LUME 1
•
CAMBRIDGE UNIVERSITY PRESS

PURE
MATHEMATICS
1
S. L. PARSONSON
CAMBRIDGE UNIVERSITY PRESS
CAMBRIDGE
LONDON · NEW YORK · MELBOURNE

Published by the Syndics of the Cambridge University Press
The Pitt Building, Trumpington Street, Cambridge CB2 1RP
Bentley House, 200 Euston Road, London NW1 2DB
32 East 57th Street, New York, NY 10022, USA
296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia
© Cambridge University Press 1970
Library of Congress catalogue card number: 70-100026
ISBN: 0 521 07683 8
First published 1970
Reprinted 1972 1974 1975 1977
PRINTED IN GREAT BRITAIN

Contents
Preface page ν
1 Numbers and inequalities
2 Vectors and vector geometry
3 Coordinates
4 Polynomials
5 Functions and inequalities
6 The trigonometric functions
7 Probability in finite outcome spaces
8 Finite series and the binomial theorem
Revision exercise A
9 Mathematical induction
10 Expectation
11 Further vectors
12 Further trigonometry
13 Matrices 1
14 Matrices 2
15 Linear equations
16 Discrete probability distributions
Revision exercise В
Bibliography
Answers
Index
1
24
45
68
78
87
103
130
151
155
169
189
215
238
258
281
301
328
339
343
375

г?

Preface
The present book is the first of a two volume course covering those parts
of modern Ά' level pure mathematics syllabuses not normally included
in standard calculus texts. The reason for omitting calculus is simply to
keep the book within manageable proportions and it is felt that there are a
number of excellent modern calculus books available. Although no formal
calculus is done here, it is expected that a student will be studying calculus
concurrently with this text; thus, for example, it is assumed in Chapter 5
that the reader can differentiate simple algebraic expressions and sketch
the graphs of rational functions. In Chapter 16, the exponential function
is used.
It is hoped that the order of presentation of topics in this book will offer
an effective teaching programme, but variations can be made at the
discretion of the teacher. Thus, for example, the chapters on probability can
be deferred while the two chapters on trigonometry could be taken in
conjunction. Again certain chapters contain work which might be deemed
suitable for a second reading; for example, Sections 4 and 5 of Chapter 10.
In some ways Chapter 1 offers the most difficult problem of presentation
in the whole book: it is necessary later to be able to refer to rational and
irrational numbers, and yet to devote too much time in the early stages
to such topics may not be desirable. The author hopes that he has found
an acceptable compromise but some might yet find parts of Chapter 1 too
formal, in which case they are strongly advised to leave the detailed study
of Sections 2, 3 and 4 for a second reading.
To learn mathematics, constant practice is necessary, some of it
repetitive. The book is therefore liberally supplied with exercises for the student.
Most of the questions represent fairly straightforward applications of the
bookwork, although in the Miscellaneous Exercises, included at the end
of all but two of the chapters, will be found some rather more searching
questions. Furthermore, questions marked Ex. occur in the text; it is
hoped that most, if not all, of these will be attempted by the pupil as he
covers the associated bookwork—certainly those marked with an asterisk
should be regarded as obligatory.
The syllabuses for M.E.I. Ά' level, the Joint Matriculation Board
syllabus in Mathematics (Advanced) and Further Mathematics (Advanced)
and the University of London Revised Syllabus in Mathematics (Advanced)

PREFACE
have been particularly kept in mind in the writing of this book and the
planning of the next. Other boards are in the process of devising new
syllabuses and draft copies published by them indicate that this book will
probably cover the necessary work. The School Mathematics Project have
produced their own text-books but the present book may be used to
supplement these if desired.
It is a pleasure to record my thanks to Mr M. J. Rawlinson who read
part of the text and made a number of valuable suggestions; to Mr A. J.
Moakes who read the entire book and whose detailed criticisms have done
much to remove obscurities and improve the presentation; and to my wife,
who also read the entire book and who lent invaluable assistance in
checking answers. I am grateful for permission to reprint examination questions
from the following Boards: Oxford and Cambridge (O&C, M.E.I.,S.M.P.),
the Joint Matriculation Board (J.M.B.), the Cambridge Local Examination
Syndicate (Cambridge), University of London (London), and the Oxford
Delegacy of Local Examinations (Oxford); also to the Clarendon Press
for permission to use Oxford Scholarship questions (O.S.) and to the
Cambridge University Press and University Registry for the use of
Cambridge Scholarship (C.S.) and Mathematical Tripos (M.T.) questions.
S.L.P.

ι. Numbers and inequalities
1. THE INTEGERS AND THEIR
REPRESENTATION
The process of counting is deeply rooted in the history of mankind and its
development is obscure. Many systems have been evolved but the Arabic
(originally Hindu) which we now employ is almost universal.
The entities used for counting are the whole numbers, от positive integers,
and zero. Ten ciphers are used to represent the whole numbers: 0, 1,2, 3,4,
5, 6, 7, 8, 9. The introduction of the cipher zero was an advance of
considerable significance and it gave to the Arabic system its great flexibility
and versatility by making possible a system of place value recording;
place value calculation (by the abacus) had already been in use for a long
time. A base of ten is most frequently used for continued counting; thus,
the one hundred and forty second integer is written 142, which stands for
Ixl02+4xl0 + 2xl.
The base, 10, of this method of enumeration has no special significance
and any other number would do as well. Indeed, were ten not so deeply
rooted in us for physical reasons, other bases would no doubt be
preferable; for example, eight (23) or twelve (22 χ 3). Had primitive man ignored
his thumbs we could well have inherited a more efficient system.
Other bases have been used. A base which has acquired special
significance recently is the number two. Counting from 1, with 2 as base, the
first eight integers are written: 1,10,11,100, 101, 110, 111, 1000. The value
of this, the binary system, is clear: it requires only two ciphers, 0 and 1.
It is thus eminently suitable for recording numbers in two-state systems as,
for example, in electronic digital computers. Conversion from one base
to another is easily effected. For example
75 (written in the scale of ten, or the denary scale)
= ^гчохгнохгн^гчохгч^г^!
= 1001011 (written in the binary scale).
The same result may be arrived at more quickly by continued

NUMBERS AND INEQUALITIES [1
division by 2, the remainders recorded being the required digits (in
reverse order)
75
37
18
9
4
2
1
0
1
1
0
1
0
0
1
Ex. 1. Explain why the method of
tinued division works.
le scale to another by con-
Ex. 2. Evaluate 11011 χ 1011 in the binary scale. Perform several other
multiplications of this type and check your answers by conversion to the denary scale.
Ex. 3. Evaluate 1110011ч-11011 in the binary scale. Perform several other
divisions of this type and check your answers by conversion to the denary scale.
Exercise 1 (a)
Solve the following equation for x, working in the binary scale throughout:
(i) 101x+ll = 1101;
(ii) llx-11111 = 1011;
10x + lt
(iii)
(iv) ll(10x-101)
111;
= 10x+101.
2. Express the denary number 275 in the scales of 2, 3 and 12. (For the last
part, you will need to supplement the digits 0,1,..., 9 by t = ten and e = eleven.
Why?)
3. With the notation of Question 2, evaluate 4te χ19ί in the scale of 12.
4. With the notation of Question 2, evaluate e89t + 2ee in the scale of 12.
η from the binary form 1100110111
5. Explain the following method of
to denary form 823
1
1
100
4
110
6
111
7
512 + 256 + 48 + 7 =
Octal
Denary
6. Show that 1331 is a perfect cube whatever the base *, provided that* is greater
than3.

1] THE INTEGERS
7. The number χ lies between 100 and 999 (denary scale) and the number у is
formed by writing the digits of χ in the reverse order. Prove that χ ~ у is divisible
by 99, where χ ~ у means the difference between χ and y.
8. Show that the difference between any number with four digits (in the denary
scale) and the number with these digits reversed is divisible by 9, and that, if the
two middle digits are the same, the difference is also divisible by 37.
2. THE INTEGERS (CONTINUED)
We have already tacitly assumed that the positive integers obey certain
laws of combination, which are summarized below. These laws are seen
to hold irrespective of the base employed for representing the numbers:
the integer seven remains the integer seven, whether it be written 7 or 111
(binary) or 21 (ternary); consistency is all that is demanded.
(i) If a = b, then a + c = b + c and ac = be.
(ii) The commutative laws: a + b = b+a; ab = ba.
(iii) The associative laws: a + (b + c) = (a+b) + c;a(bc) = (ab) с
(iv) The distributive law: a(b + c) = ab + ac.
(v) The additive and multiplicative identity laws:
a + 0 = a; a\ = a.
Considerably later historically, zero and the positive integers were
augmented by the negative integers; the word 'integer' will in future be
taken to mean a positive or negative integer, together with zero. To enable
negative integers to be combined, a further rule is required:
(vi) The additive inverse law: a + (-a) = 0.
Example 1. Prove, using only the laws (i)—(vi) above, that ( — a)(—b) = ab.
First observe that
aa + aO = a(a + 0) by (iv)
= aa by (v)
= aa + 0 by (v),
a0 = 0
on adding - aa to both sides and using (ii) and (i). Thus, any integer
multiplied by zero gives the answer zero.
Next, consider
ab + [a(-b) + (-a)(-b)] = a[b + (-b)] + (-a) (-b) by (iii), (iv)
= a0 + (-a)(-b) by(vi);
= 0 + (-a)(-b) by result
proved above;
= (-a)(-b) by(v).
3

NUMBERS AND INEQUALITIES 1]
But
ab + [a(-b) + (-a)(-b)] = ab + [a + (-a)](-b)
= ab + 0(-b)
= ab + O
= ab
Thus ab = (-a)(-b).
Laws (i)—(vi) above supply almost all the apparatus required for the
manipulation of integers. However, as Example 2 below shows, they need
to be supplemented by one more law:
(vii) The cancellation law: ab = ac^b = c, provided α φ 0. (The
sign => is read as' implies'; <= means' is implied by' and о means' implies
and is implied by'.)
In this section we have given only a brief survey of the logical structure
of the integers. The reader interested in acquiring a deeper understanding
of this topic and, indeed, of the other topics mentioned in this chapter
should consult one of the books mentioned in the Bibliography at the end
of the book.
Example 2. Arithmetic modulo 12 is defined as follows: any two of the integers
0, 1, 2, ..., 11 are added or multiplied together and the answer is taken to be
the remainder on dividing the sum or product by 12. Thus
8 + 6 ξ 2 (mod 12); 4x11=8 (mod 12);
and we also write 4-6 = 10 (mod 12).
Show that arithmetic modulo 12 satisfies laws (i)—(vi) above but that the
equation 4χ ^ 4 ^ u)
does not have a unique solution (and so law (vii) does not hold).
If a + b = I2c + r, then b + a = 12c+ r and the commutative rule holds.
All the other laws may similarly be verified.
But χ = 1, 4, 7, 10 all satisfy the equation Ax = 4 (mod 12).
Ex. 4. Prove that -(a-b) =-a + b. (a-bmeans a + (-b).)
Ex. 5. Prove that, if ax = a, then χ = 1, provided я Ф 0. (Notice carefully
what you have to prove; it is not sufficient merely to verify that χ = 1 satisfies
ax = a.)
Ex. 6. Show that arithmetic modulo 11 does not suffer from the same defect as
that exhibited by arithmetic modulo 12 in Example 2. Suggest why this is so.
4
by (ii) and
(iv);
by (vi);
by result
proved above;
by (v).

3] RATIONAL NUMBERS
3. RATIONAL NUMBERS
As soon as numbers began to be applied to problems more complex than
the mere counting of objects, the value of subdividing the interval between
two integers must have become apparent. For example, when a unit of
length was defined, lengths must have been met that were not an exact
integral number of units. Thus the concept of a fraction or, as we shall
prefer to call it, a rational number, was evolved.
In order to emphasize the fact that we are now dealing with a new type
of number, we shall avoid the familiar fractional form for rational numbers
at first, and instead we shall define a rational number as a pair of integers
written in a definite order thus: (p, q), where ?φ0. (We may call this an
ordered pair of integers: (2, 3) and (3, 2) represent different rational
numbers. See Chapter 4.)
(In what follows it may help the reader to see what is happening if he
bears in mind that our aim is to demonstrate that the rational number
(p, q) is what he would call p/q.)
Two rational numbers (ръ qj and (p2, q2) are said to be equal if
Р1Я2-Р2Я1 = 0; if Pil2-P2<li + 0> they are said to be unequal.
Ex. 7. Is it true that (я, *) Φ (*, a)?
Ex. 8. Prove that (t, 2), (2,4), (13, 26) are all equal. Prove more generally that
O, q) = (kp, kg), provided HO.
Since rational numbers are newly defined objects, rules for adding,
subtracting, multiplying and dividing them must be given, for these
operations have so far only been applied to integers. The rules are
Ol,<?l) + <>2,<?z) = Ol<?2+/>2<?l,<?l<?2); Ol,<?l)-0>2,<?2) = Ol<?2 ~ Pi4l, 4l4^\
Oi, <?i) x O2, <?2) = (.PiPz, <?i<?J; (Pi, 1i) + O2, <?2) = (PilbPrfi),
provided p2 φ 0, otherwise the left-hand side is undefined.
Example 3. Verify that the distributive law holds for rational numbers.
We have to show that
Oi> ?i) x [O2, Яд + Оз> 4s)] = Οι, 1i) x O2, Ч2) + (Pi, ?i) x (РаЯа),
L.H.S. =(A,?i)x(ft?8+/W2,fc?a)
5

NUMBERS AND INEQUALITIES [1
R.H.S. = (ft, fi) x (p2, q2) + (p1, qt) x (рв, ft)
= (AA,iift) + (Aft,iiia)
= {PiP24i4i+PiPi4i42,4\4i4i)
= (Р1Р2Ч3 +Р1Р3Ч2, ЧгЧгЧз), by Ex. 8, since
ϊι*0.
In a similar way, all the rules for combining integers may be verified to
hold also for rational numbers.
Ex. 9. Verify the laws (i)-(vii) for rational numbers.
Ex. 10. Verify that, iftherulefor division is ignored, rational numbers of theform
(p, 1) have properties identical to those possessed by the integers.
Ex. 11. Verify that, if (a, 1) χ (χ, 1) = (b, 1), then (*, 1) = (b, a), provided α φ 0.
Ex. 10 shows us that we may identify the integers with rational
numbers of the form (p, 1); Ex. 11 then shows us that the rational number
(b, a), when multiplied by the integer (a, 1) gives the integer (b, 1). In
more familiar language (b, a) has just the properties we associate with
the 'number' bja. We may thus regard a rational number as 'the quotient
of two integers'.
It thus follows that the statement 'л: is a rational number' is equivalent
to the statement 'integersρ and q may be found such that χ = pjq'.
The reader will be familiar with the process of expressing a rational
number as a decimal. For example,
^ = 0.024 = 0x1 + 2 + ^ + 4x^3.
Conversely, a terminating decimal can always be expressed as a rational
number in the form pjq; for example
ο·ΐ75 = -jWo- = -:-δ.
However, the terms 'rational number' and 'terminating decimal'
are not synonymous. Thus, f is certainly a rational number, but
f = 1-142857142857142857...
and the process of division cannot be brought to an exact conclusion.
However, this expression does recur; the reader will probably be familiar
with the notation. & = ,.{49057
In fact, any rational number may be expressed either as a terminating or
a recurring decimal, and conversely, any terminating or recurring decimal
represents a rational number. The proof of the italicized part of this state-
6

3] RATIONAL NUMBERS
ment depends upon the notion of a limit (see Chapter 8; geometric
sequences) but good grounds for believing it are given in the following
particular example.
Example 4. Express 2-378 as a rational number in standard form.
Let* = 2-378; then 100ο* = 2378-378
χ = 2-378
999л; = 2376 by subtraction
Ex. 12. Express the following recurring decimals as fractions:
(i) 0-7, (ii) 0-402, (iii) 6-28.
Ex. 13. Using only the digits 0 and 1 we may express any rational number as a
bicimal. For example
1101101 = lx2+l+0x^+lxi + lxi + 0xi + lxi.
Express as bicimals the decimals 3-75, 0-703125, 4-6, 0-82.
[Hint: see Ex. 1 and substitute repeated multiplication for repeated division.]
Ex. 14. Compare 0-9 (decimal) with 0-1 (bicimal).
Ex. 15. Express 0-01 and l-lOl (both bicimals) as decimals.
4. IRRATIONAL NUMBERS
It is useful to depict the integers and rational numbers as points on a
straight line. In Figure 1.1, an origin О is taken, representing the number
zero, and equal intervals are measured to the right, the end-points
representing 1, 2, 3, ... and to the left representing - 1, -2, -3, .... The
position of a point representing a rational number may be defined quite simply;
Fig. 1.1
for example, the point representing 3-28 is obtained by sub-dividing the
interval between 3 and 4 equally into 100 divisions and marking the end-
point of the twenty-eighth division. We may thus associate with the rational
numbers a definite ordering: л:is greater than у (written χ > у) if the point
representing χ lies to the right of that representing y.

NUMBERS AND INEQUALITIES [1
All rational numbers may thus be represented by points on the line and,
furthermore, however close together two points representing rational
numbers may be there will always be another point representing a rational
number lying between them. For example, between the points representing
3-286 and 3-287 lies the point representing 3-2865.
Ex. 16. Show that Цх+у) always lies between χ and y.
Our last remark shows that, however close together we choose two
rational numbers, we can always fit another rational number in between
them; surprisingly, however, we can never succeed in 'filling the line up'
with points representing rational numbers. In Figure 1.2 an isosceles
right-angled triangle OPQ has been drawn in which OQ = QP = 1.
The theorem of Pythagoras tells us that, if OP represents the number x,
then x2 = 2; our next example shows that Ρ does not represent a rational
number.
Q
Fig. 1.2
Example 5. Prove that J2 is not a rational number.
We have to prove that ^2 cannot be expressed in the form pjq where
p, q are integers. Our method of proof will consist in assuming that
72 can be expressed in the form pjq and showing that this leads to a
contradiction.
Suppose that <J2 = pjq, where p, q have no common factors.
Then p2 = 2q2; but, if the square of an integer is even, the integer itself
must be even, and so ρ = 2r where r is an integer
.*. 4r2 = 2q2 and so 2r2 = q2.
By the same token q must be even.
Thus, p, q must both be even, contradicting our initial assumption.
Numbers such as J2 which are not rational are called irrational. Since
rational numbers are represented by terminating or recurring decimals,
the decimal representation {if it exists) of an irrational number is non-
terminating and non-recurring.
Irrational numbers obey the same laws of combination as rational
numbers. The proof of this lies beyond the scope of this book, as it requires
a systematic definition of irrational numbers in terms of the rationale;

41
IRRATIONAL NUMBERS
we shall be content to assume this result, which is intuitively fairly obvious
if we consider rational (terminating decimal) approximations to irrational
numbers.
Exercise 1 (b)
1. Where does the proof of Example 5 break down if you try to prove λ/4 is
irrational?
2. By assuming that $2 = p/q, where p, q are integers cancelled down into their
lowest form, show that ξ/2 is irrational.
3. Prove that V3 is irrational.
4. Express as (i) a decimal, (ii) a bicimal, the fraction i (denary notation). If a
decimal recurs, does it bicimal representation necessarily recur ?
5. Prove that, if a rational number of the form p/q (cancelled down into its
lowest form) is expanded as a recurring decimal, the length of the block of digits
that recurs is less than q.
6. Which of the following statements are always true and which are not? If a
statement is true, prove it; if it may be false, give a counter-example (that is, an
example that illustrates its falsity):
(i) rational + rational = rational;
(ii) rational + irrational = irrational;
(iii) irrational+irrational = irrational;
(iv) rational χ rational = rational;
(v) rational χ irrational = irrational;
(vi) irrational χ irrational = irrational.
7. If a and * are integers and V* is irrational, prove that (a + V*)3 is irrational.
(The results of Question 6 may be used in this question.)
8. Prove that it is always possible to find an irrational number that lies between
two given rational numbers a and b.
9. lipi,p2,p3 are three unequal prime numbers, prove that
is a rational number, but not an integer.
Can you generalize this result?
10. You are given a ruler with only integral units of length marked on it. Show
how to construct, geometrically, the following lengths:
(i) V3; (ii) V7; Ου) VC + V2).
11. Prove that, if к is an integer and *jk is rational, then Jk is an integer. Show
further that, for к > 1, *j(k2-\) is irrational and deduce that, for к > 1,
V(fc- 1) + V№ +1) is irrational.
9

NUMBERS AND INEQUALITIES [1
5. SURDS
Although the existence of irrational numbers is mathematically significant,
from the point of view of practical arithmetic all numbers may be regarded
as rational, since any irrational number may be approximated to by a
terminating decimal. Indeed, wherever measurements are concerned, the
answers obtained must necessarily be in the form of rational numbers,
since every measuring device must eventually reach the limits of its possible
accuracy. However, just as i is simpler to handle than its decimal
approximation 0-3333, so V2 is often easier to deal with than its approximation
1-414.
An expression involving only rational numbers and their roots (not
necessarily square roots) is called a surd; thus surds form a class of
irrational numbers, though there are irrational numbers, such as π, which
cannot be expressed as surds.
Examples of surds are:
^ ft- {^2+lt\)·' ^3+V2·
Note: if χ is positive, *tyx = xllm means the positive wth root of x; by
convention, the 2 is omitted for square roots, e.g. J9 = 3. Again, 2гЦ(—х)
has no meaning, while
2m+\l(-x) = (-x)1/<2m+« = -{xfi^+v.
A great number of complicated surds may be simplified using the three
results: , ,, ,4
&№) = №·. <®β=ψ·>
(iii) Ца + ф) (Va~V&) = a~b-
Example 6. Simplify the expression V128-V32-V8 + V2·
V128-V32-V8 + V2 = V(26.2)-V(24.2)-V(22.2) + V2
= 2V2-2V2-2V2 + V2
= 8V2-4V2-2V2 + V2
= 3V2.
The process of removing surds from the denominator of surd
expressions is known as rationalizing the denominator.
Conventionally, a surd is usually expressed with its denominator
rationalized.
10

41
Example 7. Rationalize the denominators of the following surds:
... /3 .... 1 ..... J2 + 1
'л IV (11) V5W2: (111) узТ72"
УЗ_УЗУ5 = У15
= У5~У5У5~ 5 ;
__1_ 1 /V5 + V2\ V5 + V2.
u' V5-V2 V5-V2W5 + V2/ 3 '
V2+1 / V2 + l\ Л/3-У2У = (V2+1)(V3-V2)
MV V3+V2 W3 + V2/W3-V2/ 3-2
= V6 + V3-V2-2.
Ex. 17. V45 = 3V5. Simplify in a similar fashion the surds: (i) У8; (ii) У18;
(iii) V54; (iv) V250; (v) V5292.
Ex. 18. Simplify the following surd expressions,
(i) (V3+2V2)2; (ii) V216-V150 + V24; (iii) V75 + V147-V300;
(iv) (V32+V50-V98)2.
Ex. 19. Rationalize the denominators of the following expressions:
... 1 .... V2+1 ..... 1 ,· ч V3 , ч 2-V3
0)7з; Οΰ^-; (ш)2=72: Ov)^; W^.
We conclude this section with some further examples of manipulation
of surd quantities and the use of the у sign.
ExampleS. Simplify:
1 = 1-(V3-V2)
w 1+V3-V2 1-(V3-V2)2
1-V3 + V2
l-(5-2V6)
_1+V2-V3
2(V6-2)
_(l+V2-V3)(V6+2)
2(6-4)
_ У6+2+У12+2У2-У18-2УЗ
4
= 2 + уб-У2
4
(ii) Suppose 14-4У6 = {^a-l^jbf.
Then 14-4У6 = (a + 4b)-4J(ab).

NUMBERS AND INEQUALITIES [1
These two expressions will be equal if
a + Ab = 14,
ab = 6.
By inspection (or by solution of the simultaneous quadratic equations)
a = 2, b = 3.
ThuS V(!4-4V6) = 2V3-V2-
(Notice that we must choose the positive square root.)
Example 9. Solve the equation
V(3x + 4)-V(* + 2) = 2.
(i) Squaring both sides,
(3x+4) + (x + 2)-2V(3x2 + 10x + 8) = 4.
(ii) Collecting terms and dividing both sides by 2,
2x +1 = V(3x2 + l(h: + 8).
(iii) Squaring both sides and collecting terms,
x2- 6x- 7 = 0.
This gives χ = 7 or χ = — 1. However, we must check both of these
solutions in the original equation since, after step (iii), we could equally
well have been solving the equation
V(3x+4) + V(* + 2) = 2. (Why?)
Inspection shows us that only χ = 7 is a valid root of the original
equation.
Exercise 1 (c)
1. Express in the form a^b, where b has no perfect squares as factors
(i) V50; (ii) V363; (iii) V2400; (iv) V192; (v) V1452.
2. Simplify the following surds:
(i)(V5-l)2; (ii) (2V3-V2)2; (iii) (V3 + 1)3; (iv) (V3-V2)3;
(v) (V3 -1)4·
3. Simplify the following expressions:
(i) V18-V2; (ii) V80-V5; (Ш) V108-V75 + V48;
(iv)Vm (y) 714^/3.
12

4. Rationalize the denominators of the following expressions:
*Ъ *>M- «^ ^Ш *>£&
5. Simplify:
(i) V(6-2V5); (ii) V(10-V96); (iii) VO0+12V6); (iv) V(47-6V60);
(v) 3/V(7-V40).
6. Solve for л: the following equations:
(i) 3V*-V(*+5) = 3,
(ii) V(2* + 5)-V(* + 2) = 1,
(iii) V(* + 6) + V(4-*) = Vd-3x).
7. Rationalize the denominators of the expressions:
8. Theequation V(x + 4-7) + V(* + 4-9) = V33
was solved to give an answer χ = 31 (to one decimal point). Without using
tables, state why this is clearly wrong.
9. Given that V5 » 2-23607, J2 χ 1-41421, evaluate:
(i)75; (ii)V^72'
giving your answers to the greatest accuracy that you can guarantee.
10. Given that, to six significant figures, V6 = 2-44949, V3 = 1-73205, evaluate
as accurately as possible 1/(2^3 —V6), justifying the accuracy you give.
11. Rationalize the denominator of the expression #3/(^3 + 1), by using the
factorization a3+bs = ia + b){a2-ab+b2).
12. Rationalize the denominators, and simplify as far as possible the following
expressions:
6. SETS OF NUMBERS
Suppose we have a collection of numbers, the whole collection being
denoted by the letter It. Suppose, too, we ask a question which may be
answered unambiguously 'Yes' or 'No' for each number (or element)
of U. Then the collection of those elements of 11 for which the answer is
'Yes' is said to form a set S, which is a subset of the universal set It,
13

NUMBERS AND INEQUALITIES [1
written Sill. The collection of those elements of 11 for which the answer
is 'No' is called the complement of S, written S'.
It may happen that the answer to the question is 'No' for each
element of 11; in this case we say S is empty and write this as S = 0.
(0 is usually referred to as the null set, or empty set.)
As an example, suppose that 11 is the set of integers from 1 to 10. We
may write this as Ц = {1, 2, 3, .... 10}.
The question may be posed: 'Is the element χ of 11 divisible by 3?'
Those elements for which the answer is 'Yes' form a set A where
A = {3, 6, 9}.
The complement of A is given by
A' = {1, 2, 4, 5, 7, 8, 10}.
Again, the alternative question 'Is the element χ of 11 irrational?' may be
asked and unambiguously answered for each element of 11. The answer is
'No' in every case; the set defined is thus 0 and its complement is U.
In this example, the universal set 11 contains only a finite number of
elements. It may well happen, however, that 11 contains an infinite number
of elements; for example, 11 might be the set of all positive integers. The
proper subset Ε defined by the question 'Is the element χ of 11 even?'
also contains an infinite number of elements. {P is a proper subset of Q
if Ρ is contained in Q but is not the whole of Q, written Ρ c Q. Ρ я Q
means that Ρ is a subset of Q, but may be the whole of Q.)
It is convenient to develop a shorthand notation for the somewhat
cumbersome method we have used so far in defining our sets. The method
adopted is to use braces {:} with a colon (:) or vertical line (|) (read as
'such that') separating the two necessary pieces of information:
(i) within the bracket and to the left of the colon is stated the universal
set from which the elements are drawn;
(ii) within the bracket and to the right of the colon is the statement
defining which particular elements of the universal set are to be chosen.
For instance, for the sets 11 and A mentioned above we may write
A = {x e U:x is divisible by 3}.
The sign e means 'is a member of' or 'is an element of '.
Ex. 20. If U is the set of positive integers {1,2,3,..., 9} write down the elements of
the following sets:
(i) {x e VL: χ+4 e U}; (ii) {x e VL: J χ is rational}.

6]
SETS OF NUMBERS
Ex. 21. If U is the set of all positive integers {1, 2, 3, ...} describe in words the
following sets:
::i(x+l)eU}; (ii) {xeU: x-1 eU};
: &XeЩ; (iv) {xeU: x*-5x-6 = 0};
(0 {«U
(iii) {xeU
(v) {*eU:
x*-bx-l =
Ex. 22. Two sets A and Л which are both subsets of the same universal set U are
said to be equal (A = B) if they contain precisely the same elements.
Show that А = ВоАяВ and В Я А.
We may combine two subsets А я VL and В я VL according to the two
rules of union (u) and intersection (n) defined as follows:
A U В is the set of all elements of 11 that are members of either A or В
or both.
Α Π Β is the set of all elements of 11 that are members of both A and B.
From these definitions it follows at once that the operations of union
and intersection are commutative; that is,
A\J В = B\J A and A (] В = В f] A.
The operations of union, intersection and complementation may be
exhibited pictorially using a Venn diagram. In a Venn diagram, the uni-,
versal set 11 is represented by a rectangle; any subset A £ 11 is depicted
by a closed region lying within the rectangle; see Figure 1.3.
In Figures 1.4, 1.5, 1.6, A U В, А П В and A' are shown shaded:
Ex. 23. Verify, using Venn diagrams, that the operations of union and
intersection are associative; that is
A U (В U Q = (A U В) U С; Α η (Β η С) = (А П В) η С.

NUMBERS AND INEQUALITIES [1
Ex. 24. Verify, using Venn diagrams, that the operation of union is distributive
over intersection, and that intersection is distributive over union; that is
A U (В П Q = (A U В) П (A U С); Α η (В U Q = (Α η Л) U (Α η С).
Ex. 25. Verify rfe Morgan's Laws, using a Venn diagram:
(i) 04 U Л)' = / Л i'; (ii) (А П 5)' = Λ' U Л'.
(The reader may be able to identify these laws with the logic of and (n),
or (U) and not (').)
Ex. 26. If U is the set of all positive integers and
A= {xeU-.ixeU}, В = {xeU: ixeU}
describe An B, using the {:} notation.
Ex. 27. If U = {a, b, c, d, e} and A = {a, c, e} write down the subsets:
(i) A U A'; (ii) Α η A'; (iii) AuU; (iv) A n U; (v) A U 0; (vi) A n 0;
(vii) U'; (viii) 0'.
Ex. 28. If, with the notation of Ex. 27, Л = {я, 6, e), write down the subsets
(i) AnB; (ii) Λ' U B; (iii) Λ η Л'.
Verify de Morgan's Laws (see Ex. 25) in this particular case.
Three particular sets, the set of all integers, the set of all rational numbers
and the set of all real numbers occur throughout mathematics with such
unfailing regularity that it is convenient to introduce a notation by which
to refer to them.
The set of all integers is denoted by Z.
The set of all rational numbers is denoted by Q.
The set of all real numbers (that is, the union of the set of all rational
numbers and the set of all irrational numbers) is denoted by R.
With a suitable definition of each of the terms integer, rational and real
we have the following relation between them
Ζ С Q С R.
The sets of all positive integers is denoted by Z+, of all positive rational
numbers by Q+ and of all positive real numbers by R+.
Ex. 29. Enumerate the elements of the following sets:
(i) {xeZ:(x-2)(2x+l)(x*-2) = 0};
(ii) {xe Q+:(x-2)(2x+l)(*-2) = 0};
(iii) {xe β:(*-2)(2χ + 1)(χ·-2) = о};
(iv) {xeR+:(x-2)(2x+l)(x*-2) = 0};
(v) {xe R:(x-2) (2*+l) (x2-2) = 0}.
16

η INEQUALITIES
7. INEQUALITIES
We have already asserted that the integers and the rational numbers may
be ordered; that is, given two integers (or rational numbers) χ and y,
we may answer the question 'Is χ greater thany?' The same is true for the
set of all irrational numbers, or, indeed, for the union of the sets of rational
and irrational numbers (the set of real numbers).
If x, у are two real numbers, χ > у (read 'χ is greater than y') means
that the point representing the number χ lies to the right of the point
representing the number y. Similarly for χ < у ('х is less than y'). When
comparing two positive or negative numbers χ and у in this way, we see
that . ,. . _ .
χ > yox—y > 0 (i.e. x—y is positive),
x<yox—y<0 (i.e. x—y is negative),
χ = yox—y = 0 (i.e. x—y is zero).
The number zero separates the positive and negative numbers; it is itself
neither positive nor negative.
By continued use of these results we see that we may add or subtract
any number from both sides of an inequality without altering the validity
of the inequality sign.
Inequalities are of value in defining sets. Thus, for example, to take a
rather trivial case, {xeZ: 0 < x < 3) - {1,2).
A more substantial example would be
{xeR:x2 < 4}.
Solving an inequality means redefining the set of elements which satisfy
the given inequality in as simple a manner as possible.
Ex. 30. What does 'solving an equation' mean?
Example 10. If χ is a real number, solve the inequality 3-х > 2.
The question asks us to redefine the set
{xeR:3-x> 2}
as simply as possible.
SinCe 3-*>2
=> 3 > 2+x (by adding χ to both sides)
=> 1 > χ (by subtracting 2 from both sides)
the solution set may be rewritten
{x&R:x < 1}.

NUMBERS AND INEQUALITIES [1
The rule for redefining a set by multiplying or dividing both sides of an
inequality by a number is slightly more complicated. If x—y and ζ are
both positive, then {x-y) ζ is positive, thus
χ > у and ζ > 0 => xz > yz.
If, on the other hand, x-y is positive but ζ is negative, then (x-y) ζ is
Thus, if a set is defined by an inequality χ > у, the same set is defined by
xz > у ζ if ζ is positive,
or by xz < yz if ζ is negative.
A similar result holds for division.
We now append an alternative solution to Example 10:
3-х > 2
=> -3+x < -2, multiplying both sides by -1,
=> χ < 1, adding 3 to both sides,
and we have, as before, the solution set
{x&R:x < 1}.
Ex.31. Solve the inequalities
(i) 3 + x < l;(ii) 2-3* > -l;(iii)3-4x> 1.
The reader should have observed that in, for example, the result
χ > у and ζ > 0 => xz > yz
the two-way implication was not used: it is not valid to deduce from
xz > yz that χ > у and ζ > 0. However, inequalities involving products
may be solved by observing that, if the product is positive, then the factors
are either both positive or both negative. Another useful observation is
that a squared number is always positive or zero.
Example 11. Solve the inequality (x + 2) (x-1) > 0 (x real).
If(x + 2)(;t-l) > Othen
either (i) (x + 2) > 0 and (x-1) > 0,
or (ii) (x + 2) < 0 and (x-1) < 0.
The first pair of inequalities has as solution set
{xeR:x > -2} η {xeR:x > 1} = {xeR:x > 1}.

7] INEQUALITIES
The second pair of inequalities has as solution set
{xeR:x<-2}[){xeR:x< 1} = {xe R: χ < -2}.
The complete solution is thus the union of these two sets
{x e R: either χ < -2 or χ > 1}.
Alternatively, the argument may be presented very clearly in the
following tabular form. The critical values of χ for which (x + 2) {x— 1) changes
sign are χ = —2 and χ = 1. We divide the possible values of χ into the
three ranges χ < — 2, — 2 < л: < 1 and χ > 1 and consider the signs of
(x + 2), (x-l) and hence of (x + 2) (x-l) in each interval.
(x+2Kx-\)
x<-2
~
+
-2<x<\
+
-
x> 1
+
+
+
Example 12. Prove that the expression у = x2 + x +1 is positive for all real
values of x, and find its minimum value.
It will be useful in the solution of this example to employ the notation
> : χ it у means 'x is greater than, or equal to, y\
N°W 2, ,1
у = x2 + x+l
= (* + i)2+l,
.·. y-ϊ = (*+i)2,
.·. y-l > 0 or у > %.
Thus у is always positive and has a minimum value off (when χ = -|).
Given two positive numbers χ and у we define
(i) the arithmetic mean {a.m.) as A = \{x+y);
(ii) the geometric mean (G.M.) as G = -J{xy).
We conclude this chapter by proving a famous theorem (which may be
extended, see Miscellaneous Exercise 1, Question 13).
Theorem 1.1. The a.m. of two positive numbers is at least as great as their
G.M.
Proof. Since the two numbers are positive, we may write them as
u2 and v2.
Let A = (u2 + v2)/2, G = uv; we have to prove that
A- G > 0,
19

NUMBERS AND INEQUALITIES [1
thatis u2 + v2 > 0
L.H.S. = \{u2 + v2-2uv)
= ¥u-vf
> 0, unless и = v.
We have thus proved that A > G unless u2 = v2 in which case A = G.
Example 13. 200 m of wattle fencing are to be bent to form three sides of a
rectangular enclosure, the fourth side being a straight hedge. Find the length
of the rectangle if the area to be enclosed is to be a maximum.
Let xmbe the length, у m the breadth, A m2 the area of the enclosure.
Then we have ... „ ... .... .
(ι) x+2y = 200, (n) xy = A.
Both χ and 2y are positive; thus, by the theorem just proved, their a.m.
is at least as great as their G.M. That is
\{x + 2y) > J(2xy).
Using (i) and (ii), this gives m ^ ^(^
i.e. A ^ 5000.
Equality occurs only if χ = 2y, i.e. χ = 100, у = 50. Thus the maximum
area occurs when the length is 100 m.
Exercise 1 (d)
1. Verify, using a Venn diagram, the results:
(ί) ΑΌ(Α'Ό BY = An (B' U A),
(ii) (AU B)n(A'nC)' = AU(Bn O,
(iii) [A' U {Β' η СОГ = {А П В) и {А П С).
2. If R is the set of real numbers and
A = {xeR:x> 3}, B={xeR;x<4}
describe, using the {:} notation
(i) AD B; (ii) Α η Β; (iii) A' U B.
3. Taking the universal set, U as the set of all integers, describe using the {:}
notation
(i) the set of all negative integers;
(ii) the set of all positive integers divisible by 6;
(iii) the set of all integers, excluding 0, ±1, ±2.
4. A = {xeR: -1 < χ < 3}, В = {xe R: 2 < χ < 4}, C= {xeR:x>3}.
Find expressions for the following sets, using the {:} notation:
(i) A n B;(ii) A' n С; (iii) (A U Β) η C';(iv) iU(fin C);(v) (A' U Β') η С;
(vi) Α' υ (Β' η Q; (vii) /ninC.
20

7] INEQUALITIES
5. A = {xeR:x> 0}, В = {xe R: χ < 1}, С = {xe R: -1 < χ < 2},
D = {xeR: -2 < χ < 1}.
Find expressions for the following sets, using the {:} notation:
(i) Α η Β Π С η D; (ii) Α' η Л П С Π D';
(iii) 04 U Β') η (С η DO; (iv) (Λ' П В') и (С П Ζ>).
6. Solve the following inequalities:
(i) x + 2 < -3; (ii) 3*-l > 5; (iii) 4x-3 < 3x + 4; (iv) 2x > x;
(v) x2 > x.
7. Solve the following inequalities:
(i) (* + 3)(*-l)> 0; (ii) (2x-l)(3x + l)> 0; (iii) (x-3) (2x + 3) < 0;
(iv) x2-6x + 9 < 0;(v) x2-4x-5 < 0.
8. Solve the inequality
(2x + l)(x-2)(x + 3) > 0.
9. Prove that the following expressions are positive, for all real values of x,
and find their minimum values:
(i) x2 + 2x + 2; (ii) x2-6x + 12; (iii) 2x2-2x + l.
10. Find the signs, for all values of x, of:
(i) 2x-x2-2; (ii) 2x2-x+l.
11. For what values of χ is the expression
(*-l)(* + 2)(*-3)(* + 4)
positive?
12. If xy = 25, find the least possible value for χ+у, if both χ and у are positive.
13. If xy = 18, find the least possible value for 2x+y, if both χ and у are positive.
14. If x+y = 2, find the maximum value of xy.
15. If 2x + Ъу = 120, find the maximum value of xy.
16. Some netting is required to make three sides of ^ ρ Β
a rectangular chicken-run, the fourth side being an ι 1 1
existing wall. Find the least length of netting needed
for an area of 50 m2.
17. Figure 1.7 shows part of the framework of a kite. | | |
ABCD is a rectangle; Ρ and Q are the mid-points of О Q с
AB, CD respectively. Find the maximum area of
ABCD that can be made if 4 m of wood are available. FiS· J ·7
What are the dimensions of the kite necessary to attain this area?
Miscellaneous Exercise 1
1. Express 120 as a binary number.
Find the least number of stamps required so that any value from ip to 60p
(in steps of £p) may be selected, and give their values. (You may assume stamps
of any value are available if required.)
21

NUMBERS AND INEQUALITIES [1
2. Show that it is impossible to choose values of a and b so that the number
written as ab in the scale of ten is equal to the number written as ba in the scale
of twelve.
Show, however, that ab (scale of ten) = ba (scale of seven) is possible, and
give an example.
3. Prove the well known result that the remainder on dividing a number by
9 is the same as the remainder on dividing the sum of its digits by 9.
Show that this result may be generalized as follows: if a number is divided by s
then the remainder is the same as the remainder on dividing by s the sum of its
digits, when it is expressed to the base s+1.
4. Prove that, if a number is divided by 11, the remainder is either the same as
the remainder on dividing the difference between the sums of the digits in the even
and odd places, or else the sum of the two remainders is 11.
Generalize this result along the lines of Question 3.
5. Prove that, if the digits of an integer (expressed to the base ten) are
rearranged in any way to form another integer, the difference between the two
integers is divisible by 9.
6. If abc+abc = cba in the scale of five (when abc here means 52a + 5b + c) find
values for a, b, с
Show that this question is always soluble provided a base of Зи-l is used,
where и is a positive integer.
7. Prove that V3-V2 is irrational.
8. Solve the equation
V(4x-2)+V(*+l)-V(7-5x) = 0. (O&C)
9. Verify that the expression
a2 + b2 + c2-2bc-2ca-2ab
is equal to (a + b - c)2 - 4ab. Hence, or otherwise, prove that the expression is
6qUal l° (α+Α + у) (μ-β-Ύ) ("-Α + 7) (*+β-Λ
where a. = *ja, β - 4Ь, У = Vе·
Hence, or otherwise, find one solution of each of the equations
(i) V(*-6)+V(*-D = V(3*-5);
(ii) V(6-x)-Vd -*) = V(5-3x). (O & C)
10. If a > b and с > d, prove that ac + bd > bc + da.
What happens to the third inequality if a < b, с < dl
11. If the universal set is taken as the set R of all real numbers, the three sets
А, В and С are defined as follows:
A = {xeR:x> 2}; В = {xe R: 1 < χ < 4}; С = {xe R: χ < 3}.
Express, in terms of any or all of А, В, С (and using the notation of union,
intersection and complement) the following sets:
(i) {xeR:2 < χ < 3}; (ii) {xe R: χ < 1};
(iii) {xe R: 3 < χ < 4}; (iv) {xe R: 1 < χ < 2 or 3 < χ < 4}.
22

7] MISCELLANEOUS EXERCISE 1
12. Show that the expression x2 + 8xy-5y2-k(x2+y2) can be put in the form
a(x + by)2 when к has either one or other of two values. Find these values and the
values of a and b corresponding to each value of k.
Prove that when the variables χ and у are restricted by the relation
x2 + y2 = 1,
but are otherwise free, then
-7 < x2 + 8xy-5y2 < 3. (O&C)
13. a, b, c, d are four unequal positive numbers. By using Theorem 1.1 and
considering first the pair of numbers i(a + b) and Цс + d), and then, separately,
the pairs a, b and c, d, prove that
Ua + b+c+d)> (abed)*.
Deduce, by considering the four unequal numbers a, b, с and j(a + b + c),
that &a + b + c)> (abc)K
What happens to the last inequality (i) if a = b = c; (ii) a = b Φ с?
Suggest a generalization of the results proved above.
14. If x + 2y+3z = 1, where x, y, ζ are positive numbers, find the maximum
value of xyz.
If u+v = 1, where u, ν are positive numbers, find the maximum value of u2v.
15. A cylindrical vessel, with one end open, is made from a given piece of
material. Show that its volume is greatest if the height and radius are equal.
16. a, b, с are non-zero rational numbers. Show that, if
and if none of a*, 6*, c* is rational, then a*, 6*, c* are each rational multiples
of the same irrational number.
17. Show that, if p/q is a good approximation to V2, then (p2 + 2q2)/(2pq) is a
better one. Starting with ρ = q = 1, show that V2~ 577/408, and estimate the
accuracy of this approximation.
23

2. Vectors and vector geometry
1. SCALAR AND VECTOR QUANTITIES
Many physical quantities are completely specified by their magnitude
alone. For example, if the mass of the box is m kg and we are told that
m = 8, we know the mass of the box precisely. Not all physical quantities
are so easily described as this; for example, if the position of the point Ρ
relative to the fixed point О is denoted by r, then the statement 'r = 8 m'
is not sufficient for us to locate P: it may be anywhere on a sphere, centre
О and radius 8 m. In order to determine the position of Ρ we need to be
given its direction, specified in any suitable way.
The position of Ρ relative to О is an example of a displacement; that is, a
line segment whose length, direction and sense are given. Experimental
evidence reveals that many physical quantities obey the same mathematical
laws as do displacements; such quantities are called vector quantities. We
shall give a precise definition of a vector quantity in the next section,
observing here merely that vector quantities require a direction and sense,
as well as a magnitude, for their specification. Physical quantities which
require only a magnitude (that is, a pure number) to describe them are
called scalar quantities. Here are a few examples of each type:
Vector quantities: displacement, velocity, acceleration, force,
momentum, electric intensity.
Scalar quantities: mass, time, temperature, energy, electric charge,
electrostatic potential.
2. VECTORS AND THE TRIANGLE RULE
Since all vector quantities obey the same mathematical laws as
displacements we shall concern ourselves exclusively with displacements in building
up the relevant mathematics. It should be borne in mind, however, that
the theory being developed is applicable to a wide range of physical
quantities.
Before embarking upon a study of operations involving displacements
we must define what we mean by the equality of two such quantities. A
displacement is typified by its magnitude, sense and direction and so it
seems reasonable to regard two displacements of the same magnitude which
are parallel in the same sense to be equal. Thus, in Figure 2.1, we have
24

2] VECTORS AND THE TRIANGLE RULE
AB = CD = EF = GH = .... We shall use a single symbol to represent
any member of the class of such equal displacements
a = AB = CD = EF = ....
Such a representative of a whole class of displacements is called a
vector. In other words, a vector is a mathematical entity, to which a whole
class of geometrical objects correspond. Vectors are usually printed in bold
face type: a, b, c, .... In manuscript they should be indicated by a wavy
line beneath the letters, a, b, c.
The equality of two vectors a = b, means that a and bare both
representative of the same class of parallel and equal displacements.
Fig. 2.1 Fig. 2.2
The magnitude of a vector, a, written |a|, is the magnitude of any one of
the displacements of which it is representative. It is most important to
remember that |a| = |b| does not imply that a = b; put another way,
two unequal vectors may have the same magnitude.
Two displacements AB and ВС may be combined together, or added,
according to the triangle rule, AB + BC = AC (see Figure 2.2).
Two vectors may similarly be added: to find a + b we draw any
displacement AB represented by a; we then select the unique displacement ВС,
represented by b which has as its initial point the point В and the vector
a + b will be representative of that class of displacements of which AC is a
typical member.
We have said that vector quantities are physical quantities that may be
completely represented by a displacement. Since displacements obey the
triangle rule of combination, so must vector quantities generally. Indeed,
we are now in the position to make a formal definition:
A physical quantity is a vector quantity if
(i) it has magnitude, direction and sense;
(ii) it obeys the triangle law of addition.^
t The reader should observe that certain vector quantities must have their position
specified too. For example, to describe a force we need to know not only its magnitude
and direction, but also its line of application.
25

VECTORS AND VECTOR GEOMETRY [2
The observation that any particular physical quantity is indeed a vector
quantity is the result of experimental evidence, which must include a
verification of the triangle rule.
Ex. 1. Can you suggest any addition to the list of vector and scalar quantities
given in Section 1 ?
Ex. 2. Can three displacements be added together? Does it matter in what
order it is done?
Ex. 3. A rotation may be given a magnitude (size of angle turned through),
direction (axis of rotation) and sense (positive being direction for which rotation
is anticlockwise). Show that rotations are not vector quantities. (This example
shows that physical quantities exist that are neither scalar nor vector quantities.)
3. OPERATIONS WITH VECTORS
Since all parallel and equal line segments represent the same vector a
we may, when verifying the various rules of vector algebra geometrically,
choose as representative those line segments which are most convenient.
Rule 1. Vector addition is commutative, a+b = b + a.
If AB = DC = a and ВС = AD = b, ABCD is a parallelogram and
a+b = AC = b+a.
Rule 2. Vector addition is associative. (a+b)+c = a + (b + c).
If AB = а, ВС = b, CD = c,
then (a +b) + с = AC + CD = AD,
and a + (b + c) = AB+BD = AD.
We may thus drop the brackets when adding three vectors together, and
write a+b+c without ambiguity.
If AB represents the vector a then BA will be said to represent the vector
26

3]
OPERATIONS WITH VECTORS
-a. We shall write a + (-a) = 0 for all a and call 0 the zero vector.
(Note that the zero vector, unlike all others, has no direction.)
If AB = a and AC = b then, since CA+AB = CB, the line segment
CB represents a-b (see Figure 2.5). (Notice that we have used the com-
mutativity of vector addition here.)
To develop the algebra of vectors further, we need a new definition, the
multiplication of a vector by a number {scalar).
If A: is any positive number, кл is the vector of magnitude к |а | in the same
direction as a and with the same sense.
Fig. 2.5
Fig. 2.6
If к is any negative number, кл is the vector of magnitude -к\л\ in
the same direction as a and with the opposite sense.
Thus, multiplying a vector by a number simply stretches or contracts
the vector along its length. Notice that -a = - la.
Rule 3. Multiplication by numbers is associative.
k(h) = (kl) a.
This is immediately apparent, since a, /a and (к!)я all have the same
direction, and the magnitude and senses of Щл) and (кГ)л are both (kl) | a |.
Rule 4. Multiplication by numbers is distributive over vector addition.
к(я + Ъ) = кя+кЪ.
Representing a by AB and b by ВС, кл by AB' and kb by B'C, ACC is
a straight line by similar triangles and the result follows, since the
corresponding sides are in proportion (Figure 2.6).
Rule 5. Multiplication by numbers is also distributive in the following
sense. ,. ,
(к + Т)л = кл + 1л.
The vectors on either side of the above identity have the same magnitude,
sense and direction, as can be seen immediately from a diagram.
*Ex. 4. Prove that O.a =
zero vector?
0. What does this tell us about the magnitude of the

VECTORS AND VECTOR GEOMETRY [2
Ex. 5. If a is a vector of magnitude 4 units due north and b is a vector of
magnitude 1 unit due east, describe the vectors:
(i) 2a; (ii) -3b; (Hi) a-4b; (iv) 2(a+b); (v) 2a-6b.
Ex. 6. If a = 3x+ y, b = x— 2y, find x, у in terms of a, b, justifying your
argument in terms of Rules 1-5.
Ex. 7. If a is a vector of magnitude 1 unit due north and b is a vector of
magnitude 1 unit in the direction N 60° E, describe the following vectors (using a scale
drawing or trigonometry, if necessary):
(i) 2a+b; (ii) -(2a-b); (iii) b-a; (iv) -a-2b.
Ex. 8. If a is a vector of 2 units due north in a horizontal plane, b is a vector of
3 units due east in a horizontal plane and с is a vector of 1 unit vertically up out
of the plane, describe the vectors:
(i) a-2c; (ii) -b-3c; (iii) 3a+2b+6c; (iv) 3a-2b-6c.
Ex. 9. ABCD is a parallelogram with А В 2 units and ВС 1 unit. If a is a vector of
1 unit in the direction AB and b is a vector of 1 unit in the direction of ВС, and
if Ε is the mid-point of CD, find, in terms of a and b, the vectors AC, DB, AE,
EB.
4. COMPONENTS OF A VECTOR
It is often useful to express a given vector г as the sum of a number of
vectors in specified directions
г = я1 + а2+... + я„.
If this is done, аь л^, ..., ая are called components of r.
Given a vector г and specified directions two questions immediately
arise:
(i) Can г be split into components in these directions?
(ii) If the answer to (i) is 'Yes', in how many alternative ways can this
be done ?
We shall answer these questions for a vector г and three non-parallel,
non-coplanar directions in three dimensions by proving two theorems of
types that constantly recur in mathematics. The first theorem is an example
of an Existence Theorem: it will answer question (i) above, telling us that a
solution does exist; the second theorem is a Uniqueness Theorem: it will
answer question (ii) above, telling us that the solution which we know
exists is the only possible one, that is, that it is unique. The significance
of these two theorems will become apparent as we proceed.
Theorem 2.1 (The Existence Theorem). Given a non-zero vector г and three
non-coplanarf, non-parallel vectors a, b, c, there exist numbers Α, μ, ν, such
that ■> , u ,
г = ΛΛ + μο + νΐ.
t That is, displacements represented by a, b, c, cannot all be chosen to lie in one plane.
28

4] COMPONENTS OF A VECTOR
Proof. Represent the vectors a, b, c, r, by line segments OA, OB, ОС,
OP respectively. Through Ρ draw a plane parallel to the plane OBC to
cut the line OA at A' (this may always be done, by the data). Let P' be the
point in the plane OBC such that OP' = A'P.ThroughP'draw a line parallel
to CO to cut OB at B' (this again may be done using the data). The
constructions are illustrated in Figure 2.7.
r = OA'+A'P
= Aa + OP'
= Aa + OB' + B'P'
= λя + μЪ + ι>c.
Fig. 2.7
We have thus proved our Existence Theorem by showing that, with the
given data, a set of components may be constructed. To demonstrate the
truth of the Uniqueness Theorem we employ a common device in
mathematical arguments: we assume that it is false and show that this leads to a
contradiction (see Chapter 9).
Theorem 2.2 (The Uniqueness Theorem). The solution shown to exist in
Theorem 2.1 is unique.
Proof. By Theorem 2.1 a solution exists, say
r = Л^ + ^Ь + ^с.
Assume that a different solution also exists
г = A2a + /t2b + i>2c
where, say, Aj Φ Α2. Then
Aja-b/tib + ^c = A2a+/i2b + i>2C·
and so (A2 - Aj) a = (μ1 -μ2)Ъ + (vx -v2)c.
But (μ1 — μ2)Ъ + (v1-v2) с is a vector in the plane determined by b and с
2-2 29

VECTORS AND VECTOR GEOMETRY [2
and any non-zero multiple of a cannot lie in this plane, by the data. Thus
Aj = A2, which contradicts our initial assumption.
Theorems 2.1 and 2.2 have been proved for three dimensions; they hold
equally well in two dimensions, for a vector г lying in the plane
determined by a and b, where а ф kb.
An expression such as λл + μЬ + vc is called a linear combination of the
vectors a, b, с The theorems we have just proved may be restated in the
form 'Given three non-parallel, non-coplanar vectors a, b, с any non-zero
vector г may be expressed uniquely as a linear combination of a, b, c'.
If numbers Α, μ, ν (not all zero) exist such that
λя+μЪ + vc = 0
the vectors a, b, с are said to be linearly dependent; if no such numbers
exist, a, b, с are said to be linearly independent.
*Ex. 10. Prove that, if О, А, В, С are coplanar, then the position vectors a, b,
с are linearly dependent and, conversely, if a, b, с are linearly dependent, then
0. А, В, С are coplanar.
*Ex. 11. Prove the Existence and Uniqueness Theorems for components in two
dimensions.
Ex. 12. ABCD is a parallelogram, and Ε is the mid-point of CD. AB = a,
AD = b, AE = x, BE = y. Express χ and у in terms of components in the
directions a and b and also express a and b in terms of components in the directions χ
and y.
Ex. 13. Do either the Existence or the Uniqueness Theorems hold in three
dimensions if four directions a, b, c, d are given ?
Exercise 2 a
1. If ν = a + 2b, w = 2a—b, express the following vectors in the form
Aa+/ib.
(i) v + w; (ii) 2v + 3w; (iii) v-3w; (iv) 2(v-w).
2. If u = a + 3b, ν = 2a — b, express a and b in terms of u and v.
3. If u = a + b + c, ν = a + 2b + c, w = a—b—2c, find, in terms of a, b, c:
(i) u + v + w; (ii) u-v-w; (iii) 2u + v-w; (iv) u + 2v + 3w.
4. If u = a + b-c, ν = 2a-b + c, w = 3a+2b + c, find a, b, с in terms of
5. If i and j are perpendicular vectors of unit magnitude, i pointing due east and
j due north, find the magnitudes and directions of the following vectors:
(i) 2i; (ii) -3j; (iii) 3i + 4j; (iv) i-j; (v) -3i+4j.
30

4] COMPONENTS OF A VECTOR
6. If ρ is a vector of magnitude 2 pointing due east and q is a vector of magnitude
5 pointing due south, find the magnitudes and directions of the following vectors:
(i) 6p + q; (ii) 6p-q; (iii) -p-q.
7. If ρ is a vector of magnitude 1 pointing due east and q is a vector of
magnitude 2 pointing north-west find (by accurate drawing, if you wish) the vectors
(i) 2p + q; (ii) p-q; (iii) -p-q; (iv) -4p + 2q.
8. χ is a vector of magnitude 1 pointing due east; у is a vector of magnitude 2
pointing south-west; ζ is a vector of magnitude 20 pointing due north. Express
ζ in the form Ax+/iy, giving the explicit numerical values of λ and μ.
9. If a is a vector of magnitude 1 pointing due north, b is a vector of magnitude
1 pointing N 20° Ε and с is a vector of magnitude 2 pointing N 70° W, express с
in the form с = Ла+/Л.
If d is a vector of magnitude 2 pointing N 10° E, find approximate values for
λ', μ' where d = A'a+/i'b.
10. If i, j, к are vectors of unit magnitude pointing respectively due east, due
north and vertically upwards out of the plane containing i, j, find the magnitude
of the vector u = 4i+ j + 8k.
If ν = 4i—j + 8k and w = 4i—j —8k, find the angles between the vectors
(i) u, v; (ii) v, w.
11. Figure 2.8 represents a lattice of congruent parallelograms; OA = a,
OB = b, AB = с Express OP in terms of
(i) a, b; (ii) a, c; (iii) b, c;
Fig. 2.8
and do the same for the vectors OQ and OR. Express PQ in terms of a, b, and
RQ in terms of b, с Find two different expressions of the form oa+^b+yc for
the vector OQ, with none of α-, β, γ zero. Show that, if one of these expressions
is subtracted from the other, the known relation b = c + a is obtained.
12. In Figure 2.8 ОС = u, OD = v, QP = w. Obtain expressions for OP, OQ,
OR in terms of u, ν and hence obtain w in terms of u and v.
Why is it not possible to express CD in terms of the two vectors PQ, AR?
13. ABCDEF is a regular hexagon, and AB = а, ВС = b. Find CD, DE, EF,
FA in terms of a and b.
14. ABCDA'B'C'D' is a cuboid whose base ABCD is a square of side 2 units.
The sides AA', etc., are vertical and of magnitude 4 units. Ε is the mid-point
31

VECTORS AND VECTOR GEOMETRY [2
of AB, G is the mid-point of B'C and F is the mid-point of CC a, b, с are
three vectors, each of magnitude 1 unit in the directions AB, AD, AA' respectively.
Find, in terms of a, b, с the vectors DE, AF, EF, GF, GE.
15. With the data of Question 14, express ED', EG and EF in terms of
components in the directions a, b and с and also express a, b, с in terms of components
in the directions ED', EG and EF.
16. OABCO'A'B'C is a rectangular box with square ends О ABC and O'A'B'C.
OO', AA, BB\ CC are parallel edges. If АО = ОС = 1 unit, OO' = 2 units,
and if i, j, к are unit vectors (that is, vectors of unit magnitude) along О А, ОС,
OO' respectively, find, in terms of i, j, k:
(i) ОС; (ii) OB'; (iii) C'A; (iv) OD,
where D is the point on CB' produced such that C'B' = B'D;
(v) OM, where Μ is the mid-point of AA'; (vi) MD.
17. О ABC is a regular tetrahedron of side a; p, q, г are unit vectors (see Question
16) along О A, OB, ОС respectively. Find in terms of p, q, r:
(i) AB; (ii) OD, where ABCD is a rhombus.
18. ABCD is a square andP, Q are the mid-points of ВС, CD respectively. Find,
in terms of u = AP and ν = AQ,
(i) AB; (ii) AD; (iii) BD.
19. With the notation of Question 16, OB' = u, AC' = ν, ΟΉ = w; find, in
terms of u, v, w:
(i) OA'; (ii) ВС; (iii) ΟΉ'; (iv) ВС.
20. With the notation of Question 19, if CA' = x, find two distinct expressions
for CB' in terms of u, v, w, x.
21. ABCDA'B'CD' is a cube, with faces ABCD, A'B'CD' and edges AA', etc
Find, in terms of AD', AC, AC, the vectors D'B and A'C.
22. ABCDEFGH is a regular octagon. If AB = а, ВС = b, find, in terms of a, b,
the displacements CD, DE, EF, FG, GH, HA.
5. APPLICATIONS TO GEOMETRY
So far we have developed the algebra of vectors using geometrical
arguments ; we now reverse the process and show that the algebra of vectors
may usefully be employed to deduce geometrical results. The vector
treatment of geometrical problems has the great advantage that it is equally
applicable to two or three dimensions.
To describe a geometrical configuration consisting of a number of points
А, В, С, ... we must be able to locate each point. This may be done by
taking a fixed point О (called the origin) and referring to a point A by
the line segment OA. If a is the vector representative of all the line
segments equal and parallel to OA, then a is called the position vector of the
point A with respect to the origin O. Thus, given an origin O, we may
32

5]
APPLICATIONS TO GEOMETRY
refer to all points in the plane by their position vectors relative to this
origin O.
First, we establish a result known as the Section Formula. This theorem
enables us to write down the position vector of any point on a given line
and so is of importance in setting up a vectorial description of a geometrical
configuration.
Theorem 2.3 {The Section Formula). IfAPB is a straight line, with
APjPB = λ/μ,
and if the position vectors of Α, Ρ, B, relative to any origin O, are a, p, b
respectively, then ,,
Proof. We have (Figure 2.9)
AB = b-a
and AP = p-a.
But AP=Y^-AB
λ+μ
and so (Α +μ) (ρ - a) = A(b - a),
or (λ+μ) ρ = Ab + /<a
and the result follows.
Notice that the proof holds equally well for positive or negative values
of the ratio Α/μ.
Notice also the important special case in which A = μ: the mid-point
of AB has position vector |(a + b).
Fig. 2.9
Fig. 2.10
Example 1. Show that the medians of a triangle ABC are concurrent and
find the position vector of the centroid (meet of the medians) in terms of the
position vectors a, b, c, referred to some origin O.
Let the mid-points of ВС, С A, AB be L, Μ, Ν respectively and call their
position vectors 1, m, η (Figure 2.10).
33

VECTORS AND VECTOR GEOMETRY [2
Then, by the Section Formula
l = Kb + c), m = i(c + a); η = K*+b).
Any point on AL has a position vector
λΐ+μα
λ+μ '
jA(b + c) + /<a
be· λ + μ *
For varying values of A and μ we obtain different points of the line AL;
in particular, if we choose A = 2, μ = 1, we obtain a point whose position
vector is symmetrical in a, b, с and so lies equally well on BM or CN.
Thus, the medians of a triangle are concurrent at the centroid, G, whose
position vector is given by
Notice, incidentally, that we have shown that AGjGL = 2, etc.
Example 2. Show that the mid-points of the sides of a skew quadrilateral
form the vertices of a parallelogram.
Let ABCD be the skew quadrilateral and the mid-points of AB, ВС, CD,
DA be P, Q, R, S respectively.
Take any point О as origin, let a denote the position vector of A with
respect to О and similarly for the other points.
Then we have, on using the Section Formula,
p=i(a+b), q=i(b + c), г = i(c + d), s = i(d + a)
and so p-q = Ця-с)
and s-r = Ka_c)
Thus QP = RS and so PQRS is a parallelogram (one pair of sides equal
and parallel).
Ex. 14. Show that the joins of mid-points of opposite edges of the tetrahedron
ABCD are concurrent at a point G.
Ex. 15. Show that, in a tetrahedron ABCD, the joins of vertices to the centroids
of opposite faces are concurrent at the same point G as that obtained in Ex. 14.
*Ёх. 16. What condition must the numbers α, β possess for the points with
position vectors a, b and aa+/?b to be collinear?
34

5] APPLICATIONS TO GEOMETRY
The Section Formula may be restated in a form that gives a test for the
collinearity of three points whose position vectors are known.
Theorem 2.4 {condition for three distinct points to lie in a line).
Three points А, В, С have position vectors a, b, c. Then (i) if А, В, С lie
on a straight line, there exist numbers cc, β, γ, not all zero, such that
aa + ^b + yc = 0
and cc + β + γ = 0.
Conversely, (ii) if there exist numbers α, β, у not all zero such that
aa+^b + yc = 0
cc + β + γ = 0
С are collinear, suppose ABjBC = λ/μ. Then
. /<a+Ac
i.e. μа-(μ + λ)Ъ + λc = 0.
Take α = μ, β =-{μ + λ),γ = A and the result follows,
(ii) Given that numbers α, β, у exist such that
aa+^b + yc = 0
and α+β + γ = 0
we have aa+/?b = (cc + β) c, by substitution.
Now we know that α, β, γ are not all zero and we may assume, without any
loss of generality, that у φ 0. It follows that α + β Φ 0 and so
_ aa + lb
° ~ α+β '
i.e. с is the position vector of the point dividing AB in the ratio β:χ,
from which it follows, that А, В, С are collinear.
Example 3 (Desargues's Theorem). Two triangles ABC, A'B'C' (not
necessarily in the same plane) are so positioned that AA', BB', С С all pass
through a point V. ВС, В1 С meet at L; CA, С A' at M\ AB, A'B' at N.
Prove that L, Μ, Ν are collinear.
35
and
then А, В, С are collinear.
Proof, (i) Given that A, B,
by the Section Formula

VECTORS AND VECTOR GEOMETRY [2
Denoting the position vectors of the various points in the usual way,
we have, using Theorem 2.4:
ίν + aa + a'a' = 0,
1 1+α + α' = 0;
jv+^b+^'b' =0,
1 l+/?+/?' = 0;
Fig. 2.11
From the second and third pairs of equations
/gb-yc _ /ff'b'-y'c'
β-Ύ ~ β'-Ύ'
and, by the Section Formula, these must both represent 1. Thus
(/?-7)l = fib-yc.
Similarly, (γ —a)m = yc —aa and (a —/3)n = aa —/3b.
i(/S-r)l + (r-a)m + (a-/S)n = 0,
Thus ]
I (/3-r)+(r-«)+(«-/3) = о
and so L, Μ, Ν are collinear.
Ex. 17. If ABC, A'B'C are two triangles not in the same plane and such that
AA\ BB', CC all pass through a point V, prove Desargues's Theorem by showing
that the meets of corresponding sides of the two triangles must all lie on the line
of intersection of the planes ABC, A'B'C
Ex. 18. Prove the Converse of Desargues's Theorem.
36

6] EQUATIONS OF LINES AND PLANES
6. THE VECTOR EQUATIONS OF LINES AND
PLANES
Throughout the present section we shall adhere to the generally accepted
custom that Ρ denotes a variable point whose position vector is r.
A locus is a set of points (in a plane or in space) subject to some
condition; examples are a straight line, a plane, a sphere, the interior of a
sphere, etc. In this section we shall confine our attention to lines and planes.
The equation of a line is the algebraic condition that is satisfied by the
positions vector г of a general point Ρ lying on the line. Similarly for the
equation of a plane. Later we shall meet loci that are defined by inequalities;
the interior of a sphere would be such a locus.
A straight line is completely specified if two points A, В of the line are
given. Suppose that an origin О is taken and that the position vectors of
A and В relative to О are a and b respectively. The equation of the line
AB is the condition satisfied by the position vector г of a general point
on AB. (Of course, if a different origin O' were chosen, a different equation
would be obtained: the equation of any locus depends upon the choice of
origin.) Now, since Ρ lies on AB, or AB produced (Fig. 2.12), AP = AAB,
where A is a number. For different points on the line, different values of A
are taken. This equation may be rewritten in terms of position vectors as
r-a = A(b-a)
which represents the equation of the line AB.
Fig. 2.12 Fig. 2.13
A plane is completely specified by three (non-collinear) points А, В, С
(see Chapter 3, Section 5). For a general point Ρ of the plane, AP is a vector
lying in the plane of AB and AC and so we may split AP into components
in these two directions (remember that А, В, С were non-collinear and so
two directions are defined). Thus АР = ААВ+/гАС (Fig. 2.13) and so,
in terms of position vectors relative to some origin O,
r-a = A(b-a)+/<(c-a)
37

VECTORS AND VECTOR GEOMETRY [2
which may be rewritten as
г = -(λ + μ-1) я + λЪ+μc.
Our final theorem of this chapter is essentially a re-phrasing of the
result above. It is structurally very similar to Theorem 2.4 and the reader
is advised to refer back to that theorem before continuing.
Theorem 2.5 {condition for four points to lie on a plane). Four points A, B,
C, D have position vectors a, b, c, d.
Then
(i) If А, В, C, D lie on a plane, there exist numbers oc, β, γ, δ, not all
zero, such that . „. . , »■,
aa+^b + yc + tfd = 0,
cc + β + γ+δ = 0.
Conversely,
(ii) If there exist numbers oc, β, у, δ, not all zero, such that
aa+^b + yc + tfd = 0,
cc + β + γ + δ = 0
then A, B, C, D are coplanar.
Proof, (i) Given that А, В, С, D are coplanar, d satisfies an equation of
the form ,, , ,. , ,, ,
г =-(λ + μ-l)а + λЪ+μc,
i.e. -(A+/i-l)a + Ab+/<c-d = 0
and the first result follows if we set
α=-(Λ + /ί-1), β = λ, γ = μ, 8 =-Ι.
(ii) Given that numbers ас, β, γ, S exist such that
aa+^b + yc + tfd = 0,
cc + β + γ+δ = 0.
Wehave ~(β + Ύ+δ) a+^b + yc + id = 0,
i.e. β(}> - a)+y(c - a) = £(a - d),
or MB + у AC = δΌΑ.
But AB, AC both lie in the plane ABC and so therefore does δΌΑ. It
follows that D lies in the plane ABC and the proof is complete.
*Ex. 19. Show that the equation of the line through A parallel to the direction
defined by the vector u is г = a + Au.

6] EQUATIONS OF LINES AND PLANES
*Ex. 20. Show that the equation of the plane through A parallel to the directions
denned by the two vectors m, η is г = а + Ат+/ш.
Ex. 21. Show that the mid-points of two pairs of opposite edges of a tetrahedron
ABCD are coplanar.
Example 4. ABCD is the base of a cube whose vertical edges are AA', BB',
CC, DD'. X is the point of trisection of BB' nearer Β, Υ is the point of
trisect ion ofCC nearer С and Μ is the mid-point of ВС.
If D'M cuts the plane AXY at Z, find the ratio in which Ζ divides D'M.
Take side of cube as 6 units and call vectors of unit magnitude in the
directions AB, AD, AA' respectively i, j, k.
Fig. 2.14
Working with A as origin, and denoting the position vector of X by x,
etc., we have
χ = 6i + 2k,
у = 6i + 6j+4k
and so the position vector of any point on the plane AXY is given by
г = λχ+μγ
= 6(Λ +μ) i +6μ\ + 2(Λ +2μ) к.
Again, m = 6i + 3j,
d' = 6j + 6k
and so D'M = 6i-3j-6k.
39

VECTORS AND VECTOR GEOMETRY [2
The position vector of any point on D'M is given by
г = AD'+D'P
= <1' + Л>'М
= 6j + 6k + v(6i-3j-6k)
= 6w + (6-3i>)j + (6-6i0k.
Thus, the point of intersection is given by
6ri + (6-3i0 j + (6-6i0k = 6(A+/<)i + 6/<j+2(A+2/<)k.
But i, j, к are non-coplanar vectors so, by the Uniqueness Theorem,
(ν = λ+μ,
3-3ί> = λ+2μ
and thus v = f.
Thus D'Z:ZM = 4:3.
Exercise 2 (ft)
1. ABC is a triangle. U lies on AB produced so that AB = \AU, V lies on AC
so that AV = 2VC. Find, in terms of the position vectors a, b, с of А, В, С,
the position vectors of U, V and the mid-point of UV.
2. Draw a diagram showing the relative positions of the points whose position
vectors are a, b and 3a-2b.
Prove that these three points are collinear.
3. If G is the centroid of the triangle ABC, prove that
GA+GB + GC = 0.
Suggest a generalization of this result.
4. ABC is an equilateral triangle of side 3 cm. P, Q lie on ВС, С A respectively
and are such that AQ = CP = 2 cm. R lies on AB produced so that BR = 1 cm.
Prove that PQR is a straight line.
5. О ABC is a parallelogram and the position vectors of А, В, С relative to О
are respectively a, b, с Mis the mid-point of ВС. Write down, in terms of b and
c, and hence in terms of a and c, the position vector of a general point X on the
line OM. Deduce that, if X lies on AC, then X is the point of trisection of AC
nearer С and also the point of trisection of OM nearer M.
6. OAB is a triangle; Μ is the mid-point of AB and Τ is the point of trisection
of OB nearer B. TM produced meets О A at X. If OA = a and OB = b, write
down the position vectors of Τ and M, and hence of a general point Ρ of TM.
Deduce the position vector of X and find a relation between the points О, А
andX
40

6] EQUATIONS OF LINES AND PLANES
7. The position vectors of the vertices А, В, С of a triangle are respectively a,
b, с Μ is the point of trisection of AC nearer A, N is the point of trisection of AB
nearer B. Write down the position vectors of Μ and N and hence of general
points of BM, CN. Deduce the position vector of X, the intersection of BM and
CN.
8. ABCDA'B'CD' is a parallelepiped, with parallel edges AA', BB\ CC, DD1'.
U is the point of trisection of AA' nearer A, V is the point of trisection of С1У
nearer C" and W is the mid-point of B'C. With A as origin, the position vectors
of B, D, A' are respectively b, d, a'. Write down the position vectors of U, V, W
and a general point of DD'. Locate the point at which the plane UVW cuts DD'.
9. О UVW is a tetrahedron, the position vectors of U, V, W being u, v, w. A is
a vertex of the parallelogram О UA V, В is the mid-point of VW, С is the reflection
in the origin of the point of trisection of UV nearer U. Find, in terms of u, v, w,
the position vectors of А, В, С and hence find where the plane ABC cuts О W.
10. ABCD is a tetrahedron, Ρ is the mid-point of AB, Q is the mid-point of AD
and R is the point of trisection of AC nearer C. Taking A as the origin, write
down the position vector of the general point X of the plane PQR in terms of
the position vectors of В, С and D. Write down also, in terms of the same frame
of reference, the position vector of the general point Υ of the line DM, where
Μ is the mid-point of ВС
The plane PQR cuts the line DM at Z; find the ratio ZM/MD.
11. Four points P, Q, R, S in a plane through the origin О have position vectors
OP, OQ, OR, OS given by
2i + 3j, 3i + 2j, 4i + 6j, 9i + 6j
respectively, where i and j are given non-parallel vectors. Express the vectors
PR and QS in terms of i and j.
Show that the position vectors OA and OB of the points A and В on PQ and
RS respectively, and such that PA/PQ = a and RBjRS = b, are
(2 + a)i+(3-a)j and (4 + 56) i + 6j,
respectively. Hence determine the position vector with respect to О of the point
of intersection of the lines PQ and RS. (J.M.B.)
12. Two vectors are represented by OP, OQ and R divides PQ in the ratio
*:W. Show that mOP + nOQ = (m + n)OR.
The points D, E, F divide the sides ВС, СА, AB of a triangle in the ratios 1:4,
3:2, 3:7 respectively. Show that the sum of the vectors AD, BE, CF is parallel to
CX, where Xdivides AB in the ratio 1:3. (J.M.B.)
13. ABCD is a skew quadrilateral and a plane cuts AB, ВС, CD, DA at W, Χ, Υ, Ζ
respectively. Prove that AW BX CY DZ =
WB'XC'YDZA
Suggest a generalization of this result.

VECTORS AND VECTOR GEOMETRY [2
14. ABCD is a parallelogram and О is a point in the same plane. OD cuts AB
at Ρ and ВС at Q; OB cuts CD at R and Ζ>Λ at 5. Prove that PS and βΛ are
parallel.
15. О ABC is a square of side 2a. i, j are unit vectors along О А, ОС. The mid-point
of AB is L; the mid-point of ВС is M; OL, AM meet at P; ЯР meets О A at JV.
Show that the segment OP can be measured by the vector A(2ai + aj) and also
by the vector 2αΐ+μ(2α} — αϊ). Hence determine Λ and μ. Prove that ON = $OA.
(O&C)
Miscellaneous Exercise 2
1. ABCDE is a regular pentagon. AB = aandAE = b. Show that
CD = (b-a)/(l+2cos72°)
and express ВС and ED in terms of a and b.
2. ABC is a triangle. L divides ВС in the ratio 1:2, Μ divides С A in the ratio
1:2 and N divides AB in the ratio 1:2. Prove that the triangles ABC and LMN
have the same centroid.
Suggest a generalization of this result and prove your assertion.
3. Referred to some origin O, the position vectors of A, B, C, D are a, b, c, d
respectively. Express in terms of a, b, c, d the displacements BD, AC. What can
be said about the quadrilateral ABCD:
(i) Ifa + c = b + d;
(ii) if |a-c| = |b-d|;
(iii) if both (i) and (ii) hold?
4. The triangles ABC, A'B'C have centroids respectively at G and G'. Prove
that AA' + BB' + CC = 3GG'.
5. P, Q are the mid-points of the sides ВС and CD respectively of the
parallelogram ABCD. Prove that:
AB + AC + AD =f (AP + AQ).
6. ABCD is a tetrahedron and B', C, D' lie on AB, AC, AD produced. The
centroids of the triangles BCD', B'CD', B'C'D are Gu G2, G3 respectively and the
centroids of the triangles B'CD, BCD, BCD' are Hu H2, H3 respectively. If
the centroids of the triangles BCD, G&Gs, НгЩН3 are F, G, Η respectively,
prove that FGH is a straight line.
7. ABCD is a parallelogram and points X, У are taken on the diagonal BD such
that BX = YD. Prove vectorially that AXCYis a parallelogram.
8. ABC is a triangle and Υ, Ζ are points on AC, AB respectively such that ZY
is parallel to ВС. В Υ and CZ meet at P. Prove vectorially that AP produced
bisects ВС.
9. ABC, A'B'C are two skew lines (that is, no plane contains both lines). If
AB:BC= A'B'.B'C prove that the mid-points of the lines AA', BB', CC
are collinear. Is the converse true?

6] MISCELLANEOUS EXERCISE 2
10. Prove that, if ABC is a plane through the origin О from which position
vectors are measured, and if А, В, С have position vectors a, b, c, then there
necessarily exits a relation of the form
pst+qb+rc = 0 (/ι,ί,ΓφΟ)
where 0 is the zero vector.
The lines АО, BO, CO meet ВС, CA, AB in L, Μ, Ν respectively. Prove that
the position vector of L is
-f b + ^-c.
q + r q+r
where the magnitude and sense of each line segment is taken into account. (M.E.I.)
(The result proved in Question 10 is known as Ceva's Theorem: it and its
converse are very useful for proving concurrency theorems.)
11. If a transversal cuts the sides ВС, С A, AB of a triangle ABC at L, Μ, Ν
respectively, prove that
BL.CM.AN
magnitudes and senses of each line segment being taken into account.
(This result is known as Menelaus's Theorem: with its converse it is useful for
proving collinearity properties.)
12. ABCD is a plane quadrilateral. AB and DC meet at P; ВС and AD meet at Q.
Prove that the mid-points of AC, BD and PQ are collinear.
13. P, Q are variable points on two skew lines. Find the locus of the mid-point
ofPQ. Can you generalize this result?
14. ABCD A'B'CD' is a parallelepiped, with ABCD, A'B'C'U congruent
parallelograms and AA', etc., edges. The tetrahedron ACB'D' is inscribed in the
parallelepiped. Prove that AC passes through the centroid X of B'CD' and deduce
that the joins of the vertices of the tetrahedron to the centroids of opposite faces
are concurrent at a point G. Determine the ratios AG: GX: XC.
15. Two vectors a and b, such that b is not a multiple of a, are given in a plane.
Two other vectors с and d are defined by the equations
с = 7ia + y2b,
d = ^a + ^b.
Prove that any vector aa+/Sb can be expressed in terms of с and d provided
УгК-УгК Ф 0. Find the coefficients of с and d.
Explain the geometrical significance of the condition yx S2 - γ2 8г Ф 0. (О & С)

VECTORS AND VECTOR GEOMETRY [2
16. Let О, А, В, С be four distinct points in three-dimensional space, no three
of which are collinear. The position vectors of А, В, С with respect to О are
a, b, с respectively and Xis the point given by χ = Aa+/A> + vc. Prove that:
(i) If X is a point of the line AB, then Α +μ = 1, ν = 0.
(ii) If X is a point of the plane ABC, then λ + μ + ν = 1.
(in) X is in the interior of triangle ABC if and only if λ+μ+ν = 1 and
λ>0, μ > 0, ν > 0 and indicate on a sketch the regions in the plane ABC in
which X lies for other combinations of the signs λ, μ, v. Obtain an expression for
the position vector χ of a general point X in the interior of the tetrahedron
О ABC and find the values of λ, μ, ν corresponding to the centroid. (M.E.I.)

3· Coordinates
1. UNIT VECTORS
A unit vector is a vector of unit magnitude. Thus, a unit vector is a sort
of 'signpost' giving a direction and sense; any vector may be split into
the product of a number (its magnitude) and a unit vector with the required
sense and direction. Thus we may write
г = rt,
where r = |r| and ί is a unit vector in the direction of r. (The notation
employed here is useful and should be adopted by the reader: given any
vector x, its magnitude may be written simply as χ while a unit vector
with the same sense and direction as χ is written x.) Unit vectors are useful
in that they enable us to deal separately with the magnitude and direction
of a given vector.
Now suppose we choose two directions in a plane and specify them by
unit vectors i and j. By the Existence and Uniqueness Theorem for a
plane (see Chapter 2) any displacement AB in the plane, represented by
d may be split into components in the i and j directions
d = pi + qi
where ρ and q are uniquely determined. The vector d, which is
representative of a whole class of displacements, of which AB is a typical member,
may be written in the alternative form
■Ю-
Q the vector i and l_\
Thus, (I represents the vector 2i-j, the vector i and I _ I the
vector - j. Of course, the representation of a vector in this new notation
depends upon our original choice of base vectors i and j. Again, there is
no necessity for the base vectors to be unit vectors, though they will almost
invariably be so. To illustrate these last two remarks, consider two unit
vectors i and j and the vector a where
a = 3i-j.
Now a = (i+j)+2(i-j)
and so we have .
a = I _ I with i, j as base vectors,
45

COORDINATES
with (i + j), (i - j) as base vectors.
■®
Notice that (i + j) and (i-j) are not unit vectors. (For example, if i and j
are perpendicular, neither (i+j) nor (i-j) is a unit vector.)
Similarly, in three dimensions, if non-coplanar base vectors i, j, к
are chosen, any vector d may be split into three components in the i, j, к
directions . ...
d = pi+q} + rk
and d may be represented in the alternative notation as
■Θ
relative to the base vectors i, j, k.
We shall now adopt the convention that i, j, к represent unit vectors
each one of which is perpendicular to the other two. Furthermore, positive
senses are determined by the following rule: if the thumb, first and second
fingers of the right hand are splayed out at right angles to one another, the
thumb points in the positive j direction, the first finger in the positive j
direction and the second finger in the positive к direction, i, j, к are then
said to form a right-handed orthogonal triple of unit vectors. (Orthogonal
means 'at right-angles to one another'.) Similarly, but more simply, one
may define a right-handed orthogonal pair of unit vectors i, j for a plane.
Example 1. If л = i + 2j + 3k, express л as a column vector
(i) with i, j, к as base vectors;
(ii) with i + j, i — к, к — 2j as base vectors.
(i) With i, j, к as base vectors,
|2|.
■(!)■
(ii) Write u = i + j, ν = i-k, w = k-2j.
Solving for i, j, к we have
i = i(2u+v+w); j = |(u-v-w); к = §(2u-2v+w).
Thus, a = i+2j + 3k
= (|u+iv+iw) + (fu-fv-|w) + (2u-2v + w)
= ^u-Jv + fw
46

1] UNIT VECTORS
and we have, with u, v, w as base vectors,
I
Ex. 1. Rewrite 3i + 4j-k and (i + j) - 2(j - 2k) in the column vector notation,
with i, j, к as base vectors.
Ex. 2. Express 3i-j-k as a column vector with (j + k), (k + i), (i + j) as base
vectors. Where do you use the Uniqueness Theorem in your solution?
Ex. 3. Evaluate a + b, 2a-3b and 3(a-2b) where
•(ΐ)
(·)■ ·■{:
3-Q
what are the values of χ and yl Explain how you make your deductions.
2. COORDINATES
Suppose we now have some geometrical configuration in a plane which we
want to describe algebraically. Let us choose any point О of the plane as
origin; then all the points of the plane may be specified by their position
vectors with respect to O. To give more detailed information, these position
vectors may be split into their components in the direction defined by the
right-handed pair of orthogonal unit vectors i, j.
Thus, if r = op
we may write г = xi+yj.
χ and у are called, respectively, the χ and у
coordinates of the point Ρ and are usually
written as (x, y). The lines through О in the i
and j directions are called the χ and у axes j ψ |
respectively, and may be denoted by Ox and Oy.
In Figure 3.1 the following points have been '
plotted:
A(l, 3); 5(3,-1); C(-l, -2); Z>(-3,0).
Note that the χ coordinate is always written first.

COORDINATES [3
The reader should note the distinction between (1,3) and I I: the
first gives the χ and у coordinates of the unique point A, while the second
denotes the vector (referred to i, j as base) representative of the class of
displacements of which OA is a typical member.
Coordinates may similarly be defined in three dimensions, though now
we require a right-handed triplet of orthogonal unit vectors i, j, к to
define the directions of the coordinate axes OX, ΟΥ, ΟΖ through the
origin O. Thus, if Ρ is the point (2,-1, - 3), the position vector of Ρ is
2i —j —3k. Note again that the coordinates are given in the order x, y, z.
Ex. 5. Draw a sketch to denote the approximate positions of the points:
/4(1,0), B(-2, -1), C(l, -3), Z>(-2,0), E(-l, -2).
Ex. 6. Write down the position vectors of the points
A(l, -2), 5(3,4).
Deduce the coordinates of the mid-point of AB. Can you state a general rule for
finding the coordinates of the mid-point of a line ?
Ex. 7. Write down the position vectors of the points
Ail, -1), «5,-5).
Use the Section Formula to deduce the coordinates of the two points of trisection
of AB.
Ex. 8. What are the coordinates of the mid-point of the line joining
ДО,-1,2) and 5(2,3,2)?
Ex. 9. If ABCD is a parallelogram, what relation must hold between the position
vectors a, b, c, d? Find the coordinates of the vertex D of the parallelogram
ABCD whose other vertices are given by
/4(1,2), 5(4,3), C(3,5).
Ex. 10. The coordinates of the points А, В, С are as follows:
/4(-l, 1, 2), 5(1, 0, -3), C(0, 2, 4).
Find the coordinates of the fourth vertex D of the parallelogram ABCD and
of the fourth vertex Ε of the parallelogram ACBE.
Ex. 11. Find the coordinates of the centroid of the triangle ABC of Ex. 9.
Ex. 12. A rectangular box ABCD A'B'C'D' has base ABCD and the edges AA',
BB', CC, DD' are all vertical. Μ, Ν, Ρ are the mid-points of ВС, СС and
A'B' respectively. AB has magnitude 3 units, ВС 2 units and CC 1 unit. If
A is taken as origin and unit vectors i, j, к are taken along AB, AD, AA'
respectively, find the position vector of the centroid of triangle MNP and deduce its
coordinates.
48

3] DISTANCES IN TERMS OF COORDINATES
3. DISTANCES IN TERMS OF COORDINATES
Before proceeding with our study of coordinates we shall develop an
economic notation whose value should be readily apparent to the reader.
Tt would be convenient to refer to all points by using, say, the letter P, to
all χ coordinates by x, to all у coordinates by у and to all ζ coordinates
by ζ but to do so without further clarification would clearly lead to
appalling confusion. We may, however, differentiate between the various points
and their coordinates by the use of suffixes : thus we may call two points Рг
and P2 and take as their coordinates (хг, у1г гг) and (x2, y2, z2) respectively.
One advantage of such a notation is obvious: without further explanation
we know that Pn stands for a point, that x„ is its χ coordinate and so on.
(The reader must guard against confusion between suffixes and indices:
x2 simply means the χ coordinate of the point P2, whereas x2 represents
the result of multiplying the number χ by itself.) The choice of a good
notation is often more than half the battle in the solution of a mathematical
problem; as he gains experience in its use, the reader will come to
appreciate the deeper significance of the suffix notation. One evident further
advantage may, however, be noted here: the use of suffixes effects a
considerable economy in notation if we have to deal with a large (or
indeterminate) number of points. (It should be noted that not all sets may be
enumerated, that is, counted in the form 1, 2, 3, .... In such cases, the
suffix notation as we have presented it breaks down. Consider, for example,
the problem of naming all the points on the line segment joining /4(0, 0)
and 5(1,0).)
Now consider a plane and two points Рг and P2 lying in it whose
coordinates relative to perpendicular axes through some origin О are (хъ уг)
and(x2,y2).
\'P^ dfi
o\
Fig. 3.2
From Figure 3.2 we have
P^ = P2Q + QP1
= (*!-**) i+(ft - Л) J·
Thus, applying Pythagoras's Theorem to the right-angled triangle Р2Рг Q,
ΛΑ = ν[(*ι-*2)2+ϋΊ-Λ)2]·
49

COORDINATES [3
For three-dimensional coordinates the same argument applies but
Pythagoras's Theorem must be applied twice. In Figure 3.3, the feet of the
perpendicular from Р1г P2 to the plane Oxy are Q1} Q2, and RQ1; Q2R
are respectively parallel to Oy and Ox. Thus we have
Ρ2ΡΧ = P2Q + QP1
= QaR + RQi + QPi
= (^ι-^ί + ϋΊ-^) J + Czi-^k
and so PXP2 = J[(Xl-x2y + (уг-у^Нъ~*№
on applying Pythagoras's Theorem to triangles PXP2Q and Q1Q2R.
Fig. 3.3
Example 2. (i) The distance between /4(3, — 1) and B(— 1, 2) is
V[(3- -l)2 + (-l -2)2] = 5.
(ii) The distance between /4(1, 0, -2) and B(-l, 2, -3) is
VK1- -l)3 + (0-2)2 + (-2--3)*] = 3.
Ex. 13. A, B, C, D, E, Fhave coordinates as follows:
Д-1, -2), Я(-3, 2); C(-l, -2, 4), D(0, -4, 3);
E(a, -a, 0), Д-А:, -A:, A:).
Find the lengths of AB, CD and EF.
Ex. 14. Show that ABCD is a rectangle, А, В, С, D having coordinates:
/4(1,2), Ж5,-1), C(8,3), Z>(4,6).
Ex. 15. What is the fourth vertex of the parallelogram ABCD where
A is (1,1, 2), Л(2, 0, -1), C(3, 3, 0) ?
Ex. 16. Find the circumcentre of the triangle
/4(0, -1, 1), B(\, 0, -1), C(-1,1, 0).
What is the circumradius ? (The circumcentre of a triangle is the centre of the
circle which passes through the three vertices.)
50

DISTANCES IN TERMS OF COORDINATES
Exercise 3(a)
. Evaluate as single column vectors:
(i)u)ii); (">2(i+3(J)^
Ч:)+4В; _ <w(i)+3(:i)+2(;);
MiHH-i)- _
ίπ -(Э'ЧЗ-е-В·
find, in column vector form, the vectors:
(i) 2a-b; (ii) a+b-3c; (iii) (3a + b)-2(b+3c).
3. If a, b, с are defined as in Question 2, solve for χ the equations:
(i) 2x = a-b; (ii) 3(x + a) = b-2c
giving your answers in column vector form.
4. Again using the notation of Question 2, solve for x, y, ζ the following systems
of linear equations, giving your answers in column vector form:
= 2a + 2b,
® Γ~, η к + 2Ь' ® {2x+y+z = a + 3b + 2c,
x + 2y = b-c;
l»-«-i·» - 4a-4c.
f x + y-z =
) Ьх+y+z =
lx-y-3z =
5. Two vectors p, q (where рФ4д) are given in a plane. Referred to p, q as
base, the vectors a, b, c, d are given by:
Express a, b, c, d as column vectors, taking p—2q, p + 2q as base.
6. Three non-coplanar vectors u, v, w are given. Referred to these vectors
as base vectors, the vectors, e, f, g, h are given by:
-0-г-0--0-к-4П-
Express e, f, g, h as column vectors taking ν+w, w + u, u + v as base vectors.
7. Write down the coordinates of the mid-points of the lines joining the following
pairs of points:
(i) (2, 4), (4, 6); (ii) ( - 2, 4), (4, 2); (iii) (-1, - 3), (2, - 4);
(iv) (3, 1, -5), (2, -4, 0); (v) (a + b, a, a-b), (a-b, -a, -a-b).
51

COORDINATES [3
8. Write down the coordinates of the two points of trisection of the lines joining
each of the following pairs of points:
(i) (1, 4), (4, 10); (ii) (-2, 1), (1, 8); (iii) (a+2b, 6-2a), (a-b, a + b);
(iv) (1, 2, -1), (2, 1, -5); (v) (a, 2a, b), (a-b, b-2a, a+b).
9. Find the lengths of each of the line segments of Question 7. (Leave your
answers in the form Jm.)
10. А(Ъ, 0), B(4, -1), C(6, 2) are vertices of a parallelogram ABCD. Find the
coordinates of D. Find also the coordinates of Ε if ACBE is a parallelogram.
11. Prove that the triangle whose vertices are A(-\, 2), B(3, 5), C(-4, 6) is
isosceles. What is its area?
12. А(Ъ, 1, -1), B(\, 2, -2), C(0, 0, 2) are vertices of a parallelogram ЛЯС£>.
Find the coordinates of D. What are the coordinates of the meet of the diagonals ?
13. Find the area of the triangle whose vertices are A(\, -3), B(2, 5), C(-4, -3).
14. Show that the line joining A(-1, 4, -3) and B(5, -8, 6) meets the χ axis.
If the point of intersection is C, find the ratio AC/CB.
4. COORDINATE GEOMETRY IN A PLANE;
STRAIGHT LINES AND THEIR GRADIENTS
Throughout this section we shall confine our attention to a plane, so that
only two coordinates are required to specify a point.
The reader will recall that the vector equation of a straight line is the
algebraic condition that must be satisfied by the position vector г of a
point Ρ if Ρ is to lie on the line. In the same way, the Cartesian equation
of a straight line is the algebraic condition that must be satisfied by the
coordinates x, y, of a point P(x, y) of the line (and, conversely, no point
Ρ not on the line has coordinates satisfying the equation).
The equations of lines parallel to the coordinate axes may readily be
obtained. For example, the reader should have no difficulty in seeing that
the line through (-3, 1) parallel to the χ axis is у = 1; similarly, the line
through this point parallel to the у axis is x + 3 = 0. In particular, the
equations of the χ and у axes are respectively у = 0 and χ = 0.
A straight line is determined completely by two points on it. Suppose
now that we wish to find the Cartesian equation of the straight line
through P-Sx-l, л) and P2(x2, y^, where xx =t= x2 and уг Ф y2 and so the
line is not parallel to the axes. Let the position vectors of Рг and P2 be тг
and r2 respectively. Then (see page 37) the vector equation of the line
where rx = Xii+Jij and r2 = х21+У21·
52

И COORDINATE GEOMETRY IN A PLANE
This may be rewritten in column vector notation (with i, j as base)
■©♦
\У2-У1/
By the Uniqueness Theorem in two dimensions this gives the equations
jx-x1 = λ(χ2-χ1),
Ь-У1 = КУл-уд-
Eliminating A between these two equations we obtain the following result.
The Cartesian equation of the line joining the points Рг{хъ уг) P2(x2, y2) is
^xx = Lzy1_
*2~*i У2-У1
(The reader must note carefully the distinction in this equation between
x, у on the one hand and xlt ylt x2, y2 on the other: x, у are the coordinates
of a general point Ρ on the line; xlt ylt x2, y2 are the coordinates of two
specified fixed points of the line. If we take U as the set of all points
(x, y) in the plane, the equation (1) is the defining condition for the set of
points comprising the line; that is, the set
I "' У-Уг yt-yii
The same remarks hold for the vector equation г = тх + А(г2 —rj), in which
г15 r2 are the position vectors of two given points.)
Equation (1) may be rewritten in the form
where η
The number m is called the gradient of the line. The sign of the gradient
tells us which way the line is sloping (see Figures 3.4 and 3.5). In case (i),
хг — х2 and уг—y2 have the same sign and m > 0; in case (ii) χΎ — χ2 and
У1—У2nave opposite signs and m < 0. In both cases, the positive value of
m is equal to tan a, where a is the acute angle made by the line with the
χ axis. Equation (2) gives the form for the straight line when a point on the
line and its gradient are known.f
t The reader who has already met trigonometric ratios of obtuse angles will realize
that if the straight line AB meets the xaxis at P, then the gradient of AB is tan ΔχΡΑ in
all cases, where ΔχΡΑ is measured in the positive (anticlockwise) sense from the χ axis
(see Chapter 6).

COORDINATES
о\ ^
О] _\
Equation (2) may be rewritten as
= mx + c.
(3)
Here the constant с represents the intercept cut off on the у axis, as may
easily be seen by setting χ = 0.
Having derived these various forms for the equation of a straight line,
we may easily deduce the result that any equation of degree 1 in the two
variables x, у represent a straight line (hence the word linear usually
applied to such equations).
Consider the general linear equation in the two variables x, y:
ax + by + c = 0.
If a = 0 or b = 0 this represents a line parallel to one of the axes. If
b Φ 0, the equation may be rewritten in the form
and comparison with (3) shows that this is the equation of a line with
gradient — ajb and making an intercept —cjb on the у axis.
Example 3. Find the equation of the line L^ through (2, — 1) with gradient
— \ and the equation of the line L2 joining the points (0, 7) and (-1, 4).
What are the coordinates of the intersection ofL^ andL^ ?
Draw a sketch showing the relative positions ofl^ andL2 and the
coordinate axes.
Lj has equation y+1 = —Цх-2),
i.e. x + 2y = 0,
54

4] COORDINATE GEOMETRY IN A PLANE
L2 has equation . _fl = TZPj'
i.e. 3x-j> + 7 = 0.
Since all points lying on A satisfy the x + 2y = 0 and all points lying on
L2 satisfy Ъх-у + Ί = 0, the point of intersection of the two lines must
satisfy both equations simultaneously.
Solving
we have
fx + 2y =
Ux-y =
χ = -2,
= 1
and so the point of intersection of Lj and
£,is(-2,l).
(In this case the sketching of the lines Lj
and Z^ is straightforward: we know two points
on L2, and Lj clearly passes through the origin.
In general, to sketch the position of a straight (—2,1)"
line whose equation is given, first find the ~~'
coordinates of the points where it cuts through ц
the coordinate axes.)
Ex. 17. Draw a sketch to show the positions of the straight lines:
(i) 3x-2y-6 = 0; (ii) 2x+y + 3 = 0; (in) 2x-5y = 0.
Ex. 18. Find the equations of the lines through (2, —3) with gradients:
(i) 2; (ii) -i.
Ex. 19. Find the equations of the lines joining the pairs of points:
(i) (2,-3), (3,2); (ii) (0,1), (2,-1);
(iii) (-3, -1), (-1, 2); (iv) (1, 2), (-3, 2).
Ex. 20. What are the gradients of the following lines:
(i) 2x-y-3 = 0; (ii) x+y+l = 0; (iii) 3x + 4y+2 = 0?
Suppose now we have two lines, L^ and L2, the gradients of which are
respectively m1 and m2.
(i) If Li and L2 are parallel, the angles that they make with the χ axis
are equal and
m1 = m2.
(ii) If Lj and L2 are perpendicular then either they are parallel to the
coordinate axes or they have gradients of opposite sign: in the second case,
55

COORDINATES [3
if m1 = + tan <xj and m2 = + tan oc^ then mxm2 = —tan ax tan a2 = — 1,
i.e.
WjWa = —1.
*Ex 21. With the notation above, prove that, if Wi = гщ, Lx and L2 are parallel.
*£x. 22. Again with the notation above, prove that, if т^пц. ——\,Ь^ and L^ are
perpendicular.
Example 4. iW ί/ге equations of the lines Ly, through (1, -1) parallel to
Ъх-у + Ί = 0 and L2, through (2, -3) perpendicular to 2x + 5y+\ = 0.
Find also the equation of the line L^ joining the origin to the intersection
ofLx andL2.
The gradient of 3x-y + 7 = 0 is +3 and so Lj has equations
y+l =3(x-l),
i.e. 3x-y-4 = 0.
The gradient of 2x + 5y+1 = 0 is —| and so the gradient of a
perpendicular line is +f. Thus, the equation of L2 is
y + 3 = i(x-2),
i.e. 5x-2j-16 = 0.
(With practice, the reader will soon be able to derive such equations
more rapidly than is done here. For example, in the second case, any line
perpendicular to 2x + 5y+l = 0 is clearly of the form 5x-2y = к and,
since the required line passes through (2, -3), к = 5 (2)-2 (-3) = 16.)
Now consider the equation
(3x-y-4) + A(5x-2y-l6) = 0,
where A is any number. This is a linear equation and so represents a
straight line; furthermore, the point common to Lj and L2 clearly satisfies
this equation. Thus
(3x-y-4) + A(5x-2y-l6) = 0
56

4] COORDINATE GEOMETRY IN A PLANE
represents a straight line through the intersection of Lj and L2. If it passes
through the origin, (0, 0),
-4 + A(-16) = 0,
A=-i
and the equation simplifies down to
lx-2y = 0
which is the required equation of L3.
Ex. 23. Find the equations of the lines Lu through (4, —10) parallel to Ъх—у = 0
and L2, through (1, 2) perpendicular to2x—y—\ = 0. What are the coordinates
of the intersection of Lx and L^ ?
Ex. 24. Find the line joining the point (—1, 2) to the meet of the two lines
x+3y-l = 0,x-4y + 2 = 0.
Ex. 25. Find the equation of the line through the meet of the two lines
3x-y+l = 0, 4x-3.y+2 = 0
perpendicular to the line Ъх-у-1 =0.
Ex. 26. Find the intercept cut off on the transversal 5x—2y+ 3 = 0 by the two
parallel lines 3x-y + 1 = 0 and 3x-y+4 = 0.
Exercise 3(b)
1. Find the equations of the lines through the stated points with the given
gradients:
(i) (0, -2), 2; (ii) (-1, -1), -2; (iii) (3, -1), -}; (iv) (a, b), -alb.
2. Find the equations of the lines joining the following pairs of points:
(i) (1, -2), (2, 1); (ii) (-2, -3), (-1, 4); (iii) (2, -1), (-1, -1);
(iv) (a, b), (2a, -Щ.
3. Write down the gradients of the following lines:
(i) 2x-3.y+l = 0; (ii) Зх+б.у+2 = 0; (iii) 5x-y + l = 0;
(iv) (a + b) x+(a-b) y+ab = 0.
4. Find the equations of the lines through (3, — 1) (a) parallel and (b)
perpendicular to:
(i) 2x-y+3 = 0; (ii) 5x+4.y + 3 = 0; (iii) x+7 = 0.
In case (i), what is the distance between the two parallel lines ?
5. Find the points of intersection of the following pairs of lines:
!4x+3y-6 = 0, lax+by + a2 = 0,
(i) I (ii) I
U-.y-5 = 0; (bx-ay+b2 = 0.
6. The two lines 3x- 5y-7 = 0, 4x+ 4y+5 = 0 meet at the point A. Find the
equations of the line through A
(i) which passes through the origin;
(ii) is parallel to 7χ - у + 2 = 0;
(iii) is perpendicular to 3x+5.y-l = 0.
57

COORDINATES [3
7. Find the orthocentre of the triangle whose vertices are (0, 1), (1, 2), (4, 3). (The
orthocentre is the meet of the altitudes of a triangle.)
8. The mid-points L, Μ, Ν of the sides, ВС, С А, АВ of a triangle ABC have
coordinates (2, 1), (3, -3), (4, -5). Find the coordinates of А, В, С
9. The coordinates of the vertices of a triangle are (6, 0), (-1,1) and (5, -7).
Find the coordinates of the centre of the circumcircle.
10. The coordinates of the vertices of a triangle are (1, -4), (3, -2) and
( — 11, 12). Find the coordinates of the centre of the circumcircle and
determine the coordinates of the points where this circle cuts the axes.
11. Show that, for all values of λ, the line whose equation is λχ+y = 1—2A
always passes through the point (-2, 1). What is the equation (in terms of A)
of the perpendicular line through (1, 2)? Show that, whatever the value of λ,
the intersection of these two lines always satisfies the equation
x2 + f + x-3y = 0.
What locus does this last equation represent?
12. О ABC is a rectangle in which О А = ЪОС. Μ is the mid-point of ОС and L
is the point of trisection of CB nearer C. OL, AM meet at X. By setting up axes
along О A and ОС and assigning suitable coordinates to the various points,
determine the ratio in which X divides OL.
13. Prove that, for all values of A, the line
(1-A)x + Ay = 3-7A
passes through a fixed point. What are the coordinates of the point?
14. Find the coordinates of the centre of the circumcircle of the triangle ABC,
where А, В, С have coordinates (-1,-3), (-2,-2), (5,5) respectively.
Prove that the point P(6, 4) lies on the circumcircle and prove further that the
feet of the perpendiculars from Ρ on to ВС, CA, AB are collinear.
15. Find the equation of the line joining A(a, 0) and B(0, b). A', B' are the feet
of the perpendiculars from A, B to a variable line through the origin. If A'P, B'P
are respectively parallel to the у and χ axes, what can be said about the position
ofP?
16. What are the equations of the reflections of the line 2x - Ъу+6 = 0 in
(i) the χ axis; (ii) the у axis;
(iii) the line lx-Ъу = 0; (iv) the line 2x-3y = 3?
5. COORDINATE GEOMETRY IN SPACE:
THE PLANE AND STRAIGHT LINE
The effective use of Cartesian coordinates in three dimensions requires
rather more vector technique than we have at present at our disposal and
so a more detailed study will be deferred until a later chapter (Chapter 11).
58

5] COORDINATE GEOMETRY IN SPACE
We shall content ourselves for the moment with demonstrating the general
form of the Cartesian equations of lines and planes in space.
We may define a plane as a set of points in space which
(i) contains at least three non-collinear points;
(ii) has the property that, given any two points Rx and R2 of the set,
all points of the line РгР2 belong to the set.
From this definition, three non-collinear points
clearly define a plane. If тг, г2, r3 are respectively the position vectors of
Pj, P2, P3, we have seen that the plane РгР2Р3 is (see Chapter 2.6)
r=r1+A(r2-r1)+H'3-'i)·
In column vector form this may be written
lX\ (Xl\ (Х*~ХА (χ3-χΑ
which, by the Uniqueness Theorem, yields three equations:
χ = Xl +λ(χ2 - Xl) +μ(χ3 - χ-,),
У = Ух + ЧУг ~ Уд +/*0>з - Уд,
ζ = ζί + λ{ζ2-ζί) + μ{ζ3-ζ1).
Solving the first two equations simultaneously we obtain A and μ as linear
expressions in χ and y. Hence, substituting these values of A and μ in the
third equation, we see that the Cartesian equation of a plane is linear in
x, y, z.
Now let us prove the converse result; a linear equation in x, y, ζ
represents the equation of a plane.
Consider the equation ax+by + cz+d = 0 where we shall suppose,
for simplicity, that α φ 0, b φ 0, с φ 0, d Φ 0. (The fact that one or several
of them may be zero requires a modification of the proof given below;
see Ex. 30.) The three points {-dja, 0, 0), (0, -djb.O), (0,0, -djc)
certainly all satisfy the equation and are non-collinear (since the first two
have zero ζ coordinate, for example). Thus condition (i) for a plane is
satisfied.
To demonstrate condition (ii), suppose Рг and P2 satisfy the given
equation. Then
(■t+by-i+cz-L + d = 0,
z2 + d = 0
(ax^by^cz-L + d
lax2 + by2 + cz2 + d

COORDINATES
[3
and thus, for any values of к and /
But, by the Section Formula,
_ lr1 + kr2 _ (Ιχ,+кхЛ (ly1+ky2\ (lz1 + kz2\
r- k+i -[ k+i )1+[ k+i )i+[ k+i ;k
is the position vector of a point on the line joining P1 and P2. Thus, the
coordinates of anypointon the line joiningPjPa satisfies the given equation.
In three dimensions two intersecting (that is, non-parallel) planes define
a straight line and we should therefore expect a line to be represented by
two linear equations in x, y, z.
The vector equation of the line through P, parallel to the vector
u = /i + wj + лк
is (see Chapter 2) г = тг + Au.
In column vector form this reads
(Η3+ΛΘ
and so, by the Uniqueness Theorem, we have three equations:
χ = Xj+Λ/, у = y-L + Xm, ζ = Zj + Ли
giving the equation of the line in the form
*Zii=Zza = ?Zli = A. (4)
I m η ν '
We use this form even if one or more of /, m, η are zero. For example,
the line through the point (1, -1, 2) in the direction of the vector 2i -3k
may be written as
x-l y+l z-2
x-l z-2 , , „
Τ" = Гз-' '+1=0·
Notice that the Cartesian coordinates of a second general point of the
line (4) may be expressed in terms of one variable as
(*! + A/, y1 + Am,z1 + An).
A is called a parameter for points of the line.
60

5]
COORDINATE GEOMETRY IN SPACE
The vector u defines the direction of the line and may be termed a
direction vector. Since u = /i + mj+ик, /, m, η are proportional to the
cosines of the angles that u makes with the directions i, j, k, that is, with the
directions of the coordinate axes. /, m, η are usually called direction
ratios for the line; if u is a unit vector, the constant of proportionality is
one and /, m, η are called direction cosines (see Figure 3.8, where / = cos дъ
m = cos θ2, η = cos θ3).
Fig. 3.
Example 5. Find the equation of the plane through /4(1,1,1), 5(0, 1, -2),
С (0,0, -1) and the point of intersection of this plane with the line
x-3 =y-2 = z-2
1 -2 3"
Let the equation of the required plane be
ax+by + cz + d = 0.
Since the coordinates of А, В, С all satisfy this equation,
(a + b + c =-d,
b-2c =-d,
which gives a = —3d, b = d, с = d.
The equation of the plane ABC is thus
Ъх-y-z-l = 0.
Any point of the given line has coordinates
(3+A, 2-2A, 2 + 3A).
This lies on the plane if
3(3+A)-(2-2A)-(2 + 3A)-l = 0
giving A = - 2
and so the point of intersection has coordinates
(1, 6, -4).

COORDINATES [3
Example 6. Two straight lines are given parametrically by
L1:x=l-4A, у = l+λ, ζ = l+λ
and L2: x = 2μ, у = Ι —μ, ζ = 2 + μ.
Prove that L± andL2 intersect, and find the equation of the plane containing
the two lines.
If we can choose Α, μ such that the three equations
1-4Λ = 2μ,
1+Λ = l-μ,
l+λ=2+μ
are simultaneously satisfied, then L·^ and L2 intersect. By inspection,
^ = i> /* = — 2 fulfils the required condition (giving the point of intersec-
tion(-l, if)).
Eliminating λ (a) between the χ and у coordinates of a general point of
Li, and (b) between the у and ζ coordinates, we see that the two planes
x + 4y = 5, y-z = 0
both contain Ly. It follows that, for all values of k, the equation
(x + 4y-5) + k(y-z) = 0 (1)
represents a plane containing L·^.
By setting μ = 0, we see that the point (0, 1, 2) lies on L2 (but not on
Ц). This lies on the plane (1) if
(0 + 4-5)+A:(l-2) = 0,
i.e. k=-l.
The equation of the plane containing L·^ and L2 is thus seen to be
x + 4y-5-l(y-z) = 0,
i.e. x + 3y + z-5 = 0.
Ex. 27. What is the equation of the plane
(i) containing Ox and Oy\
(ii) through (1, -1, 2) perpendicular to Ozl
Ex. 28. Describe the position of the plane ax+by = 0.
Ex. 29. Why was condition (i) included in the definition of a plane?
*Ex. 30. Show that three non-collinear points may be found on
ax + by + cz + d = 0
62

5] COORDINATE GEOMETRY IN SPACE
in the following cases:
(i) d = 0, α φ 0, b φ 0, с Φ 0; (ii) d φ 0, a = b = 0, с Ф 0;
(iii) rf = α = 0, b φ 0, с φ 0; (iv) a = 0, b Φ 0, с φ 0, d ΦΟ.
Ex.31. Find the equation of the plane through the line of intersection of the
planes x-y + 2z+l =0, 2x+y-z + 2 = 0 which contains the origin.
Ex. 32. What are the coordinates of a general point of the line
x-3 _y + 2 _ζ + 5Ί
2 -1 ~ -3'
Where does this line cut the plane x+y + z + 2 = 0?
Ex. 33. What are the equations of the line joining (-3, 1, 1) and (2, 2, -1)?
Ex. 34. Find the equation of the plane containing the origin and the line
x-2 = ^-3 = z+4
2 3 -Γ
Exercise 3(c)
1. Find (a) direction ratios, (6) the direction cosines, for the lines joining the
following pairs of points:
(i) (1,2,3), (2, 3,4); (ii) (2,-1, 3), (1, 1, 2);
(iii) (3,1, 2), (5, -1, 1); (iv) (1, 3, 5), (5, 3, 1);
(v) (2, -3,-4), (-3,2,1);
(vi) (a + λ, 2a + 2λ, За + 3λ), (я-μ, 2α-2μ, 3α-3/ί);
(νϋ) (α2, αί>; δ2), (α2, ас, с2); (viii) (1, я, α2), (α2, α, 1).
2. Find the equations of the planes through the following sets of points:
(i) (1,0,0), (0,1,0), (0,0, 1);
(ii) (3,2,0),(0,3, -1), (1,0, -2);
(iii) (2,0,0),(1,1, -1), (6,-5,3);
(iv) (2,2,0),(-l, 1,-4), (1, 1,-1);
(v) (-1,3,1),(1,-3,-3), (3,-1,5);
(vi) (1,2,3), (2, -2, 8), (-1,2, -7).
3. Find the equations of the lines joining the following pairs of points:
(i) (0, 0, 0), (2, 1, - 3); (ii) (1,2,-1), (2, 3, - 3);
(iii) (1, 1, -1), (-1, 4, 2); (iv) (-2, 1, 3), (2, 1, -3);
(v) (1, 4, 2), (1, -1, 2); (vi) (a, 2a, 3a), (-a, a, 2a).
4. Find the equation of the plane through the line of intersection of the planes
x+y-3z = 2,2x-y-z = 1
(i) containing the origin;
(ii) containing the point (1, 1, —1);
(iii) parallel to the x axis.
5. Find the equation of the plane through the line of intersection of the planes
2x + 2y + 3z-l = 0, 3x-y-z+2 = 0
(i) containing the origin;
(ii) containing the point (1, 1, 1);
(iii) parallel to the ζ axis.
63

COORDINATES [3
о the χ axis which contains
x-l = y-3 = z+l
-1 2 ~ -3"
Find the equation of the plane containing this line and the origin.
7. Find the equation of the plane through the point (1, 1, - 5) containing the
ИШ x+l_y + 5_ ζ
8. Write down, in terms of a parameter λ, the coordinates of a general point of
the line
Where does this line cut the plane x-2y-3z-2 = 0?
9. Find the coordinates of the point where the line joining (2, 3, 1) and (4, 7, 3)
cuts the plane 2x+y-z-3 = 0.
10. Find the equations of the line joining (2,1, 1) and (1, -1, 2). Where does this
line meet the plane x—y + az = 0?
Is your result true for all a? Explain.
11. Determine the direction cosines of the line of intersection of the planes
2x-y + z-9 = 0, 4x+y + 2z-6 = 0.
Where does this line meet the plane x+y + z = 0?
12. Prove that the two lines
x-l _y+l _z + 2
1 _ 2 _ 3 '
x + l ^y z-l
1 1 1 '
are skew.
13. Prove that the two lines
x-2 = y + 4 = z-l
3 ~ 1 ~ -1
. x + l y + 4 z + 2
and ^=-Γ=^2
meet. What are the coordinates of their point of intersection? Find the equation
of the plane containing them.
14. Prove that the fr
meet. What are the coordinates of their point of intersection? Find the equation
of the plane containing them.

5] COORDINATE GEOMETRY IN SPACE
15. Find the coordinates of the point common to the three planes:
x-3y + z-6 = 0,
2x-y + 2z-2 = 0,
3x + 2y + z+2 = 0.
16. Find the coordinates of the point common to the three planes:
x-2y + z-7 = 0,
2x + 3y-4z + 26 = 0,
3x+y + 3z+7 = 0.
17. Find the coordinates of the point where the line determined by the two planes
x-3y-z+8 = 0,
x-y+z-2=0
cuts the plane containing the χ and у axes.
Miscellaneous Exercise 3
1. Find the area of the trapezium whose vertices are:
(х2,Уд, (*з,.Уз), (*2,0), (x3,0).
Deduce that the area of the triangle Pi(*b уд, А(*2, Уг), А(*з> Уз) is
ЦХ2У2-Х2У2 + Х3У1- Х1У2 + Х1У2- Х2У1)·
What is the condition that the three points РгР2Рз should be collinear ?
2. ABCD is a rectangle with AB = 2AD. Ε is the mid-point of AB and F is
the mid-point of ВС. СЕ, AF meet at X. By setting up a suitable coordinate
system, prove that DXB is a straight line and determine the ratio DX/XB.
What is the area of the quadrilateral AXCD1
3. ABCD is a rectangle and points P, Q are taken on AB, AD respectively.
The rectangle APRQ is completed. If the lines BQ, DP meet at X, prove that
X,.R, С lie On a straight line.
4. Find, in terms of a parameter w, representing the gradient, the equation of
a variable straight line through the fixed point (a, b).
If this line cuts the χ and у axes at A and В respectively, and the parallelogram
OABP is completed, prove that, whatever the value of m, the point Ρ lies on the
curve
xy = xb — ay.
Sketch this curve for the case a = b = 1 for values of χ > — 1.
5. Show that, by a suitable choice of coordinate system, the equations of two
coplanar non-parallel lines may be taken in the form

COORDINATES [3
Two fixed lines О A, OB are drawn and a variable line, passing through the
fixed point C, cuts О A, OB at Ρ and Q respectively. If the feet of the
perpendiculars from P, Q to OB, О A are U, V, prove that t/Kpasses through a fixed
point.
6. Two fixed straight lines, Lx and L2, meet at O. Through a fixed point A two
lines АРгР2 and AQXQ2 axe. drawn to cut Lx at Рг and Qu and L2 at P2 and Q,_.
Prove that whatever the position of the two lines drawn through A, the point of
intersection of P1Q2 and Ρ2βι lies on a fixed straight line through O.
7. Points Pu P2, P3 are taken on the χ axis, and Qu Q2, β3 on the у axis. Lj
is the point of intersection of P2 β3 and P3 β2,1,г of P3 Qj and Рг β3, L3 of Pj β2
and P2 Qj. Prove that L1; L2, L3 lie on a straight line.
(This is a particular case of Pappus's Theorem, which holds more generally
for two sets of three collinear points on any two straight lines.)
8. Prove that the lines χ+1 j z_3
A x-\ у-ъ z-г
are skew.
By considering the family of planes through one of the lines, prove that
there is just one common transversal to the two lines which passes through the
origin and find its equation.
9. Prove that no three of the points (1, 1, 2), (-2, - 6, 3), (- 1, 1, 5), (2,4, 2) are
collinear but that all four points lie on a plane.
10. Explain why the three planes
x-z+l = 0,
x-y-z+5 = 0,
x + 2y-z + 2 = 0
have no common point.
What can you say about the intersections of the planes:
11. ABCDA'B'C'D' is a cube, with square faces ABCD, A'B'C'D' and vertical
edges AA' etc. Μ is the mid-point of CD'. The plane AB'M cuts BD' at X
and CC produced at Y. Find the ratios BX: XD' and CC'.C'Y.
12. Prove that the three lines χ ν ζ
Li:I = 2 = 3'
Li' 2 -3-I'
r x+l у ζ

5] MISCELLANEOUS EXERCISE 3
If general points of Lt and L2 have parameters λ and μ respectively, find the
condition that the line joining the point On Lt with parameter λ to the point
on L2 with parameter μ should intersect L3.
Deduce that a unique common transversal to the three lines may be drawn
through a given point on Lv
13. In the tetrahedron ABCD, each of the faces ABC, ABD, ACD has a right
angle at A, and AB = AC = AD. X is the point on AD such that AX = 2XD
and Υ is the point on ВС such that С Υ = 2YB. The mid-point of A Y is Z. Prove
that the mid-point of DZ lies on the plane BCX.
14. A is the fixed point (a,0,0) and variable points β (Ο, λ, 0), R(0, Ο,μ)
(λ, μ positive) are taken on the у and ζ axes such that the plane AQR passes
through the fixed point Β (α, β, γ). Prove that
(α—α) λμ + αβμ + αγλ = 0.
Find the value of λ that makes the triangle OQR isosceles.
15. Explain why the equation
x2+y2 + z2 = 1
represents the surface of a sphere, centre the origin.
Find the equations of the planes through the point (2, 0, 0) which are tangential
to the sphere and parallel to the у axis.
67

4· Polynomials
1. POLYNOMIALS
Expression of the form xs_3x2 + 4x+2
and x?-x+l
are called polynomials. More generally, a polynomial of degree η is an
expression of the form
a0xn + a1xn-1 + a2xn-i + ...+an_1x+an (a0 φ 0)
involving only multiples of positive integral powers of χ and a constant
term, an. A polynomial of degree η is completely determined if its (n +1)
coefficients (including the constant term or coefficient of x°) are given and
two polynomials of degree η are said to be identically equal if the coefficient
of xr in each polynomial is the same for all values of r, and conversely.
For example, writing = to mean 'identically equal to'
ax3-x-2 = 2x3-bx2-x + c
о a = 2, 6 = 0, с =-2.
Ex. 1. If ax2 + bx + c ξ Ъх2 + lax + b, find a, b and с
A convenient shorthand is to write a polynomial as P(x); for example,
if we are considering the polynomial x3 — Ъх— 2 we could write 'Let
P(x) = х^-Ъх-Т
and subsequently refer to this polynomial simply as P(x). By P(l) we
mean the value of the polynomial when χ = 1; that is, the numerical value
attained by the polynomial when 1 is substituted for x. In the case just
quoted, P(l) = 1-3-2 = -4andP(-2) = -8 + 6-2= -4.
Ex. 2. If P(x) = xi-3x3 + 4x2-x-1, show that Д1) = 0 and find P(-1), P(0)
and P(2).
Ex. 3. Show that, if P(x) ξ Q(x), then P(k) = Q(k) for all values of k. The
converse of this proposition is also true: if P{k) = Q{k) for all values of k,
then the coefficients of xr in P(x) and Q(x) are the same for all r (see Ex. 7).
Either statement may be taken as the definition of 'identically equal to'.

1] POLYNOMIALS
A non-zero number may be regarded as a polynomial of degree zero,
or a constant polynomial. The number zero may be regarded as the zero
polynomial.^
No ambiguity will arise if we write the zero polynomial as 0. A
polynomial is identically equal to the zero polynomial if all its coefficients are
zero, and conversely.
Ex. 4. If (я-1) л^ + Сд+б) x+(a + b + c) = 0, find the values of a, b, с
Example 1. IfP{x) = ax2 + bx + c, distinguish between the identity P(x) = 0
and the equation P(x) = 0.
P(x) = 0 means that P(x) is identically equal to the zero polynomial and
so we deduce that a = b = с = 0. Thus, whatever value χ may have, say
χ = k, then P(k) = 0.
P(x) = 0 is a statement which holds only for certain values of x. In fact,
provided b" > 4ac and я Ф 0,
pr-b + J(b*-4ac)l = pr-b-J(b>-4ac)l = Q
and P(x) φ 0 for any other value of x.
Exercise 4(a)
1. If P(x) = x3 + 5x2 + 3x +1; find P(l), P(-1), Ρφ, Ρ(3).
2. If (x-2) (x-3) (x-4) = axt + bxz + cx + d, find a, b, c, d.
3. Prove that (x-af = х3-3ах2 + 3а>х-а3. For what real values of χ does
(х-аУ=-а31
4. If а(х-2У + Ь(х-2) + с = 3x2-8x-l, find a, b, с
5. Express 7x3-;c2 + 3x-4 in the form a(x- l)s + b(x- l)2 + c(x- l) + d.
6. Express 4x3 + 12л:2 + 6x in the form a(x+1)3 + b(x+ l)2 + c(x+ 1) + d.
7. Express 3λ-3 + 2λ^- Их- 10 in the form a(x+ 1)3 + Z>(x+ l)2 + c(x+ l) + rfand
also in the form а(л--2)3+/0(л--2)2 + 7(л--2) + 5.
Find the three roots of the cubic equation 3x3 + 2x2 - 1 Ix-10 = 0.
8. Express x3 + 4kx2 + 3k2x - ks in the form a(x + k)s + b(x+k)2 + c(x+k) + d.
9. If (ax2+bx + c) (x+l) = 0, prove that ax2+bx+c = 0.
t We shall not associate a degree with the zero polynomial. The reader will appreciate
that, by elementary algebra, the product of a polynomial of degree m by a polynomial
of degree и is a polynomial of degree m+n. This result holds good even if one or both
of the polynomials is a constant polynomial, but fails if one of the polynomials is the
zero polynomial if a finite degree is associated with this polynomial.
69

POLYNOMIALS [4
10. If (ax+b) (ex+ d) = (ax + c) (bx + d) prove that:
(i) if α φ 0, then Ъ = с; (ii) if b φ с, then α = d = 0.
11. If (х-с)3 = x3 + ax2 + bx-27, find α, b, с.
12. If (x-a) (x-b) (х-с) ξ (x + oi) (χ+β) (χ+γ), prove that
{a,b,c}= {-α,-β,-y}.
2. THE FACTOR AND REMAINDER THEOREMS
If P(x) = {x-a) Q(x), where P(x) is a polynomial of degree η and Q(x) is a
polynomial of degree л— 1, P(x) is said to have a linear factor. Conversely,
if P(x) has a linear factor, then P(x) may be expressed in the form
P(x) = (x-a)Q(x).
For example, 2χ3 + 5χ2_3χ_ ш s (jc + 2) (2ж» + * - 5)
and so the cubic polynomial 2x3 + 5x2-3x-10 has a linear factor (x + 2).
(It also has a quadratic factor (2x2 + x-5).)
It is important to realise at the outset that possession of a factor is not
an absolute property of a polynomial but depends on the restrictions we
place upon the coefficients. For example, if we restrict our polynomials
to have only rational numbers as coefficients, the polynomial x2-2
has no linear factors; on the other hand, if we allow our coefficients to be
real numbers, x2 — 2 = (x-J2)(x + J2). (This is often expressed by the
statement: 'x2-2 is irreducible over the rational field but reducible over
the real field'. The word 'field' has a technical meaning and will be defined
in Volume 2; all that is necessary at the moment is to realize that ' over
the rational field' means that all polynomials under discussion must have
rational numbers as coefficients.)
The theorems we are about to prove do not depend upon whether we
choose the rational field or the real field for our coefficients provided, of
course, that we are consistent. We shall not, therefore, allude to the field
from which the coefficients are drawn, but the reader may, if he wishes
interpret all the coefficients as, say, rational numbers.
It is easy to see that, if P(x) has a factor (x-a), then P(a) = 0. For
P(x) has a factor (x-a)
ο Ρ(χ) = (χ-*) Q(x)
* P(a) = 0.Q(a)
* P(<x) = 0.
A converse of this result is known as the Factor Theorem.
70

2] FACTOR AND REMAINDER THEOREMS
Theorem 4.1 (The Factor Theorem). IfP(x) is a polynomial and ifP(a) = 0,
thenP(x) has a linear factor x—a.
Proof. First observe that
xr - ar = (χ - α) (χ*-1 + αχτ~2 + ofixT~* +...+ α'-1).
Suppose P(x) = b0xn + Ьг х71-1 + b2 xn~2 +...+ bn_^ χ + bn,
Then Ρ(α) = Ь^ + Ь^-^ + Ь^а.71-2^- ...+bn_1a. + bn
and so
P(x) - P(cc) = b0(x" - a») + b^x"-1 - a""1) +... + b^x - <x)
= (x-<x)[b0(xn-1 + ...+acn-1) + b1(xn-i + ...+ccn-i)+...+bn„1].
But P(a) = 0 thus we have
P(x) s (x_a) Q{x)
and therefore P(x) has a linear factor (x-cc).
Consider now the following division process:
х*-2х-Ъ
x-2)~x3-4x2 + x+2
x3-2x2
-2x2 + x
-2x* + 4x
-Ъх+2
-Ъх+6
-4
In elementary algebra the result of this process would be described by
saying that, if x?-4x2 + x+2 is divided by x—2, the quotient is х2 — 2х — Ъ
and the remainder is — 4. There are advantages, however, in restating the
process as follows: χ3 — 4χ2+ χ + 2 is identically equal to the product of
(x-2) by the quotient (x2-2x-3), plus the remainder -4. In symbols
xs-4x2+x + 2 = (x-2)(x2-2x-3)-4. (1)
This identity holds for all values of x, whereas the division process given
above is true for all values of χ other than χ = 2.
Ex. 5. Check that χ = 2 does indeed satisfy (1).
We now prove our second result, the Remainder Theorem, which
generalizes the ideas outlined above.
71

POLYNOMIALS
Theorem 4.2 (The Remainder Theorem). IfP{x) is any polynomial oft
η ^ 1, and x — a is any linear polynomial, then P(x) may be expressed in the
form
P{x) = {x-a)Q{x)+R
where Q(x) is a polynomial of degree n— 1 and R = P{o).
Proof. As in Theorem 1, we have
P(x)-P(<x) = (x-ai) Q(x)
and the result follows immediately.
Note (i). The Remainder Theorem is often stated in the following
form: if a polynomial P(x) is divided by x-a, then the remainder is P{a).
(ii) Strictly speaking, we should prove that the expression given above
for P{x) is unique in the sense that, if P(x) = {x-a) Qi{x) + Ri, then
Qi{x) = Q{x) an^ Ri = R. The proof of this result should be supplied by
the reader.
Example 2. Find the remainder when x3 - 5x2 - χ + 2 is divided by (i) лг— 1;
{и) x + 2; (iii)2x+l.
Write P{x) = x*-5x*-x + 2.
{i)P{x)^{x-l)Q1{x) + R1;
putx=l: J?! = P(1) = 1-5-1+2 = -3;
(ii) P{x) = {x + 2) Q2{x)+R2;
put л; =-2: R2 = P{-2) = -8-20 + 2 + 2 ξ -24;
(iii) P{x) = {2x+l)Q3{x) + R3;
putx=-l: R3 = P{-\) =-i-i + i + 2 = |.
Exercise 4{b)
1. Find the remainder when:
(i) x3-x + 2 is divided by χ-4;
(ii) 3x3-5x2 + x + 2is divided by x-2;
(iii) 4x4-2x2 + x-3 is divided by x+1;
(iv) x5 — x— 1 is divided by χ + 3;
(ν) 4x3-5x2 + 2isdividedby2x-l;
(vi) 4x3 - 8x + lis divided by 2x+l.
2. Show that χ — 2 is a factor of the polynomial x3—4x2 + χ+6 and hence factor-
ize the expression completely.

2] FACTOR AND REMAINDER THEOREMS
3. Factorize the following polynomials over the rational field as far as possible:
(i) 2x3 + 7x2-5x-4; (ii) 12x3 + 5x2-19x-12;
(iii) 2x3 + 7x2-17x-10; (iv) x3 + 3x2 - 2x - 6;
(v) x3-x2-x-2; (vi) x3-7x2 + 7x + 15;
(vii) x3 + 2x2-7x-2; (viii) 8x3 + 12x2-2x-3.
4. Factorize the polynomials given in Question 3 over the real field as far as
possible.
5. If Xs - 5x2 + 7x - a has a factor χ - 2, find a.
6. If 2x3+ax2-5x-1 is divisible by 2x+1, find a.
7. If xs + Ъх2 + ax-l leaves a remainder of 3 on division by (2x +1), find a.
8. If 3λγ> + ολ:2+Ζ>λ:-2 is divisible by both x + 2 and 3x+1, find α and 6.
9. If αχ3 + 3x2 + 6x - 3 is divisible by both x-1 and 2x + 3, find a and 6.
10. By expressing x4 +1 in the form x4 + 2x2 +1 - 2x2, show that x4 +1 is
reducible over the real field.
11. Factorize x4 + 3x3 -15x2 + 9x + 2: (i) over the rational field; (ii) over the real
field.
12. ax4 + 2x3-4x2-2x + Z> has factors (x-1) and (x-2), find a and b and
factorize the expression completely.
3. THE FACTOR THEOREM (CONTINUED)
We now show how the Factor Theorem may be extended for cubic
polynomials and, in particular, how this extended result may be used to derive
certain algebraic identities. The results we prove are valid for the general
polynomial (substituting η for 3 in their enunciation) but proofs for this
require the use of mathematical induction (see Chapter 9).
Theorem 4.3. If P(x) is a cubic polynomial with leading term a0x3, and
ifP{a) = Ρ(β) = Ρ(γ) = 0 for 3 unequal numbers α, β, γ, then
Ρ{χ) = αϋ{χ-α){χ-β){χ-γ).
Proof. Since P(a) = 0, P(x) = (x-a.) Q^x), where Q^x) is a polynomial
of degree 2.
Thus, since Ρ(β) = 0, 0 = (/?-«) &(/?). But/?-α Φ 0 and so &(/?) = 0.
6i(x) therefore has a factor (χ-β).
gj(x) = (χ-β) Q2(x), where Q2(x) is a polynomial of degree 1.
By the same argument,
Q2(x) = (x — y) бз(х), where Q3(x) is a polynomial of degree 0, that is,
Q3(x) = k, where к is some number. Substituting,
Ρ(*)ΞΑ:(χ-α)(χ-/?)(χ-7).
73

POLYNOMIALS [4
Since these two polynomials are identically equal, the coefficient of x3
must be the same in each, and so к = a0 and the proof is complete.
Corollary (The Identity Theorem). If P(x) is a cubic polynomial and if
4 unequal numbers α, β, γ, δ can be found such that
P(<x) = Ρ(β) = Ρ(γ) = P(S) = 0, then P(x) = 0.
Proof By Theorem 4.3,
Ρ(χ) = αϋ(χ-χ)(χ-β)(χ-Ύ).
But P(S) = 0 and so
0 = α0(δ-*)(δ-β)(δ-Ύ).
Now δ Φ α, δ Φ β, δ Φ γ and it follows that α0 = 0. P(x) thus reduces to
a quadratic polynomial axx2 +a2x + a3 which vanishes for 4 distinct values
of x. Successive repetitions of the argument show that аг = α2 = ^з = 0
and the corollary is proved.
Ex. 6. Prove that, if P(x) is a quadratic polynomial and if 3 unequal numbers a,
A, 7 can be found such that P(a) = Ρ(β) = P{y) = 0, then Дх) = О.
Ex. 7. If P(x) and β(χ) are cubic polynomials, and if unequal numbers α, β, γ, 5
сап be found such that P(a) = Q(a), ...,Ρ(β) = Q(S), prove that P(x) = Q(x).
Example 3. Establish the identity
x2-a2 , x2-b2 , x2-c2
(a-b) (a-cy (b-c) (b-a) (c-a) (c-b)
Write „2 „2 „2 L2
P(*)
■1=0 (α Φ ft Φ c).
x2-c2
(a-b) (a-c) (b-c) (Ъ-аУ(с-а) (с-Ъу
Ъ.а±
b+b-
P(a) = 0+^h+^ + l
-b + c
b-c +
Similarly P(b) = P(c) = 0.
But P(x) is a quadratic polynomial and so must vanish identically.
Exercise 4(c)
1. Prove that
(*-6) (x-c) (x-c) (x_-a) (x-a) (x-b)
\a-b)(a-c) (b-c)\b-a) (c-a) (c-b)

3] THE FACTOR THEOREM
2. Prove that
(x-b)(x-c)(x-d) (x-c)(x-d)(x-a) (x-d)(x-a)(x-b)
(a-b)(a-c)(a-d)+(b-c) (b-d)(b-a) + (c-d)(c-a) (c-b)
(x-a)(x-b)(x-c) =
(d-a)(d-b)(d-c) l-V-
3. Prove that
a(x-b) (x-c) b(x-c) (x-a) c(x-a) (x-b) _
7fl=Wla-c) (b-c)(b-a) + (c-e)(c-6) ~* =
4. Prove that
xjx-a) xjx-b) x(x-c) x_ =
bcic-a) (a-b) ca(a-b)(b-c) + ab(b- с) (с - я) abc ~
5. By considering the quadratic polynomial ax* + bx + c, show that, if я ф 0,
" + land a + b + c=0,
an2 + bn + c = 0,
a + bn + cn2 = 0,
then a = b = с = 0.
6. Prove that the cubic polynomial P(x) defined by
= ajx-β) (*-y) (x-δ) b(x-a) (χ-γ) jx-S) c(x-a) (χ-β) (χ-δ)
K ' - (α-β) (α-γ) (α-δ) (β-α) (β-γ) (β-δ) + (γ-α) (y-β) (γ-δ)
rf(x-g)(*-fl(*-r)
"*" (δ-α)(δ-β)(δ-γ)
is the unique cubic polynomial such that Ρ(μ) = α, Ρ(β) = b, Ρ(γ) = c, Ρ(δ) = d.
Find the cubic polynomial P(x) such that
P(-l) =-з, p(0) = 2, P(l) = 5, P(2) = 12.
7. Find the cubic polynomial P(x) such that
ДО) = 1, Д1) = - 1, P(2) = -1, P(3) = 13.
8. Find the cubic polynomial P(x) which takes the value - 5 when χ has the
values - 1, 0, 1 and is such that P(2) = 1.
9. State and prove the identity theorem for quartic polynomials.
Miscellaneous Exercise 4
1. We shall say that a polynomial P(x) 'has the property R' if the remainder on
division of P(x) by the fixed factor x-a is it. Prove that
(i) if Рг(х) and P2(x) both have the property R, so also has the polynomial
P(x) where P(x) = P^+P^x);
(ii) if P3O) has the property R and P4(x) has the property S, show that the
polynomial Q(x) = P3(x) P4(x) has the property RS.
What can be said about the polynomial P(x) if it has the property 0?
75

POLYNOMIALS
2. Prove that, if P(x) is any polynomial of degree η > 2, then P(x) may be
expressed in the form „, . , . , ,. л, . „
P(x) = (x-a) (x-Z>) Q(x) + Rx + S,
where it, 5 are numbers and Q(x) is a polynomial of degree n — 2 (all numbers
and coefficients taken from the same field—say the rationals).
Express 5x3 + 9x2-8x-ll in the form
(x-l) (x + 2) Q(x) + Rx + S
and show you can check your values for R and S by putting χ = 1 and χ = - 2.
3. Find the remainder on dividing the following polynomials by x2 — x — 6:
(i) 7x3-x2-30x-30; (ii) x*-l.
4. The remainder when the polynomial P(x) is divided by (x — d) (x-b), where
α φ 6, is expressed in the form A(x — a)+B(x—b). Find explicit values for A
and Я.
Prove that, if identically equal remainders are obtained on division by
(x-a) (x-b) and (x-a) (x-c),
then (b - с) P(a) + (c-a) P(b) + (a-b) P(c) = 0.
What can you deduce about the remainder when P(x) is divided by
(*-6)(*-c)?
5. By using the Remainder Theorem or otherwise, find the factors of
α2φ + с) + Ъ\с + α) + с\а + b) + labc.
Hence or otherwise show that, if a + b + c = 0, then
я3 + 63 + с3 = Здбс.
Find (i) the factors of
(/> + 2<? - 3/·)3 + (-3;?+<? + 2/·)3 + (2/7-3<? +/-)3,
(ii) the prime factors of „„„ „„, „„ ,_ , ., ч
733-473-263. (Cambridge)
6. If ox2 + 2bx + с can be written in the form
Α(.χ-<χ)* + Β(χ-β)\
prove that ааР + Ь(ас+Р) +с = 0.
Is it true that, if numbers <x, β can be found such that
α<χβ + Ηα+β) + ο = 0
then ax2 + 2bx + ccan be written in the form Α(χ-ά)2+Β{χ-β)21
7. Prove that, if lx2 + mx + n is equal to zero for three distinct values of x,
then/ = m = η = 0.
Hence, or otherwise, prove the identity
a2(x-b) (x-c) b*(x-c) (x-a) c\x-a) (x-b) _ 2
(a-b)(a-c) + (b-c)(b-a) + (c-a)(c-b) ~ '

3] MISCELLANEOUS EXERCISE 4
Prove also that
(x + af (x + ЬУ (x + c)3 _ „
(а-Ъ)(а-сУ(Ъ-с)ф-а)+(с-а)(с-Ъ) = 3* + й+6 + с·
(O&Q
8. Factorize:
(i) bc(b-c) + ca(c-a) + ab(a-b);
(ii) a{b-c)2 + b(c-a)2 + c{a-bY + %abc;
(iii) а(6-с)3 + 6(с-а)3 + ф-6)3-
9. Given that p, q, r are distinct values of χ which satisfy the equation
— + Л + — =L
χ—α χ—/3 χ—γ
prove that, for all values of χ other than χ = α, χ = ^, χ = γ
_£_ + _*_+_£_ = ι _ (*~Ж*-д) (*->-)
*_α+*-/9 *-r~ (χ-α)(χ-/?)(χ-7)·
By considering the value ofP(x) ξ (χ—ρ) (x—q) (x—/·) for different values of x,
prove that, if a, b, с are all positive and a. < β < y, then/?, q, r lie one in each of
thesets {;c еД: α < * </?}, {χ e R-,β < χ < γ}, {χ e R: χ > γ}.
Prove alsi
id that
э that . w w .
a (α-#(α-γ)
β δ с (p-q)(p-r)
{ρ-αγ (ρ-β)* (p-r)1 (ρ-α)(ρ-β)(ρ-γ)'
(Ο & С adapted)
10. When divided by (x2 + x+l), a cubic polynomial F(x) leaves a remainder
(2x+3). When F(x) is divided by x(x + 3), the remainder is 5(x + 1). Find F(x).
(Cambridge)
11. /(x), g(x) and G(x) are cubic polynomials in x. The polynomials h(x) and
#(x) are respectively the remainders left when Дх) is divided by g(x) and by
G(x). Prove that, if αχ3 and Ax3 are respectively the terms of highest degree in
g(x) and G(x), then
[aG(x)-Ag(x)]f(x) ξ aG(x) A(x) - Ag(x) tf(x). (Cambridge)

5· Functions and inequalities
1. FUNCTIONS
The concept of an association between the elements of two sets is of prime
importance in mathematics. For example, the table of sines in a book of
mathematical tables exhibits an association between the set A of angles
between 0° and 90° and the set В of real numhers between 0 and 1. We
could exhibit the fact that sin 30° = 0-5 by writing the ordered pair
(30°, 0-5). The word 'ordered' used here reminds us that the order in
which we write the elements down within each pair must be taken into
account; that is, the ordered pair {a, b)is not the same as the ordered pair
(b, a).
We can put this concept of an association between the elements of two
sets in a more precise way if we introduce the idea of the Cartesian product
of two sets A and B. The Cartesian product of A and B, written Α χ Β is
the set of all ordered pairs {a, b) where a e A and be B. For example, if
A = {1, 2, 3}, В = {3, 4, 5},
then
A xB = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}.
Ex. 1. If the set A has m elements and the set В has η elements, how many
elements has the set Α χ Β ?
A relation between two sets A and В is a proper subset of Α χ Β. Thus,
a relation is a set of ordered pairs {a, b) containing at least one pair, but
not all the pairs, of Α χ Β. Of course, relation is a very wide term as defined
here. For example, both {(1, 3)} and {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5),
(3, 3), (3, 4)} define perfectly respectable relations between A and B. Of
particular significance are the (1-1) relations (read as 'one-to-one'
relations) which associate with each element of A just one element of В in
such a way that no element of В arises from more than one element of A;
in our example above, such a relation could be {(1, 3), (2, 4), (3, 5)}.
In this chapter we shall be concerned with perhaps the most important
of all types of relation, the function. A function f from A to B, written
f:A^-B, is a subset of Α χ Β with the property that each element of A
appears just once. Another word for function is mapping: A is mapped byf
into B. A function may be looked upon as something which associates
78

1] FUNCTIONS
with each element of A a unique element of B. In Figure 5.1 a function is
pictured diagrammatically; notice that the arrows, indicating the paths
by which elements of A are to find their assigned destination in B, start
from each point of A. Notice too that there is no restriction on the number
of elements of A for which be Bis the image. (The image of the element
a e A is the element which a is mapped into by/.)
Fig. 5.1
If we have a function/: A^-B the set A is called the domain of/and the
set of image points (a subset of В in general, but possibly the whole of B)
is called the range of / The set В is called the codomain of /. The image
of the element a e A is written Да), thus, Да) is an element of В or, in
other words, (a,f(a)) is an element of Α χ Β.
Example 1. The function f: R+ -> R+ is defined by
Ax) = ih (xeR+)·
Find:
(i) the range off;
(ii) Д2) andfU2);
(iii) the value of χ which maps into \.
(i) For 0 < x1 < x2,
1 > Г^>Г7— >0·
1+*! l+x2
Thus, since 1/(1 +x) can take any positive value less than 1 by a suitable
choice of positive x, the range of/is the set {y e R+: 0 < у < 1}.
(ii) Д2) = 1/(1+2) = ^;/(V2) = 1/(1+V2) = V2~ 1 · (Notice that both
2 and V2 belong to Л+ and sofl2),fQ2) are defined.)
(iii) If l/(l+x) = i, x = 3.
If there is no doubt as to the domain, a particular function is often
defined by phrases such as 'the functionf{x) = хг — Ъх-Т or 'у is a
function of χ such that у = хг-Ъх-2ь or even 'the function χ*-3χ-2'.
From the context we are presumed to appreciate that χ is, say, a real
79

FUNCTIONS AND INEQUALITIES [5
number. Even so, the phraseology is loose: the notation f(x) (or y) stands
for the image of χ under /, not for the function itself.
*Ex. 2. Comment upon the statement
represent the same function'.
Ex. 3.f:R^-Ris defined byf(x) = x3- 3x2- 5x + 2. Evaluate ДО), Д- 1), Д2),
ДЗ). Which elements of R map into the number 2?
.Ex. 4. Answer the same question as in Ex. 3 for the function g: Q^-R defined
byg(x) = x3-3x2-5x + 2.
A useful pictorial representation of certain functions /: R-> R (or
subsets of these sets) may be obtained by taking the χ axis to represent the
domain and the у axis to contain the range. If у is the image of the number
χ under/, that is, if у =/(x),we represent the element (x,y) of/as the point
with Cartesian coordinates (x, y). The set of all such points is called the
graph of the function. For example, suppose f: R^ R is defined by the
equation Дх) = x2-5x + 4. The graph of this function for the part of the
domain -1 < χ < 6 is shown in Figure 5.2. This graph may alternatively
be referred to as 'the curve у = x2-5x + 4\
Certain relations which are not functions may also be represented
graphically; for example, consider the subset of RxR consisting of all
ordered pairs of the form (x, y) where xe R, ye R such that y2 = x.
That part of the graph for χ < 6 is shown in Figure 5.3.
Fig. 5.2 Fig. 5.3
Ex. 5. Explain why the relation depicted in Figure 5.3 is not a function.
Suppose now we have a function /: A-+ B; under what circumstances
will the set of all ordered pairs (f(a), a), aeR, constitute a function
F: В -> A ?

1]
FUNCTIONS
Viewed pictorially, the question may be rephrased: if all the arrows in a
diagram such as Figure 5.4 were reversed, under what circumstances would
we still have a diagram depicting a function?
Fig. 5.4
The question is easily answered if we recall what is demanded of such
a function F: it must associate each point of В with a unique point of A.
Thus, the range of/ must be the whole of В (since each point of В has an
image under F) and/must be (1-1) (since each point of В has a unique
image under F). If/satisfies both of these conditions F is defined as a
function: it is called the inverse function off: A^ В and is written
f-^.B^A.
Ex. 6. Show that no inverse function exists for the function /: Л -> R where
f(x) = x2, but that an inverse function does exist for the function g: R+ -> R+,
where g(x) = x2.
Exercise 5(d)
1. The function f:R^ R is defined by f(x) = v*—*- 1. Find ДО), Д1), /(2),
Л3),л-1)·
2. The function #:Д+ -> Я is defined by
Findtf(l),*(2),tf(3).
3. The function /: R -> Я is defined by
1-х if * < 0,
0 if χ = 0,
* if * > 0.
(This is normally written/(x) = |*|; read '/(*) equals the modulus of *'.) What is
the range of/? Sketch the graph of/.
A second function g: R -> it is defined by #(x) = x— |*|. What is the range
of gl Sketch the graph of g.
81

FUNCTIONS AND INEQUALITIES [5
4. The function/: R -> R is defined by
[θ, χ = 0.
Write down the values of/(-1) and/(l). Sketch the graph of/
5. A £ R and В Я R. The set / of all ordered pairs (x, y) is formed, where
χ e А, у е В and у = $x. Show that /is «oi a function if
(i) A = R+, Β = β; (ii) Λ = β, Л = R.
Suggest sets A and В for which /is a function.
6. A = {xeR:l ^ χ ^2} and/: Λ -> R is defined by/(x) = 2/x. The function
#: /4 -> Λ is denned by #0) = 1 +.y2. A new function h: A^> R is defined by
h(x) = #(/(*)] (this is usually written h = g of). Find
(i) A(f); (ii) the range of A.
7. Sketch the graph of the function/: R -> R where/(x) = |*+ 11.
8. Sketch the graph of the function g: R^- R where
*(*)= |*+l| + |* + 2| + |x + 3|.
9. A = {xeR.O < χ <3} and/: A -> R is defined by/(x) = x2-3x+2. Find
(i) /(1),Λ2);
(ii) the range of/;
(iii) the subset of A whose elements have the image 1 under/;
(iv) the subset of A whose elements have the image 2 under/
10. A = {x e R: - V6 =£ χ =£ V6) and /: A -> R is defined by /(*) = x*-6x.
Find the range and sketch the graph of/.
11. If /: R -> Λ is defined by /(x) = (*- l)3, show that f-^.R^-R exists and
determine its form explicitly.
12. Determine the numbers which are invariantt under the mapping f. R^- R
where /is denned by
»л _/(*' + *+Dfr+3)-1 (* + -3),
/W_\ 0 (*=-3).
Show how to illustrate invariance graphically for a general function g: R^- R.
13. /: R -> Λ is a given function, A <= R and the set of images of all the elements
in /4 is denoted by B. If С is the set of numbers whose images belong to B,
prove that A <= С and that the inclusion may not be strict.
14. /: R -> R is a given function, 21 <= Λ and the set of images of all the elements
in JSTis denoted byf(X)· If A <= R, В <= Risit necessarily true that
(i) AA П Β) = ΛΑ) η f(B); (ii) ΛΑ υ Β) = ΛΑ) Uf(B) ?
15. Sketch the graph of the function /: R+ -> Λ defined by Дх) = χ2. Show that
the function,?: R+ -> Λ, where ^(x) = **, is the inverse of/. Sketch the graph of g.
What connection exists between the graphs of/and g?
f An element α is invariant under the mapping/if/(a) = a.
82

1] FUNCTIONS
16. If χ is a real number, [x] denotes the greatest integer less than or equal to x.
(For example, [-2*] = -3, [Ц] = 4, [-1] = -1.) The function /: R -> Ζ is
denned by/(χ) = [χ]; sketch its graph.
A function/is called^ez-zW/c if there exists a number к such that f(x+k) = f(x),
for all χ in the domain; к is then called the period off. Show that the function
denned by f(x) = x— [x] is periodic and find its period.
What feature does the graph of any periodic function possess?
2. INEQUALITIES
The function, /, defined by
has for its domain the set R with the single point χ = \ deleted; if we call
this set D, we may write D = R-{f}. The inequality
defines a set А с д where
IxgD:
Ъ-2х
<i;
that is, A is the subset of elements of D which map into elements less
than 1. By 'the solution' of the inequality (1) is meant the determination
of the simplest expression for the set A.
It is always good practice, in solving an inequality, to express one side
of the inequality as zero. Thus
IxeD:
1
-1 ■
IxeD-
.2(*-l) ;
3-2*
" 3-2x
To determine the values of χ satisfying the inequality
: 0
2(*-l) i:
3-2ΛΓ
it is best to express the working in tabular form (see Chapter 1.7). The
critical values of χ e R are χ = 1 and χ = f.
3-2*
x-l
7,-lx
χ < 1
-
+
-
x = 1
0
+
0
.«<i
+
+
+
,>\
+
-
-

FUNCTIONS AND INEQUALITIES [5
Thus A = {xeD:x < l}u{xeD:x > %}.
Figure 5.5 shows a graph of the function/defined above; the domain of
/is denoted by the whole of the χ axis with the point χ = f deleted, the
range off is represented by the у axis, with у = 0 deleted. The subset A
of D is denoted by that part of the χ axis drawn in a thick line. (Note
carefully that l$A,i$A; indeed, \ φ D.)
Fig. 5.5
Example 2. Solve the inequality
* < 2
(*-3)(*+lp 5
and illustrate your answer by means of a sketch of the curve
y=(x~^'3)(x+iy
The function/defined by
/(X) = (x-3)(x+l)
has as its domain the set D where
D = j?-{-i,3};
that is, the set of real numbers, less the two numbers -1 and 3. (D may
be written alternatively as the complement of the set
{xeR:x=-l or χ = 3}.)
84

INEQUALITIES
_ 5х + 2х2-4л:-6
(x-3)(x + l)'5~ 5(x-3)(x + l)
2x2+x-6
_5(х-3)(х+1)
_(2х-3)(х + 2)
5(дс-3)(дс + 1)·
We thus seek a simpler description of the set A, where
| (2Х-ЗИХ + 2) |
The critical values of χ e R are χ = - 2, χ = -1, χ
As above, we set out the work in tabular form.
= 1, χ = 3.
χ <-2
-
-
-
-
+
*=-2
0
-
-
-
0
-2 < χ <-l
+
-
-
-
-
_,<*<!
+
-
-
+
+
3
x~ 2
+
0
-
+
0
\<x <3
+
+
-
+
-
x> 3
+
+
+
+
+
Thus A = {xe D: -2 ζ χ < -1} U {xe D: \ ^ χ < 3}.
Figure 5.6 shows the graph of/, the set A is represented by that part of
the χ axis drawn in thick lines.

FUNCTIONS AND INEQUALITIES [5
Ex. 7. Verify the solution of Example 2 by evaluating /(-3), Д- Ц), ДО), Д2),
Д4). Explain why these are appropriate values to use for checking.
*Ex. 8. Show that the two expressions
^-3)(χ+1) and (*-3)(*-*)(*+l)(*+2)
have the same sign for all values of χ φ — 1 or 3. Hence solve the inequality of
Example 2.
Exercise 5 (ft)
Solve the following inequalities, illustrating your solutions by means of a sketch.
x-4 6(3*+1)
x2-4 ; * (x-l)(x-2)(x+3) ;
86

6. The trigonometric functions
1. RADIANS
We shall assume in this chapter that the reader is familiar with the usual
measurement of angles in degrees and also with the definitions of the sines,
tangents, etc., of acute angles in terms of the sides of a right-angled
triangle. We shall also assume that arc-lengths in circles may be satisfactorily
defined and measured and, in particular, that the circumference and area of
a circle of radius r are respectively 2m and яг2.
An alternative measure of angle may be defined as
follows. Figure 6.1 shows a circle of radius r, centre O.
The magnitude of the angle Δ.ΑΟΡ may be defined as
the ratio of the arc-length, s, of AP to the radius, r.
In the particular case in which s = r, the angle Δ.ΑΟΡ
is called a radian. Thus, an arc-length of 2r subtends an
angle of 2 radians at the centre of the circle. (We may
write this ' Δ.ΑΟΡ = 2rad', or, since a ratio of two lengths is a pure
number, simply' Δ.ΑΟΡ = 2' provided no confusion is likely to arise.)
*Ex. 1. Show that a right-angle is \тт rad. Convert 180° to radians and use your
result to express in radians the following angles: 30°, 60°, 45°, 75°, 135°, 225°,
330°.
*Ex. 2. Express in degrees the following angles: \π rad, |тг rad, \n rad, \π rad,
■Α-π rad, %n rad, -&т rad.
Fig. 6.1
Fig. 6.2
We shall adopt the usual convention that angles are measured as positive
in the anticlockwise sense, the sense being denoted, when necessary, by an
arrow. Figure 6.2(i) shows an angle of -§я, Figure 6.2(ii) an angle of
V-я, both measured from О A.
One advantage of measuring angles in radians is that the expressions for
87

THE TRIGONOMETRIC FUNCTIONS [6
arc-lengths and the area of a sector take on a particularly simple form.
Indeed, if in Figure 6.1 L A OP = θ rad, we have
s = rd;
also, assuming that the area of a sector is proportional to the angle
subtended at the centre,
area of sector AOP = m\d\2n)
= irW.
(A further advantage of using radians will become apparent later when the
reader learns to differentiate the trigonometric functions: if the angles are
measured in radians, the derivatives of these functions take on a
particularly simple form.)
Exercise 5(a)
1. Express in radians the following angles: 20°, 45°, 105°, 150°, 210°, 270°, 315°,
330°, 450°, 570°.
2. Express in degrees the following angles: i?rrad, $π rad, |7rrad, -fen rad,
in rad, f π rad, -far rad, 5π rad, γ^ττ rad, ^π rad.
3. Express in radians, to 3 d.p.: 25° 30', 71° 15', 146° 12'.
4. Express in degrees, to 1 d.p. : 0-48 rad, 0-65 rad, 2-18 rad.
5. Taking π= 3141593, show that 1 rad « 57-296°. (Use a calculating machine
if you have one available.)
6. If the earth is taken to be a sphere of radius 3960 miles, find the length in
feet of arc which subtends an angle of 1 minute at the centre of the earth.
7. A circle has radius 10 cm. Calculate the length of the smaller arc cut off by
a chord of length 16 cm.
8. The chord PQ subtends an angle jtt at the centre, O, of a circle of radius
10cm. Find:
(i) the length of arc PQ;
(ii) the area of the triangle OPQ;
(Hi) the area of the minor segment cut off by PQ.
9. A chord of a circle of radius r cm subtends an angle of 2Θ degrees at its centre.
Obtain expressions for the length of the chord and the length of the minor arc
which it cuts off.
If the arc has length 3 cm and the chord 2-7 cm, prove that
7T0 = 200sin 0. (O&C'O')
10. The section of a tunnel consists of the major segment of a circle standing
on a chord of length 2 m. If the greatest height of the tunnel is 3 m, calculate the
radius of the circle.
Prove that the angle subtended by the chord at the centre of the circle is
approximately 1-287 rad, and hence find the area of the cross-section of the
tunnel, correct to the nearest square metre. (O&C'O' adapted)
88

1] RADIANS
11. Two circles of radii 6 cm and 8 cm respectively have their centres 10 cm
apart. Calculate the length of their common chord, and the area common to the
two circles. (O & С Ό')
12. Thevertexofahollowconeis^andfiCisadiameterofthebase;ВС = 6cm
and AB = AC = 5 cm. The cone is unrolled into a sector of a circle. Find :
(i) the angle of the sector;
(ii) the area of the curved surface of the cone;
(iii) the shortest distance on the surface of the cone from С to the mid-point
of AB. (O&C'O')
2. THE TRIGONOMETRIC FUNCTIONS
Figure 6.3 shows a circle of unit radius; OA = i, OB = j
(where as usual, i and j are perpendicular). The angle Θ,
measured in radians from OA as initial line, defines a
unique unit vector OP. Now OP may be expressed
uniquely in terms of the unit vectors i, j (see Chapter 2)
OP = xi+y\. Fig. 6.3
Thus, given any θ e R, we determine uniquely two numbers χ e R, у е R,
which may be taken as the images of two functions R^> R, called
respectively the cosine and sine functions. We write χ = cos Θ, and у = sin Θ;
the reader will have no difficulty in appreciating that our definitions lead
to the familiar sine and cosine of acute angles if 0 < θ < %n. We have to
be a little careful if we measure θ in degrees: strictly speaking, the sine and
cosine functions would then be different functions R^y R. Of course,
given an angle, the image of the appropriate sine function in either case
would be the same number: for example with angles measured in radians,
sin^Tr = 0-5; with angles measured in degrees, sin 30° = 0-5. We adopt
the same name for the two functions, as not to do so would lead to
confusion; if extra clarity is required, we may always write 'sin^rad' or
'sin0°'. (In this chapter, 'sin Θ' will mean 'sin Θ rad' and similarly for the
other trigonometric functions; if we wish to measure θ in degrees, we
write' sin θ°\)
Since -1 < χ < 1 and -1 < у < 1, the range of both the sine and
cosine functions is the set {x e R: -1 < χ < 1}.
The reader will no doubt suspect that a tangent function will now be
defined by tan θ = yjx. However, we must be careful here, for yjx is not
defined if χ = 0; this occurs whenever θ is an odd number of right-angles,
or, to be more precise, when θ = (2fc + l) \ττ, where к is any integer.
We must therefore restrict our domain to R η A', where
A = {0eR,keZ: Θ = (2Α: + 1)^}.
89

THE TRIGONOMETRIC FUNCTIONS
With this restriction, the tangent function R Π A
tan θ = yjx.
R is defined by
η θ = yjx.
Similarly, if В = {θ e R, к e Ζ: θ = for}, the cotangent function
Rf] B' -+R is defined by cot θ = xjy.
If the reader finds the notation of the previous two paragraphs difficult
to digest, he may simply remember that
(i) tan θ = yjx provided χ φ 0; tan θ is undefined if χ = 0.
(ii) cot θ = xjy, provided у φ 0; cot θ is undefined if у = 0.
In similar fashion, the secant and cosecant functions are defined by
(iii) sec0 = \jx, provided χ φ 0; sec θ is undefined if χ = 0.
(iv) cosec θ = Xjy, provided у φ 0; cosec θ is undefined if у = 0.
In terms of the sets A and В defined above, the domain of the secant
function is R η A'; the domain of the cosecant function is R η Β'.
*Ex. 3. Prove that the range of both the tangent and cotangent function is the
set R.
*Ex. 4. Prove that the range of the secant and cosecant function is the complement
of the set {x e R: -1 < χ < 1}.
The following table of the images of the sine, cosine and tangent
functions for some commonly occurring angles is worth memorizing:
sin
COS
tan
0
0
1
0
W6
1/2
V3/2
1/V3
W4
I A/2
1/V2
1
ir/3
V3/2
1/2
,/3
τ,β
1
0
Undefined
The corresponding images of the secant, cosecant and cotangent
functions may be obtained from the following important identities, which are
valid for all angles θ at which the functions are defined:
(Proofs follow immediately from the definitions. The sign = emphasizes
that these are identities, true for all values of θ for which the functions are
defined.)
Example 1. Find the value of:
(i) sin§7r; (ii) cosf^; (iii) tan^; (iv) с
(vi) cot (—ίπ).
(v) sec (-1-я-);

>] THE TRIGONOMETRIC FUNCTIONS
In each case we find:
(a) the correct sign to attach (by considering signs of χ and y),
(b) the acute angle which defines the same numerical values of χ and y.
(i) For θ = f π, у > 0 and so sin f π > 0.
Also the acute angle which defines the same numerical values of χ and у
s \ττ (see Figure 6.4(i)).
Thus, sin |тг=+ sin \n =+i.
Fig. 6.4
(ii) θ = \n, χ < 0 and the corresponding acute angle is \n (Figure
6.4 (ii)).
Thus, cos f π = - cos \tt =-\.
(iii) For θ = Ιπ, χ > 0, у < 0 and the corresponding acute angle is
£ττ (Figure 6.4 (iii)).
Thus, tan Ιπ = - tan \π = -1.
(iv) For θ = %π, у > 0 and the corresponding acute angle is |тг
(Figure 6.4 (iv)).
Thus, cosecfrr = +cosec^7r = +^/2.
(v) For θ = - J π, χ < 0 and the corresponding acute angle is \π
(Figure 6.4 (v)).
Thus, sec (-|л-) = -sec \n =—yj2.
(vi) For θ = -\π, χ < 0, j > 0 and the
corresponding acute angle is %π (Figure 6.4 (vi)).
Thus, cot (-in) = -cotin = -V3.
Figure 6.5 gives a useful method for seeing which
sign to attach according to the quadrant in which the
angle falls. A stands for all, S for sine, Γ for tangent
and С for cosine. In the A quadrant, all
trigonometric ratios are positive; in the S quadrant, only Fig. 6.5
4 PPM 91

THE TRIGONOMETRIC FUNCTIONS [6
sines (and cosecants) are positive; in the Τ quadrant, only tangents (and
cotangents) are positive; in the С quadrant, only cosines (and secants) are
positive. Notice, as an aid to remembering the diagram, that the letters
(read anticlockwise) spell the word 'cast'.
Example 2. Solve for θ the equation
2sin20 + sin0-l = 0
giving all solutions in the interval 0 < θ < 2π.
The left-hand-side factorizes: (2 sin 0-1) (sin 0+1) = 0 and so is
satisfied by values of θ such that either
(i) sin θ = +i, or (ii) sin θ =-1.
The relevant solutions of (i) are θ = \ττ, θ = \π; and of (ii) θ = \π.
Thus, the required set of solutions of the given equation is
Since OP is a unit vector, x2+y2 = 1 for all Θ. Thus,
sin20 + cos20 = 1, for all0.
Division by cos2 θ gives
1 + tan2 θ = sec2 Θ, for all θ Φ (2k +1) π/2.
Division by sin2 θ gives
1 +cot2 θ = cosec2 Θ, for all θ φ for.
These three identities are true for all values of Θ, provided the functions
mentioned are defined.
Π
•vj
Pi
Ό
Fig. 6.6
Suppose OP is the unit vector defined by the angle θ and OQ the unit
vector defined by the angle \π — θ (see Figure 6.6). If OP = x\+y\, then
OQ = y\ + x\. It follows that
sin (i?r-0) ξ cos Θ, cos (in-θ) = sin Θ, tan ϋπ-θ) ξ cot θ,
cosec Цл-в) = sec θ, sec ($π- θ) ξ cosec θ, cot ϋπ — θ) = tan θ
92

2] THE TRIGONOMETRIC FUNCTIONS
(provided, of course, that θ lies in domains for which both sides are
defined).
*Ex. 5. Find similar simplifications for the values of the trigonometric functions
for the angles
(i) in+θ; (ii) π-θ; (iii) π+θ.
N.B. Be careful about the signs!
Example 3. Solve the equation sin Θ + cos 2Θ = 0, giving all values of θ
between 0 and 2π.
Since cos (%π + Θ) = - sin Θ, the equation may be rewritten
cos 2Θ = cos(i?r + 0).
Now
cos θ = cos φ if θ = ±φ or θ = 2π±φ or θ = 4π±φ ....
Thus the original equation is satisfied for values of θ given by
20 = ±βπ + 0); 2θ = 2π±{\π + &);
2Θ = 4π±(^τ + θ); 20 = 6π±(%π + θ); ....
Trial of these solutions shows that the values of θ lying in the given range
аГе f1 7 1 1 1
Example 4. Prove that the identity
1 tanfl + cotfl
sin θ + cos θ ~ sec θ + cosec θ
holds, provided θ Φ \кл, θ Φ ηπ — \π.
The restrictions on 0 ensure
(i) {θ Φ ^Атг) that tan Θ, cot β, sec Θ, cosec β are defined;
(ii) {θ Φ wr- \π) that sin (9 + cos θ φ 0 and sec θ + cosec 6» Φ 0.
With these restrictions
_ /sing cosfl\ // 1 1 \
R,HS- Icos θ+ sin 0//lcos0 sing/
ξ (sin2 0 + cos2 0)/(sin 0 + cos Θ)
= l/(sm0 + cos0)
= L.H.S.
Example 5. Eliminate θ between the equation
χ = a cos Θ, (1)
j = й sec 6» +с tan 6», (2)
4-2 93

THE TRIGONOMETRIC FUNCTIONS
χ = a cos θ and у = b secd + c tan#
=> xy = a cos θ (b sec 0 + с tan Θ),
=> xy = a£ + acsin#
<s> acsin0 = ху-ой.
But ac cos θ = ex.
Squaring and adding , , , , , . ,.,
м ^ ь a2c2 - c2x2 + (xy-ab)2.
1. Write down the values of cos §π, tan |π, cosec £π, cot jtt, sin Jw, sec jtt,
cos fw, tan %π, cosec ( - ^π), sec ( - |π).
2. Write down the values of cos 315°, tan 135°. cosec 330°, sin(-135°),
cot(-120°), sec 240°, sin 480°, sec (-210°), cot (-60°), sin 1020°.
3. Use your tables to find the values of sin 215°, cos 128°, tan (-40°), cosec 161°,
sin (-200°).
4. Solve the following equations for 0, giving all values lying in the interval
0 < θ < 2π:
(i) sin0=-i; (ii) tan 0+1=0; (iii) sec 0 + 2 = 0;
(iv) cot 0 = V3; (v) 4 sin2 0 = 3; (vi) 2 cos (0-i?r) + V3 = 0;
(vii) sin 0 + cos 0 = 0; (viii) 2 sin 30 = 1; (ix) sec (0+K> = 1;
(x) 3 sec2 ((?+» = 4; (xi) cosec2 (0 + Ю = 1; (xii) tan2 20 = 3.
5. Solve the following equations for 0, giving all values lying in the interval
-я- s: 0 < я-:
(i) 2 sin2 0 + sin Θ = 0; (ii) 2 cos2 0 + 3 sin 0 = 0;
(iii) cos20 = sin0; (iv)2tan0 + si "
(v) sin(i7r-0) + cos0
(vii) tan0 + cotG-7r-0) =
(ix) cot20 + cosec0+l =
(vi) cos 0 = 2 cot 0;
(viii) tan 0+cot 0 +sec 0 =
(x) sec 30 + cosec 0 = 0.
6. Find the maximum and minimum values of:
(i) 3/(2 +sin 0); (ii) (l+cos20)2; (iii) sin2 0 + 2 sin 0+2.
7. No tables to be used in this question.
(i) If \ή < χ < π and sin x = i, find cos χ and tan x.
(ii) Ιί\π<χ<π and tanx=-i, find sin χ and cos*,
(iii) If я- < χ < \n and sec χ = - 3, find cos χ and tan x.
(iv) If in < χ < π and sin χ = i, find tan χ and secx;
(v) If \π < χ < 2π and sec χ = 4, find cosec χ and cot x.
8. Prove that, provided sin 0 φ 0.
l-cos0 = 1
sin 0 — cosec 0+cot 0'
94

2] THE TRIGONOMETRIC FUNCTIONS
9. Prove that, provided sin θ Φ 0, cos θ Φ 0,
tan θ + cot θ = sec θ cosec θ.
10. Prove that, provided all the values of the functions are defined,
cosec2 θ = 1 + cos θ cot θ cosec Θ.
11. If sin θ = f find, without using tables, the possible values of sec (9 +cosec Θ.
12. If tan = i find, without using tables, the possible values of 2 cos Θ + cot Θ.
13. Use your tables to find all values of Θ, lying in the interval 0° < θ° < 360°,
which satisfy the following equations:
(i) 2 sec2 Θ0 = 5-tan 0°; (ii) 3 cos2 Θ" = 7 cos 0°-2;
(iii) 2 sin2 (9°-sin (9° cos (9° -cos2 (9° = 0;
(iv) cot2 θ° = cosec 0°; (v) 16tan20° = 9;
(vi) 3 sec2 Θ0 = 2 cosec 0°; (vii) tan 0° = 2(sec θ° + cos 0°);
(viii) sec20° = l + 2tan0°.
14. Eliminate θ between the following pairs of equations:
(i) χ = 2cos0, y= 3(l+sin0);
(ii) χ = cosΘ — sinΘ, у = cos0+sin0;
(iii) χ = 2 cos 0 — sin в,у = cos# + sin0;
(iv) χ = 3 tan 0, у = 4 sin 0;
(ν) χ = cosec 0-1,.y = cos 0+1.
15. A particle oscillates along the χ axis in such a manner that its coordinates at
time t seconds after the start of the motion are (sin ifot+e), 0).
(i) Where is the particle at the start of the motion?
(ii) Between what two points does the particle oscillate?
(iii) How long does the particle take to move from one extreme point to the
other?
16. If a particle is projected with a velocity К cos ai+ К sin aj under gravity
then its position vector at time t is given by
г = (Kicosa) i + (Kisin a,-\gt2)].
If г is written in the form г = xi+я,
prove that у = χ tan a.
2\K2/
Deduce that, if x, y, g, V are known, there are in general two values of a in
the interval 0 «S a < %π which satisfy this equation. Under what circumstances is
there only one such angle?
17. Prove that the equation „ , ,
x2+.y2 + z2 = a2
represents the surface of a sphere.
Show that the point Ρ whose position vector is
г = a cos θ cos φϊ + α sin θ cos φ} + a sin фк
lies on the sphere, for all values of θ and φ.

THE TRIGONOMETRIC FUNCTIONS [6
Is the following converse result true? Values of θ and φ can be found such that
any point on the surface has a position vector of the form
г = acos0cos0i + asin0cos0j + asin0k.
18. Discuss the possibility of solving the following equations for Θ:
(i) sin θ = a+ I/a; (ii) 2ab cos θ = я2 + Z>2;
(iii) cos2 0+4 = sin 0 + 4 cos 0.
19. What are the maximum and minimum values of the expression 4/(2 + sin x) ?
Without attempting to solve the equations exactly, state how many values of χ
in the range 0 < χ < 2π satisfy
(i) the equation cos x(2 + sin x) = 4;
(ii) the equation tan x(2+ sin x) = 4.
3. THE GRAPHS OF THE TRIGONOMETRIC
FUNCTIONS
Since revolutions through angles 2π, Απ,... about О brings the unit vector
OP into coincidence with its original position, it follows that, for any
integer к and any of the six trigonometric functions/: R-+ R
A2kn + x) = /(*).
Thus, a trigonometric function is a periodic function, period 2π (see
Exercise 5(a), Question 16).
Figure 6.7 shows the graphs of the sine function (continuous line) and
cosine function (dotted line), for values of the domain —2π < χ < 2π.
Fig. 6.7
Ex. 6. How do the graphs of sin χ and cos χ illustrate the identities
sin (%π—χ) = cos x, sin (%π + χ) = cos x,
sin &т-х) = -cos x, sin (jtt + x) ξ -cos x?
Ex. 7. Sketch roughly the graphs of 2 sin x, sin 2x, [sin x\.
Figure 6.8 shows the graph of the tangent function (continuous line)
and cotangent function (dotted line) for values of the domain
— 2π<χ<2π. Recall that the tangent function is undefined at values
of χ which are odd multiples of %π but, by taking χ sufficiently close to such
96

3] GRAPHS OF TRIGONOMETRIC FUNCTIONS
values, we make |tanx| arbitrarily large. A similar remark holds for the
cotangent* function at values of χ which are even multiples of \n. (A line
such as χ = \n is called an asymptote for the curve у = tan x.)
Fig. 6.8
Ex. 8. How does Figure 6.8 illustrate the identities ϊ&η.(\π-χ) = cot x,
tan(i7r+x) ξ —cot x, tan(f-77- — x) = cot x, tan(j7r+x) = —cot xl
Ex. 9. Sketch roughly the graphs of tan 3x and |tan x\.
Figure 6.9 shows the secant function (continuous line) and cosecant
function (dotted line).
Ex. 10, How does Figure 6,9 illustrate the identities
sec (in- x) = cosec x, sec (\n + χ) = -cosec x,
sec (Ιπ— χ) = —cosec χ, sec Цп+х) = cosec χΊ
Ex. 11, Sketch roughly the graphs of sec 2x and |cosec x\.
97

THE TRIGONOMETRIC FUNCTIONS
[6
4. INVERSE TRIGONOMETRIC FUNCTIONS
The sine function has no inverse, because if we take any number ζ in the
range — 1 < ζ < 1, we can find any number of values χ such that
sin χ = z. However, if we restrict the domain of the sine function to
-\ττ < χ < \tt (the range remaining -1 < у < 1)
then the inverse function does exist, for the mapping becomes one to one,
The inverse function so defined is called the arcsine function, Thus, arc-
sine is a function from
{xeR: -I ^ χ ^ 1} to {yeR: -%п < у < \ή)
with the property that arcsin χ = у if sin у = χ, and conversely, (Notice
that we retain χ as an element of the domain.) Another notation for arcsin χ
is sin-1 x; the slight drawback to this notation is that it suggests that the
unrestricted sine function has an inverse, which is, of course, false,
The graph of the arcsine function may be obtained directly from the
graph of the sine function for the restricted domain — \ή < χ < \π by
interchanging the χ and у axes. If we retain χ as an element of the domain,
so that у = arcsin x, the graph is as shown in Figure 6.10.
•\-
2
-1
Fig. 6.10 Fig. 6.11
The arccosine function may be similarly defined, with domain
{xeR: -1 < χ < 1} and range {0 < у ^ π}.
If у = arccos x, then cos у = χ. The graph is shown in Figure 6,11,
Again, the arctangent may be defined, with domain R and range
{-\π < у < %π} (notice the strict inequality), If у = arctan x, then
tan у = χ. The graph is shown in Figure 6,12,
Ex. 12. Suggest appropriate domains and ranges for arccotangent, arcsecant and
arccosecant.
Ex. 13, Write down the values of arcsin \, arccos 1/^2 and arctan (-1).

INVERSE TRIGONOMETRIC FUNCTIONS
Example 6. Simplify cos (arcsin x).
Let arcsin χ = у; then —\π < у < \ή and sin у = χ
cos2 у = 1 — sin2 у = 1-х2.
Thus cos у = ± V(l -x2); but — |-7г < j < \ή and so cos у > 0; it follows
that cos (arcsin x) = V(l -*2)·
Exercise 6{c)
1. Show that the function/denned by
f(x) = cos л: + sin χ
is periodic and sketch its graph.
2. Draw the graph of the function/defined by
f(x) = sin χ + sin 2x
for values of χ in the interval 0 < χ < 2π. Solve the equation
sinx + sin2x = 0,
giving values of χ for which 0 < χ < 2π.
3. Draw the graph of у = tan x°, from χ = - 20 to χ = 70, plotting points at
intervals of 10°.
Using the same axes and intervals, draw also the graph of
у = cos (x+10)° from χ = 20 to χ = 50.
Read from your graph the value of χ when tan x° = cos (x+10)°.
(O&C'O')
4. It is required to find an angle χ such that
sin л:
= τ*
where χ is measured in radians.
Draw on the same diagram the graphs of у = sin л: and у = f χ for values of
χ from 0 to \ή, taking 1 in, (or 2 cm) to represent τιπ on the scale for x, and
0-2 on the scale for y.
99

THE TRIGONOMETRIC FUNCTIONS [6
Estimate from your graphs the required angle, giving your answer in degrees
to the nearest degree. (O & С Ό')
5. Solve graphically the equation tan 2x° = 2 cot x°, giving all solutions χ such
that 0° < x° < 180°.
6. Draw the graph of у = 1 — 2 sin 60x° for values of χ from 0 to 6, using 2 cm
as unit on both axes. By drawing a suitable straight line on the same diagram,
read the solutions, within the given range of values of x, of the equation
2(1 - 2 sin 60x°) = x. (O & С Ό')
7. Solve graphically the equation
sin χ = cos2;t+l
for values of χ for which 0 < χ < п.
8. Using the same axes sketch the graphs of: (i) sin x, (ii) 2 sin x, (iii) 1 + 2 sin x.
9. Using the same axes, sketch the graphs of (i) sin x, (ii) -sin x, (iii) 2-sin x.
10. Sketch the graph of 2(1 -cos x).
11. Sketch the graph of 1 -tan x.
12. Sketch the graph of 1 + |sin x\.
13. Sketch the graph of sin (х + &т).
14. Sketch the graph of 1 +2 cos (χ-^π).
15. Find the values of:
(i) sin (arccos i); (ii) cos (arcsin -£);
(iii) tan(arccot V3); (iv) sin(arctan—1);
(v) cos(arccot — 1/V3); (vi) sec (arccos y);
(vii) cosec (arcsin i); (viii) cot (arcsin -iV3)·
16. Simplify the expressions (i) tan (arcsin x); (ii) sin (arctan x).
17. Sove for χ the equation tan (arccot x) = 2 sin (arccos x).
18. Simplify the expression sec {arccos [tan (arccot *)]}.
What restriction must be placed upon the value of χΊ
19. If arcsin χ = я·-arccos у, find a relation between χ and y.
20. Prove that:
(i) arcsin χ + arccos χ = %n;
(ii) arcsin [cos (arcsin x)] + arccos [sin (arccos x)] = in.
(O & С modified)
Miscellaneous Exercise 6
1. A chord PQ of a circle of radius r subtends an angle 2Θ rad at the centre of
the circle and 2Θ < n. Find an expression for the area of the smaller of the two
portions into which PQ divides the circle.

4] MISCELLANEOUS EXERCISE 6
If this area is a fraction 1 /2n of the area of the circle, prove that sin 2Θ = 20 — 1,
Obtain graphically an approximate solution in radians of this equation.
(O&C'O')
2. If a circle of radius a is drawn with its centre on the circumference of another
circle of radius a, find the area common to both circles,
3. Find the maximum and minimum values of the expression
sin2 0 + 6 sin 0 + 4.
4. Solve the equation
12 cos2 θ = 6 + sin0
giving all values of Θ, in radians, lying between 0 and 2π,
5. A globe representing the earth consists of a sphere of radius 10 cm, Find the
length of the circumference of the small circle representing the 30° parallel of
latitude,
Two points A, B on the 30° parallel of latitude differ in longitude by 90°,
Calculate:
(i) the length of the straight line joining A and B;
(ii) the angle subtended by AB at the centre of the globe;
(iii) the great circle distance between A and B, in cm, correct to 3 s.F.
(O&C'O')
6. Show, by sketching the appropriate graphs, that the equation
x+cosx = 1
has only one root, What is this root ?
7. Prove that the identity
sin2 χ cos2 y—cos2 χ sin2 у = sin2 x—sin2 у
is true for all values of x, y. Is the identity
sin2 χ cos2 y—cos2 χ sin2 у _
sin2 x— sin2 у ~
also true for all values of χ and yl Give reasons for your answer.
8. Solve the simultaneous equations
sin(x + .y) = УЗ,
cos(2x-x) = -У2
giving all pairs of values of x, y, lying between 0 and n.
9. Eliminate θ between the two equations
χ = tan Θ—sin Θ,
у = tan<9 + sin<9.
10. Sketch the graphs of у = x and у = cos χ and hence sketch the graph of
у = x + cosx.
Sketch the graphs of:
(i) у = |x+cos x\; (ii) у = \x\ +cos x; (iii) у = x+ |cos x\.
101

THE TRIGONOMETRIC FUNCTIONS [6
11. Draw an accurate graph of the function у = \n sin χ between χ = 0 and
χ = π. Draw in the same diagram the lines у = mx for m = 1, τ, τ, f, 1 ·
Determine the values of χ where these lines cut the graph of у = \ττ sin χ
giving your answers in the form kn, where к is correct to two decimal places.
Use the values obtained to draw a separate graph of
, sin* ,
у = \π between χ = \π and χ = π.
12. Show by means of a rough graph that the equation cos χ = х/2тт has
2n +1 positive roots when л is a positive integer. Show also that, if r is a positive
integer less than n, the root nearest to 2rn is 2m+a, where a, is the smallest
positive root of the equation
cos* = - + -—.
η 2nn
Draw a careful graph of у = cos χ for - \ή < χ < \ή and from it find the five
positive roots of the equation
4π cos χ = x.
Give your answers in the form kn, where к is correct to two decimal places.
(O&C)
13. Show that θ = 2mr + a, and θ = (2n +1) π-a,, where η is an integer, both
satisfy the equation sin θ = sin a. Prove, furthermore, that any solution of the
equation must have one or other of these forms.
θ = 2m + a or θ = (2n + 1) π- a is called the general solution of the equation
sin θ = sin a,. Find forms for the general solution of
cos θ = cos a and tan θ = tan a.
14. Find the general solutions of the equations
(i) tan λ: = cot(x + in); (ii) 2 cos2 χ -cos χ - 1 = 0.
15. AB is a chord of a given circle, centre О and radius a, subtending an angle
θ at 0 (θ < π). Prove (i) that the area of the triangle OAB is \аг sin Θ, and (ii)
that the area of the minor segment of the circle of which AB is the chord is
\α\θ-ύηθ).
Prove also that the area of the part of the circumcircle of the triangle АО В
lying outside the given circle is
16. In a circle centre O, two radii OP, OQ contain an angle θ radians (θ < π).
If the area of the sector OPQ is A and the length of the chord PQ is 2c show that
1 — cos θ = сЮ\А. Draw on the same diagram on squared paper the graphs of
the functions у — 1 — cos θ and у — сЮ/А in the particular case where
A = 7rcm2, с — 1-3 cm
for values of θ between 0 and ή. (Take 1 cm to represent -&п on the χ axis and
4 cm to represent one unit on the у axis.) Read off from your graphs the value
of θ between 0 and π which satisfies the equation 1 - cos θ = c29/A in this
particular case. Use your value of θ to find the radius of the circle in this case.
(Cambridge)
17. The function/: R+ -> R is defined ЬуДх) = χ sin \/x. Sketch the graph of the
function.
102

7· Probability infinite outcome spaces
1. ARRANGEMENTS AND SELECTIONS:
COUNTING LARGE NUMBERS
Before discussing the question of probability, we shall introduce a few
techniques whereby the procedure for counting up large numbers may be
made more efficient. Our reason for doing so is that, in probability, we are
frequently faced with the problem of deciding in how many ways an event
can occur; since the numbers involved might be large, it is desirable to
arrive at them by the simplest method available.
Consider first the following problem: in how many ways may the four
aces from a pack of cards be arranged in a row on a table ? By direct counting,
it is not difficult to arrive at the correct answer, which is twenty-four.
However, it is simpler to argue thus: consider four spaces on the table; the first
space may be filled in any one of four ways: for each of these four ways,
the next space may be filled in three ways and so on, The total number of
ways is thus 4x3x2x1 = 24.
The result of multiplying together all the integers from 1 to л is written
η! (read 'factorial η'). Thus, the number of ways of arranging the four aces
may be written 4!.
Ex. 1. Verify that 2! = 2,3! = 6,4! = 24. Write down the values of 5!, 6!, 7!.
The result above may be generalized in an obvious way: the number of
ways of arranging η unlike objects in a row is η!,
Now suppose we alter our question: how many different rows of four cards
may be made, given a standard pack of fifty-two cards! Direct counting
is now out of the question—it would take far too long, But an identical
argument to that employed above shows us that the number of ways is
52x51 χ 50x49 = 6497400. We may write this in the alternative form
52!
48!'
Again, our argument may be generalized in an obvious way: the number
of ways of arranging r objects chosen from η unlike objects is
nx(n-l)x(n-2)x ... x(n-r+l) =(-~ry,·
103

PROBABILITY IN FINITE OUTCOME SPACES [7
This is often referred to as the number οι permutations of η unlike objects
r at a time and is written nPr. Thus
(Note that 0! is defined to have the value 1 and so this formula remains
true when r = n.)
Ex. 2. Evaluate 4P2, SPS> 6P6.
Ex. 3. How many different' words' can be made using the five letters of the word
'after' ? How many of these begin with A and end with R?
We may modify our question again and ask: 'how many different hands
of four cards may be dealt from a standard pack Τ The difference here is
that we are not concerned with the order in which the cards are dealt:
we only want to know the number of selections (sometimes called
combinations) of four cards that can be made from fifty-two cards. Suppose for the
moment that this number is N. Then each of these N different hands may
be rearranged in 4! = 24 ways. There are then 24N different arrangements
of four cards. But we already know that the number of different
arrangements is 52 χ 51 χ 50 χ 49. Thus
= 270725.
24
Once more, our argument may be generalized: since there are nPr
possible arrangements of η objects taken r at a time, the number of
selections, or combinations, of η unlike objects, taken r at a time, is
nPr = П\
r\ r\{n-r)V
The number of combinations of η unlike objects, taken r at a time, is
written either as nCr or I I. The latter notation is used almost universally
now, although the former has its merits: we can read „C, as 'n choose r',
reminding us that we are choosing, or selecting, rather than arranging.
We shall retain „Cr in this chapter, but revert to ( I in later chapters,
when the need arises. 7_s
Щ^г)У
(Notice that again, since 0! has been defined as 1, „C„ = 1 by the formula,
which accords with the commonsense result that there is just one way of
selecting η objects from n.)
104
0-*-

1] ARRANGEMENTS AND SELECTIONS
Ex. 4. Evaluate SC2,6C3.
*Ex. 5. Evaluate nC0 by the formula, and interpret your result.
*Ex. 6. Evaluate nCr and nCn_r by the formula and interpret your result.
Ex. 7. In how many ways may a cricket team of eleven boys be chosen from
fourteen boys available? If three particular boys are certain to be chosen, how many
ways are there of completing the team?
Example 1. A committee of three is to be chosen from four men and three
women. If at least one man is to be included, how many possible selections are
therel
First solution. There are three possibilities:
(i) Two women included: number of ways of selecting the women
= 3C2 = 3; for each of these choices there are 4Cj = 4 ways of selecting
the remaining committee member, who must be a man.
Total (i): 3x4 = 12.
(ii) One woman included: number of ways of selecting the woman
= 3Q = 3; for each of these choices there are
г 4х3 ή
ways of selecting the remaining committee members.
Total (ii): 3x6 = 18.
(iii) No woman included: the committee is selected entirely from men,
4C3 = 4 ways.
Total (iii): 4.
The total of possible committtees is thus
12 + 18+4 = 34.
Second solution. There are
_7x6x5
,Сз"1х2хЗ ^
possible committees in all. Just one of the committees consists entirely of
women, and so 35- 1 =34 contain at least one man.
It is often necessary in probability questions to know in how many
ways it is possible to arrange η objects in a row, given that r are alike of
one kind, s are alike of a second kind and so on. We start by distinguishing
between the like objects by attaching suffixes to them: for example, if we
have r letters a, s letters b, ... then we write our η letters as
<h, Ун аз> ···> ar, K, b2, b3, ..., bs, ....
There are η J ways of arranging these η letters. But, in any one of these
105

PROBABILITY IN FINITE OUTCOME SPACES [7
arrangements, the letter a^, α2, a3, ..., ar may be rearranged amongst
themselves in r! ways; similarly the letter b1; b2, b3 bs may be arranged in si
ways and so on. Thus, if we drop the suffixes, the number of
distinguishable arrangements becomes
rlsl...'
Hence we have the following important rule.
Given η objects, r alike of one kind, s alike of another kind and so on, the
number of arrangements of the η objects in a row is
For example, the number of different arrangements of the letters of the
word SELECTIONS IS
2^j = 907200.
Ex. 8. Tn how many ways may four letters Ρ and six letters Q be arranged in a
row?
We complete this section by mentioning one further technique for
counting which is frequently useful in probability: to find the number of ways
of arranging η objects, r at a time, if repetitions are allowed. We shall
illustrate the method adopted in Example 2.
Example 2. In a simplified football coupon there are ten matches whose
results (home win, 1; away win, 2; draw, χ) are to be forecast. In how many
ways may the coupon be completed!
The first result may be forecast in three different ways (1, 2, x); for
each of these choices there are three different choices for the second
match (1, 2, χ) and so on. The total number of ways of completing the
coupon is thus 310 = 59049.
Ex. 9. A multiple choice paper consists of ten questions, to each of which is
attached five possible answers, labelled A, B, C, D, E. A candidate selects one
of these answersfor each question; in how many ways may he complete the paper?
Exercise 7(a)
1. Five boats are entered for a race. Assuming that they all finish and that there
are no dead heats, in how many possible orders can they pass the finishing line?
2. In a form of thirty boys, a first and a second prize are to be awarded; in how
many ways can this be done?
106

1] ARRANGEMENTS AND SELECTIONS
3. You have a form on which you have to give your first six choices for
university, in order of preference. If you have a list of twelve universities offering the
course you want, in how many ways can you complete the form?
4. A fruit machine has three windows, in each of which appears independently
one of six pictures. How many different arrangements (taking the order into
account) are possible?
5. How many five-digit numbers can be formed from the digits 1, 2, 3, 4, 5,
using each digit once? How many of them are even? How many are even and
greater than 30000?
6. A committee of four is to be selected from six Labour and six Conservative
M.P.s. How many possible committees are there? In how many will the
members of the Labour Party have a majority?
7. In how many ways may a tennis team of six members be selected from fifteen
available players? In how many ways may a first and second team be chosen?
8. Twelve people are to divide up into three sets of four players for a whist drive.
In how many ways can this be done?
9. Of ten electric light bulbs, three are faulty but it is not known which. In how
many ways may three bulbs be selected? How many of these selections will
include at least one faulty bulb?
10. How many different bridge hands (thirteen cards) are there which contain
(i) all four aces, (ii) three aces and one king? (Leave your answer in factorial
form.)
11. Criticize the following attempted solution of Example 1:
Ά man must sit on the committee, and he can be chosen in four ways. The
remaining two members may now be chosen arbitrarily from among the six
remaining people: this may be done in 6C2 = 15 ways. Thus, the total number of
possible committees is 4 χ 15 = 60.'
12. Prove that the numbers of distinguishable arrangements of η objects in a
row, if two are alike and the rest different, is \n\.
What is the number of distinguishable arrangements if three are alike and the
rest different ?
How many different telephone numbers can be made using all the digits of
the number 4225267?
13. How many different arrangements are there of the letters of the word
queue? In how many of these arrangements do the letters QU appear together,
in that order?
14. Find in how many ways a batting order (eleven men) may be made if Smith
is to bat before Brown.
15. A diagonal of an и-sided polygon is a line joining two non-adjacent vertices.
How many diagonals does an и-sided polygon possess?
16. In how many ways can two Is, two 2s and two 3s be thrown with six dice?
107

PROBABILITY IN FINITE OUTCOME SPACES [7
17. Given η unlike objects, find the number of ways of dividing them into three
unequal groups of sizes p, q, r where p+q + r = n.
In how many ways can they be divided into two groups of size ρ and one of
size n—2p,if no attention is paid to the order of the groups?
18. A pair of integers is selected from the set of positive integers 1, 2, 3, ..., n.
In how many ways may this be done? [In each pair the order of the integers is
immaterial, e.g. (2, 3) and (3, 2) count as one pair only.]
If the integers in each pair are multiplied together show that, in the case when
η is odd, the number of products which will be odd integers is \{n2 — 1). If
η is large show that this number is approximately one-quarter of the total number
of products. (Cambridge)
2. RANDOM EXPERIMENTS AND
OUTCOME SPACES
The subject of probability deals with 'experiments' which may have a
number of possible outcomes; more specifically, it seeks to assign a
numerical measure to the likelihood of obtaining various possible results
if such an experiment is conducted, and thereby enable us to analyse the
situation mathematically. Experiments of this nature, in which the results
obtained depend upon chance, may be called random experiments.
We may take, as three typical examples of random experiments:
(i) A coin is tossed and the result (head or tail) noted.
(ii) From the very large output of a machine producing electrical
components, a sample of twenty components is drawn and each component in
the sample is tested in turn to decide whether or not it is faulty.
(iii) A person is tested for blood group {A, B, AB or O).
Each of the above experiments has the property that its outcome may
be one of a (finite) number of possibilities. A set whose elements represent
all the various distinct possible outcomes of a random experiment is called an
outcome space for that experiment and its elements are called elementary
events. Notice particularly the words 'all' and 'distinct' in this definition:
all the possible outcomes must be represented in the set and no elementary
event can correspond to more than one possible outcome of the experiment.
(The word 'space' is used because the outcomes of an experiment are
frequently represented as points in geometrical space;' outcome set' would
perhaps be a preferable term but we shall follow the customary usage.
Alternative terms in use are 'sample space' and 'possibility space'.)
For example, denoting by r the numbers of defectives found in a sample
in experiment (ii) above, a possible outcome space would be
Si ={reZ:0 < r < 20}.
It should be noticed that we have talked of ал outcome space rather than
108

2] RANDOM EXPERIMENTS AND OUTCOME SPACES
the outcome space: any other set whose elements represent distinct
possible outcomes and which exhausts all the possibilities will do, e.g.
S2 = {G, F, P},
where G denotes a good result (no defectives);
F denotes a fair result (one or two defectives);
Ρ denotes a poor result (more than two defectives).
Generally speaking, it is best to choose as an outcome space one that
gives as much detail as possible about the result obtained, but there are
exceptions to this and other criteria may be adopted.
Ex. 10. Two coins are tossed and the result (in terms of heads and tails) noted.
Suggest three possible outcome spaces.
Ex. 11. A card is drawn from a pack and its value and suit noted. Criticize the
following outcome space:
S = {card is an ace, card is a heart, card is neither a heart nor an ace}.
Ex. 12. Suggest an outcome space for Ex. 11 if the experiment is concerned only
with the drawing of a heart or an ace from a pack of cards.
Ex. 13. A count is made of the numbers of girls and boys in a family. Under
what circumstances would the set
S = {there are more girls than boys, there are more boys than girls}
constitute an outcome space?
3. PROBABILITY DISTRIBUTIONS
We now assign to each elementary event st of our outcome space S a
positive fraction^ (i.e. 0 < pt < 1) called the probability that the outcome
of our experiment will be s{. Furthermore, for consistency, we shall make
the sum of all the probabilities/?» over the entire outcome space 1. Thus,
the probabilities associated with an outcome space S = {sly s2 s„} are
real numbers such that
0 < Pi < 1, (1)
Pi+P2+Ps+-+Pn = 1. (2)
Such a set of probabilities is said to constitute a probability distribution
for the given outcome space.
It will be observed that so far we have attached no meaning to the
numbers p{—we have only placed certain restrictions upon their possible
values. Provided we observe these restrictions we have a mathematically
meaningful system. However, it is desirable that a mathematical system
should have some relevance to the physical world. If a mathematical
system describes some physical situation, we are said to have created a
109

PROBABILITY IN FINITE OUTCOME SPACES [7
mathematical model of that situation. In using a probability distribution
as a mathematical model of a situation we shall demand that the probability
Pi shall be, in some sense, a numerical measure of our degree of belief
that the experiment will result in the outcome j».
It should be carefully noted that the assigning of a probability
distribution to an outcome space S constitutes an assumption about the
experiment under consideration. Certain' natural' ways of assigning probabilities
are discussed below. In many cases, a 'natural' way of assigning
probabilities will appear so obvious that the tacit assumptions made may be
overlooked. It is a sound point of self-discipline to pause to consider the
assumptions made when embarking upon any question in probability.
We now give two examples of ways in which probabilities may be
assigned to the elementary events of typical outcome spaces. The final
justification for such probability distributions is that calculations based
upon them are supported by empirical evidence.
(i) Symmetry among the possible outcomes may make it reasonable
to assume that all the elementary events have equal probability. For
example, in the case of a die, it may reasonably be assumed that no one
face is more or less likely to appear uppermost if the die is thrown than
any other. Then sinceрг +p2 +... +pe = 1,
Pi = Pi = Рз = Pi = Ps = Pe = 6-
If the same probability is assigned to each elementary event of an outcome
space S, the resulting probability distribution is said to be uniform and the
elementary events are said to be equiprobable.
(ii) Previous repetitions of an experiment (conducted under constant
conditions) show that outcome A occurs a per cent of the time, В occurs
b per cent of the time, etc., where A, B, ... are quite distinct. Then our
accumulated experience suggests that the probability of securing outcome
A is aj 100, that of securing В is й/100, etc. As an example, suppose that a
large number of samples of size 20 are taken from the output of a machine
manufacturing electrical components and that, on average, 0-94 per
sample turn out to be defective. Then we may reasonably take the
probability of any components being defective as 0-94/20 = 0-047. Notice that
such an assignment of probabilities satisfies the requirements (1) and (2)
on p. 109.
Ex. 14. Of the outcome spaces you suggested for Ex. 10 which do you think may
reasonably be supposed to be uniform?
Ex. 15. Two dice are thrown and the total score is noted. Criticize the assigning
of a uniform distribution to the outcome space
S = {2, 3,4, ...,12}.
110

3] PROBABILITY DISTRIBUTIONS
Ex. 16. Suggest an outcome space for the experiment of throwing two dice
that may be given a uniform distribution.
Ex. 17. From the weather records of a certain town taken over the past thirty
years, on the average five days in November have been recorded as foggy. Is
it justifiable to assume that the probability of a November day in that town
being foggy is i?
Ex. 18. The number of boys and girls in a family of five children are noted.
Justify the assumption that the outcome space
S = {there are more girls than boys, there are more boys than girls}
has a uniform distribution (i.e. ρ = \ for each of the two elementary events).
4. PROBABILITIES OF EVENTS
Any subset Ε of the outcome space S of an experiment is called an event.
(Notice that the use of the word in 'elementary events' is consistent with
this definition, since {s} с S.) If we have assigned a probability
distribution to our outcome space it becomes meaningful to consider the
probability of the event Ε (by which we would mean, in a practical example, a
measure of our degree of belief that the experiment will result in one of the
elementary events belonging to E). A moment's consideration should
show that the following definition is plausible, at least in simple cases.
Given a sample space S with associated probability distribution, and an
event Ε с S, then the probability of the event E, written Pr (E\S) (read: 'the
probability of event Ε given S'—more explicitly, 'the probability of Ε
given the probability distribution of the outcome space 5") is defined as
the sum of the probabilities of all the elementary events belonging to E.
More formally, if S = {sly s2, ..-, s„} andp{ is the probability associated
with s{, and if Ε = {sri, sr2, ..., srm} then
Pr (E\S) = pn +pn +p„ + — +prm-
Example 3. If a card is drawn from a well-shuffled pack, what is the
probability that it will be an acel
A suitable outcome space, S, is the set of 52 elementary outcomes
corresponding to the fifty-two different cards in the pack.
To continue, we must make an assumption about the probability
distribution. Since the pack is well shuffled, we shall consider this as a case of
fifty-two equiprobable events and so associate with each element of S the
probability -fe. The event Ε is the subset of S consisting of the drawing of
the ace of spades, hearts, diamonds or clubs.
Then Pr (E\S) = -h +W + A +-s\- = -h-
Ill

PROBABILITY IN FINITE OUTCOME SPACES [7
Example 4. What is the probability of securing a hand of thirteen cards all
of one suit in a game of bridge!
We take as our outcome space the set consisting of all possible distinct
bridge hands, i.e. a set with 62C13 elements.
If the cards have been well shuffled we may assume that these hands
are equiprobable and so the probability of getting a specified hand is
13! 391/52!.
Of these hands, four consist of one suit only. The required probability is
thus 13! 39! 4/52!, which works out to be roughly 1-6 x KH1.
*Ex. 19. How would you interpret the probabilities obtained in Examples 3 and
4?
Ex. 20. What is the probability of obtaining one head and two tails if three coins
are tossed?
Ex. 21. What is the probability of obtaining a total of more than 10 from the
throw of two dice?
Ex. 22. There are one thousand tickets issued in a lottery and prizes are awarded
for twenty of them. What is the probability of any specified ticket securing a
prize?
Ex. 23. What is the probability that a bridge hand contains (i) just one heart
(a singleton heart); (ii) a singleton in just one suit? (Leave your answers in
factorial form.)
Probability is occasionally formulated in terms of odds rather than as
a fraction. If the probability that a horse will win a race is estimated as $,
then the probability that it will not win is £ and the odds against its winning
are 3 to 1 (sometimes written as '3 to 1 against')· Again, if the probability
that another horse will win a race is §, then the odds infavour of its winning
are 2 to 1 (or '2 to 1 on').
Ex. 24. The odds on horse A to win a race are 3 to 1 on and on horse Л 4 to 1
against. Write down the probabilities
(i) that A wins;
(ii) that Л wins;
(iii) that A does not win.
Exercise 7(b)
1. Consider families consisting of six children, all of different ages. The 'type'
of family is defined by an ordered set of the form {B, B, G, B, G, B} the elements
of which represent the sex of each child, starting with the eldest. How many
different types are there? In how many different types are there three boys
and three girls? What is the probability that a family of six children will consist
of three boys and three girls?
112

4]
PROBABILITIES OF EVENTS
2. In question 1 an outcome space
S = {0, 1,2, ...,6}
is given, where the elements represent each of the possible number of boys in a
family. Assign a probability distribution to this space.
3. A number is chosen at random from among the integers 1, 2, 3,..., 20. What is
the probability that
(i) it is a multiple of 3 or 7?
(ii) it is a multiple of 3 or 5 ?
4. If a committee of four is chosen at random from ten women and ten men,
what is the probability that there will be two women and two men serving?
5. If, in Question 4, the male chairman is certain to be re-elected, what is now
the probability of equal numbers of men and women ?
6. Two numbers are selected at random from the integers 1 to 10. (You may
assume that all numbers are equally likely to be selected and that the same
number may be selected twice.)
The elements 0, 1,2, ..., 9 of the outcome space represent the magnitude of the
difference between the two numbers. Assign a probability distribution to this
space which you feel represents a suitable mathematical model for the
experiment.
7. If two numbers are selected at random from the numbers 1 to 10, what is
the probability that the larger number will be greater than 8 ?
8. If two cards are drawn from a pack of fifty-two cards, what is the probability
that
(i) they will both be spades,
(ii) they will be of the same suit ?
9. Two people are asked independently to write down an integer between 1 and
10 (inclusive). The sum, s, of the two numbers is then calculated. Write out an
outcome space for the possible values of s and assign a probability distribution,
on the assumption that each person is equally likely to select any one of the ten
numbers.
What is the probability that s will be prime?
10. In a mixed bag of screws there are twice as many large as small. 10 % of the
large screws are defective and so are 5 % of the small screws. Assuming that every
screw has an equal probability of being selected, what is the chance of picking
a defective screw ?
Comment upon the assumption made.
5. THE ADDITION LAWS
We are now in a position to prove some simple theorems concerning the
probabilities of compound events. Throughout we shall suppose that the
outcome space S and the probability distribution are given.
113

PROBABILITY IN FINITE OUTCOME SPACES [7
Theorem 7.1. If' ΕΎ, E2 are two events, then
Pr (Ег U E2\S) = Pr (^|5) + Рг СВД-Рг (Ег η E2\S).
Proof. The sum of the probabilities of the elementary events in E1 и E2
is equal to the sum of the probabilities in Ег and E2, less the sum of the
probabilities in Ex Π Ε2, since this has been included twice. The result now
follows from the definition of the probability of an event.
Theoreml .2. Ifthe events Ег,Е2 are mutually exclusive {i.e. ifЕгГ\ Е2 = 0),
then
Pr (Ег U £2|S) = Pr СВД + Рг (E2\S).
Proof: Pr(E1 П E2\S) = Pr(0 \S) = 0, by the definition of the
probability of an event. The result now follows immediately from Theorem 7.1.
Theorem 7.3. If Ε is any event
Pr(E'\S) = l-Pr(£|S).
Proof. Since Ε U E' = S and Ε Π Ε' = 0, this is a special case of
Theorem 2.
Theorem 7.4. If Elt E2, ..., En are mutually exclusive events
{Et[\E,= 0, idpj)
then Pr(U^,|s) =Pr(JE,1|S) + Pr(JE,2|S) + ...+PrCE,n|S).f
A formal proof of this theorem may be had by employing mathematical
induction (see Chapter 9). However, its truth is intuitively fairly obvious
if we consider a Venn diagram and note that, since no two of the Et
intersect, the sum of the probabilities in UE{ is obtained by adding up the
probabilities for Ex then for E2 and so on.
Example 5. Two dice are thrown. What is the probability of scoring either a
double, or a sum greater than 9?
We take as our outcome space the set of 36 pairs (i,j), where i,j run
independently from 1 to 6. We make the assumption that the events are
equiprobable and so attach a probability to each elementary event of -Зд6-.
Wrlte E, = {(1, 1), (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)},
E2 = {(4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6)},
t The notation U means the union of the η sets Eu E2, ...,E„; that is.
Εχ U Ег U ... U Е„.
114

51 THE ADDITION LAWS
and so Ε1{\ΕΛ = {(5, 5), (6, 6)}.
We thus have
Pr (£, |S) = Pr (E, \S) = -h; Pr (Ε, η Ε,\S) = &
and so Pr (£j и E2 \S) = i + έ - "A. by Theorem 7.1,
= A-
Example 6. Two cftce are thrown. What is the probability of not getting a
double ?
Making the same assumptions as in Example 5 and using the same nota-
ti0n' Pi(Ei\S)=l-Pr(E1\S), by Theorem 7.3,
= 1-*
= i
Ex. 25. In a class of boys, one-third have black hair and one-quarter have brown
eyes. What deductions can you make?
Ex. 26. If a card is drawn from a well-shuffled pack what is the probability thai
it is either an ace or a king?
Ex. 27. If a card is drawn from a well-shuffled pack, what is the probability that
it is an ace or a heart?
Ex. 28. If a number is selected at random from the integers 1,2,..., 30 what is the
probability that it is
(i) divisible by 2;
(ii) divisible by 3;
(iii) divisible by 6;
(iv) not a multiple of 2 or 3 ?
Ex. 29. If one thousand tickets are issued in a lottery in which there are two
first prizes, eight second prizes and ten third prizes, what is the probability of not
securing a prize with one ticket ?
6. THE MULTIPLICATION LAWS
We shall now consider how we must modify our estimate of the
probability of an events if we are given information in addition to our outcome
space and its allied probability distribution. Suppose that Ε and F are
two events (with Pr (F\S) φ 0) and we require to estimate the probability of
Ε knowing that F occurs. Essentially, we have a new outcome space F
and we have to determine Pr (E\F). (We define Pr (^l-F) to mean
Pr (Ε η F\F).)
115

PROBABILITY IN FINITE OUTCOME SPACES [7
Now since F is the new outcome space, the new probability distribution
for F must have the property that the probabilities of the elementary
events contained in F sum to 1. Making the plausible assumption that the
relative proportion of the weights attached to the elementary events of F
remain unaltered, this is equivalent to scaling up the probabilities in .Fby
a factor 1/Pr (F\S). The probability that Ε will occur is thus Pr (Ε η F\S),
scaled up by this factor ljp(F\S). We are thus led to make the following
definition:
Fv(E\F)
Fr(Ef)F\S)
?r(F\S) ·
Example 7. A bag contains twenty balls, ten of which are red, eight white
and two blue. The balls are indistinguishable apart from the colour. Two balls
are drawn in succession, without replacement. What is the probability that
they will both be red?
We take as our outcome space S the set of 20 χ 19 = 380 possible
selections of two balls. Since the balls are indistinguishable apart from
colour, we may impose a uniform probability distribution upon S.
Now define Кг as the event 'the first ball picked is red', R2 as the event
'the second ball picked is red' and similarly Wlt W2, Въ B2.
We require to find Pr (R-, η R2\S).
Now Pr (R^S) = ^ and Pr (R2\Rj)—the probability that a red ball is
selected from a bag now containing nine red, eight white and two blue
balls is -j%.
Thus, pr № n ^ |S) = pr №|jy pr ^ |S)
= &·**
= &·
Example 8. With the data of Example 7, what is the probability that we obtain
a blue and a white ball (in either order) ?
Adopting the notation of Example 7, we are required to calculate the
probability of obtaining one or other of the two mutually exclusive events
Βτ η W2 and W1 П B2; that is Pr {(B1 η W2) U (й^ П B2)\S).
Now
Pr {(B1 η W2) U (Wx η BJIS} = Pr (Bx η *ВД + Рг (й^ П B2\S),
since the events are mutually exclusive using Theorem 2;
= Pr (H^J.Pr (Я^ + Рг (52|й^).Рг (W^S)
= -h.
116

6] THE MULTIPLICATION LAWS
The arguments employed for the solution of the preceding examples may
be represented diagrammatically by a probability tree (see Figure 7.1).
Starting at О on the extreme left, the end-points of the first three branches
represent all the possible (mutually exclusive) results of the first draw.
The end-points of the next three sets of branches represent all the possible
Fig. 7.1
outcomes of the second draw, the various sets corresponding to the
different draws on the first round. The various probabilities are then attached
to the branches as shown and the probability of any desired path from
О to some end-point on the right may be found by multiplying the
successive branch probabilities together.
For example
Pr^n ад = ■&*# = ft,
Pr (B2\S) = Pr (R, η B2\S) + Pr (^ П B2\S)+Vr (Btf B2\S)
(events mutually exclusive)
= Pr СВД.Pr №|jy + Pr (^|S).Pr (B2\ Wy
+рг(ад.Рг(ад)
= «.ft+A.ft+A-iV
= ΤΓο·
(Can you see a simpler method of arriving at this last probability?)
117

PROBABILITY IN FINITE OUTCOME SPACES [7
Probability trees are particularly helpful if a number of probabilities
have to be read off. If only one probability is required, a simplified tree may
often help. For example, in Example 7 we are essentially concerned with
two alternatives: red and not red. Figure 2 shows a suitable tree for this
example.
<20
,0
<z
Pr (R, η R2\S) = Pr №|5).Рг (R^R,)
A particularly important case of the multiplication law arises in the case
of independent events. Intuitively, two events are independent if neither
has an effect upon the other. To make this notion mathematically precise
we define two events A and В to be independent if
Pr {A n B\S) = Pr (A\S).Pv (B\S).
Comparison of this definition with that given above for conditional
probabilities shows that the assumption of independence is equivalent to
the assumption that
Fv (A\S) = Fr (A\B).
In words, the occurrence of В does not lead us to review our estimate of
the probability of A.
Ex. 30. Prove that, if Pr (A n B\ S) = Pr (A \ S) Pr (B\ S), then Pr (A \ S) = Pr (A \ B).
Example 9. A coin is tossed and a die thrown. What is the probability of
securing a head and a six!
On the assumption that the two events are independent, the required
probability is clearly \ χ \ = τι.
118

61 THE MULTIPLICATION LAWS
Ex. 31. A bag contains two white balls and six red balls; a second bag contains
four white balls and four red balls. The balls are indistinguishable apart from
colour. What is the probability of obtaining two white balls if
(i) one ball is drawn from each bag;
(ii) two balls are drawn from the first bag;
(iii) two balls are drawn from the second bag?
Ex. 32. Criticize the following argument:
' A ring of anti-ballistic missiles is estimated to give a probability of 0-2 of
destroying any incoming missile. A set of five such rings will therefore render
a city immune to missile attack.'
Ex. 33. 'In a form of twenty-four boys, six failed their English examination and
four failed their Mathematics. The probability that a boy, selected at random
from the form, failed in both English and Mathematics is thus & χ -& = ■&.'
Do you think this argument is essentially correct?
Ex. 34. A coin is spun twice and a die is thrown twice. Find the probability of
obtaining at least one head or at least one six (or possibly both).
Exercise 7(c)
1. What is the probability of not throwing a six with four throws of an
unbiased die? What is the probability of throwing at least one six?
2. On average, 2 % of the electric light bulbs of a certain type prove to be
faulty: what is the probability that a batch of twelve such bulbs will be free
of defectives?
3. If four people are chosen at random, find the probability that no two of them
were born on the same day of the week. (M.E.I. Ό')
4. A bag contains a dozen apples, of which three are bad. If two apples are
withdrawn at random, find the probability that
(i) both are good ;
(ii) both are bad;
(iii) one is good and one bad. (M.E.I. Ό')
5. Show that there is a better chance of throwing at least one 6 with four throws
of a single die than there is of throwing at least one double 6 with twenty-five
throws of a pair of dice. (This is a modification of a famous problem in the
history of probability, first proposed by a gambler, the Chevalier de Mere and
transmitted by him to the French mathematician Fermat, who managed to solve
it.)
6. A bag contains two red, three white and four blue balls. If two balls are drawn
in succession without replacement, what is the probability of drawing a red and a
blue ball, in either order? What is the corresponding probability if the first ball
is replaced before the second is drawn ?
7. A bridge hand consisting of thirteen playing cards contains two aces. If
five cards are drawn at random from the hand, find the probability that the five
will contain neither of the aces. (M.E.I. 'C)
119

PROBABILITY IN FINITE OUTCOME SPACES [7
8. A teacher has twelve pupils in his form, eight boys and four girls. On a school
outing he can provide transport in his car for three pupils and decides to draw
lots for the seats. Construct a tree to show on its branches the probabilities as to
the sex of the winning pupils in the different stages of the draw. Hence, or
otherwise, find the probability that those winning will contain
(i) two boys and one girl,
(ii) at least two girls. (M.E.I. Ό')
9. Given five different letters and their respective envelopes, in how many ways
can one letter be placed in each envelope if this is done at random?
Find in how many cases only two of the letters will be in their correct
envelopes.
What is the probability that at least three of the letters are in their correct
envelopes? (M.E.I. Ό')
10. What is the probability that, after a pack of cards has been well shuffled,
two aces will lie at the top ?
11. A bag contains three red, four white and five blue balls, indistinguishable
apart from their colour. If two balls are drawn successively, without
replacement, estimate the probability of obtaining two balls of different colours.
12. A machine has N components, each of which has a probability of 2/3 of
breaking down. The machine will function provided at least one of its
components is functioning. What value should be chosen for N if the machine is to be
99% efficient?
13. A pack of fifty-two ordinary playing cards is shuffled and a card is
withdrawn. D denotes the event that the card is a diamond, К that it is a king, R
that it is red. Prove by calculating the appropriate probabilities that D and К
are independent, that К and R are independent, but that D and R are not
independent.
What is the value of Pr [(Z)> nK)U(Dn #')]? (M.E.I.)
14. Three cards are drawn from a pack of fifty-two cards and, when they have
been replaced and the pack shuffled, a second set of three cards is drawn. Find
the chance that the six cards drawn should include at least one ace. (M.E.I.)
15. A bag contains five white and three red balls. Balls are drawn in succession
and are not replaced. Show that the chance that the first red ball will appear
at the fifth draw is 3/56. (M.E.I.)
16. Three bags, А, В, С, contain respectively three white and two red balls, four
white and four red balls, five white and two red balls. A ball is drawn unseen
from A and placed in B; then a ball is drawn from В and placed in C. Find the
chance that if a ball is now drawn from С it will be red. (M.E.I.)
17. How many times should an unbiased die be thrown if the probability that
a six should appear at least once is to be greater than 9/10? (M.E.I.)
18. A batch of fifty articles contains three which are defective. The particles are
drawn in succession (without replacement) from the batch and tested. Show that
the chance that the first defective met will be the /-th article drawn is
(50-/·) (49-/0/39200. (M.E.I.)
120

6] THE MULTIPLICATION LAWS
19. Two men A and В play a game in which A should win eight games to every
seven won by B. If they play three games, show that the probability that A will
win at least two games is approximately 0-55. (Cambridge)
20. The chance of any one engine of a four-engined aeroplane failing on a long
journey is 5 %. If only one engine fails the chance of the aeroplane completing
the journey is 80 %; if two engines on opposite wings fail, its chance of
completing the journey is 50 %. It cannot fly with two engines out of action on the
same wing. Find the chance that the aeroplane will complete the journey.
(O&C)
21. (i) From a bag containing five red, four white and three green balls, three
are drawn together at random. Find the chance of their being
(a) all of different colours,
(b) all of the same colour.
(ii) Two six-faced dice whose faces bear the numbers 1 to 6 respectively are
thrown together. Find the chance of
(a) the total score being exactly 8,
φ) the total score being greater than 8. (O & C)
7. EXTENDED USE OF THE MULTIPLICATION
LAWS
Note. To gain precision we have so far always specified the conditioning
set, even if it is the entire sample space (whose associated probability
distribution is known). In this latter case, however, little harm is done by
dropping the reference to S; thus, if the outcome space is S and an event
Ε is considered as a subset of S, it is customary to refer to Pr (E\S) simply
as Pr (E). Provided no ambiguity arises we shall adopt this simplified
notation.
For example, if A, В are independent, we shall write
Pr(4 η 5) = Рг(Л).Рг(5).
More generally, if А, В are any two events in an outcome space S
Pr (Α η Β) = Pr (A\B).l?r (В).
In pursuing applications of the multiplication law, we first notice that,
since the operation of intersection is commutative, that is, since
Α η Β = Bf) A
we have, for any two events A, B
Pr (Α\Β).Ρτ (B) = Vr(A(\B) = Pr (Β η A) = Pr (B\A).Pr (A).
This form of the multiplication law is often useful, for example, when it
is desired to calculate Pr (A\B) and the three probabilities Pr (A), Pr (B),
Pr (B | A) are known or easily found.
121

PROBABILITY IN FINITE OUTCOME SPACES ['
Example 10. Two bags contain coloured balls as shown in the table below.
BagI
Bag II
Red
3
0
White
2
4
Blue
1
2
A die is thrown; if a I or 6 appears, then bag I is chosen, otherwise bag II.
A ball is then drawn from the selected bag. If the result of the throw of the die
is unknown, but merely that a white ball is drawn, what is the probability that
it came from bag II
We begin by making the assumption that the balls are indistinguishable
apart from colour and that we have an equiprobable outcome space.
Let I denote the event 'bag I is selected',
II denote the event 'bag II is selected',
W denote the event' a white ball is selected'.
ThenPr(I) = i;Pr(II) = f.
W may occur in either of two mutually exclusive ways: bag I is selected
and a white ball is drawn; bag II is selected and a white ball is drawn.
Thus we have
7r{W) = Pr(^nl) + Pr(^n II)
= Pr (W|I).Pr (I) + Pr (W|lT).Pr (II)
Pr(I|PT)
_ Pr (I n W)
~Pr(0O
Pr(^nl)
Pr(0O
Pr(^|l).Pr(I)
Pr (W)
which is the required probability.
The technique mentioned above may be generalized to yield a result
commonly known as Bayes's Theorem. (The Reverend Thomas Bayes
(1702-61) was one of the earlier writers on the subject of probability. The
theorem which bears his name was published posthumously in 1764.)
122

7] EXTENDED USE OF THE MULTIPLICATION LAWS
Before considering the general theorem we shall solve an example for
a special case (n = 3).
Example 11. Three machines produce the same type of electrical component,
20 % of the total output coming from machine A, 50 %from machine В and
30 %from machine C. Tests conducted in the past show that 5 % of the
components from A and 1 %from each of В and С prove faulty. A component
selected at random from the total output is proved to be faulty. What is the
probability that it came from machine ΑΊ
Call A the event 'component comes from machine A',
В the event 'component comes from machine B',
С the event' component comes from machine C,
F the event 'component found faulty'.
We have to determine Pr (A\F).
Now, from the data, we may assign the following probabilities:
Pr (F\A) = xj^; Pr (F\B) = y^; Pr (F\C) = χ^;
Pr (A) = -fts; Pr (B) = yVo; Pr (C) = -f^-.
Now Pr (A\F) = Pr (A n F)/Pr (F) and so the problem reduces to
determining Pr (A n F) and Pr (F), given the numerical values above. But
Fr(Af\F)= Vr(F(\A)
= Pr(F\A).J>r(A)
= τοπ·
Again F = (F() A) U (Ff] B) U (F(] C) and these three events are
mutually exclusive (no component can come from more than one machine).
ThUS Pr (F) = Pr (F(] A) + ?r (F(] 5) + Pr (F(] C)
= Pr (F\A). Pr (A) + Pr (F\B). Pr (B) + Pr (F\ C). Pr (C)
-Th + iU + rAo
Thus 50°" PrC^^f
= f
Ex. 35. Of two pennies, one is double-headed, the other normal. If one of the
pennies is selected at random and tossed twice, what is the probability of
obtaining two heads? If one of the coins is selected at random, tossed twice and gives
two heads, what is the probability that it is double-headed?
Ex. 36. Find the probabilities that the faulty article of Example 11 came from
(i) machine B, (ii) machine C.
123

PROBABILITY IN FINITE OUTCOME SPACES [7
Ex. 37. With the data of Example 11, find the probability that a component
selected at random and shown to be not faulty comes from machine A.
Ex. 38. Two articles come from one of the machines of Example 11, but it is
not known which one. On testing, both are found to be faulty. Find the
respective probabilities that they come from machines А, В and C.
Notice that in Example 11 we had an outcome space partitioned into
three mutually exclusive and exhaustive events А, В, С, and were required to
find the probability of A conditional upon the occurrence of some event F.
More generally, we define a partition of a set S as a set of subsets
{НЪН2,Щ, ...,Hn)
such that the intersection of any two of the H's is empty and the union of
the H's is S, that is
Given a set S, the set {Hlt Я2, ..., #„} of subsets forms a partition of
Sif
(i) Hif) H} = 0,1 «ξ i <j «ξ n;
(ii) ΰ# = S.
Theorem 7.5 (Bayes's Theorem). Given apartition Hu Я2, ..., Hn of an
outcome space S and an event Ея. S,
Vx{H}\E)
Рг(ЭД)Рг(Яг·)
Рг(^|Я1).Рг(Я1) + Рг(^|Я2).Рг(Я2)+...+Рг(^|Яп).Рг(Яп)·
Proof. Since the H{ are exhaustive,
Ε = {Ε η Щ U {Ε П Я^ U ... U {Ε η Яп).
But the Hi are mutually exclusive and it follows that the Ε η Щ are also
mutually exclusive. Thus, by Theorem 7.4
Pr(£)= Pr(£n Я^ + Рг^п Я2) + ...+Рг(£п Нп)
= Pr (ВДЭ.Рг (Я^ + Рг (£|Я2).Рг (Я2)+ ... +Рг (£|Яп).Рг (Яп),
but
Рг(Я,п£)
Рг(Я3-|£)- Fr(E)
_Рг(Е[)Щ
Рг(£)
_Рг(£|Я^.Рг(Я,)
Рг(£)
Рг(ОД).Рг(Я,)
= Рг(£|Я1).Рг(Я1) + Рг(£|Я2).Рг(Я2) + ... +Рг(£|Яп).Рг(Яп)·
124

7] EXTENDED USE OF THE MULTIPLICATION LAWS
The value of this theorem lies largely in its application to inferential
problems in statistics. A number of hypotheses H{ are made (hence the
conventional use of the letter Η in the theorem) and the probability
of one hypothesis Д, given the occurrence of some event Ε is determined in
terms of the probabilities of the Щ without the prior knowledge of the
event E. (This is sometimes referred to as the determination of the a
posteriori probability of the event H} given the apriori probabilities of the
events H{, 1 < i < n.) The major practical difficulty in its application lies
in the estimation of the apriori probabilities Pr (Щ.
In solving problems involving Bayes's Theorem it is advisable always
to work from first principles as in Example 11, and not to quote the
theorem in its general form.
Exercise 7(d)
1. One of four dice is known to be biased, showing a 6, on average, three times
as often as each of the other scores. A die is chosen at random and thrown three
times and three 6s appear. What is the probability that it is one of the unbiased
dice?
2. A factory manufactures three different qualities of light bulb, А, В and С in
the ratio 1:2:3. The bulbs are indistinguishable in external appearance but
extensive tests indicate that, on the average, 1 % of type A, 4 % of type В and
5 % of type С are below the advertised standard. A batch of six, all of the same
type is sent in error without a distinguishing label. If none of these turns out to be
defective, estimate the probability that they were of type A.
3. Of a large sample of items, 40 % were produced by machine A and 30 % by
each of machines В and С The machinery was very unreliable, machine A
producing 10% defective items, machine В 20% defective items and machine С
30 %. If an item, selected at random from the sample, proves to be defective,
what is the probability that it came from machine A ?
4. A card is missing from a pack of 52 cards. If this is the only information you
have, what is the probability that the missing card is a spade? The pack is well
shuffled and the first card is removed and proves to be a spade. What would your
assessment of the probability that the missing card is a spade be now? The card
removed is now replaced and the pack shuffled. The top card again proves to be
a spade. What is your assessment of the probability now?
5. Four bags, I, II, III, IV, contain white and black balls as shown in the
following table.
Bag I II III IV
Number of white balls 12 3 4
Number of black balls 9 8 7 6
A die is thrown; if a one appears, bag I is chosen, if a two or three, bag II,
if a four or five, bag III and if a six, bag IV. A ball is then drawn at random from
the bag selected.
5-2
125

PROBABILITY IN FINITE OUTCOME SPACES [7
If you are told that a black ball has been drawn, what should your
assessment be of the probability that it came from bag I ?
6. A certain rare disease from which one in ten thousand of the population suffers
is diagnosed by a test which reveals the presence of the disease in 95 % of the
cases of those tested who actually have the disease. However, it also incorrectly
yields a positive reaction in 1 % of the cases of those who are not suffering from
the disease. If a person selected at random from the population shows a positive
reaction, what is the probability that he is actually suffering from the disease?
7. One of four pennies is known to be double-headed, the other three being
normal. One of the pennies is selected at random and tossed three times. If the
result is three heads, what is the probability that the coin tossed is the double-
headed penny?
8. An engineer is responsible for servicing three small computer installations
А, В, С The probability that А, В, С will break down on any day are 2го, го, too
respectively. The probabilities that the three computers will not be in operation
after 7 p.m. are ^, §·, $ respectively. On one day the engineer receives a message
that one of the computer installations has broken down at 8 p.m. but the
message does not specify which computer. Calculate the probabilities that the
breakdown is at А, В, С and so decide the first installation which should be
checked by the engineer. (M.E.I, adapted)
9. The probability that a radio message sent from a ship is not picked up by a
land-station is p. The land-station replies to all messages received and its
transmission back has probability q of not being picked up by the ship. Find the
probability that the land-station failed to receive the message if the ship gets
no reply.
Miscellaneous Exercise 7
1. A number is drawn at random from the set of thirty-seven numbers
0, 1,2, ...,36,
and is then replaced.
(i) What is the chance that the number drawn will be odd?
(ii) What is the chance that in two consecutive draws the numbers drawn will
be the same ?
(iii) What is the chance that in one hundred consecutive draws the number 0
will not be drawn ?
(iv) What is the chance that in five consecutive draws none of the numbers
will be odd?
Give the answers to (i) and (ii) as vulgar fractions, and evaluate the answers
to (iii) and (iv) approximately as decimals. (Cambridge)
2. An encyclopaedia consisting of n(>4) similar volumes is kept on a shelf
with the volumes in correct numerical order; that is, with volume one on the
left, volume two next and so on. The volumes are taken down for cleaning and
replaced on the shelf in random order. What are the probabilities of finding
exactly я, (я—1), (я—2), (я—3) volumes in their correct order on the shelf?
(Cambridge)
126

7] MISCELLANEOUS EXERCISE 7
3. You have available two pairs of dice. The dice of one pair are both true but,
in the other pair, one is true and the other is known to be biased in some way.
Which pair would you choose if you were playing a game in which to win you
had to throw (a) a score of 7; (6) a double?
4. Two persons X and У play a game in which X deals from a standard pack of
fifty-two cards. Each player receives three cards and the game is won by X if he
holds more red cards than there are black cards held by Y.
By considering the number of black cards in play or otherwise, calculate the
probability that the dealer wins a game. What odds should be fixed by the dealer
in order that he may expect just to show a profit? (M. E.I. Ά')
5. Four cards, the aces of hearts, diamonds, spades and clubs are well shuffled
and then dealt two to player A, the other two to player B. A is then asked whether
at least one of his two cards is red. He replies in the affirmative. In the light of this
information we wish to calculate the probability that he holds both the red aces.
Consider the argument:
'We know that he has one red ace; without loss of generality we may suppose
that it is the heart ace. Among the other three cards there is no reason why one
more than another should be the diamond ace: one only out of three equally
likely possibilities gives A both the red cards: the required chance is thus $.'
Criticize this argument and produce a correct argument and answer. (C.S.)
6. A bag contains twelve coloured balls: 4 yellow, 3 white, 2 black, 2 red and
1 green. A prize is offered for obtaining 2 red balls in two draws, there being an
option of either replacing or not replacing the first ball. Which would you choose ?
Generalize your result to the case of a bag containing N balls of s different
colours where there are m, balls of colour /, and prove your conjecture.
7. Raffles of two types, A and B, are held. In raffles of type A there are 2N
tickets, of which In carry prizes; in those of type В there are N tickets of which
η carry prizes. Which would you prefer: two tickets in a raffle of type A, or two
tickets in separate raffles of type ΒΊ
8. A six-digit telephone number (i.e. any number between 100000 and 999999)
is dialled at random. Show that the probability that one or more digits are dialled
in succession (as in 322916 or 322216 or 322911 but not as in 329216) is about
41 %. (Cambridge)
9. An experiment consists in drawing balls one at a time from a bag containing
eight white and two black balls initially. Balls drawn from the bag are not
replaced. A note is kept of the number r (1 =ζ r =ζ 9) of the draw which first yields
a black ball. If the experiment is repeated many times what is (i) the most
probable value of /-; (ii) the probability that r exceeds 5 ? (Cambridge)
10. Articles are distributed among three boxes in such a way that each article
has the same chance of being placed in any one of the boxes. Find the probability
that at least one of the boxes remains empty when r articles have been
distributed.
Deduce that, if r articles are similarly distributed among five boxes, the
probability that exactly two boxes remain empty is бР'"1 - 2r + V/S*-1. (Oxford)
127

PROBABILITY IN FINITE OUTCOME SPACES [7
11. A bag contains m oranges and η lemons: sampling is at random without
replacement. Obtain the probabilities of the following events:
(i) the first draw is an orange;
(ii) the first two draws are oranges;
(iii) the second draw is a lemon, given that the first was an orange;
(iv) the third fruit is an orange. (Cambridge)
12. In a game of bridge each of four players is dealt thirteen cards after the pack
has been shuffled. One of the players has four diamond cards and he can see four
more diamonds in the exposed dummy hand, which has been put down on the
table by one of the players. What is the chance that the five remaining diamond
cards are split three in one hand and two in the other?
If, in the first three rounds of play, each player has used exactly one diamond
and two other cards, what is the chance that one of the players has no diamond
cards remaining in his hand? (M.E.I.)
13. Half the population of the city of Ekron are Philistines and the other half
are Canaanites. A Philistine never tells the truth but a Canaanite speaks
truthfully with probability f and falsely with a probability f. What is the probability
that a citizen encountered at random will give a correct answer to a question?
Tabulate (as fractions, decimals or percentages) the probabilities that 0,1,2, 3
men out of a sample of three citizens taken at random from Ekron will affirm a
proposition (i) when it is true, (ii) when it is false.
The proposition has a prior probability of $ of being true. What is the
(posterior) probability of its being true conditional on its being affirmed by only one
out of three ? (Cambridge)
14. A farmer keeps two breeds (A, B) of chicken; 70% of the egg production is
from birds of breed A. Of the eggs laid by the A hens, 30 % are large, 50 %
standard and the remainder small; for the В hens the corresponding proportions
are 40 %, 30 % and 30 %. Egg colour (brown or white) is manifested
independently of size in each breed: 30 % of A eggs and 40 % of Л eggs are brown. Find
(i) the probability that an egg laid by an A hen is large and brown;
(ii) the probability that an egg is large and brown;
(iii) the probability that a brown egg is large;
(iv) which size grade contains the largest proportion of brown eggs;
(v) whether colour and size are manifested independently in the total egg
production. (Cambridge)
15. A and В are two events and B' is the complementary event to B. Show that
Pr (A) is between Pr (A\B) and Pr (A\B'). (M.E.I.)
16. Prove that, if A and В are independent events, so also are A and B'. If
Alt A2, As are three mutually independent events such that
Pr (At nA'jnA'1)=pi (i φ у ф к),
then Pr04i) = ptKpi + λ),
where λ is a root of a certain cubic equation, which should be found.
17. A motorist driving along the main street of a town encounters two sets of
traffic lights 660 feet apart. Both sets of lights change every 30 seconds and the
second set are timed to change to 'Go' 10 seconds after the first set. The motorist
128

η MISCELLANEOUS EXERCISE 7
drives at 30 m.p.h. but if forced to stop at the lights, he accelerates uniformly from
rest at 4 ft/s2 until he attains that speed. Estimate the probabilities of his finding
(i) both sets of lights in his favour;
(ii) both sets of lights against him;
(iii) only one set of lights against him.
Suggest two ways in which the relative timing of the two sets of lights could
be adjusted to ensure that the probabilities would be the same if travelling in
either direction and calculate the probabilities.
(Assume throughout that his progress is not interfered with by other vehicles,
and also, that he drives at 30 m.p.h. right up to the lights and then stops
instantaneously if they are red at that instant.) (SMP)
18. Three players, Χ, Υ, Ζ, play a game and at each round of the game they have
an equal probability of winning that round. The player who wins each round
scores one point and the game is won by the first person to score a total of
3 points. X wins the first round; calculate the probability that Twins the game.
(M.E.I.)
19. The basic premium under a certain motor insurance policy is £A p.a. The
scale of No Claims Bonus provides 25 % discount after the first year without
claim, 33J% after two consecutive years free of claim and 50% after three or
more consecutive years free of claim. In addition a preferred policy holder's
discount of 20 % is given on the basic premium (which may already be discounted
by the No Claims Bonus) for four consecutive years without claim. If a preferred
policy holder's discount has been granted it is not lost by reason of a claim.
Suppose that for a given driver there is a constant probability q of no claim in
a given year. Evaluate the probabilities for the possible premiums in the sixth
year of the policy, in terms of q. (M.E.I.)
20. n2 balls, of which η are black and the rest white, are distributed at random
into η bags, so that each bag contains η balls. Determine the probability that at
least one bag contains no black ball.
129

8. Finite series and the Binomial Theorem
1. SEQUENCES, SERIES AND THE
Σ NOTATION
In Chapter 4, we saw that a function maps each element of the domain
into an element of the range. A function whose domain is the set of positive
integers, Z+ = {1, 2, 3, 4, ...}, is called a sequence and the elements of the
range are called terms of the sequence, f
For example, we could have as the terms of a sequence the numbers
1, 3, 5, 7, 9, ...,
where 1 (the first term) is the image of the integer 1; 3 (the second term) is
the image of the integer 2; 5 (the third term) is the image of the integer 3
and so on (see Figure 8.1).
12 3 4 r
13 5 7 2r-l
Fig. 8.1
It is common usage to refer to the range of a sequence simply as ' a
sequence'; thus we should say that the numbers 1, 3, 5, 7, 9, ... 'form a
sequence'. Other examples of sequences are
1, 2, 4, 8, 16, ...,
1, 4, 9, 16, 25, ...,
1, Ц, 2, 21, 3, ...,
1,V2, 2, 2V2, 4, ....
If we call the first term of a sequence щ, the second term щ and so on,
the rth term (general term) of the sequence will be denoted by ur. A sequence
itself is often written {ur} (provided this notation causes no confusion with
the set consisting of the single element ur). For example, the first few terms
t It is possible to extend the definition of a sequence to include Ζ as the domain, but
we shall restrict ourselves to Z+ in this book, or, occasionaly to Z+ U {0}, that is, the set
of natural numbers with zero.
130

1] SEQUENCES AND SERIES
of the sequence {r3-l} are 0, 7, 26, 80, 124, .... Again, the sequence
1, 3, 5, 7, 9, ... could be written {2r- 1}—but see Ex. 5, which illustrates
the danger of assuming that the form of the /th term is determined by the
first few terms of a sequence.
Sequences are sometimes defined by giving, say, the first term and a
relation between the (r- l)th and rth terms. For example, the sequence
defined by the relation _ , ι (i\
and the first term щ = 1
has, for its second term 2.1 + 1 = 3, for its third term 3.3 + 1 = 10,
for its fourth term 4.10+ 1 = 41 and so on.
*Ex. 1. What are the first four terms of the sequence given by
и, = /■«,_!+1,
(i) if«! = 0; (ii) ifUl = 2?
A relation such as (1), connecting general terms of a sequence, is called a
recurrence relation. As another example of a recurrence relation we might
have , ™
щ = иг_х + иг_г. (2)
To define this sequence {ur} we need two initial terms (why?). Thus, if
щ = 1, щ = 1
then (2) defines the sequence
1, 1,2,3,5,8, 13,....
Ex. 2. Write down the first four terms of the sequences defined by the following
recurrence relations and initial terms:
(i) ur = 2иг_ъ u1 = \; (ii) ur = «r_!+1, «i=l;
(iii) ur = rur_x + r\ Ul = 0; (iv) щ - Ur-г = i-O'r-i - и,-.),
иг = 1, и2 = 2;
Ex. 3. Write down the first six terms of the following sequences:
(i) {Ъг}; (ii) {3r-5}; (iii) {2-'}; (iv) {r(r*-1)};
(v) {l+(-l)'}; (vi) {l/Wr+l)]}; (vii) {f-»};
(viii) {r+(- 1)4/·+1)}.
Ex. 4. Suggest possible fifth and sixth terms for the following sequences, whose
first four terms are given, indicating how you arrived at them:
(i) {1,4,7,10,...}; (ii) {1, -hi, -*,·..};
(iii) {1,2,4,7,...}; (iv) {0,3,8, 15, ...};
(v) {1,3,6,11,...}.
131

FINITE SERIES AND THE BINOMIAL THEOREM [8
*Ex. 5. Why was the word 'suggest' used in Exercise 4?
Work out the first four terms of the sequence {/-4 - 10/·3 + 35/·2 - 50/·+25}.
Work out the fifth term of this sequence.
Can you see an easy way of constructing the general term of a sequence which
has the property that its first five terms are zero, but its sixth term is non-zero?
Given a sequence it is often necessary to obtain the sum of its first η
terms. If the first η terms of a sequence are written down and connected by
+ signs, we are said to have a series. For example
1+3 + 5+7 + ...+(2и-1),
l+i+i+i + ... + 1/л
are the series obtained respectively from the sequences {2r— 1} and
{iW·
If η is known, such sums may be found simply by obtaining the
necessary terms and adding them one by one, but the process is naturally
laborious. In this chapter we shall explore more sophisticated methods for
summing certain simple series, but, before doing so, it is desirable to
introduce a shorthand notation for sums, the Σ notation (read 'sigma
notation').
We write щ + щ + щ +... + un as Σ щ. Thus, for example,
r=l
Σ/·= 1+2 + 3 + 4= 10;
Σ (3r-l) = 2 + 5 + 8 + 11+14 = 40.
r=l
It is possible to assume that uT is defined by values of r φ Ζ+:
Σ0·2 + 1)= 1+2 + 5 + 10= 18;
Σ (r2+r + l)= 1+1+3+7 = 12.
r=-l
The general rule for finding Σ uT is to determine the value of uT for r = m
and all subsequent r, in steps of 1, until we arrive at un and then to sum all
the terms so obtained.
(0 If!;
(iv) Σ(-1)'ι·;
r=l
(vii)Jo/^.
132
(ii) ΣΟ-»-,);
(ν) Σ (ή + 2);
(iii) Σ/·(2'-1);
(νί) Σ (/·2-1);
г=-2

1] SEQUENCES AND SERIES
Ex. 7. Evaluate:
(ί) Σ&-+2); (ii) L·*; (iii) Σ ή;
(iv) Σ(ι--1)»; (ν) Σ ^; (vi) Σ(-1)·>»;
/1 1 \
(vii) Σ 2'; (viii) Σ
ЙГ
Ex. 8. Write in Σ notation:
(i) 1 + 2+3 + 4; (ii) 1-2 + 3-4;
(iii) 2+4 + 6+8; (iv) 1+2 + 4 + 8 + 16;
(v) l+i + i+i+i + i; (vi) 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6.7;
(vii) 1-1 + 1-1 + 1-1 + 1-1; (viii) 1 +2 + 3+4+... + n.
*Ex. 9. Prove that, if {«,} and {vr} are two sequences, then
Σ («, + »,) = Σ uT+ Σ vT.
*Ex. 10. Prove that, if {«,} is any sequence and a a fixed number, then
Σ auT = α Σ иг.
♦Ex. 11. Interpret and evaluate Σ с, where с is fixed. (That is, find the sum of the
first η terms of the constant sequence defined by ur = c, all r.)
2. ARITHMETIC SEQUENCES
Sequences such as {1, 3, 5,7, ...},{\\,2\, 3|, 5, ...},{4, 1, -2, -5, ...}are
called arithmetic sequences, their distinguishing feature being that the
difference between any pair of successive terms is constant. If we call this
common difference d, and the first term a, then the arithmetic sequence
takes the form {a + (r- 1) d).
Ex. 12. Write down the /-th terms of each of the following arithmetic sequences:
{1,4,7,...}, {1, 2£, 4, ...}, {2,0,-2,...}.
Ex. 13. Find to which integer r the following terms correspond in the given
arithmetic sequences:
(i) 91 in {1, 3, 5, ...}; (ii) 139 in {3, 7, 11, ...);
(iii) -253 in {2, -1, -4, ...}; (iv) 61-7 in {3-2, 4-7, 6-2, ...};
(v) -67in{7i,6i, 5|,...}.
133

FINITE SERIES AND THE BINOMIAL THEOREM [8
Ex. 14. Taking what you consider to be the simplest continuation in each case,
write down, in Σ form, the sum of the first η terms of each of the following
series:
(i) 1.4.7 + 4.7.10+7.10.13+10.13.16+...;
(ii) 2.3.5 + 4.6.9 + 6.9.13 + 8.12.17+...;
(iii) 21 + 44 + 6' + 810+...;
,., 1.3 3.7 5.11 7.15^ .
0ν)ΊΓ + —+-r + —+ ...;
1 2 3 4
W 1.2.3 3.5.10 5.8.17 7.11.24'
The sum to η terms of the arithmetic series (sometimes called the
arithmetic progression and abbreviated A.P.) is easily obtained by the
following argument.
Consider first the sum of the first η positive integers, that is Σ r- Writing
this sum as S we have
S = 1+2 + 3 + .. .+n
and S=n + (n-l) + (n-2) + ... + l.
Adding: 2S = Oz + l) + Oz + l) + 07 + i) + ...+()7 + i)
= n(n+l).
Thus 5 = 1я(и + 1).
Now, for the general arithmetic series,
| [a + (r- 1) d] = Σα+άΣ r- | d (see Ex. 9 and Ex. 10)
= na + idn(n + l)-nd
= M2a + (n-l)d].
This method is easily applicable to any arithmetic series and depends only
upon memorizing the single important result
Σ^ = Μη + 1).
Example 1. The tenth term of a certain arithmetic series is -29 and the
twentieth term is -69. Find the sum of the first twenty terms.
Let the first term be a and the common difference d. Then
a + 9d=-2% a + l9d = -69,
giving a = 1, d = - 4. The rth term of the arithmetic series is thus
134

2] ARITHMETIC SEQUENCES
7 + (/·— 1) ( — 4) = 11—4/- and the sum of twenty terms is given by
20
Σ (11-4/-)= 220-2.20.21
= -620.
Exercise 8(a)
1. Evaluate:
(i) Z(r* + r); (ii) Σ -; (iii) Σ (l+r + r*).
r=l j=l Г r=-2
2. Write in Σ notation:
(i) 1.3 + 4.6 + 7.9+... (to я terms);
(ii) 1.2.3-2.3.4+3.4.5-4.5.6+... (to>zterms);
(iii) l2-32 + 52-72+ ... (to η terms).
3. Find the first four terms of each of the sequences defined by the following
recurrence relations and initial terms:
(i) ur = 3η^-ι+1, «i= 1;
(ii) «, + 2«Γ_!-3«Γ_2 = 0, «! = 1, «2 = 1;
(iii) щ = «r2_1; Ul = 2;
(iv) 2игиг_1-и2_1 = 3, «ι = 1.
4. If {иг} is an arithmetic sequence with common difference d and first term a,
show that «r satisfies the recurrence relation
uT = iir^ + d.
If sn = Σ и,, determine a recurrence relation involving only sn, sn_u j„_,
and </. What isij?
5. Find the rth term of the following arithmetic sequences:
(i) {3, 12, 21, ...}; (ii) {i, i, i, ...}; (iii) {12, 4, -4, ...}.
6. Find the number of terms in the following arithmetic series:
(i) 1+3 + 5 + ... + 101; (ii) 2+9 + 16+... +485;
(iii) 5 + 2 + (-l) + ...+(-82).
7. Sum each of the series in Question 6.
8. The fourth term of an arithmetic series is -14 and the tenth term is —44.
Find an expression for the rth term and hence obtain the sum of the first twenty
terms.
9. The first term of an arithmetic series is 9 and the sum of the first ten terms
is 450. Find an expression for the rth term and deduce an expression for the sum
to η terms. Check your final result by setting (i) η = 1; (ii) η = 10.
10. The ninth term of an arithmetic series is 24 and the sum of the first nine terms
is 126. Find an expression for the rth term and deduce an expression for the sum
to η terms.
135

FINITE SERIES AND THE BINOMIAL THEOREM [8
11. How many terms of the series 1 + 8+15+ ... are required to give a sum of
5500?
12. How many terms of the series 12 + 6 + 0+... are required to give a sum of
-3348?
13. Show that n n η
Σ(/·+1)2 = Σ/·2 + 2 Ъг + п
and deduce that Σ/· = \η(η+1).
14. If Σ sT = n2 + n, prove that {sT} is an arithmetic sequence, and find the first
three terms.
15. Prove that the sum of η terms of any arithmetic series may be written in
the form an2 + bn+c. Prove the converse of this result: that if
Σί, = an2 + bn+c,
then {sT} is an arithmetic sequence.
16. If a2, b2, c2 are the first three terms of an arithmetic sequence, prove that
(6 + c)"1, (c + a)"1, (a + b)-1 are the first three terms of another arithmetic
sequence (provided b + c + 0, c + αφθ, a + b * 0).
17. If 3 + x+y+29 is an arithmetic series, find χ and y.
If a+x1 + x2+... + xn + b are the (я+ 2) terms of an arithmetic series, find xT
in terms of a, b, r and n.
18. An arithmetic sequence whose first term is 8 has a common difference of i.
Find N, given that the sum of the first 2JV terms is equal to the sum of the next
N terms.
3. GEOMETRIC SEQUENCES
Sequences such as {1,2,4, 8, ...}, {3, - 6,12, - 24, ...}, {1-5, 0-75, 0-375...},
in which each term is a fixed multiple of the preceding term are called
geometric sequences. If we call the first term of the sequence a, and the fixed
multiple, or common ratio, ρ then the geometric sequence takes the form
Ex. 15. Which of the following sets of numbers could be the first three terms of
geometric sequences?
(i) {1,-5,25,...}; (ii) {2,2i,2tf,...} (iii) {1, 4, 9,...};
(iv) {1, -1-1,1-21, ...}; (v) {7, -1, -\,...}; (vi) {1, -1,1, ...}.
Ex. 16. Find to which value of r the following terms correspond in the given
geometric sequences:
(i) 256 in {2, 4, 8, ...}; (ii) 1458 in {2, 6, 18,...};
(iii) 10-30 in {10, 1, 01, ...}.
136

3] GEOMETRIC SEQUENCES
A geometric series {geometric progression, G.P.) is a series obtained from
a geometric sequence. The sum of the first η terms of such a series may be
obtained as follows:
Let S = Σ ар7-1.
If ρ = l.thenS = па.
If Ρ φ l.then c , , . , , n ,
S= a + ap + ap^ + .-.+ap71-1
pS = αρ+αρ2 + ...+αρη-1+αρη
.'. (l-p) S = a —apn.
Since 1 —ρ Φ 0 we may divide by (1 —p) to obtain the following important
reSullt: » /i „ч
l-p
Σ^-^^Ь^ O0*1)·
Since, for \p\ < 1, limp™ = 0 (consult a calculus text-book for the
definition of limits and the proof of this result), we may take
Дор^-а/а-р)
less than any positive number, however small, by taking a sufficiently
large n, provided \p\ < 1. Thus, if we write
lim Σ ар*'1 as Σ αΡτ~χ
we have Σ αΡτ~
(The sum to infinity of a geometric series.)
Example 2. Find the sum of money that must be invested to yield an
income of £50 at the end of that year, and at the end of each of the
subsequent nine years, reckoning on a return of 5% compound interest on money
remaining invested. {Assume that interest is payable once a year.)
For every £1 invested, its value at the end of one year is £1-05, its value
at the end of two years is £1·052 and so on. Thus, to obtain £50 at the end
of the first year, we must invest £РЪ where
>! x 105 = 50.
Again, to obtain £50 at the end of the second year, we must invest £P2,
Whefe P2xl05' = 50
and so on.
137

FINITE SERIES AND THE BINOMIAL THEOREM f8
The total sum to be invested is given by
P1+Pa + ... +P10 = 50[1 05-1 +1-05-2 +... +1-05-10]
= 1000[1-1/10510]
= 1000[1-1/1-6289]
= 386-2.
Ex. 17. Comment upon the number of decimal points used in the calculations in
Example 2 and on the accuracy achieved.
Exercise 8(b)
1. Find the sum of the following geometric series, simplifying your answers as
far as possible, but without evaluating terms of the form />":
(i) 1+4+16+... (12 terms); (ii) 1-2 + 4-8 + ... (20terms);
(iii) V2 + 2 + 2V2 + ... (16 terms); (iv) l+i + i + ... (50 terms)
2. The third term of a geometric sequence is 5 and the eighth is 160. Find the
first term and the common ratio.
3. Find two possible geometric sequences whose fourth term is 9 and whose
eighth term is 144.
4. Which is the first term of the geometric sequence 1, j, i, ... which is less than
io-8?
5. How many terms of the geometric series 1 +i+i+... must be taken to give
a sum greater than 2-10-6?
6. How many terms of the geometric series 1+% + tg + .-- must be taken to
give a sum greater than 4—10-3?
7. If x— 4, x, x+6 are the first three terms of a geometric sequence, find χ and
the fourth term.
8. How many terms of the series 1 +5 + 25 + ... must be taken to give a sum
greater than 1010?
9. A piece of paper is 01 mm thick (so that when it is folded over, the resulting
thickness is 0-2 mm). If a very large sheet of paper could be folded fifty times,
what would be the resulting thickness? (This is a variant of a famous problem;
another form it can take is to find how much corn is required if one grain is to be
placed on the first square of a chess board, two on the second, four on the third
and so on. The answers, when first met, are surprising.)
10. The sum of the first η terms of the sequence {sT} is i(3" -1). Prove that the
sequence is geometric, and write down the first three terms.
138

GEOMETRIC SEQUENCES
11. If £250 is invested at 4 % compound interest, reckoned annually, what will
be its value after six years ?
12. What sum of money must be invested to give an annuity, starting ten years
after investment, of £100 p.a. for five years, reckoning on 4 % compound interest
(calculated annually) on money remaining invested ?
13. Sum to infinity the following geometric series:
(i) 1+-ιττ + τππ+·..; (η) 3 + H + J + ...;
(iii) 1-i+J—....
14. A recurring decimal may be regarded as the sum of an infinite geometric
series; for example _i -i -i
021 = ϊδο+ιοοοο+ιοοοοοο+""
Use this approach to express 0315 as a fractionp/q.
15. If the population of a country increases by 3 % each year, how many years
will it take for the population to double?
16. If in = Σ ар'-1, evaluate Σ sr.
17. If {«,} is a geometric sequence with common ratio p, find a recurrence
relation involving ur and ιιτ_λ. If sn = Σ ur, show that sT satisfies the recurrence
relation
Sn-psn-i = a.
What is i,?
18. If au д2, д3, я4,... are terms of a geometric sequence with common ratio/-,find
the sum of η terms of the series
(aj - Да)2 + (да - α3)2 + (α3 - β^)2 + ....
4. FURTHER SUMMATION; THE USE OF
DIFFERENCES
By factorizing the algebraic expression r{r+1) (r + 2)-(r- 1) r{r+\) we
seethat r(r+l) s £[,(,+1} (r + 2)_(r_i)r(r+i)].
Now suppose we let r take successively the values 1, 2, 3, ..., η in this
identity. Writing the resulting equalities one below the other we have
1.2 = HI.2.3-0.1.2],
2.3 =i[2.3.4-1.2.3],
3.4 = i[3.4.5.-2.3.4],
{n-\).n = \[{n-\)n{n+\)-{n-2){n-\)ri\,
n.(n+l) = i[n(n + 1) (n + 2) - (n-I) n(n +1)].

FINITE SERIES AND THE BINOMIAL THEOREM [8
If all the terms on the left-hand side are added together we have precisely
Σ /■(/" +1); if all the terms on the right-hand side are added together, we are
left with -i 0.1.2 at the beginning and %n(n+l) (n + 2) at the end, all
the others having cancelled out in pairs. Thus
jtr(r+l) = in(n+l)(n + 2).
Ex. 18. Show that
1 1_
~~r r+1
and deduce the result Σ ——— = 1 -.
r=ii-Cr+l) л+1
More generally, if the rth term of the sequence {sr} can be expressed as
the difference between the (r+ l)th term and the rth term of a new
sequence {ur}, that is, if we can find a new sequence {ur} with the property
that, for all r,
then Σ sr = Σ («r+1 - «г) = "и+1 - «ι·
In the example given above, sr = r(r+1) and uT = i(r — l) r{r+1);
in Ex. 18,
sr = λ and ur = -Ijr.
This technique for summing series is called the method of differences; its
use depends, of course, upon our ability to discover the new sequence {ur}.
*Ex. 19. Show that
r(r+Y)(r + 2) = MKM-l)(M-2)(r + 3)-(r-l)r<r+l)(r + 2)]
and deduce that
Σ/·(/·+1)0-+ 2) = Ыя+1)(я + 2)(я+3).
*Ex. 20. Show that
rjr
-1 = !Г_! L__1
+ l)(r +2) 2L/C- + 1) (r+l)(r + 2)J
and deduce that
Ά 1 1 1
r=1K»-+l)(i4-2)"4 2(л+1)(л + 2)·
♦Ex. 21. Prove that
Σ/-0-+1) (/- + 2) (/- + 3) = Ы«+1) (л+2) (л+3) (л+ 4)
and evaluate L(r+l)(r + 2)(r+ 3) (r + 4).
140

4] FURTHER SUMMATION
*Ex. 22. Prove that
Ά 1 = J_ 1
r_i r(r+ l)(r + 2) (/- + 3) 18 3(л+1)(л+2)(л + 3)
It is sometimes worthwhile to manipulate sr into a slightly different form
before applying the method of differences.
Example 3. Evaluate n 1
r{r + 2) r(r + l)(r + 2) (/■+l)(/4-2)"rr(/- + l)(/4-2)·
Then
v^__ = f; 1 +v J
r-ir(r + 2) r=i{r+\){r + 2yr^ir{r+\){r + 2)
= Si lr + Ί ~7+2} + iSx \J{r + V)~(r + l)(r + 2)j
= Ь"^+2]+2Ь"(й+1)(й + 2)]
4 2(л + 1)(л + 2)·
(Note the check by putting л = 1.)
The method of difference is thus seen to constitute a powerful technique
for summing series. The reader should commit the method to memory,
especially the particular approach used for the series given at the
beginning of this section and in Ex. 19 and Ex. 20. We shall complete this
section by using the results already proved to deduce two more important
sums:
(i) j>2 = K"+ О (2л +1); (π) if = \n\n + lf.
Proof of (i): Since
r2 = r(r + l)-r,
Д'2=|К' + 1)-Д>-
= Mn + l)(n + 2)-in(n + l)
= £л(л+1)(2л + 1).

FINITE SERIES AND THE BINOMIAL THEOREM
Proof of (ii):
r(r+l)(r + 2) = r* + 3r* + 2r
and so r3 = r(r + l)(r + 2)-(3r2 + 2r)
= г{г+\){г + 2)-Ъг(г+\) + г.
Thus Σ rs = | r(r+l) (r + 2)-3 Σ /■(/■ + !)+ | '
= *л(л + 1)(л + 2)(л + 3) -л(л + 1)(л + 2) + *л(л + 1)
= £л2(л + 1)2, on simplification.
(Observe that Σ /-3 = ί Σ Α Λ
Example 4. Evaluate n
γΣ (/■+!)(«+/■).
Д(/■ +1) (л + г) = Σ/'2 + г(л +1) + л)
= Σή + (η+1)Σ? + η\ byExs. 9, 10, 11
= £л(л + 1)(2л + 1) + -£л(л + 1)2 + л2
= |л(5л2 + 15л + 4).
1. Sum the following series:
(i) l2 + 22 + 32 + ... + 122; (ii) 13 +23 + 33+... +123;
(Ш) 32 + 42+52 + ... + 202; (iv) l2+32 + 52 + 72+... + (2/z-l)2.
2. By expressing r(r+1) (/-+ 3) in the form /■(/■+1) (/·+ 2) +r(r+1), sum the series
Σ/·(/·+1)0-+ 3).
Find the value of
3. Find
4. Find
5. Find
142
Σι-(τ+1)(τ + 4).
Σ(ι* + ι- + Ζ).
ДКл-г)1.

4] FURTHER SUMMATION
6. Find Σ(/·2 + 2τ)
(i) by expressing it as Σ r2+ Σ lr\
(Η) by expressing it as Σ (/· +1)2- Σ 1.
7. Show that, for;· > 0,
___r = 1 1_
(r+l)(r+2)(r+3) (r + 2)(r + 3) fr+l)(r + 2)(r+3)·
Hence evaluate £__£__.
8. Evaluate ^5 __^__.
9. Evaluate Д___^_.
10. Find the sums to я terms of the following series:
(i) 1.4 + 4.7 + 7.10+10.13 + ...;
(ii) l2.2 + 22.3 + 32.4 + 42.5 + ...;
(iii) 1.2.3 + 4.5.6 + 7.8.9 + 10.11.12+...;
(iv) l2.2.3 + 22.3.4 + 32.4.5+42.5.6+....
11. Show that
x(x+ 1) (2x+1) ξ Ax{x+ 1) (x + 2) (χ + 3) + β(χ- 2) (χ-1) x(x+1)
for certain constant values of A and B, and find these values. Hence find the sum
of the first η terms of the series
1.2.3 + 2.3.5 + 3.4.7 + 4.5.9 + .... (O&Q
12. Prove that
(2я+1)(2и + 3) _ (2я-1)(2я+1)_ 2(2я+1)
(n+D&i+Z) nin+\) ~n(n + l)(n + 2Y
Find the sum of the first η terms of the series
ГЬ + 2Х4 + 3-Ь+- (0&C>
13. Deduce Σ τ·2 from the identity (2r+l)3-(2r-l)3 = 24r2 + 2.
14. Evaluate Σ - „ ■■
15. Prove that the sum of the squares of the first η even numbers exceeds the
sum of the squares of the first η odd numbers by n(2n + 1). Hence, or otherwise,
find the sum of the squares of the first η odd numbers. (O & C)

FINITE SERIES AND THE BINOMIAL THEOREM [8
5. THE BINOMIAL THEOREM
By ordinary multiplication of algebraic expressions
(x + a)1 = x + a
(x + af = x2 + 2xa + a2
(x + af = xs + 3x2a + 3xa2 + as
(x + af = х* + 4х3а + 6х2а2 + 4ха3 + а*
{x + af = x5 + 5xia+10x3a2 + l0x2a3 + 5xai + a5
Notice the following features possessed by these expressions:
(i) the number of terms in the expansion of (x + a)n is η +1;
(ii) the degree of each term in the expansion of (x + a)n is n;
(iii) The coefficients of the terms in the expansion of(x+a)n are given by
the nth line of Pascal's triangle (Blaise Pascal, 1623-62); see Figure 8.2.
1 1
1 2 1
13 3 1
14 6 4 1
1 5 10 10 5 1
Fig. 8.2
Each line of Pascal's triangle starts and finishes with a 1; terms in
between are obtained by adding together the two terms on either side of it in
the row above.
*Ex. 23. Supply the sixth, seventh and eighth lines of Pascal's triangle.
Ex. 24. Check that the expansion of (x + a)6 satisfies conditions (i), (ii) and (iii)
above.
Ex. 25. Show that, if we add the 'zeroth' line, 1, to Pascal's triangle, the
expansion of (x + a)° satisfies conditions (i), (ii) and (iii).
By direct verification as above, we may show that conditions (i), (ii) and
(iii) hold for all suitably small values of η; the Binomial Theorem can be used
to show that they are true for all positive integers n. However, before
proceeding to the general theorem, we shall solve an example by means
of Pascal's triangle.
144

5] THE BINOMIAL THEOREM
Example 5. Obtain the expansion of(2x—3y)s. Hence obtain an approximate
value for 199-75, and comment upon the accuracy of the result.
From the fifth line of Pascal's triangle
(2x-3yf = (2хУ + 5(2хУ(-Зу) + Ю(2хУ(-Зу)* + 10(2х)*(-ЗуУ
+ 5(2x)(-3yy + (-3yf
= 32x5-240xiy + 120xY-1080x2ys + 8Wxyi-243f.
If we put χ = 102, у = 10-\ then (2x-3yf = 199-75. Hence,
199-75 = 3 ·2 χ 1011 - 2-4 χ 109 + 7-2 χ 106 ...
and the last three terms will not affect the first six significant figures.
Direct calculation of the line above shows us that
199-75 = 3-17607 χ 1011, correct to six significant figures.
Ex. 26. Write down the expansions of
(i) (Зх-уУ; (ii) (.2х + уУ; (ίίί) (2χ-1)'; (ίν) (3x + 2yf.
The Binomial Theorem states that, for any positive integer n,
(x + a)n = xn + ("\ x^a + Q x^a2 + ··· + (") *η~ΤαΤ + ■ · · + β"
= Σ (") xn~rar. (Recall that (?\ = l\
Proof. (x + a)n = (x + a) (x + a) (x + a)... (x + a). [n brackets in all]
Since, in forming the product on the right-hand side, one term is chosen
from each bracket, each term in the expansion must be of the form
xn~TaT (choose (n — r) terms x, and r terms a). Now the term xn~TaT may be
obtained by selecting any r brackets and choosing the a term and then
choosing an χ from each of the remaining (n — r) brackets. But there are
I I different ways of doing this and so the coefficient of xn-rar is I 1. The
proof of the Binomial Theorem is now complete.f
We make the following observations:
(i) Since r runs through all integral values from 0 to n, there are (n +1)
terms in the expansion.
(ii) The degree of each term is (n-r) + r = n.
t An alternative proof of the Binomial Theorem is given in Chapter 9.3.
145

FINITE SERIES AND THE BINOMIAL THEOREM
"(/■■
("-1)! [J-+r\
-l)l(n-r-l)l\_n-r+r\
= (")■
the coefficients in the expansion of (x + a)n are connected to those of the
expansion of (* + a)™-1 in precisely the same way as was previously
observed for Pascal's triangle.
M (":;h;'K)
is of some importance and will be used later; it is generally known as
Vandermonde's Theorem.
Example 6. Obtain the first three terms in the expansion of (1 - 2*)20 in
ascending powers of x.
(1-2*)» = (l)» + (^) (1)»(-2χ)4-(*) (l)18(-2*)2
= l-40* + 760*2....
Exercise 8(d)
1. Write down the expansions of
(i) (x-yf; (ϋ) (2χ+№;
(iii) (1+2*)'; (iv) (2x + 6yf.
2. Find the coefficient of *2 in the expansions of each of the following expres-
(i)'(3*+l)4; (ii)(2-*)'; (iii)(l+2*)12; (iv) (3-4*)°.
3. Write down the first three terms in the expansions of each of the following
expressions in ascending powers of *:
(i)(l+2xf; (ii)(l+!*)"; (iii) (2-*)'»; (iv) (3*-*)'.
4. Write down the general term in the expansion of

5] THE BINOMIAL THEOREM
Hence determine the constant term in this expansion (that is, the term not
involving x).
5. Determine the constant term in the expansion of
/ , M6
6. Find correct to five significant figures, the values of
(i) (1-01)6; (ii) (0-998)8; (iii) (9-99)s; (iv) (2-98)s.
In each case, justify the accuracy of your result.
7. Find, without using tables, the value of
(2 + V5)4 + (2-V5)4.
8. Write down the coefficient of a3x7 in the expansion of (a + x)10. Find the
coefficient of а3Ь*с3 in the expansion of (a + b + c)10.
9. Expand in ascending powers of x, as far as terms in x2, each of the following
expressions:
(i) (1+x + x2)4; (ii) (1-x+x2)6; (iii) (1 +2x-x2f; (iv) (2-x-2x2)6.
10. If a device measures linear dimensions with an error of less than \%, find
the possible percentage margin of error if it is used to measure the volume of a
cube.
11. Find the sum of the coefficients of the powers of χ (including x°) in the
expansion of (l+x)n.
If(l+x + x2)4 = ao + аг χ + α2χ2+... + αβ Xе, evaluate
(0 Σα,; (Η) Σ(-Ι)'*.
12. Prove Vandermonde's Theorem by considering the identity
(1+x)"*1 = (l+x)(l+*)".
Prove that
, +2
-1/
and suggest a generalization of this result.
Miscellaneous Exercise 8
1. Find (i) the sum, (ii) the product, of the first η terms of the geometric sequence
{1,3,9,...}.
2. The first term of a geometric series is 18 and the sum to infinity is 20. Find
the common ratio and the sum of the first six terms. Find also in its simplest form
the ratio of the /Jth term to the sum of all the subsequent terms of the infinite
series. (J.M.B.)
147

FINITE SERIES AND THE BINOMIAL THEOREM [8
3. In the expansion in powers of χ of the function
(1+х)(я-6х)12
the coefficient of x8 is zero. Find in its simplest form the value of the ratio a/b.
(J.M.B.)
4. Write down the sum of the geometric series
l+x + x2 + x3+...+x".
Deduce the sum of the series
l+lx + Sx' + .-.+nx"-1
by differentiating, with respect to x, the expression for the sum of the original
series.
Find the sum of the series
\2 + 22х + Ъ2хг + ...+п2х"-\
5. Use Vandermonde's Theorem to evaluate
6. Numbers C1; C2, C3,... are defined as follows:
C, = 2, C„+1 = 2(C, + C2+... + C„)
for η = 1, 2,.... Find C2, C3 and C4 and prove that the numbers C2, C3, C4,...
form a geometric progression.
Find the sum of η terms of the series
C1+C2+C3 + .... (O&C)
7. The numbers я0, аъ ..., a„; c0, съ ..., οη_λ are the coefficients in the two
expansions ,, , . , , , , ,
(l+x)" = a0 + a1x + a2x2 + ...+anxn,
(1 + x)"-1 = с0 + с1л: + с2х2 + ... + с,!_1л:'!-1.
Prove that:
(a) a0 = c0 and a„ = c„_j;
(b) Сг-г + Сг = Or for 1 m r *ζ n-1;
(c) rcir = «cr_j for I *Z r *ζ n.
Use these results to show that
л. α- '"+1
^-^_!+...±йо = Or+1
ior 0 < r < n-\. (O&Q
8. Integers a, b, d are connected by the relation a = b + d. By using the binomial
expansion of (6+d)n, where я is a positive integer, show that a" — Ь^ф + nd) is
exactly divisible by d2.
Replace b by a-d in this result, and hence show that if a is the first term,
d the common difference and / the nth term of an arithmetic progression, then
an — (a—d)n~11 is exactly divisible by d2.
Show that 510—214 is exactly divisible by 9. (Cambridge)
148

5] MISCELLANEOUS EXERCISE 8
9.Provethat χ +2 + 3 + ...+я = Ып + 1).
Prove also that
12-22 + 32-42 + ...+(2л + 1)2 = (я+1)(2я+1).
Show that, when m is any positive integer, odd or even, the sum of the first m
terms of the series
ist-.ir-HwOw + l). '" (O&Q
11. Show that
04-,)·
' considering the identity
(1+χ)2™(1-χ)2™ξ (1-х2)2™,
ethat limy I2m\2 /2w\2 /2w\2 ,
(θ)-(ΐ)+(2)-· + (2,)=(-
l} (W!)2
12. Prove that, when я is a positive integer and α φ 1,
1+(1+я) + (1+я + я2)+...+(1+я + я2
Show that the sum may also be written
л + (л-1)я+(л-2)я2 + ... + яя-1
and deduce from this the sum to η terms of the series
l+2Z> + 3Z>2 + ...+(>z-l)Z>"-2 + >zZ>"-1. (O&Q
13. Find, in binary form, the square of the number whose expression in binary
notation consists of the r digits 1.
By putting m = n, deduce that the sum of the product in pairs of the first η
integers is , , . ,. ,„
A«(«2-l)(3« + 2).
15. Verify that r\r +1)2 - (/· -1)2 r2 = 4r3
and deduce that 13 + 23 +... + rfi = \n\n +1)2.
Prove that l3 + 33 + 53+...+(2n-l)3 = я2(2л2-1)
and find the sum of the cubes of all odd numbers less than 50 that are not
multiples of 5.

FINITE SERIES AND THE BINOMIAL THEOREM [8
16. Prove that 12 + 22 + 32+...+л2 = \η(η + \)(2η + ΐ).
A non-degenerate triangle is to be made from three rods chosen from 2n
straight rods whose lengths are 1, 2, 3, ..., 2n units respectively. If the length of
the longest side is 2r units, show that (/- — l)2 such triangles can be made.
Deduce that the total number of possible triangles whose longest side is an
even number of units is ^_ ц ^ _ J} (Jmr)
17. The salary scale for certain employees begins at £630 a year and rises to a
maximum of £855 by fifteen annual increments of £15 each. After the maximum
has been reached the annual salary remains constant. Find expressions for the
total amount received by an employee in η years (i) when η < 16; (ii) when
n> 16.
Find the value of the second expression when η = 24.
It is proposed to revise the salary scale so that it begins at £720 a year and
rises to a maximum by 10 equal annual increments. If each annual increment is
£d find an expression for the total amount to be received in salary by an employee
in η years when η > 11.
Find the integral value of d which would make the total amount to be received
in twenty years on the new scale as nearly as possible equal to the total amount
received in twenty-four years on the old scale. (Cambridge)
18. The series of natural numbers is grouped as follows:
(1), (2,3,4), (5,6,7,8,9), ...
(i.e. each bracket contains two integers more than the preceding bracket).
(i) Find the total number of integers in the first (n— 1) brackets.
(ii) Show that the first number in the nth bracket is n2-In + 2.
(iii) Show that the sum of the numbers in the nth bracket is n3 + (n -1)3.
(iv) If the first number in the nth bracket is denoted by a and the first number
in the (n+ l)th bracket is denoted by b show that the sum of the numbers in the
nth bracket is exactly divisible by (b - a) and that the quotient is an odd number.
(Cambridge)

Revision exercise A
1. If the expressions
x3-(a + 2)x+2b and 2x3 + ax* - 4x - b
have a common factor x + 3, find the values of a and b, and find a second
common factor. (O&C'O')
2. Find the mid-point of the line joining the points (-4, 2) and (2, -6), and
show that the equation of the perpendicular bisector of this line is 3x-4y = 5.
Prove that the point (3, 1) is one vertex of a square of which the points (—4, 2)
and (2, — 6) are opposite vertices, and find the coordinates of the fourth vertex.
(O&C'O')
3. You are given that the equation
Зх*-9* + 5 = 0
has a root approximately equal to 1. By substituting χ = 1+A, and neglecting
A2 and higher powers show that χ = lg is a closer approximation to this root.
4. (i) The fourth term of an arithmetic series is 55, and the tenth term is 45.
Calculate the sum of all the positive terms.
(ii) Find, correct to one decimal point, the sum of the first fifteen terms of the
geometric series χ +(1.02) + (1.02)2 + (1.02)3 + .... (0 & c <0>)
5. In how many distinct ways may the letters of the word syzygy be arranged?
In how many of these do the three Ys appear together ?
(What does the word syzygy mean ?)
6. Without using tables, prove that tan 60° = V3.
A rectangle ABCD lies in a horizontal plane and DE is a line of length h
drawn vertically upwards from D; EA, EB, EC are joined. If the lines AE, CE
make angles of 60°, 45° respectively with the horizontal, express AD and CD in
terms of h, and hence calculate the angle which BE makes with the horizontal.
(O&C'O')
7. Find the equation of the line of gradient — 1 which passes through the point
of intersection of the lines x + 3y- 5 = 0 and 2x-y-7 = 0.
8. If x2 + 2x+2 is a factor of x4 + ax2 + b, find a and b. Express x*+ 16 as the
product of two quadratic factors whose coefficients are real numbers.
9. ABC is a triangle: points Q, R are taken on AC, AB respectively so that
AQ = 2QC, 2AR = RB.
Express AB, AC in terms of RQ, ВС.
10. If \π < χ < π and sin χ = §-, find (i) cos x, (ii) tan x, (Hi) cosec x.
151

REVISION EXERCISE A
11. Write down the expansion of (1 + xf in ascending powers of x. If
(l+x + x2)s = l+ax+bx*+...,
find a and b.
12. Given that (2x+1) is a factor of the expression
6лг> + ял:2 + 6л:+1
find a, and hence find the other two linear factors.
13. Three integers (not necessarily distinct) are chosen at random from the set of
integers {1, 2, 3, ..., 10}. What is the probability that their sum is 10? What is
the probability that their sum is 10 or less?
14. By considering a suitable rhombus of side 1 unit, show that, if sin θ = a,
then sin 20 = 2oV(l-a2)·
Deduce that _
Sinl5° = ^V2"·
15. Show that 4r3 = [r(r+l)Y-[r(r-l)f.
Hence evaluate Σ r3.
16. How many terms are there in the expansion of (α + 6 + c)6?
Can you generalize your result for (a+ b + c)nl
17. Show that it is not possible to find a number which is written abc in the
scale of 4 and bca in the scale of 5.
18. Two points, Ρ and Q, on the circumference of a circle of radius 10 cm are
such that PQ subtends an angle of 1 radian (57° 180 at its centre. State the length
of the minor arc PQ of this circle, and calculate the length of the chord PQ.
If the above circle is drawn on the surface of a sphere of radius 15 cm, find
the angle which the chord PQ subtends at the centre of the sphere, and hence
find the length of the minor arc Ρ β of the great circle of the sphere which passes
through Ρ and β. (O&C'O')
19. I take my wife out to buy her a new hat. There is a probability of те that
I shall approve of the first hat that we are shown, but if I approve, there is only a
probability of jo that my wife will agree with my choice. If we both like the hat,
I purchase it. If I dislike the first hat that we are shown, there is a probability of
τ that my wife will like it. If she does like it she naturally overrules my choice
but, when I hear the price, there is a f probability that I shall veto the purchase.
What is the probability that I buy her the first hat we are shown ? If I do buy her
the hat, what is the probability that we both like it ?
20. Prove that, if я is an integer, none of the numbers 7я + 3, 7я + 5, 7я + 6 can
be a perfect square.
21. Find the equation of the plane through the point (1, 1,1) containing the line
x-3 _y^X_ J_
5 ~ 2 ~ -4'
Find also the direction cosines of the line of intersection of this plane with the
plane Oxy.
152

REVISION EXERCISE A
22. The set S consists of all points in a plane whose coordinates relative to a
given pair of perpendicular axes are both integers (in the units used). If А, В, С
are members of S, prove that the area of the triangle ABC is a rational number
of square units.
23. Eliminate θ between the pair of equations
a sec θ + b cosec θ = c,
a' sec Θ — Ъ' cosec θ = с'.
24. Solve the inequality
and illustrate your answer by sketching the graph of
- *-l
У ~ x2-3x-4'
25. Evaluate Σγ(γ+1) (2/- + 3).
26. Л = {xeR:-3 < χ ^ 2}, В = {xe R: -2 ^ χ < 3},
С={д;ей:1<4
Express, in the form {:}, the sets
A(]B, Aft C, A'(\B, A ft В ft C, A\)(Bft C").
27. Show, by drawing a rough sketch, that the equation
χ = 2 cos л:
has just one positive solution.
By drawing an accurate graph, estimate this value of χ as accurately as you
can.
28. A five digit number (that is, a number lying between 10000 and 99999
inclusive) is selected at random. What is the probability that it contains either the
digit 8 or the digit 9 (or both)?
29. Show that the plane containing the points with position vectors
i-2j, i + j+k, 2i-j + 2k
has vector equation
г = i-2j + A(3j + k)+/<i + j + 2k).
Find the position vector of the point of intersection of this plane with the line
г = 3i + 5j-5k + K2i + 3j-k).
30. a, b are rational approximations to the irrational numbers -JA, ^B. An
approximation to j(AB) is calculated by using the formulae:
(i) JiAB) » ab; (ii) J(AB) » %(a + bf -(A + B)];
(iii) J(AB) и ЩА + В)-(а-ЬП
The errors in the three cases are respectively Elt Ег, Е3. Find a relation connecting
Ei, E2, E3.
153

REVISION EXERCISE A
31. A bag contains three white and four red balls, another bag contains four
white and three red balls. One of the bags is selected at random and two balls
are drawn from it. If both balls are red, what is the probability that, of the five
balls remaining in the selected bag, two are red?
32. A and В are two points with position vectors a, b relative to a given origin O.
Prove that the equation of the internal bisector of the angle АО В is
r = A(a + 6).
(Recall that a is a unit vector in the direction of a.)
What is the equation of the external angle bisector ?
Find the position vector of the incentre of the triangle ABO in terms of
a = |a|, 6 = |b|, с = |AB| and the unit vectors a, b. (The incentre of a triangle
is the point of intersection of the bisectors of the angles of the triangle.)
33. Evaluate the expressions
(0 Vfl+VU+VO + ···)}]; (n) Vf2 + V(2+V(2+ ·.·)}]·
If η is an integer, when is an expression of the form
V[k+V{«+V(«+-)}]
a rational number ?
34. Two cards are missing from a pack. If three cards are drawn at random from
the remaining fifty (without replacement) and all are found to be aces, what is the
probability that, if a fourth card is drawn, it will also be an ace ?
154

9· Mathematical induction
1. A NOTE ON MATHEMATICAL PROOFS
In a mathematical proof we proceed from one step to the next by processes
of deductive logic; that is, each step is logically implied by the previous
step or steps of the proof. In other words, a mathematical argument has
a direction. The first step must depend upon certain premises or known
theorems which may or may not be stated explicitly. For example, in the
traditional form of proof for a geometrical problem, the premises (data)
are usually stated before the proof commences and the argument leads
from this statement to the final deduction, using previously proved results
of elementary geometry. If, however, we are asked to prove that the
probability of a double with a throw of two unbiased dice is |, we must commence
by asserting that the set consisting of the thirty-six pairs of numbers
(1,1), (1,2), ...,(1,6), (2,1),(2,2), ...,(6,6) constitutes a possible outcome
space and that we assume that each of these elementary events has a
probability of -£$. ('The dice are unbiased.') We are then able to deduce
the required probability.
Since a mathematical argument has a direction, we shall frequently
meet the situation in which a sequence of statements p, q, r are strung
together in the form ' statement ρ implies the statement q which, in turn,
implies the statement r and so on'. The notation
ρ => q (read 'p implies q')
has already been used; its use in mathematical proofs can lead to
conciseness of expression.
Example 1. Prove that x2+y2 + z2-yz-zx-xy ^ Ofor real x, y, z.
Ε = x2+y2+z2-yz-zx-xy
^2E = (y-zf + iz-xy + ix-y)2
=> IE > 0, for real x, y, z,
=> Ε ^ 0, for real x, y, z.
The reader must always be on his guard, when writing ρ => q, that this
is indeed what he means: a genuine deduction from the immediately
preceding statement is implied. One of the principle sources of error among
б PPM 155

MATHEMATICAL INDUCTION [9
beginners attempting to prove mathematical results is to confuse the
implication ρ => q with the implication q => p. ρ => q means that, whenever
ρ is true, q is true; it does not tell us anything about the truth of the
statement ρ if q is true. (We shall not discuss the meaning of the
statement/7 => q in the case when ρ is false; the bibliography refers to books in
which problems concerning the use of implication are discussed at length.)
Ex. 1. 'If -1 > 0, then (-1)2 > 0, which is true, and so -1 > 0.' Comment.
Ex. 2. A class is asked to prove that, if a/b = c/d, then
(a + b)j(a - b) = (c + d)/(c - d) (a, b, c, d unequal and non-zero).
Criticize the following proof:
=> ac + bc-ad-bd = ac+ad-bc-bd
=>2bc = 2ad
=> a/b = c/d
which is true and so the original proposition is true.'
Ex. 3. Criticize the proof of the identity cosecM tanM -1 = tanM given below:
'If cosec2 A tan2 A-1 ξ tan2 A, then
1 sinM
-r-rz —τ—. = 1 + tan2 A = sec2 A
sin2 A cos2 A
which is manifestly true, since
= sec A.'
cos A
If the implication ρ => q is reversible, that is, if ρ => q and q => ρ both
hold, then we may write
poq (read 'p implies and is implied by q').
Example 2. Solve the equation J(2x +1) + Jx = 5.
V(2x+1) + V*=5
=>2x+l+x + 2VM2x+l)] = 25
о 2VM2x+l)]=-(3x-24)
=> 4(2x2 + x) = 9x2 - 144л; + 576
о х2-148л:+576 = 0
о (х-4)(х-Ы4) = 0.
156

1] A NOTE ON MATHEMATICAL PROOFS
Thus V(2x+1) + V* = 5 => (x-144) (x-4) = 0, but two of the steps of
the argument are not reversible, and so it is not possible to infer that
(х-144) (x-4) = 0 => V(2*+l)+V* = 5· To express it another way, if
χ = a is a root of V(2x + 1) + V* = 5, then it is certainly a root of
(x-144) (x-4) = 0,
but, if χ = α is a root of (x-144) (x-4) = 0 it may not be a root of
V(2x +1) + y/x = 5. To complete the solution of the given equation we must
substitute back the two possible solutions χ = 4 and χ = 144. It is then
seen that χ = 4 is a root of the original equation, but that χ = 144 is not.
Ex. 4. In Example 2, how do you know that χ = 4 is the only root of the equation
V(2x+1) + V* = 5?
Ex. 5. Solve Question 6, Exercise 1 c, using the implication signs, and explaining
carefully which steps are not reversible.
If a result is stated in the form ρ => q, the converse result (if it holds)
is q => p. For example, if ρ and q are defined as follows:
ρ is the statement 'the triangle ABC has AB = AC,
q is the statement 'the triangle ABC has LB = LC\
then a well-known theorem of elementary geometry asserts that ρ => q.
The converse theorem, q => p, is also true and the two theorems may be
combined together in the single two-way implication ρ ο q.
It is by no means always the case that, if a theorem is true, then its
converse is also true. Indeed, if a theorem takes the form
ρ and # => r
a converse is not clearly defined. In the case in which a theorem ρ => q
and its converse q => ρ both hold, we may use the two-way implication
-o- in its formulation; alternatively, we may use the phrase 'if and only if
—sometimes abbreviated to 'iff'.
In proving the truth of a two-way implication it is vital to remember
that two separate proofs are needed. For example, referring again to the
proposition about the triangle ABC mentioned above, we
(i) assume AB = AC and deduce that
LB = LC (AB = AC => LB = LC
or alternatively, AB = AC only if LB = ZC);f and
t From the purely linguistic points of view, the words 'only if are somewhat
ambiguous; mathematically, however, no ambiguity can arise if we define 'p only if q' to
mean lp => q\
6-2 157

MATHEMATICAL INDUCTION [9
(ii) assume ΔΒ = AC and deduce that
AB=AC {Z-B = LC => AB = AC
or, alternatively, AB = AC if LB = Z-Q.f
Ex. 6. The triangle ABC is right-angled at A if and only if ВС2 = СА2+АВ*.
What would you assume if asked to prove the Only if part of this proposition?
Ex. 7. Correct the following statement: "The integer N (expressed in the denary
scale) is divisible by 5 if and only if the units digit of N is 5.'
Ex. 8. Correct the following statement: 'Two vectors a and b are equal if and
only if |a| = |b|.'
Ex. 9. Is it true to say that, if χ > 0 then ax > x2 only if a > χΊ
It is often necessary to disprove an implication; that is, to show that
the truth of statement ρ does not imply the truth of statement q. In this
situation we write ρ φ> q (read 'p does not imply q'). Since ρ => # means that,
in all cases in which ρ holds, q holds too, to show that ρ ή> q we have
merely to exhibit one case in which ρ holds and q does not hold (a counter
example). For example, if we have
ρ is the statement 'л: is of the form (6n + 1) π/3, η integral',
q is the statement 'sin л: = sin 2x'
it is a fairly straightforward matter to prove that, if ρ is true, then q is
true; that is, p=> q. To disprove the converse, q => p, we have simply
to find a counter example, e.g. χ = 0, which certainly satisfies q, but is not
of the form (6n ± 1) тг/3.
Ex. 10. Prove that χ = (6« ± 1) π/З => sin χ = sin 2x.
Ex. 11. If ρ is the statement '« is an odd number' and q is the statement 'an
integer к can be found so that η = 4k +1' prove that q => ρ and disprove the
converse result ρ => q.
To complete this section we mention one final method of proving that
the implication ρ => q holds. If a statement ρ is modified by the addition
of the word 'not', a new statement 'not p', written p' (sometimes ~ p)
is obtained. For example, if ρ is the statement 'the integer η is divisible
by 3' p' is the statement' the integer η is not divisible by 3'. p' is called the
negation of p.
t Another way of expressing implications is by using the phrases 'necessary condition'
and 'sufficient condition'. A necessary condition for ρ is q means that.? => q; a sufficient
condition for ρ is q means that q => p. A necessary and sufficient condition for ρ is q means
that ρ oq.
158

11 A NOTE ON MATHEMATICAL PROOFS
The implication ρ => q is equivalent to saying that we cannot have ρ
true without q being true or, using negatives, if q' is true, then p' is true;
in terms of implication, q' =*■ p'. The argument also works in reverse and
thus the two implications ρ => q and q' => p' are equivalent. Expressed
more succinctly . . . , ,.
(p=>g)o (я => p)-
The mathematical value of this equivalence is that it is often easier to
prove the implication q' => p' rather than the implication ρ => q.
Ex. 12. In Ex. 11, prove that p'=>q'.
Example 3. To prove that there is an unlimited number of primes.
Let q be the statement Ί has no prime factor' and r be the statement
'there is an unlimited number of primes'. The statement # is true, since the
smallest prime number is 2; we shall show that q => r, and so r is true.
Now the negation of r is the statement /■': 'there is a finite number of
primes'.
r' => there is a greatest prime number, ρ say;
=> (k+l) is not prime, where к is the product of all the primes,
к = 2.3.5 ρ;
=> (k + l) has some member of the set {2, 3, 5, ...,p} as a factor;
=> both к and (k+l) have some member of the set {2, 3, 5, ..., p) as a
common factor;
=> 1 has some number of the set {2, 3, 5, ...} as a factor;
Thus r' => q' and so q => r.
A number of important points have been glossed over in the discussion
of implication in this section. The reader should consult one of the books
mentioned in the bibliography for a more thorough and rigorous treatment
of the subject.
Exercise 9(a)
1. Disprove the converse of the result ' if a quadrilateral is a rectangle then its
diagonals are equal'.
2. All angles θ of the form θ = кп (к an integer) have the property that
sin θ = tan Θ. Is the converse result true?
3. If a, b e Z+ and if the operation о is defined by
t ab-a-b
йо6 = ^ТГ'
prove that aob = boaoa = b.

MATHEMATICAL INDUCTION [9
4. Solve the equation J(x-4) + 2 = V(*+12)> stating which steps in your
solution are reversible and which are not.
5. Solve the equation J(2x+1)-Jx = 1, stating which steps in your solution
are reversible and which are not.
6. Solve the equation J(2x-l) + J(x + 3) = 9, stating which steps in your
solution are reversible and which are not.
7. Explain the fallacy in the following argument:
'n2 +1 has no real factors. Put η = 3: thus 10 is a prime number.'
8. Prove that the difference of the squares of two odd numbers is divisible by 8.
State and prove a converse theorem.
9. Prove that (x + h) is a factor of ax2 + bx + с if and only if ah2-bh + c = 0.
10. Prove that, if α φ — 1, then the simultaneous equations in x, y:
(ax+y^b,
(x + ay = b
can be solved for χ and y.
What happens if a = 1 ?
Is it true that the equations cannot be solved if α = — 1 ?
11. Two well-known theorems of number theory are:
(i) any odd prime of the form 4n +1 is expressible as the sum of the squares
of two integers;
(ii) any odd prime of the form An + 3 is not expressible as the sum of the
squares of two integers.
Combine these two theorems into a single theorem by using the phrase 'if and
only if.
12. a, b, c, d are four unequal, positive numbers. Prove that
a-c Ца2 + сг\ ., , , .. а с
b^d=J[v^d2) rf«"d only if ъ = ,
13. If P(x) is a polynomial with a repeated linear factor, that is, if
P(x) = (x-hYQ(x),
where Q(x) is a polynomial, prove that (x— h) is also a factor of P'(x). (jP'(x) is
the derivative of P(x).)
14. Prove that, if 0 < χ < 1,
then (X-xf <^f^.
4-х
15. ABC is any triangle and Μ is the mid-point of ВС. Prove that
AB2 + AC2 = 2AM2 + BM2 + CM2 (Apollonius's Theorem).
Is it true that, if Μ is a point on ВС such that
AB2 + AC2 = 2AM2 + BM2+CM2
then Μ is the mid-point of BC1
160

U A NOTE ON MATHEMATICAL PROOFS
16. If ρ is a prime number greater than 3, prove that either /7+1 or p—\ is
divisible by 6.
Show that the statement 'if χ is a number such that x +1 or x— 1 is divisible
by 6, then л: is a prime' is false.
Show further that the statement 'if л: is divisible by 6, then at least one of
x+1 or x— 1 is a prime number' is also false.
17. Prove that a sufficient condition for the integer χ to be divisible by 17 is that
there exists an integer η such that
x= 35я-18".
Prove further that this is not a necessary condition for χ to be divisible by 17.
2. MATHEMATICAL INDUCTION;
AN INTRODUCTORY EXAMPLE
Consider the series
Suppose we define the sequence {sn} in such a way that sn represents the
sum of the first η terms of this series. Then by direct computation,
■Sl =2. S2 = f, S3 = %, S4 = $.
The form taken by the first few values of sn suggests that
"и + 1
(*eZ+).
Let us define a new sequence {/„} in such a way that
f = —
Then we wish to prove that the two sequences {sn} and {/„} are the same,
in the sense that sn = fn for all η e Z+.
As a first step, we observe that
fn+1 = ^-±i , by the definition of/„
= ^+ϊ+(7Γ+Τ)"(^+Τ)
О)
, by the definition of sn. (2)
161
> + l)(« + 2)

MATHEMATICAL INDUCTION [9
(1) and (2) define recurrence relations (see chapter 8.1); with the initial
terms they enable us to determine uniquely any term of either of the
sequences {sn}, {/„}. But we further observe, in this case, that
(i) the two recurrence relations are the same;
(ii) Sl = i =U
Since the recurrence relations (1) and (2) develop successive terms of the
sequences {sn} and {/„} in the same way, and since the two sequences
have the same starting point, we deduce that sn = /„ for all n.
The observation that (1) and (2) represent the same recurrence relation
enables us to deduce that if sn = fn, then j„+1 = fn+1, or, expressed
alternatively, - .
Sn=fn^ Sn+1 = fn+1.
The fact that jj = /i then enables us to infer the equality of sn and/„ for
allneZ+;for г г г
Si = fi=>S2=f2=>S3 =f3... ·
Ex. 13. Define sequences sn and/„ as follows
s„ = 1.2+2.3 + 3.4+...+л(л+1),
Λ = Κ»+1)(Ί+ί-
Show that (i) st = /i, (ii) s„ and /„ satisfy the same recurrence relation. Deduce
thatjn =/„.
3. THE METHOD OF MATHEMATICAL
INDUCTION
Suppose that pn is some statement about the positive integer n. For
example (see example 5 below), we might have
pn: 'the expression 13й — 6™~2 is divisible by 7'.
Then, if we can show
(i) that pm is true for some specific integer m (in our example, m = 2
would do); and
(ii) that pk =>Pic+i for all к ^ т (in our example, рк^ is the statement
*13*+1-6*-1') we have
Pm => Pm+l => Pm+2 => ■■·
and the truth of the statement pn may be asserted for all η > m. Such a
process is called proof by mathematical induction.
The method of mathematical induction is used to prove a conjecture
about the positive integer n. The conjecture itself is usually arrived at by
inductive argument from a few simple cases—hence the use of the word
'induction', though the method of proof is deductive.
162

31 METHOD OF MATHEMATICAL INDUCTION
To summarize: given the statement p„ about the positive integer л,
to prove the truth of pn we
(i) verify that the induction starts; that is, we find an integer m for
whichpm is true;
(ii) prove the implication/^ => pk+1 for a general positive integer к *s m.
*Ex. 14. Show that n2 + n+41 is a prime number if η = 1, 2, 3. Prove that the
statement 'n2 + n + 41 is a prime number for all positive integers я' is false.
(n2 + n + 41 is, in fact, prime for η = 1, 2, 3, ..., 39 but it is false when η = 40
and whenever я is a multiple of 41: pk φ> /%+ι·)
*Ex. 15. If /?„ is the statement '5л2— 5я + 2 is divisible by 10', prove that
Pk =*■ Л+1- Prove also that /?„ is false for all positive integers n. (The induction has
no starting point.) Hint. To prove the falsity of pn, observe that Pk^Pk+i and
ihusp'k=> p'k+1.
Example 4. Prove that
1.11 + 2.21 + 3.3! + ...+л.л! = (л + 1)!-1.
Write sn = 1.1!+2.2! + 3.3! + ...+л.л!.
We have to prove that sn = (n +1)! -1; let pn be the statement
Ч, = (л + 1)1-Г.
(i) px is true, for sx = 1.1! = 1 and (1 + 1)!-1 = 1: the induction
starts,
(ii) To prove pk => pk+1:
Sk+i = sk + (k+l) (k+1)!, by the definition of sk;
**l+i = (*+l)!-l+(*+l)(*+l)!, Ъурк;
=>ί*+ι = (*+1)![1+*+1]-1
=>Ъ^ = (* + 2)!-1
and thus pk => pk+1 and the result is true for all л > 1, by mathematical
induction.
Example 5. Prove that 13й— 6й-2 is divisible by 1 for all positive integers
greater than m, where m is a certain integer, to be specified.
Write un = 13й- 6й-2; we have to prove that un is divisible by 7 for all
л ^ m, where m is some integer, to be found.
(i) щ is a fraction, and the concept of divisibility does not apply.
u2= 132-6° = 168 = 7.24 and щ is therefore divisible by 7: the induction
starts.
Let/7„ be the statement 'и„ is divisible by 7'.
163

MATHEMATICAL INDUCTION
[9
(ii) Toprove/»»=>A+i:
Ujc+1 = ΐ3*+ι_6*-ι and uk = 13*-6*-2
^uk+1-6uk = (13.13*-6.6*-2)-(6.13*-6.6*-2)
=> uk+1-6uk = 7.13*
=> "fc+i = 7.13* + 6ks
=> Mfc+i is divisible by 7, by pk, since the sum of two integers
divisible by 7 is itself divisible by 7.
Thus, 13n-6"-2 is divisible by 7 for all η > 2.
Example 6 {the Binomial Theorem). Prove that
(х + аГ = (J) *«+ф x^a+Q *-V + ... + ^) ж»-*+... + £) ««
/or positive integral n.
(i) (x + α)1 = x+a = II x¥+ xV and the induction starts.
(ii) Consider the product
(x+a) [*"+(") xn~ia+W\ xn-2a2+... + r) xn~r ar + ...+an~\.
Each term in the product is of degree (n+1), the first is xn+1, the last is
an+1 and the term containing xn+1-rar is
and the proof is complete, by mathematical induction.
164

31 METHOD OF MATHEMATICAL INDUCTION
It should be noted in passing that the proof of the Binomial Theorem
given in Chapter 8.5 strictly speaking requires induction, to justify the
process of selection from each bracket. The student is recommended to
supply the necessary details.
In our next example, we make an assumption (sometimes called
inductive hypothesis) of a slightly different form to those of the preceding
examples. The reader should observe that, to start the induction, we have
to verify that the result holds for two successive values of n.
Example l.Ifa sequence un is defined by the recurrence relation
un+1 = 3un-2un_1 (η Ζ 2)
and the initial values иг = 0, щ = 2, prove that
un = 2"-2.
(i) Since 2J-2 = 0 and 22-2 = 2, the result holds for η = 1 and
л = 2.
(ii) Suppose
uk = 2*-2 and Mfc_! = 2*-!-2 (к 3* 2).
Then,
Ub+i = 3ks —2ks_i, by the given recurrence relation;
= 3(2*-2)-2(2*_1-2), by the assumption made above;
= 6.2*-1-6-2.2*-1 + 4
= 4.2*-1-2
= 2*+1-2.
Thus uk = 2*-2 and u^ = 2к~г-2 ^ uk+1 = 2k+1-2.
But the result is true for η = 1 and η = 2, hence it is true for η = 3 and so
on.
*Ex. 16. If {«„} is a sequence defined by the relation un = Σ u„ η ^ 2 and the
initial value ux = 4, prove that un = 2" (η ^ 2). (Make the assumption that, for
к > 2, us = 2s for all s in the interval 2 < s < k.)
Example 8. Prove Theorem 7.4, using mathematical induction.
Theorem 7.4 states that, if Elt E2, ...,En are mutually exclusive events,
(EtnE1= φ, i φ./), then
PriG^ls) =_|;Pr№|S).
165

MATHEMATICAL INDUCTION [9
(i) The result is certainly true if η = 2, by Theorem 7.2, and the
induction starts.
(ii) Write F = "\3\ and suppose that Pr(F|S) ^^PrCE^S).
Now F П Ek = φ, since the events are mutually exclusive. Thus
Pr(F U Ek\S) = Pr(F|S) + Pr№t|S), ЬУ Theorem 7.2
= Σ ?r(Ei\S) + ?T(Ek\S), by the assumption made above
= ДРг(ВД
and the theorem is proved, by mathematical induction.
Exercise 9(b)
Throughout this exercise, η represents a positive integer. Prove the results 1-14.
using the principle of mathematical induction.
1. 1+2 + 3+4 + ...+л = $л(л + 1).
2. 1+3 + 5 + .. .+(2л-1) =я2.
3. 12 + 22 + 32 + 42 + ...+я2 = *и(л+1)(2и + 1).
4. 13 + 23 + 33 + ...+я3 = 1η\η + ΐ)\
5. 1.1+3.2+5.4+...+(2и-1). 2я"1 = 3+2я(2л-3).
6. 3.1! + 7.2! + 13.3! + ...+(я2 + я + 1)я! = (я + 1)2 я!-1.
9. я6 - η is divisible by 30.
10. 12" + 2.5""1 is divisible by 7.
11. 2to + 32"-2 is divisible by 5.
12. 52n + 22n-2.3"-i is divisible by 13.
13. З.Л + З'-'.И"-1 is divisible by 8.
14. (1 +x)n > 1 +nx, provided χ > -1, η > 1.
15. The terms of the sequence {uT} are all positive and s„ = Σ иг. Prove that
(1 +«,) (1 +и2) (1 +«з)... (1 +йп) > 1 +ίη for 77 > 2.
166

3] METHOD OF MATHEMATICAL INDUCTION
16. Prove that 5" < n\ for all sufficiently large n. Add such precision as you can
to the phrase 'sufficiently large n\
17. Prove by mathematical induction that the sum of the angles of a convex
я-sided polygon is (2л—4) right-angles.
18. Prove that an я-sided convex polygon has \η(ιι— 3) diagonals.
19. A straight line separates a plane into 2 regions; two straight lines separate
the plane into 4 regions and so on. If я straight lines (no three concurrent, no two
parallel) separate the plane into un regions, prove that
«n = i(«2 + «+2).
20. If я is a positive integer, prove that
52п+2-24л-25
is divisible by 576. (Cambridge)
21. If Λ„) = ι + ^ + Ιι+... + ^
prove that Σ [(3/·2 + 3/-+ 1)Д/·)] = (n + 1УДп)-Ып +1). (Cambridge)
22. Prove that
jbfr+l) (r+3 ... (r + ft-1) = «"+Ц-<" + *>.
23. If дъ д2, д3,..., д„ are all positive, prove that
(д! + д2+...+д„) (- + -+... + -) ^ л2.
24. Prove that, for any positive integer n,
2n 2n(2n-2) 2n(2n~2)(2n-4) , ч п
2^Ί+(2^ϊ)-(2^3-) + (2.-1)(2.-3)(2.-Τ) + - (t° Я t6rms) = 7л-
(O&Q
25. A motorist estimates that, by travelling along a main road at a certain steady
speed, the probability that the next set of traffic lights will be green if the last
set was green is \ and that the probability that the next set of lights will be
green if the last set was red is i. He sets out one day to test his theory. Prove that,
if the first lights he meets are green and if
pn = Pr (the Hth set of lights is green when he reaches them)
then ^n = !+i(i)"-1 (n> 1).
26. A man repeatedly tosses an unbiased coin, scoring 1 for each head and 2
for each tail. If pn is the probability that his score will ever be n, prove that
27. η bags, numbered 1 to n, each contain one white and one black ball. A
ball is taken at random from bag one and placed in bag two; a ball is then taken

MATHEMATICAL INDUCTION [9
from bag two and placed in bag three and so on, until finally a ball is taken from
bag n. Prove that the probability that this ball is white is
!(·♦£)
given that the first ball drawn was white.
28. The sequence {«„} is defined by the recurrence relation
u„-5un_1 + 6un_2 = О (n > 3)
and the initial values иг = 7, и2 = 17. Prove that
un = 2-+Ч-3".
29. The sequence {«„} is defined by the recurrence relation
un+2 + un+1-2un = 0 (Ol)
and the initial values иг = 4, и2 = — 2. Prove that
и* = 2-(-2)".
30. The sequence {«„} is defined by the recurrence relation
un+3 + 2un+2-un+1~2u„ = 0 (я ^ 1)
and the initial values иг = -1, и2 = 7, и3 = -7. Prove that
ип = 2 + (-1)«(1+2").
31. (The Fibonacci Series.) The sequence {«„} is defined by the equations
«i = «2=l, «„«=«„ + «„_! (n>2);
prove that
«η = ±{*·-Ρ) where a = !±^5 and β = ±=£
are the roots of the quadratic equation x2 = x + 1.
32. Prove that, for positive integral n, 32n —5" is divisible by 7 if and only if η
is even.
168

ίο. Expectation
1. RANDOM VARIABLES
In Chapter 7 we considered random experiments described by outcome
spaces. With each elementary event we associated a number, its
probability, which denoted our degree of belief that the experiment would result
in that particular events occurring. The elementary events themselves
may be described in various ways: in some cases, it is natural to denote
them by a number, e.g. for the fall of a die we could denote our elementary
events by the numbers 1,2,3,...,6; but sometimes no such natural
numerical description exists. For example, for a single toss of a coin, the two
element set {heads, tails} is the natural choice of outcome space, the elements
being labelled by the descriptions 'heads' and 'tails'. In order to make a
mathematical analysis of random experiments it is helpful to describe the
possible outcomes numerically even in those cases where no such 'natural'
description exists. Thus, in the case of coin spinning, the event' tails' could
be denoted by the number 0, the event 'heads' by the number 1. Such
values are called values of a random variable.
Let us pause here to recapitulate. Suppose we have a random
experiment whose possible outcomes are the η elementary events s-y, s2, ..., sn.
Then we may take as our outcome space the set
S = {s-y, s2, ..., sn}.
We now attach two numerical 'labels' to each element sf of S:
(i) the probability, pt, of that event occurring;
(ii) the corresponding value xt of the random variable (see Figure 10.1).
To help fix ideas, consider the following examples:
1. An unbiased die is thrown and the score noted. The outcome space
with the associated probabilities and values of a possible choice of random
variable are shown in Figure 10.2. If the experiment is repeated a number
of times, then the sum of the values of the random variable obtained gives
us our aggregate score.
2. Two unbiased dice are thrown in an attempt to score a double.
Since we are interested in just two possible outcomes, we may take our
outcome space, S, as shown in Figure 10.3. Associated probabilities together
with a possible choice for values of a random variable are as indicated.
In this case, if the experiment is repeated, the sum of the values of the
random variables obtained gives us the number of doubles thrown.
169

EXPECTATION
Л = 6 Л==6
{Тр
Fig. 10.3
3. A simple coin-tossing game is played as follows: A pays 5 pence for
the privilege of tossing two unbiased coins. If he gets two heads, В pays
him 5 pence and his stake money is returned; if he gets a head and a tail
(in either order), his stake money is returned; if he gets two tails, В keeps
the stake. A possible choice of outcome space, with associated probabilities
attached, is shown in Figure 10.4. In this experiment we are primarily con-
Л-4
= 4
1 J »2 Ι «Ι ί *4 !
Ш\ IhtI [та] ϊττ]
i A A A

1] RANDOM VARIABLES
cerned with the financial outcome and our choice of random variable
reflects this concern. If the experiment is repeated a number of times,
the sum of the values of the random variable obtained denotes A's net
gain in pence.
With these examples behind us we now make the following definition:
a variable whose value is a number determined by the outcome of a random
experiment is called a random variable.
The discerning reader will have observed that a random variable is a
function from a given sample space into a set of numbers, the values of
the random variable being the images under this function. The words
'random variable', however, are often loosely used to denote the image—
a vice to which we ourselves shall succumb from time to time for the sake
of brevity.
In the same way that a function is denoted by / and the image of ζ
under/by/(ζ), so a random variable may be denoted by Zand its
associated value for the z'th element of the outcome space by x{.
Ex. 1. A man pays 10 pence to throw two dice. For any double he receives his
stake momey back, together with a prize: one pound for a double six and 20
pence for any other double. Suggest a suitable outcome space, probability
distribution and random variable for this experiment.
Ex. 2. A man insures his life for £1000, paying a premium of £X. Suggest a
suitable choice of random variable to describe the situation.
2. EXPECTATION
A mass of numerical data often has an indigestible appearance and conveys
very little, unless subjected to considerable analysis. A device commonly
used to give an overall impression of the data is to determine their average
value. For example, if the heights of a hundred boys are measured, the
complete set of results may usefully be characterized by giving their average
height.
We shall now investigate what meaning can be attached to the phrase
'the average value of the random variable X associated with a given
probability distribution'. If an experiment is repeated N times, where N
is a large number, our interpretation of the probability p, as a measure
of our confidence in securing the outcome s^ leads us to expect roughly
ρλ Ν occurrences of s1} with similar results for s2, s3, etc. Thus we should
anticipate a score of xx on/^iVoccasions, x2 on/?2iVoccasions and so on,
giving an approximate average value of our random variable X for all N
171

EXPECTATION [10
repetitions as
p1Nx1+p2Nx2+...+pnNxn
-— N ——- = p1x1+p,x,+...+pn Xn
= ΣΡίΧί-
The larger the value of N, the more confidence we should place in the value
of Σ ρ{ X{ as an estimate of the average. We are thus led to formulate the
following definition:
The expectation ${X) of the random variable X is defined by the
equation n
S{X)= ^PiXi-
i=l
Example 1. An unbiased die is thrown; what is the expected score!
Here/?» = \ for each elementary event, and the associated values of the
random variable X are the integers 1, 2, 3, 4, 5, 6. Thus
£{X) = Σ £
= 3-5.
It is perhaps superfluous to remark that no one sufficiently familiar
with an unbiased die would expect a score of 3-5 on any one throw. 3-5
simply represents the best estimate available, prior to the actual experiment,
that we can make of the final average score if the die is thrown a number of
times; the larger the number of throws, the better we expect our estimate
to be. (The reader is strongly recommended to test the accuracy of this
forecast if ever he finds time lying heavily on his hands by throwing a die,
say a hundred times, and computing his average score.)
Example 2. Two players, A and B, play the following game with three coins:
A pays a stake of 10 pence and tosses the three coins in turn. If he obtains
three heads, his stake is returned, together with a prize of '30 pence; for two
consecutive heads, his stake money is returned, together with a prize of
10 pence. In all other cases, В wins the stake money. Is the game fair!
We must first consider what is meant by the question 'is the game fair?'
Intuitively it seems reasonable to label a game as 'fair' if, in the long-run
neither side anticipates any considerable financial gain. Mathematically,
a game between two players is fair if the expectation of gain for either
player is zero.
172

2] EXPECTATION
To continue with the solution of the problem, we choose as our outcome
space the four-element set
S = {(HHH), (HHT), (THH), (anything else)}.
On the assumption that the coins are unbiased and that the results of the
tosses are independent, Pr (HHH) = (£)3 = -|; similarly,
Pr (HHT) = Pr(THH) = % and Pr (anything else) = 1 -| = ·§.
As our random variable we take the net gain, in pence, for A on each
particular elementary event: хг = +30, x2 = +10, x3 = +10, xt = -10.
Summarizing in tabular form, we have:
HHH
i
30
HHT
έ
10
THH
i
10
Anything
1
-10
£{X) = £.30+i.l0+|.10 + !(-10)
= 0.
The game is therefore, according to our definition proposed above, fair.
Ex. 3. An experiment with three possible outcomes su s2, s3, has probability
distribution {j, f, ^} and associated random variables 3, 2, 1. Determine the
expectation.
Ex. 4. A man pays 1 penny to throw three unbiased dice. If at least one six
appears he receives back his stake money together with a prize consisting of the
number of pennies equal to the number of sixes thrown. Does he expect to win
or lose?
Ex. 5. An experiment can result in three possible outcomes, whose probabilities
are k, i, i. A random variable is assigned whose values are respectively x2, -x,\.
Show that, if the experiment is repeated a number of times, the player may
possibly finish with a negative score but can anticipate an aggregate score which
is positive. Can his expectation be zero ?
173

EXPECTATION
Exercise 10(a)
1. The values of a random variable, together with their associated probabilities
for four different experiments, are given in the tables below. Calculate &(X)m.
the four cases.
(i)
m
(iii)
*i
P;
*/
Pi
*/
P;
0
A
-2
A
1
A
1
A
-1
!
2
A
2
i
0
A
3
i
3
i
1
A
4
!
4
A
2
A
5
A
5
A
6
A
*/
£,·
1
0
2
1
3
0
4
5
0
6
7
0
8
i
9
0
10
i
2. In Question 1, a new random variable У is constructed so that Υ = 2X— 1.
If the probability distributions remain the same, calculate S(Y) in the four cases.
Can you generalize your result in any way ?
3. A player pays a certain sum of money to spin two coins. For two heads
he receives back Юр, for two tails he receives 2p, for a head and a tail he receives
nothing. In all four cases he forfeits his stake money. What should the stake
money be for the game to be fair?
4. Two dice are thrown; find the expectation of the higher score showing (or
the score of one of them, if they fall alike).
5. If the probability that a man aged sixty will survive another year is 0-9,
what premium should he be charged for a life insurance policy of £1000? (If he
survives the year, he receives no money back.)
6. ΑΊ and X2 are two random variables, each with values 0, 1, 2, 3, ..., 9, and
each possessing a uniform probability distribution. Evaluate
(OrfUi-.*·); (ii) ЩХг-XS.
7. Two bags each contain ten coloured discs as shown.
Red
4
5
Green
3
3
Blue
3
2

2]
EXPECTATION
A player stakes a certain sum of money for the privilege of drawing two discs,
one from each bag. For two discs of the same colour his stake is returned and, in
addition, he is awarded a prize of Юр for two reds, 20p for two greens and 25p
for two blues. For two discs of different colours he loses his stake.
Show that, if the stake money is 8p, he can anticipate gaining in the long run,
but that with the stake at 9p he should expect to lose.
8. The game of Question 7 is repeated, but the player now tosses a coin to decide
which bag he must choose from: if he tosses a head, he chooses bag I, if a tail,
bag II; he then draws a disc at random from the chosen bag, notes its colour and
replaces the disc. He repeats the process again and is paid prizes as in the
previous question. Determine the minimum stake (to the nearest penny) required to
ensure that the player will show a loss in the long run.
9. The game of Question 8 is repeated but the discs are not replaced between
draws. Determine the minimum stake (to the nearest penny) required to ensure
that the player will show a loss in the long run.
10. The game of Question 7 is repeated, but the player is now required to place
his stake after the result of the first draw from bag I is known. If he draws a red
disc first time he pays 2p, if a green, 9p, if a blue, 12p. Show that, in the long run,
the player expects to win.
Show further that the above stakes are the fairest available that still give an
advantage to the player, in the sense that, if any one of the stakes were increased
by lp, the bank would then expect to win.
11. A man pays a stake to throw two dice. If he scores a total of 3 or 11, he
receives 40p. For a total of 5 or 9 he receives 20p, and for a total of 7, Юр (in
each case the stake money being returned too). He loses his stake money for any
even score. Show that he expects to win if the stake money is 21 p, but that, if the
prizes for scoring 5 or 9 and 7 are reversed, he would then expect to lose.
12. Two identical bags contain respectively (i) four fivepenny pieces and twelve
tenpenny pieces, (ii) nine fivepenny and seven tenpenny pieces. You are allowed to
select a bag and draw a coin at random from it. If the coin you draw is a five-
penny piece, what would be a fair price for you to offer for the bag you did not
select?
13. The path in Figure 10.5 represents a simple maze along which a rat is made
to run. It starts at S and has to finish at F. If it makes a mistake at A by turning
along AA' it will return to A and be forced by the construction of the maze, to
turn towards F, and similarly at each of the other junctions. The probability of
taking either of the two paths available at each junction is \. Find the expected
number of mistakes the rat will make in running from S to F.
А' В' С D'
ι //// g
«♦А В С D _M ,
Start Finish
Fig. 10.5
14. What is the expected number of moves that can be made by (i) a bishop,
(ii) a knight, placed at random on an empty chess board?
175

EXPECTATION [10
15. Two dice are thrown in One turn', each turn costing 5p. If a prize of 40p is
given for a double six and a prize of 20p for any other double (together, in both
cases, with the stake money), determine the loss to a person playing the game
One hundred times.
16. A man puts three coins into a bag, deciding at random for each coin
separately whether it is to be a fivepenny or a tenpenny piece. Calculate the expected
total value of the coins he puts in his pocket.
17. A man puts three £5 notes into one envelope and three £1 notes into a
similar envelope. Each year at Christimas, he chooses one envelope at random and
gives his nephew a note from it. As soon as either envelope is emptied by his
taking the last note from it, the process ends.
(i) State the different totals which the nephew may have received when the
process ends;
(ii) for each of these totals calculate the chance of its occurrence;
(iii) deduce that the nephew's expectation of gain is £12-375.
(Cambridge adapted)
3. STANDARD DEVIATION; EXPECTATION OF
FUNCTIONS OF RANDOM VARIABLES
The expectation £(X) of the random variable X is often called the mean
of X and is frequently denoted by the letter μ (or, occasionally, when it is
necessary to avoid ambiguity by referring to the random variable under
consideration, μχ). μ gives us a forecast of the average value obtained for
the random variable if the experiment is repeated a great number of times;
it does not, however, give us any indication of how the individual results
will be spread out.
By way of illustration, consider two unbiased dice, one of which has its
faces printed in the usual way, but the other has three faces showing one
dot and three faces showing six dots. It is readily verified that the
expected score for the second die is 3-5, just as it is for the first but clearly,
if each die is thrown one hundred times, the separate scores in the two
cases will present a very different appearance: the spreads of the
distributions are quite different. To measure the spreads of these distributions it
might seem plausible to consider a new random variable Υ whose value for
a particular face equals the difference between the score showing and the
mean (commonly called the deviation from the mean) and to calculate the
expectation of Y. However, a calculation in the above example shows that
such a quantity is quite unsuitable.
For the first die, Υ takes the values {-2-5, -1-5, -0-5, 0-5, 1-5, 2-5}
with corresponding probabilities {\, \, \, \, \, £}. For the second die, the
values of Гаге {-2-5, 2-5} with corresponding probabilities {\, £}.
176

31 STANDARD DEVIATION
For die I, £{Y) = i(-2-5-1-5-0-5+0-5 +1-5 + 2-5) = 0.
For die II, £(Y) = i(-2-5 + 2-5) = 0.
A moment's consideration should show that our obtaining the answer
zero in both cases is no coincidence. Indeed, for any distribution, the
expectation of the deviation from the mean is zero. Why?
To avoid the cancelling out of the positive and negative scores a plausible
precaution would be to choose as our new random variable not Y, the
deviation from the mean, but Z, the squared deviation from the mean and
this does, indeed, prove a very satisfactory measure of spread.
Let us see how such a new choice of random variable works in our
example of the two dice. For die I, Z takes the values {6-25, 2-25, 0-25}
each with associated probability J, while for die Π, Ζ takes the value
{6-25}, with associated probability 1.
For die I, S{Z) = K625 + 2-25 + 0-25) = 2-92.
For die II, S{Z) = 6-25.
The precise meaning of these two numbers in probabilistic terms will be
more fully explained in the next section (see Chebyshev's Theorem); for
the moment it suffices to point out that the larger answer obtained for the
second die corresponds to the fact that the scores obtained from this die
are, on the average, farther from the mean than they are for the normal die.
Summarizing the results, we make the following definition: given a
random variable X, with mean μ, the variance, σ|, of X is the expectation
of the squared deviations from the mean. That is, if a new choice of
random variable, Z, is made, where zt = (хг-/<х)2, then
σ% = S{Z)
The positive square root of the variance, σχ, is called the standard
deviation of X; it has the advantage of being measured in the same units
as X. The standard deviation (or, equivalently, the variance) has been
shown, in an intuitive sense, to give a measure of spread; its deeper
significance will be appreciated by the reader only as he gains a fuller grasp of the
subject, when it will be seen that standard deviation plays a central part in
the development of probability theory.
There is no difficulty in extending the concept of expectation to other
functions of a random variable. Indeed, given a function/that maps the
random variable X into another random variable Y, where y{ = f{xv),
177

EXPECTATION [10
we may define the expectation off(X), £[f(X)], by
with this notation σ| = <?[(Χ-μχ)2].
Ex. 6. Find the mean and variance of the random variable X whose probability
distribution is shown in the table below:
X
Ρ
0
1
1
I
2
k
3
i
4
έ
Ex. 7. Find the variance for the number of heads showing if three unbiased coins
are tossed.
*Ex. 8. A random variable has mean μ and variance σ2. If a constant с is added
to each value of the random variable, describe (without embarking upon any
detailed working) what difference this will make to (i) the mean, (ii) the variance.
4. SOME THEOREMS CONCERNING THE MEAN
AND STANDARD DEVIATION OF A RANDOM
VARIABLE
Throughout this section we shall assume that, for some random
experiment, an outcome space S with known probability distribution is given
and a random variable, X, assigned, where £(X) = μ and «?[(X-/<)2] = σ2.
Our first theorem enables us to determine μ and σ in terms of the
expectation of Χ- α and {X- a)2. Its value lies in the fact that these latter
quantities are frequently a great deal less troublesome to calculate than
are μ and σ2 directly from the definition.
Theorem 10.1. If a is any number, then
(i) £[{X~a)] = μ-α; (ii) £\{X~af\ = σ2 + (/*-α)2.
Proof, (i) £[(X-a)] = ΣΡί(χ-α)
= %PiXi~%Pia
= μ~αΣΡί
= μ— a, since Σ A = 1 ■
178

41 SOME THEOREMS
г = 1
= Σ л(*< - μ?+Άμ - α) Σ λ(*< - /*)
+(/ι-β)!.Σί(,
since 2{μ—ά) and {μ—a)2 are common factors and thus may be taken
outside the summation; thus
<?[(X-af] = σ2 + (/ί-α)2, since |й = 1
and ΣιΡί(Χί—μ) = 0 (from Theorem 10.1 (i) putting a = /<).
Example 3. Find the mean and variance of the numbers 1, 2, ..., 20, assuming
that they are uniformly distributed.
We use Theorem 10.1, with a = 10,
^[(Ζ-10)] = -/0-[-(9 + 8 + ... + 1) + (1+2+...+9)+10]
= 0-5,
/ί = 10-5;
^[(Z-10)2] = -2J0-[2(l2 + 22 + ...+92)+102]
= -Ай.9.10.19+100]
= 33-5,
σ2 == 33-5-(0-5)2
= 33-25.
The next theorem determines the mean and standard deviation of a
linear function of the random variable. Intuitively, the results are easy to
understand: multiplication of every value of Ζ by A clearly results in a
proportionate increase in the measures of the average (mean) and spread
(standard deviation), while a shift of origin shifts the average by the same
amount but has no effect upon the spread.
Theorem 10.2. If λ, a are any numbers:
(i) the mean of the random variable Υ = λΧ+α is λμ + α,
(ii) the standard deviation of the random variable Υ — λΧ + α isAa.
Proof, (i) S[XX+ a)] = Σ л(Лх< + a)
= ΑΣίΛ + ι.ΣΑ
= λμ + α.
179

EXPECTATION [10
(ii) The mean of Υ is λμ + α, by (i) and the variance of Υ is given by
^{[γ-ίλμ + α)]*} = <?{[(λΧ+α)-(λμ + α)γ}
= Σ/^Α^-Α/*]2
г = 1
i = l
= AV2.
Our final theorem of this section is a result due to the great Russian
mathematician P. L. Chebyshev (1821-94). Its importance lies in the fact
that it gives us an interpretation for standard deviation in terms of
probabilities ; that is, it gives us, in terms of A, an upper limit for the probability
that a value of the random variable lies further away from the mean than
A standard deviations, whatever the associated probability distribution.
(Of course, if we are given some information about the probability
distribution we can usefully refine the inequality, but this does not detract
from Chebyshev's result viewed as a general theorem.)
Theorem 10.3 {Chebyshev's Theorem). If λ is any positive number,
Ρτ(|*,-/ί| > Ασ) < 1/A2.
Proof. Define the subset A of the outcome space S by
Α = {χί:\χί-μ\ > λσ} = {χ{:(χ^μγ > Α2σ2}.
Now S = A U A' and A n A' = 0. Denoting by Σ summations over
A
those values of i for which *,■ e A, and similarly Σ and Σ, we have,
A' S
'! = Σ^-/·)!
= Σ/ΜΛ-/*)2+ Σρ£χ{-μ?
A A'
>ΣΡί(Χί-μ)2, since Σ/>*(**~/*)2 > 0,
A A'
> Σ Α·λ2σ2, by the definition of A,
A
= AVUft,
Since AV2 > 0 we may divide both sides of the inequality by AV2 :
}2> Σλ = Ργ(*,:|χ*-αΙ >*°-)·
180

4] SOME THEOREMS
Ex. 9. A probability distribution has mean 3 and variance <r2 = 1 -2. Give an
approximate limit for the probability
Pr(|*-3| > 2σ).
The probability distribution for the random variable X(X = 1, 2, ..., 5) is
given in the following table:
* 1 2 3 4 5
Ρ U> i I U \i о
Find the mean and standard deviation. What is Pr (|*-3| > 2cr)? Comment
upon the results obtained and any discrepancy you observe.
Ex. 10. Find an expression for £[(Χ-μ)3] in terms of £(X3), μ and <r.
5. PROBABILITY GENERATING FUNCTIONS
We now introduce a technique of considerable value in dealing with
probability distributions of an integral variable. Given a random experiment the
outcome space for which has associated integral random variable X, the
expectation of the function tx is called the probability generating function
(p.g.f.), G(t), for the distribution. Thus
G(t) = £ψ\ =(|д<".
For example, suppose a card is drawn at random from a well-shuffled
pack and that we score the face value of the card drawn (an ace scoring 1
and a picture card 10). The probabilities of scoring 1, 2, 3, ..., 9 are each
тз, while the probability of scoring 10 is 4/13. Thus
G(t) = ^(t1+t2 + t3 + ... + ts) +-& г10.
The quantity t has no particular significance: it is simply used as a
'carrier' for the values x{ of the random variable. Since the coefficient of
tx<- is р^ the probability of each value of the random variable may be read
off if we know the form of the p.g.f. Thus, if we are able to deduce the form
of the p.g.f. we have a concise summary of the probability distribution.
Theorem 10.4 gives a method of building up the p.g.f. for a complicated
distribution but, before we embark upon its enunciation and proof, we must
make the following definition:
Two random variables X and Υ defined on the same sample space are
said to be independently distributed if
Pr (X = Xi and Υ = уд = ?τ(Χ = χ,) Pr (Υ = у,).
Our next theorem shows that, if X and Υ are independently distributed,
then the new random variable Ζ = X+ Υ has as its generating function
181

EXPECTATION
[10
the product of the generating functions of X and Y. For simplicity, we
shall consider the particular case in which X can take the η +1 values
0, 1, 2, ..., η with Pr (X = г) = Pi and that Υ can take the (n + 1) values
0, 1, 2, ..., и, with Pr (Y =j) = q,-. Then Ζ can take the 2n +1 values 0, 1,
2, ..., In, with Pr (Z = k) = гъ say.
Theorem 10.4. У/" X and Υ are independently distributed random variables
taking the values 0, 1, 2, ..., η and ifthep.g.f.s of X and Υ are Gx(t), Gv(t)
respectively, then the random variable Ζ = X+Y has p.g.f. Gz(t) where
Gz(t)=GJ.t)Gy(t).
Proof. rk = Pr(Z = k)
= Pr(Ar=0andr= A:) + Pr(Z= 1 and Υ = Jk-l) + ...
+ Pr(Ar= Arand Г= 0)
but Ga,(0 G„(0 = Oo +Л '+Λ'2+- +/>» tn) (ς0 + ς^ + ς2^ + ...+ςη tn)
= РоЯо + Oo ?i +Л ?o) ' + (Λ ft +A ft + A ft) t2 + ■ ■ ■
= Gz(t).
Before giving an example of the application of this theorem, we prove
one further result to illustrate the value of developing the theory of p.g.f .s:
we show that they may be used to calculate the mean and standard
deviation of a distribution.
Theorem 10.5. If the random variable X, whose p.g.f. is G(t), has mean μ
and standard deviation σ, then,
(i) G(l)=l;
(ii) σ(1) = μ;
(iii) G"(l) = σ2+/ί2-/ί.
Proof, (i) С?(0=Д^Г
.·. G(l)=ipr
= 1;
(ii) G'(t)
182
= grPrt™
= КГРг
= «?(*)
= μι

5] PROBABILITY GENERATING FUNCTIONS
(iii) G>(t)= ±r(r-\)pTtT-*
■'■ с?"(1)=Д^-1)л
= £(X2)-£{X)
= (σ2+μ2)-μ, by Theorem 10.1, with a = 0.
Example 4. Two unbiased dice are thrown. Find the expectation of the total
score, and its variance.
Let the random variable X denote the score from the first die; the
random variable Υ the score from the second die. Then we have to calculate
the expectations and variance of the random variable Ζ = X+ Y. Now
Gx(0 = «'1 + '2 + -+'6),
Gy(t) = Ut1 + t2 + ...+t6),
л Gz(t)=-h(t1+t2+-+t*)\
assuming that the dice fall independently. Thus
G'z(t) = -hit1+t2 + ...+t6)(l+2t + 3t2 + 4t3 + 5г4+6г6),
G"z(t) = -1h-(t1+t2 + -+t6)(2+6t+l2t2+20г3+30г4)
+is(l+2t + 3t2 + 4t3+5ti + 6t&y,
Л G'z(l) = 7,
G'z{\) = 47f.
By Theorem 10.5, μ = 7
μ2 + ο^—μ = 47f,
μ = 1, σ2 = 5f.
Ex. 11. Three coins are tossed and the number of heads noted. Find the p.g.f.
for the resulting distribution and determine μ and σ2.
6. EXPERIMENTS WITH INFINITE
OUTCOME SPACES
Experiments with an infinite number of possible discrete outcomes are
easy to visualize. For example, suppose a coin is spun until a head appears.
Take as random variable X, the number of tails obtained before the first
183

EXPECTATION [10
head is tossed; then Ζ has possible values 0, 1, 2, 3, ... and
Рг(Х = г) = ^.
Notice that £ Pr (X = r) = ^T = 1»
which is as it should be. Indeed, all the results of Chapter 7 hold for such
outcome spaces; detailed proofs will be omitted, because they depend upon
the manipulation of convergent series. However, assuming the obvious
generalizations of the definitions and theorems, the calculations involved
are not difficult. For instance, in the example above
But fi0*r=(l~*)"1 (W <i}
and so Σ гхг~г = (1 -x)~2,
assuming that the infinite series may be differentiated term by term (a
result always true, in fact, for convergent power series). Thus
Σ rxr+1 = x2(l-x)~2
and, putting χ = \, S{X) = 1.
Example 5. A and В alternately throw a die, the game terminating in a win
for A if he throws a I or a 6, or in a win for В if he throws a 2, 3, 4, 5. Find
the probability that A wins and, if he wins, find the average number of throws
he takes, given that A commences the play.
Pr {A throws 1 or 6 on a particular toss) = },
Pr (B throws 2, 3, 4 or 5 on a particular toss) = -|,
Pr (A wins on (2r+ \)th. play)
= Pr {A and В fail alternately on first 2r plays)
χ Pr {A throws 1 or 6 on (2r+ \)th. play)
= ®r (*)"(*).
Л Pr {A wins) = ΣβΥ Ш = ]4| = *·
We now have to find the expected number of throws taken by A, given
that A wins. We take as our random variable X where
X = r if A wins on his rth turn.
184

61 INFINITE OUTCOME SPACES
Since the game can go on indefinitely, X may take any positive integral
value 1, 2, 3, ... (i ly-ii
Pr (X=r\A wins) = У-3^Ц—^
= my-1
and ^(1) = Σ ir©'"1
(on using the result proved above: Σ rx7'"1 = (1 -x)-2).
Ex. 12. A man tosses a coin until a head appears and is paid a number of pence
equal to the number of tosses he makes. What is his expectation?
Exercise 10(b)
1. X is a random variable with mean μ and standard deviation σ. Write down
the means and standard deviations of the following random variables:
(i) -X; (ii) X+l; (iii) ЪХ-\; (iv) (Χ-μ)/σ.
2. Calculate the mean and variance for each of the following distributions:
(i)
(ϋ)
X
Ρ
X
Ρ
0
i
1
ft
1
i
2
A
2
A
3
ft
3
ft
4
ft
4
*
5
ft
5
1
6
ft
X
^
-3
ft
X
Ρ
50
ft
-2
ft
100
i
-1
i
150
1
0
A
200
i
1
ft
250
ft
2
ft
3
ft
4
ft
3. Calculate the η
s and variances for X2 in each part of Question 2.
4. A cubical die is so weighted that the probability of obtaining any face is
proportional to the score showing on that face. Find the mean and variance of
the score obtained.
5. If the random variable X сгп take the values 1, 2, 3 n, all values being
equally likely, calculate the mean and variance for X.

EXPECTATION [10
6. А, В and С repeatedly throw a die, A starting, then B, then C, then A again
and so on. The winner is the first to throw a six. What are their respective chances
of success ?
7. A throws a pair of unbiased dice, В a pair of dice of which one is unbiased
and the other is such that the probability of a six is p. If they throw in turn and
the winner is the first to throw a double six, find p, given that, when A has the
first throw, the game is fair.
8. An unbiased coin is tossed η times. If a head appears, + 1 is scored, if a tail,
-1 is scored. Prove that the p.g.f. for the random variable X, where x, = total
score after / turns, is given by
G(i) = (2i)-n(i2+l)n.
9. A die is thrown η times. An odd face makes the score showing, an even face
scores zero. Determine the p.g.f. and hence the mean score for η throws.
10. Two dice are thrown together and the scores added. What is the chance that
the total score exceeds 8 ?
Find the mean and standard deviation of the total score. What is the standard
deviation of the score for a single die? (Cambridge)
11. A card is drawn at random from a standard pack and scores the face value
of the card (with ace one and picture cards 10 each). Find the mean and variance
of the score.
If the card is replaced, the pack well shuffled and a second card drawn, find
the probability that the total score for both draws is 12.
12. Two unbiased dice are given, one of which has the faces numbered in the usual
way 1, 2,..., 6, but the other has two faces numbered 1, two numbered 3 and
two numbered 5. Both dice are thrown and the total score, X, is recorded. Find
the mean μ and standard deviation, σ, of X.
Find also
(i) Ρτ(\Χ-μ\ > or); (ii) Pr(|JT-H > 2σ).
Miscellaneous Exercise 10
1. Two bags contain red and white discs as shown in the table below:
Red White
5
10
15
10
One of the bags is selected at random and a disc drawn from it proves to be red.
If the red discs are now valued at £1 each and the white discs are valueless, what
would be a fair price to pay for the remaining discs in the selected bag ?
2. (The St Petersburg Paradox.) A coin is spun. If a head is obtained first time
you are paid lp; if you get a tail followed by a head you receive 2p; for two tails
followed by a head 4p, the next prize being 8p and so on. Show that, however
much you are prepared to pay to play the game, your expected profit will be
positive.
186

6J MISCELLANEOUS EXERCISE 10
Criticise any assumptions you have made and indicate what further
knowledge you would require before offering a more realistic 'fair price' for the game.
If the banker against whom you are playing starts with a capital of lOOp, what
would be a fair price for you to offer him before playing the game ?
3. A and В alternately throw an unbiased die, A having the first throw. The game
terminates when a six is thrown. В agrees to pay A £1 if A throws a six on his
first throw, £3 for a six on his second throw, £5 for a six on his third throw, and
so on. What would be a fair price for A to offer to play the game?
4. Two men play a game with two dice. A has a true die, whereas В has a die
which is biased so that each of the even faces is twice as likely to occur as each
of the odd faces. The two players throw their own die and A wins from В the
sum of the numbers thrown when the sum is even and the numbers are unequal.
Л wins the sum from A when it is odd; and the game is drawn when the two
numbers are equal.
Calculate the expectation of the game to A.
5. X and У are two random variables defined over the same probability
distribution. A new random variable Ζ is constructed, where Ζ = X+ У Prove
that the mean of Ζ is the sum of the means of X and Y.
Is it also true that the variance of Ζ is the sum of the variances of X and У?
6. If X is a random variable with mean μ and standard deviation σ prove that
the standard deviation of the random variable {Χ-μΥ is
{*[(*-/04-«*}*■
7. η points are marked on a line at 1 cm intervals. If two of the points are chosen
at random, find the probability that they will be r cm apart. Deduce that the
expected distance apart of the two points is j(n+1).
8. A and В play a game in which A's chance of winning is p, while B's is q, where
p+q = 1. They have a contest, the winner being the first to score two consecutive
successes. Prove that the expected number of games is
{2+pq)0—pqr\
9. X\s a random variable such that Pr (X < 0) = 0 and the mean and standard
deviation of X are respectively μ and σ. Prove that, for any к 5= 1,
Pr (X < A/t) > ^ .
Deduce Chebyshev's result, that
Ρτ(μ-λσ < X < μ+λσ) 5= 1--^·
10. The random variable У has a distribution such that Pr (У = r) is given by the
coefficient of θτ~' in p\\ — dq)-' for r > t and zero otherwise, where 0 < ρ < 1,
q = 1 —p. Find the expectation of У (M.E.I.)
11. Xis a random variable that can take all positive integral values 0, 1, 2, 3, ....
If
pr = Pr {X = r) and qr = Σ ρ,
s=r+l
7 "μ 187

EXPECTATION [10
and if the p.g.f. for X is G{t) and
H{t)= Zqrt\
prove that H{t) = {1~^°}.
Prove, furthermore, that H{\) = μ where μ is the mean of X, and determine
the value of #'(1).
12. A die is thrown η times. If S denotes the total sum obtained, prove that
£{&) = 7л(21л+5)/12.
13. η identical cards are numbered 1, 2, 3, ..., n, and a random sample (with
replacement) of size к is drawn. If the number on the highest card is taken as the
value of the random variable X, prove that
and deduce that, for large n, the mean value of X is approximately
nk
k + V
14. A and В play a game of golf. At each hole A's chance of winning is a and
B's is b. Find the chance, h, that the hole is halved, in terms of α and b.
Show that the chance of A being r up after η holes is the coefficient of χ to the
power (n+r) in the expansion of (ax2 + hx + b)n.
lfa=b = h, find the chance that A wins 5 and 4, i.e. that he was either 5 up
with 5 to play and then halved the fourteenth hole, or was 4 up with 5 to play
and won it.
Leave your answer in terms of powers of 3 and binomial coefficients.
(M.E.I.)
15. A man consistently stakes a fixed proportion of his available capital on a
fair bet, that is, a bet with zero expectation of gain. If he wins, his existing
capital is increased by a %, but, if he loses, he decreases it by b %. Show that he
must expect to lose in the long run, whatever values a and b might take.
16. Two numbers -Sfand У between 1 and 100 (inclusive) are selected at random,
all possible pairs {X, Y) having equal probabilities. Let Ζ denote the maximum
of X and Y. What is the probability that Ζ < 50? By use of the formulae
Ег = 1л(и+1)
and Σ/-2 = %n(n+l)(2n+l)
or otherwise, show that the mean of Ζ is just over 67. Find a median of Z.
(A median of Ζ is any number ζ such that Pr (Z < 0 5= i and Pr (Z 5= 0 > i-)
(C.S.)
17. Is it possible to devise a coin tossing experiment with a single unbiased coin
to give a probability of τ for some event ?
If you are given a coin which is biased in some unknown way, show how to
devise an experiment in which the probability of success, to be defined, is $.
188

ii. Further vectors
1. THE DOT (SCALAR, INNER) PRODUCT
OF TWO VECTORS
We have defined the combination of two vectors by the triangle rule, an
operation which we called addition and for which we used the sign +. The
choice of the word 'addition' was not entirely arbitrary, for it transpired
that the addition of vectors bore resemblances to the addition of numbers,
in the sense that certain algebraic properties of addition in the two cases
were strikingly similar.
We shall now define a new way of combining two vectors, which we
shall call dot multiplication (otherwise variously called scalar
multiplication, or the formation of the inner product) and denoted by .. The use of
a familiar name and notation is again dictated by the fact that dot
multiplication and the ordinary multiplication of two numbers display a strong
resemblance to one another. However, the reader should again be on his
guard and realize that the two operations are quite different as they
operate on different types of quantities. Having issued this caveat, we now make
our definition of the dot product of two vectors a and b:
a.b = |a||b| cos 6»,
where θ is the angle between the vectors a and b. Note that the
multiplication on the right-hand side of this equation is the ordinary multiplication
of three real numbers, |a|,|b|, and cos Θ, of which the first two are positive
(or zero).
Before proceeding we must clarify one apparent ambiguity in the
definition given above: the angle θ between two vectors is chosen as that angle
Fig. 11.1
189

FURTHER VECTORS [11
(between 0° and 180°) through which one of the vectors must be rotated to
bring it into coincidence in the same sense as the other vector (see Figure
11.1). It follows that the sign of the dot product is positive if the angle thus
defined between the two vectors is acute and negative if the angle is obtuse.
We have said that the word 'multiplication' is used in this context
because of the similarity shown between dot products and ordinary
products. This similarity we now endeavour to display, but to give added force
to our warning against pressing the analogy too far, we start by observing
two laws of multiplication of ordinary numbers that do not hold for
dot products.
Dot multiplication does not give closure; that is, a.b is not a vector.
Dot multiplication is not associative; indeed, (a. b). с is not even defined,
for (a.b) is a number, not a vector. (On the other hand, we can talk about
(a.b) c, which represents a vector in the direction of с and whose
magnitude is |a| |b| cos θ times that of c.)
However, there is a sort of associative rule that holds: if a, b are two
vectors and к is a number,
Ar(a.b) = (A:a).b = a.(jfcb).
The proof of this follows immediately from the definitions and may be
left as an exercise for the diligent reader.
Another reassuring note may be struck by observing that the
commutative law holds for dot products
a.b = b.a.
Again, the proof of this is immediate from the definition and may, with
even more confidence, be left as an exercise for the reader.
The final rule for dot products, that of distributivity over vector addi-
is more subtle and the proof is not immediately obvious. However, the
effort expended in demonstrating this result is amply rewarded, for with
it we shall have set up all the apparatus necessary for an effective algebra
of vectors.
We begin by defining the projection of a vector a in a direction specified
by a unit vector u. Suppose the displacement OA represents a; draw the
plane π through О perpendicular to u. Then, if D is the foot of the
perpendicular from A to π,
a = OD+DA
= OD+/7U
(see Figure 11.2). Further, the expression of a as the sum of two vec-
190

tors, one in the plane л
that
THE DOT PRODUCT
e in the direction u is unique. For suppose
= OO+pu
= OD'+^'u,
Fig. 11.2
where D, D' both lie in π. Then, by subtraction
0 = OD-OD'+/>u-p'u
= O'O+(p-p')u
and so DD' = (p-p') u.
But this is impossible unless D and D' coincide and ρ = ρ', for DD' and
(p-p') u lie in different planes and so cannot possibly be equal, unless
each is the zero vector.
The number ρ that arises in this construction is called the projection of a
in the direction u. We now prove the fundamental result for projections:
if the projection of яг in the direction u is^ and the projection of a2 in the
direction of u is p2, then the projection of aj + a2 in the direction u is
Pl+P2-
Suppose, with the obvious notation, that
aj = ODi+^u,
a2 = OO2+p2u;
then aj + a2 = (ODj + OD2) + (px +p2) u.
But ODj + OD2 is certainly a vector in the plane π, and so, by the
uniqueness of the expression for aj + a2 as the sum of two vectors, one in π and
the other in the direction of u, the projection of aj+a2in the direction of u
isp-L+Pz.
Having defined the projection of a vector in a given direction we may
obtain a simple geometrical interpretation for the dot product of two
vectors a and b. Represent a, b by the displacements OA, OB; let С be the
foot of the perpendicular from A to OB, and D the foot of the
perpendicular from В to О A. Then ОС is the projection of a in the direction of b, and
OD is the projection of b in the direction of a (see Figure 11.3).
191

FURTHER VECTORS
Now a.b = OA(OB cos Θ)
= OA.OD
and a.b = OB(OA cos Θ)
= OB. ОС, Fis· n·3
and thus the value of a. b is the product of the magnitude of either of the
vectors with the projection of the other in its direction.
Our proof of distributivity for dot multiplication over vector addition
has been long delayed but our result, which we shall state as a theorem,
now follows very readily and the results we have proved concerning
projections have an interest and importance in their own right and were
doubly worth pursuing.
Theorem 11.1. For any vectors a, b, c,
a.(b+c) = a.b+a.c.
Proof. Let the projection of b and с in the direction a be px and p2
respectively. Then
a.(b + c) = \я\(р1+р2)= |«|a+I«Ia
= a.b + a.c
(by distributive law for ordinary numbers).
Ex. 1. А, В have position vectors a, b referred to some origin O. If О A is
perpendicular to OB, show that a. b = 0. Is the converse result true ?
*Ex. 2. If a is the position vector of the point A, interpret the scalar product a. a.
*Ex. 3. Show that (a-b) .(a-b) = a.a + b.b-2a.b.
If a, b are the position vectors of the points A, B, interpret this equation (i) when
О A and OB are perpendicular; (ii) when О A, OB are not perpendicular.
Example 1. Prove that, if the sum of the squares of two opposite edges of a
tetrahedron is equal to the sum of the squares of
another pair of opposite edges, then the remaining /1\
pair of opposite edges are perpendicular. /\ \^
We are given that BZ_ I _ _ _\ D
AB2 + CD2 = ВС2 + DA2 XL^^^'^
and have to prove that С
ACLDB. Fig" UA
Take any origin О and let the position vectors of A, B, C, D be a, b, c, d.
Then, in terms of these position vectors, we are given that
(b-a).(b-a) + (d-c).(d-c) = (c-b).(c-b) + (d-a).(d-a).
192

1] THE DOT PRODUCT
Now
(b-a).(b-a) + (d-c).(d-c) = (c-b).(c-b) + (d-a).(d-a)
ob.b-2a.b + a.a + d.d-2c.d + c.c = c.c-2b.c + b.b + d.d-2d.a + a.a
ob.c + d.a = a.b + c.d
o(c-a).(b-d) = 0.
But c-a Φ 0, b-d Φ 0,
.·. AC1.BD.
Exercise 11 (a)
1. Showthata.a-b.b = (a-b).(a+b).
Interpret this equation geometrically and discuss the particular case |a | = |b|.
2. If a, b are non-zero vectors such that |a + b| = |a — b|, show that a and b
are perpendicular.
3. A is a fixed point with position vector a and с is a constant vector. If the
variable point Ρ has position vector r where г satisfies the equation
(r-a).(r-a) = c.c
what can be said about the position of P?
4. Prove that b + c = 0 and |a| = |b| o(a-b).(a-c) = 0.
Interpret this result as a geometrical theorem.
5. OXYZ is a rhombus of side α; Ρ is any point of О У (or О У produced). Prove
that OP. YP = XP2-a2.
6. ACBD is a straight line and О is a point not on the line. ΔΑΟΒ, /-COD are
both right-angles. If OA = a, OB = b, AC = x, CB = y, BD = z, find the
position vectors (relative to O) of С and D.
V*™**»1 CB.BD _ OB-
ACAD OA2'
7. ABCD is a tetrahedron in which ABLCD and ACL BD. Prove that ADA. ВС
and that AB' + CD2 = Ασ+BD2 = Αΰ2 + Βσ.
8. If each edge of a tetrahedron is equal to the opposite edge, prove that the
line joining the mid-points of any two opposite edges is at right-angles to each
of these edges.
9. Prove that, if two of the joins of mid-points of opposite edges of a
tetrahedron are at right-angles, the remaining edges are equal.
10. ABCD is a skew quadrilateral (that is, the vertices А, В, С, D do not all lie
in the same plane). The mid-points of AC, BD are P, Q respectively. Prove that
AB2 + BC2 + CD2 + DAi = AC2 + BD2 + 4PQ2.
193

FURTHER VECTORS [11
11. ABCD is a tetrahedron; X, Y,Z are the mid-points of AB, AC, AD and
P, Q, R are the mid-points of CD, BD, ВС. Prove that
AB^+DC + IPX2 = AC2 + BD2 + 2QY2= AD2 + BC2 + 2RZ\
12. ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that
AB* + Aa + AD* = GB2 + Ga + GD2 + 3GA2.
2. THE DOT PRODUCT (CONTINUED);
COMPONENT FORM
If i, j, к form a right-handed set of orthogonal unit vectors, the following
equalities immediately follow from the definition of the dot product:
i.i = j.j=k.k = l, (1)
j.k = k.i = i.j = 0. (2)
Theorem 11.2. If л = ai\ + a2\+a3k and b = b1i+b2j+b3k,
then a.b = сцЪ^Л-аф2 +a3b3.
Proof: a.b = (a1i + a2i + a3k).(b1i + b2j+b3k)
= a1b1i.i+a2b2}.} + a3b3k.'k+a1bii. j+o^i.k
+а2Ь-1']Л + а2Ь3\.Ъ. + а3Ь1Ъ.Л + а3Ь21к..\
= a1b1+a2b2 + a3b3 by virtue of equations (1) and (2).
From Theorem 11.2 it follows that
|a|2 = a.a = al+a\+a\
, A. A . a.b aJ)-,+αφ2 +αφ3
and that ™°-]j[\W = M+dl + d!b№+l*+l*>·
Example 2. А, В, С have Cartesian coordinates (2, 3, 4), (-2, 1, 0), (4, 0, 2)
respectively. Find:
(i) cos ABAC;
(ii) the unit vector perpendicular to the plane ABC.
χ = AB =-4i-2j-4k,
у = AC = 2i-3j-2k
giving x.у =-8 + 6 + 8 = 6.
But x.y = |x||y| cos Z-ЯЛС
= V36V17 cos ABAC.
194

Now suppose u = li + mj + nk is a unit vector perpendicular to the plane
ABC. Then
2] THE DOT PRODUCT
Thus 6 = 6^17 cos ABAC,
.·. cos ABAC = 1/V17.
zk is a unit vei
Ju.x = 0,
lu.y = 0,
f-4l-2m-4n=0,
·· 1 2l-3m-2n = 0,
— 4m — 4n = 0
and so /:т:и = 1:2: —2, on solving for the ratios l:m:n. Thus
u = Ai + 2Aj-2Ak
where A must be so chosen that |u| = 1,
1,e' A2 + 4A2 + 4A2 = 1,
i.e. A = |
and so u = ii + fj-fk.
Ex. 4. If a = i-3j + k,b = 2i + 2j-k, find a .b.
Ex. 5. Show that the two skew lines
(x-l _y+l _ z+1
e perpendicular.
\x+l
I 1
Ex. 6. Find the unit vector perpendicular to the plane A (3, 1, 1), В (2, 5, 0),
C(-l, -1, 15).
Ex. 7. Deduce the two distance formulae of Chapter 3 using dot products.
Ex. 8. Prove vectorially that the angle between two diagonals of a cube is
arccosi.
3. FURTHER COORDINATE GEOMETRY OF
TWO DIMENSIONS
Throughout this section we shall suppose a right-handed set of axes Ox,
Oy is given, their directions being defined by unit vectors i, j. Л(Х> У-д,
Р&ъУъ) ··■ are given points with position vectors rl5 r2...; P(x, y) is a
general point with position vector r.
195

FURTHER VECTORS
[И
A line is specified given a point Рг lying upon it and its direction; let
us suppose its direction is defined by the unit vector u = /i+mj. Here
/ = cos 0J and m = cos θ2, where θχ and θ2 are the angles made by the
vector u with the axes Ox and Oy respectively (senses being taken into
account) (see Figure 11.5).
The line through Ρχ(χΎ, yj in the direction u thus has the equation
(see Figure 11.5, where ΘΎ φ \ή). Since cos θ2 = sin θ1} this equation may
be rearranged in the form
У-У1 =
ri(x~xi)
or у—У\ = tan Θ^χ—Xj)
agreeing with the form given in Chapter 3. (Notice, however, that the letter
m was used differently on that occasion.)
Given two lines, one through Рг in the direction щ = /ji + mj, the other
in the direction u2 = /2i+w2j, the angle φ between the two lines may be
found by computing the dot product Uj.u2
«l-«2 = l«ll |U2|C0S ?5
= cos φ (since щ, v^ are unit vectors)
and thus cos φ = l-J2+mxm2. (3)
For example, the lines 2x-y+l = 0, 3x + 2.y-3 = 0 may be put in the
χ _ y—\ x—l _ y_
1 " 2 ' -2 "3
or, equivalently, by making the denominators components of unit vectors
(P+m2 = 1)
χ _ y—\
l/V5-2/V5;
-2/V13 3/V13'

3]
FURTHER TWO-DIMENSIONAL GEOMETRY
The angle between the lines is thus given by
1 2 , 2 3
cos^=-V5-Vi3+75-Vi3
4
"V65'
Notice in particular that, from equation (3) the two lines are perpendicular
ol1l2 + m1m2 = 0.
Provided m1 φ 0, m2 Φ 0 this second condition may be written
But, as was noted above, /j/wj and l2jm2 are the gradients of the two lines;
we have thus regained, by an alternative method, the result of Chapter 3.4.
Fig. 11.6
It is frequently required to find the perpendicular distance from a point
to a line. Suppose we are given a line through P1(x1, уг) in the direction
u = /i + mj and a point P2(x2, y2). Let Q be the foot of the perpendicular
from P2 to the given line. Then the unit vector in the direction of QP2 is
- mi + /j (consider the dot product with u). Thus
P2Q = P^.C-mi + Zj)
= ([χ*-χι]ΐ+\η-3>Δϊ).(-ή+ΐί)
= -m(x2-x1) + l(y2-y1).
Although this gives the required perpendicular distance, it may not be a
very convenient form to handle: the equation of a line is generally given in
the form ax+by + c = 0 and we shall now deduce, as a theorem, the
perpendicular distance from the point P2(x2, y2) to this line.
197

FURTHER VECTORS [11
Theorem 11.3. The perpendicular distance from the point Ρ^х2, у2) to the line
ax + by + c = 0 is
ax2 + by2 + c
V(a2 + 62) '
Proof. The given line may be rewritten in the form
x+cja _ у
b -a
or, making the denominators components of a unit vector, in the
equivalent form , ,
x+cja у
blJ(a*+b*) -a/V(a2+62)*
Thus the unit vector defining the direction of the line is
b a .
U V(a2 + 62)! J(a* + b2)h
and a perpendicular unit vector is
,_ a . b .
U V(«2 + *2) V(a2+*2)J'
IfPifa, yx) is some (arbitrary) point of the given line, the perpendicular
distance, d, from P2 is given by
d = PjPa.u'
= a(*i-x2) . Ky-i-Уг)
V02 + Z>2) +V(a2 + *>2)'
But axx + byx =—с, since Рг lies on the line and hence
, = ax2+by2 + c
V(«2 + ^2) *
For example, the perpendicular distance from the point (1, 2) to the line
2*-5J+l=0is 2-10 + 1 _ 7
V29 V29'
(The significance of the negative sign is explained below; the magnitude of
the perpendicular distance is 7/^/29.)
Suppose now that the line joining OP2 cuts the line
ax+by+c = 0
at Pi (see Figure 11.7). Then, if OP1:P1P2 = 1 :A, the coordinates of Рг are
/ *2 _У»_\
\1+λ· 1+A/·

FURTHER TWO-DIMENSIONAL GEOMETRY
Fig. 11.7
But Pj lies on the line ax + by + c = 0 and thus
._^,
:= 0
1+AT1+AT
or ax2 + by2 + c = -Ac.
From this equation it follows at once that, if ax2 + by2+c and с have
the same sign, A is negative and, if they have opposite signs, A is positive.
Since с may be written as a.0 + 6.0 + c, we may summarize the result as
follows.
If, in the expression ax + by + c, the substitution of the quantities
(i) χ = 0, у = 0; (ii) χ = x2, у = y2 gives two values of the same sign, then
P2 and О lie on the same side of the line (for, in this case, Рг divides OP2
externally and so A < 0); if on the other hand, it yields values of opposite
signs, P2 and О lie on opposite sides of the line.
Ex. 9. Find the perpendicular distances of the point (2, — 1) from the lines
3x + 4y + 3 - 0 and 4x-3y-6 = 0. What can you deduce about the position of
(2, -1) relative to the two given lines?
Ex. 10. Sketch the two lines 3x+y-3 = 0,3x+y-6 = 0 and shade in the region
determined by the four inequalities л: > 0, у > 0, Зх+у-3 > 0, Зх+у-6 < 0.
Ex. 11. Find the equation of the bisector of the acute angle between the lines
χ - 0 and 5x+12j>-60 = 0. Find the incentre of the triangle formed by the
lines χ = 0, у = 0, 5x+ 12j>-60 = 0.
Ex. 12. a, b, care constants. Show that, as θ varies, the straight line
xcos#+.ysin0+acos0 + Z>sin0 + c = 0
touches a certain circle. Find the centre and radius of the circle.
Ex. 13. Find the equations of the lines parallel to Sx + 6y + 3 = 0, and distant
1 unit from it.
Exercise 11 {b)
1. Determine the projection upon (a) the χ axis, (6) the line 3x—4y+l = 0 of
the line segments joining the following pairs of points:
(i) (2,3), (4, 4); (ii) (-1,2), (1,5);
(iii) (-1, 2), (2, -3); (iv) {a,b\ {a + b,a-b).
199

FURTHER VECTORS [11
2. Find the cosines of the acute angles between the following pairs of lines:
(i) x-y + l = 0, 3x-y + 2 = 0; (ii) 3x + 2y+l = 0, x-2y-3 = 0;
(iii) 4x-y+l = 0, 3x + 4y = 0; (iv) ax + by = 0, ax-by = 0.
3. Determine the perpendicular distances between the following points and lines.
(Leave your answers in surd form.)
(i) (0, 0), x+4y-1 = 0; (ii) (-1, 2), 2x-3y-4 = 0;
(iii) (-2, -3), x-3y + 5 = 0; (iv) (h, -k), hx-ky + 2hk = 0.
4. Find the distances between the following pairs of parallel lines:
(i) x-2j>+1 = 0,x-2j> + 3 = 0; (ii) 2x+y-3 = 0, 4x+2y+3 = 0;
(iii) 5x-y-l = 0, 5x-y + 3 = 0; (iv) ax + by + a = 0, ax+by-b = 0.
5. Find the equations of the pairs of lines, parallel to the following lines, and
distant three units from them:
(i) 3x-4j>+1 = 0; (ii) 8x+6y + 5 = 0;
(iii) 5x+12j> + 2 = 0; (iv) 3x+2y+l = 0.
6. Draw rough sketches and shade in the areas determined by the following
inequalities:
(i) x> 0,y >0,y <2,2x+3y-8 <0;
(ii) χ > 0,y > 0, x-y-l < 0, Зх+у-7 < 0;
(iii) 2x-y > 0,x-2y < 0,2x-y + 2 > 0, x-2y+4 > 0;
(iv) \x+y\ < u\x-y\ < i;
(v) \x+y-l\ < 1; ^χ-^-ΙΙ < 1.
7. Find the equations of the bisectors of the acute angles between the following
pairs of lines:
(i) χ = 0,3x-4y = 0; (ii) 3x-4y-l = 0,4x-3y-2 = 0;
(iii) 5x+12j> + 3 = 0,6x-8y+3 = 0.
8. Find the equation of the image by reflection of the line x-y = 4 in the line
2x+y = 1. (London)
9. ABC is a triangle; A is the point (0, 3) and В is the point (-5, -2). The
orthocentre (meet of the altitudes) of the triangle is the point (- 1, 1). Find
(i) the coordinates of C; (ii) the area of the triangle;
(iii) the tangent of the angle ACB.
10. Show that the reflection of the point (α, β) in the line у = тх is the point
/(l-ffl2)a + 2W/g 2wa-(l-QA
\ 1 + w2 ' 1+w2 /'
A is the point (j>, q), В is the reflection of A in the line у = χ and С is the
reflection of Л in the line у = - χ. Find the coordinates of C.
Show that С is the reflection of A in the Xmzpx + qy = 0. (London)
200

3]
FURTHER TWO-DIMENSIONAL GEOMETRY
11. A triangle ABC lies wholly within the first quadrant and has an area of
Ц sq. units. The equation of one side is 2x- 5y + 23 = 0 and the vertices A and
В are (1, 5) and (3, 4) respectively. Find the equations of the other two sides, the
angle ABC, and the coordinates of the orthocentre of the triangle. (London)
12. Hf(x,y) = ax + by + c, discuss the changes in value and sign of f(X, Y),
when the point Ρ with coordinates (X, Y) moves in the x, у plane, in relation to
the line whose equation is/(χ, у) = 0.
Determine the smallest value attained by the expression
when χ and у vary subject
22a: + 11?-21,
э the simultaneous restrictions
3x + 4y> 12,
2x-y > 2.
4. FURTHER COORDINATE GEOMETRY
OF THREE DIMENSIONS
As a straight line in two dimensions may be completely defined by a point
and a gradient, so a plane may be defined, given a point lying on it and
the direction of its normal. Suppose that a plane passes through the point
A (position vector a) and is perpendicular to the directions defined by the
unit normal n. Then, if Ρ (position vector r) is any point of the plane,
AP = r-a
(Figure 11.8). But
AP.n = 0,
Λ (r-a).n = 0,
i.e. r.n = a.n. (4)
Equation (4) represents the vector equation of
the plane through A with unit normal vector n.
If Cartesian axes are taken in the directions i, j, к
and η = /ji + y + ^k, a = а^ + а^ + азк, г = xi+yj + zk, then (4) takes
the Cartesian form , , , , , , , , , , ,,4
1гх + 12у + 132 = 1^ + 12(12 +l3a3. (5)
Thus, equation (5) represents the Cartesian equation of the plane
through А{аъ а2, a3) with normal whose direction cosines are 1Ъ /2, /3.
Conversely,
r.n =p (^constant) (6)
represents a plane with unit normal vector n. For r.n is the length of the
projection of OP in the direction n, and, if this is constant for all Ρ, Ρ
clearly lies on a plane perpendicular to n. Thus, (using components),
ax + by + cz = d
201
Fig. 11.8

FURTHER VECTORS [11
is the equation of a plane. The unit normal vector is
(л/(а2+й2 + с2)) *+ Ца2 + й2 + с2))J+ W(a2 + ft2 + c2)) k
and it follows from equation (6), that the perpendicular distance from the
origin to this plane is the numerical value of
d
V(a2 + Z>2 + cT
Ex. 14. Write down the unit vectors perpendicular to
(i) x-2y+2z = 0; (ii) Ъх-y+z = 6; (iii) 3x + 2y+z = 2.
Ex. 15. Write down the equation of the planes through the given points and
perpendicular to the given vectors:
(i) (1,2,0), i-3j + k; (ii) (2, -1, -l),i-j + k;
(iii) (3,-4,2),2i+3j-5k.
Theorem 11.4. The perpendicular distance from the point Рг{хъ ylt Zj) to the
plane ax+by + cz + d — 0 is
ax^ + by-L + cz-L + d
V(a2 + 62 + c2) *
Fig. 11.9
Proof. Let P2{x2, Уъ zz) be any point of the plane ax + by + cz + d = 0 and
let N be the foot of the perpendicular from Рг to the plane (Figure 11.9).
If η is the unit vector perpendicular to the plane, then
Ρ1Ν=Ρ2Ρ1.τι
= [(χι-χύ'ι+ Οί-λ) i+ {ζι-ζύ Ч
' LV^ + ^ + c2){+ V(a2 + *2 + c2)J + 7(а2 + й2 + с2)кJ
_ а{хг - x2) + Ну,. -y2) + c(zx - z2)
V(a2 + ft2 + c2)
202

4] FURTHER THREE-DIMENSIONAL GEOMETRY
But ax2 + by2 + cz2 = —d, since P2 lies on the plane. Thus
PXN
ax1 + by1 + cz1 + d
Again, ax1 + by1 + cz1+d may be positive or negative, depending upon
which side of the plane Рг lies. The result is completely analogous to that
for the straight line.
We conclude this chapter by giving some worked examples of coordinate
geometry in two and three dimensions.
Example 3. Find the complete locus of a point which moves so that its
perpendicular distance from 8x —y + 18 = 0 is twice its perpendicular distance
from 7x + 4y + 6 = 0. (O & C)
LetP(x, y) be any point of the plane. If dlt d2 represent the perpendicular
distances from Ρ to the two lines
8£-jH-18 _ 7x + 4y + 6
dl_± V65 ' 2 V65 ■
The required locus is the totality of points Ρ which satisfy dx = 2c4; that is
if л 8х-.у + 18 Ux + Sy+12\
This is equivalent to
\(х'у)' V65 ~ V65 /
= {(х,у):2х + Зу-2 = 0} U {(дс, у) :22х + 7у + 30 = 0}.
The locus is thus the pair of straight lines
2x + 3y-2 = 0, 22x + 7y + 30 = 0.
4. Prove that the lines
x-3=y-2
1 1
ϊ - ο - ι
are skew, and find the perpendicular distance between them.

FURTHER VECTORS [11
The given lines may be written
x-3 y-2 z-4 ,
~T~ - ~Γ - ~T~ = л'
χ v+1 z-1
and so a general point Ρ of the first line has coordinates (A + 3, A + 2, ЗА + 4)
and a general point Q of the second line has coordinates (/*, - 1, μ +1).
The two lines are clearly not parallel, since their direction cosines are
different. Thus they are either skew or else they meet in a point. If they
meet in a point, we must have
A + 3 = μ,
λ+2 =-1,
ЗА + 4 = μ + l,
for some Α, μ. But solution of the first two equations gives A = — 3, μ = 0
and these values clearly do not satisfy the third equation. The lines are
therefore skew. Now
OP = (A + 3)i + (A + 2) j + (3A + 4)k,
OQ = /ti—j+O + 1) k,
.·. QP = (A-/*+3)i + (A+3) j + (3A-/* + 3)k.
But the directions of the two given lines are given by the vectors
i+j + 3k
and i + k.f
Thus QP is perpendicular to both lines
o(A-/i + 3)+(A + 3)+3(3A-/< + 3) = 0
and (A-/t + 3) + (3A-/t + 3) = 0.
Solving these two equations gives A = -1, μ = 1 and so the coordinates of
the end-points of the common perpendicular to the two lines are (2, 1, 1)
and (1, -1, 2) and the length of the common perpendicular
= Vt(2-l)2 + (l + l)2 + (l-2)2]
= V6.
t Since we are going to use the condition for perpendicularity, in which the right-
hand side of the dot product is zero, there is no need to make these vectors unit vectors
since this would only introduce a factor that would divide out.
204

4] FURTHER THREE-DIMENSIONAL GEOMETRY
Note: A more elegant method for finding the direction of the common
perpendicular to two skew lines will be described when the vector product
is introduced.
Example 5. Find the image by reflection of the point (11, -13, 8) in the plane
2x-3y+z+l = 0.
The vector 2i-3j+k is perpendicular to the given plane and so the line
through (11, -13, 8) perpendicular to the plane has equations
x-U =y+13 = z-8_
The coordinates of a general point Ρ of this line are
(2Λ+11, -3Λ-13, A + 8)
and Ρ lies in the plane
o2(2A+ll)-3(-3A-13) + (A + 8)+l = 0
о 14А + 70 = 0
о А = -5.
Thus, the reflection of (11, - 13, 8) in the plane is given by A = -10, i.e.
(-9, 17, -2).
Exercise 11 (c)
1. Prove that the vectors:
a = i+2j + 4k, b = 2i-3j + k
are perpendicular.
Find the angle between the vectors
c = 3i + j + 2k, d = i-2j-3k.
2. Find the equations of the planes through the given points which are normal
to the given vectors:
(i) (0,0,0), i-j + k; (ii) (1, 2, -3), 2i+j-3k;
(iii) (-1, 2, -4), 3i-k; (iv) (2,4, -1), 6i-3j-2k;
(v) (2, 3, 4), 4i + j - 3k; (vi) (a, b, c), ai + b\ + ck.
3. Find the cosines of the angles between the following pairs of lines. (Leave
your answers in surd form.)
W 2 -3 -1' 1 0 -2'
205

FURTHER VECTORS
W 1 _1 4 ·
(Ш>вг=
"»ί-ί-^τ· --'--δ·
4. Find the cosines of the angles between the following pairs of planes. (Leave
your answers in surd form.)
(i) 3x-y + z = 0, χ = 0; (ii) 2x-y = 0, x+y + z = 0;
(iii) 2x-j»-z+3 = 0, 3x-y + z-l = 0;
(iv) x+j>-4z+l = 0, 2x-3y + 4z+5 = 0;
(v) ax+by = 0, aj>-fe = 0;
(vi) ax + by-cz = 0, bx-cy + ax = 0.
5. Find the cosines of the acute angles between the following lines and planes.
(Leave your answers in surd form.)
© v-v-v-
w -1 0 2 '
x-2 j>-1 ζ
W 2 2 ~-Г
.. . jc-3 y-4 r+3
w 1 0 -1'
, -4 x~a _ У-Ь _ z-c
x+y + z = 0;
lx-y-z = 0;
3x-2y + 2z+4 = 0;
4x-3y-4z = 0;
y-z=0;
px+qy+rz = 0.
6. Find the unit vectors normal to the planes through the following sets of points,
and deduce the equations of the planes:
(i) (1,1, -1), (2, 3,1), (5, -1,-13);
(ii) (2, 6,1), (0, 3,1), (4, 0, -2);
(iii) (1,4,1), (2, 7, 2), (-3,0,-1);
(iv) (1,0,-1), (0, -4, -1), (3, 2, 2);
(v) (1,1, 2), (2, 2,3), (-2,1,11);
(vi) (0, ф, b/c), (с/а, 0, -а/с), ф/a, a/b, 0).
206

4] FURTHER THREE-DIMENSIONAL GEOMETRY
7. Find the coordinates of points where the line
x-2 _ y-3 _ z+1
1 2 ~ -2
meets the planes
(i) x-y-z = 0; (ii) 2x-y-z-6 = 0;
(iii) 3x+4y+2z-9 = 0; (iv) 2x+3y + z = 0.
8. Find the reflections of the following points in the given planes:
(i) (-l,7,S),Sx-y-z-10 = 0;
(ii) (5, -9, -6),2x-y + 3z-8 = 0;
(iii) (6,13, -3),4x-y-3z+l9 - 0;
(iv) (3,5,8),3x-j;-z+26 = 0;
(v) (-5, 11, 6), 5x-6y-2z-27 = 0;
(vi) (0,0,0), ax + by + cz + d= 0.
9. Find the feet of the perpendiculars from the origin to the lines
U 2 1 -1' K} 1 -2 1 '
(ϋο=-τ = ^ = νί αν)
x-3 _ j>-8_ z-3
10. Find the foot of the perpendicular from (7, -
11. Where does the line
x-2 = y-J z-2
1 ~ 3 " -1
cut the plane x—^+z = 0?
Find the image by reflection of the line in the given plane.
12. Find the image by reflection of the line
x-10 = y-4 = z-2
9 -1 ~ 0
intheplane3x-j»+2z-2 = 0.
13. Find the image by reflection of the line

FURTHER VECTORS [11
14. Prove that the following pairs of lines are skew:
χ _ у-Ъ _ _z_
ϊ ~ ~T~ ~ ~i':
x-5 _ j>-8 _z-1
3 ~ 7 ~ -1"
Find the direction ratios of the common perpendicular and determine the
shortest distance between the two lines.
15. ABCD is a rectangle and О is a point on the normal at С to the plane of
this rectangle. AB = a, AD = b, and CO = h. Ρ is a point on АО and the line
through Ρ in the plane AOB which is perpendicular to АО meets AB at M.
If AP = x, show that
PM = xj(b*+h*)/a; AM = xj(a2 + b2 + h*)la.
Prove that the cosine of the acute angle between the planes OAB and О AD is
a6/V[(e2 + A2) (b2 + A2)]. (London)
16. Define the scalar product of two three-dimensional Euclidean vectors u, v.
Deduce an expression for the angle between the vectors
u = «1i+«2j+M3k,
ν = Cii+cjj+сзк,
in terms of ult u2, u3, vlt v2, v3.
A regular tetrahedron has vertices O, A, B, C, where О is the origin, and A,
В, С have position vectors with respect to О given by
OA=-i+j,
OB = ai+ b],
ОС = pi+q] + rk.
Find the numerical values of a, b, p, q, r given that a > 0 and /· > 0. (SMP)
17. Show that the line L given by
x+l = y-1 = z+l
5 ~ 3 ~ 2
is the intersection of the planes
3x-5j>+8 = 0 and 2y-3z-5 = 0.
Show that every plane containing the line L can be expressed in the form
λ(3χ-5γ+8) + μ(2γ-3ζ-5) = 0.
How should Α, μ be chosen in order to ensure that the plane is perpendicular
totheplane 5x-y+2z=-21

4] FURTHER THREE-DIMENSIONAL GEOMETRY
Hence or otherwise obtain the equation of the orthogonal projection of the
line L on the plane
y 5x-y+2z =-2,
expressing your answer in vector form, χ = ia+ b. (M.E.I.)
18. Prove that the equation of the plane which cuts off intercepts a, b, с on the
axes of x, y, ζ respectively is
The foot of the perpendicular from the origin to a plane isP(2, -1, 2). Find
the equation of the plane. If the plane meets the axes of x, y, ζ at А, В, С
respectively, prove that AP is perpendicular to ВС, and find the angle between AP and
CP. (J.M.B.)
19. Two planes, itx and щ, have equations
2a:+j>+z=1 and Ъх+y-z = 2,
respectively. Prove that the plane π3, which is perpendicular to nlt and contains
the line of intersection of щ and щ has the equation
x-lz = 1.
Points Ρ and Q lie on the planes щ and π3 respectively and the line PQ is
perpendicular to Щ. If the coordinates of Ρ are (-2, 4, 1), find the coordinates
of Q. Determine the angle between the line PQ and the perpendicular from Ρ to
the line of intersection of the three planes. (J.M.B.)
20. The straight line whose equations are
-2 1 2
meets the plane x+ 2y-2z = 8 at B, and A is the point (2, 0, - 1) on the line.
The foot of the perpendicular from A to the plane is С Find
(i) the coordinates of В and C;
(ii) the length of AC.
Show that the sine of the acute angle between В A and ВС is %.
The line А С is produced to D so that А С = 2CD. Find the coordinates of D.
(J.M.B.)
Miscellaneous Exercise 11
1. The perpendicular distance from the origin to a straight line / is of length
ρ and makes an angle a, with the positive χ axis. Prove that the equation of the
line may be taken in the form
χ cos a.+y sin a = p.
What is the equation of the parallel line through the point iVX, y^ ? Deduce
the perpendicular distance from Px to the given line.
2. Find the equations of the straight lines which bisect the angles between the
straightlines 3x_4j;_11 = o, 12*+5,-2 = 0. (O&C)

FURTHER VECTORS [11
3. Find the equation of the bisector of the acute angle between the lines
4x + 3y-12 = 0, 12* + 5y- 60 = 0. (O & C)
4. Sketch the triangle formed by the lines
3x-4y-4 = 0, 12x-5j> + 6 = 0, 7x + 24j;-56 = 0,
and verify by calculation and reference to your sketch that the point (1,1) is the
centre of the inscribed circle. (O & C)
5. A straight line is drawn through AQi, k) and P(x, y) so that AP makes an
angle θ with the positive direction of the χ axis, and AP = r. Prove that
Three vertices of a square are E(2, 2), F(-2, 2) and G(-2, -2); a straight
line of gradient J is drawn from A{- 3, - 1) to meet FG at Ρ and EF at Q. Use
the formulae in the first part of the equation to find the length of Ρ β.
Find also the radius of the circle with centre at the origin to which APQ is a
tangent. (O & C)
6. Points A and В lie on the same side of a line /, С is the optical image of A in
/, and ВС meets / at P. Prove that / is a bisector of ΔΑΡΒ. Given that / has
the equation x + 3y = 5, and that the coordinates of A and Л are (1, -2) and
(—11, 2) respectively, verify that these points lie on the same side of /. Prove
that the coordinates of С are (3, 4), and find the coordinates of P. (O & C)
7. Prove that the equation of the circle, centre (12, 13), radius 7, is
x2+y2-24x-26y + 264 = 0.
Two sides of a triangle are
5x-12j;+5 = 0, 12x-5j>+12 = 0
and the incentre is the point (12, 13). Prove that the third side touches the circle
x2+y2-24x-26y+264 = 0. (O&C)
8. Prove that the perpendicular bisectors of the sides of a triangle are
concurrent (at the circumcentre) and that the altitudes of a triangle are concurrent (at
the orthocentre).
9. Prove that the circumcentre S, the centroid G and the orthocentre Я of a
triangle ABC have position vectors (referred to any origin O) which satisfy the
relation
h + 2s-3g = 0.
Deduce that SGH is a straight line and determine the ratio SG: GH.
Prove further that the mid-point, N, of the line SH is the centre of the circle
through the mid-points of the triangle ABC. What is the radius of this circle?
10. О ABC is a tetrahredron in which О A is perpendicular to ВС and OB is
perpendicular to AC PQR is a triangle such that A is the mid-point of QR, В
is the mid-point of RP and С is the mid-point of PQ.

4] MISCELLANEOUS EXERCISE 11
Taking О as the origin, express the position vectors of P, Q, R in terms of those
of А, В, С Hence prove that OP = OQ = OR.
If D is the foot of the perpendicular from О to the plane ABC, prove that D
is the circumcentre of the triangle PQR.
11. ABCDA'B'C'D' is a cube with edges AA', BB\ CC, DD' and diagonal AC.
Show that B'C is perpendicular to the plane ABC. Find a line in the figure
perpendicular to the plane ACC. What is the angle between the planes ABC
and ACC"]
12. In a tetrahedron PQRS, the edges PQ, RS are perpendicular to the faces
PRS, PQS respectively; L is the mid-point of PS and Μ is the mid-point of QR.
Prove that PQ} + RS* = QR2-PS\
PM= SM= iQR,
4LM2 = PQ2 + RS2. (O & Q
13. Prove that there is one and only one line which joins two given skew lines
and is perpendicular to each of them.
Two fixed skew lines AL, BM have a common perpendicular AB, and the angle
between them is Θ. Prove that
LM2 = AL2 + BM*-2AL.BMcose.
Prove that, if θ = in and the points L and Μ vary on the fixed skew lines so
that LM is constant, the locus of the mid-point of LM is a circle. (O & C)
14. The common perpendicular of two skew lines / and /' meets them at A and A'
respectively. Points Ρ and P' are taken on / and /' respectively so that AP+ A'P'
is constant, where the sense of AP and A'P" is taken into account.
Show that the locus of the mid-point Μ of PP' is a straight line m, and
describe its relation to / and /'.
If AP— A'P' is constant, show that the locus of Μ is another straight line m',
and describe the relation of m and m'. (O & C)
15. Find the reflection of the line
3 1 ~ -1
in the plane x-y-z + 2 = 0.
16. Prove that, given two skew lines and a point О not lying on either of them,
that just one transversal may be drawn through О to cut each of the lines.
Prove that the lines _
x-4 y + 3 z + 2
are skew and find the equation of the common transversal that passes through
the origin.
17. Obtain the equation of a plane in the vector form
n.r = ρ (η a unit vector),
explaining precisely what you mean by each of the three symbols n, r, p.
211

FURTHER VECTORS [11
Prove that the length of the perpendicular to the plane from the point S with
position vector s is |n.s—p|; and find the position vector t of the mirror image
Τ of S in the plane—that is, of the point Τ such that TS is perpendicular to the
plane and bisected by it. (M.E.I.)
18. Referred to a given system of rectangular coordinates in space of three-
dimensions, the points А, В have coordinates (1, 0,0), (-1, 0,0) respectively.
A variable point Ρ has coordinates (x, y, z). Write down the direction-cosines
of PA, PB and prove that, if θ is the angle between them, then
2 θ = (x2+y2 + z2-l)2
[(x2 + y* + z* +1) - 2x]. [(x2 + y* + z* +1) + 2x]'
Deduce that, if Ρ is restricted to lie in the plane ζ = 0, then
C^+y-l)2 = [(x2+j>2-1)2 + 4j>2]cos20
and find the equations of the two circles on which Ρ must lie if the angle θ is kept
constant. (M.E.I.)
19. Two planes x-3y + 2z = 2,
2x-y-z= 9,
meet in the line /. Find the equations of
(i) the plane through the origin which contains /,
(ii) the plane through the origin which is perpendicular to /.
Find also the coordinates of the reflection of the origin in /. (C.S.)
20. Find the value oik such that the line joining the points (- 2, k, - 9), (2,1, 7)
intersects the line joining (- 2, -4, 4), (7, 2,1).
What are the coordinates of the point of intersection ?
21. A regular tetrahedron ABCD has the face ABC in the xy plane, the origin
is the centre of that face, the vertex A is at the point (1, 0, 0) and the vertex D
is on the positive half of the ζ axis. Find the coordinates of В, C, D and of the
centre of the tetrahedron, and the direction ratios of the normals to the four
faces.
Hence or otherwise show that, if any line makes angles α, β, γ, δ with the faces
of a regular tetrahedron, then
sin2 a + sin2 /3+ sin2 γ+ sin2 δ = f.
22. Prove that the equation of the straight line through the given point A,
position vector a, and in the direction of the unit vector b is
г = a + ib,
where t is the distance from A of the variable point Ρ of the line whose position
vector is r.
Prove also that the equation of the plane through the point С (position
vector c) whose normal is in the direction given by the unit vector d can be
expressed in the form
r.d = c.d.
212

4] MISCELLANEOUS EXERCISE 11
The plane through the point C(l, 2, 4) has normal in the direction
d = Ai + ^j + Hk.
Find the length of the shortest distance from the point A (3, 4, 5) to the plane.
(M.E.I.)
23. To find the position of an underground rock layer a number of vertical
borings are made at points on horizontal ground which form a coordinate grid.
Results are as follows, the unit of distance both horizontally and vertically
being 300 m:
Grid-point (0,0) (2,0) (0,2) (2,2)
Depth 0-270 0-225 0162 0117
Show that these results are consistent with this part of the rock layer being a
plane. Find the (x, y) equation of the line in which this plane when produced
would meet the ground, and show that the plane would be inclined at about 3i°
to the ground.
What would you conclude if a boring at (1,1) gave a depth of about 0-18?
(M.E.I.)
24. Calculate the shortest distance between the line of intersection of the planes
x-8y + 2z+9 = 0, x-2y-z+6 = 0
and the line of intersection of the planes
2x+y + 8z-l2 = 0, x-y + z-6 = 0;
and show that the line which cuts both these lines at right-angles passes through
the point (-4,12,8).
25. Find the coordinates of the mirror image of the point (p, q, r) in the plane
ax + by + cz + d = 0.
A ray from the origin is reflected successively in the planes
x+y-z+l = 0
and x-y + 2z-l = 0,
and then passes again through the origin. Find the points at which it meets the
two planes.
26. Prove that the lines
^±1 _ l±l - £±3 *-l_J, + 3_ ζ
2 ~ 3 ~ 1 ' 3 ~ 1 ~ -2
do not intersect.
Find the equation of the plane through the origin which contains the first
line, and find also the direction cosines of the line through the origin which meets
both lines. (Oxford Mod.)
27. The line x\l = y/m = zjn is reflected in the plane ax + by + cz + d — 0. Show
that the equation of the resulting line is
(fl2 + 62 + c2)x + 2arf {a2 + b2 + (f)y + 2bd
(cP + b + c2)l-2a(al+bm + cri) ~ {ai + bi + <F)m-2b(al+bm + cri)
= (fl2 + 62 + c2)z + 2crf
~ (а2 + Ъ2 + с2)п-2с(а1+Ът + спУ
213

FURTHER VECTORS [11
Hence, or otherwise, find the equation of the plane such that the angle between
it and the plane lx + my+nz = 0 is bisected by the plane ax + by+cz = 0.
(Oxford Mod.)
28. Write down the equations of the axes Ox, Oy, Oz. What is the equation of the
plane containing Ox and the point (a, b, c) ?
The roof of a rectangular house consists of four inclined planes, each sloping
upwards at an angle of 45° to the horizontal. What is the angle between two
adjacent faces of the roof?
29. ABCDA'B'C'D' is a cubical box, with faces ABCD, A'B'C'D' and edges AA',
BB', CC, DD'. Ε is the point on BB' such that BE = \EB'\ F is the mid-point
of CC. Find the angles between
(i) the line А'В and the plane AEF;
(ii) the plane А'ВС and the plane AEF.
30. Two lines through the origin have unit direction vectors /ji + wj + ^k and
/2i + w2j+H2k. Prove that the locus of points equidistant from the two lines is a
pair of planes and deduce the equations of the straight lines which bisect the
angles between the given lines.
31. Two straight paths on a plane hillside are at right-angles, and make angles θ
and φ respectively with the horizontal. If the hillside itself makes an angle a, with
the horizontal, prove that .,„,.,.
sin2 θ + sin2 φ = sin2 a.
Prove also that the acute angle between the projections of the paths on a
horizontal plane is . ^ ,4
arccos (tan θ tan φ).
32. A line of slope, in an easterly direction, of a plane hillside is inclined at an
angle a to the horizontal. A line of slope of the hillside in a southerly direction
is inclined at an angle β to the horizontal. Prove that the actual inclination, Θ,
of the hillside to the horizontal is
θ = arctan (tan2 a, + tan2 #)*.
A vertical pole of height h is placed on top of the hill. Show that the angle φ
subtended by it at a point distant a down the line of greatest slope through the
foot of the pole is given by
Y ~ asec0+Atan0"
Find φ, if h = 16, a = 36, a = 30° and β = 45°. (London)
33. Let G be the centroid of the acute-angled triangle ABC of circumradius R.
Showthat AG°+B(?+CG>>2R>.
34. If О is the point on AB such that АО = 20B, and if Ρ is any point, prove that
APt+lBPt-SOF2
is independent of the position of P.
What is the locus of a point which moves so that the sum of the squares of its
distances from the vertices of an equilateral triangle is constant ?

12. Further trigonometry
1. FORMULAE FOR COMPOUND AND
MULTIPLE ANGLES
In many applications of trigonometry it is essential to be able to deal with
expressions such as sin(^+5) or sin2^ in terms of the trigonometric
functions of the simpler angles A and B. Angles such as A + B are called
compound angles; in particular, those like 2A are called multiple angles.
Ex. 1. By taking A = В = 30°, show that sin (A + Β) φ sin A + sin В in general.
Is it possible to find values of A and В such that sin A + sin В cannot be expressed
as the sine of a single angle?
We shall now establish the fundamental compound angle formulae for
sin (θ + φ), cos (θ + φ). The proof depends upon the fact
that we may express a unit vector OP in the form —
cos θ\ + sin Θ],
where i, j are two perpendicular unit vectors and θ is the
angle between i and OP measured anticlockwise from i;
see Figure 12.1 which represents a circle of unit radius, „
with Fl& Ш
OA = i, OB = j, OP = p, where ρ = cos 0i + sin 0j.
Similarly, if OQ makes an angle 90°+ 0 with i (that is, if LPOQ = 90°,
measured anticlockwise), then
OQ = q = cos(9O° + 0)i + sin(9O° + 0)j
= —sin <9i + cos θ\.
These expressions for p, q hold for angles of any size and either sense.
Theorem 12.1. For any two angles:
(i) cos (<? + 0) = cos <9cos0-sin<9sin0;
(ii) sin (θ + φ) = sin θ cos φ + cos θ sin φ;
(iii). cos ((9-0) ξ cos θ cos φ + sin θ sin φ;
(iv) sin {θ — φ) = sin θ cos φ-cos θ sin φ.
Proof. Take points APRBQ as a unit circle, centre O. OA = i, OB = j;
OP = p, OQ = q; OR = r. ΔΑΟΡ = Θ, LPOR = φ (see Figure 12.2).
215

FURTHER TRIGONOMETRY [12
The proof of (i) and (ii) is effected by expressing the unit vector г (a) in
terms of i and j directly and φ) in terms of ρ and q and hence in terms of i
and j. пи
Thus (а) г = cos {θ + φ) i + sin {θ + φ) j
and ф) г = cos φγ> + sin 0q.
But ρ = cos <9i + sin Θ]
and q = — sin #i + cos Θ]
(see remarks preceding this theorem).
Substituting for ρ and q in φ) this gives
(с) г = cos 0(cos 0i + sin <9j) + sin 0(-sin (9i + cos Θ])
= (cos θ cos φ — sin θ sin 0) i + (sin θ cos φ + cos 0 sin φ) j.
But, in a plane, the expressions for the components of a vector г in the
two directions i and j are unique (see Chapter 2) and so, comparing (a)
and (c),
(i) cos {θ + φ) ξ cos θ cos φ — sin θ sin 0;
(ii) sin (θ + φ) = sin θ cos 0 + cos θ sin 0.
Formulae (iii) and (iv) now follow immediately, on writing — φ for φ and
recalling that sin (-φ) = -sin φ, cos {—φ) = cos 0,
(iii) cos {θ — φ) = cos 0 cos φ + ύηθ sin 0;
(iv) sin {θ — φ) = sin 0 cos φ — cos б sin 0.
Ex. 2. Express sin 15° and cos 15° in surd form by writing 15° = 45° -30°.
Ex. 3. Express cos 165° and sin 105° in surd form.
Ex. 4. If sin χ = tt and sin у = f find sin (x+y) and cos (x+y)
(i) when χ and j> are both acute angles;
(ii) when* is acute and j> is obtuse;
(iii) when χ and у are both obtuse angles.
Ex. 5. Express cos(^ + 5)-cos(^-5)in terms of sin A and sin Л.
The results of Theorem 12.1 are important and should be committed
to memory. They give rise immediately to the following results which we
shall also state in the form of a theorem:
Theorem 12.2. For any angles θ, φ for which all of the expressions appearing
(vi) tan (0-0) =
216

I] COMPOUND AND MULTIPLE ANGLES
(vii) sin 20 ξ 2 sin 0 cos 0;
(viii) cos 2Θ = cos2 0-sin2 θ = 2 cos2 »-1ξ1-2 sin2 Θ;
2tan0
(ix) tan 2
l-tan2(
Proof, (v) The right-hand side is not defined if θ or φ = (2k + \)%n\
neither side is defined if θ + φ = (2k +1) \π. For any other values of θ and φ
/a , л\ sin θ cos φ + cos θ sin φ , . . ,.λ , ....
tan (θ + φ) = !-, ■ a ■ · by formulae (ι) and (n);
v T' cos 0cos0-sin0sm0 J ww»
divide top and bottom of the fraction on the right-hand side by cos θ cos φ
(a non-zero expression, by the restrictions on θ and φ).
(vi) As for (v) by using (iii) and (iv);
(vii) set θ = φ in(ii);
(viii) set θ = φ in (i), and recall that sin2 0 + cos2 Θ = 1;
(ix) set θ = φ in (ν).
*Ex. 6. What values of θ and φ must be excluded in (a) formula (vi); and (b)
formula (ix) ?
Example 1. Find an expression for cos 75° in surd form.
Method (i). „„ . ,... ,,„.
w cos 75° = cos (30°+ 45°)
= cos 30° cos 45°-sin 30° sin 45°
Method (ii).
Thus
(We reject the negative square root since cos 75° > 0.)
Example 2. Prove the identity
V3 ι
" 2 V2"
V3-i
2V2 '
;os 150° :
-iV3
4-2V3
8
cos 75°
1 1
"2V2
= 2 cos2 75°-1,
= 2 cos2 75°-1,
= cos2 75°.
V3-i
2V2 ■

FURTHER TRIGONOMETRY
The r.h.s.
L.H.S. =
=
is defined,
1 si
by the restriction
η 2Θ
cos 20 cos 20
1-sin 20
cos 2 0 '
(-4^й> (sin'* + ,
cos2 0 -sin2 (9 v
cos2 0 + sii
1 -sin 2Θ
ι2 0-2 sin 0 cos 0
cos 20
(sin2 0 φ cos2 0, since 0 φ (2л +1) £тг)
The formulae we have proved are also useful in dealing with
combinations of inverse trigonometric functions, as is shown in the next example.
Example 3. Find the value of arctan 2 +arctan 3.
Write χ = arctan 2; у = arctan 3, then
\π < χ < %π, \π < у < \π and so \π < χ+y < π,
f tan j
Since
Thus
fr<
x+y < π,
" 1-tanx
2 + 3
1-6
= _i.
this gives
x+y
arctan
= π + arctan (
= π-1π.
2 + arctan 3 =
tan у
-i)
fr.
Example 4. Find angles 0 between 0° and 360° satisfying the equation
5 cos 0-6 sin 0 = 3.
Observe that 5 and 6 are two sides of a right-angled triangle whose
hypotenuse is V61. Thus, dividing both sides of the equation by ,/61,
cos α cos 0—sin α sin 0 = -^y,
where tan α = f, i.e. α = 50° 12'. Also, using tables,
vfi = cos67°25'

COMPOUND AND MULTIPLE ANGLES
No.
3
61
v'61
cos67° 25'
Log.
0-4771
1-7853
0-8926
T-5845
0-sin 50° 12' sin Θ = cos 67° 25',
cos (50° 12' + 6) = cos 67° 25',
50° 12'+ 0= 67° 25' or 292° 35'
0 = 17° 13' or 242° 23'.
Examples. Find maximum and minimum values for the expression
у = sin 20 + 4 cos 20.
This may be solved by a process similar to that
employed in Example 4. Thus
, = V17(^sin2tf + Acos2,)
= V17 sb (20 +a), Fig. 12.4
where tan a = 4. But — 1 ^ sin φ ^ 1,
.·. -V17 <J < V17·
Observe that, since tan 76° x 4, у attains its maximum when
20 + 76° χ 90°,
i.e. θ χ 7°, and its minimum when 20 + 76° χ 270°, i.e. θ χ 97°.
Ex. 7. Find the maximum and minimum values of the following expressions:
(i) sinx-3cosx; (ii) 2 cos χ + sin χ; (iii) 5 + 3 sin χ + 4 cos χ;
(iv) 1 — sinx-cosx; (v) 1/(2— sin * + cos x).
Ex. 8. Solve the following equations, giving all solutions lying between 0° and
360° inclusive:
(i) cosx — sin л: = 1; (ii) 3 cos x+4 sin χ = 5;
(iii) sin χ + 2 cos χ = 1; (iv) 12 sinx — 5 cos χ = — 4.
Ex. 9. If tan Л = i, and tan В = |, find tan (A + B) and cot (Л-Я).
£*. 10. If x+y = in and tan χ = i find tan y.

FURTHER TRIGONOMETRY [12
Ex. 11. If cos χ = i, find cos 2x. If, furthermore, 0 < χ < π, find sin 2x.
Ex. 12. Find the maximum and minimum values of 1 + sin χ cos x.
Ex. 13. Find the values of χ in the interval 0 < χ < 2π which satisfy the
equation cos 2x = 3 sin x— 1.
Ex. 14. Find the value of tan &r without using your tables.
Ex. 15. Find sin (2 arcsin x) and cos (2 arccosy).
Ex. 16. By writing χ = arctan \ andj> = arctan f and considering tan (x+y),
show that arctan i+arctan ^ = arctan f.
Ex. 17. Express arcsin f + arcsin A in the form arcsin x.
The multiple angle formulae can be used to derive three very useful
expressions, for sin 0, cos 0 and tan 0, in terms of tan \θ. Their value
derives from the fact that, with their help, a trigonometric expression may
be written as a rational function^ of a single variable t = tan \θ. The reader
will find that they have applications to problems in calculus as well as
ordinary trigonometry.
Theorem 12.3. If tan \θ = t, then
(i) άηθ**;
00
(iii)
Proi
cos θ
tan0
>/· 0)
fiO
(iii)
t A
~l+t2'
-Д-, (M*i).
sin θ = 2 sin £0 cos £0
_ 2 tan \θ
~ sec2i#
2i
1 +12''
cos 0 = cos2 -^0 —sin2 \θ
_ l-tan2i0
sec2 ^0
1-i2
l+i2;
tan 0 = T372. by dividing (i) by (ii).
rational function is defined as the rai
tio of two polynomials

COMPOUND AND MULTIPLE ANGLES
Example 6. Prove the identity
l+sin0
cos θ
1 +
l+t2
_(!±0!
tan@ir+№ (?*(2n+l)№,
1 _ - , ^n multiplying numerator and denominator by (1 +t2),
= rzv 1 + '*° (^*(2«+i)i^),
tanjg + tanjg
l-tan|7rtani^
" 1-f*
etani?r Φ 1.
-.= 3>
Example 7. Sofoe ί/ze equation of Example 4, tuzng ί/ге substitution
tan £0 = i.
5 cos 0-6 sin 0 = 3,
. 5(1 -f2) 12i_
·' l+f
5 —5i2—12i = З + Зг2,
4t2+6t~l = 0,
, _ -6+v(36+i6)
8
= K-3±Vi3)
= 0-151 or -1-651,
.·. \θ = 8° 35' or 12Г 12',
0 = 17° 10' or 242° 24'.
Note. Some discrepancy arises between the results obtained here and in
Example 4. Rounding-off explains the error in the larger answer; the
smaller answer is seriously affected in Example 7, since the subtraction
— 3 + V13 loses a significant figure.
Ex. 18. Express sin x + cos χ in terms of t, where t = tan \x.

FURTHER TRIGONOMETRY [12
2. SUM AND PRODUCT FORMULAE
The expansions derived for cos (A ± B) and sin {A ± B) lead to a further
series of identities between trigonometric functions which prove of great
value in the manipulation of trigonometric expressions. The identities
which we are about to deduce may be divided into two groups:
1. The expression of products of trigonometric functions as sums.
2. Conversely, the expression of sums of trigonometric functions as
products.
We shall state and prove these two sets of identities as two theorems.
Theorem 12 A.
(i) sin A sin В з i[cos (A - B) - cos {A + B)];
(ii) cos A cos В з £[cos (A + B) + cos (A - B)];
(iii) sin A cos В з i[sin (A +B) + sin (A-B)];
(iv) cos A sin В з £[sin (A +5)-sin (A-B)].
Proof, (i) R.H.S. = i[(cos A cos 5+sin A sin B)
— (cos A cos B— sin A sin B)]
= £.2 sin A sin -S
= L.H.S.;
(ii), (iii), (iv) are all proved in a similar fashion.
Theorem 12.5.
rs ■ . χ, ^ . A+B A-B
(ι) sin Λ+sin 5 3 2 sin—=—cos—=—;
,..ч · , · D „ Л+Я . A-B
(n) sin Л — sin В 3 2 cos—=—sin—=—;
Proof.
... . ΓΑ+Β Α-ΒΊ . ΓΑ+Β Α-ΒΊ
(ι) R.H.S. = sin j^—2 '—2—J "*"Sln \_~~2 2~~J*
= sin A+sin В
= L.H.S.
(ii), (iii), (iv) are all proved in similar fashion.
222
by Theorem 12.4(iii)

SUM AND PRODUCT FORMULAE
Example 8.IfA + B+C= 180°,prove that
cos A +cos 5+cos С — 1 +4 sin у sin j si
L.H.S. = cos A + [cos 5+cos C]
=
=
=
=
1-2
1-2
1-4
R.H.S.
sin у sin у—cos
. АГ B+C
sin-y cos—2
sin у sin
5 . С
2S1H2
5-СП
2 "J
B-
-cos—2
Example 9. Sofoe i/ze equation
sin Зх + sin л: = cos л:
gz'wng a// roots in the range 0° < χ < 360°.
sin 3x + sin л: = cos л:
о- 2 sin 2x cos χ = cos χ
ocosx(2sin2x-l) = 0
=> either cos χ = 0 or sin 2x = £.
(1) cos χ = 0 gives χ = 90° or 270°.
(2) sin 2x = i gives 2л; = 30° or 150° or 390° or 510°
and χ = 15° or 75° or 195° or 255°.
Thus χ = 15° or 75° or 90° or 195° or 225° or 270°.
Note. A very common error made in equations of the type in Example 8
is to forget the solution cos χ = 0.
Ex. 19. Express in factor form:
(i) sinx + sin3x; (ii) cos 4x+cos 2x;
(iii) cos 3x — cos x; (iv) sin x+cos x;
(v) COS {χΤΤ— Χ) — COS (ϊ7Γ + Χ).

FURTHER TRIGONOMETRY [12
Ex. 20. Prove the identity sin χ + sin 2x + sin 3x = sin 2x(2cosx+l). Hence
find all values of χ in the interval 0 =g χ < 2π satisfying the equation
sinx + sin 2x + sin Ъх = О.
*Ex. 21. Solve the equation sin χ + sin 2x = 0 by expressing the left-hand side in
factor form, giving all solutions in the interval 0 =g χ =g 2π. Solve the equation
again in the following alternative ways:
(i) by rewriting it as sin 2x = sin (-*), etc.;
(ii) by rewriting it as sin χ + 2 sin χ cos χ = 0.
Exercise 12(a)
1. Evaluate in surd form:
(i) sin 75°; (ii) cos 105°; (iii) tan 105°; (iv) cot 75°.
2. Prove that sin ЗА = 3 sin A-4 sin3 A and that cos ЗА = 4 cos3 A- 3 cos A.
By considering the equation sin 2A = cos ЗА, express in surd form:
(i) sin 18°; (ii)cosl8°; (iii) sin 54°; (iv) tan 108°.
3. Prove that the following identities hold for all angles A, B for which the
expressions appearing are defined:
(i) sin (A + В) sin (A - Β) ξ cos2 В - cos2 A;
(ii) cos (A - B) - sin (A + Β) ξ (cos A - sin A) (cos В - sin B);
..... cos (Л + B) + cos (Л - B)
{П1) sin(A + B)-sb(A-B)=COtB;
(iv) sin04+.B)sin Л ξ cos A-cos (A + В) cos В;
, л „. sec Λ sec Л cosec Л cosec В
(ν) sec (Λ + В) = ;
cosec A cosec В—sec A sec Л
(vi) tan Z4(cot A -tan A) = 2;
(vii) cos 4Λ ξ 4(cos4 Л + sin4 A) — 3;
(viii) cot A -cot 2Λ = cosec 2Λ;
sin A + sin В
cos Λ + cos В
sin Λ + sin 2.
cos Λ +cos 2Λ + cos ЗА ~
4. If А, В, С are the angles of a triangle, prove the following identities:
(i) sin Λ + sinjB + sin С = 4cos — cos — cos —;
(ii) cos2v4 + cos2fi+cos2C+4coSv4cos.Bcos C+l ξ 0;
224

2] SUM AND PRODUCT FORMULAE
(iii) cos2 A +cos2 В + cos2 С ξ 1 - 2 cos A cos Л cos C;
(iv) tan Λ + tan В + tan С = tan Л tan Л tan С (provided that the triangle is
not right-angled).
, giving values between 0° and 360° in-
(i) 2 sin χ = sin (χ+ 45°); (ii) 2 cos χ = cos (x + 60°);
(iii) sin(30° + x) = cos (15° -x);
(iv) cos (x- 30°) + sin (x- 20°) = cos x.
6. Solve the following equations for x, giving values of χ between 0° and 360°
inclusive:
(i) cosx + sinx + V2 = 0; (ii) sinx + V3 cosx-1 = 0;
(iii) 3 sin x-4 cos x + 2 = 0; (iv) 3 sin x + 2 cos x-3 = 0.
7. Solve the following equations, giving all solutions between 0° and 360°:
(i) sin 2x + sin2 χ = 0;
(ii) cos2x—3cosx = 4;
(iii) cosx + cos2x + cos3x = 0;
(iv) cos x— sin χ = cos 2x.
8. (i) Prove that, when cos φ = — \, the value of the expression
sin θ + sin {Θ + ф) + sin {θ + 2ф)
is zero, whatever the value of Θ.
(ii) Prove that if
ιη(χ + α),
= \ sin a,
stating what restrictions you impose on the value of sin α for a solution to be
possible. Hence find the values of χ between 0° and 360° which satisfy the equation
5 tan χ = tan (x + 30°). (O & C)
9. Prove the identities:
... sec2fl + 2tanfl_
(0 (cosTTiin^r2 = SCC °;
(ii) 4 cos θ cos (Θ +120°) cos (0 -120°) ξ cos 30. (O & C)
10. Prove that cos [(n + 2)6} = 2 cos θ cos [(л +1) 0] - cos И0.
Hence express cos 30 and cos 40 in terms of cos 0, and prove that
cos 50 ξ 16 cos6 0-20 cos3 0 + 5 cos 0. (O & Q
,, „ , sin 50+ 2 sin 30 +sin 0
11. Prove that ъ ^ = cot 0
cos 0—cos 50

FURTHER TRIGONOMETRY [12
and hence, using the formula sin 30 = 3 sin 0 — 4 sin3 0, show that cos 36° is
a root of the equation 8χ*_8χ2 + χ+ί = 0. (0&C)
and, without using tables, deduce that
tan22i° = V2-1
(ii) Prove that cos 30 = 4cos30 — 3 cos0 and deduce that
cos3 θ + cos3 (Θ + 60°) + cos3 (0 +120°) = f cos (0 + 60°) + i cos 30. (O & C)
13. (i) If a = sin 0+ cos ф and b = cos 0+ sin 0, prove that
cos (θ-φ) = 2αΖ>/(α2 + Ζ>2)
and hence, or otherwise, find tan (θ—φ) in terms of a and b.
(ii) Express 11 sin2 χ + 12 sin χ cos χ +6 cos2 χ in the form a + b sin (2x-0)
where a, 6 and 0 are constants to be determined.
Hence prove that
2 =g 11 sin2 x+ 12 sin χ cos x + 6 cos2 χ =S 15. (Cambridge)
14. Prove that sec χ + tan χ = tan(i?r+ix)
and express in a similar way (i.e. as the tangent of an angle)
(i) sec χ -tan χ; (ii) cosecx-cot x. (Cambridge)
15. Prove
2 sin 0(cos 2Θ + cos 40+cos 60) = sin 70 - sin 0.
Deduce that cos ^π + cos fπ + cos fw = — \.
Show also that cos ^π+cos ^π + cos %π = + \. (Cambridge)
16. Express the function cos χ+ 2 sin χ in the form R sin (x + a) where it is
positive and 0° < α < 360°. State the values of R and a. Hence, or otherwise,
find the values of χ in the range 0° to 360° inclusive which satisfy the equations
(i) 2cosx + 4sinx= 1; (ii) cos x(cos χ + 2 sin x) = 1. (Cambridge)
17. If и - cos 0+ sin 0, ν = cos 0- sin 0, prove that
cos 20 = uv, sin 20 = u2-1, и2 +1>2 = 2.
Prove that, if 0 is a root lying between —180° and 180° of the equation
cos20+asin20 = cos0-sin0,
then it is either 0° or 90° or it is a root of the equation
(a + l)cos0+(a-l)sin0 + a = 0.
Solve the equation
cos 20 + 7 sin 20 = cos 0 - sin 0
for values of 0 between -180° and 180°. (O & C)
226

2] SUM AND PRODUCT FORMULAE
18. Find all the pairs of values (θ, ψ) lying between 0 and 2π that satisfy the
equations cos θ + cos φ = cos ^
sin θ + sin φ = sin in. (O & C)
19. Prove, by induction or otherwise, that
cosa+cos (a+/?) + cos (μ + 2β) + ... +cos [oc+(n- 1) β]
= cos [α+Кя-1) β] sin £л/в cosec \β,
provided that β is not a multiple of 2π.
A regular polygon has η sides of length a; the vertices of the polygon are
Vu V2,..., Vn. Show that
(V1V2)" + (V1V3y+... + (V1 VnY = ina*cosec2 (π/ri). (Cambridge)
3. THE SOLUTION OF TRIANGLES
One of the most important applications of elementary trigonometry is to
the solution of triangles; that is, the determination of the remaining sides
and angles of a triangle some of whose sides and/or angles are given. For
consistency, we employ the following notation: ABC is a triangle with sides
ВС = а, С A = b, AB = с and the radius of the circumcircle of Δ ABC is R.
A triangle may be solved uniquely in the following three cases:
(i) given the three sides;
(ii) given two sides and the included angle;
(iii) given two angles and a side.
A triangle may also be solved (but not necessarily uniquely) in the
following case:
(iv) given two sides and a non-included angle.
The Cosine and Sine Rules of elementary trigonometry are employed in
the solution, the Cosine Rule in cases (i) and (ii), the Sine Rule in cases (iii)
and (iv). Before illustrating their use, we give the proofs for the benefit of
the reader unfamiliar with them.
The Cosine Rule.
In the Δ ABC, a2
b2
c2
227
Fig. 12.6
b2 + c2~2bc cos A,
c2 + a2~2cacosB,
a2 + b2~2abcos С

FURTHER TRIGONOMETRY
[12
Proof. Take A as origin and let the position vectors of В, С be r1; r2
respectively. (Thus \тг\ = с, |г2| = b and 1^ —r2| = a.) We have
ki-r2|2 = (^-г^.^-Гг)
= Га.Гг + г^^-г^.Гг
= #4 с2-26c cos A.
The other two forms are proved similarly, using a new origin.
Ex. 22. Prove the vector identity
2Γι.(Γ2-Γι) = Γϋ.Γϋ-Γχ.Γι-Ο-ϋ-Γύ.Ο·,!-!·!)
and show that, with the notation used above, this reduces to the second form of
the Cosine Rule.
The Sine Rule.
In the AABC a = b = с =
sin A sin В sin С
Proof. Draw the diameter BD.
\iLA is acute (Figure 12.7(i)), LBDC = LA.
If LA is obtuse (Figure 12.7 (ii)), LBDC = 180°- Ζ A.
In both cases
^- = 2*
sin A
and the result follows by symmetry.
(i) (ϋ)
Fig. 12.7
Ex. 23. Show that the area, Δ, of the Δ ABC is given by the formula
Δ = \bc sin A.
Write down the two similar expressions for Δ and deduce the Sine Rule in the
form
a _ b _ с
sin A sin В sin С
Example 10. In Δ ABC, ВС = 6 cm, CA = 4 cm, AB = 5 cm. Find the
angles of the triangle.
228

3] SOLUTION OF TRIANGLES
Clearly we must use the Cosine Rule to begin with. We select the angle B,
since this is the smallest angle, and the tables of cosines are marginally
more accurate for smaller than larger angles.
16 = 25 + 36-60 cos LB,
cos LB = 0-75,
LB = 41° 24'.
By the Sine Rule
sin A
~6~~
sin LA =
LA =
sin 46° 24'
2
= 0-9919(5),
= 82° 42'.
By the angle sum property of a triangle
LC-
= 55° 54'.
Example 11. Discuss the solution of a triangle in which А, а, с are given.
(This is often called the ambiguous case, since it is sometimes possible
to draw two triangles.)
в В
A X A Cx X C2
(0 Oi)
Fig. 12.8
Construct the point X such that LAXB = 90°, LB AX = LA.
if c sin A
с sin A > a, sin С = > 1,
a
and no triangle can be drawn (Figure 12.8(i)).
If с sin A = a, LC = 90° a unique right-angled triangle can be drawn.
If с sin A < a < c, sin С = с sin A/a < 1 and two angles С may be
found, one acute and one obtuse (see Figure 12.8 (ii)); two triangles can be
drawn.
Finally, if a > c, then A > С and the obtuse value for С must be
rejected (see Figure 12.8(ii) again).
*Ex. 24. Interpret the four congruency conditions for triangles in terms of the
solution of triangles by trigonometry.

FURTHER TRIGONOMETRY [12
4. SOLUTION OF PROBLEMS, PARTICULARLY
IN THREE DIMENSIONS
We recall that the angle between two planes is equal to the angle between
the normals to the two planes and that the angle between a line and a plane
is equal to the angle between the line and its projection in the plane. It is
important to remember, too, that bearings are always given in a horizontal
plane.
Many problems are best done by setting up coordinate systems and
applying the methods of Chapters 2 and 11. The reader must learn to
develop a flexible attitude towards problem solving and be prepared to try
several techniques in a search for the simplest approach.
Example 12. A hillside faces due north and is inclined at 20° to the horizontal.
A path up the hill has a bearing of 120°. Find the angle the path makes with
the horizontal.
Fig. 12.9
ABCD is a vertical rectangle; ADEFa horizontal rectangle. The hillside
is represented by BCEF; AF points due north, FE due east. FC represents
the path.
We have LAFB = 20°, LAFD = 60°, and we have to calculate LDFC.
Let LDFC = θ, ΑΒ = h units.
In AAFB,
In AAFD,
230
F D
Fig. 12.10
AF = h cot 20°.
FD = AF sec 60° = h cot 20° sec 60°.

41 SOLUTION OF PROBLEMS
In ADFC, tan θ = hjDF = tan 20° cos 60°
= 0-1820,
θ = 10° 19'.
Example 13. Prove the theorem of Apollonius that, in any triangle ABC with
median AM, Авг + AC2 = 2AAP + IBM2.
An aircraft flying on a constant course and at constant height with speed
V is observed from a station on the ground at times 0,t,2t to have elevations
α, β, γ respectively. Prove that
V = AV(cot2 a-2 cot2 /? + cot2 γ)Ιφ.
If the bearings of the aircraft from the station at times 0 and 2t are θχ
and θ2 respectively, determine the course of the aircraft.
В Μ Α
Fig. 12.11 Fig. 12.12
B' VI Μ' VI A1
Fig. 12.13
Let LAMB = θ (Figure 12.11).
In AAMB, AB2 = ^M2+5M2-2^M.5Mcos Θ.
In AAMC, AC2 = AM2 + MC2-2AM.MC cos (18O°-0).
But BM = MC and cos θ = -cos (180°-0) whence the theorem of
Apollonius, by addition.
AB represents the course of the aircraft, Α, Μ, Β being its position at
times 0, t, It. The projections of Α, Μ, Β on the ground are Α', Μ', Β',
231

FURTHER TRIGONOMETRY
and О is the observation point. We are given that AM — MB =
that ΔΑΟΑ' = <x, ΔΜΟΜ' = β, ΔΒΟΒ' = у.
In ΔΟΑΆ, OA' = h cot a.
Similarly OB' = h cot у; ОМ' = /г cot /?.
In Δ A O'B', h2 cot2 α + /г2 cot2 у = 2/г2 cot2 β + 2¥*ί2
by Apollonius's Theorem, whence
F = /zV(cot2 a-2 cot2 /? + cot2 у)/?л/2.
Furthermore, if 0 is the bearing of the course of the aircraft
LA'OB' = вх-вг; LB'A'O = φ~θ1
and we have, by applying the Sine Rule in ΔΟΑ'Β'
2Vt _ h cot у
sin(^-^2) ~ sin (0-0)'
/г cot у sin (#! — 02)
sin (0-00 =
A,/2V(cot2 a-2 cot2 /? + cot2 y)'
cotysin^!-^) Ί
,V(2 cot2 a - 4 cot2 /? + 2 cot2 y) J
■D
Exercise 12(b)
1. Two points Л and С on the bank of a straight river are 120 metres apart.
It is observed that a point A on the opposite bank is such that the angle ABC
is 72° 15' and the angle ACB is 38° 30'. Find the width of the river, correct to the
nearest metre. (O & С Ό')
2. In AABC, s = Ца + Ь+с); deduce from the Cosine Rule that
If the area of the triangle is Δ, use this result to prove
(i) sin I = Ур'У"0] ; GO Δ = V№"a) (s- b) (s- c)].
3. In the AABC, a = 14, b = 15, с = 13. Calculate
(i) the area of the triangle;
(ii) sin A;
(iii) the radius of the circumcircle. (O & С Ό')
4. Prove that,
0)
(ii)
232
til
. A~
tan —
in any triangle ABC,
sin A — sin В
sin A + sin B'
В a-b С
- = a^bCOt2-

41 SOLUTION OF PROBLEMS
In a triangle ABC, a = 13-41 cm, b = 9-63 cm and LC = 34°. Find the size
of the angles LA and LB.
If in a triangle a = 5, b = 4 and cos (A — B) = ii, prove that cos С = i
andthatc=6. (O & C)
5. A straight river is 80 m wide. A man on one bank observed that the angle of
elevation of the top of a tree directly opposite him is 18° 20'. He walks 60 m
along the bank. Find the angle of elevation of the top of the tree from his new
position. (O&C'O')
6. In a triangle ABC, a = 20 m, b = 28 m, с = 32 m. Prove that LB = 60°.
From the points А, В, С which are on level ground, the top of a flagstaff has
the same angle of elevation, 30°. Calculate the height of the flagstaff.
(O&C'O')
7. A snow-slope is a plane inclined to the horizontal at an angle of a. A man on
skis traverses this slope in a straight line which makes an angle β with the
horizontal. Show that the angle θ which his path makes with the line of greatest
slope of the plane is given by the equation
cos θ = sin β cosec a.
If a = 25° and β = 20°, calculate the size of the an gle Θ. (O&C'O')
8. Two adjacent sides of a roof, whose horizontal bases meet at right-angles,
slope at 30° and 45° to the horizontal. At what angle do the roofs intersect?
9. A right pyramid, vertex O, stands on a square horizontal base ABCD;
AB = 2a and the height of the pyramid is h. Express the sines of the following
angles in terms of A and a:
(i) the inclination of OB to the horizontal;
(ii) the inclination of a slant face to the horizontal;
(iii) the angle between the faces OAB, ODC;
(iv) one-half of the angle between the faces OAB, OBC. (O & C)
10. ABCD is a regular tetrahedron and Mis the mid-point of the edge CD. Find
the angle between the plane ABC and
(i) the plane ABM; (ii) the line AM.
11. From the top of a cliff the angle of depression of a ship, steaming on a
constant course at 12 km per hour, is 15° and its true bearing is 75° north of west.
Two minutes later, the angle of depression is 12° and the true bearing is due west.
Find
(i) the height of the cliff (to the nearest metre);
(ii) the ship's course (to the nearest half-degree).
12. A and В are two points on one bank of a straight stretch of a river, Ρ is a
chimney exactly opposite A and 20 m from the other bank. The angles of
elevation of the top of the chimney from A and В are α and β respectively. Calculate
the width of the river to the nearest metre, given that
a = 45°, β = 30° and AB = 200 m.
233

FURTHER TRIGONOMETRY [12
Prove that, if the angle of elevation of the top of the chimney from a point C,
midway between A and B, is γ, then
4 cot2 7 = 3 cot2 a+cot2 β
whatever the height of the chimney, the width of the river, and the distance
AB may be. (O & Q
13. In the quadrilateral ABCD, AB = 13 cm, ВС = 20 cm, CD = 48 cm,
LBCD is 90° and ABAC = ADBC. Without using tables
(i) prove that cos ABAC = ts;
(ii) prove that cos LACB = f;
(iii) find the area of the quadrilateral by adding the areas of the triangles
ABC and ACD. (Cambridge)
14. A right pyramid has vertex V and rectangular base ABCD. AB = 4 cm,
ВС = 6 cm and the height of the pyramid is 8 cm. Find
(i) the angle a sloping edge makes with the base;
(ii) the angle the face VAB makes with the base;
(iii) the angle between two adjacent sloping faces.
15. From a point Ρ in a horizontal plane a man observes the summit S of a
mountain to be due north at an elevation Θ. When the man has walked a
distance la on a bearing α east of north to a point Q in the horizontal plane, he
observes that the elevation of S from Q is again Θ. If h is the height of S above
the horizontal plane containing Ρ and Q, show that h = a tan θ sec <x. When
the man has walked a further distance a in the same plane and in the same
direction to a point R he observes that the elevations of S from R is φ. Show that
cot2 φ = (3cos2a+l)cot20
and that the distance RS is
a(sec2 α sec2 θ + 3)}. (Cambridge)
16. An observer situated at a point О in a horizontal plane observes two other
points Ρ and Q. The point Ρ is in the horizontal plane containing О on a bearing
α west of north from O. The point Q is situated due north of О at an angle of
elevation β as observed from O. If OP = OQ = r, show that the length / of
the straight line PQ is given by
/2 = 2r2(l-cosacos/?).
Hence, or otherwise, show that cos LPOQ = cos α cos β.
If α = 60°, β = 30° find the length of the arc PQ of the circle which passes
through Ρ and Q and has its centre at O. Give the answer in terms of r correct
to two decimal places. (Cambridge)
17. ABC is an equilateral triangle of side 1 m marked out on level ground.
Three vertical posts are driven in at the vertices; AP is of height a m; ВQ is of
height b m; CR is of height с m (a > b; с > b). The line PQ meets the ground at
U; the line RQ meets the ground at V. Prove that
BU = JL·, BV= *

41 SOLUTION OF PROBLEMS
If / = 2, a = 6, b = 3, с = 4, calculate the length of the perpendicular from
Л to the line UVand hence find the inclination of the planePQR to the horizontal.
(O&C)
18. A tower stands on a level plane. The inclinations of its top from three points
А, В, С in the plane are oc, β, γ. The points А, В, С are in a line which does not
go through the foot of the tower and AB = p, ВС = q. Prove that the height h
of the tower is given by
/z2[p(cot2 a-cot2/0)+9(cot2 α-cot2β)] = pq(p+q). (O & C)
19. In the tetrahedron О ABC, ВС = а, СА = b and AB = c;
LBOC = LCOA = LAOB = 90°.
Find the angle between the planes OAB and ABC.
Miscellaneous Exercise 12
1. Prove that arctan | + arctan { = }π and that
arctan i + arctan £ + arctan | = \π.
2. Solve the equations:
(i) arctan i+ arctan χ = arctan f;
(ii) 2 arcsinOc-x) = 3 arccos (x+y) = π.
3. Two circles with centres О and С meet in Ρ and Q. The radii of the circles are
a and b and the angle CPO is a. Prove that the angle between the common
tangents of the circles is Θ, where
(a-b)2 cot2 iff = 4ab sin2 ice. (O & C)
4. Find the set of values of χ which satisfy the inequality
2 sin x+sin 2x > 0. (SMP)
5. Under what circumstances is it true that
arctan χ + arctan ν = arctan; ?
1-xy
Find values of x, у for which the relationship above is (i) true, (ii) untrue.
6. Prove that, in any triangle ABC,
_a 6__ с
sin A sin В sin С '
In the triangle ABC the angle Л is a right-angle and О is a point inside the
triangle at which all the sides subtend the angle 120°. If θ is the angle CBO, prove
that .„
e_ c+*J3
If the angle С = 30°, show that CO = 2AO. (O & C)
7. Show that 4 arctan |— arctan гт^ = ίπ (a result commonly known as
Machiris Formula).

FURTHER TRIGONOMETRY [12
8. Find all pairs of angles x, у such that 0 < χ < π, Ο < у < π which satisfy
the simultaneous equations
sin χ sin у = i(V3— 1); cos χ cos у = i(V3 + 1).
9. A man stands facing the rectangular front of a building and is in the same
plane as one of the ends of the building. The elevation of the nearer top corner
A is α and of the further top corner В is β. The man walks towards the vertical
edge through В until the elevation of В is also a, and he finds that he has walked
a distance a. Show that the height of the face of the building is
a sin a sin /3/sin (α—β)·
Show also that its length is afsin (pc+fi)/sin (а-Д)]*. (O & Q
10. Find the maximum and minimum values of the expression α sin (9+6 cos Θ.
By making the substitution t — tan \θ, deduce the condition for the existence
of real roots of the quadratic equation
(b+c)t*-2at + (c-b) = 0.
11. An aircraft is observed flying on a constant course γ east of north at a
constant height. When its true bearing is θ west of north, the angle of elevation
is a, and when its true bearing is φ east of north, the angle of elevation is β. Prove
that γ is given by . , ,.„*/,
_ sin0 tan «+sing tan/g
~ cos φ tan a-cos θ tan β'
Prove that, if θ = φ, the angle of elevation δ, when the true bearing is north,
is given by Ш1 s = i(tan а + шд sec A (O & Q
3 tan x — tan3 χ
12. Prove that tan 3x
1-3 tan2л:
provided both sides of the identity are defined.
Deduce that the three roots of the cubic equation
ί3-3ί2-3ί+1 = 0
are tan -fen, tan fen, tan |π.
13. Sketch the graph of the function defined by
f(x) = |sinx + cos*|.
For what values of χ in the interval 0 =g χ =g π does |sin x+ cos
14. Find the maximum and minimum values of the expression
sinx(sinx+cosx).
15. Eliminate χ and у between the equations
sinx+sin^ = a,
cos x+cos у = b,
x+y = a,.
236

41 MISCELLANEOUS EXERCISE 12
16. The wave-train well away from a ship is modelled by the equation
z = asin[(x+.y-tf)/p]
where ζ is the height of the sea's surface above the mean horizontal plane in
which (x, y) are Cartesian coordinates, a is the maximum height of the waves, с
is the fixed speed, and ρ a fixed length. Sketch a diagram showing an airman's
view of the waves, indicating the lines of the crests and troughs and the direction
in which the waves are travelling; and prove that the wave velocity and distance
between successive waves are сД/2 and ττρ^Ι respectively.
The wave train of another ship, given by
ζ = α sin [(χ—у— ct)/p]
is superimposed on the other one. Prove that on the lines у = (N+ i) πρ, where
N is an integer, the sea is undisturbed. (SMP)
17. Find the range or ranges of values of с such that the simultaneous equa-
sec<9 + cosec$S = с
are satisfied by real values of θ and φ.
Obtain the general solutions of these equations when с = б£.
237

13. Matrices 1
1. INTRODUCTION
Any pair of simultaneous equations in two unknowns, χ and y,
{ax + by = c,
dx + ey=f
is completely specified if we know
(i) the coefficients on the left-hand side, which we may write as the
rectangular array
СЭ
and (ii) the two numbers on the right-hand side, which we may write as
the rectangular array
Ex. 1. Solve the simultaneous equations whose coefficients are given by the
rectangular array .
and whose right-hand sides are given by the rectangular array
The answer is the pair of values 4 (for x) and - 3 (for y), which can be expressed
as an array
l-з)'
as we saw in Chapter 2.
Such rectangular arrays of numbers are called matrices; in the example
above they exhibit the known quantities in a pair of simultaneous
equations but they may be used in other contexts too, to display information.
Suppose 23 boys take examinations in Mathematics (M), English (E)
and French (F). In each examination there are five grades A, B, C, D, E.
Then the fate of the boys may be summarized in the following 3x5 matrix
238

(read 'three by five matrix'; that is,
columns): . „
INTRODUCTION
matrix with three rows and five
С D Ε
7 6 3\
8 6 2.
8 7 0/
Ex. 2. In the example just quoted, describe what information is represented by
each of the following :
(i) the 3 χ 1 matrix (or column-vector)
(iii) the 1 x 5 matrix (or row-vector)
(3587 0).
More generally, any rectangular array of m rows and η columns of ele-
ments is called an r
nxn matrix:
Mi <ha
/ <h\ <ha
^13
<hb
... aln
... din
\aml am2 атЪ
■ amn
We may, if we so wish, denote the whole matrix by a single capital letter A
(printed in bold-face type; in script, write a capital A and underline). The
element а^ in the z'th row and yth column, is called the i, jth element or
entry of the matrix. A real matrix is a matrix all of whose elements are
real numbers.
2. LINEAR TRANSFORMATIONS
If we are given a pair of axes in a plane, any point Ρ may be located by
its coordinates (x, y), so that OP = xi+y}. Suppose we now associate
with each point P(x, y) a unique point P'(x', y') such that
For example, the point P(l, 2) gives rise to P'(5, -1); the point Q{-2, 3)
gives rise to g'(4, - 5). Such an association of points is a mapping, or func-

MATRICES 1 [13
tion, of the set of points in the plane into itself and is often referred to as a
transformation of the plane into itself. We may specify our mapping by the
2x2 matrix
(! -Э
which we call the matrix of the given transformation. For the present,
such a matrix is to be regarded simply as an inert array of coefficients
defining a transformation. In Section 3 we shall consider combinations of
transformations and the resulting matrices; in Section 4, rules for combining
matrices will be developed and the inert arrays which we have at present
will come to life. Finally, in Section 5 matrices, with their new-found
vitality, will be used to illuminate the concept of a geometrical
transformation.
We may also regard the transformation as mapping the vector OP into
the vector OF and the vector OQ into the vector OQ'. If we call the
transformation Γ, then we write
Γ(ΟΡ) = OP'
and T(OQ) = OQ'.
Notice carefully that Γ(ΟΡ) is a vector: the position vector ofP'. Г(ОР) is
called the image of OP; P' is the image of Ρ and may be written T(P).
The transformation defined above has an important property: it maps
the vector OP+OQ into the vector sum of Γ(ΟΡ) and T(OQ). Let us
demonstrate this property first with the points Ρ and Q above :
OP = i + 2j, OQ=-2i + 3j.
Let OR = OP+OQ
so that R is the point (-1, 5) and OR = -i + 5j. By the definition of T, R'
has coordinates (-1 +2.5, -1-5), i.e. (9, -6) and so
OR' = 9i-6j.
But OP' + OQ' = (5i-j) + (4i-5j)
= (9i-6j)
= OR'.
Now let Ρ be any point (Als kj, Q any point (/z2, k2) and let R^, k3) be
the point such that 0R = OP + OQ,
or h3i + k3\ = (A1 + A2)i + (fc1 + A:2)j.

2] LINEAR TRANSFORMATIONS
then T(OR) = T(h3i + k3j)
= (h3 + 2k3) i + {h3 - k3) j, by the definition of T,
= (h1 + h2+2k1 + 2k2)i + (h1+h2-k1-k2)i
= Pi + 2kd i + (Ai - kd j] + P2 + 2*2) i + (A2 - *2) j]
= Γ(ΟΡ) + T(OQ), again using the definition of T.
The result we have just proved for the transformation Τ can be
generalized. If S(/z4, fc4) is the point whose position vector is given by
OS = AOP+/<OQ,
where Α, μ are any numbers, so that /z4 = λ^+μη^ kt = λ^+μΐ^, then
r(os> = T{Ki+Ki)
= (/z4 + 2jfc4) i + (A4 - Jt4) j, by the definition of T,
= (ЛЙ!+/</г2 + 2A^ + 2/4*2) i + (Λ/zj +/i/z2 - A/q -/<fc2) j
= λ[(η1 + 2^) i + ihi-кг) j]+/*[(A2 + 2*2) i+ (*,-**) j]
= λΤ(ΟΡ)+μΤ(00).
The property which we have just demonstrated for the transformation
holds for a wide class of transformations, called linear transformations.
A linear transformation Γ is a transformation such that the image of the
vector Ax +/*y is the vector sum of A times the image of χ and μ times the
image of y. In symbols, Γ is a linear transformation if, for any vectors x, у
and any numbers Λ,„ τ{λχ+μ^ = λτ{χ)+μΤ^
*Ex. 3. By writing χ in the form x+ 0, show that, for any linear transformation
T, ДО) = 0.
An example of a mapping, T, that is
particularly easy to visualize geometrically is that in
which the image P' of Ρ is obtained by a half-
turn about the origin (see Figure 13.1). The
coordinates of P'(x', /) in terms of P(x,y) are given
by
ПАУ)
*> Fig. 13.1
and thus the matrix of this transformation is

MATRICES 1
[13
To see that Γ is a linear transformation, consider the points P(//1; kj
and QQh, k2) and suppose that
OR = AOP+/<OQ,
where OR = h3 i + k3 j = (A^ + /i/z2) i + (Akx+/tfc2) j,
then T(OR) = lX.hsi+k3j)
= -h3i-k3\, by the definition of T,
= - (Λ/zj+A) i - (A&!+M2) j
= A(-A1i-fc1j)+H-A2i-/c2j)
= A7T[OP)+/<r(OQ).
Thus, Γ is a linear transformation.
*Ex. 4. Write down the matrix for the transformation
x' = kx,
у = ky.
This transformation is called the enlargement transformation, or dilatation
transformation. Draw a diagram and consider the effect on the points (2, 3),
(-1,2) when к is (i) 2, (ii) \. Can you suggest a reason for the name?
*Ex. 5. Prove that the enlargement transformation is linear.
*Ex. 6. The function/:R -* R defined by f{x) = ax+b is a linear function (that
is, its graph is a straight line). Show that, regarded as a mapping of points of the
χ axis into points of the χ axis, it is not a linear transformation.
Ex. 7. The function /: R -+ R defined by f(x) = x2 maps the χ axis into itself.
Prove that/is neither a linear function nor a linear transformation.
Example 1. The transformation Τ maps the vector OP into the vector OP'
where
|OP| = |OP'| and LPOP' = <x,
a fixed angle measured in the anticlockwise sense. Find the matrix for Τ
and verify that Τ is a linear transformation.
Suppose OP makes an angle β with the χ axis
(see Figure 13.2). Writing |OP| = |OP'| = r we
have
x' = r cos (α +β) = (r cos β) cos α - (r sin β) sin a,
y' = r sin (ρί + β) = {r cos/?) sin a + (r sin/?) cos α С ^
and the transformation is given by Fig. 13.2
x' = л: cos α-y since,
y' = χ sin α + j cos a.
242
■ϊ(χ,ν)

2] LINEAR TRANSFORMATIONS
The matrix of the transformation is therefore
(cos α — sin a\
sin α cos a/
To prove that Τ is linear, consider P(/z1; kj, Q(h2, k2), R(h3, k3) where
OR = AOP+/<OQ,
Г((Ж) = T{h3\ + k3\)
= Qi3 cos <x-k3 sin a) i + (h3 sin a + ka cos a) j, by the equations
for Τ defined above,
= [(λ/jj +/</z2) cos α _ (Mi +M2) sm °0 >
+ [(Mi +Мг) sin α + (Afcj +/<fc2) cos a] j
= A [(/zj cos a - fcj sin a) i + (/zj sin a + кг cos a) j]
+/*[(/z2 cos a - fc2 sin a) i + (h2 sin a + k2 cos a) j]
= λΤ(ρΡ)+μΤ(00)
and the result is complete.
3. LINEAR TRANSFORMATIONS
AND THEIR MATRICES
Suppose we have a linear transformation Τ whose matrix A is given by
Since i is the position vector of the point 7(1, 0) and j is the position vector
of the point /(0, 1), we have
r(i) = fli + 4 T(j) = bi+dj.
Recalling that vectors may be written in column form, this result shows us
that the first column of A, regarded as a vector, is the position vector of the
image of I and the second column of A, regarded as a vector, is the position
vector of /.
Ex. 8. Prove the converse of the result above that, if
T(i) = ai+cj and T(j) = bi + dj,
then the matrix of Γ is A where
243

MATRICES 1 [13
(The results above show us that we are entitled to talk of A as the
matrix of a given linear transformation Τ only if we take i, j as our base
vectors: to put it another way, if we change our coordinate system, we
change the equations which define the linear transformation. We shall
continue in this chapter, unless specifically stated otherwise, to assume that
our base vectors are i and j.)
Since any vector χ may be expressed in the form
χ = Ai+/ij,
knowledge of the effect of Γ upon i and j enables us to predict the effect of
Τ upon x; for _, ч _,v , „
T(x) = Τ(λι+μι)
= λΤφ+μΤφ.
Now suppose we have two linear transformations S, Τ whose matrices
are respectively A and W where
If we define a new transformation U, called the sum of S and Τ by the
equation ^ = 5(χ) + Γ(χ)
it may be shown that
(i) U is a linear transformation;
(ii) U has matrix , , , , .
v ' /a + w b+x\
[c+y d+z)·
For we have
(i) i/(Ax + /ty) = S(Ax+/ty) + Τ(λ\ + /ty), by definition of U,
= λΞ(\)+μΞ(γ) + λΤ(χ)+μΤ(γ), since S, Τ are linear,
= А(5(х) + Г(х))+К5(у) + Г(у))
= λυ(χ)+μυ(γ), by definition of U,
and U is a linear transformation.
Again, (ii) to find the matrix of U, we have simply to find i/(i) and C/(j).
U® = S(\) + T®, by definition of U,
= (ai + c]) + (wi+y'})
= (a+w)'\ + (c+y) j.
Similarly Щ) = φ + χ) i + {d + z) j
and the result (ii) follows.
244

MATRICES OF LINEAR TRANSFORMATIONS
and Ρ is the point (1, 1) sketch in a diagram the effects of S, Τ and U upon the
point P.
*Ex. 10. The transformation L is defined by
L(x) = cT(x).
(L is sometimes written as cT.) Prove that L is a linear transformation and that
its matrix is
*Ex. 11. The transformation Mis defined by
M(x) = S[T(x)],
that is, χ is first transformed under Τ into 7Щ, and Γ(χ) is then transformed
under S into M(x). (M is called the product of S and Τ and may be written as
ST.) Prove that Mis a linear transformation and that its matrix is
(aw+by ax+bz\
cw + dy cx+dzj'
+ dy
All the transformations mentioned so far have transformed points in a
plane into points in the same plane. It is not difficult to generalize the
concept of transformations of points in space to points in space. For example,
the equations , _
define a transformation whose matrix is
Ί 0 0\
0 1 0 .
*0 0-1/
Ex. 12. Give a geometrical interpretation for the transformation Τ whose
matrix is
/1 0 0\
0 1 0.
\0 0 -1/
Exercise 13 {a)
1. The transformation Τ is defined by the equations

MATRICES 1
[13
What is the image of the point P(-1, 3)? What point Q gives rise to the point
G'(3,5)?
2. A transformation is represented by the matrix
(A :S)·
Find the image of the point (-1, 2) and also the point whose image is (14,9).
What points are invariant under this transformation? (A point is invariant
under a transformation if it maps into itself.)
3. What points are invariant under the transformation whose matrix is
(? :?)'
4. A transformation whose matrix is
(if)
is called a shear. Consider its effect upon the square with vertices (0, 0), (1, 0),
(1,1), (0, 1) and suggest a reason for the name.
5. Prove that the shear transformation defined in Question 4 is linear.
6. Describe the following transformation in geometrical terms:
x' = y,
/ = 0.
Draw a sketch to illustrate the transformation; mark in a number of points and
their images.
Prove that the transformation is linear.
7. Answer the same questions as in Question 6 for the transformation
x' = 2y,
y' = 2x.
8. Sketch a diagram to illustrate the effect of the transformation which has matrix
(ίί)
on the four points P(0, 0), β(1, 0), R(l, 1), 5(0, 1).
Determine the area of the image quadrilateral P'Q'R'S'.
What can you say about the particular transformation in which ad-be = 0?
9. Prove that the translation transformation defined by
*' = x,
y' = y+l
is not a linear transformation.
246

3] MATRICES OF LINEAR TRANSFORMATIONS
10. Find the matrix of the linear transformation which maps the points U, J
whose position vectors are given by
-©·-©
into the points U', J' whose position vectors are given by
■■-G-'-C)·
11. Write down the matrix for the shear, S, given by
x' = x + ky,
У' = У,
and the matrix for the reflection R, given by
*' = x,
V = -У-
Two new transformations Γι and T% are defined as follows. To find Тг(Р),
find R(P) = Q\ say, and then find S(Q'); to find Тг(Р), find S(P) = Q", say,
and then find R(Q"). Prove that 7Ί and T2 represent different transformations.
4. THE ALGEBRA OF MATRICES
The results we proved in the last section for the sums, numerical multiples
and products of linear transformations and their associated matrices
suggest that definitions should be given for the sums, numerical multiples
and products of matrices so that, for example, the sum of two matrices
A and W is the matrix of the sum of the two transformations whose matrices
are respectively A and W. In this section we shall give some definitions for
combining and manipulating matrices which are indeed motivated by the
corresponding rules for linear transformations (proved only in the plane,
but holding more generally). In Section 5 we shall show how the definitions
of this section tie up with our previous work on linear transformations.
We begin by defining equality of matrices. Notice that such a definition
is required: it is intuitively fairly clear that
G ?) - Π Э
are different matrices, but are
/0 2\ ./0 2 0\
(5 1) and (5 1 0)
different?
247

MATRICES 1 [13
(i) Equality of matrices.
Since column vectors are 3 χ 1 matrices (in three dimensions) we must
ensure that our rule for the equality of matrices includes as a special case
the rule for equality of vectors.
Two matrices A and В are equal if and only if
(a) they have the same number of rows and the same number of columns;
(b) the i,jth element of A equals the i, jth element of B, for all possible
values of i and).
Example 2. If
\1
then a = -2, b = 4, с = 2, d = 1, e = 0. The two matrices
/2 a 3\/c -2 3\
\l Ob) \d e 4/'
a /0 2 0\
and (5 1 0)
are not equal.
(ii) Addition of matrices.
Two matrices are said to be conformable for addition if they have the
same number of rows and the same number of columns.
By analogy with the corresponding rule for column vectors, we make the
following definition: if two matrices are conformable for addition, A + B
is the matrix whose i, jth element is the sum of the i, jth elements of A andB.
If A and В are not conformable for addition, A+Bis not defined.
Example 3. If /1-3
A = (4 0 3
m
/1+2 -3 + 2 2-3\ /3 -1 -1\
4 + 1 0-3 3 + 4 = 5 -3 7 1.
\l+3 -1-3 2+1/ \4 -4 3/
•B
then A + С is not defined.
(iii) Multiplication of matrices by numbers.
Again working by analogy with column vectors, if A is any matrix and к
is any number, kA is a matrix, conformable for addition with A, whose i,jth
elements is к times the i, jth element of A.
248

4] THE ALGEBRA OF MATRICES
Example 4. If
ЧЫЬЧВ -2Ч:Л)·
We may now define the subtraction of two matrices A and В which are
conformable for addition by the equation A-B = A + (-l)B.
Example 5. With the matrices A and В of Example 3
/-1 -5 5\
A-B =33 -1 .
\-2 2 1/
(iv) Zero matrices.
A zero matrix is a matrix all of whose elements are zero. Provided no
confusion is likely to arise, a zero matrix may be written 0 but note that
there are many different zero matrices; for example,
/0\ /0 0\ /0 0 0\
[θ/ \0 0/ \0 0 0/
are all zero matrices, and are all different,
(v) Multiplication of matrices.
Two matrices A and В are said to be conformable for the product AB if the
number of columns in A is the same as the number of rows in B. We now
define the product AB of two matrices A and В which are conformable for
the product AB as follows: the i,Jth element of AB is obtained from the ith
row of A and the jth column of В by multiplying together the corresponding
elements and adding. If А, В are not conformable for the product AB, then
AB is not defined.
Example 6. If
/1 3 2\ /-1 1 -1\
A = 5 -1 3 and В = 2 0 2 1,
\l 1 2/ \ 0 3 1/
then
/(1) ( -1) + (3) (2) + (2) (0) (1) (1) + (3) (0) + (2) (3)
AB= (5)(-l) + (-l)(2) + (3)(0) (5)(l) + (-l)(0) + (3)(3)
\(1) (-1) + (1) (2)+(2) (0) (1) (1)+(1) (0) + (2)(3)
(l)(-l) + (3)(2) + (2)(l) \
(5)(-l) + (-l)(2) + (3)(l)
(1)(-1)+(1)(2) + (2)(1) /
/ 5 7 7\
= -7 14 -4 .
\ 1 7 3/
249

MATRICES 1
but CA is not defined.
1 0 then AC
-(■! "3
Ex. 13. If
find
(i) A + B;
Ex. 14. If
find
(i) AB;
*Ex. 15. If
verify that
(i) A + B =
(iii) A(BQ =
Hi
(ii) 2A-
A=|
(ii) BA.
*-G-
B + A;
= (AB)C;
Show also that AB φ ΒΑ.
2
0
2
-3B;
I
t
"?
1
В =
^-δ i-
(iii) AB;
■-G 4
-e ?)- --G
(ii) (A + B) + C =
-3
(iv) BA.
i)
?)·
= A + (B+C);
(iv) A(B+C) = AB+AC.
In Ex. 15 above, certain of the basic laws of algebra have been shown to
hold good for the particular matrices А, В, С With the definitions of
equality, addition and multiplication we have given, the following laws
may be shown to hold good for all matrices А, В, С, provided all the sums
and products are defined:
(i) Matrix addition is
(a) Commutative: A + B = B + A;
(b) Associative: A + (B+C) = (A + B) + C.
(ii) Matrix multiplication is
(a) Associative: A(BC) = (AB) C;
φ) Distributive over addition: {*^ Ζ fc++f^
However, matrix multiplication is non-commutative; indeed, two
matrices А, В may be conformable for the product AB and yet BA is
not defined. This is not to say that matrix multiplication is never
commutative.
&..6.,f a-(J J) and ,-(-· J).
verify that AB = BA.
250

4] THE ALGEBRA OF MATRICES
A unit or identity matrix is a square matrix in which the elements of the
leading diagonal (top left to bottom right) are all unity and every other
element is zero. Thus, the 2x2 identity matrix is
U
while the 3 χ 3 identity matrix is
f. ϊ Λ
\0 0 1/
Provided that no confusion is likely to arise and the correct size (2 χ 2,
3x3, ...) is obvious from the context, an identity matrix is denoted by the
letter I.
Ex. 17. Show that, if A is any 2 χ 2 matrix, В is any 3 χ 2 matrix and I is the 2 χ 2
identity matrix, (i) AI = IA = A; (ii) BI = B.
Ex. 18. Show that, if I is the 3 χ 3 identity matrix and A is any matrix with three
columns, AI = A.
5. LINEAR TRANSFORMATIONS
AND THEIR MATRICES (CONTINUED)
The rules for matrix addition and multiplication given above enable us
to obtain a deeper insight into the machinery of linear transformations.
If Ρ has coordinates (x, y) and its image, P', has coordinates (x\ y') then
the position vectors (in column vector form) of Ρ and P' are respectively
If we have a transformation Ρ to P' defined by the matrix
then x, у, х', у' are related by the equations
x' = ax + by,
У = cx + dy
and these may be written, on using the definition of multiplication of
matrices, in the form of the matrix equation
or, more briefly, p' = Ap. (1)
9 ppM 251

MATRICES 1 [13
Furthermore, A always represents a linear transformation, for
A(Ax+/*y) = AAx+/<Ay.
A second linear transformation p' to p" with matrix В has the form
p" = Bp' (2)
and so, combining (1) and (2),
p" = B(Ap)
= (BA)p.
Thus, if the linear transformation whose matrix is A is followed by the
linear transformation whose matrix is B, the combined effect is a linear
transformation whose matrix is BA.
In three-dimensional space, precisely similar results hold. Indeed, if
p, p' are the position vectors (in column vector form) of the point Ρ and its
image P', and if A is a 3 χ 3 matrix, then the linear transformation defined
by A takes the matrix form p' _ др
which is precisely equation (1), although ρ', Α, ρ have different meanings.
Example 7. The linear transformation Ρ to P' of points in three-dimensional
space is defined by the matrix:
-V, i)
Show that, under this transformation,
(i) any point Ρ is mapped into a point lying in a certain plane;
(ii) all points of the line
x-l _y+l _ z-1
1 1 -2
are mapped into the same point.
(iii) Find the set of points that are mapped into the origin.
(i) The transformation is given by the equations
x' = 2y + z,
y' = 3x+y + 2z,
ζ = Ъх-y + z.
But y'-z = x' and so all points map into the plane x-y + z = 0.
(ii) A general point Ρ of the given line has coordinates (Λ + 1, Λ-1,
-2A+2). Thus, the position vector of the image of Pis given by
ii!) (Ш
and so all points of the given line map onto the point (0, 6, 6).
252

5] LINEAR TRANSFORMATIONS (CONTINUED)
(iii) If the image of the point P(x, y, z) is the origin, then x, y, ζ satisfy
the equations , . „ , „
M (a) 2y+ ζ = 0,
(b) 3x+ y + 2z = 0,
(c) 3x- j> + ζ = 0.
Planes (a) and (ft) meet in the line χ = j> = -\z which clearly lies also in
plane (c). Thus, all points with coordinates of the form (A, A, -2A) map
into the origin.
1. If
evaluate
0) A + B;
2. If
evaluate
(i) A+B;
3. If
evaluate
Exercise 13(b)
*-(-Μί)--(-1-5 2).
(ii) A-B; (iii) 2A-B; (iv) 2A+3B.
1-Х 2 1\ /12 1\
A=3 0 3 and В = 1 1 3 ,
\ 1 2-1/ \2 -1 2/
(ii) 2A-3B; (iii) 3A-B; (iv) 4A-2B.
-М^-И'-П-
(i) A-3B + 2C; (ii) 2A(B-Q; (iii) A2-AB + AC.
Verify that (A + B)2 = A2 + AB + BA+B2; can this be put in the simpler form
A2 + 2AB + B2?
4. А, В, С are all 3 x 3 matrices. Remove brackets in the following expressions:
0) A(B-
5. If
-2C); (ii) (A + B)(A-Q; (iii) (A-B-C) (A + B + C).
-H-e-i »)·
find AB and BA.
6. If
find
(i) AB;
(iv) B2;
7. If
/1 0 2\ /3 1 -2\
A= 0 -1 -1 , Β = 1 -1 -2 ,
\2 11/ \l 1 1/
(ii) BA; (iii) A2;
(v) (2A-B)2; (vi) (A + 2B) (A-B).
/2 1 -1\ /0 2 1\
/ 2 1 -1\ / 0 2 1\
= ( 1 1 2 , B= ( 1 3 0 ,
\-l 2 1/ \-2 -1 2/

MATRICES 1
find
(i) AB; (ii) BA; (iii) A2;
(iv) B2; (v) A2-4B2; (vi) (A-2B) (A+2B).
evaluate
(i) AC; (ii) ВС;
Verify that (A+2B) С = АС+2ВС.
3 4\ /-1 0 1\ /
1 -2 , B= 2 1 0 , C =
2 3/ \ 0 2 3/ \-
-Hi
find
(i) AB; (ii) BA; (iii) ABA; (iv) BAB.
10. A matrix Μ is said to be transposed into the matrix M' if the first row of Μ
becomes the first column of M', the second row of Μ becomes the second column
of M', and so on. Write down the transposes of the matrices:
M:
■&т-Ш-
Calculate the matrix products M'M and TM; show also that (TM)' = M'T".
If the elements of Μ are the Cartesian coordinates of a point P, what
information is provided by the element of M'M?
If the matrix Τ describes a transformation of the point Ρ of three-dimensional
space, interpret geometrically the equation:
(TM)' (TM) = M'M,
and find all appropriate values of a, b and с (SMP)
11. Show that under the linear transformation defined by the matrix
= (2 2 l)
\8 0 -1/
0 -1/
any point Ρ in space is mapped into a certain plane, and find the equation of this
plane.
Show further that all points of the line
x-l y-\ _£=3
1 -5 _ 8
are mapped into a certain point, and find the coordinates of this point.
12. Show that, under the linear transformation defined by the matrix
4πί)'
any point Ρ in space is mapped into a certain plane, and find the equation of this
plane.

5]
LINEAR TRANSFORMATIONS (CONTINUED)
Show also that all points of the line
x-1 _y-2=z-l
1 -1 -1
map into a certain point and find the coordinates of this point.
13. The linear transformation Τ of three-dimensional space into itself maps
the point Ρ into the point F and the point Q into the point Q'. The points Ρ and
Q are distinct. Show that
(i) if P' and Q' are distinct, then all points of the line PQ map into points
of the line P'Q'.
(ii) if P' and Q coincide, then all points of the line PQ map into P'.
14. The linear transformation Τ of the plane into itself has matrix
и: -D-
Show that Τ maps the interior of the triangle 0(0, 0), P(2, 0), g(l, 1) into the
interior of another triangle O'FQ' and find the coordinates of the vertices
O', F, Q'.
Prove also that the image of the centroid of the triangle OPQ is the centroid
of the triangle O'F Q'.
Miscellaneous Exercise 13
map a set of parallel lines into another set of parallel lines when ad φ be?
What happens to the transformation defined by A when ad = bcl Is it still
one-one? (A transformation is said to be one-one if each image F arises from a
unique P.)
2. What are the position vectors of the images of the points whose position
vectors are i, j, к under the linear transformation whose matrix, A, is given by
(a-i. аг д3
δι h из
Ci C2 C3
Describe geometrically the linear transformation whose matrix is
(ο ι o).
\0 0 l/
3. Describe the effect of the linear transformations whose matrices are:
<» (! J), <«> U "'); CD (? ?)■
For each transformation determine the set of vectors which are transformed into
the zero vector.
255

MATRICES 1 [13
4. A is the matrix of the transformation that rotates all points in the plane
through an angle α; Β is the matrix of the transformation that rotates all points
in the plane through an angle β. Calculate BA and simplify the resulting matrix.
What result does this illustrate?
5. Can you find a point Ρ with position vector χ which is mapped into itself by
the tranformation whose matrix is A, where
Can you determine any real values of λ which would enable you to find non-zero
vectors χ such that Ax = λχ? If you can, find both λ and the corresponding vectors.
6. Answer the same questions as in Question 5 for the linear transformations
with matrices
ЧП): «(-3 1
7. The transpose of the matrix A, denoted by A', is defined as the matrix whose
z'th row is the same as the /th column of A. For example, if
чм ?)■-(< J i)·
then / 1 1\ /2 4 3\
*'-(-, 4) -»-(, 2 -5).
If С and D are both 2x2 matrices, prove that (CD)' = D'C.
8. A symmetric matrix A is a matrix such that A' = A. If A and В are two
symmetric matrices, find a further condition that A and В must satisfy to ensure
that AB is a symmetric matrix.
9. A square matrix is said to be diagonal if its only non-zero elements lie on the
leading diagonal (that is, top left to bottom right). For example, the following
two matrices are diagonal:
-?)· 6 i ?)■
Prove that, if Db D2 are two 3x3 diagonal matrices, then
(i) DjDa is a diagonal matrix; (ii) D^ = D^;
(iii) Di = Di.
If A is a 3 χ 3 matrix and D any diagonal 3x3 matrix, what с
about A if AD = DA?
10. A linear transformation is given by the matrix
(111)·

5] MISCELLANEOUS EXERCISES 1
Show that
(i) all points are mapped into points on a certain plane;
(ii) all points on the line
x-1 y-\ z-1
1 ~ 3 ~ -5
are mapped into the same point.
11. Let А, В, С be real 2x2 matrices and write
[A,B] = AB-BA, etc.
Prove that
(i) [A,A] = 0;
(ii) [[А, В], С]+ [[В, С], А]+ [[С, А], В] = 0;
(iii) [A, B] = I => [A, B™] = wB™-1 for all positive integers m.
At each step you should state clearly any properties of matrices which you use.
The trace 7>(A), of a matrix
A = К °A
is defined by 7>(A) =
Яц+approve that:
(iv) 7>(A+B) = 7>(A)+7>(B);
(v) ЖАВ) = 7>(BA);
(vi) 7>(I) = 2.
Deduce that there are no matrices satisfying [A, B] = I. Does this in any way
invalidate the statement in (iii)? (M.E.I.)
12. If Μ denotes the matrix /1 0\
and I denotes the matrix /j q\
\0 1/
prove that Μ2 = 3Μ-2Ι.
Prove further that, if η is any positive integer,
M" = (2я-1) М-2(2"^- 1) I. (M.E.I, adapted)
13. Prove that a linear transformation Τ of three-dimensional space into itself
maps the interior of a tetrahedron into the interior of the image tetrahedron.
Prove further that the image of the centroid of the given tetrahedron is the cen-
troid of the image tetrahedron.
257

ι φ Matrices 2
1. THE INVERSE OF A MATRIX
The reader has seen that a definition of multiplication of matrices may be
formulated which has some of the properties of multiplication of real
numbers. Notably, if the matrices are square and of the same order (say
three, for the sake of argument), then the matrix I, where
/1 0 ov
1=01 0 ,
\o 0 1/
has some of the properties of the number 1.
Ex. 1. The linear transformation Γ maps points of the x-axis according to the
rule ,
Ί\χ) = kx.
Interpret Tin the case where к = 1.
Ex. 2. The linear transformation Tmaps points in a plane into points in the same
plane according to the rule
Г(Х) = Ax where A is a 2 χ 2 matrix.
Interpret Τ in the case where
*-n·
Ex. 3. Interpret the matrix /1 0 0\
1=0 1 Ol
\0 0 1/
as the matrix of a linear transformation.
The question thus naturally arises: 'Is there a matrix analogue to the
reciprocal of a number ?'; that is,' given a square matrix A, does there exist
a matrix В such that BA = I?'
We begin our investigation by actually constructing such a matrix В
for a given 2x2 matrix A. Suppose
Hb3)·
We effect the construction in two stages: we first form a matrix Cj such that
258

1] THE INVERSE OF A MATRIX
the product QA has as its first column ι ι,
that CaCQA) = I. By associativity, it follows that CaQ = B.
Since the element au of A is already 1, we have only to assume that QA
makes the element of the second row and first column of the product zero.
Choose
(J?)·
Q =
Next choose
Q =
Thus if we take
= (_* J); then QA =
(J _J); then C2(QA)
В = C,Q = (_2 1) we have BA =
We write В = A-1 and call A-1 the inverse of A.
(Notice that we have not yet justified the use of the definite article' the':
in fact, В is unique, although the intermediate matrices Cj and C2 are not.
Notice also that we have shown that BA = I but not that AB = I although
this latter equation does indeed hold, as we shall soon show.)
In this example, the choice of Q and C2 was determined by trial and
error. Whilst this is easy for 2x2 matrices, in more complicated cases
we shall need a more general method. This is given by considering the so-
called elementary row operations on a matrix.
We shall consider the 3 χ 3 matrix
-- a2l
w
«22 «23 )
«32
although our results will easily be seen to generalize to the nxn case (and,
in particular, to cover the 2 χ 2 case).
(i) The first elementary row operation: the interchange of two rows of Α..
Consider the matrix
0 0\
0 1 ,
1 0/
«12 «1
EjA = \a31 a32 a3
«22 «23/
and the effect of pre-multiplication by Ej has been to interchange row 2 and
row 3 of A. A can be any 3x3 matrix; taking A = I, since Ejl = Ej, Ej is
formed by interchanging row 2 and row 3 of I. If r„ r't represent the z'th
259

MATRICES 2 [14
rows of A and EjA respectively, the effect may be symbolically expressed
by К = rlt r2 = r3, r3 = r2.
Similarly /0 0 1\
E2 = 10 1 0
\l 0 0/
has the effect of interchanging the first and third rows of
A {r[ = r3> r2 = r2> r3 = rv)
Ex. 4. Show that Ei is the matrix of the linear transformation which reflects
points in the plane у = ζ.
Ex. 5. Interpret E2 as the matrix of a linear transformation.
(ii) The second elementary row operation: the multiplication of a row of A
by a non-zero constant с
Consider I с 0 0\
Q = |0 1 0),
\0 0 1/
(ca11 ca^ ca13\
a21 a22 Ой J
a3i a32 a3J
and the effect of pre-multiplication by Q has been to multiply the first row
of A by с (ri = crlt r'2 = r2, r3 = r3). Notice again that Q has been
obtained by performing the required elementary row operation on I. (The
reader may wonder why the restriction с Ф 0 has been imposed; so far as
the result is concerned, this is of course unnecessary, but multiplication of a
row by zero must be specifically excluded when we use elementary row
operations to determine inverses.)
Ex. 6. Interpret Q as the matrix of a linear transformation.
(iii) The third elementary row operation: addition to any one row a
constant multiple с of another row.
Consider /1 0 0\
M! = (0 1 0),
\0 с 1/
/ On an a13 \
MjA = Ι α21 α22 α23 I
\a31 + ca21 a32 + ca22 а^ + са^/
and we have a matrix whose first two rows are those of A, but whose third
row is obtained by adding to each element of the third row of А, с times
the corresponding element of the second row {r[ = rlt r2 = r2,r3 = r3 + cr2).
260

1] THE INVERSE OF A MATRIX
Ex. 7. Interpret Mj as the matrix of a linear transformation.
Matrices such as Ε, Μ and C, which effect elementary row operations
are called elementary matrices. Let us now reconsider the reduction of
to a unit matrix in terms of pre-multiplications by elementary matrices.
We wish to effect the transformations
и; и; ι
The first transformation may be obtained by leaving the first row as it is
and subtracting from the second row twice the first row
0i = rltri = r2-2/j)
that is, by multiplying by the elementary matrix
Operating thus on A we obtain
(J?)·
We now leave the second row as it is and subtract from the first row three
times the second row {r[ = гг - 3r2, r2 = r2); that is, we multiply by
and the reduction is complete.
If we attempt a reduction of any square matrix A to unit form by a
succession of row operations represented by the elementary matrices
Xj... X„, then γ γ α τ
ХИХИ_1 ··. X2XjA = I
and X„X„_!... XuXJ = XnXn.1...X2X1 = A"1.
Thus, in a practical reduction we perform the successive row operations on
two series of matrices, one side starting with A and finishing with I, the
other side starting with I and finishing with A-1 (see Example 1).
Ex. 8. If A is the matrix of a linear transformation T, and if A-1 exists, interpret
A-1 as the matrix of a linear transformation.
Example 1. If A= (-1 2)*
find В such that BA = I.

MATRICES 2
-1 2\
3 4}
-5 10\
0 10/
/ о
г2 + 3Г1 \ 1
( ο ίο)
r2
r2
Φ0
(1
(Ϊ
и
n
1)
s
Thus B= [\ з5 .
\lT7 To/
Notice that, in the third transformation, we multiplied the first row by 5
to avoid introducing fractions for a little longer—an artifice worth
remembering. With practice, the reader will learn to contract some of the
working, though it is best to write it out in full at first. It is always advisible
to check one's working by computing BA.
Ex. 9. Find, using elementary row operations, matrices В such that BA = I in
each of the following cases:
Example 2. If
/1 2 5\
A =
find В such that BA = L
262
m

THE INVERSE OF A MATRIX
/1 2 5\ / 1 0 0\
(2 3 4 I 0 1 0
\l 1 2/ \ 0 0 1/
/1 2 5\
0-1-6 r't = r2-2
\l 1 2/
/1 2 5\ / 1 0
(O -1 -6) -2 1
\0 -1 -3/ ri = rt-rx \-l 0
/1 2 5\ / 1 0 0\
(0 -1 -6) (-2 1 0)
\0 0 3/ r3' = r3-r2 \ 1 -1 1/
/1 0 -7\ ri = r! + 2r2 /-3 2 0\
(θ -1 -6) (-2 1 0)
\0 0 3/ \ 1—1 I/
/1 0 -7\ /-3 2 0\
(O -1 0 r,' = r2 + 2r3 0-1 2
\0 0 3/ \ 1—1 1/
/ι ο o\ ri = ri+ira /-f -i f\
0 -10 0 -1 2
\0 0 3/ \ 1 — 1 1/
/1 0 OV Z-f-i }V
0 1 0) ri =-r2 0 1 -2
\0 0 3/ \ 1 -1 1/
о o\ /-!- -i i\
10 0 1-2
0 l/ ri = ir3 \ i -* i/
■Г:.
1 -2
Ex. 10. Find, using elementary row operations, matrices В such that BA = I in
each of the following cases:
/6 8 5\ /4 7 3^
(i) A= 3 5 3 ; (ii) A= 2 5 2;
\2 3 2/ \5 13 5/

MATRICES 2
(iii) A
/4 3 5V /4 1 IV
= 4 2 3 ; (iv) A= 3 2 3 .
\6 3 5/ \4 3 4/
Having determined the inverses of some specific matrices, we next
consider whether all square matrices A have an inverse. The perceptive
reader may have observed that the process outlined above breaks down if
a complete row of zeros is obtained. In fact, if this occurs at any stage, A has
no inverse. Matrices for which no inverse can be found are called singular
matrices; matrices which have an inverse are called non-singular matrices.
The proof of a necessary and sufficient condition for singularity is deferred
to the next section.
Before leaving the consideration of elementary row operations it should
be pointed out that entirely analogous elementary column operations exist
and post-inverses (that is, matrices С such that AC = I) may be found by
post-multiplication by elementary matrices.
Exercise 14(a)
1. Find matrices В such that BA = I in each of the following cases:
■G3'
\5 4/'
(iii) A =
Ov) A= L _ ; (v) A= '
VA-1 h V
2. Find matrices С such that AC = I for each of the matrices A of Question 1.
3. Show that the matrix /3 2\
1б 4/
is singular.
4. Find matrices В such that BA = I in each of the following cases:
\4 1 4/
/3 4 1\ /3 4 5\ /3 6 2\
) 2 3 1 ; (v) 4 3 11 ; (vi) 1 4 2 .
\3 7 2/ \l 0 3/ \2 4 2/
Verify in each case that AB =
5. Show that the matrix
is singular.
264

1] THE INVERSE OF A MATRIX
6. A matrix of the form
\o ο ι/
leading diagc
is non-singu
■0 ··(■)■
with zeros in every entry below the leading diagonal and 1 as each element of that
diagonal is called an echelon matrix.
Prove that every echelon matrix is non-singular and that, if A is echelon and
BA = I, then В is also echelon.
7. If
find В such that BA = I.
show that the system of equations
(6x + 9y + 8z = 1,
3x + 5y+4z = 3,
2x + 4y + 3z = 2
may be written in the matrix form
Av = w.
Pre-multiply both sides of this equation by В and hence solve the equations.
8. Solve the equations (3x + 2y + 3z = 17,
\lx-3y + 4z = -7,
[2x- y+3z= 1
by the method of Question 7.
9. For what value of a is the matrix
-3 -1
singular?
10. If AX =
i

«ι2\
«22/
e
α
о}
wj
and
к,
u12
z + a21H
Z+iZ22W
o,
1,
MATRICES 2 [14
2. DETERMINANTS
Recall that a square matrix A is non-singular if a (pre- or post-) inverse
may be found; otherwise it is singular. We now establish a simple criterion
for singularity for 2 χ 2 matrices.
Theorem 14.1. If A is the matrix
/«n
W
then A is non-singular о a11a22-a^a^ φ 0.
Proof, (i) Suppose A is non-singular.
A is non-singular => a matrix В =
exists such that BA =
^ {^x + a^y =
[a^x + a^y =
!x(a11a22-a12a21) = a^,
у(а11а22-а12а2г) = -a12,
z(a11a22-a12a2i) = -a21,
w(a11a22-a12a21)= au.
But the matrix A is non-singular and thus Α φ 0, from which it follows
that at least one of the elements of A is non-zero and hence that
«11 «22-«12 «21 * 0·
(ii) Suppose ДцОгг-«i2«2i + 0·
Consider
/ α22 -α12\ /оц α12\ = /aua22-a21a12 0 \
\-«2l «ll/ \«21 «22/ I 0 «11 «22- «21 «12/
then
«11«22-«12«21 * 0 ^ l ( ^ "Μ («" aA = I
ii"22 12 21-r a^a^-a^a2X\-a^ au) \a21 a22J
/ «22 -«i2\
«11 «22-«12 «21 I-«21 «ll/
=> A is non-singular.
The quantity a11a22—a12a21 is called the determinant of A and is written
Ι α" αΐ2 I or, more shortly, det A (or |A|).
where В =

2] DETERMINANTS
Thus the result of Theorem 14.1 may be stated as:
Ά is a non-singular 2x2 matrix -o- det Α φ Ο'.
If the elements of A are real numbers then det A is a real number and
we have a mapping A -> det A from the set of all 2 χ 2 matrices with real
elements to the set of real numbers. Similarly, we have mappings A -> det A
from the set of all 2 χ 2 matrices with rational/integral elements to the set
of all rationals/integers.
The rale for determining the value of det A must be carefully memorized:
products of the elements are formed diagonally (the leading diagonal first)
and then subtracted. Thus
Il 21 =_2. I 2 _11 =_7. I cosa sina| = ι
I 3 4 I I - 3 - 2 I ' I - sin a cos a|
Ex. 11. Evaluate:
I 3 1 I I 1 -2 I I 1 -2 Ι
(i)|o i|; co|2 t|; cd|3 ,[.
The construction of the determinant of a 2 χ 2 matrix has thus given us a
test for singularity. We now define determinants of higher orders; although
the definition will no doubt appear complicated, the motivation for their
introduction should be clear—we shall use them to extend the result of
Theorem 14.1.
If A is the 3 χ 3 matrix /au a12 a13\
I «21 «22 «23).
\«31 «32
then the determinant of A, written
I «11 «12
«21 «22 «23 |
I «31 «32
is defined by
. . A I «22 «23 I | «21 «23 L I «21 «22 |
det A = an\ —«12 +«13
I «32 «33 I I «31 «33 I I «31 «32 I
= Дц «22 «33 - «11 «23 «32 - «12 «21 «33 + «12 «23 «31 + «13 «21 «32 - «13 «22 «31 ·
It is simply verified by direct calculation that det A may be expressed in
the two alternative forms
, . A I «12 «13 I . I «11 «13 I „ I «11 «12 I
det A = —α2ι Ι+α22 —«23 >
I «32 «33 I I «31 «33 I ! «31 «32 I
or, more briefly, det A (or |A|)

MATRICES 2
«12 «13
I «22 «23 I
«11 «12
I «21 «22 I
These three forms give the expansion of the determinant by rows. The
determinant may also be expanded by columns; for example, expanding
by the first column
det A = Оц
«22 «23
-«21
«12 «13 L «12 «13
+ «31
«32 «33 I «22 «23 I
I «32 «33
The term I а& а2з Ι
I «32 «33 I
is called the minor of au; similarly, the minor of a12 is
I «21 «23 I
I «31 «33 I
(see expansion by first row) while the minor of a32 is
I «11 «13 I
I «21 «23 I
(from the expansion by the third row). To find the minor of any term,
write down the determinant and cross-out the row and column containing
the term under consideration; the required minor is the 2 χ 2 determinant
that remains. See Figure 14.1, where the minor of a23 is being sought: it is
«11 «12
ОГ («11^32-«12«3l)·
«u a13 a13
• —аа_—ашг-.а^-
To find the sign to be attached to each minor in the expansion of a
determinant, draw the chess-board pattern, as shown in Figure 14.2,
starting with + along the leading diagonal. Thus, if we were expanding by

2] DETERMINANTS
the second row, or by the third column, we should include the term
| «11 «12 I
~«23
I «31 «32 I
A minor, together with its correct sign as given in Figure 14.2, is called a
cofactor, the cofactor of the term ai} being written At}. Thus
«11 «13
= «ii«33-«i3«3i;
«11 «13
= «13 «21 "«11 «23·
I «21 «23 I
In terms of cofactors, the expansion of the determinant may be written
det A = a11A11+a12A12 + a13A13 (first row),
or det A = α12Λ12+«22Λ2+ «32^32 (second column), etc.
Determinants of orders 4, 5, 6, ... may be defined similarly. Thus, for
the 4 χ 4 determinant ,
«11 «12 «13 «14
«21 «22 «23 «24
«31 «32 «33 «34
«41
«43 aM
each cofactor is a 3 χ 3 determinant, obtained by rejecting the row and
column in which the corresponding element stands and attaching the
required sign, obtained from the chess-board pattern, to the remaining
determinant. Thus
I «11 «13 «14 I I «11 «12 «14 I
A22 = α31 α33 a34 ; Лз = - «21 «22 «24 ·
I «41 «43 «44 I I «31 «32 «34 I
The expansion of this determinant by the third column is
«13^13+ «23 Лз + «33^33+ «43^43·
Ex. 12. Evaluate the determinants:
-1 1
1 1

MATRICES 2
3. PROPERTIES OF DETERMINANTS
We shall now develop some simple properties of determinants, restricting
ourselves to the 3 χ 3 case for simplicity, although the results and proofs
all extend to η χ η determinants.
First observe that the interchange of rows and columns does not affect
the value of a determinant: for we may expand in the first case by the first
row and in the second case by the first column and, in either case, the
corresponding cofactors remain the same. Thus, any property possessed
by the rows of a determinant is possessed equally well by the columns.
For brevity, we shall denote the determinant
I «ii «12 «13 I
«21 «22 «23 Ьу Δ.
I «31 «32 «33 I
Property 1.
I Оц + x (h2+y a13 + z
«21
«22
«23
For
«11 «12 «13
«21 «22 «23
+ «21 «22 «23
«31 «32 «33 I I «31 «32
= (an + x) An + (a12 + y) A12 + (a13 + z) A13 = R.H.S.
Property 2. If two rows (or columns) are identical, det A = 0.
For, if two rows are the same, expanding by the third row the
corresponding cofactors are all zero.
Property 3. If an elementary row operation is performed on A, the
determinant of the new matrix is equal to the product of det A and the determinant
of the corresponding elementary matrix.
Case (i): the interchange of two rows (columns).
If, for example, we interchange rows 2 and 3, on expansion by row 1 the
cofactors all change sign but remain the same in magnitude; but
II 0 0 1
0 0 1 =-1.
Ι ο ι о I
The result may similarly be verified for any other interchange of two rows
(columns).
Case (ii): the multiplication of a row (column) by a non-zero constant.
сОц ca12 ca13
«21 «22 «23
= с det A,
by expanding by the first row.
270

PROPERTIES OF DETERMINANTS
But the corresponding elementary matrix is
Ic 0 0\
0 1 0
\o ο ι/
с
0
0
0
1
0
0
0
1
Case (iii): the addition to any one row (column) a constant multiple of
another row (column).
Suppose, for example, that we add to row 1 с times row 2
On
(къ + саы a^ + ca^
<h\ (hs, <hi
= det A + c a21 <къ «23
I a31 a32 a33 I
by properties 1 and 3 (case (ii))
= det A, by property 2.
But the corresponding elementary matrix is
\o 0 1/
0 10
0 0 1
= 1,
on expanding by the first column.
Other possibilities are dealt with in a precisely similar manner.
Property 3 may clearly be extended to a sequence of elementary row
(column) operations on A.
Property 4. det (AB) = det A det B.
Write
then
detAB
I OilKl+ 0·\Φ·ί\ + ^13^3
= аяЬц + α22 b21 + а2з b3.
Ι а31йи + а32*21 + азз*г
= ^ώ,
ап b12 + a12 έ22+α13 b32 an b13 + a^b^ + α13 b33
02! b12 + α22 b22 + α23 b32 α21 b13 + α^2 Ъ23 + %, b33
α31 b12 + α32 έ22 + Озз b32 a^ b13 + ^ b23 + a33 b33
But this may be expressed, by an extension of property 1, as the sum of
twenty-seven determinants, twenty-one of which are zero. (For example,
one of the vanishing determinants would be
I <hibu (hib-a, (hbK
<hsb33
-- bnb12b3i
<h\ an <ha
<hl «21 ^23
= 0 by property 2.)
a21bu a21bn
I ^31*11 a31b12
We are left with six determinants which do not have two identical columns.
However, using properties 1 and 3 we get
detAB = det A(bnb22b33 — b11b23b32 + b12b23b31 — b12b21b33 + b13b21b32
-b13b22b31) = det A det B.
271

MATRICES 2
[14
One simple but useful result that follows immediately from property 2
can now be derived. It will be recalled that a11A11 + a12A12 + a13A13 = det A
and similarly for other rows and columns. Now suppose we multiply each
element of some row (or column) by the cofactors of the corresponding
elements of another row (or column); for example
Now
a11A21 + a12A22 + α^Αχ,
anA21 + a12A22 + a13A23
«ιι <hs an
«31 «32 «33
I «11 «12 «13 I
A similar result holds for any other combination of rows (or columns).
Such an expansion is called an expansion by alien cofactors and we have
demonstrated that expansion by alien cofactors gives the value zero.
Example 3. Evaluate the determinant
18 4 12
9 3 3
I 5 15 10
12 13
= 4.3.5 3 1 1
I 1 3 2
18 4 12
9 3 3
5 15 10
(removing factors)
1
= (-60)
= 1020.
3 4 |
(-17)
(expanding by second
row)
Example 4. Express the determinant
as the product of factors.
272

3] PROPERTIES OF DETERMINANTS
If we set b = c, the given determinant is zero (two columns identical) and
so (b-c) must be a factor. By symmetry, (c-a) and (a-b) must also be
factors. But the expansion of the determinant clearly gives an expression
of degree 4, in a, b, с and so the remaining factor must be linear and
symmetrical, i.e. (a + b + c).
-- X(b-c) (c-a) (a-b) (a + b + c).
By comparing the term in ab3 on both sides, the value of λ is clearly 1.
2 1 -
1
4 2 1
-5 -1 -2
8 13 3
1 14 8
19 55 22
Ex. 13. Evaluate the following determinants:
16 15
With the properties of determinants that we have just proved we are
almost in a position to extend our theorem on the necessary and sufficient
condition for a matrix to be non-singular to matrices of order higher than
two. However, before we do so, we shall introduce a further concept, the
adjugate (or adjoint) of a matrix, since there is an intimate connection
between the adjugate and inverse of a non-singular matrix.
Given any square matrix A, where
(an a12
Chi «22
«31 «32
the adjugate (or adjoint) of A, written adj A, is defined by
adj A =
- u,
Us
that is, the adjugate is obtained by substituting for each element its co-
factor and transposing (see Miscellaneous Exercise 13, Question 7).
Theorem 14.2. For any square matrix A
A (adj A) = (adj A) A = (det A) I.

MATRICES 2
[14
Proof. (3x3 matrices)
Ml Ou Ou\ /Λΐ Λΐ Λΐ\ /Δ 0 0\
A(adjA)= α21 α22 αΔ Iai2 Λ22 AsA = 0 Δ 0 I,
\α31 α32 α33/ \Α13 Λ23 A J \0 0 Δ/
since each term in the product is either an expansion of the determinant,
or an expansion by alien cofactors;
= ΔΙ, where Δ = det A.
A precisely similar result holds for (adj A) A.
Corollary. If det Α φ 0, det (adj A) = (det A)2 (3x3 matrix).
Proof. Take determinants of both sides of A (adj A) = (det A) I and
divide both sides by det A.
It can be shown that, in fact,
det A = 0 => det (adj A) = 0
and so the restriction that det A must be non-zero may be dropped.
If det Α φ 0 we have now obtained an inverse for A, namely
adj A . adj A T
-j-^-г, since A.-j-^-r = I.
det A det A
We note further that multiplication of this inverse with A is commutative.
To obtain a complete picture about inverses we need to clear-up two final
points:
(i) Can a matrix A have an inverse even if det A = 0?
(ii) If det Α φ 0, can we find an inverse of A other than adj A/det A?
Theorem 14.3. If A is any 3x3 matrix, we have:
the matrix A is non-singular о det Α φ 0.
Proof (ϊ). To prove the implication => we show that det A = 0 => A is
singular (see Chapter 9).
det A = 0 ^ for any matrix B, det (AB) = det A det В = 0
*AB + I, since det I = 1
=> A is singular;
(ii) det Α Φ 0 ^ §^4 exists
det A
=> A is non-singular, by Theorem 14.2.
274

3] PROPERTIES OF DETERMINANTS
Theorem 14.4.
A-i _ adJ A
A ~ det A
is the unique inverse of the non-singular matrix A.
Proof. Suppose L is a left inverse other than A-1; that is, suppose that
LA = I, L Φ A"1. Then
L = LI = ЦАА-1) = (LA) A"1 = IA"1 = A"1,
which contradicts the assumption thatL Φ A-1. Similarly for right inverses.
Exercise 14(b)
1. Evaluate the following determinants:
«I!! I; oo I _f -T N
Oii)
Ι 6 5 Ι
(iv)
2. Evaluate the following determinants:
(·)
(iv)
(νϋ)
(x)
-4 7
3 1
5 -9 -
4 3 1
3 4 2
10 8 3
1 3 5
2 7 1
5 16 19
112 129 1
67 78
99 114
4
2
-4
04
62
92
7 17 16
13 33 30
10 21 23
3. Determine which of the following matrices are non-singular:
(9 6 10\
(iii) 17 9 29
/10 7
(ii) 23 14
\4 1
/-1 ί
(iv) 3 1
\ 11 -1

MATRICES 2 [14
4. Find the adjugate of each of the following matrices and, in each case, evaluate
the product of the matrix with its adjugate. Write down the determinant of each
matrix.
f ' 5):
\5 13 9/
/1 3 IV
) 3 10 5 ;
\l 4 5/
'4 11 5\
1 4 21.
1 2 1/
5. Find the inverses of the following matrices by the adjugate-determinant
method:
(п>Ш
6. Find the values of a which make the following matrices singular:
/1 3 7\ / 7 13 1\ / 1 -3
(i) 4 2 5 ; (ii) -1 a 3 ; (iii) -2 6
\5 5 β/ \ 3 7 2/ \ 4-11
7. Factorize the following determinants:
( ' - l}
\-l 4 0/
) (2 6
\l 4
/1 2 7\ /2 0 0\
= (0 1 2j, B= jl 1 01.
\l 2 8/ \0 2-1/
Solve for X the matrix equation
4 4\
8 8/
/0 2
B= 1 1
\3 -2

PROPERTIES OF DETERMINANTS
Solve for X the matrix equation
AX-B =
/2 1 -4\ / -4 2 1\
= (1 1 3), B= 11 -5 -3l.
\0 2 -1/ \-16 7 5/
Miscellaneous Exercise 14
-1 if \ 0 1/ \0 1/ \1 1/
Κι)·
Deduce that the rows of a 2x2 matrix can be interchanged by operations
which add multiples of one row of the matrix to the other, together with
operations changing the sign of a row.
Find a 2 χ 2 matrix X such that
la b\ v (b a\
2. Do 2 χ 2 matrices А, В with integer entries, exist such that
(а) AB = 0, ΒΑ Φ 0?
(б) AB = ΒΑ = 0, Α Φ 0, Β φ 0, Α Φ Β?
(с) AB = ΒΑ Φ 0, Α, Β φ Ι, ΑΦ Β?
In each case, if your answer is 'yes', justify it by giving examples of suitable
Α, Β. (Μ.Ε.Ι.)
so that the transpose a' is given by
·(:)
write down the 3 χ 3 matrix aa' and prove that the determinant of this matrix
has value zero.
Obtain the analogous result when
(a 0\
(SMP)
■B
4. If A is any 3x3 matrix and Pisa non-singular 3x3 matrix, prove that
det(A-Al) = det(B-AI), where Β = Ρ1 ΑΡ.
277

MATRICES 2
[14
5. S is a skew-symmetric matrix if S + S' = 0; A is an orthogonal matrix if
AA' = I. If S is a given skew-symmetric matrix and A, B given orthogonal
matrices, prove the following results:
(i) det A = ± 1;
(ii) AB is orthogonal;
(iii) (I-S) α + S)-1 is orthogonal.
6. If
prove by induction that
Prove also that the nth power of the matrix
»_/i ь\ ., /i
Do these results hold if η = -
7. Prove that
I b2c2 + a2d2 bc+ad 1 I
c2a2 + b2d2 ca+bd 1
I a2b2+c?d2 ab + cd 1 |
Also prove that
= (b-c) (c-β) (fl-b) (.a-d)(b-d)(c-d).
= (aj-x) (я2-*) (Дз-л) (я4-*) (l + * Σ -
■(Hi)
and det МФО, prove that, if λ is a root of the equation
Ι Α + λ Η
where A is the cofactor of a, etc., then det Μ/λ is a root of the equation
(x+a)(x+b) = h2. (O&C)
9. Factorize the determinant
Ι 1 + я2 + я4 1 + αί> + α2ί>2 l + ac+aV I
\l + ab + a2b2 l+62 + 64 1 + Z>c+Z>2c2 . (О & Q
I l + ac+a*c2 l + bc + b2c2 l + crHc4 |
278

10. Prove that
yz
(y+z-x
11. If
MISCELLANEOUS EXERCISE 14
zx xy = - 16xyz(y-z) (z-x) (x-y).
(z + x-yY (x+y-z)1 I (O&C)
0 0 rf \
/0 6 0
1 c / ,
/
where a, b, ...,/arerealanda > b > c, prove that the equation
det(A-*I) = 0
has real roots. (O&C adapted)
12. If A and В are square matrices such that AB = 0, prove that either A = 0
or В = 0 or det A = det В = 0.
13. Two matrices A and В are said to commute if AB = BA. Let
prove that every matrix В that commutes with A can be expressed in the form
В = AA+/tI, where A and μ are scalars and I is the unit matrix. Obtain
expressions of this form for A2 and A"1. (M.T.)
14. Prove that, if a is a 3 χ 3 matrix, with adj a = A and det a = a, that
aA = Aa = αϊ.
Assuming that α Φ 0, prove that:
(i) det A = a2;
(ii) X = A is the only solution of the matrix equation aX = αϊ;
Oii) adj A = aa.
Which, if any, of the above results, are still valid if the restriction on α is
removed ?
15. If A is a 3 χ 3 matrix with αη φ 0, and if the cofactor of every element of
A is zero (that is, if adj A = 0), prove that A has the form
(an а1г а13 \
kan ka12 ka13 J.
lau 1агг laj
16. What is meant by the statement that multiplication of real numbers is
associative and is distributive over addition?
Prove that, if я,,·, b{j cw (/,/' = 1, 2, 3) are real numbers, then
3/3 \ 3 /3
Σ Σ aiabap\ cPj = Σ aia Ι Σ *α/,<
β = \ \α = 1 / α = 1 V = l

MATRICES 2 [14
Deduce that multiplication of 3 χ 3 real matrices is associative and that, if A
is a 3x3 real matrix, then the notation A3, A4, As,... may be interpreted
unambiguously.
If the 3 χ 3 real matrices В and A2 +A+I are non-singular (I is the identity
matrix) why would the notation A/B be ambiguous, but
17. The non-singular matrix В has the property BB' = B'B, where B' is the
transpose of B. Prove that B'B"1 = B-'B'. Prove also that, if С = B-'B', then
CC is the identity matrix.
■H i i)

15. Linear equations
1. LINEAR EQUATIONS IN TWO UNKNOWNS;
INTRODUCTION
Given perpendicular coordinate axes Ox, Oy in a plane, any linear
equation connecting the two variables χ and у is the equation of some straight
line in the plane; that is, the linear equation defines the set of points
(x, y) comprising the line. Two such sets of points will have as their
intersection a single point, in general, whose coordinates may be obtained by
solving the equations simultaneously, or, in other words, by requiring the
defining properties of the sets of points comprising the two lines to hold
simultaneously.
For example, the pair of straight lines
(2x + 5y= 1,
1 x-3y = 6
intersect at the point (3, -1), a result derived by solving the simultaneous
equations in the usual way.
Consider now the pair of straight lines
(2x + 3y = 1,
(2x + 3y = 2.
It is immediately obvious algebraically that these equations have no
solutions. Geometrically, they represent parallel straight lines, with no
finite point of intersection.
Again, consider the 'pair' of straight lines
| x + 2y = 1,
(2x + 4y = 2.
The second equation is a thinly disguised re-write of the first so any point of
the single line represented has coordinates satisfying the equation. In fact,
setting у = A and solving the first equation for x, we see that (1-2A, A)
is a solution, for all values of A. Since we may assign values of the single
parameter A arbitrarily, our equations are said to have one degree of
freedom (or to be a one parameter system). Geometrically, we have a pair of
coincident straight lines.
The reader may feel this last case somewhat frivolous, and its extensive
281

LINEAR EQUATIONS [15
discussion pedantic. However, if the coefficients are unknown, this
possibility must be remembered.
To summarize; given two equations in two unknowns, one of three
possibilities occurs:
(i) The equations have a unique solution. (The two lines are distinct
and intersect.)
(ii) The equations have no solution. (The two lines are distinct and
parallel.)
(iii) The equations have one degree of freedom. (The two lines coincide.)
Example 1. Solve the equations
[х+У = 2,
\ax + 2y = b.
Eliminating v, we have ,„ „ ,
b·^ (2-a)x = 4-b.
Case (i). If α Φ 2, we have the unique solution
4-b b-2a
X = 2^a' У = ^а-·
Case (ii). If a = 2, we have 0.x = 4-b.
Sub-case (iia). If b Φ 4, no solution for χ exists.
Sub-case (ii*). If b = 4, χ may take any value, у then being determined
from the equation x+y = 2.
Thus, the complete solution of the equations takes the form
4-b b-2a
αΦ2:χ = ϊ-α, У=^-а,
a = 2, b Φ 4: no solution;
a = 2, b = 4: one-parameter solutions (λ, 2 —A).
The reader would be wise to consider the meaning of this solution
geometrically.
The reader is probably so accustomed to solving pairs of simultaneous
equations that it may not have occurred to him to ask what happens if
he is given three (or more) equations in two unknowns. From geometrical
considerations it should be fairly clear that, in general, no values of χ
and у satisfy all three equations simultaneously. However, consider the
system ( x + 2y = 5,
3x- y=l,
{ x-5y= -9.
Solving the first two equations we get χ = 1, у = 2 and these values cer-
282

11 TWO UNKNOWNS; INTRODUCTION
tainly do satisfy the third equation. Geometrically, these three equations
represent three straight lines through the point (1, 2) (see Figure 15.1).
Ъх-у = 1 *
Fig. 15.1
It will be recalled that, if
^x+bjy + C! = 0, 02X + b2y + c2 = 0
are two non-parallel straight lines, then any line through their point of
intersection has the form
μ(α1χ + ό1γ + α1) + λ(α2χ+^γ + €2) = 0.
Thus, in the particular case we are considering, any straight line through
the point (1, 2) has an equation of the form
/4х + 2у-5)+ЦЗх-у-1) = 0.
Setting λ/μ = - \ we obtain the third equation. This example illustrates the
general result, that, if three equations in two unknowns have a solution,
then any one equation is a linear combination of the other two (see
Chapter 2.4).
2. LINEAR EQUATIONS IN TWO UNKNOWNS;
CONTINUED
We consider pairs of equations of the form
ioax + аау = Ьъ)
{a^x + a^y = bj
Writing
the equations (1) may be rewritten in the matrix form
Ax = b. (2)
IO PPM 283

LINEAR EQUATIONS [15
If we regard A as the matrix of a linear transformation T, equation (2)
tells us that the image of the point P, with position vector I I, is the
of simultaneous equations may be reinterpreted as the problem of finding
the point Ρ whose image, under a given linear transformation, is a given
point B.
If A is non-singular, that is, if det Α φ 0, we have, for any point B,
Ax = b
ο Α1 Αχ = A-Jb
ο χ = A-Jb
and we see that the unique point P, whose position vector is A_1b, has
image В under the linear transformation T.
Ex. 1. Show that A-1 is the matrix of a linear transformation S.
[You must show that, for every point Q with position vector y, the image S(y)
exists and is unique and that 5(Ay+μζ) = A5(y) +μδ(ζ).] The linear
transformation S so defined is called the inverse of the transformation T; S and Γ satisfy the
relations S[T(x)] = x; T[S(y)l = у for all χ, у.
As an example of the solution of equations whose matrix, A, is non-
singular, consider
Write
then det A =
and we have
\3x
19 Φ 0 and
A-1 =■
G
-Ю-
~0-
-e>-
-2y =
+ 5y-
D·
■0
■H
Η
= 4,
= 1.
b =
5 2
2 3
5
-2
-D-
0
)
I)
The solution may be stated in the form χ = ff, у = -γ-9- or, in terms of the
linear transformation Twhose matrix is A, the unique point (y§, -Д) maps
into (4, 1).

2]
TWO UNKNOWNS; CONTINUED
Ex. 2. Solve the equations Ъх-2у = 4,
2х + Ъу = 1
by using the inverse matrix method.
Suppose now that det A = 0. Before discussing the existence of a
point Ρ whose image is В under the linear transformation T, consider the
effect of Τ upon the unit square OIMJ. For a general transformation
(0 00
Fig. 15.2
(det Αφ 0) ΟΓ Μ'J'is a parallelogram whose area is det A (see Chapter 13)—
depicted in Figure 15.2, but, if det A = 0, the parallelogram collapses and
O, I', J', M' are collinear as shown in Figure 15.3.
*Ex. 3. If det A = 0, show that every point Ρ of the plane maps into a point of
the line ΟΓ.
Since, in this latter case, Τ maps every point Ρ into a point of the line
ΟΓ, a point В not lying on the line cannot be the image of any point under
T: the equations Ax = b are inconsistent.
However, if В does lie on ОГ, we may write the position vectors of/',
/', В as r, тт, ст respectively; that is
Aj = r, Ai = тт, b = ст.
Then, P(x, y) is a point which maps into В
о ст = A(xi+yj)
о ст = x(Ai)+XAj)
о ст = (mx+y)i
о Ρ lies in the straight line y + mx = c,
and infinitely many points map into B, by the reversibility of the argument.
Ю-2 285

LINEAR EQUATIONS [15
To summarize for the case det A = 0:
(i) the equations Ax = b have no solution (are inconsistent; equations
with one or more solutions are said to be consistent).
о there is no point which maps into В under Τ
о В does not lie on the line ОГ;
(ii) the equations Ax = b have infinitely many solutions
о there are infinitely many points Ρ which map into В under Τ
о В does lies on the line ОГ.
Ex. 4. Show that the equation ax = b is uniquely soluble only if α φ 0, in which
case, the unique solution is χ = ar^b.
Discuss the case a = 0, showing that failure to obtain a unique solution can
occur in two different ways.
Ex. 5. What can you say about the equation Ax = b if T(i) = T(J) = 0? (Γ is
the linear transformation with matrix A.)
Exercise 15(a)
1. Solve, using the inverse matrix, the following pairs of simultaneous equations:
\Ъх- у = 1, \Ъх + 2у = 1, j70x+105y = 12,
(0 1 (и) 1 (ш) 1
\2х + Ъу = 19; l2x + 7y = 29; I 5x- 7y = 5.
(i) ax + a2 = Ьг-Ьх; (ii) ax + b = bx + c.
3. Solve for x, у the following pairs of equations, commenting upon any special
cases that arise for particular values of the coefficients a:
. (Ъх+ау =-2, ί x + 2y = 4, .... (3x+ ay = 1,
w \3x + 2y = 3; K } \2x + ay = 2a; ^ } W+12.y = 2.
4. Solve for x, у the following pairs of equations, commenting upon any special
cases that arise for particular values of the coefficients a, b;
( ax + by = 2, ίαχ + Зу = 6; ... iax + Zy = 1,
1 \l2x-4y = a + Z>; W \ x-by = 2; wy \bx + ay =-1.
5. Discuss the solution of the following simultaneous equations for various
values of α: ,
x+ y = h
\2x + ay = 5,
U*+ у =-3.
6. Discuss the solution of the following simultaneous equations for various
values of the coefficient a:
(ax + 3y =
fr
6,
2y = a.

2] TWO UNKNOWNS; CONTINUED
7. Discuss fully the solution of the following simultaneous equations for the
unknownsx,y: , , .
J ( ax+ by = с,
\агх+Ьгу = с2.
8. Under what conditions do the homogeneous equations
(a^x+b^y = 0,
\a2x+b2y = 0
have a non-trivial solution (that is, a solution other than x, у = 0) ? Solve the
equations fully in the case where non-trivial solutions exist.
9. Solve for x, у the equations:
f χ cos θ +y cos φ = cos a,
\ χ sin θ+y sin 0 = sin a.
10. Sketch the set, E, of points in the plane whose coordinates satisfy the
inequality x+y > 0. The linear transformation, T, has matrix A where
Sketch the image set T(E).
11. Γ is a linear transformation whose matrix is
■6 :!)·
If Ε is the set of points whose coordinates satisfy the inequality у > 0, sketch the
image set T{E).
3. SYSTEMS OF LINEAR EQUATIONS IN THREE
UNKNOWNS; INTRODUCTION
Given perpendicular axes Ox, Oy, Oz in space, a linear equation
connecting the three variables x, y, ζ is the equation of some plane. Two linear
equations represent two planes which may be parallel (same unit normal
vector) but otherwise intersect in a straight line. In general,
three planes intersect in a unique point and thus, in general,
three equations in three unknowns possess a unique solution;
see Figure 15.4.
For example, the three planes
x+ y+ ζ = 0,
x + 2y + 3z =-3,
x+ y- z = 4, Fig· 15.4
have in common the single point (1, 1, —2)—a result readily derived by
successively eliminating two of the variables.

LINEAR EQUATIONS [15
However, just as in the case of two lines, complications may occur.
Consider the equations , ,
[ x+ У + Зг = 1,
| x-2y+ z= 1,
I2x- y + 4z = 2.
If we write
Lx = x+y+3z-l, L2 = x-2y + z-l, L2 ξ 2x-y + 4z-2,
then it is readily seen that L3 = Lj+Z^; thus, these three equations
represent three planes through a common line (see page 62) and so
possess infinitely many solutions (one degree of
freedom); see Figure 15.5.
Again, the system
(2x+y + 3z = 6,
x+y + 4z = 1,
x- z= 1,
has no solution, for by subtracting the first two equations we obtain
x-z = 5, which is inconsistent with the third equation. Geometrically,
x — z = 5, which is a plane through the line of intersection of the two
planes 2x+y+3z = 6 and x+y + 4z = 1,
is parallel to the third plane x-z = 1 and thus the line of intersection of
the first two planes is parallel to the third plane. A similar result would
hold whatever line of intersection was chosen and so the three planes form
a triangular prism of uniform cross-section; see Figure 15.6.
Fig. 15.5
Fig. 15.6 Fig. 15.7
Another possibility is illustrated by the system
f x+ y+ ζ = 1,
\2x + 2y + 2z= 3,
I x+ y + 3z= 1,
which clearly has no solution, the first pair of equations being inconsistent.
Geometrically, the first two planes are parallel to one another but not to
the third plane; see Figure 15.7.

3] THREE UNKNOWNS; INTRODUCTION
Again, we may have equations in the form:
(x+y + z = 1,
jx+j + z = 2,
Ух+y + z = 3.
The equations are manifestly inconsistent and thus have no solution.
Geometrically, they represent three parallel planes.
Finally, we may have two or three equations which reduce to the same
equation, in which case we have one or two degrees of
freedom—geometrically, two or three coincident planes. (Or, possibly, two coincident
planes and the other plane parallel—giving no solution.)
To summarize: given three equations in three unknowns one of four
possibilities may occur:
(i) The equations have a unique solution (no two of the planes are
parallel and the planes do not have a common line of intersection).
(ii) The equations have no solution (either at least two of the planes are
parallel or the three lines of intersection are parallel but not coincident or
two of the planes coincide and the third is parallel).
(iii) The equations have a one-parameter solution (the planes have a
common line of intersection).
(iv) The equations have a two-parameter solution (the planes all
coincide).
The situation is more complicated than in two dimensions, as was to be
expected but, if the reader keeps the various geometrical possibilities in
his mind, he should avoid confusion.
Example 2. Solve
(1)-(2) gives
(1) - (3) gives ■
If b Φ 5, z =
and substitution
χ =
1 completely the equations
(Ъх+у+ ζ = a, (1)
\2x + y- ζ = 4, (2)
[5x+y + bz = 1. (3)
x +2z = a-4,
-2x + (l-b)z = a-l,
(5-b)z= 3(a-3).
3(a-3)
5-6
back gives
-a-2-ab + 4b _ l5 + 5a-12b + 2ab
5-
corresponding to the case of unique solution.
If b = 5, α φ 3 there is no finite solution for ζ—we have a triangular
prism of planes (clearly no two of the planes are parallel).

LINEAR EQUATIONS [15
If b = 5, a = 3 the equations are consistent: putting χ = λ in (1) and
(2) we have 2y = 7-5A, 2z = - l-λ and the line of intersection of the
first two planes is , ,
χ _y-j z + Jr _
2~ -5~ -\~ μ·
By substitution, the general point (2μ, \ — 5μ, -\-μ) of this line lies on
plane (3) for all μ. Since no pair of the planes coincide, we have three
planes with a common line of intersection and the general solution may be
taken as „ - , ,
χ = 2μ, у = ϊ-5μ, ζ = -\-μ
(or any comparable form).
If we are given four equations in three unknowns, we have, geometrically
four planes and these will not, in general, have a common point. However,
it is not impossible for four equations in three unknowns to be consistent;
consider, for example, the four planes given by the equations
x+ y+ ζ = 0,
2x- 3y+ ζ =-3,
x+ 3y+ ζ = 2,
3x+Uy+4z = 9,
all of which contain the point (1, 1, - 2). In practice, the best way of
tackling such a system is to solve three of the equations and check whether or
not the fourth equation is satisfied.
Ex. 6. Show that the system of equations
Г x+ y+ ζ = 2,
-5y+ z=-l,
j x+2y + 3z =-1,
[3x-3y + 2z = 2
is consistent.
4. SYSTEMS OF LINEAR EQUATIONS IN
THREE UNKNOWNS; CONTINUED
As in the case of two unknowns, we may write the system
саих + а12у+а1Ъг = Ьгл
la^x + ^y+a^z = ьЛ (З)
\аЪ1х+аъгу+аъъг = b3,l
in the matrix form Ax = b. (4)
290

4] THREE UNKNOWNS; CONTINUED
Regarding A as the matrix of a linear transformation T, equation (4)
tells us that the image of the point P, whose position vector is
(i)·
is the point B, whose position vector is
(i)
Again, the problem of solving a set of simultaneous equations may be
interpreted as the problem of finding the point Ρ whose image, under a
given linear transformation, is a given point B.
If det Α φ 0 we have, as in section 2,
Ax = b
о A-JAx = A-4>
ο χ = Α-4>;
the unique point P, whose position vector is A_1b, has image В under the
linear transformation T.
Ex. 7. Show that A-1 is the matrix of a linear transformation S—the inverse
of the transformation T.
As an example of the solution of equations whose matrix, A, is non-
singular consider
Oc+ y+z = 4,
\l 2
and we have
(by the usual inversion
process)
:|)'*-(H-3
(Э-ШН
4
The solution is χ = 1, у = 1, ζ = 2 or, in terms of the linear
transformation whose matrix is A, the point (1, 1, 2) maps into the point (4, -2, 1).
291

LINEAR EQUATIONS [15
The analysis of the case in which det A = 0 follows the same pattern as
the two-dimensional case and is pursued in the following series of exercises.
Ex. 8. Show that, under the linear transformation Τ whose matrix is given by
ia-ί a2 a3\
A = lbx b2 b3\
Vi c2 cj
the images of the points /, /, K, whose position vectors are i, j, k, are /', /', K',
whose position vectors are respectively the first, second and third columns of A.
Ex. 9. Show that, under T, the unit cube, three of whose edges are OI, OJ, OK,
transforms into a parallelepiped (a figure all of whose faces are parallelograms).
Ex. 10. Show that the volume of the parallelepiped obtained in Ex. 8. is det A.
(This result is necessary for the remaining exercises in this section, but its proof
is rather hard and may be omitted at a first reading. The best proof depends
upon the triple scalar product, which will be met in Book 2.)
Ex. 11. If det A = 0, show that О, Г, J', A"'are coplanar. What does this tell us
about the columns of A?
Ex. 12. If the point D, with position vector d, is the image of a point Ρ under
a linear transformation whose matrix is singular, what can be said about
(i) the point D; (ii) the vector d?
Ex. 13. Show that, if det A = 0, and if D lies in the plane OI'J'K', then all the
points of a certain line map into D.
Ex. 14. In the extreme case in which О, Г, J', K' are collinear, what can be said
about the columns of A? If Ax = d has a solution in this case, what can be said
about (i) the point D; (ii) the vector d?
In an actual example, det A should be calculated and the various cases in
which det A = 0 should be treated by reverting to the original equation.
Matrix methods are admirably suited to studying the structure of systems
of equations but rarely constitute an efficient procedure for their solution
in individual cases.
Example 3. Discuss the solution of the system of linear equations
x + ay+ ζ = b,
ax+3y+ ζ = 1,
5x + Sy+3z = 1
for various values of a and b.
/1 a 1\
A=la 3 II and so det A =-3α2+13α-14.
\5 8 3/
292

4] THREE UNKNOWNS; CONTINUED
Thus det A = 0 if a = 2 or J.
Case (i). α φ 2, α φ }.
The equations have a unique solution for all values of b.
Case (ii). a = 2.
The equations reduce to
x + 2y+ z = b, (1)
2x + 3y+ ζ = 1, (2)
5x + 8y+3z = 1. (3)
Eliminating ζ between (2) and (3), and between (1) and (2),
Г x+ у =2, (4)
\ x+ у = 1-6. (5)
Thus, the equations are consistent if and only if b = -1, in which case we
have a one-parameter solution л: = λ, у = 2-λ, ζ = λ-5 (on using
equations (4) and (1)).
Case (iii). a = y.
The equations reduce to
(3x + 7y+3z = 3b, (6)
\7x + 9y+3z = 3, (7)
[5x + 8y+3z= 1. (8)
(6) + (7)-2(8) gives 0 = 36 + 3-2.
Thus, the equations are consistent if and only if b = - $, in which case we
have a one-parameter solution χ = λ, у = 2(1 - λ), ζ = ^(11λ-15) (from
equations (7) and (8)).
5. HOMOGENEOUS EQUATIONS
We now consider the system of linear equations
(апх + а12у+а13г = 0,
a21x + a22y+a23z = 0,
a31x + 03^+033 ζ = 0,
or Ax = 0.
Such a system is said to be homogeneous. If A is regarded as the matrix of a
linear transformation Τ we seek these points which are mapped into the
origin under T. It is easy to see that the origin maps into the origin under
any linear transformation; that is, that χ = 0 is always a solution for a set
of homogeneous equations—we call this the trivial solution—but it is of
more interest to look for non-trivial solutions χ φ 0.
293

LINEAR EQUATIONS
Ex. 15. Show that the linear transformation Γ whose matrix is
'-G -! -!)
maps all points of the line
nto the origin.
What does this tell us about the system of homogeneous equations
[ x + 3y+ ζ = 0?
We conclude this chapter by proving a necessary and sufficient condition
for the existence of non-trivial solutions of the homogeneous equations
Ax = 0.
Theorem 15.1
Ax = 0 has a non-trivial solution o- det A = 0.
Proof, (i) If det Α φ 0, A-1 exists and χ = 0 is the only solution.
Thus, if Ax = 0 has a non-trivial solution, then det A = 0, and we
have proved that a necessary condition for the existence of a non-trivial
solution is det A = 0 (the implication =>).
(ii) If det A = 0 and all the cofactors are zero then, by the result proved
in Miscellaneous Exercise 14, Question 15, the rows of A are in proportion
and the three equations reduce to a single equation, which has a
two-parameter solution.
If det A = 0 and at least one cofactor, say А1Ъ is non-zero, then
a11A11+a12A12+a13A13 = 0 (detA = 0),
α21 Λΐ + α22 Ли + α23 Лз
α3ΐΛΐ +Я32Л2 +^33^13
We have thus constructed a solution
x = ^11» У = Лг, г = A13
which is non-trivial, since An φ 0.
The condition det A = 0 is sufficient to ensure the existence of a non-
trivial solution (the implication <=).
Exercise 15 (b)
1. Solve the equations (3x+ y+ z= -2,
\2x+2y + 3z = 8,
I x + 3y + 2z = 6
(i) by successive elimination; (ii) by using the inverse matrix.
294
= 0/1
= 0j
expansion by alien cofactors.

5] HOMOGENEOUS EQUATIONS
2. Calculate the inverse of the matrix
C-i-i-i)
and hence solve the equations
i-3x+Uy + 7z = a,
\ - x + 4y+3z = b,
[ 2x- 7y-5z = с
(i) when a = 4,b = l,c = -2; (ii) when β = 1, b = l, с = -1;
(iii) when α = 1, * = -6, с = 0.
3. Solve the equations ( x — 2y+ ζ = 1,
hx+2.y-3z = 7,
l.5x-2j,- ζ = 9
and interpret your result geometrically.
4. Show that the matrix /3 -1
A = (2 3
\l -15
is singular.
Discuss the solution of the equations
(3x- y+ 2z = 4, \Ъх- у+ 2z = 4,
(i) \2x+ Ъу- z=\, (ii) \2x+ Ъу- ζ = 1,
I x-l5y+10z = 8; [ x-l5y+l0z = 9,
and interpret your results geometrically.
5. Show that the equations (3x-y+ ζ = 3,
]αχ+^+ ζ = 4,
[8x-.y+3z = *
are consistent and have a unique solution, provided α φ 2.
Discuss the solution of the equations for the case a = 2.
6. Discuss the solution of the equations
Г x+j, + z=l,
i ox + ay+2z = 1,
U* + 3.y+(a+l)z = *
for various values of a, b.
Interpret your various results geometrically.
7. /3 -1 1\ /2 -1 -1\
A= 2 1 -1 , B= 3 -1 2 .
\1 -4 -l/ \5 -1 -2/
A maps the point Ρ into the point Q and В maps the point Q into the point R,
whose coordinates are (-5, 0, 0). Find the coordinates of P.
295

LINEAR EQUATIONS [15
8. Discuss the solution of the equations
(3x + 2y+az = 2a,
\ax + y + 6z = 15,
[ x + 2y + 3z = S
for various values of a.
Interpret your results geometrically.
9. Find the value of a for which the homogeneous equations
(6x + 2y- z = 0,
\Ъх + ау-2г = 0,
[3x-2y+ z = 0
have a non-trivial solution, and find solutions for this case.
10. Prove that, if det A = 0 and if Axj = b and Ax2 = 0, then Xi + Ax2 is a
solution for the equation Ax = b, for all values of λ.
Find a solution of the equations
x+ y+ z = a,
Ъх-5у- z = b,
2x-8y-3z = с
when a = b = с = 0 and hence write down the general solution of the equations
when a = 1, b = 3, с = 2.
11. A linear transformation Г has matrix
■6 J I)'
Show that Τ maps all points of the line
into the origin, and find the line of points that maps into the point (6, 6, 0).
What is the relation between these two lines?
Iax bx сЛ
= я2 h сг\.
\<h Ih cj
Ax is the cofactor of дь etc., and det A = Δ. If
a^x + b^y + dz = dlt
a2x + b2y + c2z = a\,
a3x + b3y + c3z = d3,
prove that Δχ = d1A1 + d2A2 + d3 A3.
If Δ = 0, what does the value of the expression dx Ax + d2A2 + d3 A3 tell you about
the solution of the given system of equations ?
296

5] HOMOGENEOUS EQUATIONS
13. Prove that, if
= [a2 b\ °cX d = IdX χ = (y)
\a» b3 c3J \dj \zj
and if det Α Φ 0, then the unique solution of the equation
Ax = d
is given by
<?i ii Cj αϊ i/i Ci αϊ Z>i dj.
\ d2 b2 c2\ = a2 d2 c2 \ = \ a2 b2 d2
| d3 b3 c3 I I a3 d3 c3 I I a3 b3 d3 I
14. The linear transformation, T, has matrix A where
■(-1J -5-
Ε is the set of points in space whose coordinates satisfy the inequalities χ > 0,
у > 0, ζ > 0, x+y + z < 1. Determine the image set T(E).
Miscellaneous Exercise 15
1. Find the inverse of the matrix
Given the equations Г 2*!—4x2+ x3
+ 3x3 ■■
(-1 i -!)·
(2лг!-4л:2Н
-X!+2x2-
find the solutions
(i) when a = 1, b = 1, с
2. Let A be a 3 χ 3 real matrix and b a three-rowed column vector. It is proposed
to solve the equation _
and a particular solution χ = Xo is noted. Prove that any other solution may be
written in the form
X = Xo + U,
where u is a solution of the equation Ax = 0. Prove conversely that any vector
Xo + u is a solution of the equation, where Xo is a fixed particular solution of the
equation and u is any vector such that Au = 0.
Interpret geometrically the equation and its solutions if A is singular (i) when
Xo exists, and (ii) when there is no such particular solution Xq. (M.E.I.)
297

LINEAR EQUATIONS
M-(j
form the product MM' and show that M' = 49M"1.
Without multiplying out, state the product M'M, giving reasons for your
answer. Hence, or otherwise, find the solution of the equations:
2*!+3x2+ 6x3 = 1,
6*! +2x2-3x3 = 1,
3*!-6x2+ 2*3 = 2. (M.E.I.)
4. The simultaneous equations
Зх + у = 5
may be written in matrix form as
(1 -Do-® °tAx=a
Carry out numerically the procedure of the following three steps:
(i) A'AX = A'B;
(ii) (A'A)-1 A'AX = (A'A)-1 A'B;
■0
(iii) IX = = (A'A)-1 A'B.
Verify that the values of x, у so found do not satisfy all the original three
equations. Suggest a reason for this.
Under what circumstances will the procedure given above, when applied to
a set of three simultaneous equations in two variables, result in values which
satisfy the equations? (SMP)
5. By systematic elimination, find values of А, В, С in terms of a, b, c, such that
( x+ y+ z = 1,
\ ax+ by+ cz = a + b + c,
[a2x + b2y+c2z = bc + ca + ab,
if and only if ix+y+ ζ = 1,
{ Ay + Bz=b+c
[ Cz = a2 + b2.
Solve the equations, assuming a, b, с are all different.
Describe geometrically the configuration of planes, in three-dimensional
Euclidean space, with equations:
x+ y+ z= 1,
x + 2y+2z = 5,
x + 4y + 4z = 8. (SMP)
298

5] MISCELLANEOUS EXERCISE 15
6. Solve the simultaneous equations
x+ y + ζ = 3,
x+2y + 3z = 6,
x+3y + kz = 4 + k
(i) when/: φ 5;
(ii) when к = 5, giving the general solution. (SMP)
7. If A is the matrix Iй I and X is the non-zero matrix I I, show
Iй I and X is the non-zero matrix I* I,
that, if AX = AX, where λ is a number, real or complex, then
X*-(a + d)X+(ad-bc) = 0.
Show that A itself satisfies this quadratic equation, in the sense that
A2-(a + d)A + (ad-bc)\ = 0,
/1 0\
s the matrix
(Itl-
8. By considering det (A - XT), extend the result of Question 7 to a 3 χ 3 matrix A.
9. Factorize the determinant I 1 1 1 I
I a2 b2 c2 I'
Show that, if no two of a, b, с are equal, the equations
x+ y+ ζ = a,
ax+ by+ cz = ab,
a2x+b2y+c'z = abc,
have unique solutions for x, y, z, and find them.
Discuss the special cases
(i) a = b Φ с; (ii) b = c + a; (iii) a = b = с. (М.Т.)
10. Prove that the only value of λ allowing a real non-trivial solution of the
simultaneous equations _ .
y + z = Xy,
4x+2y-z = Xz,
is λ = 1. (M.T.)
11. If А, В are 3 x 3 matrices and χ a three-rowed column vector such that there
exist numbers λ, μ such that
Ax = λχ, Bx = μχ,
prove that there exists a number ν such that
ABx = »x.
Prove further that, if x1; x2, x3 are three linearly independent column vectors
with this property, then A and В are commutative for multiplication.

LINEAR EQUATIONS [15
12. Write the two sets of three equations
Яа*1+Я*2*2 + Я>3*3 = С, ЬцУ!+bi2y2+biay3 = Xt (ί = 1, 2, 3)
in matrix form, and prove that they can be solved uniquely for х1г x2, х3, у1г у2, y3
if and only if det AB φ 0, where A and В are the 3 x 3 matrices of coefficients
in the two sets of equations.
Show that the equations
Χ-ίΛ-2χί = 2, Хг + Хг + Хз = 1, Зх2-Х3 = к,
Зу1+У2 + 4у3 = Χι, -y-ί + 2у2-Зу3 = хг, У!+5у2 -2у3 = х3
are inconsistent if к φ 7, and find the most general solution for х^.-.уз if
к = 7. (M.T.)
13. Prove that, if λ: φ 0, the system of equations
2x+ у = а,
x+ky- ζ = b,
y + 2z = с
has a unique solution (x, y, z) for every choice of (a, b, c).
Show also that, when к = 0, the system is consistent if and only if (a, b, c)
satisfy a certain linear relation, and find this relation.
Verify that the system is consistent when к = 0 and (a, b, c) = (1, 1, -1)
and find an expression for the general solution of the system in this case.
(M.T.)
14. Find, for all values of the parameter λ, the number of solutions of the
eqUati°nS x + 2y+Xz=0,
2x+3y-2z = λ,
λχ+ y+ ζ = 3. (M.T.)

16. Discrete probability distributions
1. INTRODUCTION:
THE UNIFORM DISTRIBUTION
In Chapter 7 we discussed the concept of an outcome space for a random
experiment. To each elementary event we ascribed a probability and the
entire set of probabilities was described as г. probability distribution for the
outcome space. In this chapter we shall enumerate various possible
probability distributions that find frequent use as mathematical models for
random experiments. We shall generally define our outcome space in terms
of an associated random variable Ζ (see Chapter 10), and we shall thus be
able to talk about the mean and variance of the distribution; however, it
must be remembered that, for example, the mean is defined as ${X) and
so, if a new random variable, Y, is chosen, the mean naturally changes, too.
Unless there is any possibility of confusion, we shall refer to the mean and
variance of X as μ and σ2 respectively.
Suppose we have an outcome space consisting of η elementary events,
with associated random variable, X, where
X= {1,2, 3,..., л}.
Perhaps the simplest assumption we can make about the η possible
outcomes is that each one is as likely to occur as any other. Thus we take
Pr(Z=r)=l/n. (1)
Equation (1) defines the uniform distribution for η possible outcomes.
The uniform distribution is, in a way, fundamental, for we may often
subdivide the elements of an outcome space in such a way that the new
outcome space may reasonably be given a uniform distribution. This is not
always possible, however; a simple counter-example might be the fall of a
biased coin.
The mean, μ, of the random variable X with a uniform distribution is
given by
μ = <?(Χ)
= !("+!)· (2)
301

DISCRETE PROBABILITY DISTRIBUTIONS [16
Again, the variance, σ2, is given by
= £[Χ*]-μ*
= Κ"+ΐ)(2«+ΐ)-Κ"+ΐ)2
= -h(n2-l). (3)
The reader will find a large number of examples on the uniform
distribution at the end of Chapter 7.
Ex. 1. A man spins a coin and throws a die. For a head he scores —1, for a
tail +1, and this he adds to the score showing on the uppermost face of the die.
The values of the random variable, X, where
X= {0, 1,2, ...,7}
represent his total score. Why do you think a uniform distribution would be
an unsuitable mathematical model for this eight-point space? Suggest a way of
subdividing the sample space so that a uniform distribution would be suitable.
2. THE BINOMIAL DISTRIBUTION
The simplest possible type of random experiment is one that has just two
possible outcomes, which we may conveniently designate success and
failure. An experiment of this type is called a Bernoulli trial. (J. Bernoulli,
1654-1705, one of a family of distinguished mathematicians, whose
famous work on probability, the Ars Conjectandi, was published
posthumously in 1713.). Some examples of Bernoulli trials are
(i) a coin is tossed : does it show heads or tails?;
(ii) a person is tested for disease: has he got the disease or not?;
(iii) the height of a person is measured: is it less than six feet or not?;
(iv) a marksman fires at a target: does he bit or miss the bull?
If the probabilities of the two possible outcomes of a Bernoulli trial are
ρ (success) and# (failure) (where p+q = 1) and η independent repetitions
of the trial are made, we may define the random variable X as the number
of successes obtained in the η trials and calculate p(r) = Pr (X = r),
r = 1,2,3, ...,n.
Method I. Suppose first we assign some specific order in which we should
obtain the r successes: for sake of argument, suppose the first r trials
result in success and the final (n — r) trials in failure. Then, by independence,
302

2] THE BINOMIAL DISTRIBUTION
the probability of obtaining this sequence is prqn~r. But there are Iй)
different sequences possible. Thus
p(r) = -Pr(X=r)=typ'q'". (4)
Method II. The p.g.f. for a single Bernoulli trial is given by
S{f) = p.t1+q.t0
= pt+q.
Thus, the probability generating function for η independent Bernoulli
trials is given by G(t) = (pt+qy.
Thus p(r) = coefficient of tT in the expansion of (pt+q)n
The probability distribution defined by equation (4) is called the
binomial distribution. It is easily verified that (4) does indeed define a valid
probability distribution. For Pr (X = r) > 0 and also
№
ргдп-r = (j)+qYj by the Binomial Theorem;
= 1, since p+q = 1.
It finds a wide field of application as a probability model but the reader
must always take care to remember the binomial distribution applies only
if we have η independent repetitions of a Bernoulli trial. A binomial
distribution is known completely if the two parameters η and ρ are given; the
phrase 'the binomial distribution arising from η repetitions of a Bernoulli
trial with probability of success/'' is often abbreviated to B(n,p).
Example 1. Six unbiased dice are thrown. What is the probability of securing
three or more sixes!
The dice may be assumed to fall independently and so the random
variable X (= number of sixes showing) has a binomial distribution with
probability \ of success on each of the six Bernoulli trials.
Thus
Pr(Z> 3) = l-p(X < 3)
* 006.

DISCRETE PROBABILITY DISTRIBUTIONS [16
We next derive expressions for the mean, μ, and variance, σ2, of the
number of successes in a binomial distribution. We shall give two methods,
the first one using the definition (4) directly, the second employing the
probability generating function (and using Theorem 10.5).
Method I. By definition
* /л-1\ r n r ■ /n\ n\ nln-\\
= np%l \rZi)pr'1^n~r
= npip+q)71'1, by the Binomial Theorem,
= np, since p+q = 1.
Again
σ2 = <?[(Χ-μ)2] = <?[Χ2]-μ\ by Theorem 10.1,
= Σ''2( I Prqn~T — n2p2, using the first part,
= Σ nr Г_ J prqn~r-n2p2
= ^n{n-\)p2^rZ^pr-2q^
+ ^np^~_l}jPr-1qn-r-n2p2,
on writing r = (r-l) + l,
= n{n—\)p2 + np — n2p2, since p+q = I,
= npq.
Thus we arrive at the important result: the mean and variance of the
number of successes in a binomial distribution B(n, p) are given by
μ = np, σ2 = npq. (5)
Method II. Since the p.g.f. for the binomial distribution is given by
G(t) = (pt+qY
we have
G'(t) = np{pt+qY-\ G\t) = n{n-l)p2(pt+qy-2 (n>2).
Putting t = 1, μ = np, σ2+μ2—μ = n(n—l)p2 and the results follow.
304

2] THE BINOMIAL DISTRIBUTION
Example 2. A very large number of balls are in a bag, one-eighth being black
and the rest white. Twelve balls are drawn at random. Find
(i) the probability of drawing three black balls and nine white balls;
(ii) the probability of drawing at least 3 black balls;
(iii) the expected number of black balls in the sample;
(iv) the most likely number of black balls in the sample.
Drawing a ball from the bag and noting its colour may be regarded as a
Bernoulli trial (a black ball drawn being a success, ρ = £). Since a ball is
not replaced before its successor is drawn, the twelve Bernoulli trials
constituting our sample are not strictly independent; however, since the
number of balls in the bag is very large, we may assume that Pr (black ball
drawn) remains sensibly constant at ^; that is, each trial is independent of
what has previously occurred. Thus the random variable
X = number of black balls contained in a sample of 12
has a binomial distribution 5(12, |).
к 0-129;
* 0-182;
(iii) S(X) = np
= 1-5.
(iv) Write p(r) = Pr (X = r) = (^) p^~r.
= 12! rl(12-r)! β\
(r+l)!(ll-r)! 12! \i)
Yl-r
7(r+l)·
Thus ^±1) < j if r > ι
P(r)
and so /7(1) = щщ p(0) = y^(O)
is the maximum of the p(r); that is, the most likely number of black balls

DISCRETE PROBABILITY DISTRIBUTIONS [16
[Notice that the ratio deduced in Example 2(iv) enables us to express
p(r) in terms of p0. Thus:
Α = ΨΡο,
a = Ha = Mpo,
Рз = \%Рг = ΤιΐΡο, etc.
This is a particularly useful device for calculating binomial probabilities
if a hand calculating machine is available.]
Ex. 2. Find the values of Pr (X = r) in terms of Pr (X = 0) for the binomial
distribution B(4, £).
Ex. 3. If fifteen dice are thrown what is
(i) the expected number of sixes showing;
(ii) the most likely number of sixes showing ?
Ex. 4. Calculate the mean and variance for the binomial distribution B(9, |).
Ex. 5. In a family of four children, what is the probability of there being two
boys and two girls ? What is the probability that the eldest two are boys and the
youngest two girls ? What are the odds against having all four children of the
same sex?
Exercise 16(a)
1. A coin is tossed four times. Find the probability that heads appear
(i) at the first two tosses, followed by two tails;
(ii) just twice in the four throws;
(iii) at least twice.
2. 10 % of the very large number of articles produced by a machine are faulty.
What is the probability that a random sample of ten articles will
(i) be free of faulty articles;
(ii) contain more than two faulty articles?
3. From a packet containing a large number of seeds, 40 % of which are
advertised to give red flowers and the others white, 10 plants are produced. What is the
probability
(i) that all the plants have red flowers;
(ii) that all the plants have white flowers;
(iii) that half the plants have red flowers and half white?
4. 10 % of the very large number of articles produced by a machine are faulty.
If articles are taken at random and tested, how many articles will be tested, on
average, before the first faulty article is found ? What is the probability that the
testing procedure will have to go on longer than this before the first faulty article
is found?
5. (i) In a trial, eight coins are tossed together. In one hundred such trials how
many times should one expect to obtain three heads and five tails ?
306

2] THE BINOMIAL DISTRIBUTION
(ii) If 8 % of articles in a large consignment are defective, what is the chance
that a sample of thirty articles will contain fewer than three defectives ?
(O&C)
6. A battery of four guns is firing on to an enemy emplacement. It is reckoned
that each gun should score on the average one direct hit in every five shots,
and that three direct hits are needed to destroy the emplacement. If each gun
fires one shell, calculate the probability that the emplacement will be destroyed.
With new gun crews it is reckoned that two of the guns should score one direct
hit in every three shots and that the other two guns should score one direct hit
in every four shots. If each gun now fires one shell, calculate the probability that
the emplacement will be destroyed. (Cambridge)
7. Nine unbiased dice are thrown. Find p(r), the probability that r sixes appear,
and hence determine the value oip(r+\)jp(r). Find
(i) the expected number of sixes;
(ii) the most likely number of sixes;
(iii) the probability of obtaining more than one six.
8. (i) In a binomial distribution where the probabilities of the occurrence of an
event are given by the terms of the expansion of (p + g)m, the mean μ and the
standard deviation σ of the distribution are given by the formulae
μ= mp; σ = J(mpq).
Prove these formulae in the case where m = 3.
(ii) Two men A and В play a game in which A should win eight games to
every seven won by B. If they play three games, show that the probability that
A will win at least two games is approximately 0-55. (Cambridge)
9. Playing a certain One-arm bandit', which is advertised to 'increase your
money tenfold', costs 5p a turn; the player is returned 50p if more than eight balls
out of a total of ten drop in a specified slot. The chance of any one ball dropping
is p. Determine the chance of winning in a given turn, and for ρ = 0-65, calculate
the mean profit made by the machine on five hundred turns. Evaluate the
proportion of losing turns in which the player comes within one or two balls of
winning (j> = 0-65). (Cambridge)
10. Prove that, in the binomial distribution B(2n, %), the probability of scoring
an even number of successes is \.
11. An experiment consists of tossing an unbiased coin twelve times and
counting the number of heads obtained (X). If the mean and variance of X are μ
and σ2 respectively, find the value of
(i) Ρτ(\Χ-μ\ > σ); (ii) Ρτ(\Χ-μ\ > 2σ).
3. THE GEOMETRIC AND NEGATIVE
BINOMIAL DISTRIBUTIONS
We must now consider two further probability distributions associated
with the independent repetitions of Bernoulli trials. However, instead of
repeating the trial a fixed number of times and asking how many successes
307

DISCRETE PROBABILITY DISTRIBUTIONS \\6
have been obtained, we reverse the process by asking how long we must go
on repeating the trial until a stipulated number of successes is obtained.
First, suppose we wish to calculate the probability that the first success
occurs at the rth repetition of a Bernoulli trial, the probability of success
in any one trial being p. (As in Section 2, we assume that each trial is
independent of what has previously occurred.) We take as our random
variable X, where
X = number of trials up to and including the trial which results in the
first success.
There is no upper limit to the value X may take: we can theoretically
continue obtaining failures for ever, although the probability of doing so
steadily decreases.
Now Pr (X = r) = Pr (initially (r — 1) failures, followed by a success),
Le· Pr(X=r) = <r-1P· (6)
Equation (6) defines the geometric distribution for the random variable
X= {1, 2, 3, ...}. It is easily seen that this is a valid probability distribution,
for
= 1.
The mean, μ, and variance, σ2, of the random variable Ζ for a geometric
distribution may be derived without undue difficulty: we give two possible
methods.
Method I. S(X) = Σ rpqr-\
qS(X) = ^rpq^ = ^(r-\)pq^.
r=l r=1
Subtract: pS( X) = ρ + Σ pq^
= 1, since p + q — 1,
■·· μ = W = \\p.
Again £{X*) =jtrW~1.
^(χ2) = Σ'2μγ = Σ('-1)2μ'--1·

3] BINOMIAL DISTRIBUTIONS
Subtract: P#(X*) = Ρ + 2 Σ rpqr~x - Σ РЧГ~Х
r=2 r=2
= ρ + 2(β-ρ)-^
= p + 2{(llp)-p}-(\-p)
= (2/p)-l,
= 1-1-1
/>2 /> />2
= ?/P2·
Thus, for the geometric distribution,
μ = -, σ2
/>
Method II. The p.g.f. for the geometric
G(t) = g(r)
=£pqr-1rr
= ptl(l-qt).
Thus G\t)=pl(\-qt)\ G"(t) = 2pql(l-qty
and results (7) follow on substituting t = 1 and using Theorem 10.5.
The geometric distribution is a suitable probability model in a number
of practical examples; for example, in inverse sampling, articles are tested
until a faulty article is found.
Example 3. 5 % of the output of a certain machine is faulty. Articles are
taken at random from the output and tested, the process stopping when the
first faulty article is obtained. Find
(i) the expected duration of this sampling process;
(ii) the probability that the process will terminate before the expected
value is reached.
We shall assume that the total output of the machine is large and so the
individual Bernoulli trials of testing articles are independent, each with
probability of success (that is, of finding a faulty article) equal to -fa.
Adopting the geometric distribution as a suitable mathematical model, we have
(i) mean length of sampling process = l/(-2ir) = 20;
309
distribution is given by

DISCRETE PROBABILITY DISTRIBUTIONS [16
(ii) Pr (X < 20) = 1 -Pr (X > 20)
= 1-Σ (НГЧА)
r=20
= 1-[(М)19Ш]/[1-Ш
к 0-623.
The concept underlying the geometric distribution may be generalized
by asking how many Bernoulli trials will be required up to and including
the Mi success. Taking as our random variable X = {1, 2, 3, ...} we
observe that, if X = r, the rth outcome must be the Mi success, probability
p, and the previous (fc— 1) successes are binomially distributed B(r— l,p).
ThuS f/r-l\ \
Pr (X = r) = p[\^_\) /,*-y-«-<*-«j (/>, r ^fc)
(8)
Equation (8) defines the negative binomial distribution; to appreciate the
reason for this name and also to facilitate the verification that this is indeed
a valid probability distribution, we first observe that
X= X^Xz+... + Xi+... + Хъ
where Xt is the number of trials after the (г— l)th success up to and
including the rth success. Thus, the p.g.f. for a negative binomial distribution
is the Mi power of the p.g.f. for a geometric distribution, i.e.
G{i) = P4X\-qty\
Writing G(t) = а^ + а^^ + а^ь+Ч...
we have for r ^ k, Pr (X = r) = ak+r,
Furthermore,
and it follows that
310
ДРг (ЛГ = r) =βΣ β*4*
= G(1)
G'(t) = Λρ*ί*-1/(1-ϊ0*+1»
* 2 kq
μ = -, σ = -τ·

3] BINOMIAL DISTRIBUTIONS
Ex. 6. Devise an experiment for which you feel the negative binomial
distribution would constitute a suitable mathematical model.
Ex, 7. Dice are thrown in succession until two sixes have appeared; find
(i) the probability that two dice in all are thrown;
(ii) the probability that more than four dice are needed;
(iii) the expected number of throws required.
4. THE POISSON DISTRIBUTION
Another distribution of common applicability associated with an infinite
outcome space and random variable
X= {0,1,2,3,...}
is the Poisson distribution, defined by the equation
Pr(Z=r) = e-a.^ (e>0).
Since Σ Pr (X = r) = e-a Σ -} (10)
= e~aea
= 1
and Pr (X = r) > 0, for all r, equation (10) does constitute a valid
probability distribution.
We shall show that the Poisson distribution arises as the limit of a
binomial distribution in which n^oo andp^O, in such a way that np = a
remains constant. Thus the Poisson distribution will form a suitable
probability model, not only to situations where the binomial distribution
applies directly and η is very large and ρ very small but also to situations
where the binomial distribution is not directly applicable. For example,
consider requests for trunk call connections made at a telephone exchange.
Suppose that, for the period 7.00 a.m. to 8.00 a.m. they average out at
three and suppose further that the request takes a negligible time to make.
Dividing the hour up into a large number of short time intervals (say 720
of 5 seconds each), the probability of there being a call in any one of the
intervals may be taken as 2b", on the assumption that the calls arrive
independently throughout the hour, and the probability of two calls in
the interval is negligible. Thus we have 720 independent repetitions of a
Bernoulli trial ('Is there a call or not in a given 5 second interval?') with
small probability of success ρ = j-Jj, and the Poisson distribution may
be regarded as a good mathematical model to employ.
We must now verify our original assertion that a binomial distribution
311

DISCRETE PROBABILITY DISTRIBUTIONS
[16
B{n, p) tends to a Poisson distribution as η -> oo and ρ -> 0 in such a way
that np = a, a > 0. First, for B(n,p)
Pr(Z=r)=(^V"-r
= ^i)(^bJ^±lV(1_^(1_,r
= Ι »M */ I—«l^j^V/»-.
since л/7 = a.
Since r is a fixed number, the product of the r factors in the numerator of
the first fraction tends to 1 as η -+ oo, and (1 — p)~r tends to 1 as ρ -+ 0.
Furthermore,
and we have Pr (jf = r) -> ^p
as required.
The Poisson distribution is completely described if we are given the
value of a; thus, a Poisson distribution is a one-parameter distribution and
a Poisson distribution with parameter a may be referred to as P(a).
The mean, μ, and the variance, σ2, of the Poisson distribution P(a) may
be found by the direct computation of $(X) and ${X*). Thus
μ = <?(Ζ)
= ^r eracf
= a.
Similarly, σ2 = «?(Х2)-/*2
= α as above.
For the Poisson distribution P(a),
μ = a, σ2 = a, (11)
i.e. the mean and variance of the Poisson distribution P(a) are both equal
to a.
t For the proof of this and other results concerning the exponential function, see one
of the books on calculus mentioned in the bibliography.
312

4] THE POISSON DISTRIBUTION
Ex. 8. Prove that, if the random variable X = {0, 1, 2, ...} has equal mean
and variance, this does not necessarily mean that X has a Poisson distribution.
{Hint: consider a suitable uniform distribution.)
Example 4. Traffic accidents are reported in a certain town at an average
rate of four in a week. Estimate the probabilities:
(i) of a given week being accident free;
(ii) of there being three or fewer accidents in a given week;
(iii) of there being more than four accidents in a given week.
To produce reasonable estimates of these probabilities we have first to
set up a mathematical model of the situation. The assumption that
accidents occur independently of one another is arguable but, without more
detailed data, it would be difficult to make a more plausible assumption.
With this hypothesis, a Poisson distribution for the number, X, of accidents
per week would appear to be a reasonable model, for the reason outlined
earlier. Thus we have
(i) Рг(Х=0) = е-*
к 0-018;
(ii) Рг(Х<3) = е-*(1+4 + ^+:£)
и 0-433;
(42 43 44\
1+4 + 2! + з, + 4|)
к 0-371.
Exercise 16(b)
1. The random variable X can take Values 0, 1, 2, 3 Given that X has a
Poisson distribution, mean 2, calculate the probabilities
Pr (X = 0), Pr (X = 1), Pr (X = 2), Pr (X = 3),
Pr (X = 4), Pr (X = 5), Pr (X > 5).
2. Samples of forty articles at a time are taken periodically from the continuous
production of a machine and the number of samples containing 0, 1, 2, ...
defective articles are recorded in the following table:
No, defective per sample 0 1 2 3 4 5 6 Total
No. of samples |~~^ ^ ^ ~ ~ \ 0 | 100
Find the mean number of defectives per sample.
313

DISCRETE PROBABILITY DISTRIBUTIONS [16
Assuming that this is the mean of the population and that the Poisson
distribution applies, find the chance of:
(a) a sample containing four or more defectives;
(b) two successive samples containing between them four or more defectives.
(O&C)
3. The following table shows the results of recording the telephone calls handled
at a village telephone exchange between 1.00 p.m. and 2.00 p.m. on each
of a hundred weekdays (e.g. on thirty-six days no such calls were made):
Calls 0 I 2 3 4 or more
Days |36 35 22 7 0
Assuming that calls arrive independently and at random, estimate
(i) the mean m of the corresponding Poisson probability distribution;
(ii) the probability that if the operator is absent for ten minutes no call will
be missed;
(iii) the probability that if the operator is absent for ten minutes two or more
calls will be missed. (Cambridge)
4. In an examination 60 % of the candidates pass but only 4 % obtain distinction.
Use the binomial distribution to calculate the chance that a random group of
ten candidates should contain at most two failures.
Use the Poisson distribution to calculate the chance that a random group of
fifty candidates should contain more than one distinction. (Cambridge)
5. The road accidents in a certain area occur at an average rate of one per two
days. Calculate the probability of 0,1, 2 6 accidents per week in the district.
What is the most likely number of accidents per week ?
How many days in a week are expected to be free of accidents ? (M.E.I.)
{Note. This question is best attempted using a hand calculating machine, if
one is available.)
6. Explain briefly what is meant by a Poisson distribution and show that for
such a distribution the mean is equal to the variance.
In a bakery 3600 cherries are added to a mixture which is later divided up
to make 1200 small cakes.
(i) Find the average number of cherries per cake.
(ii) Assuming that the number of cherries in each cake follows a Poisson
distribution, estimate the number of cakes which will be without a cherry and the
number with five or more cherries. (O & C)
7. 10 % of the output of screws from a machine are faulty. If screws are taken
at random from the output until two faulty screws are found, what will be the
average number of screws tested ? What is the probability that precisely this
number of screws will be needed ?
8. 20 % of the butterflies in a district are of type A. If a random sample of size
10 is taken, find the probability
(i) that there will be just two butterflies of type A in the sample;
(ii) that the tenth butterfly caught will be the second one of type A.
314

41 THE POISSON DISTRIBUTION
9. Mass-produced articles are taken at random from a batch and tested until a
faulty article is found. If the twenty-first article proves to be the first defective,
is this at variance with the assumption that 10 % of all the articles are faulty?
10. If X is distributed according to a Poisson distribution with mean A, write
down the probability that X = r.
How is this probability modified if the values X = 0 are unobservable ? Prove
that the mean is now λ/(1 -e~A), and find the second momentt about X = 0.
(Oxford)
11. Evaluate Pr(^= r+l)/Pr(X = r) for the Poisson distribution P(a). If
the numbers of misprints on pages of an uncorrected proof have a Poisson
distribution with mean 2-7, what is the most likely number of misprints to be
found on a given page ?
12. The average proportion of bad eggs in an egg packing station is one in
2000. The eggs are packed in boxes containing six eggs each.
(i) Evaluate to two significant figures the probability that a box contains
one bad egg.
(ii) A housewife complains if she obtains two or more boxes, with one bad
egg each per hundred boxes. What is the probability that she complains.?
(M.E.I.)
13. During World War II, 537 flying bombs fell on south London. The
distribution of the number of hits in 576 areas, each of 0-25 km2, is given in the
table. Compare the actual frequency with the theoretical frequency obtained by
assuming that the aim was effectively random and followed a Poisson distribution
with the same average number of hits.
No. of hits 0 1 2 3 4 5 6 or more
Frequency 229 211 93 35 7 1 0
(M.E.I, adapted)
5. SAMPLING INSPECTION SCHEMES
Suppose a machine produces a large number of articles and it is required to
keep a check upon the number of articles produced which do not attain
some prescribed standard; for convenience, we shall describe such articles
as 'faulty'. The most obvious method is to check each article individually,
but such a process suffers from several drawbacks; for example
(i) the process of testing might be very costly;
(ii) the process would almost certainly be time consuming;
(iii) by the very nature of the test, the article tested might be destroyed
(consider, for example, the testing of photographic flash bulbs).
A more economic approach is to take a random sample from the total
output, to test each article of the sample and to deduce, from a
probabilistic argument, the quality of the whole output. The process of selecting
a random sample is not as straightforward as it might appear; we shall,
t The second moment about X = 0 Is SiX*).
ττ PPM 315

DISCRETE PROBABILITY DISTRIBUTIONS
[16
however, here assume that such a sample has been drawn, and concentrate
upon the second part of the problem: inferringthequalityofthepopulation.
A simple approach would be to take a sample of size n, test each article
and reject the batch as sub-standard if, say, more than m of the articles
prove faulty. We begin by assuming that we have some percentage, lOOp
say, as an upper limit to the number of faulty articles which may be allowed
before the manufacturing process is stopped and corrected for any fault.
Example 5. Samples of size 20 are taken from a large batch of articles
produced by a machine; if more than two faulty articles are discovered the
batch is withdrawn, otherwise it is accepted. What is the probability that, if
the machine is producing 5 % of faulty articles, the batch will be accepted!
If the proportion of faulty articles rises to 10 % of the total output, what is the
probability the batch will now be rejected!
Since the batch is large, we may assume that the removal of twenty
articles does not sensibly alter the proportion of defectives and so the
binomial distribution 5(20, 0Ό5) may be taken as a suitable mathematical
model for the experiment, where our random variable is the number, X,
of defectives in a sample of size 20.
(i) If Pr {article faulty) = -2\, then
Pr (not more than 2 faulty articles in 20)
χ 0-925-
(ii) If Pr (article faulty) = -&, then
Pr (more than 2 faulty articles in 20)
a 0-649.
Thus, if only 5 % of the articles produced are faulty, the chance of
accepting the batch is about 92^%, whereas, if the number of faulty articles
increases to 10 %, there is a 65 % chance of rejecting the batch.
Ex. 9. Comment upon the efficiency of the test given in Example 5. Suggest
ways in which you feel it could be improved.
The testing procedure exhibited above is an example of a single sampling
inspection scheme. Any sampling inspection scheme is open to two types
of error:
(I) A batch with an acceptable number of faulty articles may be
rejected.
(II) A batch with aq unacceptable number of faulty articles may be
accepted.
316

51 SAMPLING INSPECTION SCHEMES
If α = Pr (error of Type Ι), β = Pr (error of type II), then, in Example 1
above, α и 0075, β χ 0-351, if we regard 5 % faulty as acceptable, but
10 % faulty as unacceptable.
It can be shown that, if the size of the sample is held fixed, but the
standard of rejection is altered, either α increases and β decreases or vice versa.
Thus, there is an unavoidable margin of error in any inspection scheme;
the decision whether to minimize α or β depends upon such external
arguments as whether it is economically more desirable to withdraw good
batches or allow sub-standard batches on to the market. Of course, if the
sample is increased, more accurate information about the population may
be obtained and both α and β may be decreased.
Ex. 10. What objections are there to increasing the sample size in order to reduce
aand/??
If the probability, p, that an individual article is faulty is now regarded
as a variable, and the testing procedure is defined (that is, the sample size
and number of faulty articles required for rejection are given), the
probability Ρ of accepting the batch is a function of p. The curve obtained by
plotting Ρ against ρ is called the operating characteristic for the testing
plan. For example, with the test outlined in Example 5, the operating
characteristic has the following shape:
"°K

DISCRETE PROBABILITY DISTRIBUTIONS [16
From this graph, the probability of accepting (or rejecting) a sample
given the proportion of defectives in the population may be read off.
An increase in the sample size causes the operating characteristics to fall
off more steeply, thereby improving the test.
Other schemes are possible, besides single sampling schemes. For
example one can devise double sampling schemes as outlined in the next
example.
Example 6. A sample of size 8 is taken from a large batch of articles
produced by a machine. If the sample contains no faulty article the batch is
accepted, if more than one, it is rejected; if, however, the sample contains
a single faulty article, a second sample of eight articles is taken. The batch is
now accepted if this second sample is free of faulty articles, otherwise it is
rejected. Find
(i) the probability of accepting a batch containing 1 % of faulty articles;
(ii) the probability of rejecting a batch containing 10 % of faulty articles;
(iii) the average size of sample tested in case (ii).
PI
1-0 L
0-8
0-6 l·
0-4 V
Q-2h \
0-2 04 0-6 0-8 1-0 ρ
Fig. 16.2
Since the batch is large, the binomial distribution constitutes a suitable
mathematical model. Thus, if ρ is the probability of a randomly chosen
article being faulty, the probability, P, of accepting the batch is given by
318

61 HYPOTHESIS TESTING
i>=(i-/08+[8(i-/0»(i-/08
= (i -/О8+8/41-/О15;
(i) ifp = 1/100, Ρ и 0-99;
(ii) if/7= 10/100, 1-Рй 0-55;
(iii) a first sample of size 8 is always taken; the probability that a second
will be required is 8(0·9)7(0-1) χ 0-382 and so the expected sample size is
8 + 8x0-382» 11.
As in the case of single sample inspection schemes, the operating
characteristics curve can be drawn in this case too; see Figure 16.2.
Ex. 11. What advantage does a double sampling scheme possess over a single
sampling scheme? Are there any disadvantages?
6. HYPOTHESIS TESTING
Probability theory finds a further application in the testing of statistical
hypotheses. Consider the following problem.
Example 7. A coin is tossed ten times and shows nine heads. Is this sufficient
evidence to support the claim that the coin is biased!
In order to attack this problem, we adopt the approach made in testing
scientific theories: we make a hypothesis and then consider whether or not
observational evidence supports this hypothesis. In problems of statistical
inference, our hypothesis is generally called the null hypothesis (because it
often takes the form of an assumption about the absence of bias in the
population): in this example, we shall take as our null hypothesis the
statement ' the coin is unbiased; that is, Pr (heads on any toss) = \.
Having set up our null hypothesis we may ask the question ' Is the
probability of obtaining a result as bad as or worse than, the observed result so
small that we are reluctant to ascribe it to chance?' If it is, we agree to
reject our null hypothesis in favour of some alternative hypothesis.
To complete the framework of our solution we must further decide
what numerical value to prescribe to the indefinite phrase 'so small';
following common practice, we shall define ρ = -2й0- as a small probability
(the so-called '5% significance level'); in other cases it might be more
reasonable to take ρ = -j-^ (' 1% significance level').
Returning to the given example, our null hypothesis, H0, is given by
Ho* = h
we may take as our alternative hypothesis
Η,-.ρ + i.
319

DISCRETE PROBABILITY DISTRIBUTIONS [16
Our question now takes the form ' What is the probability of obtaining
a result as bad as, or worse than, nine heads with an unbiased coin?' Since
we are comparing the hypothesis H0:p — \ with the alternative hypothesis
Η-ι'.ρ Φ i, we are simply testing for absence of bias and thus we have no
reason to distinguish between an apparently abnormal number of heads
or tails; it follows that a result 'as bad as, or worse than, nine heads' is to
be interpreted as the result 'nine or ten heads or tails'.
We adopt as our probability model for the situation the binomial
distribution. Thus we have
Pr (9 or 10 heads or taih/H^ = 2[φ10 + 10(i)9(i)]
Since this is less than yg·, we infer that the result is significant at the 5 %
level: we have reason to believe that the coin is biased.
Sometimes a test of a statistical hypothesis is constructed before the
observational evidence is obtained. For instance, in our last example, we
might have compared the null hypothesis:
H0:p = i
with the alternative hypothesis:
Η,-.ρ Φ i
and so decided what results we should require before accepting or
rejecting #0.
In Example 7, our null hypothesis took the form of a statement about the
probability of an event; more generally, a null hypothesis is a statement
about the probability distribution of a population. We conclude this chapter
with a further example, in which our null hypothesis is a statement about
the mean of a random variable.
Example 8. Batteries are sent out by manufacturers in batches of five
hundred. On average, 2-4 batteries per batch are faulty. How many batteries
would we need to find faulty in a particular batch before we could adduce
evidence that more faulty batteries were appearing than could be ascribed
to pure chancel
Since we are here concerned only with the question of finding more than
the expected number of faulty batteries, we take as our null hypothesis
#0: mean number of batteries per batch = 2-4,
and as our alternative hypothesis
Нг: mean number of batteries per batch > 2-4.
We adopt as a mathematical model of the situation the Poisson
distribution (there is a large number of batteries in every batch and the prob-
320

6] HYPOTHESIS TESTING
ability of any particular battery being faulty is small), with random variable
X = r when there are r faulty articles in a batch.
Thus we have Pr (X = r\ H0) = <r2 4 χ ^^
and we wish to find the least integer η such that
Pr(Z> n\H0) < 0-05.
Equivalently we have to find the least integer η such that
Pr(Z< n\H0) > 0-95.
By direct computation
(2-42 ?-43 2-44\
1+2-4 + ^- + ^- + ^-)
и 0-0907(1 + 2-4 + 2-88 + 2-304 + 1-382)
« 0-90,
Pr (X < 5|#0) = Pr (X < 4|Я0) + Рг (X = 5|#0)
и 0-96.
Thus, η = 5: if we were to find 5 or more faulty batteries in a batch, we
should have reason for rejecting our null hypothesis #0 in favour of our
alternative hypothesis Hv
Exercise 16(c)
1. A machine is believed, on average, to produce 0-1 % of faulty articles.
Estimate, to 2 d.p., the probability of finding a batch of five hundred articles free of
defectives
(i) by using a binomial distribution;
(ii) by using a suitable Poisson distribution.
A batch of five hundred is tested and found to contain two faulty articles.
Comment.
2. A large batch of manufactured articles is accepted if either of the following
conditions is satisfied:
(i) a random sample of 10 articles contains no defective article;
(ii) a random sample of 10 articles contains one defective article and a second
random sample of ten is then drawn which contains no defective articles.
Otherwise the batch is rejected.
If, in fact, 5 % of the articles in a batch to be examined are defective, find the
chance of the batch being accepted [(0-95)10 = 0-5987]. (O & C)
3. From a batch of manufactured articles a sample of ten is taken and each
article is examined. If two or more articles are found to be defective the batch is
321

DISCRETE PROBABILITY DISTRIBUTIONS [16
rejected; otherwise it is accepted. Show that, if ρ is the proportion defective in a
batch and Ρ its chance of being accepted,
Ρ = (1-/»)·(1+9/ή.
Find an expression for Ρ if it is now decided to modify the scheme so that when
one defective is found in the sample a second sample of ten is taken and the batch
rejected if this second sample contains any defectives.
In the second case, what will be the average number sampled per batch over a
large number of batches when ρ = 0-05 ? (M.E.I.)
4. In a certain inspection scheme a sample of fifty items is selected at random
from a very large batch and the number of defectives is recorded. If this
number is more than three the batch is rejected; if it is less than three the batch is
accepted. If the number of defectives is exactly three a further sample, this time
of twenty-five items, is taken and the batch is rejected if there is more than one
defective in the second sample but accepted otherwise.
If the proportion of defective items in the batch is 1 % determine the values of
the following probabilities;
(i) that the batch is accepted as a result of inspection of the first sample;
(ii) that a further sample has to be taken and the batch is accepted as a result
of inspection of that sample;
(iii) that the batch is rejected.
Take (0-99)48 as 0-6173. (M.E.I.)
5. State the formula for the probability that a variable following a Poisson
distribution of mean m takes the value r. Prove that the variance of r is m.
Past experience has shown that the number of defective items produced in a
shift by a certain machine is a Poisson variable of mean 4. A new employee in his
first shift produced six defectives. Is this clear evidence that he is operating the
machine inefficiently? (Cambridge)
6. The probability P(r) that there will be r damaged tomatoes in a crate can be
taken as _„ ,
PM = ^\
where m is the expectation of r. Over a large number of crates the value of m
has been found to be 10. In the first crate from a new supplier the value of r
was 4. Test whether this is significant evidence that the value of m for this
supplier is less than 10; explain carefully the logic of your argument.
(Cambridge)
7. A die is thrown six times. If a score of six is made on three of these occasions,
have you any reason for believing that the die is biased?
8. A cubical die with faces marked 1 to 6 is thrown η times. Show that on the
hypothesis that the die is unbiased the chance that the face marked 4 will appear
uppermost not more than once is p, where
322

6] HYPOTHESIS TESTING
If η = 40 and the face marked 4 comes uppermost exactly once, test whether
the hypothesis that the die is unbiased is contradicted
(я) at the 1 % significance level;
(b) at the 0-1 % significance level. (Cambridge)
9. A bag contains ten balls, each of which is either black or white, but otherwise
the balls are indistinguishable one from another. Three balls are drawn without
replacement, and all are found to be white. Test the hypothesis
H0: there is an equal number of black and white balls in the bag.
The three balls are now replaced and three more are drawn; these are found to
be two white and one black. Test the hypothesis Щ again.
10. A sampling inspection scheme is operated by taking a random sample of size
10 from each large batch of a product. The batch is rejected if more than one
defective is found and otherwise the batch is accepted. Plot the operating
characteristic of the given plan.
Explain the applications of operating characteristics when choosing a suitable
plan. (M.E.I.)
[You are advised to use a hand calculating machine for the first part, if you
have one available.]
11. A double sampling inspection scheme is devised as follows: a sample of size
15 is drawn from a batch of articles; if the sample contains none or one faulty
article the batch is accepted, if more than three, it is rejected. If it contains two
or three faulty articles a second sample is taken; the batch is now rejected if this
second sample contains more than one faulty article. If the batch contains
100/7 % faulty articles, find
(i) the probability that the batch will be accepted;
(ii) the value of ρ which gives the largest expected sample size;
(iii) the size of this largest expected sample.
12. An assembled instrument contains two critical components A and B, Sample
tests show that we may expect one in ten of A and one in eighteen of В to be
defective. Estimate and compare the costs of the following inspection plans,
per hundred fully tested instruments:
(i) to test every component before assembly at a cost of 2p for each A tested
and 3p for each В tested; or
(ii) to test every instrument after assembly, if this test adds nothing to the cost
but making good a defective instrument costs on average 24p. (M.E.I.)
Miscellaneous Exercise 16
1. A pack of cards is cut and the suit of the exposed card noted; the pack is then
well shuffled and the process repeated. Find
(i) the probability that a spade will appear for the first time on the fourth cut;
(ii) the average number of cuts required before the first spade appears;
(iii) the average number of cuts required to expose cards of all four suits.
2. Show that the probability of r successes in η independent trials is the coefficient
of tr in the expansion of (q+pt)n, where ρ is the chance of success in a single
trial and q = 1 —p. Prove that the mean number of successes is np.
323

DISCRETE PROBABILITY DISTRIBUTIONS [16
Samples, each of eight articles, are taken at random from a large consignment
in which 20 % of the articles are defective. Find the most likely number of
defective articles in a single sample and the chance of obtaining precisely this
number.
If a hundred samples of eight are to be examined, calculate the number of
samples in which you expect to find three or more defective articles. (O & C)
3. Two coins are identical in appearance, but one is unbiased while the other
gives, on the average, heads three times as often as it does tails. One of the coins
is taken at random and tossed four times. If two heads and two tails appear,
what is the probability that it is the unbiased coin?
4. Explain briefly what is meant by a Poisson distribution of rare events and its
relation to the binomial distribution. Prove that the mean of the distribution is
equal to its variance.
A shopkeeper's sales of washing machines are four per month on the average.
Assuming that the monthly sales fit a Poisson distribution, find to what number
he should make up his stock at the beginning of each month so that his chance
of running out of machines during the month will be less than 4 %. (O & C)
5. A machine depending for its energy upon four complexes of solar cells, each
complex functioning independently of the others, will work provided one of the
four complexes is working. Each complex has probability ρ of failing. The
machine is redesigned to have six complexes and will function if two of the six
complexes are working. Is the new design an improvement on the old?
6. A and В play N games, each of which must result in a definite win to one or
other player, ,4's chance of success in any one game is p. For his nh win, A
receives from В £r, for his ith loss, he gives В £s. Find A's expected gain.
7. A coin is tossed repeatedly an even number, In, times. Show that, whatever
the bias (provided Pr (Η) Φ 0 or 1) the probability of obtaining the same
number of heads as tails decreases as η increases.
8. In lawn tennis a set is won by the first player to win six games, except that
if the score reaches 5-5 the set is won by the first to lead by two games. Two
players have chances respectively ρ and q of winning in any game (p+q = 1);
games may be treated as independent. Find the chance that a set lasts exactly
2n + 2 games (n > 5). (Cambridge)
9. The number of eggs laid by an insect has Poisson distribution with mean μ.
If the probability that an individual egg survives is p, show that the number
of eggs surviving has Poisson distribution, and determine its mean value. (You
may assume that the survival of an egg is independent of the fate of the other
eggs laid.)
10. An experiment has probability^ of success. In л independent trials, pn is the
probability of an even number of successes (p„ = 1). Prove that
Pn-(l-2p)Pn_1=p.
If ДО = Σ^ηίη, prove that
/(0 = [1 - (1 -P) 0/{(l - 0 [1 - (1 - 2p) t]}.
Deduce that pn = Ш + (1 - 2p)n]-
324

61 MISCELLANEOUS EXERCISE 16
11. Two machines, A and B, produce large numbers of articles, with 10 % of
those from A and 20 % of those from В defective. Machine В produces 50 % more
articles than machine A. A batch of ten taken from one of the machines contains
two defective articles. What is the probability it came from A ? (Give to 1 d.p.)
12. In a simplified probability model of the service in a barber's shop, it is
supposed that all haircuts take exactly six minutes and that a fresh batch of
customers arrives at six-minute intervals. The number of customers in a batch is
described by a Poisson probability function, the mean number being three. Any
customer who cannot be served instantly goes away and has his hair cut elsewhere.
The shop is open for forty hours a week. Calculate the theoretical frequencies
with which batches of 0,1, 2, 3, 4, 5 and more than 5 customers will arrive.
The proprietor reckons that it costs him £25 a week to staff and maintain each
chair in his shop, and he charges 25p for each haircut. Calculate his expected
weekly profit if he has
(i) three; (ii) four; (iii) five chairs. (SMP)
13. A sample of η coins is drawn at random from a large collection in which a
fraction ρ are pennies. What is the probability that just r of the η coins are
pennies ?
If the probability that a penny is a Queen Elizabeth one is q, what is the
probability that there are exactly s Queen Elizabeth pennies among the r pennies of this
sample?
Write down the probability that a sample of η coins will contain s+k Юр
pieces, only s of which are Queen Elizabeth ones, and calculate the sum of these
probabilities for all possible values of k. (C.S.)
14. A bag contains a large number of red, white and blue dice in equal numbers.
If η are drawn at random, show that the probability Pin, r) of drawing exactly
r red dice is equal to the term containing ft)r (i)n~T in the expansion of (} + §)".
If r dice are thrown, find the probability Q{r, s) of throwing exactly s sixes.
If η dice are drawn from the bag and the red dice drawn are thrown, show
that the probability of throwing exactly s sixes is
nifp(n,s+t)Q(s+t,s)
and prove that this is equal to a term in a binomial expansion. Explain why a
binomial distribution is obtained. (C.S.)
15. A random sample of size 10 is taken from a batch of a thousand components
and one defective is found.
(i) What is the largest possible percentage of defectives in the batch ?
(ii) ρ is the smallest proportion of defectives in the batch such that the
probability of obtaining not more than one defective is not less than 95 %. Find
the value of ρ to two decimal places by trial of suitable values of p.
(iii) Find the most likely percentage of defectives in the batch (i.e. such that
the probability of obtaining one defective in the sample is a maximum).
(M.E.I.)
325

DISCRETE PROBABILITY DISTRIBUTIONS [16
16. In sampling inspections of batches of manufactured articles a random sample
of twenty is taken; if none or one defective occurs in the sample the batch is
accepted, if three or more defectives occur the batch is rejected. If two defectives
occur a second random sample of twenty is taken and if in the combined sample
of forty less than four defectives occur the batch is accepted; otherwise it is
rejected.
Assuming that the proportion/? of defectives in a batch is sufficiently small for
the Poisson distribution to apply, show that the chance Ρ of the batch being
accepted is given by
Ρ = е-20" (1 + 20p) (1 + 200p2 е-20*).
(i) Find the chance of a batch which is 2 % defective being rejected.
(ii) Find the chances of batches which are respectively 5 and 10 % defective
being accepted.
(iii) Sketch the operating characteristic curve for the inspection scheme.
Determine the average sample size per batch if ρ = 0-05. (О & С)
17. The number of a certain type of organism in a given volume of water has
Poisson distribution with mean 2. A test, applied to indicate absence of the
organism has a 90 % chance of success if the organism is in fact absent, but also
indicates absence in 10 % of those cases in which they are present. If the test is
applied and indicates absence, what is the probability that the water is free of
the organism?
18. The probability that a source emits r α-particles in a given time is
proportional to μτ\ϊ\, where μ is a constant. Obtain the constant of proportionality
and calculate the mean of r.
The probability of the same source emitting s ,5-particles in the same time is
proportional to v'/sl Assuming that the two types of particle are emitted
independently, write down the probability that r α-particles and s /3-particles are
given off in this time and show that the probability that a total of η particles of
the two types are emitted is
β-(μ+") Qi + v)n/nl (Cambridge)
19. The probability that any randomly chosen rat from the colony used in a
certain laboratory will show a certain undesirable characteristic is p; it is known
that the value of ρ for the colony is either 0-4 or 0-6 and it is desired to set up a
sampling scheme to decide which value is correct. The procedure is to take rats
one at a time at random and test for the presence of the characteristic; after η
rats have been tested, let r be the number with the characteristic.
Show how to determine limits Ln and U„ such that for known η
Pr (/· < Ln\p = 0-6) к a and Pr (/· > Un\p = 0-4) « β
where α and β are specified.
Use the table of partial binomial sums below to determine Ln and Un for
η = 5,7, 9, 11, 13, 15, 17 where α and β are less than 0-1 and as close to it as
possible. Mark these values on a plot of π — r against r, and explain how such
a diagram could be used in the decision procedure.
326

MISCELLANEOUS EXERCISE 16
J£
3
4
5
6
7
8
9
10
11
5
317
87
10
7
580
290
96
19
2
Partial biiu
9
768
517
267
99
25
4
0
,,
881
704
467
247
99
29
6
1
0
mial sums
13
942
831
647
426
229
98
32
8
1
15
973
909
783
597
390
213
95
34
9
17
988
954
874
736
552
359
199
92
35
Table of ΙΟ3 χ Σ I I (0-4)s (0-6)"-. (Cambridge)
20. At a certain seed testing station it is found that a proportion 0-4 of a certain
type of seed is fertile. By accident the remaining stock of this seed (whose total
amount is very large) is completely mixed with an equal quantity of a second
type of seed which is believed to be completely infertile. If this latter assumption
is true, what is the probability that a seed taken at random from the mixture
will germinate?
Each of seven pots is planted with two seeds taken at random from the mixture.
Six pots eventually produce one or more plants each. Is this result consistent, at
the 5 % level of significance, with the infertility postulate for the second type of
seed? (Cambridge)
21. Derive the probability p(r) of obtaining exactly r successes in η independent
trials, the probability of a success being ρ at each trial. Determine the mean and
variance of this distribution.
Find an expression for [p(r)]/[p(r— 1)] and hence or otherwise find the
conditions that rm must satisfy if p(rm) is such that no other value of p(r) is greater
than/KrJ.
Show that the mode of a binomial distribution differs from the mean by less
than unity. (M.E.I.)

Revision exercise В
1. Find the equation of the line Lt through the origin with gradient \, and
also the equation of the line L2 perpendicular to Lx and passing through the point
(4, -*).
2. In a large university, one-third of the men and one-quarter of the women
read science. If four men are selected at random, what is the probability that not
more than one reads science? If two men and two women are selected at random
what is the probability that not more than one of the four reads science?
What is the probability that, in a mixed group, one man reads science and the
other man and the two women read something else?
3. A square lamina ABCD of side a is held with the corner A on a horizontal
plane. The feet of the perpendiculars from B, C, D on to the plane are В', С", D'
and the angles B'AB, D'AD are <x, β. The angle B'AD' is Θ. Prove that
(i) cos0 = -tan α tan β;
(ii) the area of the triangle B'AD' is %a-(cos2 α cos2 β- sin2 α sin2 /?)*;
(iii) the inclination of the lamina to the horizontal can be expressed in the
ОГт arccos [cos (α+β) cos (α-β)]*. (О & С)
4. If α Φ i(4n+ 1)я, find x if sin (x+oc) = cos (x-oc).
What can you say about χ if α = \(4n + 1)я?
5. Find the sum to infinity, S, of the geometric series
1+7ϊδ+ϊδ+ιονιο+ "■
If SN denotes the sum of the first N terms of this series, find the least value of
N such that S and SN are the same correct to 3 d.p.
6. A tennis match usually consists of either three or five sets, and ends when one
side has won a majority of the sets. If the probability of a side winning a set is p,
and if the result of each set is independent of any previous results, show that the
probability of a match going its full legth is 2pq in the case of a three-set match
and 6/?V in the case of a five-set match (q = 1 —p).
Show that the first probability is always greater than the second, if;? Φ 0 or 1.
7. The function/: R -> R+ is defined by
Λ*)= |*+1| + |2χ-1|+|*-2|.
Sketch the graph off and determine the least value of A*).
8. Call a matrix of the form lx Л
328

REVISION EXERCISE В
which is symmetrical about both diagonals 'super-symmetrical'. If A, B, ... are
super-symmetrical 2x2 matrices and m, n, ... are non-negative integers, prove
that A™B" ... is super-symmetrical.
Does this result hold for 3 χ 3 matrices?
9. Factorize into linear factors the expression 2x2 + 5xy — 3y2—3x+5y—2 and
describe geometrically the set of points {(*, y): 2x2 + 5xy - Зу2 - 3x + 5y - 2 = 0}.
Describe geometrically the sets of points
(i) {(x,y): x*-2y* = 0}; (ii) {(*, у) : x2 + 2y2 = О}.
10. i,j, к are unequal positive integers and x, y, ζ are real numbers. Prove that
x(j-k)+y(k-i) + z(i-j) = 0ox,y,z are respectively the zth, y'th, kth terms of
an arithmetic sequence.
11. Prove by mathematical induction, or otherwise, that
л.13+(л-1).23 + (л-2).33+... + 1.и3 = -knin+V, (л+2) (Зи2 + 6и+1).
(Oxford)
12. Find the inverse of the matrix
(J \ I
Hence solve the simultaneous equations
Г 5x + 3y + lz = a,
\ 3x+4y + 6z = b;
[-x + 2y+ z = c;
(i) when a = -1, b = 2, с = -3; (ii) when a = b = 1, с = 2.
13. What are the first three terms in the expansion of (l-2x)24 in ascending
powers of x?
Find the value of 0-9824, correct to 4 d.p.
14. Show that, whatever value is chosen for k, the equation
(3 + 5/c) x + (2-7/c) y + (5-4k) = 0
represents a straight line through a fixed point A.
Find the particular line of the system
(i) which passes through the origin;
(ii) which is parallel to the у axis.
15. The polynomial P(x) leaves a remainder of x+1 on division by x2-2 and
the polynomial Q(x) leaves a remainder of 2x-f-3 on division by x2-2. New
polynomials R(x) = P(x)+Q(x) and Six) = P(x) Q(x) are defined. Find the
remainder
(i) when R(x) is divided by x2-2;
(ii) when S(x) is divided by x2-2.
329

REVISION EXERCISE В
16. If A is the point (a cos a, b sin a) and В is the point (a cos /δ, 6 sin β), show
that the equation of the line AB is
What is the connection between α and β if AB passes through the origin?
17. Prove that the number of spheres that can be drawn to pass through three
given points and touch a given plane is 2 or 1 or 0, explaining how the three cases
arise.
18. О A, OB, ОС are three concurrent straight lines lying in one plane. Ρ is a
point outside the plane such that the angles POA, РОВ, РОС are equal. Prove
that PO is perpendicular to the plane.
19. Find the matrix of the linear transformation which reflects all points in the
plane x+y+z = 0.
20. ABCDA'B'C'D' is a parallelepiped, with opposite faces ABCD, A'B'C'D'
and edges AA', etc., AB = BB' = B'C, and the mid-points of BB', A'D', A'B'
are respectively F, G, H. Find the ratio in which the plane AD'Η divides the line
FG.
21. Two boxes each contain one hundred cards, numbered 1 to 100. A card is
taken from each box, the numbers on the two cards noted, and they are then
returned to their respective boxes. If the process is repeated fifty times, find the
probability that at no stage a pair of cards bearing the same number will be
drawn. How many draws are needed for the probability to be > £?
22. Prove that lg 15 is irrational. (lg 15 means log1015.) Is it true that lg χ is
irrational <s> χ is not a power of 10?
23. Evaluate the following determinants:
I b-c c-a a-b I I (x-p)2 iy-p)2 (,ζ-ρ)2 I
(i) c-a a-b b-c \; (ii) ι (x-q)2 (y-q)2 (z-q)2 .
| a-b b-c c-a \ \ (x-r)2 (y-r)2 (z-rf \
24. Define the projection of a vector in a given direction. ABCD is a
quadrilateral in which LA = Ζ С = 90°. The feet of the perpendiculars from B, D
to AC are X, Г respectively. Prove that AX = CY.
25. In a large crate of oranges, 100p% are bad. A random sample of ten oranges
produces two bad ones. On the assumption that this is the most likely number
of bad oranges to find in the sample, what can you say about the possible values
ofp?
26. ABCD is a square lying in the plane 2x—y — 2z = 5. If A has coordinates
(1, 1, -2) and С has coordinates (5, 5, 0), find the coordinates of Л and D.
27. The function, /, defined by
f(x) = sin χ + cos χ
has for its domain the set of real numbers {x e R: 0 < χ < π}. Find the range
of/

REVISION EXERCISE В
28. Factorize the expression ax2 + (ac + b—c) x+c(b — c) into two linear factors.
Hence factorize the expressions
(i) a + b-c + bc + ca-c2;
00 a-b + c + bc-ca-c2;
On) asb2 + a2bc + ab2-abc+bc-c2.
29. Describe geometrically the transformation of three-dimensional space
represented by premultiplying the position vector of the point (x, y, z) by the matrix
i 0 -i\
0 1 0 .
-i о i)
Which points are left unaltered by this transformation ? (O.S.)
30. When A and В play chess the chance of either winning a game is always
i and the chance of the game being drawn is always \. Find the chance of A
winning at least three games out of five.
If A and В play a match to be decided as soon as either has won two games,
not necessarily consecutive games, find the chance of the match being finished
in ten games or less. (J.M.B)
31. Solve the equations
0) 2sin2x+sin2x = 0;
00 sin* = cosQ-я-*);
giving all solutions in the interval 0 =g χ «S 2я in each case.
32. Solve the inequality x2-\ < 1
x2-4 " 5
and illustrate your solutions by means of a sketch of the curve
33. The base AB of a triangle ABC is fixed and К is a fixed point on AB. The
vertex С of the triangle moves so that the perpendicular distances of К from С А
and CB are always equal in length. Prove that, in general, the locus of С is a
circle through K.
What is the exceptional case? (Cambridge)
34. A pitcher is taken daily to a well and back. Its chance of being broken on
an outward trip is рг. If it survives the outward journey it has a further chancer
of being broken on the return. Show that on any day its change of being broken
is P, where
Ρ = Pi+Pi-PiP*·
Show that the chance that the pitcher will survive for at least η days is Qn,
where Q = (1 -рг) (1 -p^), and find the chance it will survive (n-1) days but
be broken on the nth day.
If two such pitchers are each taken independently to the well and back daily,
prove that the chance that they survive for exactly the same number of days is
Pli\ + Q). (CS.)
331

REVISION EXERCISE В
35. Find the image by reflection of the point (4, 3,1) in the plane
3x+2y+3z+l = 0.
36. A and В are two towers, В being four miles due east of A. The true bearings
of a flagpole С from A and В are a° east of north and a° west of north
respectively; the true bearings of a second flagpole D from A and В are (a+/3)° east
of north and (μ-β)° west of north respectively. Draw a sketch-plane to indicate
the positions of А, В, С, D.
Assuming that А, В, С and D are on level ground, prove t hat D is 4 sin2 β cosec 2a
miles south of С and 2 sin 2β cosec 2a miles east of C. (O & C)
37. If axi + 3xs + bx2 + cx + 2 is exactly divisible by (χ+ 2) (x2-l), find the
values of a, b, с
38. Sketch the graph of the function defined by
f(x) = x+\x\ (xeR).
In a separate diagram, shade in the set of points satisfying simultaneously the
three inequalities
y>x+\x\, y>\x\-x, x-y+l>0.
What is the maximum value of the expression
E= x+y+l
if χ and у are subject to the three given inequalities?
39. Prove that, if А, В are two non-singular 3x3 matrices, then AB is a non-
singular matrix and (AB)-1 = B_1A_1.
If / 6 3 5\ /1 0 0\ /10 0\
A= 2 5 2 , E,= 0 0 1 , Et= 0 1 0
\-3 4 -2/ \0 10/ \l 1 1/
find Α-1, ΕΓ',Ε^1.
What are the inverses of the following matrices:
/ 6 3 5\
(ii) -3 4 -2 ?
\ 5 12 5/
40. Using the formula for tan (A + B) in terms of tan A and tan B, show that,
if А, В, С are the angles of a triangle, then
tan^ + tan Л + tan С = tan Л tan .В tan C.
Calculate, in degrees and minutes, the angles of a triangle ABC if
tan A: tan В: tan С = 1:2: - 6. (Cambridge)
41. Find the equation of the plane containing the origin, the point (1, 1, 1)
and the point (3,1, — 1). Find also the direction ratios of the line of intersection
of this plane with the plane x+y-z-4 = 0.
332

REVISION EXERCISE В
42. In a game of tennis one point is scored either by A or by his opponent B.
The winner of the game is the player who first scores four points, unless each
player has won three points, when 'deuce' is called and play proceeds until one
player is two points ahead of the other and so wins. If A's chance of winning any
point is | and B's chance is i calculate the chance of
(i) A winning the game without 'deuce' being called;
(ii) a similar win by B;
(iii) 'deuce' being called.
If' deuce' is called, prove that A's subsequent chance of winning the game is f.
Deduce that A's chance of winning the game is nearly six times that of B.
(Cambridge)
43. Prove that " tn\
ДАЛ-2·
If A is a square matrix such that A2 = A, express (A + I)" in the form
α,Α+βΙ.
44. Use the method of mathematical induction to prove that
(fc+l)l (fc+2)l (fc+w-l)l = (k+n)\
1! 2! (л-l)! (k+l) (л-1)Г
45. Show that the equations
Uk + 2)x + 2y + 3z = 7,
| x + (k+2)y + z = 0,
{5x+2y + (6-k)z = 13
have a unique solution for all but three values of k.
Discuss the solution of the equations in the three exceptional cases.
46. Prove the formula cos ЗА = 4 cos3 А-Ъ cos A. (Formulae for cos 2A
and sin 2A may be assumed.)
Substitute χ = i+cos θ in the equation 8x3- 12x2+ 1 = 0 and, with the aid
of the above formula, solve the resulting equation in Θ, giving values between 0°
and 180°.
Hence find the three roots of the cubic equation in x, correct to two decimal
places. (Cambridge)
47. A1A2... An is a regular polygon of side 1 unit. Two (distinct) vertices are
selected at random. Taking as random variable X, the shorter distance measured
along the perimeter between the two points, find S(X)
(i) when η is odd; (ii) when η is even.
48. When л is a positive integer the coefficient of an~rbr in the binomial
expansion of (a+ b)n is denoted by I I. Write down an expression for I J and prove
that . . ...
"- for r= 2.3, ...,*.
By comparing the binomial expansions, or otherwise, prove that, when
> °'b> °' (а+ЬУ- α" < nb{a+ Ь)"-1. (О & С)

REVISION EXERICSE В
49. Factorize the determinants
I x a b с |
(O&C)
I я2 (я+l)2 (я + 2)2 Ι ι χ a b с
(i) b2 φ + iy ф + 2У , (ii) \a χ с b
| c2 (c+1)2 (c+2)2 | \b с χ a
\ с b a x
50. Find the image by reflection of the plane χ = 0 in the plane x+y+z = 4.
51. Using the same axes draw the graphs of у = sin x° and у = 1 — --- for
values of χ from 0 to 360. Estimate from your diagram the roots of the equation
The equation sin x° = mx+c, where m and с are constants, has roots χ = 60,
χ = 330. With the aid of your diagram estimate
(i) the other root of the equation;
(ii) the values of m and с (Cambridge)
52. Prove that, when any polynomial Дх) is divided by (x-d), the remainder
i*№
The remainder, when/(x) is divided by (x— a) {x—b) is written in the form
A(x-a) + B(x-b); prove that
b—a a—b
When αχϊ+βχ^ + γχ + δ is divided by x2— 1, the remainder is Kx where К is
a constant; when it is divided by x2 — 4, the remainder is K. Prove that
a=-/?=-iy=S= -i*. (O&C)
53. Let A(0) and B(0) denote the matrices
/cos0 sin 0\ /cos0 -sin0\
\sin# -cos0/ \sin# cos Θ J
respectively.
(i) If
(ί)-»(ί)
show that the point (A", 1") is the mirror image of {X, Y) in the line у = xtan£0.
(ii)PrOVethat AWA№) = B(0),
where 0 is some angle (to be determined), and hence, or otherwise, explain the
relation between the points (X\ V) and (X, Y) when
(ί)-»(ί)-
(iii) Prove that A(0,) A(02) A(03) = A(03) A(0a) A(0,)
and interpret this result geometrically.
334

REVISION EXERCISE В
54. One side of a hill has the form of an inclined plane, the line of greatest
slope, AB, being due north. A man starts from A and walks to B, the distance
being dx km. At В he walks for a distance d2 km in a straight line, inclined at
an acute angle θ to AB, reaching a point C. The vertical heights of В and С
above the level of A are hx km and /z2 km respectively. Prove that
cos a = ———.
кгаг
The bearing of С from A is φ east or west of north. Prove that
хапф = Ш^ь- (0&0
55. Show that, for all but two specific values of k, the equations
Ix-ky + 2z = 3,
3x+ y-2z= 1,
kx-9y+2z = 7,
represent three planes with a single common point.
Interpret geometrically the two special cases.
56. In how many different ways may a red balls, b white balls and с black balls
(indistinguishable apart from colour) be arranged in a straight line?
Find the coefficient of xlyizi in the expansion of (x+y+2z)12.
57. Solve the inequality I 5x+l I
\х*-х-6\ <
and illustrate your solution by drawing a sketch of the curve
5x+l
58. Three spheres of radii 1, 2, 3 units, touch each other externally. Prove
that a plane which touches each sphere makes an angle arcsin (J VI3) with the
plane containing the centres of the spheres. (O & C)
59. Find all values of x, у and ζ which satisfy the equations
where v-2w = a,
-u +3w = b,
2u-3v = с (C.S.)
60. If I deal cards one by one from a well-shuffled pack, find the expectation of
the number of cards I shall have to deal
(i) to secure a spade; (ii) to secure all the spades.
335

REVISION EXERCISE В
61. ABCDA'B'C'D' is a rectangular box with faces ABCD, A'B'C'D' and edges
AA', etc. AA'D'D is a square of side 1 unit, AB = 2 units and Μ is the
midpoint of CD'. By setting up a suitable coordinate system, or otherwise, find the
acute angle between the planes BMA', AB'D', correct to the nearest degree.
62. Prove that I {a-xf {a-yf {a-zf {a-wf I
Ub-xf ф-уУ (b-ζγ (b-wf\ 0
{c-xf (c-y)2 {c-zf (c-wY \ '
I (d-xy (d-уУ (d-zy (d-wy |
63. If θ denotes the angle between two intersecting lines with direction cosines
/, m, η and /', m', ή respectively, show that
cos# = U'+mm' + nn'.
Prove that the lines χ = y+3 — z+\
and * = KH-4) = H*+3)
intersect. If Ρ is the point of intersection, find the equation of the line through
Ρ perpendicular to both given lines. (J.M.B.)
64. Two jars, one white and the other black, contain a + b balls each; in the white
jar there are a white and b black balls and in the black jar b white and a black.
Single draws are made as follows: at the /-th draw a ball is drawn from the white
or black jar according as the (/·- l)th ball drawn was white or black, the colour
of the ball noted and then returned. If pn is the probability that the nth draw is
white, show that , ,. , ,
{.a + b)pn=b + (a-b)pn_1.
By means of the substitution pn = i+qn,
or otherwise, determine pn, when the jar from which the first draw is made is
(i) chosen at random, (ii) white. What is the probability in the two cases that
both jars have been drawn after three draws? (O.S.)
65. The number η whose digits in the scale of 10 are a, b, c, d in that order is the
same as the number whose digits in the scale of 9 are d, b, c, a in that order; in
other words, we have
103a + 1026 + 10c + i/= 94+92b + 9c+a
and the digits a, b, c, d all lie between 0 and 8 inclusive. Prove that there is
exactly one number л (Ф 0) with this property, and find n. (C.S.)
66. Prove that, if /A 1 0\
иА"-1 Ми-1)А"-2\
А" лА"-1
where л is a positive integer.
If A0 is defined to be I, is the result still true for any integer л?

REVISION EXERCISE В
67. The points Pt and P2 on the surface of the earth (assumed to be a sphere of
radius r and centre O) are at the same (northern) latitude A, and their respective
longitudes are Lx and L2. Show that the length С of the route from Рг to P2 via a
circle centre О (called a great circle) satisfies the equation
cos I —J = sin2 A + cos2 A cos (I^—LJ.
If the separation in longitude is 90° show that either
_ „ . /cos A\ '™° ^
С = 2r arcsin I —— I c
68. If A is a real number, show that the equations
λχ+ y+ ζ = 2,
x+Xy + ζ =-l,
x+ y + λζ =-1,
have a unique solution (x, y, z) for all but two exceptional values of λ.
Show that, for one of the exceptional values of λ, the equations have no
solution and that, for the other exceptional value of λ they have infinitely
many solutions. (O.S.)
69. Determine θ so that the line
lx+ my+ η = θ (Ι'χ+ т'у+rf)
is perpendicular to the line λx+μy + v = 0.
Prove that the perpendiculars from the vertices Au A2, As of the triangle formed
by the lines , . ,. , „ „.
их+пцу+щ = 0 (г = 1,2,3)
on to the sides B2Bz, BZBU ВгВг respectively, of the triangle formed by the lines
λιΧ+μ{ν+νι= 0 (г = 1,2,3)
are concurrent if
(/A + wj/ia) (/2λ3 + »г2/*з) (/3λ! + wa/ti)
= (/jAa + Wl/i3) (/.Л + пцрд (/3A2 + Wa/ta).
Deduce that, in this case, the perpendiculars from the vertices of Вг Вг Bs
on to the respective sides of Αλ Α2 A3 are also concurrent. (C.S.)
70. A hill 0-3 km high is in the shape of a spherical cap, with a horizontal circular
rim, the radius of the sphere being 0-6 km. A man walks up from a point of the
rim to the peak at a steady speed of 1-6 km per hour but never ascending at a
gradient of more than arcsin (4). Find the minimum time the walk can take him,
and sketch roughly a possible minimum path (as seen from above); does it
necessarily have no sharp corners ? (C.S.)
71. In a population of identical elements, an element produces no or one
progeny with equal probabilities i. If the population consists of 0, 1, 2, ...

REVISION EXERCISE В
elements with probabilitiesp0,puP2, ■ ■ · show that the probability P, of r progeny
is given by the coefficient of tr in G(i), where
When Pn = -I*
find the probability that the number of progeny is greater than or equal to N.
By considering G(t) and its derivatives, or otherwise, find r, the expected value
of /·, the number of progeny, and the expected value of (/—F)2. (O.S.)
72. А, В and С are the three angles of a triangle. Show that
I sin Л sin Л sin С I
cos A cos В cos С = 0. (C.S.)
| sin3 A sin3 В sin3 С |
73. If л is a positive integer and ρ a prime number, а„(и) denotes the greatest
integer к such that pk divides л. If η is written in the form
N
η = Еад' (0 < ar <p-\),
show that η— Σ ar
^(пП=~~-. (CS.)
74. .4.BCD is a parallelogram and Ε a point not necessarily in the plane ABCD.
Show that a2 + c2+g2 + h2 = 62 + i/2 + e2, these being the lengths shown in the
figure, and find a relation involving only a, b, c, e, f. (You may use vector
geometry.) (C.S.)
338

Bibliography
GENERAL
Allen, R. G. D. Basic Mathematics (Macmillan). A broad survey of mathematical
ideas—useful as a reference book.
Backhouse, J. K. and Houldsworth, S. P. T. Pure Mathematics (Longmans,
2 volumes). Written some years ago, with traditional 'A' levels in mind—
consequently rather old-fashioned, but workmanlike in its approach and
contains a great number of examples for the student.
Kemeny, J. G., Murkill, H., Snell, J. L. and Thompson, G. L. Finite
Mathematical Structures (Prentice-Hall). An introductory survey of some modern
topics, containing a much more thorough account of the ideas sketched in
Ch. 9.1 and also including chapters on linear algebra and probability.
S.M.P. Advanced Mathematics. (Cambridge, 4 volumes). Specifically written
for the new S.M.P. 'A' level, these books are full of new ideas, presented in
a lively and stimulating manner.
ALGEBRA
Beckenbach, E. F. and Bellman, R. An Introduction to Inequalities (Random
House.) A readable introduction to the subject.
Birkhoff, G. D. and Maclane, S. A Survey of Modern Algebra (Macmillan,
New York). Really a university text book, but beautifully written and
containing much that can be read with profit by sixth-formers. Gives a much more
thorough account of the ideas outlined somewhat sketchily in Ch. 1.
Cohn, P. M. Linear Equations (Routledge and Kegan Paul). A sound
introduction at low cost.
Durell, С V. and Robson, A. Advanced Algebra (Bell, 3 volumes). Written a
long time ago, but these books wear well and still contain a great deal of
valuable reading; they also contain an excellent selection of problems, many of
which are stimulating and challenging.
Ferrar, W. L. Higher Algebra for Schools (Oxford). A good introductory course
in traditional sixth-form algebra.
Maxwell, E. A. Algebraic Structures and Matrices (Cambridge). Carries the work
beyond 'A' level standard; contains some good exercises.
Neill, H. and Moakes, A. J. Vectors, Matrices and Linear Equations (Oliver and
Boyd). Covers the course of linear algebra required for the M.E.I. Ά' level
examinations. Approaches the subject from a rather different standpoint to
that adopted in this book.
Niven, I. Numbers: Rational and Irrational (Random House). A comprehensive
account of the work of Ch. 1.
339

BIBLIOGRAPHY
CALCULUS
Hardy, G. H. Pure Mathematics (Cambridge). A very famous book. Not easy
reading, but repays careful study. A work for the specialist.
Matthews, G. Calculus (Murray). An excellent modern text.
Maxwell, Ε. Η. Calculus (Cambridge, 4 volumes). A comprehensive text, useful
as a work of reference.
Quadling, D. A. Mathematical Analysis (Oxford). An excellent introduction to
university work: presupposes some acquaintance with the calculus.
Scott, D. B. and Timms, S. R. Mathematical Analysis (Cambridge). A university
text, but contains much good reading for the serious sixth-form student.
Siddons, A. W., Snell, K. S. and Morgan, J. B. A New Calculus (Cambridge,
3 volumes). A standard text-book; volume 3 is a particularly good
introduction to advanced work.
Toeplitz, O. The Calculus, A Genetic Approach (Chicago). A stimulating
approach to calculus by tracing its historical development.
GEOMETRY AND TRIGONOMETRY
Barton, A. Introduction to Coordinate Geometry (London). A good account of
two-dimensional coordinate geometry.
Conn, P. M. Solid Geometry (Routledge and Kegan Paul). See Conn: Linear
Equations.
Forder, H. G. Higher Course Geometry (Cambridge). Full of excellent material—
fascinating for the keen geometer.
Kazarinoff, N. Geometric Inequalities (Random House). Stimulating reading:
contains a good selection of problems, with hints on solutions.
Macbeath, A. M. Elementary Vector Algebra (Oxford). Despite its title, this is
really an introduction to vector geometry. Very clearly written.
Maxwell, E. A. Coordinate Geometry with Vectors and Tensors (Oxford). A
good introduction to three-dimensional coordinate geometry—takes the
subject well beyond 'A' level.
Nobbs, С G. Trigonometry (Oxford). A good introduction to'A' level work.
Tuckey, С A. and Rollett, Α. Ρ Trigonometry (Murray). A sound introductory
text, with many interesting digressions.
PROBABILITY
Feller, W. An Introduction to Probability Theory and Its Applications, Volume 1
(Wiley). An advanced text, but highly recommended for the serious post-
Ά' level student.
Mosteller, F. Fifty Challenging Problems in Probability (Addison-Wesley). Some
good material here for the problem addict.
Mosteller, F., Rourke, R. E. K. and Thomas, G. B. Probability and Statistics
(Addison-Wesley). A very elementary introduction to probability—wordy at
times, but thorough.
Parzen, E. Modern Probability Theory and its Applications (Wiley). A very
thorough introduction, carrying the subject well beyond Ά' level.
340

BIBLIOGRAPHY
Whitworth, W. A. Choice and Chance (Hafner). Published seventy years ago,
the bookwork is distinctly old-fashioned, but it ends with a delightful collection
of problems, some distinctly tricky. Excellent bedside reading for the inveterate
problem-solver; a key is published to prevent insomnia.
BOOKS FOR GENERAL READING,
COLLECTIONS OF PROBLEMS, ETC.
Abramowitz, M. and Stegun, I. A. Handbook of Mathematical Functions (Dover).
Over a thousand pages of mathematical tables (most to sixteen or eighteen
significant figures), formulae and graphs, together with extensive notes.
A marvellous reference book and not at all expensive for what it contains.
Ball, W. W. R. Mathematical Recreations and Essays (Macmillan). First
published in 1892 but it still provides excellent reading material.
Bell, Ε. Τ. Men of Mathematics (Pelican, 2 volumes). A racy and amusing account
of the lives of great mathematicians.
Boyer, С. В. A History of the Calculus (Dover). A readable account of the
subject.
Burkill, J. С and Cundy, Η. Μ. Mathematical Scholarship Problems (Cambridge).
A good collection of problems at university scholarship level.
Cadwell, J. H. Topics in Recreational Mathematics (Cambridge). Digressions on
the underlying mathematics in fifteen different problems: it makes very good
reading.
Courant, R. and Robbins, H. What is Mathematics? (Oxford). A famous book,
which discusses a number of mathematical situations in some depth.
David, F. N. Games, Gods and Gambling (Griffin). An excellent book on the
early history of probability.
Dorrie, H. 100 Great Problems of Elementary Mathematics (Dover). Extended
discussion of 100 problems of considerable historical interest, together with
their solutions—excellent value.
Gardner, M. Mathematical Puzzles and Diversions (Pelican). Entertaining
reading in hexaflexagons, polyominoes, the games of nim and hex, etc.
Lockwood, Ε. Η. A Book of Curves (Cambridge). A delightful collection, with
historical notes and comments on the geometrical properties. A good book for
browsing.
Maxwell, E. A. Fallacies in Mathematics (Cambridge). Will be read with interest
and enjoyment by any mathematician.
Moakes, A. J. Core of Mathematics (Macmillan). An elementary account of
mathematical structure and methods of proof: easy to read but stimulating.
Newman, J. R. World of Mathematics (George Allen and Unwin). An extensive
anthology of mathematical writings—good browsing material.
Polya, G. How to Solve It (Doubleday). An analysis of the art of problem
solving
Polya, G. Mathematics and Plausible Reasoning (Princeton, 2 volumes). The
title is self-explanatory. Compulsory reading for the serious student, while the
more frivolous will be able to skip through with profit.
Salkind, С. Т. The Contest Problem Book (Random House). A collection of
multiple choice problems, not all of which are easy.
Sawyer, W. W. Prelude to Mathematics (Pelican). A book for the layman,
341

BIBLIOGRAPHY
but can be read by the specialist with profit. Strongly recommended to all
sixth-formers.
Smith, D. E. A Source Book in Mathematics (Dover, 2 volumes). Contains
excerpts from the writings of great mathematicians of the past—all within the
understanding of the sixth-former.
Weston, J. D. and Godwin, H. J. Some Exercises in Pure Mathematics
(Cambridge). A fine collection of problems, based on the Welsh Ά' level
(alternative syllabus). Contains detailed solutions.
342

Answers
CHAPTER 1
PAGE
2 Ex. 2. 100101001. Ex. 3. 100, remainder 111.
Exercise 1 (a)
1. (i) 10; (ii) 1110; (iii) 10000; (iv) 101.
2. 100010011, 101012, lie.
3. 8e242. 4. 3e, remainder 19.
5 Ex. 7. Yes, unless a2 = b2 φ 0.
7 Ex. 12. (i) 7/9; (ii) 134/333; (iii) 283/45.
Ex.13, (i) 11-11; (ii) 0-101101; (iii) 100-1001;
(iv) 0-li010001111010111000.
Ex.15, (i) 0-3; (ii) 1-714285.
Exercise 1 (ft)
9 4. (i) 0-3; (ii) 0-Oi.
6. (i) True; (ii) True; (iii) False; (iv) True; (v) True (provided rational
number is not zero); (vi) False.
11 Ex. 17. (i) 2V2; (ii) 3V2; (iii) 3V6; (iv) 5V10; (v) 42V3.
Ex. 18. (i) 11 + 4V6; (ii) 3V6; (iii) 2V3; (iv) 8.
Ex. 19. (i) iV3; GO i(2 + V2); (iii) «2 + V2); (iv) 3 + V6; (ν) 7-4Λ/3.
Exercise 1 (c)
12 1. (i) 5V2; (ii) 11V3; СЮ 20V6; (iv) 8V3; (v) 22V3.
2. (i) 6-2V5; (ii) 14-4V6; (iii) 10+6V3; (iv) 9V3-11V2;
(v) 28-16V3.
3. (i) 2V2; (ii) 3V5; (Hi) 5V3; (iv) Vl3; (v) (3V3)/2.
13 4. (i) V5/5; (ii) V21/3; (iii) 2(V7 + 2)/3; (iv) (13 + 4V3)/11;
(v) V6 + V3-V2-2.
343

ANSWERS
13 5. (i) V5-1; (ϋ) V6-2; (iii) 3V2+2V3; (iv) 3V3-2V5;
(v) V5 + V2-
6. (i) 4; (ii) ±2; (iii) -5.
7. (i) «2+V2-V6); (H) i(VlO+V5-V2-3).
9. (i) 0-44721 (5 d.p.); (ii) 0-27395 (5 d.p.).
10. 0-98560 (5 d.p.). 11. i(3 - ^9 + ^3).
12. (i) (V(*+ D+V(*-1)}/2; (ϋ) (*+2) V(*+ Ό/(*+1);
(iii) (V(2x-a)-V(2x+a)}7(4х2-Д2)/(4х2-я2);
(iv) W(*- D-(*-2) V(*+ D-2(x-l)}/(x2-2x).
14 Ex. 20. (i) {1, 2, 3, 4, 5}; (ii) {1, 4, 9}.
15 Ex. 21. (i) Odd integers; (ii) integers 5= 2; (iii) multiples of 10;
(iv) the integer 6; (v) 0.
16 Ex.26. jxeU: | e uj.
Ex.27. ©И; (ii) 0; (iii) U; (iv) Λ; (ν) Λ; (vi) 0; (vii) 0;
(viii) U.
EX. 28. (i) {a, e); (ii) {a, A, d, e); (iii) {C}.
Ex. 29. (i) {2}; (ii) {2}; (iii){-},2}; (iv) Ш, 2};
(v) {-V2,-i, V2, 2}.
18 Ex. 31. (i) χ < -2; (ii) * < 1; (iii) * < £.
£Ьотсда7(</)
20 2. (i) Λ; (ii) {* e R: 3 < χ < 4}; (iii) {x e R: χ < 4}.
3. (i) {x e Ζ: χ < 0}; (ii) {x e Z+: x/6 e Z+};
(iii) {xeZ:x^ -3 or x? 3}.
4. (Ϊ) {xeR:2si x< 3}; (ii) {xe Д.* >3};
(iii) {xeR: -1< x< 3}; (iv) {xeR: -1 =g x« 3};
(v) {xeR:x > 3}; (vi) {xe Д: л: < -1 or χ =ϊ 3};
(vii) {хеД:3 < л: < 4}.
21 5. (i) {хеД:0< л: < 1}; (ii)0; (iii) {xe R: 1 < x < 2};
(iv) {«i: -2< x< -1}.
6. (i) {* e R: χ < - 5}; (ii) {χ eR:x> 2}; (iii) {x e R: χ < 7};
(iv) {xeR:x > 0}; (ν){*εΛ:χ<0 or x > 1}.
7. (i) {xeR:x< -3 or χ > 1}; (ii) {xe R: x< -i or χ > i};
(iii) {хеД: -| < χ < 3}; (iv) 0; (v) {xeR: -1< x< 5}.

ANSWERS
21 8. {xeR: -3 < χ < -\ or χ > 2}.
9. (i) 1; (ii) 3; (iii) \. 10. (i) Negative; (ii) positive.
11. χ < -4; -2 < χ < 1; χ > 3.
12. 10. 13. 12. 14. 1. 15. 600.
16. 20m. 17. fm2; ^=lm, AD = im.
Miscellaneous Exercise 1
I. 1111000,· Seven: ip, lp, 2p, 4p, 8p, 16p, 32p.
22 2. a = 2, b = 3. 6. a = 1, * = 4, с = 3.
8. 1-1/V7. 9. (i) 10; (ii) f. 10. Unaltered.
II. (Expressions not unique) (i) An C; (ii) A' n Л'; (iii) С п Л;
(iv) ВП(А'\) С).
23 12. к = -7, α = 8, * = \; к = 3, а = -2, Ъ = -2.
13. (i) Expressions equal; (ii) strict inequality.
14. 1/162,4/27.
CHAPTER 2
28 Ex. 5. (i) 8 units N; (ii) 3 units W; (iii) 4V2 units NW;
(iv) 2V17 units N 14° 2' E; (v) 10units N 36° 52' W.
Ex. 6. χ = (2a + b)/7, у = (a-3b)/7.
Ex. 7. (i) V7 units N 19° 6Έ; (ii) V3 units S 30° E;
(iii) 1 unit S 60° E; (iv) 41 units S 40° 54 W.
Ex. 8. (i) 2V2 units at 45° with downward vertical in plane through SN
line;
(ii) 3V2 units at 45° with downward vertical in plane through EW line;
(iii) 6V3 units at 45° to Ν, Ε and upward vertical;
(iv) 6V3 units at 45° to N, W and downward vertical.
Ex. 9. (i) 2a+b; (ii) 2a-b; (iii) a + b; (iv) a-b.
30 Ex. 12. χ = b + ia, у = b-^a, a = x-y, b = i(x + y).
30 Ex. 13. Yes, no.
Exercise 2(a)
1. (i) За + b; (ii) 8a + b; (iii) -5a+5b, (iv) -2a+6b.
2. a= (u + 3v)/7, b = (2u-v)/7.
345

ANSWERS
30 3. (i) 3a + 2b; (ii) -a + 2c; (iii) 2a + 5b + 5c; (iv) 6a + 2b-3c.
4. a = K"+v), b = (-u-4v + 3w)/9, с = (-7u-v+3w)/9.
5. (i) 2 units E; (ii) 3 units S; (iii) 5 units N 36° 52' E;
(iv) V2 units SE; (v) 5 units N 36° 52' W.
31 6. (i) 13 units S 67° 23' E; (ii) 13 units N 67° 23' E;
(iii) V29 units N 21° 48' W.
7. (i) 1-53 units N 22° 30' E; (ii) 2-8 units S 59° 38' E;
(iii) 1 -47 units S 16° 20' E; (iv) 7-4 units N 67° 30' W.
8. -20х-10л/2у.
9. λ = 2 sec 70°, μ = -2 tan 70°; λ' = μ' κ, 1-02.
10. 9units:(i) 12° 45'; (ii) 125° 28'.
11. OP:(i) a + b; (ii) 2a + c; (iii) 2b-c.
OQ:(i)2a + 3b; (ii) 5a + 3c; (iii) 5b-2c
OR:(i)3a + 4b; (ii) 7a + 4c; (iii) 7b-3c.
PQ = a + 2b; RQ=-2b + c.
12. iu + ^v, fu + Jv, u + v, w=-Ju-^v.
13. b-a, -a, -b, a-b.
14. a-2b, 2a + 2b+2c, a + 2b+2c, b-2c, -a-b-4c.
32 15. -a + 2b + 4c, a + b + 4c, a+2b + 2c;
(-3ED' + 2EG+2EF)/7, (ED'-3EG+4EF)/7, (ED' + 4EG-3EF)/14.
16. (i) j + 2k; (ii) i + j + 2k; (iii) i-j-2k; (iv) 2i + j + 2k;
(v) i + k; (vi) i+j + k.
17. (i) aq-ap; (ii) ap-aq + ar.
18. (i) (4u-2v)/3; (ii) (4v-2u)/3; (iii) 2(v-u).
19. (i) u-iv-iw; (ii)i(v-W); (iii) i(ii + w), (iv)i(v-u).
21. AC-2AD', 2AC-AC.
22. bV2-a, b-aV2, -a,-b, a-bV2, aV2-b.
35 Ex. 16. α + β= 1.
Exercise 2{b)
40 1. 2b-a,K2c + a), K-a+3b+c). 5. A(b + c), A(2c + a).
6. lb, i(a + b), 2a, О A = AX.
41 7. |a + ic,ia+|b; χ = }(2a + 4b + c).
346

ANSWERS
PAGE
41 8. Ja', |b+d + a', b+-£d + a'; the point of trisection of DD' nearer D'.
9. u + v, \(y + w), - i(v + 2u); point A' on WO produced such that
ox = $wo.
10. 3:1. 11. 2i + 3j, 6i + 4j, 6j-i.
42 15. i |.
Miscellaneous Exercise 2
1. b+ 2 cos 72° a, a+ 2 cos 72° b.
3. (i) Parallelogram; (ii) equal diagonals; (iii) rectangle.
43 14. 3:1:2.
15. {αδ2 - A)/(7i ^ - У2 <U (/07, - <*72)/(7ι <?2 - 7s <*i)! c, d not parallel.
CHAPTER 3
Ex. 4. - 2, 6.
48 Ex. 6. (2,1). Ex. 7. β, - J), (¥, - i1). Ex. 8. (1, 1, 2).
Ex. 9. (0, 4) Ex. 10. (-2, 3, 9), (0, -1, -5). Ex. 11. (f, ψ).
Ex.12, ίΐ + j + ik, (|,1, i).
50 Ex. 13. 2V5, V6, V(2a2 + 3/t2). Ex. 15. (2, 4, 3).
Ex. 16. (0, 0, 0), V2.
Exercise 5(a)
H:!)HiH (J-
.«(Ii); CO |
4· <»»- (J· '-HiW-(J·'- ©·-(-;)·
» PPM 347

ANSWERS
» 5· (Ι)· (-1)· (-!)· Θ-
411 (:)· Μ)-
7. (i) (3, 5); (ii) (1, 3); (iii) (}, -*); (iv) (i -i -j);
(ν) (α,Ο, -*).
52 8. (i) (2, 6), (3, 8); (ii) (-1, Щ, (Ο, ψ); (iii) (e+ b, b-a), (a, *);
Gv) & 1, -i), (1, f-, -V); (v) (а-i*, fa+i*, ia+*),
(a-ib, -ia+ib,ia+b).
9. (i) V8; (ii) V40; (iii) VlO; (iv) V51; (v) J(8a2+4b*).
10. (5, 3), (1, -3). 11. V.
12. (2, -1, 3), (4, i, i). 13. 20. 14. 1:2.
55 Ex. 18. (i) lx-y-Ί = 0; (ii) 2x+3.y + 5 = 0.
Ex. 19. (i) 5x-y-13 = 0; (ii) x+y-1 = 0; (iii) 3x-2.y+7 = 0;
(iv) y-2 = 0.
Ex.20. (0 2; (ii) -1; (iii) -f.
57 Ex. 23. 3x-.y-22 = 0, x+2^-5 = 0, (7, -1).
Ex. 24. 11*+5^+1 = 0. Ex. 25. x+3y-l = 0.
Ex. 26. 3V29.
Exercise 3(b)
1. (i) 2x-y-2 = 0; (ii) 2x+.y+3 = 0; (iii) x+2y-l = 0;
(iv) ax+by-(a2+b*) = 0.
2. (i) 3x-y-5 = 0; (ii) 7x-y+ 11 = 0; (iii) y+1 = 0;
(iv) 3bx+2ay-5ab = 0.
3. (0 I; GO -i; (Ш) 5; (iv) (b+a)Kb-a).
4. (i) 2x-.y-7= 0, x+2^-1 = 0;
(ii) 5x + 4.y-ll = О, 4Х-5.У-17 = 0; (iii) x-3 = 0,y+l = 0. 2V5.
5. (i) (3, -2); (ii) (-(a3 + *W+n -ab(a-b)/(a*+b*)).
6. (i) 43x+3.y = 0; (ii) lx-y-2 = 0; (iii) 10x-6^-9 = 0.
58 7. (-3, 10). 8. (5, -9), (3, -1), (1, 3). 9. (2, -3).
10. (- 5, 4), (- 5 ± 2V21, 0), (0, 4 ± 5V3).
11. x-Xy = 1 - 2λ; a circle on (- 2, 1), (1, 2) as diameter.

ANSWERS
58 12.3:4. 13.(3,-4). 14.(2,1). 15. Lies on AB.
16. (i) 2x+3y+6 = 0; (ii) 2х+Ъу-6 = 0; (iii) 2х-Ъу-6 = О;
(iv) 2x-3j>-12 = 0.
62 Ex. 27. (i) ζ = 0; (ii) ζ = 2.
63 Ex. 31. 3.y-5z = 0. Ex. 32. (2λ + 3, -λ-2, -3A-5), (1, -1, -2).
Ex.33. Kx+3) = y-1 =-«z-l).
Exercise 3 (c)
1. (i) [1, 1, 1], [1/V3, 1Д/3, 1/V3]. The direction ratios only are given in
the remaining seven parts; (ii) [1, -2, 1]; (iii) [-2, 2, 1]; (iv) [1,0,-1];
(v) [-1, 1, 1]; (vi) [1, 2, 3], (vii) [0,a,b+c]; (viii) [1, 0, -1].
2. (i) x+y+z = 1; (ii) x + y-2z = 5; (iii) 2x+y-z = 4;
(iv) 3x-y-2z = 4; (v) 5x+3y-2z = 2; (vi) 5x-z = 2.
3. (i) ix = y = -iz- (H) x-1=y-2 = -i(z+iy,
(iii) }(*-!) = -40,-1) = -^+!); (iv) %x + 2) = -i(z-3),y=l;
(v)x=l, ζ =2; (vi) &x-a)=y-2a = z-3a.
4. (i) 3x-3y + z = 0; (ii) 5x-4y-l = 0; (iii) 3^-5z-3 = 0.
5. (i) 7x + 3.y+5z = 0; (ii) 4x-4.y-5z+ 5 = 0; (iii) llx-^+5 = 0.
64 6. 3.y+2z-7 = 0; 7x-4^-5z = 0.
7. 2x+.y+2z+7 = 0.
8. (3λ+8,λ+1,1), (5, 01). 9. (1,1,0).
10. x-2 = i(y-l)= -(z-1), (3,3,0).
11. [1/V5.0, -2/V5], (1,-4,3).
13. (-7, -7, 4), x-4.y-.z-17 = 0.
14. (l,l,3),3x-4y-2z+7 = 0.
65 15. (1, -2, -1). 16. (-3, -4, 2). 17. (7, 5, 0).
Miscellaneous Exercise 3
2. 2:1, 5y. 4. mx-y+ib-am) = 0.
66 8. χ = -y = \z.
10. (i) Planes form a prism;
(ii) planes all pass through the line χ — ζ + 1 = 0, у = 4.
11.2:1; 2:1. 12. 18λ/ί+2λ-5/* = 0.
67 14. α(^+ у)/(д - α). 15. χ + ζ V3 - 2 = 0.
ra-a 349

CHAPTER 4
PAGE
68 Ex. 1. 3,6,6. Ex.2. 8, -1,5.
69 Ex.4. 1, -1,0.
Exercise 4(a)
1. 10, 2, 3-J, 82. 2. 1, -9, 26, -24. 3. 0.
4. 3, 4, -5. 5. 7, 20, 22, 5. 6. 4, 0, -6, 2. 7. 3, -7, -6, 0;
3. 20, 33, 0; -If, -1, 1 8. 1, k, -Ik2, -k\
70 11. -9, 27,3.
Exercise 4(b)
72 1. (i) 62; (ii) 8; (iii) -2; (iv) -241; (v) Ц; (vi) 4*.
2. (x-2)(x-3)(x+l).
73 3. (i) (2x+l) (x+4)(x-l); (ii) (x+l)(3x-4) (4x + 3);
(iii) (x + 5) (x-2)(2x+l); (iv) (x+3) (^-2); (v) (x-2) (xHx+l);
(vi) (x+l)(x-3)(x-5); (vii) (x-2) (x2 + 4x + l);
(viii) (2x-l)(2x + l)(2x + 3).
4. As for Qu. 3 except (iv) (x + 3) (x - V2) (x + V2);
(vii) (x-2)(x + 2-V3)(x+2 + V3).
5. 2. 6. -5.
7. -6i 8. 4, -5. 9. 2, -2.
11. (i) (x-l)(x-2)(x* + 6x+l);
(ii) (x-l)(x-2)(x + 3 + 2V2)(x+3-2V2).
12.0,4; 2(x+l)(x-l)(x-2).
Exercise 4(c)
75 6. x3-x2 + 3x + 2. 7. 2x3-5x2 + x+l.
8. x3-x-5.
Miscellaneous Exercise 4
76 2. Я =-2, 5 = -3.
3. (i) l8x + 6; (ii) 1261X + 2777. 4. P(b)/(b-a), P(a)/(a-b).
5. (b + c)(c + a)(a+b). (i) 30 + 2tf-3/-)(-3/?+9+2r)(2;>-39 + /·);
(ii) 2, 3, 13, 47, 73.
350

ANSWERS
PAGE
77 8. (i) -(b-c)(c-a)(a-b); (ii) (b + c) (c + a) (a + b);
(iii) (a + b+c)(b-c)(c-a)(a-b).
10. x^ + Sx'+Sx+S.
CHAPTER 5
78 Ex. 1. wi.
80 Ex. 3. 2, 3, -12, - 13; 0, (3 ± V29)/2.
Ex.4. 2, 3, -12, -13; 0.
Exercise 5(a)
81 1. -1, -1, 5, 23, -1. 2. Ц, 2$, 3$.
3. {y e R: у > 0}, {у e R: у < 0}.
82 4.-1,1. 6. 2i, {zeR: 2=g z=g 5}.
9. (i) 0, 0; (ii) {yeR:-h^y<2}; (iii) {(3 - V5)/2, (3 + V5)/2};
(iv) 0.
10. {yeR: -4V2< *<; 4V2}. 11. f^iy) = 1 +jA
12. i. 14. (i) No; (ii) no. 15. Reflection in χ = у.
Exercise 5(b)
15. χ < -KV13+1) or χ > -KV13-1).
16. -V3 < x < -V2 or j2 < χ < V3.
17. л: < li or * > 3. 18. χ > 2|.
19. χ < -i or * > If. 20. * < -л/2 or χ > V2.
21. x<0 or li<x<l 22. χ < -l-\ or 34 < x < 5.
351

CHAPTER 6
PAGE
87 Ex. 1
■ π π 5π 3π 5π 11π
'3'4'Ϊ2'~4'~4'~6~'
Ex. 2. 45°, 150°, 120°, 40°, 105°, 36°, 48°.
Exercise 6{ά)
π π 7π 5π Ίπ 3π 7π \\π 5π 19π
' 9'4'Ϊ2'~6~'~6~'Τ'Τ'ΊΓ'Τ'ΊΓ'
2. 30°, 60°, 135°, 75°, 210°, 810°, 54°, 900°, 50°, 63°.
3. 0-445, 1-244, 2-552. 4. 27-5°, 37-2°, 124-9°.
6. 6080 feet—the 'nautical mile'. 7. 18-5 cm.
8. (i) 10-5 cm; (ii) 43-3 cm2; (iii) 9-1 cm2.
9. 1r sin θ, 7Γ/-0/9Ο. 10. 1-7 m, 8 m2.
Э 11. 9-6 cm, 26-5 cm2.
12.(i)216°; (ii) 47-1 cm2; (iii) 6-24 cm.
Exercise 6{b)
\ 1. -i, -1, 2, -V3/3, -V2/2, 2, 0, V3/3, -J2, -2V3/3.
2. V2/2, -1,-2, -V2/2, V3/3, -2, V3/2, -2V3/3, -V3/3, -V3/2.
3. -0-5736, -0-6157, -0-8391, 3-0716,0-3420.
. ... 7л- 11Я- 3π 7тг 2тг 4тг тг 7тг
4.(0Т,—; (п)т,т, (ш)т>у; (.ν)-,-
, я- 2π 4π 5π \3π Ππ 3π Ίπ
(ν)3-,τ,τ.τ; (νΟ^,^; (vu)T,T;
, ...ч π 5π 137Γ 177Γ 25π 29π ,. ч 5π , ч п . , „
«""'Τί-ίί-ΊΪ'ΤΓ-Ίί-Τί-· «Τ· W<M».».i».an
<·?■ «Π·?·?·
«ν)--,»,»; ,ν,-ϊΛ" W-J.Ji W0 -£-££''*
(viii) no solution; (ix) --; (χ)
12' 12' 12"
' 3π 7π

ANSWERS
94 6. (i) 3, 1; (ii) 4,0; (iii) 5,1.
7. (i) -2V2/3, -V2/4; (ii) V5/5, -2V5/5;
(iii) -i, 2V2; (iv) -V15/15,-4V15/15; (v) -4V15/15,-V15/15.
95 11. -A, H. 12. 3 + 6V10/10.
13. (i) 45,123-7, 225, 303-7; (ii) 70-5, 289-5; (iii) 45, 225,153-4, 333-4;
(iv) 38-2, 141-8; (v) 36-9, 143-1, 216-9, 323-1; (vi) 30, 150;
(vii) no solution; (viii) 0, 63-4, 180, 243-4, 360.
14. (i) 9*2 + 4(y-3)2 = 36; (ii) x2+y* = 2; (iii) 2*2-2*y + 5.y2 = 9;
(iv) *2(16-j>2) = 9.y2; (ν) yil-y) (1 + *)2 = 1.
15. (i) (sine,0); (ii) (±1,0); (iii) π/ω seconds. 16. a. = \n.
96 19. 4, li (i) 0; (ii) 2.
98 Ex. 12. arccot: R -> {y e R: 0 < у < π};
arcsec: {xeR: \x\ > l}^ {y eR: 0< y^ π,y+ \ττ).
Ex.13. &r,in, -\π.
Exercise 6{c)
99 2. 0, |7Г, тт. |тг, 2тг. 3. 35-2. 4. 65°.
100 5. 35-3, 144-7. 6. 0-4, 3-32, 5-14. 7. 0-90, 2-25.
15. (i) V3/2; (ii) V3/2; (iii) V3/3; (iv) -V2/2; (v) \; (vi) 3;
(vii) 4; (viii) -V3/3.
16. (i)x/V(l-x2)l GO */V0+*2)· 17. V2/2.
18. x, |*| > 1. 19. *2+/ = 1.
Miscellaneous Exercise 6
1. 0-967.
101 2. 2a2(>-V3/4). 3.11,-1. 4. 0-73, 2-41, 3-99, 5-43.
5. 54-4 cm. (i) 12-25 cm; (ii) 75-5°; (iii) 13-2 cm.
6. 0. 7. sin * Φ ± sin у
8. (j>,3» or (Ця.з». 9. (*2-/)2= I6*y.
102 11. 0-887Γ, 0-787Γ, 0-69тг, 0-60я, 0-50я.
12. 0-467Г, 1-бЗтг, 2-317Г, 3·95π, 4тг.
13. 2ηπ + α., ηπ+α..
14. (i) i(3n+l)n; (ii) §(3я+1)я- or 2кл. 16. 0-39тг, 2-3 cm.
353

ANSWERS
CHAPTER 7
PAGE
103 Ex. 1. 120, 720, 5040.
104 Ex. 2. 12, 60, 720. Ex. 3. 120, 6.
105 Ex. 4. 10, 20. Ex. 5. 1. Ex. 7. 364, 165.
106 Ex. 8. 210. Ex. 9. 510.
Exercise 7(d)
1. 120. 2. 870.
107 3. 665280. 4. 216. 5. 120, 48, 30. 6. 495, 135.
7. 5005, 420420. 8. 34650. 9. 120, 85.
10. (i) 48!/(9!39!); (ii) 48! 16/(9! 39!). 12. \n\, 840.
13. 30, 12. 14. 19958400. 15. i«(«-3). 16. 90.
108 17. n\/(plq\r\),n\/{3(n-2p)l(pm. 18. \n{n-1).
109 Ex. 10. {##, Я71, TH, TT}, {OH, 1#, 2#}, {coins fall the same, coins fall
differently}.
112 Ex. 20. i Ex. 21. A. Ex. 22. зтт-
Ex. 23. (i) 132(39!)2/(27! 52!); (ii) 13 (39!)2/(27! 51!).
Ex.24, (i) I; (ii) i; (iii) i.
Exercise 7(b)
1. 64, 20, A.
113 2. {■ά,^,ϋ,ϋ,ϋ,ή,ή}.
3. (i) i; (ii) A- 4. Ш. 5. Ш-
6. Pr (0) = A, Pr (/·) = (10- /·)/50, r = 1 to 9. 7. &-.
8. (i) τ^; (ii) A.
9. {2, 3, 4, ..., 20}, {тк, т^, jh, .·., tSit, т№, ifo, ■·■> ue}l Ϊ&.
ю. Α.
115 Ex. 26. A. Ex. 27. A- Ex. 28. (i) i; (ii) i; (iii) £; (iv) |.
Ex. 29. fg.
119 Ex.31, (i) i; (ii) A; (iii) A. Ex.34. Ш
354

ANSWERS
Exercise 7(c)
PAGE
119 1. &&, £ft. 2. 0-784. 3. Ш-
4. 0) Α; (ϋ) А; (Ш) A. 6. *, if. 7. if.
120 8. (i) H; (ii) if. 9. 5!, 20, nfe. 10. rir.
11. fi 12. At least 12. 13. Ц. 14. 0-22.
16. if. 17. 13 times.
121 20. 0-956. 21. (i) Α; Α; (ϋ) А, те-
123 Ex. 35. |,f. Ex.36. iV, i
124 Ex. 37. -4^r. Ex. 38. й, зтг, зтг.
Exercise 7(d)
125 1. #r. 2. 0-1997. 3. Д. 4. J, A, #T. 5. ,.
126 6. !^W; 7. A. 8. W, ff, A.
9. p/(p+q-pq).
Miscellaneous Exercise 7
1. (0 Η; (ϋ) aV; 0») 0-065; (iv) 0-036.
2·^!' °' 2(^2)!' ЗОРз)Т
127 3. No difference in either case. 4. Approx. y. 5. i.
6. Replacement yj; no replacement ^g.
7. Probability of securing no prize greater for B, but this is compensated
by greater probability of two prizes.
9. (i) 1; (ii) f. 10. (2'-l)/3-i.
128 11. (i) mlim + n)· (ii) m(m- 1)/{(т + л) (м + л- 1)}; (iii) n/(m + n- 1);
(iv) m/(m+ri).
12. т¥г, A.
13. i (i) AV, ift, i¥5, тЬ; GO rb, тй, A, t^V; Й.
14. (i) 0-09; (ii) 0-111; (iii) 0-336; (iv) small; (v) no.
16. (Λ + λ)(Α + Λ)(Λ + λ) = A2.
17. it, 0, TT; exactly and oppositely synchronized; i, 0, J; J, ^, J.
18. j&.
355

PAGE
128 19. λ, (1-9)(1-<74);!λ, (1-9)9*; |λ, (1-9)9; |λ, (1-<?)<72;
}λ,(1-9)<Ζ3;ΊΛ,94.
20. 1 - (л-1) Ι (л»- лг) Ι и—VCi" - 1)!-
CHAPTER 8
131 Ex. 1. (i) 0, 1, 4, 17; (ii) 2, 5, 16, 65.
Ex. 2. (i) 1,2,4,8; (Ю 1, 2, 3, 4; (iii) 0, 4, 21, 100; (iv) 1, 2, 5, 17;
(v) 1-5, 1-4, 1-41, 1-414, (approx.).
Ex. 3. (i) 3, 6, 9, 12, 15, 18; (ii) -2, 1, 4, 7, 10, 13;
(iii) 1, 2, 4, 8, 16, 32; (iv) 0, 6, 24, 60, 120, 210; (v) 0, 2, 0, 2, 0, 2;
(vi) i, i, А, А, то A-; (vii) 1, i, 1, 4, 25, 216;
(viii) -1, 5, -1,9, -1, 13.
132 Ex. 5. 1, 1, 1, 1, 25.
Ex.6, (i) 1 + 4+9; (ii) 0+2 + 6+12; (iii) 1 + 6+21 + 60;
(iv) -1 + 2-3 + 4-5; (v) 2+3 + 10+29; (vi) 3 + 0-1+0+3 + 8;
(vii) 0+i + i+i + l
133 Ex. 7.(i) 18; (ii) 14; (iii) 100; (iv) 6; (v) lA; (vi) -15;
(vii) 31; (viii) A;
Ex. 8. (i) Σ /·; (ii) Σ (-1)^1/·; (iii) Σ lr; (iv) Σ Τ-1;
(ν) Σι-1; (vi) Σζ-e+l); (νϋ) Σ(-1)'+ΐ; (viii) Σ л
Ex. 11. en. Ex. 12. Ъг- 2, Ц3г~ Ό, 4- 2/·.
Ex. 13. (i) 46; (ii) 35; (iii) 86; (iv) 40; (v) 100.
134 Ex. 14. (i) Σ(3/·-2)(3/·+1)(3/·+4); (ii) Σ 6/·2(4/·+1); (iii) Σ (2r)3r-2;
(iv) Σ(2/·-1)(4/·-1)(3/·-1)-1; (ν) Σ(- 1)'+V{(2/-1)(3/·- l)C7r-4)}-1.
Exercise 5(a)
135 1. (i) 112; (ii) 2&; (iii) 15.
2. (i) Σ3Κ3/—2);
(ii) Σ (-1ГМг+1)(г+2); (iioJU-D'+H^-l)2.
356

ANSWERS
135 3.(i)l,4,13,40; (ii) 1, 1, 1, 1; (iii) 2, 22, 24, 28; (iv) 1, 2, 1|, lfi-
4. sn-2sn^ + sn^ = d.
5. (i) 9/·-6; (ii) i(2r-l); (iii) 20-8/·.
6. (i) 51; (ii) 70; (iii) 30. 7. (i) 2601; (ii) 17045; (iii) -1155.
8. 6-5/·, -930. 9. 8/-+1,л(4л + 5). 10. i(5r+3), £л(5л+11).
136 11. 40. 12. 36. 14. 2,4,6. 17. llf, 20j.
18. 63.
Ex. 15. (i), (ii), (iv) and (vi). Ex. 16. (i) 8; (ii) 7; (iii) 32.
Exercise 8(b)
138 1. (i) K412-l); (ii) -K220-l); (iii) (2 + V2)(28-l);
(iv) i[3 -3-»].
2. Ц, 2.
3. li,2i,4i, ... or -li, 2i, -4i 4.18.
5. 21. 6. 29. 7. 27. 8. 16.
9. 1-126 xlO8 km. 10. j, 1, 3.
139 11. £316-1. 12. £313-07. 13. (i) 1^; (ii) 6; (iii) f.
14. AS-. 15. 23i years. 16. {pa(p"-l)-an(p-1)}(/>-1)-2.
17. «r = />«,._!. 18. a2(l - /·) (1 - /·2") (1 + ζ-)"1.
140 Ex. 21. in(n+ 1) (л + 2) (я + 3) (и + 4) (я+5).
141 Ex. 22. 5L- |{(я + 1) (η + 2) (я + 3) (я + 4)}"1.
Exercise 8(c)
142 1. (i) 650; (ii) 6084; (iii) 2865; (iv) iw(4«2-l).
2. Ая(я+1)(« + 2)(Зя + 13), Ал(я+1)(я + 2)(Зя + 17).
3. Ая(я+1)(« + 2)(Зя + 1).
4. i^"3 + 2я2 + Ъп + 10). 5. tVC*2 -1).
143 6. Ып +1) (2и + 7). 7. i - i(2« + 3) {(я + 2) (я + З)}"1.
8. з7-*(3я+1){(я+1)(я + 2)(я + 3)}-1.
9. AV- А(4я +13) {(п+1) (л+ 2) (л+ 3) (я + 4)}-1.
10. (i) я(3я2 + Зя - 2); (ii) hniri + 1) (я + 2) (Зя + 1).
(iii) Ып+ 1) (Зя+ 1) (Зя- 2); (iv) ^я(я +1) (я + 2) (я+ 3) (4я +1).
357

ANSWERS
143 11. l-i;i„(„+lf(„+2).
12. i(2«+D(2« + 3){(«+l)(«+2)}-1-i
14. i-iMfl+l)}-1. 15. i«(4^-l).
144 Ex. 23. 1, 6, 15, 20, 15, 6, 1; 1, 7, 21, 35, 35, 21, 7, 1;
1. 8, 28,56,70, 56, 28,8,1.
145 Ex.26, (i) 81x4-l08x3y + 54x2y!-12xy3 + ;»>4;
(ii) 64x6 +192x5y + 240л:V +160*V + 60*V +1Ixf + у*;
(iii) 128*' - 448x6 + 672x6 - 560x4 + 280x3 - 84x2 +14* -1;
(iv) 243x6+810xV+1080xV + 720xV+240x/ + 32/.
Exercise 8(d)
146 1. (i) x6-6x&y+15x1y2-20xsys+15x2yi-6xy5 + f;
(ii) 32x5 + 40x1y+20x3f+5x2y3 + ixyi+^2y;
(iii) 1 + 14* + 84л^ + 280х3+ 560x4 + 672x6 + 448x6+128*';
(iv) 64(x6 +18x6y + 135xV + 540xV +1215лУ +1458*/ + 72V).
2. (i) 54; (ii) 672; (iii) 264; (iv) 19440.
3. (i) 1 + 16X+112*2; (ii) 1 + Ч-х + Чкх2\ (ш) 256(4- 20x+45x2);
(iv) -y> + 2\y*x-№fx\
4. (8\ 2rx*r-*,nw.
147 5. A. 6.(i)l-0615; (ii) 0-98411; (iii) 0-99501; (iv) 235-01.
7. 322. 8. 120,4200.
9. (i) l+4x+10x2; (ii) l-6x+21x2; (iii) 1 + 10x + 35x2;
(iv) 16(4-12x-9x2).
10. ±li%. 11. 2". (i) 81; (ii) 1.
Miscellaneous Exercise 8
1. (i) i(3"-l); (ii) 3*"<«-i> 2. A, 19-99998,9.
148 3. f. 4. {nx"+1-(n+l)x"+l}(x-l)-\
{„2xn+2_ (2„2 + 2„- 1) X"+i + (n+lf X--X-1} (х- 1)-з.
5. (;+;). 6. 4, 12, 36; 2.3-
149 12. {(l-^-n^a-^Jd-*)-2.
13. (/·-1) digits 1, followed by r digits 0 and finally the digit 1.
358

ANSWERS
PAGE
149 15. 627500.
150 17. (i) 15л(л+83)/2; (ii) 45(19«-40). 18720, l0dn+720n-55d; 30.
18. („-If.
Revision Exercise A
151 1. 5,3;x-l. 2. (-5, -5). 4. (i) 1110; (ii) 17-3.
5. 120, 24. 6. 40-9°. 7. 7x+7.y-29 = 0.
8. 0, 4; (x2 + 2V2x + 4) (x2 - 2V2x + 4).
9. 3RQ-2BC, 3RQ-BC.
10. (i) -V5/3; (ii) -2V5/5; (iii) Ц.
152 11.5,15. 12. 11, (3x+l)(x+1).
13. 0-036, 0-12. 16. 28, Цп+1)(п + 2).
18. 10 cm, 9-6 cm, 37-25°, 9-75 cm. 19. two, H-
21. 2x+3.y + 4z-9 = 0; 3V13/13, -2713/13,0.
153 23. (be'+b'cf (ca' - c'af = (aU + a'bf [(be' + b'cf +(ca'- c'af].
24. i(5-V97) < x< -1 or i(5 + V97) < χ < 4.
25. b0i+l)(f+2)(3/! + 7).
26. {-2 < χ < 2},{-3 < χ < 1},{2 < χ < 3},{1 < * < 2},
{-3<x<2}.
27. 1-03. 28. MM. 29. -i-j-3k.
30. 2Et = E2+Ez.
154 31. J. 32. г =/<a-b), a*(a+b)/(a+* + c).
33. (i) i(V5+l), (ii) 2; 4n + 1 is a perfect square. 34.^.
CHAPTER 9
158 Ex. 6. ЛЯС is right-angled at A. Ex. 7. Not' only if.
Ex. 8. Not 'if. Ex. 9. Yes.
Exercise 9(d)
160 4. 13. 5. 0or4. 6. 13.
10. No solution if a = - 1 unless * = 0.
15. No: e.g. right-angle at В and Μ coincident with B.
359

ANSWERS
CHAPTER 10
PAGE
171 Ex. 2. lOOO-X -X.
173 Ex. 3. l£ Ex. 4. Lose.
Ex. 5. Yes, if*
Exercise 10(a)
174 1. (i) 2|; (ii) -A; (iii) 3^;
(iii) 6A; Gv) 10.
3. 3p. 4. W.
175 8. 9p. 9. 7p.
14. (i) 8-75; (ii) 5-25.
176 15. 27ip. 16. 22ip.
178 Ex. 6. 2, li. Ex. 7. i
(iv) 5i.
5. £100.
12. £1-32.
17. *i3 8
16p | 2 3
Ex. 8. (i) μ+c;
= 1.
2. (i) 4h 00-ii;
6. 0, 3-3.
13. 2.
13 15 16 17
3 2 3 3"
(ii) σ\
Ex. 9. i, 0. Ex. 10. <?(Χ)*-3μσ*-μ\
Ex. 11. |(1 + f)3, Iii
Ex. 12. 2p.
Exercise 10(b)
1. (i) -μ, σ; (ii) μ+1, σ; (iii) 3/t-1, 3σ; (iv) 0, 1.
2. (i) 2-81, 4-41; (ii) 3-7, 2-01; (iii) 0-15, 3-53; (iv) 150, 3000.
3. (i) 12-31, 121-1; (ii) 15-7, 102-8; (iii) 3-55, 16-75;
(iv) 2-55 xlO4, 2-83 xlO8.
4.4J-.2J. 5. Hn+l),Un2-D.
6. M, H, H. 7. A. 9. (i6 + i3 + ί + 3)"/6", |я.
10. A, 7, 2-415, 1-71. 11. ff, ^, ^.
12. 6-5, 2-36. (i) |; (ii) 0.
Miscellaneous Exercise 10
1. £7-33. 2. 4p favours player, Цр favours banker.
3. Just over £3. 4. — lj^. 5. Yes, if Χ, Υ independent.
7. 2(n-r)/{n(n-l)}. 10. 0p-i. 11. «σ2+/*2-
"■НШ'Н'5К)Ш1Ю

ANSWERS
CHAPTER 11
PAGE
192 Ex. 1. Yes, provided a Φ 0, b Φ 0. Ex. 2. OA2.
Ex.3, (i) Pythagoras; (ii) Extension of Pythagoras (Cosine Rule).
Exercise 11 (a)
193 3. Sphere, centre A, radius |c|.
195 Ex. 4. -5. Ex. 6. (3i + j + k)/Vll.
199 Ex. 9. 1, 1; lies on angle bisector.
Ex. 11. ЗХ + 2.У-10 = 0, (2, 2). Ex. 12. (-a, -*), \c\.
Ex. 13. 8x + 6y+13 = 0, 8x+6y-7 = 0.
Exercise 11 (ft)
1. (i) 2, ψ, (ii) 2, ¥; (iii) 3, |; (iv) Z>, (3e-26)/5.
200 2. (i) 2V5/5; (ii) V65/65; (iii) 8V17/85; (iv) |(я2-*2)/(я2+*2)|.
3. (i) V17/17; (ii) 12V13/13; (iii) 6V10/5; (iv) (h + lcf (/z2+/t2)-}.
4. (i) 2V5/5; (ii) 9V5/10; (iii) 2V26/13; (iv) (a+b) (a2 + *2H.
5. (i) Ъх-4у + \6 = 0, Ъх-Ьу-U = 0;
(ii) 8x+6y + 35 = 0, 8x + 6y-25 = 0; (iii) 5x+12^+41 = 0,
5x+12.y-37 = 0; (iv) 3x + 2.y+ 1 ±3Vl3 = 0.
7. (i) 2x-y = 0; (ii) 7x-7y-3 = 0; (iii) 128x+16y+69 = 0.
8. x-ly-18 = 0. 9. (i) (9, -9); (ii) 52i; (iii) +h
10. (-p, -q).
201 11. x+2y-U = 0, x-y+l = 0, arccos (8V145/145), (¥, 3).
12. 37.
202 Ex. 14. (i) i(I-2J + 2k); (ii) (3i- j + к)Д/11; (iii) (3i + 2j + к)Д/14.
Ex. 15. (i) x-3y+z = -5; (ii) x-^ + z = 2; (iii) 2x+3.y-5z = -16.
Exercise 11(c)
205 1. 110-9°. 2. (i) x-.y+z = 0; (ii) 2x+y-3z = 13;
(iii) Ъх-z = 1; (iv) 6x-3.y-2z = 2; (ν) 4x+.y-3z = -1;
(vi) ax+by+cz = a2 + b2 + c2.
3. (i) 2V70/35; (ii) V28/84; (iii) 2/29; (iv) 1/2.
361

ANSWERS
206 4. (i) 3VH/H; (ii)Vl5/15; (iii)V66/ll; (iv) 17V58/174;
(v) аЫаг + Л2)-1; (vi) \ab - be - ca \ (a2 + Ьг + с2)"1.
5. (i)V7/7; (ii)Vl05/15; (iii) 1; (iv) 40/41; (v) ,/3/2.
(vi) Vffi(<?/ -ртЩМР + m* + «3)(p2 + q2 + r2)}.
6. (i)K2i-2j+k),2x-2^ + z= -1;
(ii) i(3i-2j + 6k),3x-2.y + 6z = 0;
(iii)^(i + j-4k),x+^-4z = 1;
(iv)^(4i-j-2k), 4x-y-2z=6;
(v)^(3i-4j+k),3x-4.y + z =1;
(vi) (a2i - b*j + c2k)/V(a4 + b1 + c4), a*x - V-y + c2z = 0.
207 7. (i) (2, 3.-1); (ii) (4, 7,-5); (iii) (1, 1, 1); (iv) (0,-1, 3).
8. (i) (9, 5, 3); (ii) (7, -10, -3); (iii) (-6, 16, 6); (iv) (-9, 9, 12);
(v) (15, -13, -2); (vi) (- 2ad(a* + Ρ + c*)-\ - ΉκΚ,α* + & + <?Y\
-Icdtf+tf + c2)-1).
9. (i) (1, 1, 3); (ii) (2, 3, 4); (iii) (6, 2, 4); (iv) (6, 4, 1).
10. (8, 2, 0). 11. (1, 4, 3), K*-l) = y-4 = z-3.
12. Цх-1) = - Hr- 5) = fa- 2). 13. x = y- 3 = - (z+ 6).
208 14. 3: -1:2, Vl4.
16. «V3 -1), «V3 +1), V3(i - V3)/6, V3(i + V3)/6, iA
17. 5λ = 2μ, a = 5i + 29j + 2k, b = -(i + 5j + k).
209 18. 2x-.y + 2z = 9, arccos -f. 19. (1, 5,0), arccos V(rr)·
20. (i) (4, -ι, -3), (¥, f, -¥); (ii)!;(!,!, -|).
Miscellaneous Exercise 11
1. |/7-Xj cos a—^sinal.
2. 3x + ll.y+19 = 0, llx-3^-17 = 0.
210 3. 14x + 8.y-57 = 0. 5. 3-75, 1. 6. (-4,3).
9. 1:2.
211 11. 60°. 15. x + 2 = Ky+5) = z-5.
16. \x=-\y = -z. 17. s + 2|u.s-p|(±u).
362

ANSWERS
PAGE
212 18. x*+y* = l + 2.ycot0.
19. (i) x-5y + 4z = 0; (ii) x+y+z = 0, (6, -2, -4).
20. -11,(1, -2,3).
21. Ж-*, V3/2, 0), C(-i, - V3/2, 0), £(0, 0, V2), (0, 0, V2/4); -4:0:V2,
2:±2V3:V2,0:0:l.
22. 2.
213 23. 5x+12^-60 = 0. 24. 2V6.
25. χ = [p(b* + c2 - a2) - 2a(*9 + cr+d)] (a2 + b2 + с2)"1, etc.;
(A, -A,l),(-i -if).
26. 15x-9.y-3z = 0, 1:2: —1.
27. E[(a2+Z,2 + c2)/-2a(a/+*w+w)x] = 0.
214 28. у = ζ = 0, ζ = χ = Ο, χ = у = 0; cy-fe = 0. 120°.
29. arcsin 8Д/82, arccos 4Д/82.
30. Intersection of planes with 2,(m2ns—m3n2) χ = 0.
32. 14-2°. 33. Sphere, centre the centroid.
CHAPTER 12
215 Ex. 1. Yes, e.g. A = В = 60°.
216 Ex. 2. KV6-V2), KV6+V2)· Ex. 3. -i(V6+V2), i(V6+V2).
Ex.4. (i)ff,M; («) U, -U; (in) -ft, M-
Ex.5. -2 sin Л sin Я.
217 Ex. 6. (a) 0-0 Φ (2A:+1) in; φ) θ φ \т±\тт.
219 Ex. 7. (i) ± V10; (ii)±V5; (iii) Ю, 0; (iv)l±V2; (v)i(2±V2).
Ex. 8. (i) 0, 270, 360; (ii) 53-1; (iii) 90, 323-1; (iv) 4-9, 220-5.
Ex. 9. 1,7. Ex. 10. \.
220 Ex. 11. -J, fV2· Ex. 12. Ц, \. Ex. 13. &τ, £π.
Ex. 14. V2-1. Ex. 15. 2*VG ~x\ 2/-1. Ex. 17. arcsin Ц.
221 Ex. 18. (1 + 2i- f2) (1 + i2)-1.
223 Ex. 19. (i) 2 sin 2x cos χ; (ii) 2 cos 3x cos x; (iii) -2 sin 2x sin χ;
(iv) V2 sin Й7Г+ χ); (v) s/2 sin χ
224 Ex. 20. 0, \тт, |тг, тг, §тг, |тг, 2тг. Ex. 21. 0, ψ, π, %π, 2π.
363

Exercise 12(a)
224 1. (i) i(V6+V2); (ii) -i(V6-V2); (iii) -(2+V3); (iv) 2-V3.
2. (i) KV5-D; (ii) W(10+2V5); (iii) i(V5+l);
(iv) -KV5+DV(10+2V5).
225 5. (i) 28-6°, 208-6°; (ii) 120°, 300°; (iii) 37-5°, 217-5°; (iv) 18-3°, 198-3'
6.(0 225°; (ii) 90°, 330°; (iii) 29-5°, 256-7°; (iv) 22-6°, 90°.
7. (i) 0°, 116-6°, 180°, 296-6°, 360°; (ii) 180°; (iii) 45°, 120°, 135°,
225°, 240°, 315°; (iv) 0°, 45°, 90°, 225°, 360°.
8. |sina| < |; 9-3°, 50-7°, 189-3,° 230-7°.
10. 4 cos3 Θ- 3 cos Θ, 8 cos4 Θ- 8 cos2 0+1.
226 13. (i) ± (a2-*2) (2a*)-1; (ii) a = 8i, * = 6*; tan θ = Д.
16. R = V5, a = 26-6°. (i) 140-5°, 346-3°;
(ii) 0°, 63-4°, 180°, 243-4°, 360°.
17. -97-6°, 0°, 90°, 171-3°.
227 18. (Цп, Дтг), (Дтг, Цп).
Exercise 12(b)
232 1. 76 m. 3. (i) 84; (ii) Ц; (iii) 8i
233 4. 101-2°, 44-8°. 5. 14-8°. 6. 9-3 m. 7. 36° 8. 127-8°.
9. (i)A(A2 + 2a2)-*; (ii)/г^2 + a2)"*;
(iii) 2я/г(/г2 + α2)-1; (iv) V№2 + 2a2)/V(2A2 + 2a2).
10.(i)35-3°; (ii) 64°. 11. (i) 77 m; (ii) S 46° W. 12.121m.
234 13. 529 cm2. 14. (i) 64-9°; (ii) 69-4°; (iii) 31-6°.
16. 1-12/·. 17. 1-96, 56-8°.
235 19. exccos{(ci-ai-bi + 2aW)H2l?c:i+2(?a2 + 2a2l?--ai-bl-c1)-i}.
Miscellaneous Exercise 12
2. (i) i; (ii) h ~i- A. 2m < x < (2л +1) π, n e Z.
236 8. GK Aw), (А", *я), фя, in), б", Ня). 13. 0, *π, тт.
14. i(l ± V2). 15. tan ia = a/*.
237 17. о 4.
(i) (9 = 2n7T + arccosf, 0 = 2w?r+arcsin |;
(ii) (9 = 2n7r+arccosf, ψ = (2w+l) 7r-arcsini;
(iii) θ = 2n7T+arccos|, ψ = 2w7r+arcsinf;
(iv) θ = 2n7r+arccosi, ψ = (2w+l) w-arcsin f.

CHAPTER 13
PAGE
245 Ex. 9. S(P) = (3, 5), T(P) = (3, 0), U(P) = (6, 5).
Ex. 12. Reflection in Oxy plane.
Exercise 13(a)
1. (-5, -10), (i, -i).
246 2. (-5, -4), (3, -5); (0, 0). 3. The line x-2y = 0.
6. Reflect in χ = у and then project on to χ axis.
7. Reflect in χ = у and stretch by factor 2.
8. \ad — bc\; all points map into line ex — ay = 0.
11.
П-
250 Ex. 13. (i) 2 1
Exercise 13{b)
•(-"3=
4. (i) AB-2AC; (ii) A2-AC + BA-B(
(iii) A^AB-BA+AC-CA-iB+C)2.

(iv) О О - 2 ; (ν) 2 2 -6 ; (vi) -6-2
\5 1-3/ \-l -3 19/ \- 8 -4
/ 3 8 0\ / 1 4 5\ /6 1 -1\
7. (i) -3 3 5 ; (ii) 5 4 5; (iii) 1 6 3 ;
^«(-»
-27 -40 - 1
254 8. (i) 4 1
^{_1-·_ϋ··π-3=
/0 О й\
10. (χ j> ζ), * 0 0; Τ preserves lengths; each + 1.
\0 с 0/
11. 2x+y-z = 0, (-1, 7, 5).
12. x-y + z = 0, (5, 8, 3).
255 14. (0,0), (2, 2), (-1,0).
Miscellaneous Exercise 13
2. Shear; χ, ζ coordinates fixed.
3. (i) (0, 0) only; (ii) line 2x-y = 0; (iii) all points in the plane.
256 5. (0, 0) only; no.
6. (i) (0, 0) only; λ = 10: line 7x-5y=0,X=-2: line x+y = 0;
(ii) all points of line x-y = 0; λ = 6: line 4x + y = 0.

PAGE
256 8. AB = ΒΑ.
9. A diagonal, provided non-zero elements of D distinct.
257 11. No.
CHAPTER 14
l-i о i\ ι i i -¥
(iii) 1 5 -4 ; Ov) 0-4 3
\ 0 -3 2/ \-i ! -I
Exercise 14(d)
>(14);
/ f -3 V\ / i i -i\ I f -i ¥\
(iii) -f 2 -Ϊ ; (iv) i -i i ; (v) \-i i -ψ);
0 -1
i i
■u

ANSWERS
PAGE
267 Ex. 11. (i) 3; (ii) 5; (iii) 7.
Ex. 12. (i) 2; (ii) -3; (iii) 0;
(iv) -3; (v) abc+2fgh-ap-bg*-ch\
273 Ex. 13. (i) 12; (ii) -9; (iii) 0, (iv) 1.
Exercise 14(ft)
275 1. (i) -2; (ii) -1; (iii) -1; (iv) 22.
2. (i) -30; (ii) 0; (iii) 0; (iv) 1; (v) 24; (vi) -9;
(viii) -12; (ix) 8; (x) 6.
. (iii) Singular.
16 -16,0,0;
/30-11 5\ /0-1 2\
1-10 4 -2,21,2; (iv) 1 -1 -3,1
\ 2 ^1 1/ \-2 3 5/
i i -i ·
6. (i) 12; (ii) 1; (iii) -10. 7.(i)(b-c)(c-d)(a-b);
(ii) (*-c)(c-a)(a-*)(a + * + c); (iii) (b-c)(c-a)(a-b)(bc + cc
36\
-95.
136/
/-3 33 4\
.0-8 0 .
\ 2 - 2 -l/
Miscellaneous Exercise 14
2. (a) Yes; (*) yes; (c) yes.
ι ac\ /a* ab
be); [ba b2 +
c7 \ca cb+
1 bc+2 .
c2+4/
278 6. Yes. 9. (* - cf (c - a)2 (a - bf.
368

PAGE
279 13. A2 =
CHAPTER 15
285 Ex. 2. (Й.-А).
286 Ex. 5. True for any χ provided b = 0.
Exercise 15(d)
1. (i) (2,5); (ii) (-3,5); (iii) (f, -?).
2. (i) If α Φ -b, χ = b-a; if α = -*, true all x.
(ii) If αφ b,x= ic-b) (а-ЬУ1; if a = b φ с, no solution;
if a= b = c, true all x.
3. (i) χ = (4 +За) (За-б)-1, у = 5(2-a)-1, provided α φ 2.
If a = 2, equations inconsistent.
(ii) If α φ 4, * = 0, у = 2; if α = 4, * = 2(2-λ), j; = λ, all λ.
(iii) If α φ ±6,χ= 2(6 +аГ1, > = (6 +α)"1.
If α = 6, χ = λ, у = i(l - 3λ); if α = - 6, equations inconsistent.
4. (ί)ΚαΦ-3*,Λ:=(*2 + ύ*+8)(4α+12*)-1,^=(24-α*-α2)(4α+12*)-1.
If α = -ЪЪ, b2 φ 4, equations inconsistent; if b = 2, χ = λ, у = 3λ+ 1;
if* = -2, л: = λ, у = ЗА—1.
(ii) If ab φ -3, χ = 6(1 +*) (οΛ + 3)-1, j, = 2(3-α) (аб + 3)-*.
Ifa= Ъ,Ь = — 1,х = Л,у = 2-λ; if ab = -3,α φ 3 equations inconsistent,
(iii) If α2 φ Λ2, χ = (α—Λ)-1, ^ = (b- a)-1. If a = b, equations inconsistent;
if a + b = 0, χ = λ, у = (αλ-Ι)α-1.
5. Inconsistent unless α = 3 or -1;
if α = 3, χ = - 2, j> = 3; if β = - 1, * = 2, > = -1.
6. Inconsistent unless a = 1, 2 or - 3. If a = 1, χ = V, .У = t',
if a = 2, χ = 2, j> = 0; if a = - 3, χ = 1, .у = |.
287 7. (i) α Φ 0, * Φ 0, α φ *, * = c(b-c) сгЧЬ-аУ\у = c(c-a) b-\b- аУ1;
(ii) a = 0, b φ 0, с Ф 0, equations inconsistent unless
* = c, when χ = λ, у = 1; (iii) α = 0, b φ 0, с = 0, consistent,
л: = λ, у = 0; (iv) α φ 0, * = 0, с φ 0, equations inconsistent unless
369

PAGE
287 я = с, when solution is χ = 1, у = λ; (ν) я φ 0, * = 0, с = 0,
consistent, χ = 0, у = λ; (vi) a = b φ 0, с Φ 0, inconsistent unless
я = 6 = с, when solution is л: = λ, ^ = l-λ; (vii) α = b = 0, с Φ 0,
inconsistent; (viii) я = b = с = 0, consistent, χ = λ, у = μ.
8. яА-яА = 0, χ = АяГ1, .у = -Щ1.
9. If 0 φ hr+φ (к integral), x = sin (φ-a,) cosec($S-0),
j> = sin(0-a) cosec(0—φ); if 0 = ктт + ф, equations inconsistent unless
θ and ψ both differ from α by an integral multiple of π. If θ = α + 2w7r,
0 = α+2/OT, χ = X,y = l-λ; if0 = α. + 2ητπ,φ = а + (2я+1)тг,
л; = А, у = λ-l; if θ = a + (2w + l)7T, 0 = α, + Ίηπ, χ = λ, > = λ+1;
if# = α + (2ηι+1)π,ψ = а + 2(я+ 1) π, χ = λ, у =-1-λ.
292 Ex. 12. (i) Lies in plane OI'J'K'; (ii) can be expressed in terms of any
two of i', j', k'.
Ex. 14. Columns of A proportional; (i) lies on ΟΓ; (ii) Ai'.
Exercise 15(b)
294 1. (-2,0,4).
/ 1 6 5\
295 2. 1 1 2l. (i) (0, 1, -1); (ii) (2,0, 1); (iii) (-35, -5, -7).
3. Any point of line i(x-l) = Ну+1) = i(z+2).
4. (i) Line λ, Цб-Ίλ), i(13— 11λ); (ii) inconsistent: planes form
prism in direction 5i—7j —Ilk.
5. b = 10, line; b φ 10, inconsistent.
6. я = "2, inconsistent; я φ 2, consistent only if * = я + 2, giving line.
7. (2, 1,-3).
296 8. я = 5, inconsistent; я = — f, line; unique point otherwise.
9. 4, (0, λ, 2λ). 10. (λ, λ, - 2λ), (1 + λ, λ, - 2λ).
11. Цх+6) = у = -^-(ζ-12), parallel.
297 14. Tetrahedron 7x-6.y-5z > 0, χ-2y- ζ < 0, 9x-8.y-7z < 0,
3x-4.y-3z > -2.
Miscellaneous Exercise 15
1. Ii i l), (-2, -l, l), (4,4,4)-
\0 -2 -ι/
2. (i) Line or plane; (ii) inconsistent.
370

ANSWERS
PAGE
298 3. (f, -1,4).
5. A = b-a, В = c-a, С = (b-c) (c-a);
χ = (*2 + с2)(й-*)-) (с-я)-1, у = {с2Л-a2) {a-b)-1 (b-c)~\
z= (a* + b*)(b-c)-1(c-a)~1.
299 6. (1, 1, 1), (λ, 3-2λ, λ).
9. (*-с)(с-й)(й-*), л: = я*(*-с)(с-я)-1(я-*)-1;.У = оЧа-Ь)~\
ζ — аЫ^с—а)-1; а = b Φ с, inconsistent unless α = 0; а Φ Λ = с, line;
а = b = с, plane.
300 12. - 4, 3, 2, -11 (1 + λ)/7, 5(1 + λ)/7, λ.
13. a-2b-c = 0, (λ, 1-2λ, λ-1).
14. λ φ f, -1, unique; λ = ^-, inconsistent; λ = -1, line.
CHAPTER 16
306 Ex.2. TPo,TsPo,ihPo,6hsPo. Ex.3, (i) 2-5; (ii) 2.
Ex. 4. 1-8,1-44. Ex. 5. f, ^, 7 to 1.
Exercise 16(a)
1. ii^ 2.0-348,0-071. 3.0-000105, 0-00605, 0-201.
4. 9, 0-387. 5. (i) ^2U; (ii) 0-566.
307 6. <й-, Hi. 7. (9-/-K5/-+5)-1. (i) 1-5; (ii) 1; (iii) 0-46.
9. /79(10-9p), £3-52, 46-8%. 11. (Ί) 0-388; (ii) 0-0386.
311 Ex.7, (i) A; (ii) Ш; (ш) 12.
Exercise 16(b)
313 1. 0135, 0-271, 0-271, 0-180, 0090, 0-036, 0017.
2. 1-45; (a) 0-06; (*) 0-33.
314 3. (i) 1; (ii) 0-846; (iii) 0-013. 4.0-167,0-594.
5. 0-030, 0-106, 0-185, 0-216, 0-188, 0-132, 0-077; 3, 4-25.
6. 3; 60, 222. 7. 20, 0-029. 8. 0-302, 0-060.
315 9. No: Рт(Х^ 21) « 0-121. 10. (λ* + A)/(l -e"A).
11. afr+l)"1, 2. 12. 0-0030, 0-0379.
13. 226-8, 211-4, 98-6, 30-6, 7-1, 1-5.
371

ANSWERS
Exercise 16(c)
PAGE
321 1. (i) 0-63; (ii) 0-61. Pr (2 or more) « 0-09. 2. 0-79.
3. (1 -p)w {1 +10^1 -p)»}, 13-15.
322 4. 0-9860, 0-0119, 0-002. 5. No.
6. Pr (/· < 4|w = 10) = 0-029; yes. 7. Pr (3 or more sixes) = 0-062.
8. Yes; no {ρ π 1/163).
323 9. Not significant in either case.
11. (i) (1-ρ)14(1 + ΐν{1 + 105^2(1-,ρ)13 + 455^3(1-,ρ)12};
(ii) 0-164; (iii) 17-5.
12. £2-25, £1-50.
Miscellaneous Exercise 16
1. 0) Яг; (ii) 4; (iii) Si 2. 1, 0-335; 20.
324 3. M- 4. 8. 5. Yes, if ρ < i; no otherwise.
6. }iV(iV+1) (2p-1). 8. ¥ (2/И7)" <J?+q*)-
9. μρ.
325 11. 0-3.
12. 20, 60, 90, 90, 67, 40, 33. (i) £157-50;
(ii) £167-50; (iii) £160-75.
13. f^jjrXl-p)-',
Qo*)*(i-да)"--
14. β(',*)= Q(*)*(t)"
15. 0)99-1%; (ii) 0-03; (iii) 0-1.
326 16. 0-011, 0-870, 0-516; 23-7. 17. 0-585.
18. e-'4, μ, е-^+оу^гЫУ1.
327 20.0-2; no. 21. p(n-/·+!) (r-rp)-1, rm = [р(л+1)].
372

Revision Exercise В
PAGE
328 1. x-ly = 0,4x+2y-15 = 0. 2. Щ, %, \. 4. i(4/H
5. J(10+V10), 7. 7. 3. 8. No.
329 9. (2x-.y+l) 0+З.У-2), two lines; (i) two lines; (ii) origin.
/-8 11 -10\
12.^-9 12 -9. (i)(20,20,-23); (ii) (-V, -5,
\ 10 -13 11/
13. l-48x+1104x2, 0-6158. 14. (i) 37x-27.y = 0;
(ii) 31x+27 = 0.
15. (i) 3x + 4; (ii) 5x + 7.
/ 1 -2 -2\
330 16. a. ~ β= (2k+l)n. 19. И-2 1 -2.
\-2 -2 1/
20. 3:1. 21. About f, 69.
23. (i) 0; (ii) 2(9-0 (r-p)(p-q)(y-z)(z-x)(x-y).
25. A < ρ < A. 26. (2, 5, - 3), (4, 1, 1).
27. {yeR: -1 < у < J2}.
331 28. (x + c)(ax+*-c). (i) (1+ c) (a+*-c); (ii) (c-1) (*-<
(iii) (ab + c)(a2bc+b-c).
29. Plane л: + ζ = 0.
30. -fo, UU. 31. (i) 0, in, π, in, 2π; (ii) Д-тг, Ня.
32. -2 < χ < -i or i < л: < 2. 34. β-
332 35. (-2, -1, -5). 37. 2, -4, -3. 38. 4.
CO - 2 - 2 3 ; (ii) - 2 3 - 2 .
\ 23 24-33/ \ 43 -62 45/
40. 26° 34', 45°, 108° 26'. 41. x-2y+ ζ = 0, 1:2:3.
42. (i) fff; (ii) ih\ (Ш) Ш- 43. α = 2"-l, /S = 1.
45. A: = 1, line; A: = — 2 or A: = 3, inconsistent.
46. 2Q°, 100°, 140°; 1-44, 0-33, -0-27.
47. (i) i(«+l); (ii) in\n-iy\

ANSWERS
334 49. (i) - Mb - c) (c - a) (a - *);
(ii) (x + a+b+c)(x + a-b-c)(x-a+b-c)(.x-a-b+c).
50. x-2y-2z + S = 0.
51. 53°, 155°, 344°. (i) 158°; (ii) m = -0-005, с = 1-7.
53. φ = ег-вг.
335 55. к = 5, line; /с = -7, prism. 56. 554400.
57. л: < -5, -1 < χ < Ι, χ > 7.
59. u + v+w = 0 and Зя+2*+с = 0 for consistency;
line (λ, λ+Ub-a), X-i(2a+b)).
60. ЗЦ, 49-A-.
336 61. 47°. 63. x-1 = -i(y+2) = z.
64. (i) \\ (ii) i + \(a - b)n (a + b)~"; *(* + 2a) (a + *Г2 in both cases.
65. 5567. 66. Yes, provided λ Φ 0.
337 68. λ = -2, line; A = 1, inconsistent.
69. (λΙ'+μηΐ)<№+μη,)-\ 70. 23min.
338 71. 3-N, \, |. 74. 2a2 + 2c2 +P = 2b* + 2d2 + e2.

Index
a posteriori, 125
a priori, 125
addition of vectors, 25
additive identity, 3
additive inverse, 3
adjoint matrix, 273
adjugate matrix, 273
alien cofactors, 272
ambiguous case, 229
Apollonius's Theorem, 160
arcsine, 98
arithmetic mean, 19
arithmetic sequence, 133
arithmetic series (progression), 134
arrangements, 103
associative, 3
axes of coordinates, 47
base, 1
base vectors, 45
Bayes's Theorem, 124
bearings, 230
Bernoulli trial, 302
bicimal, 7
binary system, 1
binomial distribution, 302
Binomial Theorem, 145
Cartesian equation, 52
Cartesian product, 78
centroid, 33
Ceva's Theorem, 43
Chebyshev's Theorem, 180
circumcentre, 50
codomain, 79
cofactor, 269
cofactor alien, 272
column vector, 239
combinations, 104
common difference, 133
common ratio, 136
commutative, 3
complement, 14
component form of dot product, 194
component of vector, 28
compound angles, 215
consistent, 286
coordinates, 47
cosine, 89
Cosine Rule, 227
cosecant, 90
cotangent, 90
counterexample, 9, 158
degrees of freedom, 281
de Morgan's Laws, 16
denary system, 1
Desargues's Theorem, 35
determinant, 266
deviation from mean, 176
dilatation transformation, 242
direction cosines, 61
direction ratios, 61
direction vectors, 61
displacement, 24
distributions
binomial, 302; geometric. 308; negative
binomial, 310; Poisson, 311; uniform,
110, 301
domain, 79
dot product, 189
double sampling, 318
elementary events, 108
elementary row (column) operations,
259
empty set, 14
enlargement transformation, 242
entry (matrix), 239
equality sets, 15
equality vector quantities, 24
equation of line, 37, 53, 60
equation of plane, 37, 59
equiprobable, 110
event, 111
Existence Theorem, 27
expectation, 172
Factor Theorem, 70
factorial, 103
Fibonacci Series, 168
fraction, 5
function, 78
geometric distribution, 308
geometric mean, 19

TNDEX
geometric sequence, 136
geometric series (progression), 137
gradient, 53
graph, 80
homogeneous equations, 287, 293
hypothesis testing, 318
i, j, k, 46
identical polynomials, 68
identity matrix, 251
if and only if (iff), 157
image, 79
implication =>, 4, 155
incentre, 154
inconsistent, 286
independent events, 118
inductive hypothesis, 165
inequalities, 17, 83
infinite outcome spaces, 183
inner product, 189
integer, 3
intersection, 15
invariant, 82, 246
inverse function, 81
inverse matrix, 259
inverse sampling, 309
inverse trigonometric functions, 98
irrational numbers, 8
linear combination, 30
linear dependence, independence, 30
linear equations, 54, 281
linear transformations, 241
inverse, 284; matrix of, 244; one-one,
255; product of, 245; sum of, 244
locus, 37
magnitude (vector), 25
mapping, 78
mathematical induction, 162
matrix, 239
addition, 248; adjoint, 273; adjugate,
273; conformable, 248, 249; diagonal,
256; echelon, 265; element of, 239;
elementary, 261; equality of, 247;
inverse, 259; multiplication by number,
248; multiplication of, 249;
non-singular, 264; orthogonal, 278; singular, 264;
skew-symmetric, 278; symmetric, 256;
trace of, 257; transpose of, 254; unit,
251; zero, 249
mean, 176
median, 188
Menelaus's Theorem, 43
method of differences, 140
376
minor, 268
modular arithmetic, 4
modulus, 81
moment, second, 315
multiple angles, 215
multiplication of vectors by numbers, 27
multiplicative identity, 3
mutually exclusive, 114
necessary condition, 158
negation, 158
negative binomial distribution, 310
non-trivial solutions, 287, 294
null hypothesis, 319
null set, 14
odds, 112
one-one, 78
operating characteristic, 317
ordered pair, 78
orthocentre, 58
orthogonal, 46
outcome space, 108
infinite, 183
parameter, 60
one-parameter distribution, 312; one-
parameter solution, 281
partition, 124
Pascal's triangle, 144
periodic function, 83
permutations, 104
Poisson distribution, 311
polynomials, 68
identical, 68
position vector, 32
possibility space, 108
probability, 109
addition of, 114; of event, 111;
multiplication of, 116
probability distribution, 109, 301
probability generating function, 181
probability tree, 117
projection, 190
proper subset, 14
q, e+, i6
R, R+, 16
radian, 87
range, 79
random, experiments, 108
random variable, 171
expectation of, 172; mean of, 176;
standard deviation, 177; variance, 177
rational number, 5

INDEX
rationalizing the denominator, 10
real matrix, 239
recurrence relation, 131
recurring decimal, 6
relation, 78
Remainder Theorem, 72'
row vector, 239
suffix, 49
sum to infinity, 137
surd, 10
tangent, 89
terminating decimal, 6
terms of sequence, 160
transformation, 240
triangle rule (vectors), 25
trivial solutions, 287
uniform distribution, 110, 301
union, 15
Uniqueness Theorem, 28
unit vectors, 32
universal set, 13
Vandermonde's Theorem, 146
variance, 177
vector, 25
components of, 28; dot product, 189;
multiplication by number, 27; triangle
rule, 25; unit, 32
vector addition, 26
vector equations, 37
vector quantities, 24
Venn diagram, 15
Ζ, Ζ+, 16
zero matrix, 249
zero vector, 27
St Petersburg Paradox, 186
sample space, 108
scalar product of vectors, 189
scalar quantities, 24
section formula, 33
selections, 104
sequence, 130
series, 132
set, 12
complement of, 14; empty, 13; equality
of, 15; intersection, 14; null, 13; union,
14; universal, 12
shear, 246
sigma notation, 132
sine, 89
Sine-Rule, 228
single sampling, 316
skew lines, 42
spread, 176
solution of triangles, 227
standard deviation, 177
subset, 13
sufficient condition, 158

LU Ε
CAMBRIDGE UNIVERSIT PRESS

PURE
MATHEMATICS
2
S. L. PARSONSON
Senior Mathematics Master
Harrow School
CAMBRIDGE UNIVERSITY PRESS
CAMBRIDGE
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Bentley House, 200 Euston Road, London NW1 2DB
32 East 57th Street, New York, NY 10022, USA
296 Beaconsfield Parade, Middle Park, Melbourne 3206, Australia
ISBN:0 521 08032 0
First published 1971
Reprinted 1973 1975
Printed in Great Britain
at the
University Printing House, Cambridge
(Euan Phillips, University Printer)

Contents
17
18
19
20
21
22
23
24
25
26
27
28
29
Pre/асе
Complex numbers (1)
Polynomials and partial fractions
Complex numbers (2)
Mappings in the Argand diagram
Quadratic equations and quadratic functions
The parabola and rectangular hyperbola
Polynomial equations
Vector products and their applications
Continuous probability distributions
Numerical solution of equations
The ellipse and hyperbola
Further matrices
Further coordinate geometry
Revision exercise С
Bibliography
Answers
Index
page ν
379
401
423
438
450
461
483
506
520
552
573
609
633
657
680
683
703

Preface
This book completes the course designed to cover the work required for
modern Ά' level pure mathematics syllabuses (including probability),
particularly the M.E.I, syllabus. As explained in the Preface to Volume 1,
the calculus has been excluded, since it is already adequately covered by
existing texts, but a knowledge of the subject is demanded throughout the
book, and particularly in Chapter 25, on continuous probability.
In places the subject is carried beyond the bare limits of 'A' level
requirements: most chapters contain a small proportion of work which
would probably best be left to a second reading, while Chapters 20, 24, 28,
29 contain a substantial proportion of * S' level work.
As in Volume 1, the book is liberally supplied with exercises for the
student. Most questions are straightforward applications of the bookwork,
though a few harder questions may be found in the Miscellaneous
Exercises. The questions marked Ex. occurring in the text illustrate the
associated bookwork: those marked with an asterisk should be regarded as
obligatory.
The same nomenclature as in Volume 1 has been used to indicate the
source of examination questions and I am grateful to the Examination
Board of the University of London and the Oxford and Cambridge Schools
Examination Board for permission to reproduce their questions. I should
also like to record my thanks to Dr N. A. Routledge and Mr A. J. Moakes
who read the book and made many valuable suggestions and to my wife
who once again lent invaluable assistance in checking the answers.
S.L.P.

^

\η. Complex numbers (1)
1. AN EXTENSION OF THE REAL NUMBER
SYSTEM
We have seen (Chapter 1, et seq.) that, by broadening the meaning of the
term 'number', we are able to ascribe solutions to an increasing range of
problems. To take a simple example, consider the solution of certain
equations. If we restrict 'number' to mean 'positive integer', the equation
*-2 = 0
has a root, but the equation
x+2 = 0
has not. The latter equation does have a root, however, if we postulate
the existence of negative integers. Thus, by augmenting the set Z+ of
positive integers into the set Ζ of integers we are able to solve a wider
variety of equations. Z+ is a subset of Z, and the familiar laws of algebra
governing the combination of elements of Z+, that is
(i) closure: if a, b, e Z+, then a + b, ctbeZ+;
(ii) commutativity: a + b = b + a, ab = ba;
(iii) associativity: (a + b) + с = a + (b + c), (ab) с = a(bc);
(iv) distributivity:a(6 + c) = ab + ac, (b+c)a = ba + ca;
(v) the existence of a multiplicative identity: \a = a
hold for the elements of Z. But we also have two new laws which hold for
Ζ but not for the subset Z+:
(vi) the existence of an additive identity: 0+a = a;
(vii) the existence of additive inverses: ( — a) + a = 0.
In a similar way, an equation such as
2x+l =0,
which has no solution in Z, has a solution if we augment Ζ into the set
Q of rational numbers. Again, Ζ is a subset of Q, and laws (i)—(vii) above
hold in Q, together with an additional law
(viii) the existence of multiplicative inverses: α_1α = 1 (α φ 0).
Laws (i)-(viii) define afield; that is, any set of numbers combined by the
operations of addition and multiplication and satisfying laws (i)-(viii)
above constitute a field.
379

COMPLEX NUMBERS (1) [17
Ex. 1. Show that the set
S = {x = a+b^/2: a, b rational}
constitute a field under the usual operations of addition and multiplication. (For
example, to verify (iv) you have to show that
(αι + ό1λ/2)(α2 + ό2λ/2)
is of the form a3 + ό3λ/2,
where аъ ...,b3 are all rational numbers. Check, in a similar way, the other seven
laws.)
The extension of the number system to the set Q vastly increases the
number of equations with a solution, but it is not difficult to formulate
equations in terms of the elements of Q which have no rational solution:
consider, for example, the equations
x2-2 = 0 or 2 sin χ = 1 or 10* = 5.
Many such equations acquire solutions if we augment the set Q into
the set R of all real numbers. Although it is possible to define R in terms
of the elements of Q, we have contented ourselves with the intuitive
geometrical concept of R as ' completing the number line' but, since real
numbers may be approximated arbitrarily closely by rational numbers, it
seems reasonable to assume that laws (i)-(viii) above hold in R as well.
Since we have 'filled in the gaps' of the number line it might seem
reasonable to suppose that our work is done and that any equation
involving elements of R as coefficients should have elements of R as solutions.
Such is not the case however; indeed, if we modify the three equations of
the previous paragraph only slightly to
x2 + 2 = 0 or sin л; = 2 or 10* =-5,
none of these equations has a solution in R. However, these equations have
solutions provided we extend the number systems still further. It will be
the purpose of the remainder of this chapter to introduce such an extension
in a reasonably informal manner. The resulting augmented set we shall call
the set of complex numbers and denote it by C. Since С will contain R as
a subset, it will be necessary to verify that the laws of combination (i)-(viii),
suitably redefined, hold in С also.
Up to this point, each extension of the number system has led inexorably
to a further extension. However, it has been shown (by С F. Gauss in the
Fundamental Theorem Algebra) that no further such extensions are needed,
in the sense that any polynomial equation with its coefficients in С will
have all its roots lying in С as well. But the real significance of the set С
goes deeper than simply enabling solutions to be found for any polynomial
380

2] MANIPULATION OF COMPLEX NUMBERS
equation: it reveals an underlying unity among mathematical concepts
that would otherwise remain unsuspected, for example between
trigonometric and exponential functions, suitably redefined.
2. COMPLEX NUMBERS AND THEIR
MANIPULATION
The extension of the real number system that was foreshadowed in Section
1 is effected as follows: we introduce a new symbol j and define a complex
number to be an expression of the form a+b] where a, b are any real
numbers. (The sign + is used at the moment simply to unite the two
components a and b] of the complex number: the operation + has only been
defined so far for combining real numbers.) For the moment, j is to be
regarded purely as a new symbol: an interpretation of j will appear as soon
as we have defined operations between complex numbers. Before we do
so we must define equality: two complex numbers a+b] and c + d] are equal
if and only if a = с and b = d.
Addition, subtraction and multiplication of complex numbers is defined
to proceed precisely as if each complex number were a polynomial in j,
where j is subject to the condition j2 = — 1.
Thus for example,
(3+j) + (5-2j) = 8-j
(3+j)(5-2j)=15-6j + 5j-2j2
= 17-j (since j»=-l).
Ex. 2. The result of dividing the polynomial P(x) by x2 +1 may be written in the
form P(x) = (x2 +1) Q(x) + ax + b.
Discuss the relation of this polynomial identity to the concept of a complex
number. (Put л: = j.)
Ex. 3. Simplify: (i) (2+j)-(3-j), (ii) (7-7j) + (4+3j), (iii) (2-j)(2+3j),
(iv)(l + 3j)2, (v)(l+3j)3.
In the same way that constant polynomials were identified with real
numbers, so a complex number of the form α + Oj can be identified with
the real number a. Thus we write
2+0j = 2 and 2(3-4j) = (2+0j)(3-4j) = 6-8j,
as might have been expected. 0+0j is written as 0 (the complex number
zero). Similarly, we abbreviate 0 + 6j to b].
The existence of a multiplicative inverse for any non-zero complex
number has already been postulated in Section 1. To find an explicit form for
such an inverse we use the fact that
(a + b])(a-b]) = a2-(6j)2 = a* + b\
381

COMPLEX NUMBERS (1)
1 2+] 2+j
\<-—i) —
more generally,
a + b] a—b]
Thus ^-^_ — — _--.
(я+ВД-1 = whz τΞπ = ά-iJ ^ b>not both zer°)·
Ex. 4. Show that the multiplicative inverse of (2—j) obtained above is unique;
that is, show that ,„ ... . . , ,
(Z-j)-1 = x+y]^x = f, y = i.
Division of complex numbers by complex numbers now follows:
ζ 4- w means zw~\ For example
4-j _ (4-j)(l~3j) _ l^j _ ! _is-
i+3j~(i+3j)(i-3j) ίο Λ loJ·
Ex. 5. If Zj = 2+ 3j, z2 = 3-4j, express in the form a+oj:
(i) 2г1-Зг2, (ii) (2Ζι-ζ2)2, (iii) (z1-2jz2)2.
Ex. 6. Express as complex numbers in the standard form a+by.
(i) 1/(2 + j), (ii) (2 + 3j)/(l + j), (iii) {(1 + j) (1 + 2j)}/(l + 3j).
*Ex. 7. Taking 0 = 0 + 0j as the additive identity and 1 = l+0j as the
multiplicative identity, verify that laws (i)-(viii) of Section 1 hold for complex
numbers.
The interpretation of α + Oj as a real number enables us to speak
meaningfully of the square roots of a negative real number. Thus, if a e R,
then the real number — a2, regarded as an element of C, is the square of the
complex number a] and also of the complex number — <zj.
Given a complex number ζ = a + b) we call a the real part of ζ and b
the imaginary part of z:
a = Re (z), b = Im (z).
(Note that the imaginary part of a complex number is raz/!)
If ζ is non-zero, we have seen that the reciprocal of z, \j{a + b]), is
obtained as follows:
_! = _[_ = _J_ а-Щ = fl-ij
a + b] a + b] a—b] a2 + b2'
The number ζ* = α—6j, obtained by writing — j for j in z, is called the
complex conjugate of z. The real number \z\ = V(fl2 + *2) is called the
modulus of z. Thus we have shown that, for any non-zero complex number z,

2] MANIPULATION OF COMPLEX NUMBERS
Ex. 8. Write down the real and imaginary parts of the following complex
numbers:
(i) 3-2j, (ii)(3-2j)2, (ШИЗ-г.])-1.
Ex. 9. Show that Re, Im may be interpreted as functions С -» R. Find
Re [Im (a + b})] and Im [Re (a + b})],
R being regarded as a subset of C.
Ex. 10. If zx = 2-j, z2 = 3 + 2j find:
(i) z*, (ii) z*, (iii) (Zl + Z2)*, (iv) (ζιΖ2)*, (ν) (1/zO*, (vi) (Zl + 3z2)*, (vii) (z« + z|)*.
Ex. 11. The conjugacy function/: С -» С is defined ЬуДг) = ζ*. For what subset
of С is /the identity function?
fit. 12. Find:
(i) |3+4j|, (ii) |2j|, (iii) |l/(l-j)|, (iv) |cos0+j sin0|, (v) |l-cos θ+] 5ΐηθ\.
Ex. 13. The word 'modulus' has, prior to its use in this chapter, been used in the
context of real numbers only, to mean ja2. Show that the two uses coincide if
R is regarded as a subset of C.
*Ex. 14. If zl5 z2 are two complex numbers, and z2 Φ 0, prove that:
(i) (Zl + z2)* = z* + z*, (ii) (Zlz2)* = z*zt, (iii) zlZl· = \ζλ\\
Ov) R* = ΐ = rii- (v) ζχ + zf = 2Re (Zl),
\z2/ z2 |z2|
(vi) z1-z* = 2jlm(z1).
The results proved in Ex. 14 are of great importance and should be
committed to memory.
The concept of modulus enables us, in a sense, to order complex
numbers, but this is not entirely analogous to the ordering of real numbers,
since different complex numbers can have the same modulus. It is not
possible to order the complex numbers in the same way as we order the
reals.
Exercise 17(a)
1. Ifzj = l + 2j,z2 = 2—j,z3 = 4 + 5j, express as complex numbers in the
standard form a+by.
(i) гг + гг + г3·, (ii) 3z!-z2 + 2z3; (iii) ζ^; (iv) z! + 2z3; (v) z2z3 + z3zx + z^;
(vQz^z,,; (vii)(z1-jz2)/(z2 + 2jz3); (viii) -+-; (ix) (z2 + z3)3;
(x) (z1-z2)(z2+jz3); (xi) j/tf+jzl); (xii) (г?-г2г3)/(г22-гзг1).
2. If Zj = cos <?! + j sin 0l5 z2 = cos 02 + j sin 02, show that
Zl z2 = cos (θ1 + θ'„) + j sin φχ + θ2),
and find zjz2. Evaluate z\ and (z1z2)2.
3. Solve the following equations:
(i) x2-Ax + 5 = 0; (ii) 2л:2-2л: + 1 = 0; (iii) x2-5x + l = 0;
(iv) jx2-2x-2j = 0.

COMPLEX NUMBERS (1) [17
4. If (5- 12j) = (a + b])2 where a > 0, find a, b. Find similarly complex numbers
which when squared give (i) j; (ii) 3 +4j.
5. Solve the equation x2-(4+j) лг + 5-j = 0, giving each root in the form
a + b].
6. Given that A(3 +2/) -5(1 -j)- (5 + 2j) = 0, find A and B: (i) if A, В are both
real; (ii) if A, B are conjugate complex numbers.
7. If ζ = 1 — cos Θ—j sin 0, write down the value of z*. Express z'1 as a complex
number in standard form.
8. If α is a real number and ζ is a complex number, prove that (1+az)* = 1 + az*.
Deduce that, in standard form, (1 + az)-1 = (1 + az*)/(l + 2a Re (z) + a2|z|2).
9. Given zeC, show that ζ is real oz = z*.
10. What is the conjugate of z*? By considering the product zxz* for any pair
of complex numbers zx and z2, prove that zx z* + z*z2 is real. What can you say
about the complex number ζ1ζ* — ζ^ζ2Ί
11. Show, by constructing an example, that non-real numbers a and b may be
found so that the quadratic equation
x2-ax + b = 0
has a real root.
Is it possible for the equation to have two real roots if a, b are non-real ?
12. Form the quadratic equation whose roots are:
(i) 2 + j, 2 - j; (ii) 2 - 3j, 2 + 3j; (iii) 2 - j,/3, 2 +jV3; (iv) 1 + 2j, 2 + j; (v) 4,1 + j.
Can you conjecture any general result about the coefficients of a quadratic
equation and its roots ?
13. By writing хг+у2 in the form хг — Qy)2, split x2+y2 into linear factors with
coefficients in C.
Factorize into linear factors with coefficients in C:
(i) л:2-2л:+2;00x44/·,(iii)x2+ 3y2;(iv)j(x2 + x+1)-1:(y)jx2-4xy-4]y2;
(vi)x3+l; (vii)jc*+l.
(Hint for (vii): the expression may be written (x2+l)2 — 2x2.)
14. By putting χ equal to j in the identity
χ*-3χ3-χ+2 ξ (л:2 + 1) Q(.x) + ax+b
determine the numerical values of a and b.
Find the remainder, on division by x2 +1, of (i) Xs +1; (ii) x9 +1.
15. Evaluate j" in the four cases η = Am, 4m+l, 4m+2, Am + 3 (m integral).
Find the sum of the series n
16. If (1 +j)" = a+b], prove that a2 + b2 = 2":
(i) by taking the complex conjugate of each side of the original expression;
(ii) by mathematical induction.
384

2] MANIPULATION OF COMPLEX NUMBERS
17. Evaluate the determinant
I 2 1+j 2 1
1-j -1 4-j ,
I 2 4+j 3 |
where j2 = — 1.
Explain how without evaluation, it could have been concluded that the value
of the determinant was real. (M.E.I.)
3. THE ARGAND DIAGRAM
As forecast in Section 1, we may set up a 1-1 correspondence between the
set С of complex numbers and the points in a plane by associating the
complex number x+yj with the point Ρ whose coordinates are (x, y)
referred to a given pair of rectangular coordinate axes. The χ axis
corresponds to the set R of real numbers, the у axis to the set of pure imaginary
numbers, while a general complex number a+ b], α ή= 0, 6 + 0 lies off the
axes in one of the four quadrants.
Ex. 15. Plot the points corresponding to the complex numbers l+2j, l-2j,
-j,+j. What is the geometrical relationship that exists between the points
representing ζ and z* ?
Demonstrate geometrically the result
ζ = ζ* ο zeR.
Such a geometrical representation of complex numbers is generally
referred to as the Argand diagram (J. R. Argand, 1768-1822)f or as the
complexplane.
As we saw in Chapter 3, there is a 1-1 correspondence between the set
of points P(x, y) and the set of position vectors OP = x\ + y\. Thus, the
complex number x+yj may be alternatively represented in the Argand
diagram by the point Ρ or by the position vector of P. Both representations
have their value, and we shall use both freely.
*Ex. 16. ΙΐΡ represents the complex number a + b] (P is the affix of the complex
number a + bj) show that |a + oj| = |OP|.
Consider now two complex numbers zx = at + 6ij and z2 = a2 + b2],
with affixes Рг andi>2. The sum гг + г2 = (a1+a2) + (b1 + b2)];b\it
OPi + OP2 = (β! + α2) i + fo + b2) j
t The first exposition of the geometrical treatment of complex numbers was in fact
published in 1797 by a Norwegian surveyor, Casper Wessel (1745-1818). For a
translation of his paper see The Treasury of Mathematics: 2 by Henrietta Medonick (Pelican).
385

COMPLEX NUMBERS (1) [17
and thus the position vector representing the sum of two complex numbers
is the sum of the position vectors which separately represent the two numbers
(see Figure 17.1).
Fig. 17.1 Fig. 17.2
*Ex. 17. Describe the vector representing the complex number z1 — z2.
Ex. 18. Mark in the Argand diagram the affixes of the complex numbers 2—j,
2(2—j) and |(2—j). Interpret multiplication of a complex number ζ by the
real number (i) a > 0, (ii) b < 0, in terms of an operation upon the vector OP
representing ζ in the Argand diagram.
Ex. 18 shows that multiplication of a complex number ζ by a positive
real number a is represented geometrically by an enlargement of OP
(possibly by a factor less than 1). Multiplication by a negative real number
— a both enlarges OP and rotates it through an angle π (see Figure 17.2).
*Ex. 19. Mark in the Argand diagram the affixes of the complex numbers 3 + 2j,
j(3 + 2j), — (3 + 2j), — j(3 + 2j). Interpret multiplication of a complex number ζ
by the pure imaginary number j in terms of an operation upon the vector OP
representing ζ in the Argand diagram. Show that multiplication of ζ by the pure
imaginary number bj is represented in the Argand diagram by an enlargement of
magnitude \b\ followed by (i) an anticlockwise rotation through \π if b > 0, or
(ii) a clockwise rotation through iniib < 0.
Ex. 20. Interpret geometrically the statement j2 = — 1.
To interpret geometrically the multiplication of two general complex
numbers, it is convenient to introduce a new concept, the argument of a
complex number z.
Given a complex number ζ = x+yj, represented by the vector OP, we
have χ = rcos Θ, у = r sin Θ, where r = \z\ = ■Jix'+y2) (see Figure
17.3).
The two real numbers, r and Θ, determine the complex number ζ
uniquely; conversely r is determined uniquely by z, but there is an infinite
386

3] THE ARGAND DIAGRAM
number of values of θ corresponding to z, differing from one another by
integral multiples of 2π. Each such value of θ is called an argument of z.
Just one of these values will lie in the interval —π<θζπ, and this value
is called the principal argument ofz, and is written arg z.
If arg ζ = φ, we may write
ζ = r(cos φ+) sin φ).
Fig. 17.3
A complex number may always be stated in this modulus-argument form.
For example
l+j = V2(^ + ^j) = V2(cosi77+jsini77),
l-j = V2(cos[-^]+jsin[-^]),
-l+j = V2(cos^+jsin^),
-1-j = V2(cos {-%*]+) sin [-%*]).
(Notice that we must always be careful to select the angle to lie in the range
— π < arg ζ ^ π.)
Example 1. Express in modulus-argument form the complex numbers
(i) -V3+j, (ii) 2-j, (iii) l+j cot α (-π < α < ^anda Φ 0).
(i) I — V3+j| = 2 and we have
= 2(cos|^+jsinfn').
(ii) |2-j| = V5 and we have
= 5(cos α+j sin a),
where a is the angle in the range -\n < α < 0 such that tan α = --£.
387

COMPLEX NUMBERS (1)
[17
(iii) First observe that
|l+jcota| = V(l+cot2a) = |coseca|.
If 0 < a < π, |cosec a| = cosec a and we write
1 +j cot α = cosec α (sin α+j cos a)
= cosec a (cos φ+] sin φ),
where φ is chosen so that . ^
r -η < φ «ϊ π
and sin φ = cos α
and cos φ = sin a.
Thus, for 0 < a < π
1 +j cot a = cosec a[cos (i^-a)+j sin (i?r-a)].
If — я < a < 0, |cosec a| = —cosec a and we write
1 +j cot α = |cosec a| (—sin a—j cos a)
= |coseca|(cos 0+j sin Θ),
where (9 is chosen so that
— π < θ ^ π
and sin θ= -cos α
and cos θ = sin a.
The reader should verify that this gives
1+j cot α = |coseca|[cos (-^-a)+j sin (-|я-а)]
for — я < α < 0.
£лг. 21. Find the modulus and principal argument of each of the following
complex numbers, and write each in modulus-argument form
(i) 1+V3J; (ii)2 + 2j; (iii) -1; (iv) j; (v) -j; (vi) -,/3-j; (vii) ,/3-j;
(viii) 1 + j tan a; (ix) tan α+j; (x) 1 -cos a-j sin a (use half angle formulae).
*Ex. 22. Show that arg ζ = -arg z* and hence find arg z"1 in terms of arg z.
Ex. 22. By writing ζ in the modulus-argument form r(cos θ+j sin #), discuss the
geometrical interpretations for the multiplication of the complex number ζ
by: (i) the real number a > 0; (ii) the real number a < 0; (iii) the pure imaginary
number bj, b > 0; (iv) the pure imaginary number bj, b < 0.
Returning now to the question of the geometrical interpretation of the
multiplication of two complex numbers ζ and w, let us write z, w in the
modulus-argument form:
ζ = r (cos θ + j sin Θ), w = s (cos φ + j sin 0).
388

3] THE ARGAND DIAGRAM
Then
wz = sr (cos Φ+) sin φ) (cos 0+j sin Θ)
= sr [(cos φ cos θ -sin φ sin 0)+j (cos 0 sin (9 + sin φ cos (9)]
= sr [cos (0 + 0)+j sin (0 + 0)],
and thus the product, wz, is a complex number with modulus |vc||z| and
an argument argw+argz. In words, the modulus of a product is the
product of the moduli and an argument of the product is the sum of the
arguments. Geometrically, if the affix of ζ is P, multiplication by w
(i) enlarges OP by a factor |vc|, and (ii) rotates OP anticlockwise through
an angle arg w.
Ex. 24. If ζ is any complex number and w = u+vj (и, ν real and positive), plot
in the Argand diagram, the affixes of z, jz, uz, jvz, uz+jvz. Deduce that
\wz\ = \w\\z\ and arg (wz) = arg w+ arg z.
Is it necessary to modify your demonstration in any way if either и or ν is
negative?
R(wz)
y\ / I
/ /^7m
o\ ДО
Fig. 17.4
In Figure 17.4, /, P, Q are the affixes of 1, z, w respectively. If each side
of triangle IOP is enlarged by a factor | w\ and the triangle is then rotated
anticlockwise through an angle arg w about O, I is brought into coincidence
with Q. Let Ρ be brought into coincidence with R (so that the triangles
OIP, OQR are similar). Then
\OR\ = \w\ \OP\ = \w\\z\
and 10R = arg w+arg z; thus R is the affix of the complex number
wz. (Notice that IP, which represents the complex number ζ -1, is rotated
into the displacement QR, which represents the complex number w(z — 1).)
389

COMPLEX NUMBERS (1)
Ex. 25. Form the product (1 +j) (д/3+j) and interpret your result geometrically.
What are the values of cos 75° and sin 75° ?
Now suppose that w, zx and z2 are any three complex numbers. Let
P, Q, be the affixes of z1 and zx + z2 respectively (Figure 17.5). Then OP
represents ζλ and PQ represents z2. Now enlarge triangle OPQ linearly
by a factor |vc| and rotate through an angle arg w to bring it into
coincidence with the triangle ORS. Then OR represents vezl5 RS represents wz2
and OS represents w(>i + z2). But OS = OR + RS and we have thus
verified the distributive law . .
w(zi + z2) = wz-l + wz*
.Slw^+z,)]
У.
0
lz-l= —
®-
Fig. 17.5
Ex. 26. If w φ 0, show that
and that arg
or differs from arg z—arg w by 2π.
*Ex. 27. Draw diagrams similar to that of Figure 17.4 to illustrate the
construction from the affix of the complex number z, of the affixes of the complex
numbers (i) z2; (ii) z3; (iii) z~\
Example 2. zx and z2 are arbitrary complex numbers. Give a geometrical
verification of the triangle inequalities
h-ζ,ΐ ζ \\Zl\-\z2\\.
Describe geometrically the set S = {ze C:\z-1— j| ^ 1} and show that
ifzeS, then
V2-1 ^ |z-2| ^ V2+1-

3]
THE ARGAND DIAGRAM
In Figure 17.6, OP represents zu PQ represents z2 and thus OQ
represents z-l + Zz and RP represents z1 — z2, where OPQR is a parallelogram.
B\it,inAOPQ, |OQ| ^ |OP| + |PQ| (with equality only if OPQ is a straight
line) and thus . , . „ . , , . ,
Again, in ΔίλΡΚ,
|OP| ^ |OR| + |RP| and |OR| ^ |OP| + |RP|;
thus, |Zl| ^ |г2| + |г!-г2| and \z2\ a |zi| +|zx-z2|,
which combine together to give
Ι^-ζ,ι > |N-|z2||.
Fig. 17.6
For the second part of the question, we know that \z — w\ gives the
distance between the affixes of z, w in the Argand diagram, and the set S
is thus represented by the circumference and interior of the circle, centre
(1+j) and radius 1.
We demonstrate two methods of deducing the final part of the question.
Method (i) {Analytical)
From the inequalities
\ъ-ъ\> 1Ы-Ы1, W+zt\^ Ы + Ы,
\z-1\ = |(z-l-j)-(l-j)l > ||z-l-j|-|l-j| ^ V2-1,
\z-2\ = |(z-l-j) + (-l+j)| ^ |z-l-j| + |-l+j| ^ V2+1-
Method (ii) (Geometrical)
Let С be the centre of the circle \z— 1— j| = 1 (which touches the real
axis at A, say) and let В be the affix of the number 2. Let ВС cut the circle
atP, Q (see Figure 17.7).
391
we have

COMPLEX NUMBERS (1)
[17
For given real a > 0, \z — 2| = a represents a circle, centre В and radius
a. Of all such circles, the smallest containing a point of S (that is, the
circumference and interior of the circle \z— 1 —j| = 1) has radius BP and the
largest a radius of BQ. But, by Pythagoras, ВС = •N/2andthus5i> = ^J2-l,
BQ = ^2+1 and the result follows.
Ex. 28. The triangle inequalities stated in the Example above may be deduced
without recourse to geometry. Show that
Zj z2* + zf z2 = 2 Re (zj z2*)
and deduce that |z! + z2|2 = |Zi|2 + 2 Re (z1z^)+ |z2|2.
Show also that Re (zx z2*) < | zx 11 z21
and deduce that |zi + z2| < |z!| + |z2|.
Prove the second triangle inequality in a similar fashion.
Consider now the effect of multiplying the complex numbers 0, 1,
l+j.j by a+by. we get
0, a + b), (a-b)-(a+b)], -b + aj
—the vertices of another square (Figure 17.8).
(fl-b,a+b)
(0,1)
(1
(0,0)
(1
1)
0)

3] THE ARGAND DIAGRAM
The result is strongly reminiscent of the linear transformation with
^^ /« -*\ /1 0\ JO -14
(* a)=a[0 l)+*(l θ)
= αϊ+6J, say.
The result suggests that it might be profitable to associate the number 1
with the matrix I, the number j with the matrix J and the general complex
number a + b] with the matrix al + b5.
Consider now two complex numbers a+b], c + d] and the associated
matrices , -
e d- с "3-
We have
(i) (a+bj) + (c+dj) = (a+c)+(b + d)j
(ii) (e+uj)(c + aS) = (ec-M) + @c + <fc)j
and (b aj[d cj-\bc + da ac-bd)'
The 1-1 correspondence between the set С of complex numbers on the
one hand and the set of all 2 χ 2 matrices of the form
on the other is thus seen to preserve the structures of addition and
multiplication.
Ex. 29. Confirm, by direct computation of (a + oj)-1 and the inverse of the matrix
I I that the structure of division is also preserved.
*Ex. 30. Interpret the correspondence between the complex number cos θ+j sin#
and the matrix I . . J in the light of their geometrical properties.
\ sin a cos 0)
Ex. 31. If C, R have their usual meanings and Μ is the set of all matrices of the
form I J, a, b e R, determine the nature of the function /: Μ -> R which
corresponds to the modulus function С -> R.
To summarize, we have the following representations of the complex
number a + b}:
(i) the point (a, b) in the Argand diagram;
393

COMPLEX NUMBERS (1) [17
(ii) the vector ax + b\;
(iii) the matrix al+bJ where
>-n-'-(?-i)·
Representation (i) is, as it were, a static representation: it is useful in that
it enables us to discuss sets of complex numbers in geometrical terms, but
by itself, it lacks the additive and multiplicative structure of complex
numbers.
Representation (ii) is, as it were, dynamic: addition of vectors and
addition of complex numbers have the same structure. However, multiplication
of complex numbers has to be interpreted by a new operation on vectors:
dot products of vectors on the one hand and products of complex numbers on
the other do NOT have the same structure.
Representation (iii) gives a. full structural representation of the complex
numbers in terms of matrices: addition, subtraction, multiplication and
division of complex numbers have their exact analogue in the
corresponding operations on the associated matrices.
Example 3. The affixes of the complex numbers a = 3+j and b = 1 +2j
are A and В respectively. Find complex numbers ρ, ρ', q, q' with affixes
P, P', Q, 0! such that ABQP andABQ'P' are squares.
Find also complex numbers r, r' with affixes R, R' such that ABR, ABR'
are equilateral triangles.
AB = b-a
= -2+j.
To obtain AP and AP' we must rotate AB through ± \v\ that is, we must
multiply {b-a) by ±j (see Ex. 19). Thus
AP = j(-2+j)
= -l-2j
and AP'=-j(-2+j)
= l + 2j.
Thus p = a+(-l-2])
= 3+j-l-2j
= 2-]
and p' = Ъ+] + \+2]
= 4 + 3j.
394

THE ARGAND DIAGRAM
Similarly q = b + (-l-2})
= l+2j-l-2j
= 0
and q = l+2j+l+2j
= 2 + 4j.
To rotate AB through an angle ± \π we seek complex numbers ζ such
that \z\ = 1 and arg ζ = +%π; thus
ζ = cosyn-lj sin \π
= i(i±JV3).
It follows that AR = Ц1 + j V3) (6 - a)
and thus /■ = a+ i(l +JV3) (*- a)
= 3+j+Kl+W3)(-2+j)
= 2-W3 + (f-V3)j.
Similarly AR' = $0 - JV3) (* - a)
and thus r' = 3+j + i(l-JV3)(-2+j)
= 2 + W3 + (f + V3)j.

COMPLEX NUMBERS (1)
Exercise 17 φ)
1. Determine the modulus and principal argument of each of the following
complex numbers, where necessary leaving the modulus in surd form and giving
the argument in degrees, correct to the nearest degree: (i) 3 —4j; (ii) 2 + 3j;
(iii)-l-2j;(iv)i+j; (v)-2 + 5j; (vi) -2-j; (vii) 12-5J; (viii) -9-40J.
2. Simplify the following complex numbers:
(i) (costf+jsintf)3; (ii) (cos θ+j sin 0)/(cos ф-j sin?5);
(iii) (cos β+j sin Θ) (sin Θ+j cos Θ); (iv) (cos ^π+j sin £π)2;
(ν) (cos tjtt+J sin -^tt) (cos t&t +j sin γ^π);
(vi) (cos xr?r + j sin jjtt) (cos -γψτ — j sin γ^π);
(vii) (cos^+j sin^Vicos^-J δίη^π);
(viii) (1 +cos Θ+j sin Θ) (1 - cos 0-j sin φ).
3. Determine the principal argument of the complex number
(sin β+j cos Θ) (cos φ+j sin 0)
(i) when —\n < 0 — β < |π; (ii) when |τ7 < φ — θ < |π.
4. If Ρ is the affix of the complex number z, show how to construct geometrically
the affixes of: (i)l+z; (ii)l-2z; (iii)2j + 3z; (iv) 1-jz; (v) (1-j) z; (vi)j + z2;
(vii) z + z2; (viii) (l+j)(l-z).
5. If Ρ is the affix of the complex number z, show how to construct the affixes
of: (i) z*; (ii) z-z*; (iii) 1+2/z; (iv) (1+z)*; (v) 1/(1 +z); (vi) j/(j-z);
(vii) z/(l-z);(viii)(l-z)-2.
6. If Λ Q are respectively the affixes of the complex numbers z, w show how to
construct the affixes of: (i) 2z+ve; (ii) ^(ζ + 2κ>); (iii) £(3z-ve); (iv) z+jw;
(v) (1 +j) (z + иО; (vi) (z+1) (и>+1); (vii) w/z; (viii) j(z-w).
7. Describe geometrically the set S = {z e C:|z-2-j| < 1} and prove that, for
allze5, V5-1 < |z| < V5+1.
8. Describe geometrically the set S = {z e C:|z-2| < 2} and prove that, for
allzeS, 3 =£ |z+2 + 3j| < 7.
9. Show that |z-6j| < 3^ 10 < |z-5 + 6j| < 16.
10. Show that |z-3-j| < 1^ V10-1 < \z\ < VlO+1.
11. Prove, using mathematical induction, the following results:
(i) |Zl + z,+ ...+z.| =S lzxl + Ιζ,Ι + ...+ Ιζ,Ι;
(ii) \ζΛζ,...ζη\ = |г1||г,||г1|...|г.|;
(iii) aig(z1z1zt...z^ = Σ argz*;
(iv) (z1+zt + z, + ...+z^* = Σ ζ*;
(ν) (zxz2z3...zn)* = ζ*ζ%ζ?...ζ%.
12. The function/: C-* С is defined ЬуДг) = 2z+j. Describe the effect of this
mapping geometrically and find the value of ζ which is invariant under the
mapping, again interpreting your result geometrically.

3] THE ARGAND DIAGRAM
13. Answer the same question as in Question 12 (substituting where necessary,
values for value) for the functions denned by: (i) f(z) = jz+1; (ii) f{z) = z2 + 1;
(iii) flz) = z2 +1 + j; (iv) f{z) = 2z* +1 + j.
14. Show that the points representing the complex numbers 3 + 4j, 7 + 2j, 5 + 8j
in the Argand diagram are three vertices of a square and find the fourth vertex.
Show that the interior of the square represents the set
{zeC.z = 3 + 4j+A(2-j)+/i(l+2j), real λ, μ, 0 < λ < 2, 0 < μ < 2}.
15. Prove that, if S= {zeC: \z-2\ < 1}, then, for all ζ e S, -%n < argz < £π.
16. Prove that the modulus of the quotient of two complex numbers is the
quotient of the two moduli, and that the difference of the two arguments is an
argument of the quotient.
(i) If ζ is a complex number, and = 2, prove that the locus of the
point which represents ζ in the Argand diagram is a circle, and find its centre and
radius.
(ϋ) If "86τϊ)=^·
find the locus of the point which represents z. (O & C)
17. If ζ is the complex number cos 0+j sin Θ, express 1/(1 +z) and 1/(1 -z) in
the form χ+yj.
If \z\ = 1, prove that the real part of 2z/(l -z2) is zero. (O & C)
18. If A, B, C, D are the affixes of the complex numbers a, b, c, d prove that
a + c = Ьл-do ABCD is a parallelogram.
What can you say about the parallelogram if a + bj = c + djl
19. Explain with the help of a sketch, why no complex number ζ can be found
such that argz = j* and |z-4-j| = |z-2-3j|.
20. a, b are real numbers; show that
is represented by a straight line in the Argand diagram.
Miscellaneous Exercises 17
1. By writing 1 =-j2, express ζ = (V3 + l)+j(V3-l) as the product of two
complex numbers and hence write down the values of \z\ and arg z.
2. If |z—2—j| < 2 and \w— 5 — 5j| < 1, find the maximum and minimum values
of|z-Hi
3. Describe geometrically the relation between the complex numbers 6j and
2 + 4j and the complex numbers (i) 1 + 5j; (ii) 3 + 3j.
If λ is an arbitrary real number, what can you say about the complex number
(l+A)+j(5 + A)?(over)
397

COMPLEX NUMBERS (1) [17
If B, Care the affixes of the complex numbers 6j, 2 + 4j, find the two complex
numbers whose affixes, A, A' are such that the triangles ABC, А'ВС are
equilateral.
4. zu z2, и>1, ve2 are four complex numbers and
By considering det (AB), prove that the product of two sums of four squares
is itself the sum of four squares.
5. Write down the product of the complex numbers хг+у^ and хг + уг].
Show that this product has modulus rxr2 and argument θ1 + θ2, where rlt r2 are
moduli and 0l5 #2 the arguments of the given numbers. Solve the equation
^ z-l = ^_
3 + 4] 5j 3 —4j
Show that, if (z-2)/(z-j) is real, the point corresponding to ζ = x+yj in the
Argand diagram lies on a straight line through the points 2 and j, and find the
equation relating χ and у when the ratio is purely imaginary. (O & C)
6. If the complex numbers zl5 z2, z3 are represented in the Argand diagram by
points ZltZ2,Z3, interpret geometrically (giving a justification) the modulus
and argument of the complex number (z3 — z^l{z2— zj.
If the complex numbers a, b, c, x, y, ζ are represented by points A, B, C, X, Y,
Zand if
x-c _ y-a _ z-b
b — c c—a a—b'
prove that the triangles ВСХ, С Α Υ, ΑΒΖ are similar.
Prove also that the triangles ABC and XYZ have the same centroid.
(O&C)
7. The cross-ratio of four complex numbers zl5 z2, z3, z4, written (гггг, ζ3ζύ, is
defined by
, , , \ _ (.Zi-z3Hz2-z4)
(ZlZ2'Z3Z4)-(!^J^7)·
Show that, Ί{(ζ1ζ2, ζ3ζ^ = λ, then the twenty-four possible permutations of
the numbers zl5 z2, z3, z4 give rise to six distinct cross-ratios with values λ, 1 — λ,
λ-ι, (1-λ)-1, λ(λ-1)-1, λ-\λ-1).
If a,b,c,d are four complex numbers, and the complex numbers Wi,Zi
(i = 1, 2, 3, 4) are connected by the relation
_ аг{+Ь
czi + a"
prove that (ζχζ2, ζ3ζ^ = (w^w^, w3w^).
8. A complex number zr is represented on the Argand diagram by a point Zr.
Prove that arg [{z3-ζ^Ι{ζ2-z-^j] is equal to the angle Z,,ZXZ3.
If zl5 z2, z3, z4 are distinct, and the cross-ratio {.z2-z^(Zi-z3)l{Zi~z^) (z2-z3)
(that is (z^Zi, ζχζ3)) is real, prove that, in general, Z1; Z2, Z3, Z4 lie on a circle.
In the exceptional case, what is the relation between Z1; Z2, Z3, Z4?
(O&Q
398

3] MISCELLANEOUS EXERCISE
9. The complex numbers a, b, c, x, y, ζ are represented in the Argand diagram by
points A, B, C, Χ, Υ, Z.
(i) If the triangles ABC, XYZ are directly similar (that is, corresponding
angles are equal in sense as well as magnitude) prove that
Ι ι ι ι I
(ii) If the triangles ABC, XYZ have equal a
| ι ι ι | | ι ι ι |
10. If β and у are complex numbers whose moduli are both equal to 1, prove
that (1+βγ)/(β+γ) is real.
Hence, or otherwise, prove that, if a, b, с are complex numbers whose moduli
are all equal, then (a2 + bc)/[a(b + c)] is real. (O & C)
11. The points А, В, С are the affixes of the complex numbers a, b, с in the
Argand diagram. If the circumcentre of the triangle ABC is at the origin, prove
that the orthocentre H, is the affix of the point a + b + c. Deduce that O, G, Η
are collinear, where G is the centroid of the triangle, and that OH = 30G.
12. Points A, B in the Argand diagram represent complex numbers a, b
respectively. О is the origin, and Ρ represents one of the values of j(ab). Prove that,
if О A = OB = r, then also OP = r, and OP is perpendicular to AB.
А, В, С lie on a circle with centre O, and represent complex numbers a, b, с
respectively. Prove that the point D which represents -bc/a also lies on the circle,
and that AD is perpendicular to ВС.
The perpendiculars from В, С to С A, AB meet the circle again at E, F
respectively. Prove that О A is perpendicular to EF. (O & C)
13. A complex number ζ is represented by a point Ζ on the Argand diagram.
Prove that multiplication of ζ by w is represented by taking the point Z' on OZ
(where О is the origin) such that OZ' = OZ\w\ and rotating OZ' through the
angle arg w.
If ζ is represented by a point on the circle of radius a which touches the ^-axis
at О and lies in the first and fourth quadrants, prove that
z-2a = jz tan (arg z). (O & Q
14. A point Ρ representing the complex number ζ moves in the Argand diagram
so that it lies always in the region defined by
|z-l|<|z-j| and |z-2-2j|<l.
Indicate on a sketch the region within which Ρ lies.
If Ρ describes the boundary of this region find: (i) the value of |z| when arg ζ
has its smallest value; (ii) the values of ζ in the form x+jy when arg (z— 1) = \π.
(London)
399

COMPLEX NUMBERS (1) [17
15. In the Argand diagram А, В, С and D are four points representing the
complex numbers zl5 za, z3 and z4 respectively. If О is the origin, prove that the
triangles OAB and OCD are similar if
£i_ f_3
z2 z4"
The complex number ζ is given by
= (Q±*Y
13J-4/ ■
Find, by drawing to scale, the position of the point representing ζ on the Argand
diagram. Check your result by calculation. (London)
16. The complex number ζ = x+jy = r (cos θ+j sin Θ) is represented in the
Argand diagram by the point (x, y). Prove that, if three variable points z1; z2, z3
are such that z3 = λζ2 + (1 — λ) zu where λ is a complex constant, then the triangle
with vertices zl5 z2 and z3 is similar to the triangle with vertices at the points 0,1
and λ.
ABC is a triangle. On the sides ВС, С A, AB triangles BCA', CAB', ABC'
are described similar to a given triangle DEF. Prove that the centroids of the
triangle ABC and A'B'С are coincident. (O&C)
400

18. Polynomials and partial fractions
1. THE EVALUATION OF POLYNOMIALS
(Unless specifically stated otherwise, we shall assume that all polynomials
have integral coefficients.)
We have seen in Chapter 4 that, given a polynomial P(x) the process of
dividing P(x) by x-a leads to the identity
P(x) = (x-a)Q(x)+R,
where Q(x) is a polynomial of degree one less than the degree of P(x).
The actual coefficients of Q(x) are obtained by the process of long division,
but the process may be shortened by the method of synthetic division
which consists, essentially, of comparing coefficients on both sides of the
above identity, as illustrated in the following example.
Suppose that P(x) = 2х* + Зл:8-5л: + 7 is to be divided by x+3. Then
the quotient, Q(x) will be a cubic polynomial, and we have
2л4+Зл;3 + 0л;2-5л; + 7 ξ (x + 3)(ax3 + bx2 + cx + d) + R.
Comparing coefficients
a = 2, 3a+b = 3, 3b + c = 0, 3c + d = -5, 3d+R = 7
and, working from the left, we have, by successive substitution
a = 2, b=-3, с = 9, d = -32, R = 103.
Observing that successive coefficients of Q(x) are obtained by multiplying
the preceding coefficient by — 3 and adding the result to the corresponding
coefficient of P(x), the work may be set out succinctly:
2 3 0-5 7(-3
-6 9 -27 96
2 -3 9 -32 103
giving Q(x) = 2*3-Эх2+ 9*-32 and R = 103.
Example 1. Find the quotient and remainder on dividing Ах* — 5хг + 1х + 2
byx-5.
4 0-5 7 2(5
20 100 475 2410
4 20 95 482 2412
Thus Q(x) = 4х3 + 20х3 + 95л;+482 and R = 2412.
401

POLYNOMIALS AND PARTIAL FRACTIONS [18
It will be seen that the above procedure constitutes an efficient method
for evaluating R = P(a), even if the coefficients of Q(x) are not required
explicitly.
Ex. 1. Explain how the method above may be modified for division by ax + b,
α Φ 1, by first dividing P(x) by a. Obtain Q(x) and R when
P(x) ξ 2xl-15x3-33xs-x+U
is divided by 2л: + 3. (In such cases Q(x) will not, of course, generally have integral
coefficients.)
Given a polynomial P(x) of say, degree four,
P(x) = a0xi+a1x3+a2x2 + a3x+ai,
the procedure outlined above to calculate P(b) amounts to finding suc-
a0, аф + а^, b{a0b+a^ + a2, Ь[Ь(а0Ь + а^)+а^+а3
and finally, P(b) = b{b[b(a0b + a^+aj+a3}+a4.
Thus the polynomial P(x) has been nested into the form
P(x) ξ x{x[x{a0x + ai) + a2]+a3}+al,
and P(b) calculated by substituting χ = b and working outwards from the
middle bracket. The process is often referred to as nested multiplication:
the reader familiar with the use of hand calculating machines will see that,
using the facility of back transfer, the whole computation may be effected
without having to record any numbers and clear the machine.
Ex. 2. Use a hand calculator to evaluate P(37) where
P(x) = 15χ*-61λ:3 + 46λ:2 + 18λ: + 56.
Details of the use of hand-calculating machines in evaluating
polynomials and, indeed, in a wide variety of other applications will be found
in Numerical Mathematics by A. J. Moakes.
It is often useful to express a polynomial P(x), of degree n, in the form
P(x) = A0(x-a)n + A1(x-a)n-1 + A2(x-a)n-2 + ...+An-1(x-a) + An
(A0, A1} A2, ..., An integral). Such a reduction may be effected by successive
applications of Horner's method of synthetic division. (See Example 1.)
402

U EVALUATION OF POLYNOMIALS
Example 2. Express 4л;4 - 5л;2 + 7л; + 2 in the form
A£x - 5)4 + A,(x -5)a+A2(x- 5)2+A^x - 5) + A,.
4 0-5 7 2(5
20 100 475 2410
4
4
4
20
20
40
20
60
20
95
200
295
300
595
482
1475
1957
2412
4 | 80
ThusP(x) = 4(л— 5у + Щх- 5)3 + 595(л— 5)2+1957(л— 5) + 2412.
Ex. 3. Show that, when χ χ 5,
P(x) ξ 4х4-5;с2 + 7л;+2 « 1957x-7373.
Find Д5-1) correct to 2 significant figures.
Ex. 4. With the notation of Ex. 3, what is the equation of the tangent to the
curve у = P(x) at the point (5, 2412)?
Ex. 5. With the notation of Ex. 3, evaluate P(5), P'(5), P"(5).
Ex. 6. Show how, in Example 2, the values of Ait A3, A2, Аъ А0 enable us to
calculate P(5), P\5), P"(5), P"'(5), i"v(5).
2. DIFFERENCES
Consider the quartic polynomial
/(л-) = л-4-5л;3 + л; + 2.
Tabulated overleaf are the values of f(x) for χ = -4(1)5 (that is, for all
values of χ in steps of 1 from χ = -4 to χ = 5). The first and second
columns give respectively the values of л; and f(x), while each subsequent
column gives the successive differences between entries in the preceding
column.
It is seen from this table that, although the first and second difference
columns follow no readily discernible pattern, the third difference column
contains the terms of an arithmetic sequence and the fourth differences are
constant. This is no coincidence: for any polynomial of degree n, the nth
differences are constant (see Ex. 11).
403

POLYNOMIALS AND PARTIAL FRACTIONS
X
-4
-3
-2
-1
0
1
2
3
4
/
574
215
56
7
2
-1
-20
-49
-58
A/
-359
-159
-49
-5
-3
-19
-29
-9
65
Δ2/
200
110
44
2
-16
-10
20
74
Δ3/
-90
-66
-42
-18
6
30
54
Δ4/
24
24
24
24
24
24
Δ5/
0
0
0
0
0
If we are content to assume for the moment that the fourth differences
continue constant, it is an easy matter to extend the table working
backwards from the constant fourth difference, to supply further values for
the polynomial:
x f Af Δ2/ Δ3/ Δγ

2] DIFFERENCES
Ex. 7. The cubic polynomial f(x) has the values/(0) = 2,Д1) = -8,/(2) =-20,
f(3) = —28. Form a difference table and deduce the values of Д4), Д5).
.Ex. 8. Explain the following check on a difference table : the sum of any column
of differences is equal to the difference between the first and last entries in the
preceding difference column.
Differences may be calculated for non-integral values. For example, in
the table below, differences are calculated for the polynomial
f(x) = x-x3
for values of χ = 0 (0-1) 0-6. (That is, for values of χ from χ = 0 to χ = 0-6
in steps of 01.) Since the values of f(x) are calculated to 3 decimal places
all the differences will also be to 3 decimal places and it is unnecessary to
enter the decimal point; for example, the constant difference (Δ3/), written
as - 6, is really - 0-006.
X
0
01
0-2
0-3
0-4
0-5
0-6
/
0
0-099
0-192
0-273
0-336
0-375
0-384
Δ/
99
93
81
63
39
9
Δ2/
-6
-12
-18
-24
-30
Ex. 9. Extend the table above, back from the constant difference in the Δ3/
column, to obtain Д-0-1) and/(0-7).
If we denote the successive values of the variable χ by л;0, х1г х2, ■■■,
where the step between successive values of χ is constant, the corresponding
values off(x) may be written f0,A,f2, .... Again, we write
Afr=/r+1-/r
and A% = Δ(Δ/,) = Afr+1-Afr, etc.
Similarly, if we work back from x0 through values ли, jc_2, X-a> ···to obtain
2 PPMn 405

POLYNOMIALS AND PARTIAL FRACTIONS [18
values /_i,/_2,/-3, ... of the polynomial, the successive differences are
written ΔΑ, Δ/_2, ···, and Δ2Α, Δ2Α, ···. etc.:
Δ/ Δ2/ Δ3/
Δ/-.
Δ2Α
ΔΑ Δ3/_2
Δ2Α
Δ/0 ΔΑ
Δ%
Δ/
Tf we wish to refer to an explicit value of x, we may write
4/t*) =Λ* + Α)-Λ*),
А2Дх) = Δ/(χ + /ζ)-Δ/(χ),
where /г is the difference between successive values of л; for which the
polynomial/^) is being tabulated.
*Ex. 10. If f{x) = xix-\) ix-2) ... ix-n+ 1) (and thus fix) is a polynomial
of degree ri) prove that, if /is tabulated for integral values of л:,
(i) ДДл:) = nxix-l)ix-2)...ix-n + 2),
(ii) Δ«Λ*) = и(и-1)*(х-1)(х-2)...(х-и + 3).
Suggest a form for Д'Дл:) and prove your conjecture by mathematical induction.
*Ex. 11. Prove that any polynomial fix) of degree η may be expressed in the
form
fix) = AlXix-l) ix-2) ...ix-n+l) + AzXix-l)...ix-n + 2) + ... + Anx+An+1,
where АиАг, ...,Ап+1 are numbers, and deduce, using Ex. 10, that the nth
differences of an nth degree polynomial are constant. Prove furthermore, that
Δ-/= Αχη\.
*Ex. 12. Prove, by mathematical induction, the Gregory-Newton formula
fin) =fi0) + (^Afi0) + ... + h Δ7·(0) + ... + Δ»/(0).
Example 3. The quadratic polynomial f(x) has values/(2) = 15,/(3) = 41,
/(4) = 81. Find an explicit form for f(x).
Since/(л;) is a quadratic polynomial, the second differences will be con-
406
/
*-2 /-2
x-i A
*o /o
Xl /l
*2 /2

2] DIFFERENCES
stant. Thus, we are able to build up the table of differences below to infer
the values of ДО) and Δ/(0).
Δ/ Δ«/
χ
0
1
2
3
4
/
5
3
15
41
81
Thus/(0) = 5, Δ/(0) = -2, Δ2/(0) = 14, and by the Gregory-Newton
formula (Ex. 12)
/(") = /(0) + ("J АЛО) + (у Δ*/(0)
= 5-2и + 7и(и-1)
= 7и2-9и + 5.
The Gregory-Newton formula was only proved for integral n, but
f(x) ξ 7χ2-9x +5
is the unique quadratic polynomial defined by the three values
Д2) = 15, /(3) = 41, /(4) = 81.
(We may thus be permitted to write formally
(l) = X> Q^^*-1^ (з)=*^-1)(^-2), ...
for all real x.)
Ex. 13. The quadratic polynomial f(x) has values Д1) = 0, f(2) = 1, ДЗ) = 6;
use the Gregory-Newton formula to find the explicit form of f(x).
Exercise 18(a)
1. Use the method of synthetic division to find the quotient and remainder:
(i) whenx3-5x2+x+16 is divided by x— 2;
(ii) when x4 + 3x3 - 9л:2 - 23л: +14 is divided by χ + 3;
(iii) when3x4 + x3-^2-l^-24is divided by л:+ 2;
(iv) when x1—5x— 5 is divided by л:—4;
(ν) when 7л:5 - Ъх—5 is divided by χ + 5.

POLYNOMIALS AND PARTIAL FRACTIONS [18
2. Use the method of synthetic division to find the quotient and remainder:
(i) when Ix3 - 5л:2 - 17л: is divided by 2л: + 1;
(ii) when Зх4 + 2x3 - Зх2 + Ίχ+6 is divided by Зл: + 2;
(iii) when 2л* - 9x* + 2л:3 + 7л:2 + Ίχ - 3 is divided by 2л: -1;
(iv) when Ax9· - 9л:3 - 9л:2 + χ + 2 is divided by 2л: + 1;
(ν) when 3χ?-χ*-21χ3+9χ2-1Αχ+12 is divided by Зл:-1.
3. Use a hand calculating machine to evaluate the following polynomials for
the given values of л:, giving your answers to 1 decimal place:
(i) 3·5λ:3 + 7·1λ:2 + 2·8λ:+5·9,λ:= 2-9;
(ii) 7·7λ:3-9·6λ:2-3·8λ:+17·2,λ: = 1-97;
(iii) 8-6jt*-9-2x8 + 7-3xa-5-8x+13-2,jt = 2-61.
4. Express the following polynomials in the form
a0(x-h)n + a1(x-h)n-1+...+an.1(x-h) + a„
for the given values of A
(i) x3 + x2 + 2x+l, A = 1;
(ii) 2л:3-17л:2 + 30л:, А = 2;
(iii) л:3 + 7л:2-6л:-4, А = -3;
(iv) je*-l, h= -2.
5. Use the method of synthetic division to find the quotient and remainder:
(i) when jc* + (1 + 4j) x2-2x + (\+)) is divided by (x+j);
(ii) when jc* + (1 -j) л:3 -jx2 + 3(1 + 2j) x+ (4- 3j) is divided by (x-2j).
6. By expressing P(x) = л:3- 11л:2 + 35л:- 30 in the form
а0(л:-3)3 + а1(л:- 3)2 + сфс- 3) + а3,
estimate (i) Ρ(3·1), (ii) Д301), both correct to 1 decimal place.
7. Express the polynomial P(x) ξ λ:4 + 16л:3 +100л:2 + 288л: +320 in the form
Σ аАх+4У.
What does your answer tell you about the shape of the curve
у = ^+16л:3+100л:2 + 288л:+320
at the point (-4,0)?
Solve the equation л:4+ 16л:3 + 100л:2 + 288л:+ 320 = 0.
8. Show how the method of synthetic division may be modified for division
by the polynomial л:2 + л:+2.
Find the quotient and remainder on dividing
λ:7 + λλ-λ5-3χ4-4λ:3+3λ:2+5λ:+3 by л:2+л:+2.
9. If л:2+1 is a factor o{ x'7 + Axs +x3 +Bx2 + 2x+l, find the values of A and B.
10. If л:2 + 4 is a factor oix*+Axi-x3 + Bx2 -Ax-A, find the values of A and B.
11. A cubic polynomial P(x) has values Д0) = -1, P(l) = 4, P(2) = 21,
A3) = 62. Use a table of differences to find P( -1) and P(A).
12. A quartic polynomial P(x) has values Д-2) = 11, Д-1) - 2, P(0) = 5,
P(l) = S,P(2) = 23, Find ДЗ) andP(-3).
408

2] DIFFERENCES
13. Find the quadratic polynomial f(x) defined by the values
Д1)=-15, Д2)=-20, Д3)=-23.
14. Find the cubic polynomial f(x) defined by the values
Д1) = 3, Д2) = 4, ДЗ) = 21, /(4) = 66.
15. Find an expression for Σ rl as a polynomial of degree five in n.
3. PARTIAL FRACTIONS
The process of adding together fractions with polynomials as
denominators will be familiar to the reader. For example
_2 3_
x-3 x-2
2(x-2)-3(x-3)
(x-3) (x-2)
_ 5~x
(*-3)(jc-2)·
The reverse process is called 'resolving a rational function into partial
fractions'. Such a resolution is of importance in differentiation and
integration and also in obtaining power series expansions of rational functions,
as we shall see later in this chapter.
Two questions arise—can the resolution always be effected, and, if it can,
are the partial fractions obtained unique? Detailed analysis of these
questions lies outside the scope of this book.f
13 7'
tion into arithmetical partial fractions is not unique.
Suppose we are given a rational function of the form
Q(x)
where P(x), Q(x) are polynomials in x. As a first step, if the degree of P(x)
t See, for example, Ferrar: Higher Algebra. We shall convince ourselves, in specific
examples, that the answer to both questions is a qualified 'yes'.

POLYNOMIALS AND PARTIAL FRACTIONS [18
is greater than, or the same as, the degree of Q(x), we may divide out until
this is no longer the case. For example, we should express
χ*-2χ3-3χ2 + χ + 4
We now suppose that this has been done; that is, that we have a rational
function of the form
Let us first consider the case in which Q(x) contains only non-repeated
linear factors. Consider, for example, the function
*+18
(*-3)(2x + l)·
If we can find numerical values A, B such that
x+18 ξ A(2x+l)+B(x-3) (1)
then, on division by (x-3) (2x+1), we have
(*-3)(2x+l) *-3 + 2*+l· K)
From (1), suitable values of A and В can be found by equating со-
efficients: 1=2A + B, 18 = A-3B,
and then solving the resulting simultaneous equations. A preferable method
is to put χ = 3 in (1) to find A, and then to put χ = —J to find B. The values
obtained are A = 3, В = -5 and
x+18 _J 5_
(*-3)(2x+l) x-3 2x + Y
Ex. 16. Students often object that, since (2) is meaningless for χ = - i or χ = 3,
it is invalid to substitute these values in (1) to obtain A and B. Expose the fallacy
in this argument.
Observe that, in our last example, A and В were also given by
A. 3 + 18 (-a + 18.
A "(2x3 + 1)' ΰ~ (-i-3)'
410

3] PARTIAL FRACTIONS
that is, A was found by covering up the term (s - 3) in
s+18
(jc-3)(2jc+1)
and then putting s = 3 in what was left. Similarly, В was found by
covering up the term (2л;+1) and putting χ = —\ in what was left. This
useful technique for obtaining a partial fraction expansion is usually
called the cover-up rule.
Example 4. Express in partial fractions
n 4s2- 19s + 7 ... x3 + x
W (s-l)(s-2)(s + 3)' {П) s2-4'
(i) By the cover-up rule
4s2- 19s + 7 _ 1 /4-19 + 7\ 1 /16-38 + 74
(s-l)(s-2)(s + 3) (s-l)l(-l)(4)j+(s-2)l (1) (5) )
1 /36 + 57 + 7\
+(s+3)l(-4)(-5)j
_2 3_ _5_
jc—1 s-2+s + 3"
(Note the check by putting χ = 0:
2=_2+§+£)
(ii) Since s3 + χ is a cubic polynomial and s2 - 4 is only a quadratic, we
must first divide out
x3+x s3-4s + 5s 5s
s2-4 ~ s2-4 ~ x s2-4'
,T 5s 5s
Now,
(s-2)(s + 2)
x~^2\a)+x~+2\^a)'
by the cover-up rule. Thus we have
x3 + x _ 5 5
s2-4 •X + 2(s-2) + 2(s + 2)·
Ex. 17. Express in partial fractions
... jc+11 л:+
(1) (x-l)(x + 2); (11) *=
('") 335 (fl + °); (iv> 3z

POLYNOMIALS AND PARTIAL FRACTIONS [18
*Ex. 18. A common application of partial fraction expansions is to the
evaluation of certain integrals. Evaluate, by first expressing the integral in partial
fractions:
*Ex. 19. Partial fraction expansions can also frequently be used to simplify the
work in differentiating. Differentiate
х'-Бх+б' К ' x^-Sx-b'
It is worthwhile considering the 'cover-up rule' from a rather different
standpoint. Let us suppose that we have to express
f(x) = 5Jc+l
ЛХ) (jc-3)(2jc-1)
in partial fractions.
We proceed by considering approximations to this rational function.
It has singularities at χ = \ and χ = 3; that is, f(x) is very large when χ
is near \ or 3 (but is not defined at χ = i or χ = 3). Near χ = 3, the
dominant term in f(x) is 1/(jc —3): this term is multiplied by a factor with
numerical value almost exactly equal to
5x3 + 5
2x3-1"
= 4
(the 'cover-up rule' again). We may thus say that f(x) is approximately
equal to .
j^3
near л; = 3.
The approximate shape of the graph of f(x) is shown in Figure 18.1
while the approximate shape of the graph of у = Aj{x— 3) is shown in
Figure 18.2.
Near the common asymptote χ = 3 the two curves are seen to be very
similar in shape and position.
Now consider the deviation off(x) from the approximation:
5s + 5 4
/(*)-
c-3~ (2x-l)(x-3) x-3
_5x+5-Sx+4
(2jc-1)(jc-3)
-3(x-3)
-(2*-l)(*-3)
3__
~ 2x-V

PARTIAL FRACTIONS
Thus we have finally
/(*) =
and our rational function has been expressed in partial fractions.
*Ex. 20. The fact that (x - 3) cancelled at the final stage in the example above was
no coincidence. Suppose that f(x), F(x) are polynomials and that F(x) has a
linear factor χ-a, that is,
F(x) s (x-a)g(x), ЙФО.
Show that, near χ = ex., the rational function f{x)/F(x) is approximately
№ ι
№*-«'
Я*)*(«)-*(*)Л«)
(x-a)g(x)g(a) ■
with deviation
Prove that the numerator of this fraction is a polynomial with a factor (х- а) and
deduce that cancelling is always possible.
The method of obtaining partial fractions outlined in Ex. 20 is clearly
applicable whenever the denominator contains a (strictly) linear factor.
We conclude this section with two worked examples.

POLYNOMIALS AND PARTIAL FRACTIONS
Example 5. Express
in partial fractions.
f(x) has singularities at χ = 1 and χ = -2. The results proved in Ex. 20
suggest that we consider the singularity χ = -2. Near χ =-2,/(л;) is
approximately /4(_2)2_3(_2) + 5 , J_ ^_
I (-2-1)2 / U + 2/ * + 2'
The deviation is
4s2-3;c + 5 3_ 4х2-Зл: + 5-3(л:2-2л:+1)
(*-1)2(* + 2) * + 2 (*-1)2(* + 2)
х2 + Зл: + 2
(х-1)2(* + 2)
(х+1)(х + 2)
(х-1)2(* + 2)'
and thus we have f(x) = -^ + (*_ 1)2.
The reduction can, if desired, be carried one stage further by writing
je+1 s(*-l)+2:
3 1 2
/(*) =
л+2т^-Г(л-1)2'
Example 6. Express 2,19
-(x-2)(x2 + 4)
w partial fractions.
f(x) has just one singularity, at χ = 2. Near jc = 2, f(x) approximates
t0 /22+12\ / 1 \ __2_
U2 + 4/ U-2/ jc-2"
The deviation is
л:2+12 2_ x2+12-2Q2 + 4)
(*-2)(*2 + 4) jc-2 (jc-2) (*2 + 4)
_ (jc-2)(jc + 2)
(*-2)(*2 + 4)
_ x + 2
~ *2+4'
2 x + 2
and thus we have f(x) = ——= —2~ΤΊ·
414

PARTIAL FRACTIONS
Ex. 21. Express in partial fractions
W (χ-2)*(χ+1)' (U) (х+1У(2х+1)'
Ex. 22. Express in partial fractions
,. x+2 x*+l
W (л:-1)(л:2 + 2); W л:3-!-
Ex. 23. By writing x2 + 4 = 0 + 2j) (>-2j), use the 'cover-up' rule to find the
partial fraction expansion for 2
-2)(^ + 4)·
л:2-л:-1
(i) as the sum of partial fractions with rational coefficients;
(ii) as the sum of partial fractions with real coefficients.
Exercise 18(b)
. Express in partial fractions
© ,-zfezbi №
3-7*+2*2' w *2 + 2*' v"' 2*2 + 5*-3'
2x-l , ч 2λ? + 3 , .ч 13ЛГ-19
1+Зл:
х* + х' (х-1)(х + 2)(х
20-х3
-5л: + 6'
(xii)
х2 + 2х3' ^"' (2 + x)(l-2x)2' у™' (2 + х2)(1+х)'
, ** ■ (xivl {Х~1У ■ (хх) χ3 + χ* + χ ■
v ' **(x-2) ·
2. Find numbers А, В, С, D such that
2л:2 + 8л: + 3 = Л(л:-2)3 + Я(л:-2)2 (х-1) + С(л:-2) (х-1) + Щх- 1)
and deduce the partial fraction expansion of
2λ? + 8λ: + 3
(x-2)3(x-l)·
3. Adapt the method of Question 2 to express
9л:3 - Xs + Ίχ -5
(х-1)Чх+4)
η partial fractions.

POLYNOMIALS AND PARTIAL FRACTIONS [18
4. Express
(*+!)(*+2)
in partial fractions.
Hence evaluate n
,?i (r+l)(r + 2)"
Evaluate similarly Σ — .
5. Express
(i) as the sum of two partial fractions with real coefficients;
(ii) as the sum of three partial fractions with complex coefficients.
6. Express 1/(χ»-3χ+2) in partial fractions and deduce the partial fraction
expansion of ?\{2f -3^+1).
Express x2/(Sx2-6x+1) in partial fractions.
4. POLYNOMIAL APPROXIMATIONS:
THE BINOMIAL SERIES
We saw, in Chapter 8, that, for any number χ and any positive integer η
(l+*)»s 1 + $χ+(^χ2+ - + (") *'+··■+*"■
If χ is numerically much less than one, so that, say, its rth and higher
powers are negligible, the identity above may be used to give a polynomial
approximation of degree (/■ -1) for (1 + x)n. Thus, for example, with η = 10,
(1+л;)10« 1 + 10х + 45л;2,
neglecting terms in χ of degree greater than 2 {terms of order 3).
Ex. 25. Use the quadratic approximation to (l+л:)10 obtained above to find
(0-998)10 correct to 4 decimal places.
By using Taylor's series expansionf
f{x + d)= f{a) + xf'(d) + i*«/*(e) + ■ ■ ■ ,
polynomial approximations may be obtained for many other functions.
In particular, taking a = 1, axiaf(x) = xk (к е Q) we have
/'(1) = k, fV) = k(k-l), /w(l) = k(k-l) (k-2)
f See, for example, Siddons, Snell and Morgan: Calculus III.
416

4] POLYNOMIAL APPROXIMATIONS
giving, for sufficiently small *f the binomial series expansion
*Ex. 26. Verify the following expansions:
(i) (l+л:)-1 = \-x+xt-x3 + xi...;
(ii) (1+л:)-2= l-2x+3x2-4x3 + 5x1....
(Notice that this result may be obtained by the formal differentiation of (i).)
(iv) (i+x)i/» = ι + (i*) - ^ (**)■+^ (**)»-(^^ (**)*...;
1.3.5
3!
Ex. 27. Suppose, when χ is very small, that (l+χγ can be approximated by
1 -Vax-Vbx2. Then 1 +x and {1+ax+bx2)2 are nearly equal. Deduce that, for a
'best fit', 2a = 1 and a2 + 26 = 0, giving the first three terms of expansion (iii) of
Ex. 26.
Deduce similarly a quadratic approximation to (1 + x)~* if χ is very small.
Example 7. If χ is sufficiently small for x1 and higher powers to be neglected,
find cubic polynomial approximations for
0) M-9v^9-vV (»)
(1-2*) (2-*)' W(1-*)V(1+*)'
(i) By the 'cover-up' rule
(l-2*){2-*) 3(1-2x) 3(2-*)
* ?(l+2*+4*» + 8*«)-g(l+;[+£+£)
s i + |*+¥*2+f|*3.
t The decision as to what constitutes a 'sufficiently small' χ depends, of course, upon
the accuracy demanded of the answer. In numerical approximations, answers are usually
required to a certain number of significant figures or places of decimals and the point
at which to truncate the binomial series can be found by inspection.
It can be shown that
lim l (»-0(Я-2) :(й_-н:1)х^
for any real n, provided \x\ < 1. The result is also true for χ
for χ = -1 provided η > 0. It is not true for any |x| > 1, ι

POLYNOMIALS AND PARTIAL FRACTIONS
(») Π-vWH+v^C1-*)-1 (!+*>-*
(l-x)V(l+*)
» 1+д:(1-^) + л:2(1-| + |)+Д:3(1-^ + |-А)
ξ 1+^л; + ^л;2+1а6-л;3.
Alternatively, we may write
1
i l+ax + bx2 + cx3.
(ΐ-*)να+*)Λ
Then, using Ex. 26 (v), we have
l-ljc+f*2—hx3 * (1-дс)(1+ах+йдс«+сх»),
giving a-1 = -i, b-a = f, c-6 = -re
or α = i, b = J, с = -!%
as before.
Ex. 28. Solve Example 7 (i) by the method of Example 7(ii).
Example 8. By using the binomial expansion for (1 — 2x)~i, estimate *]5 and
calculate an upper limit for the error in your approximation.
Since (1 -2x J-0-)-i = 1Д/Й) = -W5 we have (see Ex. 26 (v))
W5 = 1-(-А)+У (-А)'-^|г (-Л)'
τ 4|—riw 5[ *. ш -г···,
1+0-1 +0-015+ 0-0025 +0-00044+ 000008.
Thus У5 в 1-11802 and V5 « 2-23604.
To estimate the accuracy of our answer we have to obtain an upper limit
for the sum of the remaining terms in the series. Now the coefficient, ur,
of (Уо)г in the expansion above is given by
Thus
418
1
2r-
.3.5....
r\
=1=2-
< 2wr_!
.{lr-
1
-- <
-1)
2.

4] POLYNOMIAL APPROXIMATIONS
and the sum of the remaining terms in the binomial series above is certainly
less than
= ' ' — (Уй-)6 γ^-j (sum of infinite geometric series)
χ 1-805 χ 10-».
Since 4-4 χ 10~4 and 8 χ 10~6 are overestimates of the fifth and sixth terms
of the series above we may safely conclude that the error in taking 1 · 11802
as JV5 is less than 1-805 χ 10~5 and so our answer of 2-23604 for V5 has an
error of less than 3-61 χ ΙΟ"5.
Ex. 29. In fact, J5 = 2-236068 (6 dp.), and our limit for the error in Example 8
appears to have been rather crude. Explain why this is so. How could a more
accurate estimate of the error be obtained?
Ex. 30. To4d.p.,V5 = 2-2361, while in Example 8 above, we obtain J5 = 2-2360
(to 4 d.p.) although we subsequently show our error to be less than 5 χ Ю-5.
Explain.
The binomial series gives a polynomial approximation to (1 +x)n valid
in the neighbourhood of χ = 0. If we require a polynomial
approximation valid in some other neighbourhood, it is necessary to shift the origin,
as shown in our next example.
Example 9. Obtain a quadratic approximation to (1 +2x)_1 in the
neighbourhood of χ = 3.
(1+2χ)-ι = [7 + 2(*-3)]-1
, where у =
■\№"
since у is approximately zero
= ^-Д-(*-3) + з!з(х-3)2.
The quadratic approximation is best left in this form but it may be
reducedto *&-&* + &*
Notice that this is quite different from the quadratic approximation to
(1 +2*)-1 in the neighbourhood of χ = 0:

POLYNOMIALS AND PARTIAL FRACTIONS
Exercise 18(c)
1. Obtain cubic polynomial approximations for the following expressions in the
neighbourhood of л: = 0:
2. Obtain cubic polynomial approximations for the following expressions in the
neighbourhood of л: = 0.
3. Obtain cubic polynomial approximations for the following expressions, in
the neighbourhood of л: = О:
(i) (1 + x)*; (ii) (1 + 2*)-*; (iii) (1-х)-*; (iv) (2-х)"5; (v) (8-x)*.
4. Obtain cubic polynomial approximations for the following expressions, in
the neighbourhood of χ = 0:
W (\-х)(\-2хУ (U) (х-2)(х-ЗУ (ш) (1+хУ(2 + хУ
x(x+3) _ 1 + 4л:
(1VJ (1 + χ2) (1 + 2x)' W (1 - 2хУ (1 + 2x2)'
5. Obtain quadratic polynomial approximations for the following expressions
m the neighbourhood of χ = 0:
1+x ' w (l+x)V(l + 2x)"
6. Obtain quadratic polynomial approximations for:
(i) (1 + x)-1 in the neighbourhood of χ = 1;
(ii) (2 — x)-1 in the neighbourhood of л: = — 1 ;
(iii) V(3 + 2x) in the neighbourhood of χ = -1 ;
(iv) (2 + Зл:)* in the neighbourhood of χ = 2;
(ν) ——— in the neighbourhood of χ = -1.
7. Prove that . ... „. ,
V(l+x2) + l л:2
and deduce that, if χ is very small,
Vd+*2)+i ^ *~^2"
8. Find, correct to 5 significant figures, the values of (i) (0-998)1'3; (ii) (1-02)1'2;
(iii) (4-01)1'2; (iv) (0-799)1'3.

4] POLYNOMIAL APPROXIMATIONS
9. By substituting χ = 1/1000 in the expansion of (1 - л:)1'3, find (999)1'3 correct to
4 significant figures.
10. Prove that 19-97 < V399 < 19-98.
11. Prove that, if
2-x-2V(l-x)'
then£= 2-x+2V(l-*)-
Deduce that, if χ is small, then Ε is approximately equal to ^x2.
12. If χ is so small that x3 and higher powers of χ may be neglected, express the
fUnCti°n V(4 + x)
l-2x + (l+3x)2'3
in the form, a+bx+cx2. (O & C)
13. Express the function Ε given by
(2x+1)(1 + jc·)
in partial fractions.
Hence prove that, if л: is so large that x~i can be neglected, then
E = 5-^. (O&Q
14. Use the binomial expansion to calculate the value of (16-32)1'4 correct to six
places of decimals. (O & C)
Miscellaneous Exercise 18
1. The polynomial xi + Axs + Bxi-Ax3 + Ax!t+Bx+C is exactly'divisible by
x3+l;fmdA,B,C.
2. If (1+х+хг)п = а0 + а1х+а2хг+ ...+агпхгп, write down the values of a0
and a2n and prove that aT = а^п-г.
In In
Show that Σ aT= 3" and find Σ (-1)^·
If и is an even number, prove that Σ (-1)7+1 a2r-i = 0.
3. Find the value of the constant A for which the expression
2x+3-A(x*+x+l)
has (x—2) as a factor and find the remaining factor.
Hence, or otherwise, put into partial fractions the expression

POLYNOMIALS AND PARTIAL FRACTIONS
4. Provethat χ2 + ^+1 g {x_aHx+2b + a) + a> + 2ab+L
(x-2)(x* + 2bx+l)
into partial fractions
(i) when b = i,
(ii) wheno = 1.
5. Find А, В such that, for all values of χ other than i, -i, -f,
x+1 Ax+B A(x+1) + B
(2л:-1)(2л:+1)(2л:+3) (2x-1) (2x+ 1) (2л:+ 1) (2x+3)'
Find the sum to η terms of the series whose rth term is
(2r-l)(2r+l)(2r+3)·
6. Prove that, if - i < χ «Ξ i,
1.3 , 1.3.5 , 1.3.5.7 ,
Deduce that χ J3 13 ζ 135J
1 + 4+4.8 + 4.8.12 + 4.8.12.164
7. Sum to infinity the series:
,.4 , 1 1.3 1.3.5 1.3.5.7
(1) 1 + 3 + зТб+зХ9 + 3-Х9Л2 + -;
1 1.4 1.4.7 1.4.7.10
(11) +4 + 4.8 + 4.8.12 + 4.8.12.16+""'
,.. 3 3.9 3.9.15 3.9.15.21
} 8 + 8.16 8.16.24 + 8.16.24.32 ;
„ , , 1 1.5 1.5.9 1.5.9.13
(1V) 1 + 6 + 6Л2 + 6Л2Т8 + 6Л2Ж24 + ·-
8. The cubic polynomial P(x) assumes the values 2, 3, 2, 11 respectively for
χ = 1, 2, 3, 4. By writing
Fix) ξΑ + Β(χ-1) + C(x- 1) (x- 2) + D(x- 1) (л:- 2) (л:- 3) find Дл:).
Show how to fit a quadratic polynomial to coincide with the values of
у = sin χ at χ = 0, £π, π.
9. Find a linear approximation to J(3-x) in the neighbourhood of χ = - 1,
and interpret your result graphically.
Show that V3-9 « 1-975.
10. Find the sixth term and the rth term of the series whose first five terms are
on the assumption that the rth term is a polynomial in r of as low a degree
as possible.

19. Complex numbers (2)
1. INTEGRAL POWERS OF
COMPLEX NUMBERS
In Chapter 17 we saw that the multiplication of complex numbers was best
expressed in terms of their moduli and arguments: if ζ = r (cos 0+j sin 0)
and w = s (cos Φ+] sin φ) then zw = rs [cos (θ + φ)+) sin (θ + φ)]. In
particular, if ζ = cos 0+j sin 0 then z2 = cos 20+j sin 20, from which it
follows that z3 = cos 30+j sin 30 and so on; the general result,
z" = cos ηθ+] sin и0 (n е Z+),
may be proved by mathematical induction.
Theorem 19.1. (De Moivre's theorem for positive integral exponent.) For any
positive integer η ^ 0+. ^ ^ = ^ ^+. sin ^
Proof.
(cos 0+j sin 0)"-1 = cos (и-1) 0+j sin (и-1) 0
=> (cos 0+j sin 0)" = [cos (и-1) 0+j sin (и-1) 0] (cos 0+j sin 0)
=> (cos 0+j sin 0)m = cos «0+j sin ηθ
(cos 0+j sin 0)1 = cos 0+j sin 0
and the result holds for all positive integral n, by induction.
Ex. 1. Illustrate de Moivre's theorem using the Argand diagram for the cases
η = 2, 3, 4. What can you say about 0 if (cos 0+j sin#)" = 1 (n a positive
integer) ?
Ex. 2. Express V3 + J 'п modulus-argument form and deduce the value of
(V3+J)9·
De Moivre's theorem may be extended without difficulty to include
negative integral exponents. If we define z°to be 1, this enables us to say
that (cos 0+j sin 0)m = cos «0+j sin ηθ for all integral n.
Theorem 19.2. {De Moivre's theorem for negative integral exponents.) For any
negative integer n,
(cos 0+j sin 0)m = cos «0+j sin ηθ.
423

COMPLEX NUMBERS (2) Г19
Proof. Write η = —m; then m is a positive integer and
(cos (9+j sin &)m = cos mff+) sin mff,
by Theorem 19.1. Thus
(cos 0+j sin 0)" = . . j.
v J ' cos m(9+j sin m(9
= (cos тв-) sin m0), since |cos m(9+j sin m(9| = 1,
= cos >j(9+j sin и0.
Ex. 3. Illustrate de Moivre's theorem, using the Argand diagram, for the cases
η =—2, — 3, —4. What can you say about (i) zn+z~n, (ii) zn—z~n where
Μ = ι?
Ex. A. Evaluate (V3+j)-9.
The results proved in Theorems 19.1 and 19.2 are frequently useful in
deriving further results. We shall illustrate, in the next examples, some of
the techniques most commonly employed.
Example 1. Express cos 6(9 in terms of cos θ and sin Θ.
Since cos 6Θ +j sin 6Θ = (cos 0+j sin Θ)6 we have, using the Binomial
Theorem,
cos 6Θ +j sin 6Θ = c6 + 6jcs5'-15c45'2-20jc35'3+15c25'4+6jc5'5-ie,
where с = cos Θ, s = sin Θ. Comparing the real parts of both sides,
cos 6Θ = cos6 (9-15 cos4 θ sin2 (9+15 cos2 θ sin4 (9-sin6 (9.
Notice that, in Example 1,
(i) by comparing imaginary parts we immediately derive an expression
for sin 6(9 in terms of cos (9 and sin (9;
(ii) the expression for cos 6Θ (but not sin 6Θ) may be written as a
polynomial in either cos (9 or sin Θ.
Ex. 5. Show that
cos 6(9 = 32 cose (9-48 cos4 (9+18 cos2 (9-1.
Ex. 6. Express tan 6(9 as a rational function of tan <9.
Example 2. Express cos6 θ in terms of multiple angles.
Writing ζ = cos θ + j sin (9, we have z-1 = cos θ — j sin (9 and thus
2cos0 = z + z-\
Then 64 cos6 0 = (z + z-1)6
= (z6 + z-6) + 6(z4 + z-4) +15(z2 + z-2) + 20.
424

1] INTEGRAL POWERS
But
ζ6+ζ~6 = (cos 60 +j sin 6(9) +(cos 6(9-j sin 6(9), by de Moivre's theorem
= 2 cos 60,
and similar results hold for z4 + z~4 and z2 + z~2. Thus
32 cos6 0 = cos 60 + 6 cos 40+15 cos 20 + 10.
*£лг. 7. Suggest a quick check on the accuracy of the answer to Example 2.
*Ex. 8. By writing 2j sin 0 = z-z'1, express sin" 0 in terms of multiple angles.
Ex. 9. Evaluate cos6 0d0.
Example 3. Find the sum of the series
χ sin θ+χ2 sin 20 +x3 sin 30 + ... +xn~1 sin («- 1) 0,
w/геге χ is a real number.
The given series is reminiscent of a geometric series, but we have sines
of multiple angles, sin rff, rather than powers of sines, sinr 0. De Moivre's
theorem suggests a possible way of changing from the multiple angle form
into an exponent form. Consider the two series:
С = l+Jccos0 + ;c2cos20+...+;c'l-1cos(n-l)0,
S= xsin0 + .x2sm 20 + ...+ x"'1 sin (n-1)6».
Multiplying the second series by j and adding, this gives
C+jS = 1 +x(cos 0+j sin 0) + x2(cos 20+j sin 20) + ...
+ xn-1(cos(n-l)e+] sin (и-1)0).
Thus, writing ζ = cos 0+j sin 0,
C+jS = 1 +xz + x2z2+... +xn~1zn-\ by de Moivre's theorem
= -: , from the formula for the sum of a geometric series
1— xz
l-x*z* (1-jcz*)
1-jcz -(l-xz*)
= 1-Λ:ζ*-Λ:'Ιζπ+Λ:η+1ζπ-1|ζ|8
l-*(z + z*) + *2|z|2
1 -x(cos 0-j sin θ)-χη (cos «0+j sin и(9) +xn+\cos, (n-1)(9
= +j sin (n-1)0)
l-2xcos0 + x2
since |z| = 1.
425

COMPLEX NUMBERS (2)
Comparing the imaginary parts of both sides we then have
S =
χ sin θ — χη sin ηθ + xn+1 sin (n — 1) θ
l-2xcos0 + ;c2
Ex. 10. What is the value of С:
Ex. 11. If jjtrj < 1, find Σ λ:'si
1. Simplify
(i) (cosiTT+jsiniw)4; (ii) (cos ^π+j sin^)-3;
(Hi) (cosi^—j sin^r)6; (iv) (sin^r+j cos^n)6;
(v) (cos j7r+j sin %π)3 (cos i?r +j sin ^π)4;
(vi) (l+jtaniw)5; (vii) (1 -j cot ^n·)-";
(viii) Q+cos20+jsin20)-4.
2. Express in standard form
(i) (cos in +j sin ^)12; (ii) (1 + j)8; (iii) (V3 -j)4 (1 + JV3)e;
(iv) (cos 0+j sin (9)"/(cos φ -j sin ф)т.
3. Express in terms of sin Θ: (i) cos 4(9; (ii) sin 5(9.
4. Express in terms of cos (9: (i) cos 4(9; (ii) cos 5(9; (iii) . .
5. Express tan 40 in terms of tan Θ.
6. Prove that cos 7(9 = 64 cos7 (9-112 cos5 (9 + 56 cos3 5-7 cos (9.
Write down the seven roots of the equation
64л:7-112x5 +56л:3-7л: = 0,
and also the seven roots of the equation
64л:7-112х5 + 56л:3-7л:-1 =0.
7. Express, in terms of cosines of multiple angles:
(i) cos5 (9; (ii) sin4 (9; (iii) cos7 <9.
8. Express, in terms of sines of multiple angles:
(i) sin5 Θ; (ii) sin7 Θ; (iii) sin3 θ cos Θ.
9. Express sin5 θ cos4 θ in terms of sines of multiple angles.
10. Evaluate
e cos40d0and ι
Jo Jo
11. Sum ton terms the series:
(i) С08л: + С08 2л:+... + С08Лл:;(п) sin л: + 8Ш 2л: + ... + 81П п
12. Sum to η terms the series
cos л: + 2 cos 2л: + 4 cos Зл:+ ... +2"-1 cos nx.
426

1] INTEGRAL POWERS
13. Sum the series
l+Qcostf+Qc
n-l
14. Find Σ 0+1) sin r0.
16. By considering cos (<?i + 02 + θ3 + 0J + j sin (вг + θ2 + θ3 + 04), express
tan (θ1 + 02 + 03 + 0ί) in terms of tan 0b tan 02, tan 03, tan 04.
17. Find Σ sin(2r+l)(?.sin1!'-+1(?.
r=l
18. If ζ = cos 0+j sin (9, show that
z- 1/z = 2j sin (9 and zn- 1/z" = 2j sm л0.
Express sin5 0 in the form a sin 50+ ό sin 3(9 + с sin (9 and hence solve completely
the equation 16 sin5 0 = sin 5(9. (O & C)
2. RATIONAL POWERS OF
COMPLEX NUMBERS
In this section we shall denote a general rational number by pjq, where p, q
are integers and q > 0.
Given a real number r > 0, there is just one positive #th root of rp, that
is, there is just one a > 0 such that r* = a9. We write
a = ri' or a = %r*).
*Ex. 12. Prove that, it p, q are integral and r > 0, there is only one positive
real number a such that rv = a". (Show that, for b > 0, r* = ό5=> ό = α.)
*£лг. 13. Prove that, if? is odd there is just one real number a such that, for real
r, rv = a" and that, if q is even, there are either two or no such numbers.
Now let us consider the problem of finding a complex number w such
that zv = wq, where ζ is the complex number /-(cos 0+j sin 0), r > 0.
Suppose vc = sfcos φ + j sin φ); then
j-^cos />0+j sin pff) = s"{cos qφ +j sin #0).
(/j, q being integers, we may apply de Moivre's theorem to both sides.)
Now two non-zero complex numbers can be equal only if they have the
same moduli, and arguments differing by an integral multiple of 2π. Thus
ρθ + Tkn = дф.

COMPLEX NUMBERS (2)
Whether q is even or odd, we may take s =
,P1±7J^+..,P1±7^
ь-mi·
giving w = .
In this expression, к can take any integral value, positive or negative. Since
values of к differing by q or any multiple of q give rise to the same w, there
are precisely q different values of w given, for example, by taking
£ = 0, 1,2, ...,(9-l);
in other words, a complex number has exactly q #th roots. Since the real
numbers form a subset of the complex numbers, every real number has
q qth. roots; by Ex. 13 at least (q — 2) of these will be complex.
Ex. 14. What are the four fourth roots of: (i) 1; (ii) 4; (iii) -1 ?
If ζ = /-(cos θ +j sin Θ), where θ = arg z, we shall define zvlq to be
r*te(cos^+jsin^).
It is important to realise that zp/s is only one of the qth roots of z*. We
shall sometimes use the notation "Jz11 for zplq; in particular, 4Z = z^·
*Ex. 15. Under what circumstances do (ζ*)1'5 and zTlq represent the same
complex number?
Ex. 16. Verify that the definition given above for z*lq yields the correct value for
г*1", where r > 0 is a real number.
Example 4. Find (— 1 +j)i and the other fifth roots of the complex number
(-i+j).
Writing ζ = -1 + j
= V2(COS|77+jsini77)
and w6 = z,
we have № = 2" cos-— +jsin^-^ .
Distinct values for w are given by к = 0, 1, 2, 3, 4; к = 0 gives (— 1 +j)^.
ThUS (-1 + j)* = 2*(cos *V +J sin 2»·
The remaining four fifth roots of ( —1+j) are 2"llff(cos ^я+j sin y-J-я),
2"lA°"(cos j&r+j sin i&r), 2A"(cos f>+j sin f», and 2^(cos \v+\ sin Ίχή).
428

2] RATIONAL POWERS
Using tables we have
(-1+j)* χ 0-955 + 0-487J,
and similar approximations may be found for each of the other fifth roots
of (-1+j).
Ex. 17. Plot the positions, in the Argand diagram, of the affixes of the complex
number —— + — and of its five fifth roots.
Ex. 18. Find the three cube roots of j, and verify that their sum is zero. Plot their
positions in the Argand diagram.
3. THE nTH ROOTS OF UNITY
The equation zn = 1 has, by the results proved in the last section,
precisely η roots, for
1 = cos2fo7+jsin2fo7,
These η complex numbers are the nth roots of unity. Writing
and applying de Moivre's theorem
Ι ω* = cos hj sin 1
the >jth roots of unity may be written 1, ω, ω2, ω3, ..., ωη_1.
*Εχ. 19. Show that the nth roots of unity are represented by the vertices of a
regular и-sided polygon inscribed in the unit circle \z\ = 1.
Since ωη — 1 =0 and ω Φ 1, we have, by summing the geometric series
in the usual way, the following important result:
1+ω1 + ω2 + ω3+...+ω'1-1 = 0;
that is, the sum of the η nth roots of unity is zero.
*Ex. 20. If ρ is a prime number, and if ξ is anypth root of unity other than 1,
show that the complete set otpth roots of unity may be written as 1,1, ξ2, ..., I*-1.
Discuss possible generalizations of this result for the case where ρ is not prime.
(If you have difficulty in proving the general result, consider the particular cases
ρ = Ъ,р = 4.)
429

COMPLEX NUMBERS (2) [19
Example 5. Prove that (1 + z)n = z*=> Re (z) = -%.
(l+z)- = z-
=> (1 + z) = z£, where ξ is an «th root of unity other than 1,
(1+ζ*) = ζ*ξ*
^> (l+z)(l+z*) = zz*£*
=> l+(z+z*)+zz* = zz* (since ξξ* = 1)
=> Re (z) = -i. (See p. 383 Ex. 14(v).)
Exercise 19(b)
1. Find in standard form the three cube roots of:
(i) 8; (ii) -1; (iii) -j; (iv) (1+j)3.
2. Express in the form a+bj, giving a,bto2 significant figures in (ii)-(iv)
(0 V0-4J); (ii)Vd+2j); (Ш) V(-3-j); (iv)V(i-3i).
3. If ω is a complex cube root of unity, show that the cube roots of z3 are
ζ, ζω, ζω2.
Find the three cube roots of — 2 + 2j and deduce surd expressions for
cos (π/12) and sin (π/12).
4. Find, correct to 2 significant figures, the real and imaginary parts of (2-j)1,e.
5. Simplify:
(i) V(cos θ-i sin 0)5-=-V(cos θ+i sin β);
(ii) (sin Θ-j cos 0)1'3;
K ' л/ \cos30+jsin30/'
(iv) (1 +j cot 0)1'4;
(v) {(cos Θ+) sin Θ) (sin 0-j cos Θ)}111.
6. Plot in an Argand diagram the four fourth roots of 16 and, on a separate
diagram the six sixth roots of 64.
7. Solve the equation z*-z2+1 =0.
8. Solve the equation l + z + z2 + z3 + z4 + z5 = 0.
9. Solve the equation z3-(j-z)3 = 0.
10. Solve the equation (1 + jz)5 - (1 - jz)5 = 0.
11. Show that, if (J—z)" =(jz— 1)", then ζ must be a real number, and find all
the real numbers satisfying this equation.
430

3]
nTH ROOTS OF UNITY
12. If 1, ω, ω2 are the cube roots of unity, prove that:
(i) (α+ω-ωη(α-ω + ωη = α* + 3;
(ii) (l+jW-w2)(l-w+jW2) = 2;
(iii) (a + b) (a + bo>) (α+όω2) = a3 + b3;
(iv)(a+b + c)(a + bb) + c(jj2)(a + bb)2 + c<i)) = a3 + b3 + с3 - ЗаЬс.
13. Describe geometrically the effect of multiplying a general complex number
ζ by ω, a complex cube root of unity. Deduce geometrically that, if \z\ = 1,
then|z + wz| = 1.
14. 1, i, ξ2,Γξ3, ξ4 are the five fifth roots of unity and a, b two given complex
numbers, and if Au A2, A3, Ai; A5 are the affixes of the complex numbers
α + b, a + bi,a + bi2,a + bi3, α + όξ4 show that Λ^ Λ 3Λ4Λ5 is a regular pentagon.
s a factor of the
Show also that
Hence express Δ as a product of linear factors and also as a product of a real
linear and a real quadratic factor.
Factorize into four linear factors the determinant
4. COMPLEX POWERS OF
COMPLEX NUMBERS
{This section may be omitted at a first reading)
If e is the base of the natural logarithms, and if у is a real number, it
seems plausible to assume that, provided d" has a meaning,
On this assumption, ζ = eJ" satisfies the differential equation
dz .
dy = ^

COMPLEX NUMBERS (2) [19
But w = cos y+i sin у also satisfies this differential equation (by direct
verification). Also, when у = 0, ζ = 1 = w and thus ζ = w for all real y:
e3" = cosy+'] siny.
Since ex+ly = e^.eJ»,
it follows that e^+J" = e^cosj+j sin y). (1)
♦Ex. 21. Show that e~Sv = cos^—j sin у and deduce that cos^ = K^+e"1·),
sin^ = — (&y - e~ly). Deduce from these expressions and the infinite series for
ex the series expansions for cos у and sin y.
Ex. 22. Show that cos у = cosh j> and that j sin у = sinh \y.
*Ex. 23. Prove that e*1 = &*<?> z1—zi = 2кщ, к an integer.
Now consider the equation _ w
Given any non-zero ζ we can certainly find a w which satisfies this equation.
For example, if w = In \z\ +j arg z,
e«> _ glnl^l+jarg2
= glnlilelarg»
= \z\ (cos arg z+j sin arg z)
By Ex. 23, any other solution of the equation may be written in the form
w = In \z\ +j(arg z + 2kn). Such an expression is called a logarithm of z;
the particular expression with к = 0 is called the principal logarithm of ζ
and is written In z: , , , . .
In ζ = ln|z|+j argz.
With this definition of the logarithm of a complex number we are in a
position to define a complex power of a complex number:
zW = eWln, (z + 0).
For example, j]' = eJlnl

3] COMPLEX POWERS
while (i+j)i-i = ed-J'^d+J)
= e(l-J){lnV2+i*J>
= 6(1ηΛ/2+ίπ)+ί(ίπ-1ην2)
= V2ei*(cos (i7r-lnV2)+j sin (£π-1η V2))
χ 2-8 + 1-3J.
Ex. 25. Find expressions for:
(i) ln(-l); (ii)ln(-j), (iii)(-l)J; (iv) Kl+j)^.
Ex. 26. Verify that the definition given in Section 2 for zvlq (p/q rational) is in
accordance with the definition given here for a complex power of a complex
number.
Exercise 19(c)
1. If a, b, r, s are real numbers and
и = r&e, ν = se'e,
find:
(i) \au + bv\;
(ii) an argument of the complex number au+bv.
2. Express in the form a + by.
(i) e1-·*»!; (ii) e<"+»J>.
3. Express in the form a + by,
(i) ln(V3-j); (ii) (V3-j)J.
4. If ζ moves once anticlockwise around the unit circle in the Argand diagram,
starting at the point — 1, describe the motion of the point representing zK
5. If ζ is a complex number and sin z, cos ζ are defined by
sin ζ = ^(eJ«-e-i«), cos ζ = ^(fi>z + e-!'),
prove that:
(i) sin2 z+cos2 ζ = 1; (ii) sin 2z = 2 sin ζ cos z;
(iii) cos (in- z) = sin z; (iv) cos (ζλ + z2) = cos zx cos z2 - sin zx sin z2.
6. Show that:
(i) cos (x + jd) = cos χ cosh y-\ sin л: sinh y;
(ii) sin (лг+jd) = sin χ cosh ^+j cos χ sinh y.
7. Show that: . , ... ..
If x+jy = с tan i(u+jv), express хг+уг + сг in terms of u, υ and с If и and с
are positive constants show that the locus of the point (x, y) referred to Cartesian
axes is a circle of radius с cosech υ. (tan ζ = sin z/cos z.) (London)

COMPLEX NUMBERS (2) [1
8. Give a sketch of the representation in the Argand diagram of the two sets:
A = {ze C: \z\ = 1, -%n < Im ζ < \π};
B={weC: zeA, ezw-ez + w + l = 0}.
Miscellaneous Exercise 19
1. Solve the equation zs-zi+1 = 0 and mark the positions of the roots in the
Argand diagram.
2. Prove that, if л is a positive integer,
(cos 0+j sin 0)" = cos ηθ+i sin ηθ.
By putting η equal to 5 in this formula, or otherwise, prove that
sin^=^i. (0&Q
3. Prove that cos 70 = cos7 0(1 -21 tan2 (9+35 tan4 (9-7 tan" (9).
Find the real part of (i+Cos 0+j sin0)« „ . ,
7Γ, a ■ ■ mn· (London)
(l+cos(9-jsin(9)«
4. Express each of the complex numbers zx = (1+j)V2, z2 = 4(-l+j)V2 in
the form r(cos #+J sin 0), where r is positive. Prove that z\ = z2, and find the
other cube roots of z2 in the form r(cos θ+j sin (9). (O & C)
5. Solve the equation z4 + 2z2+3 = 0, giving the real and imaginary parts of
each root correct to 2 significant figures.
6. If α+}β = yj{(a+ib) (c+jd)} where a, b, c, d, a. and β are real, find the value
of a? in terms of a, b, с and d. (London)
7. Prove that the roots of the equation (z+ l)"-(z-1)" = 0 (n > 3) all lie on
the imaginary axis. Illustrate the result geometrically in the case η = 3.
where ω is a complex cube root of unity, prove that £>2 = — 27. (O & Q
9. Evaluate Σ cos (cc+rfi) and Σ sin(a+r/?).
r=0 r=0
10. Ifz= cos 0+jsin#, showthat z+l/z = 2 cos 0 and find the corresponding
result for z—l/z.
Prove that
cos9 0 = тЪ· [cos 80 + 8 cos 60+28 cos 40+ 56 cos 20 + 35].
Evaluate (cos8 0 + sin8 0)d0. (London)
434

3] MISCELLANEOUS EXERCISE
11. If \z\ - 1 and arg ζ = θ Φ 0, express in modulus argument form:
(i) 1+z, (ii) 1-z; (iii) У(^)·
Provethat w = J Ш
lies on a fixed straight line, whatever the value of Θ.
12. Demonstrate the following results geometrically, where ω is a complex cube
root of unity:
(i) (1 + ω) (1+«»■) = 1; (ii) ^~ = ω; (iii) Re(l+2w) = 0.
13. Find the roots of the equation z5 +1 = 0 in the form
cos^+j sin^,
where φ is to be determined.
Deduce, or prove otherwise, that the roots of
16х5-20л:3 + 5л:+1 = О
are cos H(2r+1) π] (r = 0, 1, 2, 3, 4). (London)
14. Solve the equation .,„ .. ., ..
j(l-xj)" = (1+дд)я,
and verify your solution by setting: (i) η - \; (ii) η = 2.
15. Find |w+l+j| and arg(w+l+j) geometrically, where ω is that complex
cube root of unity with positive imaginary part.
Hence, or otherwise, find surd expressions for sin -fer and cos γ^π.
16. Prove that:
(χ+γ + ζ)(χ+ωγ + ω*ζ)(χ+ω*γ + ωζ) = x3+y3 + z3-3xyz,
where ω is a complex cube root of unity.
Hence, or otherwise, solve the following problems:
(i) Prove that the product
(x3 +y3 + z3 - 3xyz) (a3 + b3 + c3 - 3abc)
is expressible in the form
A3 + B3 + C3-3ABC,
where A = ax+by + cz, В = ay + bz + cx, С = az + bx + cy.
(ii) Solve the equation , „
л:3-9л:+12 = 0. (О & С)
17. In the determinant
| a b с d
d a b с
с d a b
b с d a
the cofactors of a, b, c, daxe denoted by А, В, С, D respectively (the expansion of
Δ by its first column being аА+ЬВ+cC+dD).

COMPLEX NUMBERS (2)
By considering the product ΔΩ, where
0 0 1 |
prove that, if ω is any root of the equation x*—l =0, then a+bo) + cu>2 + d(u3
divides into Δ, the quotient being
A + B(jJ3 + C(jJ2 + D(jJ.
Hence show that
A + B+C+D = (a + c-b-d){(a-c)2 + (b-d)2}. (O & С)
18. P, Q are the affixes of the complex numbers p, q and ξ = cos 2π/η +j sin 2π/η.
Locate the affix of the complex numbers (j>-q)i·
If a, b are two complex numbers, prove that the affixes of the complex numbers
zb z2, ..., zn are the vertices of a regular л-sided polygon if and only if
(zT-a)" = b" 0 = 1,2,3,..., n).
19. If Z,, Z2, Z3 are the affixes of the complex numbers zl5 z2, z3 prove that a
necessary and sufficient condition for the triangle ZXZ2Z3 to be equilateral is that
Zl + Zl+ Z$— Z2Z3 — Z3Zi — ZXZ2 = 0.
20. Prove that, if л is a positive integer (n > 1) and
ω = cos 2π/η + j sin 2π/η,
ΐηβη1+ω + ω2+... + ω"-1 = 0.
In the Argand diagram the points Аъ А2, ..., A„ are the vertices of a regular
polygon inscribed in a circle of radius a with its centre at the origin. The complex
numbers represented by the points Ax, Аг, ..., An are zl5 z2, ..., z„. Prove that
zl + z\+... + zl = 0.
The perpendicular distances of the points Аъ А2, ..., An from any given line
through the centre are dlt 4> ···> ^n· Prove that
dl + dl+... rf2 = i«a2. (O & C)
21. If x+jy = tanh (ii+jf) where г/, и, л:, д' are real, find χ and д' in terms of и
andt\
Prove that
лЛ+>'2-2л:соШ2г/+1 = 0
and x2+y2 + 2ycot2v-l = 0.
If и and >> are regarded as variable parameters, and χ and у as Cartesian
coordinates, describe the relationship between the two families of circles.
22. Prove that .,,„.,, . , „ , . , „ . ,
sinh (Р+}ф) = sinh 0 cos 0 +j cosh a sin 0.
For all real or complex values of ζ the sum of the infinite series
«0 + α1ζ + α2ζ2 + ...
is/(ζ). Prove that if ω is a root of the equation ω2 + ω+1 = 0, then
a0 + a3x3 + aex°+...+ = ΗΛ*) +Λωχ) +/(ω2χ)}.
436

3] MISCELLANEOUS EXERCISE
By considering the series for sinh x, prove that
♦ж
23. A sequence uu и2 ··· is denned by иг = 1 and
un+1 = ошп + п+1 (и>1),
where α is independent of л. Find an expression for un when α = 1.
By induction or otherwise show that, when α Φ 1, u, is of the form
un = Accn + Bn + C,
where А, В, С are independent of n, and find А, В and С
If now α is a complex wth root of unity and л is a multiple of m, determine the
real part of u„. (O & C)
24. Prove that n_x . . .
лг2"-1 =(jc-1)(jc+1) Π x2-2xcos — + 1
ы\ я /
and devise a similar expression for x*n+1 -1.
25. Prove that
x*"-2xncosnff+l ="n ix2-2xcos (W — ) + l|.
Deduce the following results:
(i) sin/га = 2""1 Π sin (a+—);
fc=o V n)
(ii) cos noc-cos nfi = 2"_1 Π jcosa-cos (/?+ J J.
From (i) and (ii) deduce, by logarithmic differentiation, the further results:
(iii)cot«a = - Σ cot[oi+ — l, αφ—;
Пк=о \ η) η
(iv) cosec2 ηθ = — Σ cosec2 ((?+— ), 0 Φ —.
26. Prove that л:2+.у2+г2-.уг — ζλτ-λ^ has a linear factor Λ:+ω_ν+ω2ζ, where
ω is a complex cube root of unity. Deduce that, if л: Φ 3, хг + уг + ζ2 —yz - zx - xy
is a factor of (y-z)n+(z-x)n+(x-y)n.

2o. Mappings in the Argand diagram
The equation of a curve in the Argand diagram is the condition imposed
upon the complex number ζ whose affix, P, is any point of the curve.
For example, the equation of a straight line through О and containing the
point В (represented by the complex number b) is
ζ = Xb, λ real.
A straight line not containing О is uniquely defined by the foot, A, of the
perpendicular from О to the line (see Figure 20.1). Suppose A is the affix of
the complex number a and that P(z) is any point on the given line.
Fig. 20.1
Then OP = OA + AP or, in complex number notation,
ζ = a+jAa, λ real. (1)
(Recall that multiplication by j rotates the vector representing a complex
number through %π.) Taking complex conjugates of both sides of (1)
Eliminating λ this gives
z* = a*—jAa*.
a*(z-a) + a(z*-a*) = 0,
a*z + az* = c,
(2)
where с (= 2|α|2) is a real number, as the equation of the line through A
and perpendicular to О A.
*Ex. 1. By writing α = A + &jandz = χ+я, prove conversely that any equation
of type (2) above represents a straight line.

STRAIGHT LINES AND CIRCLES
Ex. 2. What is the equation of the perpendicular bisector of the line joining
the points which represent the complex numbers 0 and 2 — 3j ?
*Ex. 3. Show that
a*z + az* = 2k\a\2, к real,
represents a straight line and describe its relationship to the line
a*z+az* = 2\a\\
Interpret geometrically the quantity Re (a*z).
Ex. 4. Where does the line (1 + 2j) ζ + (1 - 2j) z* = 12 cut
(i) the real axis; (ii) the imaginary axis?
Fig. 20.2
The equation of a circle in the Argand diagram is also easily found.
If the centre of the circle is B, corresponding to the complex number b,
and if the radius is r, then, for any point P(z) of the circumference,
\z-b\=r,
or (z-b)(z*-b*) = r\
which gives, on writing the real number \b\2 — r2 as d,
zz*-b*z-bz* + d = 0 (3)
as the equation of the circle centre В and radius [|6|2— d]i.
Ex. 5. Show conversely, that for real α Φ 0 and real c, the equation
azz* + b*z+bz* + c = 0
represents a circle, provided ас < \Ь\г.
Ex. 6. What is the equation of the circle, centre 1 — 2j and radius 3 in the form
(3)?
Ex. 7. Show that 2zz* + (3-j) z+(3+j) z*+ 1 = 0 is the equation of a circle,
centre -i(3+j) and radius J2. Find the centre and radius of the circle with
equation:
(i) zz*-z(l-3j)-z*(l+3j) + 6 = 0;
(ii) 4zz*-z(2+4j)-z*(2-4j)+l = 0.
3-2 439

MAPPINGS IN THE ARGAND DIAGRAM
[20
We now consider the images of straight lines and circles under three
simple functions /, g, h from the set С into the set С given by
(I) f{,z) = w, where w = ζ+β;
(II) g(z) = w, where w = ccz;
(III) h(z) = w, where w = z_1;
α, /? being complex constants.
(I) w = ζ+β.
This is clearly a translation of the whole plane by an amount equivalent
to the position vector corresponding to the complex number β (see
Figure 20.3).
ζ+β
Fig. 20.3
Geometrically it is obvious that this function maps a straight line into a
parallel straight line, and a circle into an equal circle with its centre
translated an amount β.
Analytically, „ „
crz + az* = с
=>α*(\ν-β) + α(\ν*-β*) = с
=> a*w + aw* = ο + α*β+αβ*.
But α*β+αβ* = 2 Re {α*β); thus the right-hand side is real and the
equation represents a straight line.
Again, for a circle we have
\z—b\=r (r real and positive)
=> \w-{b+jl)\ = r,
which represents a circle of the same radius but with its centre translated
by an amount β.
Ex. 8. Explain why the straight lines a*z+az* = с and
a*w + aw* = β+α*β + αβ*
are parallel.
440

STRAIGHT LINES AND CIRCLES
(II) w = olz.
As shown in Chapter 17 this maps P^ Q where OQ = |a|0i> and
/-POQ = arga; that is, it represents an extension by the factor \a\
followed by an anticlockwise rotation of magnitude arg α (Figure 20.4).
Fig. 20.4
The effect of this mapping upon straight lines and circles may be deduced
analytically as follows:
a*z + az* = с
=> (ад)* w + (oca) w* = c|a|2
and \z-b\ = r
\w-ba\ = r\a\.
Thus straight lines are mapped into straight lines and circles are mapped
into circles under the transformation w = clz.
*Ex. 9 Show that the centre of the ζ circle maps into the centre of the w circle
under this transformation and that the ratio of the two radii is \a\; 1.
Ex. 10. Illustrate geometrically the effect of the transformation w = (1 + j) ζ
upon the circle |z—1| = 1.
(Ill) w = z-\
We must restrict the domain of this function to the whole of С with the
number 0 deleted. To describe the transformation geometrically it is
useful to define the inverse of a point Ρ with respect to a circle. Given a
441

MAPPINGS IN THE ARGAND DIAGRAM
circle, centre О and radius r, the
on OP such that О P.OP' = r2.
of the point Ρ is the point P'
*Ex. 11. Show that the transformation w = z~x maps the point Ρ into the
reflection in the real axis of the inverse of Ρ with respect to the unit circle \z\ = 1
(see Figure 20.5).
Fig. 20.5
Now consider the effect of the transformation w =
straight line and circle. The straight line
a*z + az* = с
1 upon a general
maps into
<№)'■
This represents a circle through O, if с Φ 0, and a straight line through О
if с = 0. Thus, a straight line not containing О maps into a circle through O,
while a straight line through О maps into another straight line through O.
*Ex. 12 Show that a straight line through О maps into its reflection in the real
axis under the transformation w = z_1. What lines map into themselves under
this transformation?
*Ex. 13. Show that, under the transformation w = z_1, the straight line through
A and perpendicular to О A maps into a circle with its centre on the line О A*,
where A, A* axe, the affixes of the conjugate complex numbers a, a*.
Now consider the effect of the transformation w = z_1 upon the circle
zz*-b*z-bz*+d = 0.
442

STRAIGHT LINES AND CIRCLES
The image is the set of points defined by
thatis,by dww* — b*w*— bw +1 = 0.
If d = 0 this represents a straight line, otherwise it represents another
circle. Thus, a circle passing through О maps into a straight line not through
O, while a circle not through О maps into another circle not through O.
*Ex. 14. Show that, under the transformation w = z_1, a circle, centre B, which
passes through the origin maps into a straight line perpendicular to OB*,
where В, В* are the affixes of the conjugate complex numbers b, b*.
*Ex. 15. Show that, under the transformation w = z~\ a circle, centre B, which
does not pass through the origin maps into a circle with its centre lying on the
line OB*, where В, В* are the affixes of the conjugate complex numbers b, b*.
*Ex. 16. Given a circle, centre B, not passing through O, show that the
transformation w = z_1 does not in general map В into the centre of the image circle.
*Ex. 17. Show that a diameter of a circle is mapped onto a diameter of the
image circle under the transformations w = ζ +β and νι> = olz. Show also that,
under the transformation w = z_1, a diameter through the origin is mapped
onto a diameter of the image circle.
To summarize the effect of the mapping w = z_1 upon straight lines and
circles we have (reading -> as 'maps into'):
(i) straight line through О -> straight line through O.
(ii) straight line not through О -> circle through О;
(iii) circle through О -> straight line not through О;
(iv) circle not through О -> circle not through 0.
Example 1. //" г/ге />οί/ζί ζ /ιβί on the circle \z\ = 1, find the locus of the
point w where w = j/(z+j).
Method (i)
Consider the sequence of transformations:
f1:z^u = z+j,
f3:v^W = iv,
upon the circle \z\ = 1. Their effects are shown in the following sequence
of diagrams (Figure 20.6). (For/2, recall the result of Ex. 14.)
The locus is seen to be the line w = i + Aj or w + w* = 1.
443

MAPPINGS IN THE ARGAND DIAGRAM
Method (ii)
-
^>
^>
^>
^>
4*+J) = j
w
|z| = l
IH = |j(i-w)|
N = |(i-w)l
vw* = (l-W)(l-w*)
w + vf* = 1, with the sa
e conclusion as before.
Ex. 18. If ζ moves anticlockwise around the circle \z\ = 1, starting at ζ = j,
how does w move along the line w + w* = 1 ?
Example 2. Show that the transformation
_2z-3+j
W jz-2
maps the unit circle \z\ = 1 into a circle with centre on the real axis and radius
V2.

TRANSFORMATIONS
Method (i)
Write
2z-3+j_2(z + 2j)-3-3j
jz-2 j(z + 2j)
2J z + 2j'
and consider the effect of the successive transformations
l-j)M = 3V2(cos^+jsin^M,
upon the circle \z\ = 1. (Recall the result of Ex. 15 for deducing the effect

MAPPINGS IN THE ARGAND DIAGRAM [20
of/2. Notice that, by Ex. 17, the line joining the successive images of +j
remains a diameter of the corresponding circle.)
Thus, the image is a circle, centre M(l-j)+(3+j)] = 2 and radius
il(3+j)-0-J)l=i|2 + 2j|=V2.
Method (ii)
_2z-3+j
jz-2 '
Qz-2)w = 2z-3+}
_2w-3+)
^ Z jw-2 '
Thus \z\ = 1
|jW-2| = |2W-3+;|
=> (jn, _ 2) (-)W* - 2) = (2w - 3 + j) (2w* - 3 -j)
=> ww*-2)(w-w*) + 4 = 4ww*-6(w + w*)-2j(w-w*) + 10
=> ww*-2(w + w*) + 2 = 0.
This equation certainly represents a circle. Furthermore, since the
interchange of w and w* does not affect the equation, it is symmetrical
about the real axis and thus its centre lies on the real axis. It cuts the real
axis at points given by w = w*,
i.e. x2-Ax + 2 = 0,
i.e. χ = 2 + V2,
giving a radius of л/2, as before.
Ex. 19. With the notation of Example 2, if ζ moves anticlockwiseroundthe circle
\z\ = 1, starting at ζ = j, how does w move around the circle
ww*-2(w+w*) + 2 = 0?
Example 3. The function f: С -> С w defined by Дг) = (ζ— l)2. Prow ί/ζαί
г/ге image о/ г/ге иш7 cw-c/e |z| = 1 ми^ег this mapping is a closed curve,
consisting of all points w with the property that r = 2(1 + cos Θ), where
r = \w\ and θ = argw.
Since \z\ = 1, we may write ζ = cos φ+j sin φ; then
z-1 = cos$5-l+jsini5
= -2 sin2 \φ +2j sin Ц cos y>
= 2 sin i0(-sin i^+j cos i0)
= 2 sin ^{cos (i" + W) +j sin Цтт +Щ).
Thus w = (z -1)2 = 4 sin2 10{cos (я + φ) + j sin (я + φ)) and it follows that
446

TRANSFORMATIONS
;* = 4 sin2 ^φ, θ = π + φ — lkn, where A; = —1, 0, 1, according to which
value is required to bring θ in the range — π < θ < π. Eliminating φ gives
;·= 4cos2i<9 = 2(1+ cos Θ).
In the equation ;* = 2 (1 + cos Θ), any
value of θ in the range 0 < θ < π
determines a unique value of ;* and thus a
unique complex number ;*(cos θ+j sin Θ).
As θ varies the corresponding complex
number moves around a curve Γ which is
clearly closed since ;* returns to its original
value after θ has turned through an angle
of 2π. r = 2(1 + cos ff) is called the polar
equation ο/Γ; the shape of the curve, known
as a cardioid, is shown in Figure 20.8.
Ex. 20. Prove the first part of Example 3 geometrically by constructing the unit
circle and the affixes of numbers z-1, (z-1)2.
Exercise 20
1. The function/has domain {ze C: \z\ = 1}. Find the range of/in each of the
following cases:
(i) /(z) = z + 3; (ii) /(z) = z+j; (iii) flz) = 2z; (iv) /(z) = jz;
(v)/(z)=(l+j)z.
2. In Question 1, if the point ζ is regarded as moving anticlockwise around the
circle |z| = 1, starting at ζ = 1, describe the motion of the image point w = f(z)
in each of the cases (i)-(v).
3. The function/: С -> С is defined by/(z) = z3 + z2 + 2z+1. Find the values of
ζ which remain invariant under this mapping and illustrate your answer by
reference to the Argand diagram.
4. The function /: С -> С is defined by
f(z) = az* + bz + c,
where a, b, c, e C. If Д1) = 0, ДО) = 2j, f(j) = 1 +j, show that just one ζ
remains invariant under this mapping, and find its value.
What two complex numbers map into 3 + j ?
5. As ζ moves once anticlockwise around the unit circle \z\ = 1, starting at
ζ = 1, describe the motion of the point w where:
(i) w = z2, (ii) w = z3; (iii) w = jz4, (iv) w = z~\
6. Show that the affixes of the сотрГех numbers 1, i(-l+V3J)> i( — 1 — V3J)
form the vertices of an equilateral triangle. Describe the effect of the transforma-
447

MAPPINGS IN THE ARGAND DIAGRAM [20
tion w = jz +1 + j upon this triangle, illustrating your result by a sketch of the
Argand diagram.
7. Answer the same question as in Question 6 for the transformation
w=(l+j)z + l+j.
8. Show that, under the mapping defined by w = j/(z-l), the interior of the
circle \z-1| = 1 is mapped into the exterior of the circle |w| = 1. Show that two
points remain invariant under this transformation and locate their approximate
positions on an Argand diagram.
9. Show that, under the mapping defined by w = (j-jz)/(l +z), the interior of
the circle \z\ = 1 is mapped into the half-plane Im (ve) > 0.
10. Show that, under the mapping defined by w = (j + 2jz)/(l —z), the image of
the set {z e C: \z\ < 1} is the set {w e C: Im(vf) > - i).
11. Find the equation, in the form a*z + az* = ό, of the line joining the points
ζ = 1 and ζ = j.
Find the image, under the mappings defined by w = z\(z + 1), of the set of
points represented by the interior of the triangle with vertices ζ = 0, ζ = 1, ζ = j.
In nos. 12-17 find the image of the circle \z\ = 1 under the given
transformation. If the image is a circle, find its centre and radius.
1 + z
18. Prove that, if (z - 8j)/(z + 6) is purely imaginary, the locus of ζ in the Argand
diagram is a circle with centre at the point 4j - 3 and radius 5. (O & C)
19. Show that the set of points {z} satisfying
where θ is a real number in the interval - π < θ < η and α, b are complex
numbers, is represented by the arc of a circle through the affixes А, В of α, b.
What is the condition for a point ζ to lie on the other arc of this circle?
20. Show that, if & is a real number not equal to 1, the set of points {z} satisfying
is a circle, with its centre the point С on AB such that CAjCB = k2, and that A
and В are inverses with respect to this circle.
What can you say about this circle (i) if к = 0; (ii) if к is very large?

EXERCISE 20
Draw on the same diagram the system of circles obtained for various values of
к by taking a = 1, b = -1.
What happens if к = 1 ?
21. Show that the transformation denned by w = 1/(3+j-z) maps the circle
|z-2-j| = 1 into the straight line Re (w) = i. What transformation maps the
line Re(z) = } into the circle |w-2-j| = 1?
22. Show that the transformation defined by w = 1 + z2 maps the unit circle
|z| = 1 into the unit circle \w—1| = 1. Draw a sketch of the Argand diagram,
construct the point 1 + z2 and deduce the above result geometrically.
23. Show that the transformation defined by w = 1/(1 -z)2 maps the unit circle
\z\ = 1 into the curve with polar equation 2r(l +cos Θ) = 1. Draw a sketch of
this curve, indicating on it the images of the points 1, j, —1, -j.
24. Prove that if 1
then the points on the Argand diagram defined by making A constant lie on a
circle, and the points defined by making μ constant lie on a circle.
Prove also that, whatever be the values of the constants, the centres of the two
systems of circles obtained lie on two fixed perpendicular lines.
25. In the transformation defined by w - l/(z-j), describe the motion of the
image point w if ζ moves anticlockwise around the rectangle with vertices 0, b,
b + aj, oj, starting at the origin (where a, b are positive real numbers).
26. The points Ρ and Q represent, in the Argand diagram, the complex numbers
ζ and l/(z2 + 2). The point Ρ describes a quadrant of a circle, from the origin
along the real axis to the point ζ = a, round the arc of the circle \z\ = a to the
point ζ = ja, and back along the imaginary axis to the origin. Describe the path
traced out by Q (i) when a = 1, (ii) when a = J2.
27. Find the transformation of the form
_az + b
W cz + d'
which maps 0 into j, j into 0 and 1 into — 1.
Show, that, if this particular transformation maps the complex number ξ into
the complex number i/, then it also maps η into ξ.
Can you generalize these results in any way?
28. Show that, if the image of the complex number ζ under the transformation
w = z + z-1 is real, then \z\ = 1.
Deduce a geometrical construction for determining the roots of the quadratic
equation z2 + az+1 = 0 where α is a real number in the interval -2 < a < 2.
Extend your results to deal with the quadratic equation z2 + az + b = 0, where
a, b are real numbers such that a2 < 4b, by considering the transformation
w = z + bz~\
449

21. Quadratic equations and
quadratic functions
1. THE QUADRATIC EQUATION
We have shown in Chapter 19 that it is always possible to extract the
mth root of a complex number ρ; that is, it is always possible to solve the
nation ^_p = Q
In particular, if m = 2, the quadratic equation
w2-p = 0 (1)
has roots +Jp and -φ, and the real and imaginary parts of these
two numbers may be calculated to any required degree of accuracy.
The more general quadratic equation
az2 +bz + c = 0 (a, b, с е С, а + 0) (2)
may be reduced to form (1) by the transformation
—*έ·
a process usually known as 'completing the square':
az2 + bz + c = 0
^> z2 + -az=-? (α + 0)
/ b\2 b2-Aac
-b±J(b2-4ac)
Z~ la
If b2 = 4ac, only one root is obtained, otherwise equation (2) has two
distinct roots. For uniformity it is useful, in the case b2 = Aac, to say that
(2) has two coincident roots, or a repeated roof, with this convention, every
quadratic equation has two roots.
Ex. 1. Solve the equation (3+j) z2 - 8jz - 6 + 2j = 0.
450

1] QUADRATIC EQUATION
If the two roots of equation (2) are denoted by α and β then az2 + bz + c
has linear factors (z - a) and (ζ - β) and we may write
az2 + bz + c = α{ζ-α){ζ-β),
or az2 + bz + c ξ αζ2-α(α.+β)ζ + ααβ.
Thus α+β = -δ/α (3)
and a/? = cja. (4)
Relations (3) and (4) frequently enable us to avoid the explicit solution
of a quadratic equation, especially in those cases where we are concerned
with symmetrical relations between the roots. (Particularly important
cases of this arise in analytical geometry. See Chapter 22.)
Example 1. If the roots of the equation (1 +j) z2-2jz+1 -j = 0 are α., β
find:
(i) \+\; (") "2+/?2;
(iii) χβ*+χψ, (iv) a3+/?3;
(v) ct+β*; (vi) |a-/?|.
From (3) and (4).
(i)
(ϋ)
H-
Since a
cc + β
αβ
-, β both
*+β -
χβ =
-J
-i?j-I+i
satisfy the given equation:
(l+j)"2
(l+j)/?2
-2ja+l-j=0:
-2j/?+l-j = 0:
whence, by addition,
(1+])(*2+β2)-2)(χ+β) + 2-2} = 0
Buta+/? = 1+j; thus
(1+))(*2+β2) = 2j(l+j)-2 + 2j
= -4+4j
giving χ2+β2 = 4j.

QUADRATIC EQUATIONS AND FUNCTIONS [21
[Alternatively we could write a2+/?2 = (a+/?)2-2a/?, but this approach
is less easy to generalize to higher powers—see (iv).]
(iii) φ+βα? = *β{α?+β*)
= -J(4j), by (ii),
= 4.
(iv) Since (1 + j) a2 - 2ja +1 - j = 0
we have (1 + j) a3 - 2ja2 + (1 - j) a = 0;
similarly, (1 + j) β* - 2j/?2 + (1 - j) β = 0
whence, by addition,
(1+j) (а3+Я-2Ха2+/?2) + (1-.|) ("+/?) = 0.
Thus (1 + j) (a3 +Л = 2j(4j) - (1 - j) (1 +j), by (ii),
= -10,
giving a3+ β3 = -5 + 5j, on multiplying both sides by i(l -j).
[Alternatively, α3+/?3 = (a+/?) (a2-a/?+/?)
= (a+/?){(a+/?)2-3a/?},etc]
(v) As in (iv)
(1 + j) (<* +/?4) = 2j(a3+Л - (1 - j) (a? + Л
giving
Thus,
= 2j(-5 + 5j)-(l-j)(4j)
= —14—14j
a4 + /?4 =-14.
(1+j) («·+/?«) = 2j(a*+/?«)-(l-J)(as+^
= 2j(-14)-(l-j)(-5 + 5j)
= -38j
giving a5+/?6 = -19-19J.
[Alternatively,
(а2+Л (а3+Л = afi+p* + a*fl\a+fi), etc.]
(vi)
(a-/?)2 = (a+/?)2-4a/?
= 6j,
|a-/?|2 = 6,
|a-/?| = V6.
Example 2. The roots of the equation 2z2 — 3z + 5 = 0 are α and /?. Form
ί/ге equations: (i) vWi/г rooii oc + μ,β + μ; (ii) vf/г/г roote λα, λ/?; (iii) with
roots а2, Л

1] QUADRATIC EQUATION
(i) If w = ζ + μ, then, when 2z2-3z + 5 = 0, w = α + μ or β + μ. Thus,
ct + μ and β + μ are the roots of the equation
2(\ν-μ)2-3(\ν-μ) + 5 = 0,
i.e. 2w2 - w(3 + Αμ) + 2μ2 + 3μ + 5 = 0.
Alternatively, cc + μ and/? + /t are the roots of the equation
w2 + Aw + B = 0,
where A = -(a+/?+2/t), Β = (α + μ) (/?+/*) = αβ + μ{α+β) + μ*.
Thus Л = f — 2/t, Β = \ + \μ + μ2,
leading to the same equation as before.
(ii) With reasoning similar to that of (i), we make the transformation
z-> (l/λ) w: the equation
2νν2-3λνι> + 5λ2 = 0
has roots λα, λβ.
Alternatively, the equation is
w2 +Cw + D = 0,
where С = -λ(α+/?) = -|A; D = λ2<χβ = |λ2, etc.
(iii) If w = z2, then either ζ = wl or ζ = — w%. Thus we have
2z2-3z + 5 = 0
o2z2 + 5 = 3z
■neither 2vi> + 5 = Ъ\Л or 2vi> + 5 =-3vi>£
o{2w + 5-3w^} {2и> + 5 + Зи>*} = 0
oAw2-Uw+25 = 0,
on multiplying out and rearranging terms; and this last equation has roots
cc2, β2.
Alternatively, the required equation is
w2-Ew + F = 0,
Where Ε=α2+β2={*+β)2-2*β, F = a2/?2 = (a/?)2,
giving £=(f)2-5, F=(%)2, etc.
453

QUADRATIC EQUATIONS AND FUNCTIONS [21
Ex. 2. If α, β are the roots of the equation 2z2 - 3z + 7 = 0, find:
(i) a2+/?2; (ii) a3+/?3; (iii) \α-β\; (iv) ^+^2.
£лг. 3. If α, /? are the roots of the equation 3z2 - 5z + 3 = 0 form the equation
(i) with roots 2a, 2/?; (ii) with roots a- 1, /?- 1; (iii) with roots a2, /?2.
2. QUADRATIC EQUATIONS WITH
REAL COEFFICIENTS
If in the equation , .
az2 + bz + c = 0,
a, b, с are all real (and a is non-zero) all the previous results naturally still
hold but the additional restriction placed upon the coefficients enables us
to deduce further results. In this section it will be assumed, unless explicitly
stated otherwise, that a, b, с are real numbers and α φ 0.
*Ex. 4 If fix) = ах? + Ьх + с and if/(a) > 0, /(β) < 0 (μ, β real) prove that
the equation ax2 + bx+ с = 0 has a root lying between α and β. Why is there no
corresponding theorem if a, b, с are complex ?
Theorem 21.1. If the equation
az2 + bz + c = 0 {a, b, с e R, a + 0)
has a complex root a, Im (α) Φ 0, then its other root is the complex
conjugate a*.
Since act2 + bet + с = 0, we have by taking the complex conjugate of both
sides' ew+w+c'-o·.
But a* = a,b* = b, c* = c, 0* = 0 and thus
a(ct*)2 + bct* + c = 0
and the result follows.
*Ex. 5. Explain why the proof of Theorem 21.1 breaks down if a, b, с are not
restricted to the set R. Explain also why the condition Im (α) Φ 0 was added in
the enunciation of the theorem.
*Ex. 6. Prove that a quadratic equation with real coefficients either has no non-
real root or two distinct non-real roots.
Ex. 7. If α, β are two non-real numbers with the property that
Im(a+/?) = Im(a/?) = 0,
prove that α, β are complex conjugates.
There is an analogous result to Theorem 21.1 for irrational roots of a
quadratic equation with rational coefficients.

2]
REAL COEFFICIENTS
Theorem 21.2. If the equation
az2 + bz+c = 0 (a, b, с e Q, a * 0)
/km α rooi p + q^r where p, q, r are rational (q φ 0) and φ is irrational, then
the other root of the equation is ρ - q*]r.
(Notice carefully the change in conditions for this theorem: a,b,c are
now restricted to be rational numbers.)
a(p + q*]rY+b(p+q<Jr) + c must be of the form P+Q-Jr, where Ρ and Q
are rational. Thus p<n ι — η
since ρ +qψ is a root of the given equation.
But this is possible only if Ρ = Q = 0 since ψ is irrational; thus
and it follows that p — q^r is a root of the given equation.
Ex. 8. Explain why the proof of Theorem 21.2 breaks down if a, b, с are not
restricted to the set Q. Explain also why the condition ?Ф0 was added in the
enunciation of the theorem, and where the fact that Jr is irrational is used.
Exercise 21 {a)
1. Write down the sum and product of the roots for each of the following
equations:
(i) z2-3z-7 = 0; (ii) 2z2-4z+ll = 0; (iii) (1+j) z2-z + (l-j) = 0;
(iv) (z+j)2 = (4-j)z; (ν) (z-l+j)« + z» = (2jz-l)2.
2. Write down the equations with roots:
(i) -2,3; (ii) -if; (iii)3-V5,3 + V5; (iv)2-V2,3 + 2V2; (v)3-4j,3 + 4j;
(vi) Kl-ty3), K1+JV3); (vii) l + 2j, 1 -j; (viii) V2-J, W2j.
3. Solve the following equations, giving the real and imaginary parts of the roots
correct to 2 significant figures. Check your solutions by calculating the
approximate product of the roots (using a slide rule):
(i) 2z2-3z-7 = 0; (ii)2z2-3z + 7 = 0; (iii) z2 + (l+j) z+(l-2j) = 0.
4. Find л + β, αβ, α2+/?2, (1/α) + (1//?) and α4/?+α/?4:
(i) when α, β are the roots of the equation z2-3z —9 = 0;
(ii) when α, β are the roots of the equation 3z2 - ζ - 5 = 0;
(iii) when α, β are the roots of the equation z2-jz-(l -j) = 0.
5. α, β are the roots of the equation 2z2-9z-4 = 0. Find the values of
(i) (1/α)+(1/0; (ii)(a//?2) + G%2); (iii) |a-/?|; (iv) |a2-/?2|.
6. α, β are the roots of the equation 3z2-2z-7 = 0. Find the values of :
(i) (a + k) (β + к); (ii) a3 + β3; (iii) (1 + a)-1 + (1 + β)~\
7. α, β are the roots of the equation (z-a)2 = Azb. Find:
(i) a2 + /?2; (ii) (1/α) + (1/Α· (ш) \α-β\.
455

QUADRATIC EQUATIONS AND FUNCTIONS [21
8. The equations
(b2x2 + a2y2 = a2b2,
\ у =mx + c
are solved simultaneously to give two pairs of solutions χ = Χι, у = yi ar»d
x = *2> У = >V Find iUi+*2> and iOi+^a).
9. If a, /? are the roots of the equation 2z2-z —7 = 0, form the equations with
roots:
(i)a-l,/?-l; (ii) 10a, 10/?; (iii) (1/a), (1//?); (iv) a2, β2.
10. If a, /? are the roots of the equation (1 +j) z2-2jz+(2-3j) = 0, form the
equations with roots:
(i) a-2j, /?-2j; (ii) (1 +j) a, (1 +j) β; (iii) a2, β2.
11. If a, /? are the roots of the equation 2z2 - 5z + 4 = 0, form the equations with
roots:
(i) α-/?,/?-α; (ii) α+j/?, /?+ja; (iii) α2 + α,/?2+/?.
12. The roots of the equation 4z2 + az-37 = 0 differ by 1; find the possible
values of a.
13. The roots of the equation z2 — az+ 9j = 0 are λ and jA. Find the possible
values of a.
14. The roots of the equation (2—j) z2 + (3+j) z-4-5j = 0 are α and β. Form
the equations with roots (i) (1/a), (1//?); (ii) α*, β*.
15. If the roots of the equation az2 + bz + c = 0 are α, β prove that
-aw2 + bjw + c = aQw-cc)(jw-fi).
16. If a, b are real and non-zero and z2 + az + b = 0, where ζ is non-real, prove
that |z| = 1 => й = 1 and [«[ < 2.
17. Discuss the application of the method of completing the square to the vector
quadratic equation , „ ,
4 дг.г + Ь.г + с=0 (α Φ 0).
18. The two equations
z2 + az+b = 0,
z2 + Az + B= 0
have a common root. Prove that
b(a-Ay-a(a-A) (b-B) + (b-B)2 = 0.
19. If one root of the equation az2 + bz+c = 0 is the square of the other, find a
relation connecting a, b, с
20. If one root of the equation az2 + bz + с = 0 is j times the other, find a relation
connecting a, b, с
21. Prove that, if one of the roots of the equation z2 + za+1 = 0 has unit
modulus then a is real.

3] QUADRATIC FUNCTION
3. THE QUADRATIC FUNCTION
A function/: R-> R defined by
f(x) = ax2 + bx+ с (a, b,ceR)
is called a (real) quadratic function. Since in this section we shall be
primarily concerned with inequalities, we confine our attention to real
quadratic functions and the coefficients a, b, с as well as the variable x,
will be assumed to be real.
We now derive a necessary and sufficient condition iorf(x) to be positive
for all real values of x.
Theorem 21.3. If a is not zero, then
ax2 + bx+c > 0 for all real χ ο α > 0 and b2 — Aac < 0.
Proof. Write у = ax2 + bx+c.
f , b b2 с b2\
Then y=a\x2 + -x + -r-. + — -^
[ a Aa2 a Aa2)
(( b\2 4ac-b2\
= α[{Χ + 2*) +-4Η- (1)
(i) If a > 0 and b2 < Aac, then у > 0 from (1), since
(Х + Ш > Oforalljc·
(ii) If ax2 + bx + с > 0 for all χ, then certainly α > 0, for we may choose
χ such that
(X+2a) +^F~ > α
But, if b2 > Aac, the equation ax2 + bx + с = 0 has real roots, α, β say, and
ax2 + bx+c = α(χ-χ)(χ-β),
from which it follows that γ<0ϊΐα<χ<β. But we are given that у is
always positive; thus b2 < Aac.
*Ex. 9. Give a geometrical demonstration of the"truth of Theorem 21.3.
*Ex. 10. Prove that
ax* + bx+c <0 for all real x<s>a < 0 and b2-4ac < 0.
The quantity Δ = b2 — Aac occupies a position of central importance
when considering the quadratic expression ax2 + bx+c or the quadratic
457

QUADRATIC EQUATIONS AND FUNCTIONS [21
equation ax2 + bx+c = 0; Δ is called the discriminant of the expression
ax2 + bx+c.
*Ex. 11. If the roots of the quadratic equation ax2 + bx+ с = 0 are α, β, prove
that α\α,-β)2 - Δ. If the coefficients are real, what does this tell us about the
reality of the roots of the quadratic equation?
Example 3. Determine the range of the function f: R-* R defined by
fix) = x+a
JW x2 + ax+l
for different values of a.
If x2 + ax +1 = 0,/(л;) is undefined; excluding this case, if у is a value
assumed by the given expression then
y(x2 + ax+\) = x + a,
or yx2+ (ay-1) x + (y-a) = 0.
Thu sy is a possible image of the function / if this quadraticf in χ has
real roots; that is, if , , 4, . . ч
(ay-I)2 > 4y(y-a)
or Ε ξ (a2-A)y2 + 2ay+\ ^ 0.
To determine what values of у satisfy this inequality we have, by
Theorem 21.3, to consider the discriminant of Ε and the sign of the coefficient
of y2. The discriminant of Ε is
Δ = Aa2-A(a2-A) = 16
and thus, since Δ > 0, the equation Ε = 0 has real roots α, β (where,
in fact, α = (a + 2)-1 and β = (a-2)-1). Thus we have
provided \a\ Φ 2.
The coefficient of y2 in Ε shows us that the critical values of α are +2.
0) N < 2:
For Ε to be greater than or equal to zero, у must lie between α and β;
the range is thus {y e R: *< у ζ β}.
(ii) \a\ > 2:
By a similar argument, the range is
{yeR: у ^ α or j ^ /?}.

3] QUADRATIC FUNCTION
(iii) a = 2:
Ε = Ay+1 and thus the condition Ε > 0 gives us the range
{yeR:y> -$.
(iv)fl=-2:
By a similar argument, the range is
{yeRiy^i}.
Ex. 12. Illustrate the solution of Example 3, by drawing rough sketches of the
curve у = (x+a)/(x* + ax+1) in the separate cases a = -4, a = - 2, a = 0, a = 2,
α = 4. [Express^-1 in the form ζ = .у"1 = χ + (χ+a)'1 and deduce that the graph
of ζ has turning points at x+a = ± 1; hence sketch the у curve.]
Miscellaneous Exercises 21
1. If α,, β are the roots of the equation z2-5z—11 = 0 form the equation
(i) with roots α + (βΙα),β + (α,Ιβ); (ii) with roots (α+j/?, ja+/?).
2. If a, /? are the roots of the equation az2 + bz + c = 0, find the roots of the equa-
tl0n a*z* + (lea - ό2) ζ + с2 = 0 in terms of α, β.
3. If a, b, c, d are real numbers, prove that the roots of the equation
can be complex only if b and с are of opposite signs.
4. Show that, if χ is real,
5. Show that, if χ is real, the expression
2л:2- 13л:-
cannot assume any value between 25/9 and 9.
6. Show that, if л: is real, the expression
л:2-4л:+3
л:2-6л:+8
can attain all real values.
7. If a,/? are the roots of the equation ax2 + bx+c = 0 express the roots of
the equation , , , ,. ... , .
^ ac(x2 +1) - (ό2 - lac) χ = 0
in terms of α and β.
Prove that, if л:ь х2 are roots of the equation
(л:2 + 1)(а2+1)-иал:(ал:+1) = О,
then (*?+l) (*| + 1) = шл:1л:2(х1+л:2). (О & Q
459

QUADRATIC EQUATIONS AND FUNCTIONS [21
8. x, у are real numbers.
(i) Show that, if χ and у are connected by the equation
x*+y*-2x+4y-4 = 0,
then -2 < χ < 4and -5 «Ξ у < 1.
(ii) Show that, if л: and у are connected by the equation
x2 + xy-2y2-3x+3y + 9 = 0,
then χ may have аиу value, but у cannot lie between -1 and 3.
9. Show that, when ,
Хл:-1)(л:-3) = л:-2,
л: is real for every real value of _y.
Sketch the graph of у as a function of л:, indicating the behaviour of the graph
when χ or у is large. (O & C)
10. Two quadratic equations have roots α, β and α + λ, β+λ respectively;
prove that their discriminants are equal, provided they are written with their
leading coefficients as unity.
11. If a, b, c, p, q are all real and if the solutions of the simultaneous equations
px+qy = 1, ax2+by2 = 1
are all real, prove that ab < aq2 + bp2.
12. If a, b, h, λ are all real, discuss the reality of the solution of the simultaneous
equations „ „, , . .
ax2 + 2hxy+by* = A,
x2+y2 = 1.
by putting χ = cos Θ, у = sin θ and considering a quadratic equation in tan Θ.
13. Show that, if , .4, ,.
X*-2)0-3) = лг-1,
then л: is not real when у has any value between - 3 - 2^2 and - 3 + 2^2. (O & Q
14. Show that, if p, q, r are not all equal and if
y (x-q)(x-r)'
then χ is real for every real value of у provided thai
(p-4)(p-r)<o. (o&C)
15. If α, β are the roots of „ л
x2-2ax+ab+c = 0
and y,S are the roots of x,_2bx+ab_c = Q>
provethat ^_^ (у_^ = (а_^ ^_^ = (α_^ ^_^
Hence, or otherwise, prove that, if the quadratic equations have a common root,
it is a repeated root of one of them. (O & Q

22. The parabola and rectangular
hyperbola
1. SETS OF POINTS IN A PLANE
DEFINED BY PARAMETERS
The reader has already met (Chapter 3) the parametric representation of a
line in two and three dimensions. For example, given coordinate axes Ox,
Oy in a plane, the set of points
S = {(x,y)-x = 2t-\,y = t + 2; ί real}
represents a line. The set of points may be written in the alternative form
S= {(x,y):x-2y+5 = 0},
for ~2~= y~2= U
ox-2y+5 = 0
(Notice that, to show that the parametric equations
χ = 2t-l, у = t + 2
represent the line x— 2y+5 = 0,
it is insufficient merely to verify that the point (2t-l, t + 2) lies on the line:
it is necessary also to show that any point of the line has coordinates of this
form. See Ex. 29 below.)
Ex. 1. Show that, if both χ and у are given as linear expressions in a parameter t,
then the equation connecting χ and у is linear and thus represents a straight line.
Ex- 2. Find the Cartesian equation of the line given by the parametric form
x=3i-2, y = 2t+3;
and find a parametric form for the line with Cartesian equation
Ъх-Sy-l = 0.
If the expressions for χ and у involve non-linear expressions in a
parameter t, the set of points so defined generally constitute a curve. For
example, Г = {(*,,):*= i, ,= *V real}
represents a curve, since the expression for у in terms of χ is not linear.
461

PARABOLA AND RECTANGULAR HYPERBOLA [22
Ex. 3. Show that the Cartesian equations of the curve Γ defined above is
у = x\
Distinguish between the curve Г and the curve Г' defined by
Г'= {(*,*):*=/*, * = **;/real}-
Sketch Г and Г' for -3 < χ < 3.
Ex- 4. Show that the set of points
{(x, y) e R: χ = i2, у = ί2-1; t real}
constitute a line segment. (Only in such artificial cases does a non-linear
parametric expression yield a straight line.)
The expressions for χ and у need not be algebraic: the circle with centre
(0, 0) and radius 2 may be expressed in the parametric form
χ = 2 cos Θ, у = 2 sin Θ.
Ex. 5. Find the equation of the curve with parametric representation x= α sec 0,
у = b tan Θ.
The use of a parametric representation greatly facilitates the study of the
geometry of curves with known equations. This is because the position of a
point on the curve may be specified by one variable (the parameter) rather
than two variables (jt and y) connected by an equation (the equation of the
curve).
2. TRANSLATION OF COORDINATE AXES
The equation of a given curve naturally depends upon the choice of origin
and axes. Equations can often be much simplified by retaining the directions
of the χ and у axes (defined by the unit vectors i and j) but shifting the
origin to another point O'. Such a change in coordinate system is referred
to as a translation of coordinate axes.
Suppose we are given a pair of perpendicular unit vectors i and j and a
point Ρ with coordinates (jc, y) referred to the origin О and axes Ox, Oy
in the directions i, j. Then r = OP = xi+yj is the position vector of Ρ
with respect to O. Now let O' be the point (h, k) referred to the axes
Ox, Oy (Figure 22.1). Then
a = OO' = hi + kj
is the position vector of O' with respect to O. It follows (Figure 22.1) that
s = O'P = r-a = (x-h)i + (y-k))
is the position vector of Ρ with respect to O'. Put another way, if O' is
taken as the origin, with axes O'x', O'y' in directions defined by i, j then
462

COORDINATE AXES
the coordinates (x', y) of Ρ with respect to the new axes and origin O'
are given by
-h, у' = у—к.
Ex. 6 The coordinates of A, B, C, D referred to perpendicular axes through О
are respectively (1, 2), (-2, 3), (0, 4) and (-1, -5). What are their coordinates
referred to a pair of parallel axes through O' (2, -1) ?
У
~0
yt
O'V^^^
^^P
"-^*
Example 1. Show, by a translation of coordinates to a new origin O', whose
coordinates in the original system were (2, — 3), that the equation
x2 + xy-2f-x-Uy-20 = 0
represents a pair of straight lines.
Denoting new coordinates by dashes we have
x' = x-2, y' = y+3,
^x = x' + 2, y = /-3,
and the given equation becomes
(*' + 2)2 + (*' + 2)(/-3)-2(/-3)2-(*' + 2)-14(^-3)-20 = 0)
which simplifies to „ , , , . „
F χ 2 + x у -2у2 = 0.
But the expression on the left of this equation factorizes:
(x' + 2y')(x'-y') = 0;
and thus the equation represents the pair of straight lines
x' + 2y = 0, x'-y' = 0
through O'.

PARABOLA AND RECTANGULAR HYPERBOLA
Referred to Ox, Oy, the equations are
(x-2) + 2(y+3) = 0 and (je-2) -(y+3) = 0,
or x + 2y+4 = 0 and x-y-5 = 0.
Exercise 22(a)
Find the equations of the curves given parametrically in Questions 1-6.
1. χ = cos Θ, у = 2 sin Θ.
2. χ = t*,y = i3.
3. χ = l-f,.y = 1 + ί + ί2.
4. л: = cos (9-sin θ,γ = cos(9 + sin(9.
. at2 at
5-х = ш» ^ = йтз·
6. x= t+l/t, y= l+t2.
7. Show that the parametric form
represents the curve x2+y2 = 1,
less the point (-1,0).
8. If the axes are translated to (1, -4) as new origin, find the new coordinates of
the points:
(i) (0,0); (ii) (3, -4); (Hi) (3,4); (iv) (5, -5).
9. By translating the axes to pass through (1,1) as new origin prove that the
equation *-4y*-2x+8y-3 = 0
represents a pair of straight lines, and find their gradients.
10. Prove that the equation
2х2 + Зху-2у2-Пх + 6у + 8 = 0
represents a pair of straight lines intersecting in the point (2, 3).
11. Show, by a translation of axes to pass through (p, q) as new origin, that the
equation ax*+7hxy + by* + 2gx+2fy + c = 0
is transformed into the equation
ax'2 + 2hx'y' + by'* + d = 0
if p, q are chosen so that ap + hq+g = 0,
hp + bq+f= 0.
464

COORDINATE AXES
Deduce that the equation
ax2 + 2hxy + by2 + 2gx + 2fy+ с = 0
represents a pair of straight lines only if det A = 0 where
i\}
12. Use the results of Question 11 to show that the equation
6x2 + 5xy+y2-x-l = 0
represents a pair of straight lines, and find their point of intersection.
3. THE PARABOLA
A parabola is defined as the set of points in a plane whose distances from
a fixed point, S, are equal to their distances from a fixed line / not
containing S: see Figure 22.2. For the two typical points Px andi>2,
SPi = Pi-Μί, SP2 = P2M2.
X
Mx
P, /
0\
M2
I
s\
^2\
The curve is clearly symmetrical about the line through S perpendicular
to /; this line is called the axis of the parabola. The axis cuts the parabola
at O, the vertex of the parabola.
Ex. 7. In Figure 22.2 why does SO = OTl
The point S referred to in the definition is the focus of the parabola;
the line / is the directrix. From the symmetry of the figure about the axis,
the tangent at the vertex is parallel to the directrix.

PARABOLA AND RECTANGULAR HYPERBOLA [22
Ex. 8. If the chord PQ of a parabola, when produced, cuts the directrix at K,
prove that SK bisects the exterior angle between SP and SQ. What can you say
about the angle PSZ, where Ζ is the point at which the tangent at Ρ cuts the
directrix ? What does this tell you about the angle SPM, where Μ is the foot of
the perpendicular from Ρ on to the directrix ?
Ex. 8 shows that a number of properties of the parabola may be derived
very quickly from the definition by pure geometry, but the further study
of the curve is best made by analytical methods which we shall now develop.
As has been pointed out already, the equation of a parabola depends
not only upon the relative positions of the point S and the line / but
also upon the choice of coordinate axes. Let us suppose (Figure 22.3) that
Μ
X
У
Ί
О
Р^"
S >
Fig. 22.3
SX = la. The fact that SX is an axis of symmetry suggests strongly that
this line should be taken as one of the axes of coordinates. The absence
of a perpendicular axis of symmetry leaves us without an immediately
obvious choice for second axis; however, if we take the origin to be a
point of the curve, we shall avoid a constant term in the equation. Thus we
shall take О as the origin, OS as the χ axis and the perpendicular through
О as the у axis. Now suppose that Ρ is any point on the parabola, Γ, and
that the coordinates of Ρ referred to our axes of coordinates are (x,y);
let Μ be the foot of the perpendicular from Ρ on to /.
Since XO = OS, S is the point (a, 0) and / has equation x + a = 0.
ThuS SP>=(x-af+y\
PAP = (x + a)2,
and Γ = {(*, у): (x-af+f = {x+af}
= {(x,y):y2 = 4ax}.
466

3] PARABOLA
Thus, referred to its axis as the χ axis, its vertex as the origin and S as the
point {a, 0), the equation of a parabola is
y2 = Aax. (1)
*Ex. 9. Show that χ = at2, у = lat is a parametric form for the parabola
y2 = Aax. Need a be positive in this equation ?
From the equation derived above it may be seen that all parabolas have
a similar shape but differ in size (depending upon the distance of the focus
from the directrix). The chord of the parabola through S and perpendicular
to the axis is called the latus rectum; the length of the latus rectum
determines the size of the parabola.
*Ex. 10. Show that the length of the latus rectum is 1SX.
*Ex. 11. What is the length of the latus rectum of the parabola y2 = Aaxf
Answer the same question for the parabolas y2 = —ax and x2 = Ъау.
Ex. 12. Using the same axes, plot, on graph paper, the parabola y2 = Aax, for
a = 2,l,i,Tfo, -1,-2.
*Ex. 13. By writing the equation
y2—Aax—lay-За2 = О
in the form (y — a)2 = Aa(x+a)
and translating the axes to the new origin ( — a, a), show that it represents a
parabola of latus rectum Aa. Show also that, with the original coordinate system,
the vertex is the point ( — a, a), the focus (0, a) and the directrix x+la = 0.
Ex. 14. Show that the equation
y2 -lax-lay + 5a2 = 0
represents a parabola and find its focus and directrix. Answer the same question
for the equations:
(i) y2 + 3ax+6a2 = 0; (ii) x2 + lax - Say - la2 = 0.
The chord joining the points P^at2, lat^, P2{atl, 2at2) of the parabola
is easily obtained. Its gradient, m, is given by
m ~ at\-at\
2
h+h'
and thus its equation is
y-2ah = -^-{x-atb
which reduces to 2x-{h+t^ у+2а^2 = 0. (2)
467

PARABOLA AND RECTANGULAR HYPERBOLA [22
By letting i2-> tx this leads immediately to the equation of the tangent at
ЛК,2<>: x-by+atl-0. (3)
*Ex. 15. Interpret the parameter tL geometrically.
*Ex. 16. Л is the point {atl, 2αία), P„ is the point {atl, 2at^). Show that РгР2
passes through the focus, S, of the parabola if and only if ίιί2 = — 1. (A chord
which passes through the focus is called a focal chord.)
*Ex. 17. Obtain the gradient of the tangent to the parabola y2 = Aax at the
point {at2, 2at) by differentiating the equation y2 = Aax.
*Ex. 18. If lx+my + na = 0 cuts the parabola at the points Λ(αίί, 2ah),
P2{at\, 2at2), show that ib t2 are the roots of the quadratic equation
lt2 + 2mt + n = 0.
Write down the values of m/l, n/l in terms of ii and t2 and deduce the
equation of the chord РгР2. How may this method be modified to obtain
the equation of the tangent at P(at2, 2at)?
*Ex. 19. Show that the line
(χ-χύ/l = (У-Уд1т = Я
meets the parabola y2 = Aax at points with parameters Ab λ2 which are roots of
the quadratic equation
»г2Л2 + (2тед'1-4аОЛ + 0'?-4ал:1) = О.
Deduce that, if (лг1; уг) lies on the parabola and the given line is a tangent, then
2my1-Aal = 0.
Hence show that the tangent at (лг1; у^ has equation
(x-xi)lyi = (y-y1)l2a.
The equation of the normal to the parabola y2 = Aax at the point
P{at2, 2at) (that is, the line through Ρ perpendicular to the tangent) is now
easily found. For, since the gradient of the tangent is \jt, the gradient of the
normal must be — t and thus its equation is
y-2at =-t(x-at2),
which reduces to tx+y = 2at+at3. (4)
Ex. 20. Show that, if the normal to the parabola y2 = Aax at the point
P{ap2, Zap)
cuts the parabola again at the point Q(aq2, 2aq), then q is a root of the quadratic
equation in λ «,„■»-» ·,
ρλ2 + 2λ-2ρ-ρ3 = 0.
What is the other root of this equation? Find q in terms of p.
468

3] PARABOLA
The parabola is rich in geometrical properties which may be easily
derived analytically from the results already found. We shall content
ourselves here by proving two famous results; the reader will find further
examples in Exercise 22(6).
In proving the following two properties, we shall take the parabola in
the standard form y2 = Aax by making the vertex the origin, the axis of the
parabola the χ axis, the tangent at the vertex the у axis, and writing the
length OS as a.
(i) Tangents at the extremities of a focal chord meet at right-angles on the
directrix.
The chord joining Ρχ{αί\, 2atJ and P2{at\, 2at2) is, from equation (2),
Ix-fa+Qy+lat^z = 0.
[f ΡΧΡ2 is a focal chord, (a, 0) lies on this line and thus
From equation (3), the gradient of the tangents at Ρλ and P2 are l/ix
and l/t2 and thus, since t±t2 = -1 for a focal chord, tangents at the
extremities of such a chord certainly meet at right-angles.
On solving the equations
x-^y+atf = 0,
x-t2y+ati = 0,
we obtain the point of intersection R of the two tangents in the form
(at^, a(h+ Q)· For a focal chord РхРг, this reduces to {-a, a{tx+12))
and thus R lies on the directrix. (See Ex. 22 for an alternative method of
obtaining the coordinates of R.)
(ii) If Ρ is any point on the parabola, focus S, and Μ is the foot of the
perpendicular from Ρ on to the directrix, then the angle SPM is bisected by the
tangent at P.
The reader who has successfully completed Ex. 8 will have already proved
this property by pure geometry.
In Figure 22.4 we have
tana = gradient of tangent at Ρ {at2, 2at)
= \\t, from equation (2),
tan β = gradient of SP
2at
at2 —a
It
t2-\
2(1/0
i-(i/02·
4 PPM π 469

PARABOLA AND RECTANGULAR HYPERBOLA [22
= tan 2a
and the result follows.
Property (ii) is usually known as the parabolic reflection property since
it shows that a ray of light emanating from the focus will be reflected by a
parabolic mirror along a line parallel to the axis.
Μ
y.
0
^
p^^
rf .
5
Fig. 22.4
Ex. 21. Explain the use of parabolic reflection in car headlights and reflecting
telescopes. Explain, with the aid of diagrams, the ' headlights on' and 'headlights
dipped'positions.
Parabolas frequently occur both in nature and in design. For example,
a first approximation to the path of a projectile moving under gravity is a
parabola. Again, in architecture, it is becoming quite common to find roofs
of buildings designed so that certain cross-sections are parabolic. The
curves of suspension bridges are also approximately arcs of a parabola.
We conclude this section with an example illustrating the method of
determining a locus associated with a parabola.
Example 2. PQ is a focal chord of a parabola, focus S. Show that the locus
of the mid-point ofPQ is another parabola and locate its focus and directrix.
Set up coordinate axes with the vertex, O, as origin, the axis of the
parabola as χ axis, the tangent at the vertex as у axis and call the length of
OS, a. The equation of the parabola is then
y2 = Aax.
470

3]
PARABOLA
= 0, has the form
Any line through the focus, S(a, 0), other than л
у = m(x-a).
(Notice that m is a parameter: different values of m give different lines
through the focus and all such lines, with the single exception mentioned
above, may be put in this form.)
The line у = m(x-a) cuts the parabola y2 = Aax in points with у
coordinates y1 and y2 given by the roots of the equation
that is, by my2 — Aay — Aa2m = 0.
Let (h, k) be the mid-point of this focal chord; then
к = Кл+Л) = 2a/m
by a property of the roots of a quadratic equation.
Again, since (h, к) lies on the chord,
к = m(h—a)
or h = a+(2alm2),
on substituting for к.
The locus is thus given parametrically by
x = a+(2alm2), у = (2a/m).
Elimination of m gives this locus as the curve
y2 = 2a(x-a).
(Strictly speaking, to complete the
question we should also prove that any
point of the curve y2 = 2a(x— a) is a
point of the locus. The reader may care
to supply the details.)
By translating the axes to the new
origin (a, 0) it is easily seen that this
equation represents a parabola of latus
rectum la. In the original coordinate
system, the focus is <S"(fa, 0) and the
directrix, /', χ = $a. The directrix of the
locus cuts the common axis at A where
О A = AS = SS' (see Figure 22.5). Fig- 22-5
Notes on Example 2.
(i) The use of symmetric functions of the roots of a quadratic equation
is frequently applicable to problems involving the parabola and has the
advantage of retaining symmetry in the calculation.
4-2 471

PARABOLA AND RECTANGULAR HYPERBOLA [22
(ii) An alternative method of solving the question would have been to
write Ρ as the point (atf, 2at^) and Q as the point (at\, 2at2). The coordinates
of R could then be found in terms of tx and t2, again giving the locus of
R parametrically.
(iii) If LL' is the latus rectum of the given parabola, the mid-point of
the particular chord LL' is S which must therefore lie on the required
locus, giving us a check on our working. Again, the symmetry of the locus
about the χ axis was to be expected from the symmetry of the parabola
about its axis.
Ex. 22. Show that if the tangent at the point P(at2,2at) passes through the point
R(h, k) then г is a root of the quadratic equation
aA2-kA + h = 0.
Deduce that the coordinates of the meet of tangents at Λ(α?ί, 2at^ and P2(at\, 2at£
is Riat^aifi+Q).
Exercise 22(b)
1. Find the latus rectum, vertex, focus, and directrix of each of the following
parabolas:
(i) (y-2)2 = Aix-3); (ii) (x-3)* = 2(y + l);
(iii) (y + 1)2 = -(x-1); (iv) (x+2y = 3(y + 3);
(ν) (y + 3)2 = 4(x-5); (vi) (x-2y = 80>-2);
(vii) (y-4)* + 2x= 0.
2. Find the equations of the parabolas having the given points as foci and the
given lines as directrices:
(i) (2, 1), χ = 0; (ii) (-3,2),y = 3.
3. Find the equations of the parabolas: (i) with focus (4, 3) and vertex (1, 3);
(ii) with focus (-1, - 7/4) and vertex (-1, -1).
4. Find the equations of the tangents to the following parabolas at the points
given:
(i) y* = 2{x+1), (7, -4); (ii) 0> + 2)2 = x, (9, 1);
(iii) y*-3y-5x = 0, (2, 5); (iv) л:2-2х-3y-21 = 0,(-3, -2).
5. Show that the line x+y-2 = Oisa tangent to the parabola x2 + Sy = 0 and
find the point of contact.
6. Show that the line 5x+2y + S = 0 is a tangent to the parabola
л:2-2л:-4.у = О,
and find the point of contact.
7. Any element Ρ of the set S of points in a plane has a position vector of the
f0m r = U-Di + i%
where г is a scalar parameter. Show that S defines a parabola and determine the
position vectors of its vertex and focus.
472

3]
PARABOLA
8. A ball is thrown obliquely into the air in such a way that its position vector
at time ns given by r = 10гН-(Ш-Ш*) к,
where i is a horizontal unit vector and к is a unit vector pointing vertically
upwards. Show that the path of the ball is a parabola, and determine the distance
of its directrix from the origin.
9. Any element Ρ of the set S of points in a plane has a position vector of the
form r = cos20.i-(l+sin0).j,
where θ is a scalar parameter. Show that S forms part of a parabola with its axis
parallel to i and sketch the set of points S.
Questions 10-15 refer to Figure 22.6. О is the vertex, S the focus of the parabola,
AM is the directrix, ANPM is a rectangle. The tangent at Ρ cuts the axis at Τ
and the normal at Ρ cuts it at G. PT and SM intersect at Z. You may use any
method which you regard as suitable.
Fig. 22.6
10. Prove that SZ and PT are perpendicular and that Ζ lies on the tangent at
the vertex. Suggest how, with the aid of a set square, this property may be used
to draw a parabola.
11. Prove that ST = SP and deduce that SPMTh a rhombus. (Notice that this
gives the parabolic reflector property.)
12. Prove that ST = GS and deduce that NG is of fixed length.
13. Prove that NT is bisected at O.
14. Prove that SZ2 = SO.SP.
15. Prove that TZ = ZN and that GP = 2SZ.
473

PARABOLA AND RECTANGULAR HYPERBOLA [22
16. PQ is a focal chord of a parabola. If the tangents at Ρ and Q meet at Γ and
the normals at Ρ and Q meet at [/ prove that TV is parallel to the axis of the
parabola.
17. If R is the mid-point of the focal chord PQ of a parabola and V is the foot
of the perpendicular from R on to the directrix, prove that RV = \PQ.
18. Ρ is a variable point on a parabola, focus 5, and R is the mid-point of PS.
Prove that the locus of R is a parabola and find its axis, focus and directrix.
What checks can you apply to your result?
19. The chord PQ of a parabola subtends a right angle at the vertex. Find the
locus of the mid-point of PQ.
20. In Figure 22.6 the line UOV is perpendicular to the tangent PT, meeting
PT at U and the parabola at О and V. Prove that UO.OV is constant for all
positions of Ρ on the parabola.
21. Ρ is a variable point of the parabola y2 = Aax. Find the locus of the
midpoint of OP and sketch in the same diagram the given parabola and the locus.
22. With the notation of Figure 22.6, К is the mid-point of PG. Prove that SK
is parallel to TP and find the locus of K.
23. Prove that the equation of the tangent at the point (at2, lot) on the parabola
A and В are any two points on the parabola y2 = Aax, and the tangents at
A and В meet at P. The line PM is parallel to the axis of the parabola and meets
the line AB at M. Prove that Μ is the mid-point of AB.
If the parameters of the points А, В are t and 2i respectively and the tangents
meet at P, find the coordinates of Ρ and show that it always lies on the parabola
y2 = 9ax/2. (O & C)
24. Show that any line through the fixed point (ha, ka) may be written in the
form , , , ,
y—ka= m(x—ha),
where m is a parameter.
Deduce that, if the chord PQ of the parabola y2 = Aax passes through a fixed
point, then the tangents at Ρ and Q meet on a fixed line.
25. РгР2 and Qr Q2 are parallel chords of a parabola. If Qr Q2 produced cuts
the tangents at Д and P2 at ^ and R2 then QiRt = Q2R2·
26. Prove that the equation of the chord joining the points (at2,2at) and
(aT2,2aT) on the parabola y2 = Aax is
2x-(t+T)y + 2atT= 0,
and deduce the equation of the tangent at the point whose parameter is t.
The parameters of the three points А, В and С on the parabola are tu t2 and
t3. The tangents at A and В meet the tangent at С at Q and Ρ respectively; the
tangents at A and В meet at R. Prove that if Ρ is the mid-point of QC, then
2ί2 = ί!+ί3.
Prove also that RP is parallel to А С. (О & Q
474

3] PARABOLA
27. Prove that the equation of the normal at the point P(at2,2at) to the parabola
y* = 4ax'is tx + y= at3 + 2at,
and that it meets the parabola again at QiaT2, 2aT) where Τ = -t- lit.
The tangents at Ρ and Q meet again at R; prove that, if Ρ is a variable point
on the parabola, the locus of R is
y2(x + 2a) + 4a2 = 0. (O&C)
28. The tangents at two points Ρ and Q on the parabola y2 = Aax meet at R.
Find the equation of the locus of R in each of the following two distinct cases:
(i) when the sum of the ordinates of Ρ and Q is 4a;
(ii) when the sum of the squares of the ordinates of Ρ and Q is 4a2. (O & Q
(The ordinate of a point is the у coordinate.)
29. The chord PQ of the parabola y2 = Aax subtends a right angle at the vertex,
Ρ and Q being the points (at2,2at) and (aT2,2aT). Prove that tT+ 4 = 0 and that
the locus of the point of intersection of the normals at Ρ and Q is
уг = 16a(x-6a). (O&Q
30. Show that the differential equation
d2y/dx2 = a,
where α is a constant, defines a family of parabolas (that is, any solution of the
given equation represents a parabola). If a = 6, find the equation of the parabola
of the family which has its vertex at the point (f, - f).
31. Prove that, in general, a unique parabola with its axis parallel to a given
direction may be drawn through three given points. What exceptional cases
may arise?
Find the equation of the parabola with axis parallel to the у axis and passing
through the points (-1, 5), (0,2), (1, 3).
32. A man rows a boat across a river flowing at a constant speed v. If his speed
in still water is also ν and he always aims at a point on the further bank exactly
opposite his starting point, show that his path is a parabola.
4. THE RECTANGULAR HYPERBOLA
The parabola proved particularly amenable to analytical treatment
because of its simple parametric form when referred to suitable axes.
Another curve with a simple parametric representation is the rectangular
hyperbola. We shall see later that the rectangular hyperbola is a special
case of a more general type of curve, namely, the hyperbola, and, for this
reason, we shall postpone a geometrical definition (see Chapter 27) and
content ourselves with a purely analytical treatment by defining a
rectangular hyperbola as the set of points
{(x,y):xy = c2},
475

PARABOLA AND RECTANGULAR HYPERBOLA [22
where the coordinates are referred to perpendicular axes Ox, Oy. (c is a
constant which appears here as c2 since it will later be interpreted as a
length. It is not necessary for the constant to be positive and the set of
also represents a rectangular hyperbola.)
*Ex. 23. Show that χ =
rectangular hyperbola.
ct, у — cjt is a suitable parametric representation of the
Fig. 22.7
The shape of the curve is shown in Figure 22.7. It is seen to consist of
two branches (notice that, if (A, k) lies on the curve, so also does (- h, — k)).
The point О is called the centre of the rectangular hyperbola (see Ex. 24)
and the curve is symmetrical about the lines x—у = 0 and x+y = 0.
When χ = 0, there is no corresponding value of у and vice-versa;
in fact, as x-^0 through positive values, j-^αο and as x^O through
negative values, у ->—oo. Thus, the line χ = 0 is an asymptote and
similarly so is у = 0.
Ex. 24. Show that, if any line through О cuts the rectangular hyperbola at Ρ
and Q, then О is the mid-point of PQ. (This corresponds to a property of the
centre of a circle.)
Ex. 25. Trace the position of the point P{ct, c/t) as t varies from -oo to +oo.
The chord joining the points P^ch, cjt^) and P2(ct2, c/t2) on the
rectangular hyperbola has equation
с φ2-Ί) , ,ч
У— = — ,—^ч (х— cU)
476

4] RECTANGULAR HYPERBOLA
which reduces to ± ^ .. , „ ч
The tangent at Д may be obtained immediately by letting t2 -> ^:
♦Ял:. 26. Obtain the gradient of the tangent at Pt by differentiating the equation
xy = c2.
*£Ά\ 27. Obtain the equation of the chord РгРг by the method of Ex. 18.
Ex. 28. Show that the equation of the normal to the curve at Px is
t\x-y= ci/i-l)//i.
Problems concerning the rectangular hyperbola are solved by methods
reminiscent of those employed for the parabola. We conclude this chapter
with a worked example.
Example 3. A rectangular hyperbola has centre O. Through a fixed point A
lines are drawn to cut the hyperbola at Ρ and Q. Show that the mid-point,
M, ofPQ lies on another rectangular hyperbola and find its centre and
asymptotes.
Set up coordinate axes with О as origin and the asymptotes of the given
rectangular hyperbola as the axes of coordinates, labelled in such away that
one branch of the hyperbola lies in the first quadrant (positive x, positive;;).
Let A be the point (α, β) and let PQ be any chord through A, where Ρ is the
point (ctv c/ij and Q is the point (ct2, cjt2). Then PQ has equation
x + ht2y = (fti + h)
(as shown above) and thus, since PQ contains the point A,
a+ ht^ = c(?! +12). (1)
Let (h, k) be the mid-point of the chord PQ; then
2h=c(t1 + t2), 2*=C(I+})=^,
t,t2
_<h + t2)
2lc
Hence, using (1) we see that h, к are connected by the equation
which may be rewritten as
(Α-*α)(*-*/ί) = *α£

PARABOLA AND RECTANGULAR HYPERBOLA [22
Thus {h, k) lies on the curve
(x-fr)(y-W)=W (2)
which represents the required locus (or, more accurately,1 contains the locus:
see Ex. 29).
By translating the coordinate axes to pass through the new origin
(ia> iA)> this is seen to represent a rectangular hyperbola, centre (^α, %β),
with axes parallel to the axes of coordinates. Thus, Μ lies on a rectangular
hyperbola, centre the mid-point of О A and with asymptotes parallel to
those of the given hyperbola.
The reader should compare the choice of parameters in Example 3
with that in the similar problem of Example 2.
Ex. 29. The locus sought in Example 3 may be definitely only part of the curve
(x— ia) (y- \β) = ia/3. Show this by taking A as the origin.
Exercise 22(c)
1. The tangent to a rectangular hyperbola at the point Ρ meets the asymptotes
at Q and R. Prove that Ρ is the mid-point of QR.
2. An axis of symmetry cuts a rectangular hyperbola at A, A'. P is any point of
the hyperbola and N is the foot of the perpendicular from Ρ to AA'. Prove that
PN2= AN.NA'.
3. With the notation of Question 2, if О is the centre of the hyperbola and G is
the point of intersection of the normal at Ρ with the line AA', prove that
ON ■= NG.
4. With the notation of Questions 2 and 3, if Ζ is the foot of perpendicular from
О to the tangent at P, prove that OZ.OP - ОАг.
5. With the notation of Questions 2 and 3, if the tangent at Ρ meets AA' at T,
prove that ON .ОТ = OA\
6. Show that the equation
(x-a)(y-b)= c2
represents a rectangular hyperbola and find its centre and asymptotes.
Find the centre and asymptotes of the rectangular hyperbola
ху-хЛ-Ъу-Ί - 0.
7. A variable line passes through the point (1, 1) and meets the χ and у axes at
P, Q respectively. The rectangle OPRQ is completed. Prove that R lies on a
rectangular hyperbola, and find its centre and asymptotes.
8. The tangent to a rectangular hyperbola at the point Ρ meets the asymptotes
at Q and R. If О is the centre of the hyperbola, prove that QR = 2θΡ and that
the area of the triangle OQR does not depend upon the position of P.
478

4] RECTANGULAR HYPERBOLA
9. Ρ is a variable point on a rectangular hyperbola and Q is the foot of the
perpendicular from Ρ on to one of the asymptotes. If R divides PQ in a fixed ratio
show that R lies on a rectangular hyperbola with the same asymptotes as the
original hyperbola.
10. Find the equations of the normals to the hyperbola xy = c2 which are
parallel to the line x— Ay = 0.
11. The tangents at Ρ, β to a rectangular hyperbola meet at R. If О is the centre
of the hyperbola, show that OR bisects PQ.
12. Л, P2, P3, Pi are four points on a rectangular hyperbola. Prove that, if Р±Р2
is perpendicular to P3Pi, then each of the four points is the orthocentre of the
triangle formed by the other three.
13. The normal to the rectangular hyperbola xy — c2 at the point A(cib c/O
meets the curve again at P2(ct2, c/t2). Prove that t\t2 = — 1 and deduce that the
locus of the mid-point Μ of РгР2 has equation
4x3y3 + c2(x2—y2)2 = 0.
Miscellaneous Exercise 22
1. Prove that the line px+qy = 1 is a tangent to the parabola
y2 = 4a(a-x) if a{p2+q2) = p.
A line / and a point О not on / are given; Ρ is a variable point on / and Q
divides OP in a fixed ratio. Prove that the line through Q perpendicular to OP
touches a fixed parabola with focus at О and directrix parallel to /. (O & C)
2. The coordinates of the mid-point of the line joining the points (au2,2au),
(av2, lav) are (X, Y). Express u + v and uv in terms of X and Y.
A variable chord of the parabola y2 = Aax passes through the fixed point
(b, 0). Prove that the locus of the mid-point of the chord is a parabola, and
find the coordinates of its vertex and focus. (O & C)
3. The normal at a point Ρ meets the parabola again at Q. Find the length of
PQ in terms of the acute angle θ between the normal and the axis of the parabola.
4. Two parabolas, y2 = 4a(x + b) and x2 = 4a(y + b), where a > 0, are given.
Prove that each point of intersection lies either on the line у = л: or on the line
y + x+4a= 0.
Hence, or otherwise, prove that, if the parabolas have four real points of
intersection, then b > Ъа. (О & Q
5. Lines are drawn through a variable point Ρ oi the parabola y2 = Aax, making
angles oc. and π— ос. with the axis of the parabola. The lines meet the parabola
again at Q and R. Prove that:
(i) the point of intersection of the tangents at Q and R lies on the parabola
y2 = 4a(x+4acot2a);
(ii) the line QR touches the parabola y2 = 4a(x-4a cot2 a). (O & Q
479

PARABOLA AND RECTANGULAR HYPERBOLA [22
6. Prove that the line у — mx+a/m touches the parabola^2 = Aax for all values
of m. Deduce that, if two tangents to a parabola are perpendicular, then their
intersection lies on the directrix.
7. Two directed lines meet at A, and points Ρ and P' are taken on the two lines
so that AP+AP' is constant, sense being taken into account. It is required to
prove that PP' touches a fixed parabola. Prove this by first showing that the
equations of the lines may be taken in the form
у = (x + c) tan α and у = - (x + c) tan a,
where 2a is the angle between the lines and
AP+AP' = 2c sec a,
and that the coordinates of Ρ and P' may be taken in the form
(t cos a, t sin a + с tana) and (-i cos a, t sin a— c tan a).
Deduce that PP' touches the parabola уг = Acx tan2 a. (O & C)
8. Prove that the mid-points of parallel chords of a parabola lie on a straight
line (called a diameter of the parabola) parallel to the axis.
Given a parabola traced on paper, find a construction for its focus.
9. If a point is marked on a rectangular sheet of paper, one of whose sides is AB
and the paper is then folded in such a way that AB passes through the point,
show that the crease will always touch a fixed parabola.
10. The tangents /, m from a point Ptoa parabola meet the directrix at L, Μ
respectively. The other tangents to the parabola from L, Μ meet m, /at Q, R
respectively. Prove that QR passes through the focus, S, and that the angle PSQ
is a right angle.
11. A straight line passes through the pomt P^x^y^ and has gradient m = tana.
Prove that a general point of the line has coordinates (xt+ r cos а, уг + г sin a),
where r is a parameter, and that the values of r giving the points of intersection of
the line with the parabola y2 = Aax are the roots of the quadratic equation
r2sin20+2f-O>1sin0-2acos0)+>'2-4ax1 = 0.
Deduce that, if Д is the mid-point of the chord QR, then the gradient of QR
is 2а/у! and its equation is „ . „
4 угу-2ах = yl-2axv
12. Two chords Д Qls P2 Q2 of a parabola intersect at K. Prove that the value of
thefraction (ΚΡ^ΚΩύΚΚΡ,.ΚΩύ
depends on the directions of the two chords, but not upon the position of the
point K.
[This result, known as Newton's Theorem, holds for all conies (see Chapter 27).
Use the analysis of Questions 11 to prove the result for the parabola.]
13. Follow through the work of Questions 11 and 12 to prove Newton's Theorem
for the rectangular hyperbola xy = c\

MISCELLANEOUS EXERCISE
14. Three points P(ap2, lap), Q(aq2, 2aq), R(ar2, 2ar) taken on the parabola
y2 = Aax are such that PQ subtends a right angle at R. Show that
(p + r)(q + r)+4= 0.
Show that every chord of the parabola which subtends a right angle at R
intersects the normal at R at the same point F. Show also that, as R varies, F describes
another parabola. (London)
15. The chord AB joining points A(ctu c/iO and B(ct2, clQ on the rectangular
hyperbola xy - c2 is of constant length /. Show that, as the position of the chord
varies, the centroid G of the triangle AOB, where О is the origin, moves on the
CUrVe (9xy-4c2) (x*+y2) = Pxy.
Find the area of the triangle AOB when the coordinates of G are (c, 2c).
(London)
16. Prove that the equation of the tangent PT at the point P(ct, c/t) on the
rectangular hyperbola xy = c2 is
x + t2y = let.
The perpendicular to PT from the origin О meets PT at Q and the normal at
Ρ meets the hyperbola again at R.
Prove that:
(i) as Ρ varies, the locus of Q is
(x2+y2)2 = 4c2xy;
17. Prove that the equation of the chord joining the points P{ct, c/t) and
QicT, c/T) on the rectangular hyperbola xy = c2 is
x+tTy = c(t + T).
Μ is the mid-point of PQ and PQ meets the л: axis at N. Prove that OM = MN,
where О is the origin.
The line through N parallel to OM meets the hyperbola at R and S whose
parameters are h and i2; С is the mid-point of RS. Prove that:
(i) CM is parallel to the у axis;
(ii) iti2 = - tT (O & C)
18. Prove that the equation of the tangent at the point (A, k) on the parabola
y2 = Aax and the equation of the tangent at the point (Я, К) on the rectangular
hyperbola xy = c2 are, respectively,
ky = 2a(x+h) and Хх+Яу = 2c2.
Find the coordinates of the point of intersection, P, of the parabola y2 — Aax
and the rectangular hyperbola xy = 4а%/2 and prove that the tangent to the
parabola at Ρ is the normal to the rectangular hyperbola at P.
Prove that, if this normal meets the rectangular hyperbola again at Q, then
the abscissa of Q is - 4 a. (O & C)
481

PARABOLA AND RECTANGULAR HYPERBOLA [22
19. Prove that the normal to the rectangular hyperbola xy - c2 at the point
P(ct, c/t) meets the curve again at Q{- c/t3, - ct3).
The circle on PQ as diameter meets the hyperbola at R; prove that PR passes
through the origin. (O & C)
20. Given a point Ρ on a rectangular hyperbola, prove that one and only one
real chord PQ can be drawn which is normal to the hyperbola at Q. Deduce that
there is only one (real) chord AB of the hyperbola which is normal at both A and
В and locate the two points A, B.
21. Points P, Q, R, S are taken on the rectangular hyperbola xy = c2. Prove
that PQRS is a rectangle if and only if the parameters of the four points are of the
form t, t~\ -t, -1-1. Deduce that it is impossible to inscribe a square in a
rectangular hyperbola.
22. The tangent to a rectangular hyperbola Г atthe point Ρ meets the asymptotes
at Q and R. О is the centre of the hyperbola and OQSRis a rectangle. If SQ, SR
cut Г at Q', R' prove that Q'R' touches a second rectangular hyperbola whose
asymptotes coincide with those of Г.
23. Prove that all chords of a parabola which subtend a right angle at a point
Ρ on the parabola pass through a fixed point Q.
Q is called the Frigier point of P. If a given chord subtends right angles at
two points, Ρ and P' of the parabola, prove that the join of the two Fregier
points, Q and Q', is parallel to the given chord.
24. Find the equation of the tangent to the parabola 5Ί: y2 = Aax at the point
P(at2, 2at).
If this tangent cuts the rectangular hyperbola SV· xy = k2 at real points Q, R
and Μ is the mid-point of QR show that Μ lies on part of the parabola Σ:
2y2 + ax= 0.
Sketch in the same diagram the curves Si and 52 and the set of all such points
Μ in the case a > 0. Mark in your diagram the line other than χ = 0 which
is a tangent to both 5t and 52.
482

23. Polynomial equations
1. SOME PRELIMINARY OBSERVATIONS
The results we have obtained in Chapter 18 generalize very readily to
general polynomial equations. In particular, it is possible to find
expressions for symmetric functions of the roots of polynomial equations in
terms of the coefficients. We shall pursue the question of symmetric
functions in Section 2; in this section we shall discuss a number of other results
connected with the solution of general polynomial equations of degree n.
If we are given a specific equation, say
z* + (3+2j)z2-(l-j)z + 4=0
it is not at all obvious whether or not the equation has a solution. Indeed,
two questions arise:
(i) Does every equation possess a solution ? (That is, is there a complex
number which satisfies the equation ?)
(ii) If a solution exists, can it be found by processes similar to those
employed for solving a quadratic equation ?
The answer to the first question is 'Yes' but the proof of this is rather
difficult (but not beyond the understanding of a really enthusiastic pupil: see,
for example, Hardy's Pure Mathematics, Appendix II or Courant and Rob-
bins What is Mathematics ?, Chapter 5). The result was first proved, like so
many other central theorems of mathematics, by C. F. Gauss and is
generally known as 'The Fundamental Theorem of Algebra': explicitly, it states
that any polynomial equation with complex numbers as coefficients (and
in particular, any equation with real coefficients) has a root which is a
complex number (R is regarded as a subset of С in this result). We shall
content ourselves with assuming the truth of this theorem.
The answer to the second question is 'Yes, if the degree of the given
equation is less than five' but 'No, in general, if the degree is five or more'.
The first part of this result is comparatively easy to prove and is due to a
number of Italian mathematicians of the Sixteenth Century; the second
part is very much harder and is due to two mathematicians, E. Galois
(1811-32) and N. H. Abel (1802-29). (The truly dedicated reader may care
to follow the matter up in, say, Birkhoff and Maclane, 'Survey of Modern
Algebra. Brief readable accounts of the life and work of Galois and Abel
may be found in Bell, Men of Mathematics, Volume 11.)
483

POLYNOMIAL EQUATIONS [23
Ex. 1. Although it is not possible, in general, to solve a given quintic equation
except by approximate numerical methods, this is not to say that no quintic
equation is soluble. Solve the equation z5— 32j = 0.
If P(z) is a polynomial of degree n, and if P(z) has a factor (z - cc)r (but
not (z~x)r+1) where r is an integer ^ n, the equation P(z) = 0 is said
to have a root of multiplicity r. For example, the quartic equation
(z- 1) (z + 2j)3 = 0 has a root ζ = 1 and a root ζ = - 2j, of multiplicity 3.
Ex. 2. Solve the cubic equation z3+3z2 — 4 = 0, given that it has an integral
root of multiplicity 2.
*Ex. 3. Assuming that every polynomial equation has a root, prove, by induction,
that a polynomial equation of degree η has η roots, where a root of multiplicity
r is counted as r roots.
*Ex. 4. By writing P(z) = (z-α)' Q{z) prove that, if P(z) = 0 has a root of
multiplicity r, then P'(z) = 0 has a root of multiplicity 0- 1).
Prove that the converse is false.
*Ex. 5. If P(z) ■= 0 has a root of multiplicity r > 2, what can you say about the
equation P"(z) = 0?
As with quadratic equations, additional restrictions upon the coefficients
of a polynomial equation enable us to make further assertions about the
roots. In particular, if all the coefficients are real we have the following
three important results which enable us to decide whether we have an odd
or even number of roots and to locate such roots.
Theorem 23.1. If all the coefficients of
P(z) s a0zn + a1zn-1+...+a„
are real and if the complex number zx is a root ofP{z) = 0 so also is zf.
Proof. Since „ , „ , , , _
α^ϊ + α^ϊ^+.,.+αη = 0,
we have a0{zt)n + a^zf)71'1 + ...+an = 0,
on taking the complex conjugate of each side, and noting that a* = ait since
the at are real. The result follows immediately.
Ex. 6. Given that — 1 + j is a root of the equation
z4+3z3 + 5z2+4z+2 = 0,
solve the equation completely.

1] PRELIMINARY OBSERVATIONS
Theorem 23.2. If all the coefficients of
P(z)sa0z>+o1z«-4...+a» (a0 * 0)
are real and if the equation P(z) = 0 has no real root, then P(z) is always
positive or always negative for all real z.
Proof (The theorem is intuitively obvious from graphical considerations.)
By Theorem 23.1 the roots of P(z) = 0 occurs in conjugate pairs:
Oi+JA, «,±JA. ....
But [*-(«,+j&)] [*-K-j/?r)] = (*-«,)'+#
and thus P(z) = α0[0-θι)2+Λ2] [(.z-atf+ffl -
and every square bracket is positive for all real z. The sign of P(z) is thus
determined by the sign of a0.
*Ex. 8. Prove that, if a0 and an have opposite signs, the equation P(z) = 0 has
a real root.
Theorem 23.3. If all the coefficients of
P(z) = α0ζη + α1ζ"-1 + ...+Λ„ (α0 * 0)
are real and if 'xx, x2 are real numbers such that P{x^), Р{Хг) have opposite
signs, then P(z) = 0 has an odd number of real roots between χλ and x2
(and, in particular, of course, at least one real root). For the purposes of
this theorem a root of multiplicity к is counted as к roots.
Proof. (Again, the result is intuitively obvious from graphical
considerations.)
Let ocj, a2, ..., ccr be all the real roots of P(z) = 0. Then
P(z) = ίψ-αύ{ζ-αύ ... (ζ-Ο Q(z),
where Q(z) is either positive or negative for all real ζ (by Theorem 23.2).
Thus (x1 — cc1)(x1 — cc2) ... (x1 — ccr) and (χ2—α.1)(χ2—οί2)...(χ2—α,τ) have
opposite signs and
K*l-«l) (*.-«!)] K*l-"2) (*2-"2)] .·. К*!"*,) (*,-«,)] < 0.
Thus (x1—oci) (x2— a,·) is negative in an odd number of cases and the result
follows.
Ex. 9. Prove that x3-x + 3 = 0 has a root between л: = -2 and χ = - 1.
We conclude this Section with two worked examples. The first shows
how one equation may be transformed into another in such a way that
485

POLYNOMIAL EQUATIONS [23
the roots of the two equations bear a given relationship to one another.
The second example indicates how calculus methods may be used to
solve problems on polynomial equations.
Example I. If α, β, γ are the roots of the equation
3z3-5z2 + 2z + 2 = 0
form the equation: (i) with roots α - 2, β — 2, γ — 2;
(ii) with roots 2a, 2β, 2γ;
(iii) with roots α?, β2, γ2.
(i) As in Example 2, Chapter 21, a- 2, β-2, γ- 2 are the roots of the
3(w + 2)3-5(w + 2)2 + 2(w + 2) + 2 = 0.
equation
This may be expanded directly. Alternatively, since we are essentially
making the transformation ζ -+ w + 2 we may rewrite the polynomial
3z3-5z2 + 2z + 2
in the form a(z -2)3 + b(z- 2)2 + c(z - 2) + d
and the required equation will be
aw3 + bw2 + cw + d = 0.
By Horner's Method (Chapter 18, Section 1) we have
3-5 2 2(2
3
3
3
6
1
6
7
6
13
2
4
14
18
8
10
giving the required equation
3w3+13w2+18w+10 = 0.
(ii) In a similar way, to form the equation with roots 2a, 2β, 2γ we
make the transformation ζ -+ wj2, giving the required equation:
3(iw)3-5(iw)2 + 2(iw) + 2 = 0.
or 3w3-10w2+8w+16 = 0.

1] PRELIMINARY OBSERVATIONS
(iii) If w = z2, then either z= wb or ζ = — w*. Thus
3z3-5z2 + 2z + 2 = 0
ο z(3z2+2) = 5z2-2
о either w*(3w + 2) = 5w-2
or -v^(3w + 2)= 5w-2
о {5w-2-w*(3w + 2)}{5w-2+wi(3w + 2)} = 0
о 9w3- 13w2 + 24w-4 = 0,
on multiplying out and rearranging terms, and this equation has roots a2,
β\ У2·
Example 2. Discuss the nature of the roots of the equation
3xi-4x3-l2x2 + k= 0
for different real values ofk.
P(x) = Ъх*-4х*-\2х2;
P'(x) ξ \2χ3-\2χ*-2Αχ
P'(x) = 0 when χ = - 1,0, or 2.
Write
then
and thus
Since P(x) > 0 for sufficiently large x, the values - 1, 0, 2 for χ yield
respectively a minimum, maximum and minimum of P(x). The graph of
у = P(x) thus assumes roughly the shape shown in Figure 23.1.
487

POLYNOMIAL EQUATIONS [23
Now the given equation may be written as
P(x) = -k
and the number of real roots of the equation may be deduced immediately
by considering the intersections of the curve у = P(x) and the straight
line у = —k:
к < 0: the equation has two unequal real roots and two complex
conjugate roots.
к = 0: the equation has three unequal real roots, one of multiplicity 2.
0 < к < 5: the equation has four unequal real roots.
к = 5: the equation has three unequal real roots, one of multiplicity 2.
5 < к < 32: the equation has two unequal real roots and two conjugate
complex roots.
к = 32: the equation has a real root of multiplicity 2 and two conjugate
(complex) roots.
к > 32: the equation has two pairs of conjugate complex roots.
Ex. 10. What are the roots of the equation in Example 2 in the cases
(i) к = 0; (ii) к = 5; (iii) к = 32?
Exercise 23(a)
1. Form the equation with roots 1, -2, 3~V5, 3 + V5.
2. Form the equation with roots 1 + 2j, 1 -2j, -2, 3.
3. Form the equation with a root — 1 of multiplicity 3 together with a pair of
roots +j, -j.
4. Solve the equation „„,„„.,
~* z4- 2z3 + 3z2- 4z+2 = 0,
given that it has a root of multiplicity 2.
5. Use the Remainder Theorem to solve the equation
z3-3z2-8z + 30 = 0,
given that it has an integral root.
6. Theequation 24zi+100z3+ 126z*+27z_27 = 0
has a root of multiplicity three. Solve the equation completely.
7. If oc, β, γ are the roots of the equation
z3-z-4= 0,
form the equation with roots;
... 1 1 1. _.,«+! β+1 7+1

1] PRELIMINARY OBSERVATIONS
8. If α, β, γ are the roots of the equation
z3-3z2-3z-4 = 0,
form the equation with roots:
(i) 2a, 2β, 2γ; (ii) a\ β*, y\
9. If α, β, γ are the roots of the equation
z3+jz+l+j = 0,
form the equation with roots:
0) К tf, Ь; (ϋ) (1 +JK (1 + j)A (1 + j)7·
10. If a, /?, γ are the roots of the equation
2z3-z2+4z+7 = 0,
form the equation with roots
(i) α + 2,/? + 2,γ+2; (ii) α-3, /?-3, γ-3.
11. If a, /?, γ, £ are the roots of the equation
z4-z-5= 0,
form the equation:
(i) with roots - a, - β, - γ, - δ; (ii) with roots a + 2, /?+2, γ+2, <?+2;
(iii) with roots α2,/?2, γ2, <?2.
12. If a, /?, γ are the roots of the equation
3z3-20z2+39z-18 = 0,
form the equation with roots a- 3, β— 3, γ- 3. Hence find a, /?, γ.
13. Solve the equation z4+4za + 5z2+2z_2 = α
by first increasing the roots by 1.
14. Solve the equation
z3+3jz2-5z-3j = 0
by first increasing the roots by j.
15. Show that the cubic equation
z3 + 4z2-3z+25 = 0
has three real roots, one of which is positive and the other two negative.
16. Show that the cubic equation
z3~3z+k= 0
has three distinct real roots if and only ii k is real and — 2 < к < 2.
17. Discuss the nature of the roots of the equation
2ζ3-15ζ2+24ζ+& = 0
for different real values of к.
18. Discuss the nature of the roots of the equation
ζ4-4ζ3-2ζ2+12ζ+&= 0
for different real values of к.
489

POLYNOMIAL EQUATIONS [23
19. Discuss the nature of the roots of the equation
3z4-4z3-24z2 + 48z+&= 0
for different real values of k.
20. Find a necessary and sufficient condition that the three roots of the cubic
equati°n z°+3tfz+G=0
should be real and unequal.
2. RELATIONS BETWEEN ROOTS AND
COEFFICIENTS FOR POLYNOMIAL EQUATIONS
Consider first the cubic equation
α0ζ3 + α1ζ2 + α2ζ+α3 = 0
with roots α, β, γ. We have
ας?3 + a1z2 + a2z+a3
Sflb(z-a)(z-yS)(z-y)
ξ α0ζ3-€φί+β+γ) ζ2 + α^βγ + γα, + ΰίβ) ζ-α0χβγ
and, equating coefficients of z2, z1 and z°,
cc+β+γ =-aila0,
βγ + αα, + χβ = +а2/а0,
αβγ =-α3/α0.
Similarly, for the quartic equation
a0zi + a1z3 + a2z2 + a3z + ai = 0,
with roots α, β, γ, δ we have
a0zi + a1z3 + a2z2 + a3z+ai
= α£ζ-*)(ζ-β)(ζ-γ)ίζ-ί)
s α0ζί-α(£ΰΐ,+β+γ + δ)ζ3 + αο(αβ + αγ + χδ+βγ+βδ+γδ)ζ2
-αάβγδ+γδα + δχβ+χβγ) ζ + α0*βγδ
from which, by equating coefficients,
cc+β+γ+δ = -αι/α0,
а£+ау + а*+у?у+уЮ+у* = +a2/a0,
βγδ+γδα, + δχβ+α,βγ = -α%\α*
χβγδ = +а^
490

2]
RELATIONS BETWEEN ROOTS
In words, the sum of the roots one at a time is - ajcio, two at a time is
+ a2ja0, three at a time is —a3/a0, and four at a time is aja0. For brevity,
we denote χ+β+y + S by Σα, αβ+αγ + αδ+βγ+βδ+γδ by Σα/?, etc.;
no ambiguity arises provided we know the degree of the equation under
consideration (in this case four).
Ex. 11. The roots of the equation
2z3-3z2-5z-13 = 0
are α, /?, у. Write down the numerical values of Σα, Σα/?, αβγ.
Ex. 12. The roots of the equation
z4-4jz+l = 0
are a, /?, y, δ. Write down the numerical values of Σα, Σα/?, Σα/?γ, α/?γ£.
*£лг. 13. If the coefficients of a cubic equation are all real, it follows that
Σα, Σα/? and αβγ are all real. Explain how this can be so, even if the equation
has complex roots. Elustrate your answer by considering the equation z3- 1 = 0.
Now suppose that the general polynomial equation
α0ζη + α1ζη~1+α2ζη~2+...+αη = 0
has roots ocj, a2 a„. Then it may be shown by induction that the results
proved previously in the particular cases η = 2, 3, 4 hold generally.
Σα; = — flj/flo
Σα^-α,- = +α2/α0
Σ^α,-α^ = -α3/α0
cc1cc2cc3...ccn = (-1)иак/а0.
(Let the equation with roots ocj, a2, ..., aK-1 be
a0zn~1 + b1zn-2 + b2zn-3+...+bn_1 = 0
and consider the identity
(z-aj (a0zn-1 + b1zn~2 + b2zn-3 + ... +bn_1)
ξ α0ζη + α1ζη-1 + α2ζη-2 + ...+αη.)
Ex. 14. By considering the equation zn— 1 = 0, prove that the sum of the nth
roots of unity is zero (n > 2).
Example 3. Find the quartic equation with roots 1, 3, —2, —4.
We have Σα = - 2,
Σα/? = 3-2-4-6-12+8 = -13,
Σα/?γ = 24+8-12-6 = 14,
χβγδ = 24,
491

POLYNOMIAL EQUATIONS [23
and the required equation is
z4 + 2z3-13z2-14z + 24 = 0.
Example 4. If the roots of the equation z3 —4z—6 = 0 are ot-, β, у find
(i) Σα2, (ϋ) Σα2/?, (iii) Σα3, (iv) Σα*.
We have Σα = 0, Σα/? = - 4, α,βγ = 6:
(i) Σα2 = (Σα)2-2Σα/?
= 0 + 8.
= 8.
(ii) Notice first that Σα2/? contains six terms:
Σα2/? = a^+a^+^y+y^a+r^ + y2/?.
Consider the product (Σα) (Σα/?): each term such as a2/? occurs just once,
but the terms α/?γ appears three times, as α./?γ, β.γα, and γ.α/?. Thus
Σα2/? = ΣαΣα/?-3α/?γ
= -18.
(iii) Since α is a root of the given equation, we have
α3-4α-6 = 0
and similarly for β and γ. Adding these three results
Σα3-4Σα-18 = 0.
Thus Σα3 = 18.
(iv) By an argument similar to that adopted in (iii),
Σα*-4Σα2-6Σα = 0.
Thus Σα* = 32,
on using the result of part (i).
Example 5. Solve the equation
8z3-12z2-66z+35 =0
given that the roots are consecutive terms of an arithmetic sequence.
From the equation we have
Σα = f, Σα/? = -^, αβγ = -ψ.
Since the roots are consecutive terms of an arithmetic sequence, we take
α = a—d, β = α, γ = a + d.
492

2] RELATIONS BETWEEN ROOTS
Then Σα = За, Σχβ = 3a2-d2, giving
За = I, 3a2-d2 = -ψ.
Thus a = \, d = + 3 and the roots are -2\, \, 3\.
Example 6. Solve the simultaneous equations
x+y + z = 2,
x2+y2 + z2 = 30,
x3+y3 + z3 = 116.
Let x, y, ζ be the roots of the cubic equation
t3-at2 + bt-c = 0.
Then v
a = Σχ = 2,
b = Σχγ = ${(Σχ)2-Σχ2} = K4-30} =-13.
To find c, we make use of the identity
x3+y3 + z3-3xyz ξ (x+y + z)(x2+y2 + z2-yz-zx-xy);
с = xyz = %{Σχ3 — Σχ{Σχ2 — Σγζ]}
= i{116-2[ 30 + 13]}
= 10.
Thus χ, у, ζ are the roots of the cubic equation
t3-2t2-l3t-l0 = 0.
By observation, this has a root t = — 1:
0+1)02-Зг-10) = 0,
i.e. 0+1) 0 + 2) 0-5) = 0,
and χ = — 1, у = —2, ζ = 5 (or any equivalent permutation).
Example 7. If the roots of the cubic equation z3 + 3Hz + G = 0 are α, β, у,
form the cubic equation with roots (β—γ)2, (у—α)2, (α,—β)2.
w = (β-γ)2
=> w = Σχ2-χ2-2βγ
^> aw = α(Σα)2-2αΣα/?-α3-2α/?γ (since (Σα)2 = Σα2 + 2Σα/?),
*ад= -6tfa-a3 + 2G (since Σα = 0, Σα.β = 3tf),
^> a = 3G/(w + ЗЯ) (since a3 = - 3tfa- G)
=> 27G3 + 9HG(w + 3#)2 + G(w + ЗЯ)3 = 0 (since a3 + 3tfa + G = 0)
=> κ;3+18#κ;2 + 81#2νι; + 27((72 + 4#3) = 0.
493

POLYNOMIAL EQUATIONS [23
Thus, {β—γ)2 is a root of the equation
κ;3+18#κ;2+81#ν+27((72+4#3) = 0
and, by symmetry, so also are (γ —α)2, (a—/?)2.
Note: the equation derived in Example 7 is of importance in studying the
nature of the roots of cubic equations—see Section 4.
Exercise 23(b)
1. If α, β, γ are the roots of the equation
z3-4z2 + z + 2 = 0,
find:
(i) Σα*; (ϋ) Σα3; (iii) Σα*; (iv) Σα*/?;
(ν) Σα2/?*; (vi) (α+ 2) (/?+ 2) (у + 2); (νϋ) Σ1/α;
(viii) Σα/?(α2 + /?2); (ix) Σ1/α/?; (χ) Σ(α+/?)*.
2. If α, /?, γ, δ are the roots of the equation
z4-4z3-z + 9 = 0,
find:
(ί)Σα*; (ϋ)Σα*/?; (iii) Σ1/α; (iv) Σα3;
(ν) Σα3/?; (vi) Σ(α+/?+γ)*; (νϋ) Σ(α+ 1) (β+ 1).
3. If α, /?, γ are the roots of the equation
z3+pz+q = 0,
find:
(i) Σα4; (ii) Σ1/α4; (iii) Σ(α+/?-2γ).
4. Solve the equation
-3z2-23z+12 = 0,
given that its roots are successive terms of an arithmetic sequence.
5. Solve the equation 32z3+48z2 + 6z_5 = 0>
given that its roots are successive terms of an arithmetic sequence.
6. Solvetheequation 24zs_38z2+19z_3 = 0>
given that its roots are successive terms of a geometric sequence.
7. Solve the equation
-5z + 50 =
given that one of the roots is twice a second root.
8. Solve the equations:
x+y+z = -4,
yz + zx+xy = 1,

2]
RELATIONS BETWEEN ROOTS
x+y + z = 1,
x2+y2 + z2 = 29,
x3+y3+z3 = -29.
лг+^ + z = 4,
λτ2+^2 + ζ2 = 14,
x3 + >,3 + z3 _ 34
11. The roots of the equation:
ax3+3bx2+3cx + d = 0
are in arithmetic progression. Express с in terms of a, b and d. (O & C)
12. Prove that, if the cubic equation
ax3 + bx2+cx + d= 0
has a pair of reciprocal roots (that is, α and 1/a), then
a2—d2 = ac—bd.
Verify that this condition is satisfied for the equation
6х3+Пх2~2Ах~9 = 0
and hence, or otherwise, solve the equation. (O & C)
13. If α, β, γ are the roots of the equation
z3+pz+q = 0,
form the equations with roots:
(i) α2, β2, γ2; (ii) fly, γα, α/?;
(iii) /? + γ, γ + α, α+/?; (ίν) β + γ-οί,γ + α-β, α,+β-γ.
14. If α, /?, γ are the roots of the equation
z3+pz+q = 0,
form the equations with roots:
(i) α3, β3, γ3; (ii) α(β+γ), β(γ+ά), γ(α,+β).
15. If α., β, γ are three non-zero complex numbers such that α,+ β+γ = 0,
prove that α2-βγ = β2-γα, = у2—αφ.
16. Given that α, /?, γ are the roots of the equation
л:3+.рл:2 + 9л: + г = О,
and that a = fiy—a,2,b = γα—β2,ο= α-β—y2, prove that
a = poc + q, b = pfi + q, c = py + q.
Hence, or otherwise, prove that
ax + bfi+cy = (a + b + c)(a+fi+y). (O&C)
17. If α, β, γ are the roots of the equation
x3+px2+qx + r = 0,
495
9. Solve the equations:
10. Solve the equations:

POLYNOMIAL EQUATIONS [23
find the equations whose roots are:
(i) α2-/?γ, /?2-γα, γ2-α/?; (ii) /?+γ-2α,γ + α-2/?, α+β-2γ.
What can be said about a, /?, γ if
(i) q3 = pV; (ii) 2p3~9pq+21r =0? (O &Q
18. By considering the product (/?+γ) (γ + α) (μ+β), find a necessary and
sufficient condition that two of the roots of the equation
z3+pz2+qz+r = 0 HO)
should be equal in magnitude but of opposite signs.
3. THE CUBIC EQUATION
For practical purposes an algebraic solution of the general cubic is of
limited value since, in practice, only approximate numerical solutions are
required and these are best found by the iterative methods to be described
in Chapter 26. Nevertheless, a brief survey of the method of attack seems
worthwhile on at least three counts: having seen a solution to the general
quadratic it is natural to wonder whether a similar method holds for the
cubic; such a question did occur to mathematicians of the past and the
study of the cubic is of considerable historical interest; finally, the question
whether general polynomial equations are soluble leads one to some of the
central ideas of modern mathematics.
The general cubic equation was first solved in the sixteenth century by
Tartaglia and published by Cardan in one of the most famous of all acts
of mathematical plagiarism. (See Cardan: the Gambling Scholar by O. Ore
for a biography of this unsavoury yet fascinating character.) The general
quartic soon followed (Ferrari) but, as has been pointed out already in
Section 1, the general quintic evaded all attempts at solution, for the reason
discovered independently by Abel and Galois in the nineteenth century.
An account of the Italian mathematicians of the sixteenth century whose
work we have referred to will be found in History of Mathematics by
D. E. Smith, Volume 1, Chapter 8.
We shall take the general cubic in the form
a0z8 + 3a1z2+3a2z + ai = 0. (1)
If we apply the transformation ζ -> (w—a-^ja0 we have
a0z3 + 3a1z2 + 3a2z + a3 = 0
=> alz3 + 3a1a\z2 + 3a2a0a0z + azal = 0
=> (w - aj3 + За^и; - аг)2 + 3a2a0(w - a-,) + a3 a\ = 0
^> w3 + 3w(a2a0 -af) + (2a? - ЗодС^ a2 + a3al) = 0
=> w3+3Hw + G = 0, where Я = a2a0-a\, G = Ъа^ЗооО^а^ + а^.
496

3] CUBIC EQUATION
Thus any cubic equation may be reduced to
w3 + 3Hw + G = Q (2)
and we shall, from now on, regard this as the general cubic equation which
it is our intention to solve.
Ex. 15. Show how to reduce the equations:
(i) x3- 9x2 + 22л:- 3 = 0, (ii) 2л:3 + 6л:2+ 1 Ox- 1 = 0,
to the standard form x3 + 3Hx + G = 0.
If the roots of (1) are zlt z2, z3 and the roots of (2) are κί, h>2, w3, then the
geometrical interpretation of the transformation z->(w—aj/flo is that it
transforms the triangle z1z2zs into the similar triangle \νχ\ν2\νβ which has its
centroidat the origin (и>! + и>2+и>3 = 0). (See Figure 23.2.)
Fig. 23.2
Ex. 16. Why are the triangles ZiZ2z3 and w1w2w3 similar?
Cardan's solution of (2) consists in expressing the roots as the sums of
pairs of complex numbers lying at the vertices of two equilateral triangles
with centroid at the origin. (There is no 'completing the cube' process in
general and thus it is not possible to find a further transformation that
maps the triangle и^и^и^ into an equilateral triangle.)
Since
w3-3wpq-(p3+q3) = (w-p-q)(w-a)p-u?q)(w-u)2p-<1>q)
(see Exercise 19(6), Question 12) we have
w3 — 3wpq — (p3+q3) = 0
о w = p+q or w = b)p + o)2q or w = ω2ρ + ως. (3)
Now write -pq = H, -(p3+q3) = G; (4)
497

POLYNOMIAL EQUATIONS [23
if we can find values p, q satisfying (4) then we shall have a complete
solution of the cubic equation.
From (4), p3 and q3 are the roots of the quadratic equation
ZZ + GZ-H3 = 0. (5)
Let α, β be the roots of the quadratic equation (5). Take ръ q1 as any
cube roots of α and β; then
Α+£ = α+β= -G,
p\ql = *β= -Η3.
If Я = 0, the original cubic is immediately soluble; if Я Ф 0, then
giving && = e,
where e = 1, ω or ω2, and^ and qje are suitable values to take for;? and q.
In summary, to solve the equation
ц;3 + ЗЯц; + С7 = 0 (Я + 0),
let α, β be the roots (repeated if necessary) of the quadratic equation
z2+Gz-IP = 0,
and let ρ = ca,q = fie where e = oaftl(-H). Then
p+q, ωρ + ω% aj^p + ajq (6)
are the required roots for the cubic equation.
For
(w-{p+q)){w-{o)p + u)*q)){w-{o)*p + o)q)) = w3-3wpq-(p3+q3),
and p, q have been so chosen that
pq=—H, p3+q3 =-G.
Example 8. Solve the equation z3 — 6z + 6 = 0.
We seek p, q such that „ „ ,
p3 + q3=-6,
pq = 2.
Thus, p3 and q3 are the roots of the quadratic
Z2 + 6Z+8 = 0,
498

CUBIC EQUATION
from which it follows that^3 = 2, q3 = 4 = 22.
Thus z3-6z + 6 = 0
=> ζ3 + (2*)3 + (2Ϊ)3-3ζ(2*) (2?) = 0
^> (z + 2* + 2*) (ζ+ω2» + ω2ΐ) (ζ + ω22* + ω2?) = 0
^>z=-2*-2i or -ω2*-ω22ΐ or -ω22ϊ-ω2?
^>z=-(2* + 2t) or {(2* + 2t)±j(2t-2*)V3}/2.
*Ex. 17. Show that w= Z-(H/Z) transforms equation (2) into
Z°+GZ3-H3= 0
the roots of which are ρ, ωρ, οβρ, q, o)q, ω\.
*Ex. 18. With the notation of Ex. 17, show that, to each w correspond two
values of Ζ and that to the triangle κί κί w3 formed by the roots of (2) correspond
two equilateral triangles.
Suppose now we have determined ρ and q as two complex numbers.
Then, by (6), we have
p+q, \ν2 = ωρ+ώ
w3 = rfp + uq.
(See Figure 23.3.)
Ex. 19. Show that κ>2, w3 may be found by constructing equilateral triangles on
—p, —q as base.

POLYNOMIAL EQUATIONS
[23
From now on we shall suppose that the coefficients of (2) are real: then
either one root is real and the other two are conjugate (complex) numbers,
or all three roots are real. From (5), since G, Η are real, either^3 or q3 are
both real or else p3 and q3 are complex conjugates, in which case we must
have \p\ = \q\.
Case (i). p3 and q3 both real (and unequal)—in which case we may take
ρ and q as both real and the geometrical construction (Figure 23.4) shows
that, of the solutions of (2), one is real and two are complex conjugates.
Case (ii). p3 and q3 are complex conjugates (and thus p, q may be taken
as complex conjugates). In this case, we must have \p\ = \q\ and the
geometrical construction (Figure 23.5) shows that all three roots of (2) are
real.
Fig. 23.4
Since this exhausts all the possibilities, we have arrived at the famous
paradox that, if only one root of (2) is real, then;? and q are both real, but,
if all three roots of (2) are real, then they arise from complex values of ρ
and q {the irreducible case). Put another way, if all three roots of a cubic
equation are real then the working of Cardan's solution necessarily
involves complex numbers ρ and q at the intermediate stages.
Ex. 20. Explain geometrically how (i) a root of multiplicity two, (ii) a root of
multiplicity three, arise.
e expression д = ^ _ ^ ^ _ ^ ^ _ ^2
500

3] CUBIC EQUATION
is called the discriminant of the equation
w3 + 3Hw + G = 0.
FromExample?, Δ = -27(G" + 4/7»).
Consider now the various cases that may arise (remember that we are
considering cubic equations with real coefficients).
(i) One real root a, two conjugate complex roots β+ ')"/.
A = (2/r)2(a-/?-Jy)2(/?-Jy-*)2
= _4γ2(_α2_^2_72+2α^)2
,-тС
~'\^
τ
\ I
\ 1
\Щ
\P
ys
/1
(ii) A multiple (real) root. д _ η
(iii) Three unequal real roots
Δ > 0.
Thus, the nature of the roots determines the sign of Δ.
*Ex. 21. Show, conversely, that the sign of Δ determines the nature of the roots
of the real cubic equation w3 + 3Hw + G = 0.
The case Δ > 0 leads to three real solutions via complex values of ρ
and q: the consequent working in Cardan's solution is frequently rather

POLYNOMIAL EQUATIONS [23
tedious, but can be avoided in this case by employing the trigonometric
solution illustrated in Example 10, which employs the identity
cos 3(9 ξ 4 cos3 0-3 cos Θ.
(See p. 244.)
Example 9. Solve the equation z3 — 6z + 4 = 0.
Here Я = -2, G = 4 and Δ > 0: there are three real roots. Put
ζ = 2(2)i cos Θ.
16^2 cos3 Θ-12^2 cos θ = -4
4 cos3 0 — 3 cos θ = — ^
cos30 = -Vi,
giving θ = \π or -γ^π or fytf and ζ = 2 or 2* cos -Д-7Г or 2* cos \\π.
*Ex. 22. Show that if Δ > 0, the substitution ζ = 2(-#)*cos<9 will always
yield the solution of the cubic equation z3 + 3Hz + G = 0.
Solve the cubic equations in Questions 1-5, using a trigonometric substitution
if possible and otherwise employing Cardan's solution.
1. z3-2z+4 = 0.
2. 4z3-5z + 6 = 0.
3. z3-3z-l = 0.
4. 12z3-9z-2 = 0.
5. z3-9z + 12 = 0.
6. Reduce the equation ^ fe2+ lfe_ ^ = Q
to the form z3 + 3ffz + G = 0, and hence solve it completely.
7. Show that the equation ,,_,,,_ , , „
az3 + 3bz2 + 3cz + d = 0
is reduced to the equation , ., , 4,
p(w~q)3 = q(w-pY
by the transformation w = az + b where p,q are the roots (supposed unequal)
of the equation
(ac - b2) A2 + (А/- ЪаЬс + 2b3) λ - (ас - ό2)2 = 0.
Solvetheequation za_15z2 + 57z_5 = 0. (O & Q
8. Solve the equation z3 + 5z2 + 5z + 4 = 0.
502

3] CUBIC EQUATION
9. Solve the equation z3 + 5z + 2j = 0.
10. Iip,q are real, find the turning points (if any) of the graph of у = x3+px+q.
Deduce a necessary and sufficient condition for all the roots of the equation
x3+px+q = 0
to be real.
Prove that the roots of the cubic equation det (A—Al) = 0 where
A =
В
are all real. Can two of the roots be equal?
11. (The Quartic Equation.) Show that the general quartic equation
z* + 4pz3 + 6qz2 + 4rz + s = 0
can be reduced to the form „ ,
wi + aw1 + bw+c = 0.
By identifying this equation with the equation
(w> + ay = β(νν+γ)*
show that α is a root of the cubic equation
8a3-4aa2-8ca+4ac-62 = 0.
If one of the roots of this equation is a,, and the corresponding values of/?, γ
are β а, У a, show that
w* + aw2 + bw+c = (w2-β\\ν+α0~γαβΙ) (w2+β^+α,,+ΎοβΙ).
Solve the quartic equation . „
xl + Ax-\ = 0.
Miscellaneous Exercise 23
Is or otherwise, that ;
:3 + bx2 + cx + d= 0
1. Prove, by graphical methods or otherwise, that a cubic equation (with real
coefficients) of the form
has at least one real root.
Prove that, if A is real, the equation
χ3-3χ* + 3χ+λ = 0
is satisfied for one and only one real value of л:.
2. By considering two alternative expansions of the expression
\η{(1+χχ)(1+βχ)(1+γχ)}
(ос, β, γ all real), prove that
&+βι+γι = pl-4p2q + 4pr+2qi,
where α, β, γ are the roots of the equation
x3+px*+qx + r = 0.
3. In the equation . , ,„ „ ,
xi + ax3~20x2 + bx~516 = 0,

POLYNOMIAL EQUATIONS [23
a and b are to be chosen so that two of the roots are equal and the sum of the
other two roots is zero.
Find all possible values of a and b, supposing them chosen so that the roots
of the given equation are all real. (O & Q
4. Find the value or values of a for which the roots of the equation
2х* + 6х* + 5х+а = 0
are in arithmetical progression. (O & Q
5. Given that — 2+jk is a root of the equation
2xl + 8x3+Ux2 + 4x+S = 0,
find к and the other three roots of the equation.
6. Given that α, β, γ are the roots of the equation
x3-px2+qx-r = 0,
and that sn = α.η+βη+γη, prove that
Si = P, s2 = p2—2g, s3 = i"2—qsx+Ъг.
Calculate s2 and s3 for the equation
χ3-12x4 30л:-20= 0.
Show that (s2)i and (j3)* are both near to 9 and verify that 9 approximates to a
root of the equation. (O & C)
7. If ω is a root other than 1 of the equation ω7 = 1, prove that the other roots
are ω2, ω3, ω4, ω5, ωβ.
Prove that 1+ω + ω2 + ω3 + ω4 + ω5 + ω6 = 0.
If α = ω + ω6, β = ω2 + ω5, γ = ω3+ωι, prove that the equation with roots
a>/?> 7 is „ л
'H' ' z3 + z2-2z-l = 0.
Hence, or otherwise, find the values of
(i) cos \n+cos \n + cos fπ. (ii) cos fyr cos |я cos fn-. (O & C)
8. When f{x) and^(x) are given polynomials having no common factor, prove
that the values of the constant к for which the equation
f(x)~kg(x) = 0
has a repeated root are given by/(a)/^(a) where a is a root of the equation
f(x)g'(x)~f'(x)g(x) = 0.
Hence, or otherwise, find the values of & for which the equation
л:3-Зл:2 + ЗЬ:-1 = 0
has a repeated root, and find also all the roots of the equation for each case.
(O&C)
9. Discuss the reality of the solution of the equation
2л:3 + Зл:2-36л: + А: = 0
for different real values of k.
504

MISCELLANEOUS EXERCISE 23
10. Iff[x) is a polynomial in χ which has a factor (x—oCf, prove that (л:—a) is а
factor of/"(*)· Prove conversely that, iff{x) and f\x) have a common factor
x-cc, then (лг-α)2 is a factor off(x).
Solve the equation ^^^^«^ = 0
given that it has a repeated root. (O & C)
11. If α, β, γ are the roots of the equation
x3+px+q = 0,
find the equation with roots β/γ + γ /β, γ/α + α/γ, α/β + β/α. (Ο & C)
12. The sum of the Mi powers of the roots of the equation
xt+px^+qx+r = 0
is denoted by sk. Prove that
j2 = -2p, s3 = -3q.
sn+i+psn+2+qsn+1+rs„ = 0 (n > 0),
and deduce that st = 2p*-4r, s7 = 7q(r-p%).
Given that the roots a, b, c, d of the equation are all real and that s, = 0,
prove that q = 0 and deduce that
(a+b)(a+c)(a+d)= 0. (O & Q
13. Show that, if ζ = 2(-#)*cos Θ, where ζ is a root of the cubic equation
z4-3tfz + G=0,then cos3, = V(_G2/4JP).
Deduce that, if Η < 0 and Ga+4#3 < 0 the cubic equation has three real
roots.
By using the expression ^ ^ = ^ + e_^
solve the equation when G2 + AIP > 0,
treating separately the cases ff < Q ^ ff>Q_

24. Vector products and their
applications
1. THE VECTOR PRODUCT
We have already seen, in Chapter 11, that it is frequently useful to find a
vector which is perpendicular to two given vectors; for example, to find
the equation of the plane through three given points А, В, С it is sufficient
to know the unit normal vector; that is, a unit vector perpendicular to
both AB and AC. We therefore introduce a new method of combining two
vectors a and b to form a third vector perpendicular to both of them and
called their vector product, written а л b (or a x b). It is defined by
a Ab = |a||b| sin 6>έ,
where θ is the angle between a and b (see
Chapter 11) and t is a unit vector perpendicular
to both a and b and with sense that makes a,
b, с (in that order!) a right-handed triple (see
Chapter 3). The relative positions of a, b,
and а л b are shown in Figure 24.1 where a,
а л b lie on the page and b points into the
*Ex. 1. Showthata Л а = 0. 6'
*Ex. 2. Show that а Л Ab = A(a Л b).
*Ex. 3. Show that, if a and b are parallel, then a A b = 0.
*Ex. 4. Show that, if a A b = 0, then either a and b are parallel or one or other
or both of a, b is zero.
*Ex. 5. Show that the vector product is anti-commutative; that is, that
aAb = -bAa.
*Ex. 6. Verify that iAi = jAj = kAk=Oand that
jAk=-kAj = i, kAi=-iAk = j, iAj=-jAi = k.
Ex. 7. a is a vector of magnitude 2 units pointing due east ;b has magnitude 3 units
and points N60°E. с is a vertical unit vector. Find a A b and b A a.

1] VECTOR PRODUCT
So far, apart from the result of Ex. 2, the vector product appears
remarkably dissimilar to the product of ordinary numbers. However, there
is one rule which vector products obey and which, in a sense, justifies the
use of the word 'product': it is distributive over vector addition; that is
а л(Ь + с) = (а лЬ) + (а л с).
We shall deduce this very important result in a manner reminiscent of
that used to prove the analogous result for scalar products:
a.(b + c) = (a.b) + (a.c).
In Chapter 11, we saw that the scalar product of a = OA with a unit
vector Й could be interpreted geometrically as the (scalar) projection of a
Fig. 24.2
in the direction Й and it was this fact that enabled us to deduce the
distributive law for scalar products. Consider the plane, П, perpendicular to fi
and through O; let the foot of the perpendicular from A to Π be D; then
the vector OD is called the (vector) projection of a on the plane Π (see
Figure 24.2).
It follows from the definition of the vector product that
OD= |алй|,
but OD Φ а л Й,
since OD and а л Й are perpendicular to one another.
As in Chapter 11, Section 1, the vector projection of a on Π is unique
and, ifthe projection of a! and a2 on Π are ΟΌ1 and OD2, then the projection
of OB = a! + a2 is ОС = OD! + OD2 (see Figure 24.3 and consider the
mid-point of Λ^2).
In words, the (vector) projection, in a plane, of the sum of two vectors is
the sum of the separate projections of the two vectors.
507

VECTOR PRODUCTS
[24
Now consider й л (a! + a2): from what has just been demonstrated,
|й л(а!+а2)| = |(й ла^ + СЙ ла2)|.
But й л (ах + а2) is perpendicular to the plane О ВС and thus to ОС, й л at
is perpendicular to OD^ й л a2 is perpendicular to OD2 and thus, by
rotating the parallelogram OD-^CD^ through a right angle in the plane П,
it follows that . . ч . .
й л (а! + а2) = й л а!+й л а2.
Fig. 24.3
*Ех. 8. By writing а = Ai, deduce that
аЛ(Ь + с) = (аЛЬ) + (аЛс).
Show also that (b+c) Л a = (b Л а) +(с Л а).
With the distributive rules behind us, it is an easy matter to express the
vector product а Л b in terms of components in three mutually
perpendicular directions i, j, k. If a = a!i+a2j+a3k andb = ^i+^j + ftgk, then
a Ab = Oii+aJ + азк) л (^i + ftj + ^k)
= (a2b3-a3bj i +(<%6i-ai6s) j +{а1Ьг- а2Ъд к,
on repeated application of the distributive rule and using the results proved
in Ex. 6. The result is best remembered written formally as a determinant:
j к |
bi b2 b3 I
Ex. 9. If a = i+2j + 3kandb = ί-j-k, find а Л b.
Since |a Ab| = |a| |b| sin Θ, we see that the magnitude of the vector
product gives us twice the area of AOAB (see Figure 24.4). The vector
508

1]
VECTOR PRODUCT
i(a л b) is often referred to as the vector area of AOAB. (Notice that the
vector area of AOAB is minus the vector area of AOBA.) More generally,
the vector area of the triangle ABC is i(AB л AC).
Ex. 10. OABC is a tetrahedron; prove that the sum of the vector areas of the
triangles OBC, OCA, OAB is the vector area of the triangle ABC.
Fig. 24.4
Example 1. ABCD is a plane quadrilateral whose sides CD, В A intersect
at O; if P, Q are the mid-points of the diagonals AC, BD, prove that the
area ofAOPQ is one-quarter of the area of the quadrilateral ABCD.
О d С
Fig. 24.5
Take origin О and a = OA, b = Ла, с = ОС, d = μο. We have
Ρ = i(a + c),
q = i(Aa + /*c),
and the vector area of OPQ
= ipAq
= l(a + c) л(Ла + /*с)
= 1(/<-Л)(алс)
on using the distributive and anti-commutative rules. But since ABCD
is a plane quadrilateral, its area is the magnitude of the sum of the vector
509

VECTOR PRODUCTS [24
areas of AADB and ABDC (notice that we letter the triangles in the same
sense). Thus
ABCD = i(d лЬ+b ла + а Ad)+i(d лс + с лЬ + Ь Ad)
= ^(bAa + aAd + dAc + cAb)
= MaA(d-b)-cA(d-b)]
= i(a-c)A(d-b)
= i(a-c)A(/*c-Aa)
= i(/*— λ) (a A c), and the result follows.
Ex. 11. The position vectors, relative to some origin O, of the vertices of a
convex polygon Р^РгРъ ...P„ are denoted by Pi, p2, p3,..., p„. Extend the definition
of the vector area of a triangle to the vector area of a convex polygon and show
that the vector area о{РгРгР3...Рп is
i(Pi Λ ρ2 + ρ2 Λ ρ3+ ... + Ρ„-ι Λ ρ„ + ρ„ Λ Pl).
Prove the result of Example 1 by considering the vector areas of ABCD and
OPQ.
2. PRODUCTS OF THREE VECTORS
If from two vectors, b and c, we form the vector product b л с, we may
multiply a by this new vector in two quite different ways, to give either a
scalar or a vector:
(i) the scalar triple product, a.fb л с);
(ii) the vector triple product, а Л (Ъ л с).
The component form for a. (b л с) is easily found. Recall that
X У = Х1У1 + Х2У2 + ХгУг
«li + fly + ^k and
b Ac =
а.(Ьлс) =
i
bi
Cl
ai
bi
Cl
J к
ь2 h
c2 c3
a2 a3
b2 b3
c2 c3
*Ex. 12. Prove that a.(b A c) = b.(c л а) = с. (а Л b) (that is, the scalar triple
product is unaffected by cyclic interchange of the letters) and that
a.(bAc) = (a Ab).c
(the . and A may be interchanged).
510

2] PRODUCTS OF THREE VECTORS
The scalar triple product a. (b л с) has an interesting and important
geometrical interpretation. Suppose Й is perpendicular to the plane OBC
(see Figure 24.6). Then b Л с = 5Й, where S is the area of the
parallelogram OBDC. Thus
а.(Ьлс) = (а.Й)^
= volume of the parallelepiped formed by a, b, с
Fig. 24.6
Ex. 13. Show that a.(b л с) = 0<> О, А, В, С, lie in a plane.
Ex. 14. Prove that x.(y Λ ζ) = (χ Λ у).ζ without using components (and thus
without assuming the distributive law аЛ(Ь + с) = аЛЬ + аЛс).
Ex. 15. The fact that χ. (у Λ ζ) = (χ Λ у). ζ may be used to verify the distributive
law: define d = a Λ (b + c)-a Λ b-a Л с and form the product u.d, where u
is an arbitrary unit vector. Show that u.d = 0. Why does this imply that d = 0?
*Ex. 16. Show that the volume of the tetrahedron О ABC is |a. (b Л с).
*Ex. 17. Prove that, if any one of the vectors a, b, с is a scalar multiple of one
of the other two, then a. (b Л с) = 0.
The vector triple product а л (b л с) is a little more difficult to express
in component form. First observe that, if a is perpendicular to b and also
perpendicular to c, then it is parallel to b л с and thus а Л (b л с) = 0;
let us assume that none of a, b, с is zero and that a is not perpendicular to b
and thus that a.b φ 0. Again, if b = Ac, а Л (b л с) = 0; let us assume
that b φ Ac. Since b л с is perpendicular to both b and с and since
а л (b л с) is perpendicular to b л с it follows that а Л (b л с) lies in the
plane of b and с Thus, а л (b л с) can be expressed uniquely as λЪ+μc;
we have to determine A and μ. By Ex. 17,
а. [а л(Ь л с)] = 0,
.·. A(a.b)+/i(a.c) = 0,
.·. * = -^^ (since a.b Φ 0),
μ (a.b) ^ h
511

VECTOR PRODUCTS [24
and thus а л (b Λ c) = v[(a.c) b-(a.b) c].
To find v, consider the i component of both sides:
<h{b\ c2—b2c^ — a3{b3 c-t-b-L c3)
= v[(axcx + a2c2 + a3c3) bx-(axbx + a2b2 + a3Ьэ) cj,
.·. 6i(a2C2 + a8cs)-ci(«2*2 + as*s) = ^Ь^^ + ^с^-с^а^-а^].
.·. ν = 1.
Thus we have finally
а л(Ь Л с) = (a.c)b-(a.b) с.
*Ex. 18. Prove that the vector triple product is not associative; that is
а Л (b Л с) Ф (а Л b) Л с.
Ex. 19. Prove that а Л (b Л c) + b Л (с Л а) + с Л (а Л b) = 0.
Exercise 24(a)
1. Prove that (a-b) Л (a + b) = 2a Л b and interpret the result geometrically in
terms of areas.
2. What can you deduce from the equation гЛа = гЛЬ?
3. Prove that a + b + c =0*ЬЛс= сЛа = аЛЬ and interpret the result
geometrically.
Is it true that ЬЛс = сЛа = аЛЬ* a + b + c = 0?
4. Find а Л b in the following cases:
(i) a= 3i-j + k, b= 2i+j + k; (ii) a= 5i + 4j + 3k, b= 3i + 4j + 5k;
(iii) a = i-k, b = j + k; (iv) a = 2i-j-k, b = 2i+j+k;
(v) a = ai + ck, b = b].
5. If a = 3i+j + 2k, b = i + 2j + 3k, с = i-3j-4k, find а.(ЬЛс). What can
you deduce about the points А, В, С?
6. If a = i+j, b = 2i+j, с = i + k, form the vector products а ЛЬ and
(а Л b) Л с and mark the position of the five vectors a, b, с, а Л b, (а Л b) Л с
in a rough sketch.
7. The points А, В, С have position vectors a = i-4j + 3k, b=2j-3k,
с = 3i- 6k. Find the unit vector perpendicular to the plane OAB and deduce
the perpendicular distance from С to this plane.
8. With the notation of Question 7, find the area of b.OAB and hence the volume
of the tetrahedron О ABC.
9. If А, В, С are non-collinear points with position vectors a, b, c, prove that
the vector ЬЛс + сЛа + aAbis normal to the plane ABC.
If d is the position vector of the point D and a = 5i+j + k, b = i+j + 3k,
с = 3i + 4j-k, d = 8i + 3j + 2k, find:
(i) the unit normal to the plane ABC;
(ii) the area of the triangle ABC;
(iii) the volume of the tetrahedron ABCD.
512

PRODUCTS OF THREE VECTORS
«i (h a3\
i Ui h b,\ cubic units-
I ct c2 c3 I
If d = i^i + iy + i^k, write down the volume of the tetrahedron ABCD.
What is the condition that A, B, C, D should be coplanar. Is this a necessary
and sufficient condition?
11. If a, b are perpendicular vectors and if г Л а = b, prove that
г = Ла + /«(аЛЬ),
where λ may be chosen arbitrarily. Find μ in terms of |a|.
12. If /t... m3 are real numbers such that
l\ + ll+ll = ml + ml + ml = 1 and /^Ч- /2w2 + l3m3 = 0
prove that (/2 m3-l3m^2 + (l3 тг - /ι w3)2 + (/i w2 - /2 w^2 = 1.
13. Prove that (aAb).(cAd) = (b.d) (a.c)-(b.c) (a.d). What trigonometric
identity follows if O, A, B, C, D are coplanar and the angles AOB, COD are both
right angles?
14. The vectors a = OA, b = OB, с = ОС are non-coplanar; prove that a
given vector г may be expressed uniquely in the form
г = ab Л с+/?с Л а+уа Л b
and find a, /?, γ in terms of a, b, с and r.
Ifr = OP, locate Ρ in terms of О, А, В, С: (i) if α = β = γ = 1; (ii) if α = О,
/9 = 7=1.
15. Points β, R are taken on the sides CA, AB {not produced) respectively of a
triangle ABC in such a manner that the triangles ABQ, ARC are equal in area.
Prove that RQ is parallel to ВС.
16. G is the centroid of the triangle ABC, and E, F are the mid-points of CA,
AB respectively. Prove that the quadrilateral AFGE and the triangle GBC have
the same area.
17. ABCD is a skew quadrilateral; the mid-points of the opposite sides ВС, DA
are P, Q, while the mid-points of the diagonals AC, BD are R, S. Prove that
PRQS is a parallelogram whose area is i\&ABC-AABD\.
18. ABC is a triangle and points L, M, Ware taken on ВС, CA, AB respectively
such that
BL _ CM _ ЛЛГ _
LC ~ ' МЛ ~ μ' ΝΒ~ V'
Prove that ALMN = (1 + λ/tν) ДЛВС and deduce the theorem of Menelaus, that,
iiLMN is a straight line, A^v = - 1.

VECTOR PRODUCTS
[24
19. The sides (and sides produced) of the parallelogram ABCD separate the plane
into nine regions (see Figure 24.7). Prove that, if a point О is taken within
region I, II, VI or IX the area of О AC is the sum of the areas of OAB and
О AD but that, if О is taken within any other region, the area of О AC is the
difference of the areas of OAB and О AD.
(The reader who has studied mechanics may be able to interpret this result in
terms of moments.)
Fig. 24.7
20. ABCD is a tetrahedron. The lengths of AB and CD are a and b and the angle
between AB and CD is Θ. A plane is drawn parallel to AB and CD and meets the
edges CA, CB, BD, DA at P, Q, R, S. Prove that PQRS is a parallelogram and
that its area is
where χ is the distance of the plane from AB and h is the distance between AB
and CD.
Hence, or otherwise, prove that the volume of the tetrahedron is
\abh sin Θ.
3. APPLICATIONS TO COORDINATE
GEOMETRY
Throughout this section we shall assume that an origin О and a right-
handed set of coordinate axes Ox, Oy, Oz specified by the unit vectors
i, j, к are given. The position vector of the point A... with respect to О will,
as usual, be denoted by a ....
Consider first the line through A in the direction of the unit vector u.
If R is any point on this line, then AR = λύ and thus
ARaU = 0,
giving the equation of the line in the form
(r-a) ли = О.
514

3] APPLICATIONS TO COORDINATE GEOMETRY
If г = xi+yj+zk, a = axi+OiJ+a^k, й = Γι+mj+nk, we can transform
this to (or from) the familiar coordinate form as follows:
(r-a) Ли = 0
=> [(x-ai)i + (y-a2)i + (z-a3)k] А(П + т\ + пк) = 0
o(y-a£n-(z-(Qm = {z-a^l-{x-a^n = {x-a^)m-{y-a2) I = 0
χ—αχ _ y—a2 _ z—a3
^ I ~ m ~ η "
Fig. 24.
The distance from the point В to the line, the length BN (see Figure
24-8».'s |(Ь-.)лО|.
If В is the point (bu b2, b3) this gives
BN = \[(b2-aj n-(b3-a3) m] 1 + Wb-aJ l-ih-a^ n] j
+ [(b1-a1)m-(b2-a2)[\k\.
Ex. 20. Show that the perpendicular distance from the point (3, 3, 2) to the line
x-l _y-2_z-3 .
/83
1S Jt4-
The vector product is particularly useful in giving us the vector
perpendicular to two given vectors. For example, consider the plane through the
(non-collinear) points А, В, С The vector
d = AB л AC
is perpendicular to this plane. Thus
d = (Ь-а)л(с-а)
= b лс + с ла + а лЬ
and the equation of the plane is now found easily.

VECTOR PRODUCTS
[24
Example 2. Find the equation of the plane through the points A(l, 1, —3),
B(2, -1, 10), C(-l,-l, 1).
AB= i-2j+13k,
AC=-2i-2j+4k,
AB лАС= 18i-30j-6k,
and thus the vector 3i—5j —к is perpendicular to ABC Since A lies in the
plane the equation of ABC is therefore
3(x-l) -5(y-l)-(z + 3) = 0
which gives, on simplification,
3x-5y-z = 1.
*Ex. 21. If г = xi+y) + zk is the position vector of the point Ρ show that the
volume of the tetrahedron Ρ ABC is
1(г-а).[(Ь-а)л(с-а)].
(See Ex. 15.)
Hence obtain the equation of the plane ABC of Example 2 in determinant
form, and verify that the expansion of this determinant gives the same equation
as that obtained above.
Example 3. Find the equation of the plane through the point (2, — 5, — 1)
perpendicular to the line of intersection of the planes
x + Jy + z = 0, x-5y-2z = 0.
The line of intersection is perpendicular to both of the vectors
i+7j+k and i—5j —2k
and thus its direction is given by
(i+7j+k) A(i-5j-2k) = -9i + 3j-12k.
Thus 3i — j + 4k is normal to the required plane, whose equation must be
3(x-2)-(y + S) + 4(z+l) = 0
or Ъх-y + Az = 7.
Example 4. Find the length of the common perpendicular to the two skew lines
x y+l z + 2 x-2 y-6 z-Ъ
1~ 2 ~ 4 ' -1~ 4 ~ 2 ■
The direction of the common perpendicular is given by
(i + 2j+4k) A(-i + 4j + 2k) =-12i-6j + 6k.
516

3] APPLICATIONS TO COORDINATE GEOMETRY
A unit vector in the direction of the common perpendicular is thus
fl = (2i+j-k)A/6.
The lines pass through the points A(Q, — 1, — 2), B(2, 6, 3) respectively and
we have the length of the common perpendicular equal to the projection
of AB in the direction u
= (b-a).U
= (2i + 7j + 5k).(2i+j-k)/V6
= (4+7-5)/V6
The vector product may be used to obtain an alternative method of
solving three linear simultaneous equations. Consider the three equations
3x + 5y + Sz=-l Pu
2x + 3y + 4z = 3 P2,
4x+2y + 5z =-2 P3.
These three equations represent three planes, Pu P2) P3; their solution
consists in finding the point Q common to each plane (assuming such a point
exists).
From equations P1 and P2 we obtain (3PX +P2), the equation of the plane
through the line of intersection of Ръ Р2 containing the origin. Thus
Ux + 18y+28z = 0 Pi
is a plane containing О and Q. Similarly, ДО-Д),
2x + 8j + llz = 0 P5
is a plane containing О and Q.
Thus OQ is perpendicular to both 1 li +18j +28k and 2i + 8j +1 Ik; that is
OQ = A(lli + 18j428k) A(2i + 8j + llk)
= -13A(2i + 5j-4k),
and Q has coordinates (—26λ, — 65λ, 52λ) where Λ is some scalar. Writing
— 13λ = μ, Q has coordinates (2μ, 5μ, —4μ). Substituting in Ръ we find
that μ = 1 giving Q as the point (2, 5, — 4). The solution of the given set
of equations is thus „ , .
n χ = 2, у = 5, ζ = — 4.
517

VECTOR PRODUCTS [24
Ex. 22. Use the method outlined above to find a necessary and sufficient
condition for the existence of a non-trivial solution to the set of three homogeneous
linear equations: , , ,
a-LX+Wy + CiZ = 0,
агх + Ьгу + czz = 0,
a3x+b3y + c3z = 0.
Exercise 24(b)
1. Find the equations of the planes through the following sets of points:
(i) (1, -1,1), (2, 1,2), (3,1, -1);
(ii) (1, 1, 1),(0, -3, -2), (-1,3, -1);
(iii) (1,1,5),(1, -1, -3),(2, 1,1);
(iv) (4, 0, 1), (7, 3, 5), (-10, -4, -1);
(v) (a, b, с), (За, b, 0), (4a, 2b, c).
2. Find, in each part, the equation of the plane through the given point and
perpendicular to the line of intersection of the given planes:
(i) (1, -1, 3); 4x-y + 4z = 0, 12x-5y + 8z = 0;
(ii) (-5, -4, -1), 3x+4y + 3z = 15, 9x-2y-12z = 3;
(iii) (2, 3, 4), 5x+y + 2z = 7, x+y+z = -4;
(iv) (9, 3, 1), 3x+2y+2z = 0, 5x+2^+z = 4;
(ν) (1, 1, -1), 3x+2y-4z = 0, 4x+y + 3z = 0.
3. Find the length of the common perpendicular to the following pairs of skew
lines :
ω -з~ 2 ~ 2 ' ■>-_-;-*;
(ii) x-6 = y-3 = z-l,
x-4 = y-7 =
3 ~ 5
x+l _y+l _ ζ
2' 2 -1 2 "
4. Find the length of the common perpendicular to the two lines
x-4 у z+3 . x-6 y+1 z + 5
What deductions can you make from your answer?
5. Find the equation of the plane containing the points A (6, 3, 0), B(—2, 0, — 5)
which is parallel to the vector i + j + k.
6. Find the equation of the plane containing the points A(3, 5, 10), B(4, 3, —3)
which is parallel to the vector 3i + 5j - 6k.
7. Find the equation of the plane parallel to the two lines
f(x-6)=i0>-6) = |(z-l), $x = -{y-2) = z + A
which passes through the point (2, 0, — 1).
518

EXERCISE 24
8. Given the plane P: 2x+3y-z = -1 and the line L: %x = -y = -(z-5),
find·.
(i) the coordinates of the point of intersection of Ρ and L;
(ii) a vector parallel to Ρ and perpendicular to L;
(iii) a vector in the direction of the projection of I, in P;
(iv) the equation of the projection of I, in P.
9. Show that the equation of the plane containing the points А, В, С has
equation
(г-я).(Ь-я)Л(с-я) =
Write down, in determinant form, the equation of the plane through the points
А(аъ Ch, a3), В{ЬЪ Ьг, b^), С(съ с2, с3).
10. Find a vector perpendicular to both ca + bj + ck and li + nq + nk. Hence
write down the equation of the plane through the origin containing the line
x—a _ y—b _ z—c
Use this result to find the line through the origin which intersects each of the
ППе8 K*+D = y-2 = i(z + 2),
i(*+D= Ky-4) = (г-3),
giving your answer in standard form.
11. Find the vector equation of the line through the point С (position vector c)
which intersects both of the lines
r = a!+Abi, r = a2 + /tb2.
Discuss any special cases that might arise.
12. Find the volume of the tetrahedron formed by the planes
βγ+γζ = 0, γζ + αχ = 0, αχ + βγ = 0, αχ + βγ + γζ = δ.
13. A square piece of paper ABCD of side a is folded along the diagonal BD so
that the planes of the triangles ABD, CBD are perpendicular. Find the shortest
distance between the edges A B and CD.

25. Continuous probability distributions
1. DENSITY FUNCTIONS
In Chapters 7, 10 and 16 we discussed probability distributions in discrete
outcome spaces, that is, in spaces whose elementary events we could count
by putting them into one-to-one correspondence with the set N of natural
numbers. Associated with each elementary event 5* of the outcome space
S was a number x{, usually an integer—the value of our associated random
variable X.
Random experiments occur, however, in which it is mathematically
more convenient to take as outcome space a set which cannot be counted
and as random variable a subset X of the set R of real numbers. For
example, suppose we take a point Ρ at random on a line AB of length / m.
Since we can only measure lengths to within some given limits of accuracy,
say 0-5 mm, we could take as values of a (discrete) random variable the
number of intervals of length 0-5 mm Ρ is from A, giving 201 possible values
in all (0-200). Mathematically the situation may be simpler to analyse if
we take as values of our (continuous) random variable the supposed' exact'
length of AP, measured as a real number of mm; that is
X={xeR:0< χ «ξ 100}.
If we are dealing with a subset of R, calculus methods are available and it is,
for example, easier in general to integrate a function than it is to sum a
series.
Given a continuous random variable X, it will generally be impossible
to assign a probability to a specific value of X; rather we assign a
probability to each interval
Ε = {x eX; хг < χ < x2}
as follows. Let/be an integrable function such that
(i) f(x) Ζ 0, for all χ e X;
(ii) $xAx)dx= 1;
where the notation denotes that the integration is to be performed over
the set X. Then we define
Pr(£)=£V(*)d*.
520

1] DENSITY FUNCTIONS
The function/so defined is called a. probability density function. As in
the discrete case the choice of/ constitutes an assumption about our
random experiment; that is, to choose/is to choose a particular mathematical
model. Geometrically, probabilities are measured as areas under the curve
y=f{x).
Consider again the random experiment of marking a point on the line
AB. To analyse the situation we must first set up a mathematical model;
in other words, we seek to make the vague statement ' we mark a point Ρ
at random on AB' mathematically precise by stating explicitly what we
feel 'at random' could mean.
0 50 100
Fig. 25.1
We may regard AB as an interval on the χ axis, with A as origin and В
100 units to the right. Suppose we conjecture that a point is more likely to
be chosen towards the middle of AB than near the ends of the interval.
Taking, as is customary, the whole set R as the range of our random
variable Χ,-f a possible choice of/(x) would be
(0 if x < 0
kx if 0 < χ < 50
£(100-x) if 50 < χ < 100
0 if χ ^ 100,
where к is some number. Figure 25.1 shows the graph of у =f(x).
f(x) = 0if;c<0or.x;> 100 simply means that we are restricted to a
choice of Ρ within AB.f(x) = kx if 0 < χ < 50 and f(x) = £(100-x) if
50 < χ < 100 ensures that our probability distribution is symmetrical
(reflecting a feeling that we are as likely to take χ to be between, say, 35 and
45 as we are to take χ to be between 55 and 65) and that the probability of
selecting a point Ρ within an interval of some specified length becomes
progressively greater the nearer the interval is to the centre of AB (see
Figure 25.1 again).
t For consistency we shall take R to be the range of X throughout this chapter.
521

люо
Jo
CONTINUOUS PROBABILITY DISTRIBUTIONS [25
To find к in this case, we must have
JO
f(x)dx= 1
Го люо
kxdx+\ !c(100-x)dx= 1
^ik[x^0+k[l00x-ix2]^= l
^>i.£.502 + A:.100.50—1£.150.50= 1
2500'
люо
(Figure 25.1 is not drawn to scale!) Since f(x) dx represents the area of
AAPB, integration is a heavy tool to use here—we employ it to illustrate
the general method.
Now suppose we wish to find the probability that Ρ lies between 30 and
60 mm from A. We define the event Ε by
Ε = {χ e R: 30 < χ ^ 60}
and obtain
Pr (E) = Γ 2Ao xdx + Г 25ТЙ) (100-jc) dx
J30 J50
= Ц= 0-54 (see Figure 25.2).
(Again, integration is heavy in the circumstances.)
A different mathematical model, that is, a different choice for/, leads to
a different estimate of the probability of the event Ε as we shall now show.
We shall suppose a function/defined thus:
if χ < 0
f{x) = ]fce(100-je) if 0 < χ < 100
if χ > 100

И
DENSITY FUNCTIONS
(see Figure 25.3). This distribution is again symmetrical about χ = 50
and gives a greater probability to points near the centre. To find Pr (E)
we must first calculate k.
We have
"*" кх{Ш-х)ах = 1
(Again, Figure 25.3 is not drawn to scale!) Defining Ε as before we have:
3 ,mn ч j
/•60
Jso 500000л^"" л'ил
= 6xlO-6[5Qx2-l;c3]!0°
= Λψ5- = 0-432.
Ex. 1. Why would you expect the second answer for Pr (E) to be less than the
first?
Ex. 2. Write down the value of Pr (E) on the assumption that all points of the
interval are equally likely. What form does f(x) take in this case?
Ex. 3. lif(x) = kx, 0 < χ < 100 and f(x) = 0 otherwise, find & and Pr (E).
A mode of a random variable with given density function/is a value of
the random variable for which / has a local maximum. The median, m, is
the value of the random variable defined by the equation
i-J(x]

CONTINUOUS PROBABILITY DISTRIBUTIONS
Similarly, the lower and upper quartile scores L, Q, are defined by
£j(x) dx = i, jQJ(x) dx = I
Ex. 4. Interpret the median and quartiles in terms of probabilities.
2. DISTRIBUTION FUNCTIONS
The density function, /, of a continuous probability distribution bears the
same relation to probability as mass density does to mass:mass density
is the mass per unit volume, probability density is the probability per unit
interval. It must be remembered that f(x) itself is not a probability, but
δρ = f(x) δχ is.
We now introduce a function, closely related to the density function,
whose values do represent probabilities. Suppose the domain of our
random variable is R and that we are given a density function/: R-* R. Then
weknow (i)/(*)>o,
(ii) £j(x)dx=l.
Let Ε be the set of real numbers less than, or equal to x, that is, the set of
real numbers to the left of χ on the real axis:
Ε = {у e R: у < χ}.
Then Pr (E) = ^Jlx) dx
depends upon χ and we may define a function F: R-> R where
F(x) = jX_/(x)dx=PT(E).
F is called the distribution function for the probability distribution;
F(x) is the probability that, given a probability distribution with density
f(x), a point chosen at random will be less than x.
From the original definition we have (provided/is continuous)
dF
f.
dx~
Since/(л;) > 0 for all x, F(x) increases as χ increases. Also we have
lim _F(;c) = 0, lim _F(;c) = 1,
which gives the range of F as the set {x e R: 0 < χ < 1}.
524

2] DISTRIBUTION FUNCTIONS
Again, given an event Ε = {χ e R: x1 < χ < x2} we have
Pr(E) = £7(*)dx
= F(x2)-F(Xl).
Since the sets Ε and {x eR: χ < xj are mutually exclusive, this illustrates
the result that the probability of the union of mutually exclusive events is
the sum of the probabilities of the separate events.
*Ex. 5. If Еъ E2, ..., E„ are mutually exclusive events in a continuous outcome
space, prove that
Pr^ + Pr^ + .-. + Pr^J = Pr(E1uEtu...uEJ.
Example 1. The density function, f, is defined by
10 if χ < 0
kx(l-x) if 0 ^ x< 1
0 if χ > 1.
Find
(i) the value ofk;
(ii) the distribution function F;
(iii) the density and distribution functions, g and G, of the random
variable W where ,
w = x2.
Sketch the graphs off and F.
(i) Since ГV(*) ax = 1,
we have Г kx{\ -x) ax = 1
=> k= 6.
(ii) If χ < 0, F(x) = 0; if χ > 1, F(x) = 1.
If 0 ^ χ < 1,
F{x)= J 6jc(l-*)cb:
= [3x2-2x*]0*
= *2(3-2*).
Thus F is the function F: R -> {j e Д: 0 < у ζ 1} defined by
if χ < 0
F(x)= {хг(3-2л;) if 0 < χ «ξ 1
if je > 1.
525

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
(iii) If w < 0, G(w) = 0; if w > 1, G(w) = 1.
IfO < w < 1,
G(w) = Pr {z e R: 0 < ζ < w}
= vi>(3 — 2.» (since л; is positive, x = V^)·
(0 if №<0
G(w) = J3w-2wt if 0 < w < 1
[l if w > 1.
i0 if w < 0
g(w)= 3(1-» if 0< w<l
(θ if w > 1.
The graphs of/, F are shown in Figures 25.4(i) and (ii).
(i) 0i)
Fig. 25.4
Notice in this question that, since Jx increases as χ increases, as we move
from left to right along the χ axis so we move from left to right along the
w axis. More generally, if a transformation у = h(x) of a random variable
is made, the method of Example 1 may be used directly if у either increases
or decreases steadily with χ but, if this is not the case, the domain of/
should be split into those parts for which it is an increasing function and
those for which it is decreasing. (See Example 4.)
The value of transforming from one random variable to another is
that a complicated form for the density function may be changed into a
simpler (or more familiar) density function.
F is sometimes referred to as the cumulative distribution function, to
emphasize the property it has of increasing from left to right. The graph
526
Thus
Also,
giving

2] DISTRIBUTION FUNCTIONS
of F, which usually has a shape very similar to that shown in Figure 25.4,
is called a cumulative probability curve (or, occasionally, an ogive curve).
Ex. 6. Explain how the cumulative probability curve may be used to find the
median and upper and lower quartiles.
In the case of discrete distributions, if we had a random variable X
which could take all integral values r in the range 1 ^ r ^ n, where
PT(X=r)=Pn
we defined the expectation of X by
*(X) = Д rpr.
For continuous distributions, integration takes the place of summation
and we define the expectation of the random variable X, whose density
function is fi by
S{X)=Llxf(x)dx·
${X) is usually referred to as the mean of X, and is denoted by μ.
Example 2. Find the mean of the random variable X whose density function,
f L· defined by
(0 if χ < 0
f{x) = -LjcO-jc2) // 0< *< 1
10 if χ > 1.
We have
μ = *(Χ)= J*^xf{x)dx
= Γ 4x2(l -x2) dx (since Дх) = О outside the interval
0 ^ χ < 1),
More generally, if we have a function g: R-* R, the expectation of the
function g of the random variable X is defined by
<?[g(x)]= £\(х)Лх)<1
Particularly important is the case where
g(x) = (χ-μΥ.

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
The variance of X is the expectation of the function g so defined:
σ* = £[{χ-μγ\ = Ι*\χ-μγΑχ)άχ.1
Since
/_+Γ (*-Ό'Λ*) <** = J"_+J (*-2μχ + μ*)Λχ) α*
= £J **/(*) d*-2/i £^*/W dx+μ2 J^/OO &c
= £" Χ*/[χ)άχ-2μ.μ+μ*Λ,
we have σ2+μ2= x2f(x)dx,
a result which may be compared with that on page 178.
*Ex. 7. Show that the variance of the random variable in Example 2 is given by
о-2+Йт = ΓΑχ\\-χ2)άχ.
Deduce that σ2 = J^·.
tion ah
g(x) = \χ-μ\.
*Ex. 8. The mean deviation about the mean, η, is defined as the expectation of
the function g where
Find η if
[0, x<
10, xi
Ax) = \i, -1 < χ < ι
► 1.
Exercise 25(a)
1. /(x) = &л-2 if χ > 1, Длг) = 0 if χ < 1; find the value of к and hence find
Pr (x < 2).
2. f(x) = ке~2х if χ > 0, /O) = 0 if χ < 0; find fc and hence find
Pr (1 < χ < 2).
3. If Дх) = к sin πχ if 0 < χ < 1 and is zero otherwise, find к and hence find
Pr (x > i).
4. Find the distribution function for each of the density functions in Questions
t The use of σ2 to denote variance is sometimes insufficiently explicit. If it is necessary
to refer to the random variable, X, under consideration, the associated variance may be
written σ| or, more commonly, "f{X).

Find the mean, median, mode and
function defined as in Questions 5-8.
DISTRIBUTION FUNCTIONS
for the distribution with the density
r
χ) if
if
0 if
(6x+l)/4 if
0 if
0
\π cos πχ
0
0 < χ < 2
x>2.
χ < 0
0< лг« 1
χ > 1.
if *< -i
if -*<x
if x>i.
<i
/W=W- if x>a
9. If the density function, /, of a distribution is given by
ix
i
1(3-*)
0
if 0<
if 1 <
if 2<
if χ >
find the distribution function, F, and sketch the graphs of/and F.
Find the density and distribution functions, g and G, of the new random
variable Y, where у = л:2, and sketch the graphs of g and G.
10. The density function, /, of a probability distribution is given by
if 0 < χ <. 1
: if 1 < χ < 2
if x>2.
Find the density function, g, of the new random variable, Y, where у = *Jx
Их > 0,y = Oifx <0 and sketch the graphs of/and g.
Find the mean and variance of X and the mean of Y.
11. A probability density function of a random variable Xh defined as follows:
Cx(x~l)(x-2) for 0<л:<1
f(x) = \ λ for 1 < χ < 3
lo otherwise,
where Λ is a suitable constant. Calculate the expectation μ of x. What is the
probability that χ is less than or equal to /t? (M.E.I.)
529

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
12. A continuous probability distribution has density function/, where
fix) = a+bx+cx2 for 0 < χ < 1, and f(x) = 0 outside this range.
The mean is 2/3 and the variance is 4/45. Find the values of a, b, and c.
13. The probability density function pit) of the length of life, t hours, of a certain
component is given by , ., ,n
pit) = kerkt (0 < t < oo),
where & is a positive constant. Show that the mean and standard deviation of this
distribution are each equal to l/k.
Find the probability that the life of a component will be at least t0 hours.
Given that a particular component is already tt hours old and has not failed,
show that the probability that it will last at least a further t0 hours is е-*'».
An apparatus contains three components of this type and the failure of one
may be assumed independent of the failure of the others. Find the probability
that:
(i) none will have failed at t0 hours;
(ii) exactly one will fail in the first t0 hours, another in the next t0 hours and a
third after more than 2i0 hours. (M.E.I.)
14. A random variable X has the distribution function
Fix) =0 (χ < 0),
F{x) = kx3 (0 < χ < 2),
F{x) =1 ix > 2),
where & is a constant. Find the mean, median and variance of Χ. (Μ.Ε.Ι.)
15. A random variable X has probability density given by
fix) = A/il + x*) if 0< x< 1,
and fix) = 0 otherwise.
Find A the mean and the variance of the distribution.
Find also the median and the 90th percentile.
(The 90th percentile is the value of Xior which Fix) = i%.) (O & С adapted)
3. SOME IMPORTANT CONTINUOUS
DISTRIBUTIONS
(i) The uniform (or rectangular) distribution
If a continuous random variable can assume all real values between л; = а
and χ = b and if the density function, /, is independent of position within
this interval, then the random variable is said to possess a uniform dis-
530

3] CONTINUOUS DISTRIBUTIONS
tribution. / and F are thus given, for the uniform distribution over the
interval a ^ χ ^ b, by
0 if χ < a
τ-ί- if a ^ χ υ b
b — a
0 if χ > b;
0 if χ < a
x~a -с ^ г.
-, if a 4: x < b
b — a
1 if χ > b.
*Ex. 9. Sketch the graphs of/, F as defined above.
*Ex. 10. Prove that the mean and variance of the uniform distribution defined
over the interval a < χ < b are given by
μ = Ца+Ь), σ* = &а-ЬУ.
The continuous uniform distribution arises as a direct extension of the
discrete uniform distribution. As in the discrete case, the assumption of a
uniform distribution is often not made explicit in the formulation of
problems. This is a weakness in the statement of the problem and the reader
must be on his guard. For example, such a statement as 'a point is chosen
at random on the line AB' is likely to imply, in the eyes of an examiner,
that a uniform distribution is the required mathematical model, although
there may be little physical justification for this. The dangers of a loose
enunciation of a problem are illustrated in the following very well-known
example, due to J. Bertrand.
Example 3. A chord of a circle is drawn at random. Find the probability
that its length will exceed that of the side of the inscribed equilateral triangle:
(i) if one end, A, of the chord is regarded as fixed and the other end, P,
obeys a uniform distribution on the circumference of the circle;
(ii) if the direction of the chord is regarded as fixed and the perpendicular
distance, x, of the chord from the centre obeys a uniform distribution over
the interval 0 ^ χ ^ a, where a is the radius of the circle.
(i) Let ABC be an equilateral triangle inscribed in the circle (Figure
25.5). The chord AP will exceed the length AB if Ρ lies on the arc ВС and
the probability of this occurring, on the assumption that Ρ is distributed
uniformly on the circumference of the circle, is j.
(ii) Let AB, CD be the two chords in the given direction which are
distant \a from the centre of the circle (and thus are of length а^Ъ, the side
of the inscribed equilateral triangle). If x is the distance of a random chord
531
Ax) =

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
from О in the given direction, and if χ is uniformly distributed in the
interval — a =ξ χ =ξ a, the chord will exceed а^Ъ in length (e.g. PQ, in Figure
25.6) if — £a =ξ χ =ξ \α giving a probability of \.
Ex. 11. If, in Example 3, the mid-point of the chord is taken to have a uniform
distribution over the interior of the circle, prove that the probability that the
chord will exceed the length of the side of the inscribed equilateral triangle is \.
Fig. 25.5 Fig. 25.6
Example 4. A point Ρ is taken on a line AB, of length la, the distribution
being uniform. A rectangle is then drawn with adjacent sides of lengths AP,
PB. Find:
(i) the expected value of the area of this rectangle; (ii) the distribution
function, G,for the random variable, Y, whose value represents this area.
Let AP = χ and the area of the corresponding rectangle = y. Then
у = x(2a-x).
Now, if/is the density function of X, the random variable whose value for
given Ρ is the length AP,
10 if χ < 0
l/2a if 0 ^ χ ^ 2a
0 if χ > la,
and we have
C2a 1
£(Υ) = γ x(2a - χ) άχ, by the definition of expectation,
As χ varies from 0 to а, у varies from 0 to its maximum value, a2, and,
532

3] CONTINUOUS DISTRIBUTIONS
as χ varies from a to 2a, у varies from a2 to 0. If F is the distribution
function of X,
(0 if x<0
F(x) = \x/2a if 0 ^ χ ζ la
(l if χ > 2a
and, by the symmetry of the curve у = x(2a—x) about the line χ = a
(see Figure 25.7), we have, for any point (x, y) on the curve,
Pr (z e R: ζ < у) = Pr (ζ e R: ζ < x) + Pr (z e Д: ζ > 2a-χ)
= F(x)+[l-F(2a-x)]
χ 2a—χ
2a
2a
But, since
we have
giving
Thus we have
у = 2ax—x2,
χ = a-J{a2-y) (x < a),
G{y) = Pi(zeR:z < y)
_a-J(a2-y)
a
(0 if у < 0
1-7(1-$) if 0^<Я
1 if у > а2.
The graphs of F and G are shown in Figure 25.8.

CONTINUOUS PROBABILITY DISTRIBUTIONS
Example 5. Χ, Υ are both uniformly distributed between 0 and 1. Values
x, у are chosen independently at random and a new random variable, Z,
is formed where
z = x+y.
Show that Ζ has a triangular distribution {that is, a distribution with density
function whose graph is triangular).
Like most problems concerned with independent uniformly distributed
random variables, it is best to consider this question graphically. Since
Χ, Υ are distributed independently, the choice of two numbers, χ
and y, may be represented by plotting the point R(x, y), all points
of the square О ABC (see Figure 25.9) being equally likely. If ζ = x+y,
A\ .β
m ρ с *
Fig. 25.9
for given ζ all possible points R lie on the line segment PQ, where Ρ is the
point (z, 0). For 0 < ζ < 1, PQ ос ζ; for 1 < ζ < 2, PQ cc (2-z) and
thus the density function, /, of Z, whose value for given ζ is proportional
to z, is triangular. In fact
Ό
if
if
if
if
ζ < 0
0 ^ ζ < 1
1 < ζ < 2
ζ > 2.

3] CONTINUOUS DISTRIBUTIONS
Ex. 12. Find the distribution function, F, for the random variable Ζ of Example 5
and sketch its graph.
Ex. 13. If, with the notation of Example 5, the random variable И^ is defined by
w = x—y, describe the distribution of W.
(ii) The normal distribution
Suppose that shots are fired at the centre, O, of a large circular target. Set
up coordinate axes as shown in Figure 25.10 and let P(x, y) be the point
at which one of the shots hits the target. If we suppose that shots aimed at
the centre are liable to errors, equally likely to occur above or below Ox
and to the left or right of Oy, we may reasonably assume that the
probability of x lying between χ and x + Sx, the 'x error', depends upon the
Fig. 25.10
numerical value of x; that is, the probability is of the form φ(χ2) δχ. On the
assumption of the symmetry of the errors made above, the probability of
the 'y error' is ф(у2) Sy and the probability of lying in a small area
δ A = SxSy
around Ρ is ф(х2) ф(у2) δ A, on assuming that the errors occur independently.
But on these assumptions the probability may also be written
Ф(х'2) Ф(У'2) δ A, (approx.)
where new axes Ox', Oy' are taken with Ox' passing through P, as shown
in Figure 25.10. Thus, since x'2 = x2+y2 and y' = 0 at the point P,
ф(х2) ф(у*> = ф(х'2) ф(у'2) = ф(х2+у2) ф(0)
and φ is a function with values satisfying the equation
ф(х2)ф(у2) = кф(х2+у2).
6-2 535

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
A solution of this equation is φ(χ2) = ks~Xx2 where A is taken as a
positive constant, since it is reasonable to assume that the probability decreases
the further we move from O.
The argument above (due to Thompson and Tait) suggests that the
function Дл;) = ks~Xxl could form a suitable model for the probability density
of errors occurring in observations. This indeed turns out to be the case
and we shall therefore make the following definition.
The random variable X is said to have a normal distribution if its density
function, ф(х), is of the form
The letter φ is customarily used in place of/to denote the density function
of the normal distribution; the corresponding distribution function is
The factor (2π)-% appearing in the forms for φ and Φ ensures that the
requirement r+00
)_βαφ(χ)άχ=1
is met—that is, that φ(χ) is a genuine density function. This follows from
the well-known integral
j_ooe-^d^ = V(2^)>
a result whose proof will be found in any sufficiently advanced book on
calculus. (The reader should not be too discouraged by the remark,
attributed to Sir William Thompson, about this integral: 'No-one can call
himself a mathematician to whom this result is not obvious!')
The shape of the curve of the normal density function is shown in Figure
25.11. As is to be expected, the curve is symmetrical about χ = 0, which
clearly gives the mean of the distribution:
<?(X) = 0.
Furthermore, the variance of a normally distributed random variable
X is l.for
= ,,- , е-**2 dx, on integrating by parts,
= 1, on quoting the integral result above.

3] CONTINUOUS DISTRIBUTIONS
Statistical tables (e.g. The Cambridge Elementary Statistical Tables by
Lindley & Miller) give values of Ф(л;) for values of χ from 0 to about 4.
(Since Φ(4) « 0-99997 values of Ф(л;) for χ > 4 are very rarely required.)
By the symmetry of the curve
about the line χ = 0, values of Φ(χ) for χ < 0 may be deduced immediately.
Fig. 25.11
The normal distribution is frequently used as a probability model.
Suppose that the random variable X has mean μ and standard deviation
σ; then the new random variable Υ defined by
has mean 0 and standard deviation 1. If Υ is normally distributed it is
customary to say that ' X is normally distributed about the mean μ with
standard deviation σ\ sometimes written as ' X is distributed Ν(μ, σ)', and
that' X has been standardized to the normal random variable Υ by the given
substitution'. For emphasis, Υ is often referred to as ' the standard normal
distribution JV(0, 1)'.
Example 6. If X is normally distributed N(0, I) find
(i) Pr(* < 1-4); (ii) Pr(0-8 < χ < 1-4);
(iii) Pr (* < -1-4); (iv) Pr (-1 ·4 < χ < 0-8).
// Υ is normally distributed N(2, 0-75), find
(v) Pr(j < 0); (vi) Pr (1 < у < 3).
537

CONTINUOUS PROBABILITY DISTRIBUTIONS
From statistical tables,
(i) Pr(* < 1-4) = Φ(1·4)
« 0-9192
(Figure 25.12).
(ii) Pr(0-8 < χ < 1-4) = Φ(1 ·4) - Φ(0·8)
и 0-9192-0-7881
= 0-1311
(Figure 25.13).
Fig. 25.14 Fig. 25.15
(iii) Pr (x < -1-4) = Pr (x > 1-4) by symmetry,
= 1-Φ(1·4)
« 1-0-9192
(Figure 25.14).
(iv) Pr (-1-4 < χ < 0-8) = Pr (* < 0-8)-Pr (* < -1-4)
= Φ(0·8)-[1-0(1-4)]
«0-7881-0-0808
~. oc, „ = 0-7073
(Figure 25.15).
538

3] CONTINUOUS DISTRIBUTIONS
(ν) The random variable Z, where
z=^
0-75
has standardized normal distribution. Thus
Pr(y< 0) = Pr(z < -2-67)
= 1-Φ(2·67)
« 1-0-9962
= 0-0038.
(vi) Again,
Pr(l < у < 3) = Pr(-l-33 < ζ < 1-33)
= 2Φ(1·33)-1
« 1-8164-1
= 0-8164.
Example 7. Г/ге heights of a large number ofschoolchildrenare measuredcorrect
to the nearest centimetre and the mean and standard deviation of the resulting
frequency distribution are calculated and found to have the values 122 cm
and 5-2 cm respectively. As a model of the situation it is assumed that the
heights, x, of the children are distributed normally about a mean μ = 122
with standard deviation σ = 5-2. Calculate the probabilities for each
of the class intervals χ < 105, 105 < χ ^ 110, 110 < χ ^ 115, ...,
130 < χ < 135, χ > 135.
Since the heights are measured to the nearest centimetre, the upper
limits for χ for the class-intervals above are successively
105-5, 110-5, 115-5, .... 135-5
(there is no upper limit, of course, for the last class).
We now standardize our variable, x, by the transformation
'- 5-2 '
so that у has the standard normal distribution (/* = 0, σ =1) and using
statistical tables, we find Ф(у) for the upper limit of each class-interval.
Finally, the probability associated with each class interval may be
calculated , = Ф^-Ф^).
539

CONTINUOUS PROBABILITY DISTRIBUTIONS
The working is set out in tabular form below:
Class χ у (у)
130-135
125-130
120-125
115-120
110-115
105-110
< 105
135-5
1305
125-5
120-5
115 5
110-5
105-5
10000
0-9953
0-9495
0-7517
0-3859
0-1038
0-0132
0-0007
00047
0-0458
0-1978
0-3658
0-2821
00906
0-0125
00007
1-0000
It should be noticed that a continuous distribution has been used here
as a mathematical model for a discrete situation (heights measured to the
nearest centimetre). Again, it might be objected that, whereas heights could
reasonably be expected to fall, say, within the interval 90-180 cm, the
normal distribution is defined for all real values of x; however, as has been
pointed out earlier, for the standard normal distribution,
Pr (\y\ > 4) * 0-00006
and the two tails of the distribution, beyond у = + 4, may be neglected.
As well as appearing as a distribution in its own right, the normal
distribution is often used as an approximation to the binomial distribution
B(n, p) for large n, provided neither ρ nor 1 — ρ is too near zero. In Figure
25.16, the probabilities of obtaining the scores 0,1,2, ...,20 in the

3]
CONTINUOUS DISTRIBUTIONS
binomial distribution 5(20, \) are proportional to the lengths of the
corresponding vertical lines. The outline shape is strongly reminiscent of the
normal curve. Even in a skew case, where ρ Φ \, the outline is still
approximately normal: see Figure 25.17, in which the length of the vertical lines are
proportional to the probabilities of obtaining the scores 0, 1, 2,..., 20 in
the binomial distribution 5(20, f).
In fact, it may be shown that, given a binomial distribution B(n,p)
where η is large and neither ρ nor 1 — ρ is too near zero, then the
probability of obtaining a score r, that is,
lftpV-РУ"
is approximately equal to σ_1 times the ordinate of the standard normal
curve at the point
where μ, σ are the mean and standard deviation of the corresponding
binomial distribution
μ = Пр, σ = *]{npq).
Furthermore (and more importantly for applications) we can relate
areas under the normal curve to sums of successive terms of the given
541

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
binomial distribution; in fact, the probability that the variable takes a
value in the interval
/■i ^ r ζ r2
is approximately
■т-»ж
where, as before, μ = ηρ, σ = J(npq). (The limits rb r2 are changed to
>Ί— i »2+i respectively in order to avoid complications arising out of
approximating to a discrete distribution by a continuous distribution;
for example, we wish to associate a non-zero probability with a single value,
r, of the variable.)
The reader is referred to one of the standard texts on probability for a
proof of the results outlined above. As a working rule, the approximations
give reasonable results if the lesser of the two numbers, the mean number of
successes and the mean number of failures is greater than about five.
Example 8. A coin is spun 250 times and turns up tails 139 times. Does this
provide any evidence of bias?
Take as the null hypothesis
We have to assess the probability of getting a result as bad as, or worse
than, 139 tails on the assumption that Я is true; that is, we have to deter-
mme Pr (139 or more tails |Я) + Рг (139 or more heads \H).
The calculation of this from the binomial distribution 5(250, £) would
be prohibitively laborious; we therefore use the normal approximation
to the binomial distribution as our model where
/* = 250х£= 125,
σ2 = 125xi= 62-5,
giving σ = 7-906.
We seek 2 Pr (je > 139),
where χ is distributed 5(250, £), and this is approximately
'['-•Н1Л]
*2[1-Φ(1-71)]
« 0-087.
There is thus insufficient evidence of bias at the 5 % level.
542

3] CONTINUOUS DISTRIBUTIONS
Note. Tables of the normal distribution show us that
Pr (b| S; 1-96) «0-05
and thus a normal variable greater than 1 -96 indicates significance at the
5% level. In Figure 25.18 the two shaded areas each contain 2-5% of the
total area under the curve.
One further application of the normal distribution should be mentioned:
if хъ x2, ...,xn are independent observations of any random variable X
and a new random variable Ζ is defined by
ζ = (x1+x2+... + xn)ln
then, if the mean and variance of Zare respectively μ and σ2, the distribution
of Ζ will be approximately normal, with mean μ and variance σ2/η. (The
Central Limit Theorem.) The proof of this result is beyond the scope of this
book.
(iii) The exponential distribution
The exponential distribution is defined by the density function
(0 if χ < 0,
/(Нле- if x>0,
where λ is a positive constant.
*Ex. 14. Show that, for the exponential distribution,
SiX) = l/λ, -Г(Х) = 1/λ2.
The result of Ex. 14 suggests a resemblance between the exponential
and Poisson distributions. We shall exhibit the relation between the two
distributions in the following problem. (See Exs. 16-18.)
543

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
Consider the calls received at a telephone exchange, the rth call occurring
at time tT, where the time, t, is measured from t = 0. We shall assume that
the number of calls constitutes a purely random process; that is, the number
of calls received in any interval (t, t') is independent of anything that has
occurred previously. Furthermore, we shall assume that the purely random
process is a stationary process: that is, the number of calls received in the
interval (t, t') depends only upon its length, not on its position.
Ex. 15. Discuss the validity of the assumptions made above in the light of what
you imagine would be a typical exchange.
Suppose now that
Pr {no calls are received in the interval (0, t)) = pJit),
and let us assume thatp0 is a differentiable function. Since, by the definition
of a purely random process, the probability of receiving no calls in the
interval (t, t+St) is independent of receiving no calls in the interval (0, t),
p0(t+St)=p0(t)p0(St).
Put t = St = 0: ,m r .....
and thus, since p^O) Φ 0, we must have^0(0) = 1. Again
p<lt+St) = p<lt)pa(St)
=> Po(t+$t)-Po(t)=Po(t)[p0(St)-l]
=> p0{i + St) -pJit) = ρ Jit) [p0(St)-p0(0)]
p,(t + St)-PJjt) _ p0(St)-p0(0)
Έ ~p<kt) st ·
pO{t)=pS)p'SH)
=> p0(t) = At-*, (1)
where λ is a constant (-p'0(0)) which must be positive since p0{t) < 1 and
A = 1 since;? o(0) = 1.
Result (1) now enables us to prove that if ΧΊ& the random variable whose
value is the length of the time interval up to the first call, then X has an
exponential distribution. If F is the distribution for X then, for χ ^ 0,
F(x) = Pr (length of the time interval up to the first call ^ x)
= Pr {at least one call has been made by the time χ has elapsed)
= 1-е-л*
and thus f{x) = Ae~Ax for χ > 0.
Since/(л;) = 0 for χ < 0 it follows that X has an exponential distribution.
LeUi^O:

3] CONTINUOUS DISTRIBUTIONS
Ex. 16. If pn(t) is the probability that there are η telephone calls in the time
interval (0, t) and if St is sufficiently small for there to be a negligible probability
of more than one call in the interval (i, t+St) show that
pn(t+St) = (1-λ&)Α,(0 + λΡ»-ι(0&
and deduce that —- = λ(ρη^—pn).
Ex. 17. Prove by induction that the probability, pn(t), defined in Ex. 16, is
8"еПЬу м e-^tr
Ex. 18. Show that the distribution of the number, n, of calls received during a
fixed time interval is Poisson and account for this in terms of the Poisson
distribution being a limiting form of the binomial distribution.
Exercise 25(b)
1. The random variable, X, has uniform distribution in the interval 0 < χ < 10.
Find Pr (X = x, where хг - 5x + 4 > 0).
2. The line AB has length 10 cm. A point Ρ is taken at random on AB, all points
being equally likely; what is the probability that the area of the circle of radius
AP will exceed 10 cm2?
3. The line AB has length 10 cm. An interval of length 2 cm is marked at
random on the line, the positions of the interval being uniformly distributed. What
is the probability that it will contain the mid-point of ΑΒΊ
4. A circular disc of radius 10 cm is placed on a table. Another disc, of radius
3 cm, is now placed on the table so that it is at least partially in contact with the
first disc. On the assumption that the permissible positions of the smaller disc
are uniformly distributed, what is the probability that it covers the centre of the
larger disc?
5. A point Ρ is taken at random on the side AB (and between A and B) of the
square ABCD, the positions of Ρ being uniformly distributed. If PC cuts BD at
X, find the probability (i) that BX < \BD; (ii) that BX < \BD.
6. Figure 25.19 shows a square wooden frame
ABCD with a square hole A'B'C'D' cut
symmetrically in it. AB = 50 cm, A'B' = 30 cm.
A ball of diameter 5 cm is dropped on to the
frame. Assuming that its centre falls within
the square ABCD and that it is equally likely
to meet the plane of the frame at any point
within ABCD, what is the probability that the
ball will pass straight through the hole without
touching the frame ?
7. A point A is marked on the circumference of
a circle of radius r and a chord AP is drawn at Fig. 25.19
random, the positions of the point Ρ on the
circumference being uniformly distributed. Find the expected length of the chord.
545

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
8. Determine the variance of the length AP of Question 7.
9. A point Ρ is marked on the side AB of the square ABCD, the points within
AB being uniformly distributed. Find the mean and variance of the area of the
triangle APD.
10. ABC is an isosceles triangle right-angled at B. Through A a line is drawn
at random to cut ВС at P, the angle BAP being uniformly distributed between 0
and \π. If AB = a, show that the expected area of the triangle ABP is (In 2) α2/π
and find its variance.
11. The intelligence quotients of 500 schoolchildren are assumed to be normally
distributed with mean 105 and standard deviation 12. How many children may
be expected:
(i) to have an intelligence quotient greater than 140;
(ii) to have an intelligence quotient less than 90;
(iii) to have an intelligence quotient between 100 and 110?
12. If an unbiased die is thrown 600 times, what is the probability of throwing
a six less than 90 times ?
13. X is normally distributed with mean 2 and standard deviation \. Find
numbers p, q such that Pr (x > p) = 0-2 and Pr (j? < χ < q) = 0-1.
14. A multiple choice test has 100 questions, each question having written
beside it five answers, only one of which is correct. If the pass mark is 30%,
what is the probability that a student who makes a completely random guess at
each answer will pass ?
15. Rods are manufactured with a mean length of 20-2 cm and standard
deviation 0-09 cm, the distribution being normal. If rods of length less than 20-1 cm
are rejected, what is the probability that a rod that is retained has a length in
excess of 20-3 cm?
16. Experience has shown that when a certain machine is functioning
satisfactorily it produces capacitors with capacitances which are distributed normally
with standard deviation 0-080 /tF. Under these circumstances find the mean
capacitance if 99% of the output has a capacitance of at least 2 /tF. Find also
what proportion of the output will then have a capacitance between 2-10 and
2-30 /tF.
Tests on a large batch of capacitors just produced reveal that their mean
capacitance is 2-20 /tF and that 10 % of them have capacitance below 2 /tF.
What do you deduce about the variability of the capacitance of the capacitors
in this batch? (M.E.I.)
17. The average proportion ρ of insects killed by administration of χ units of
insecticide is given by Пх-ма
p= (2 л)-1/2 е-'2'2 di,
where μ and σ are constants.
Whenx= \0,p= 0-400 and when χ = \5,p = 0-900. What dose will be lethal
to 50% of the insect population, on average?
546

EXERCISE 25
If a dose of 17-5 units is administered to each of 100 insects, how many will
be expected to die? What is the probability that just two insects will survive?
(M.E.I.)
18. Packets are advertised as containing 500 g of sugar. Tests carried out show
that 6-7 % of such packets contain a mass greater than 508 g, while 0-6 % have
a mass less than 492 g. Estimate the average mass of the contents of a packet
on the assumption that a normal distribution constitutes an acceptable
mathematical model.
19. A firm advertises that their runner bean seeds give a 95 % germination rate.
Of 200 such seeds, 14 fail to germinate. Have you cause for complaint ?
20. A man claims that he can forecast rainy or dry weather 48 hours ahead.
Careful records are kept over 80 days and his forecast is found to be correct
47 times. Is his claim justified, or is he merely lucky ?
21. A 'chance of failure' distribution is given for time t by the probability
density function , . ,., . ,, .„
pit) = (1/a) e-"a (0 < / < со).
Show that a is the mean time of failure and that the variance is a2.
Two components in a machine have failure time distributions corresponding to
means a and 2a respectively. The machine will stop if either component fails and
the failures of the two components are independent. Show that the chance of the
machine continuing to operate for a time a from the start is e~3'2. (O & Q
Miscellaneous Exercise 25
1. AB is a rod of length 2a. A point Ρ is taken on AB, the points being uniformly
distributed, and the stick is broken at P. Find the expected value of AP2+PB2.
The two parts of the rod are placed on a table in such a way that A and В
are Ол/3 apart. What is the probability that the triangle APB will be obtuse?
2. The mean deviation about μ for a distribution with density function / is
defined by the equation . m
V = J_ \χ-μ\/(χ)άχ.
Prove that the mean deviation of a random variable distributed normally about
μ with standard deviation σ is approximately |σ.
Rods are manufactured with a mean length of 18-4 cm and standard deviation
0-25 cm. If only those rods with a length greater than 18-4 cm are retained, what
is their average length on the assumption that the rods were originally normally
distributed about 18-4 cm with standard deviation 0-25 cm?
3. If Χ, Υ are independently and uniformly distributed between 0 and 2, find
Pr (xy < 1).
4. The side of a square is uniformly distributed between 0 and 1. Find the
density and distribution functions for the area of the square and sketch their graphs.
Determine the expected value of the area and its variance.

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
5. The probability density function for the life л: of a motor car tyre is given by
Ax) = Ле~л*
for χ =s 0, where λ = 0-04 and χ is measured in units of 1500 km. What is the
probability that a single tyre will last more than 30 units?
A car has four tyres in use and they are all of the same age. What is the
probability that all of them will need replacing before 30 000 km? (M.E.I.)
6. A random variable X has the probability density function
nexp{-nxl^3)}
ПХ} ^3{1+ехр[-™/(^3)]}2
for all real x. Show that the distribution is symmetrical about χ = 0. Determine
the cumulative distribution F(x) = Pr {X < x].
You are given that the variance of X\s σ2. Compare the value of F(cr), Щ&),
Ρ(3σ) with the corresponding values for the normal distribution with zero mean
and variance σ2. (Μ.Ε.Ι.)
7. A random variable Xis distributed normally with expectation μ and standard
deviation σ. Find the mode and median of the distribution.
What is the proportion of the population lying between the points of inflexion
of the curve of the probability density function? (M.E.I.)
8. The diameters of some machined components are distributed normally with
mean 5-00 cm and standard deviation 0-05 cm. Find the expected proportion
of the components which will be outside the range 4-925 to 5-075 cm and the ratio
of the expected proportion in the range 5-025 < d < 5-050 to the expected
proportion in the range 5-050 < d «Ξ 5-075.
It is desired to adjust the mean of the process so that there are, on average,
twice as many components in the range 5-025-5-050 cm as in the range 5-050-
5-075 cm. Show that this can be done if the mean is adjusted to a value between
5-00 cm and 4-95 cm, and find the value by trial to the nearest 0-01 cm.
(M.E.I.)
9. The probability density function of a distribution is given by
Ях) = Ту—ттт (x> 0,λ an integer > 0).
(Λ-1)!
Find the expectation and variance of X. И^(Х) = μ, find ${(X— μ)3}.
Sketch fix) when λ = 2. (M.E.I.)
10. A grocer sells bread and can buy batches of 120 loaves. The number of daily
customers for bread is distributed normally with mean 100 and variance 100.
The net profit on the sale of a loaf is 2-5p and the net loss on an unsold loaf is
3-5p. What is the average daily net profit to the grocer? (M.E.I.)
11. The random variable X has normal distribution. If a new random variable
Υ is denned by у = x2
show that the distribution function, F, for Υ is given by
548

MISCELLANEOUS EXERCISE 25
Deduce that Υ has α χ2 distribution with one degree of freedom, defined by the
density function /, where
12. Two chords are drawn independently at random in a circle. What is the
probability that they intersect?
13. Two men, A and B, are allowed a lunch-break of an hour. A is at liberty
to leave the office at any time between 12.00 and 12.45, while В may leave the
office between 12.45 and 2.00. What is the probability that they will both be out
together, on the assumption that the permissible starting times for both A and
В are uniformly distributed?
14. A needle is pivoted at the point (0, a) and is rotated. When it comes to rest,
the point Ρ at which its axis (produced if necessary) cuts the x axis is marked.
The random variable X is defined to have as its value the x coordinate of P.
On the assumption that the angle that the needle makes with the у axis is
distributed uniformly between ± \π, prove that X has a Cauchy distribution,
defined by the density function
л*)=й^Ь) <-»<*<«»·
Discuss the existence of S(X) and Ψ~(Χ) for this distribution.
Sketch the graph of the density function for a = 1.
15. The Laplace distribution has probability density function
fix) = Се-Л'*' (-oo < χ < сю)
where С is a constant and \x\ denotes the magnitude of x. Find Pr {|χ| < λ}
and the variance of X. (M.E.I.)
16. A famous early example of the 'Monte Carlo' method for solving problems
by random numbers was Buffon's determination of π. Small rods of length Lt
were dropped at random on to a sheet of paper ruled with thin parallel lines a
distance L2 (> LJ apart. A' success' is recorded when any part of the rod touches
a line. The ratio of 'successes' to the total number of tosses is recorded over
several thousand tosses. How may π be estimated from this ratio? (C.S.)
17. Let X be a random variable uniformly (rectangularly) distributed over the
interval 0 < χ < 1. Derive the probability density function of the following
random variables: (a) Υ = X2-l,(b)Z= sin πΧ. Find the mean and standard
deviation of Υ and Z. (C.S.)
18. The chance that the customer at the head of a queue completes service in
any interval of length St is μδί, and the chance that a new customer arrives in any
interval is ASt, arrivals and departures are independent and λ < μ. The chance
that at time t there are η customers in the queue (including the one being served)
is denoted by pn(t). Show that
pn(t) = λ8ίΡη_1(ί-δί) + (1-λδί-μδί)ρη(ί-δί) + μδίρη+1(ί-δί) (η > 1)
and obtain the corresponding equation for p0(t). (over)
549

CONTINUOUS PROBABILITY DISTRIBUTIONS [25
By considering the forms that these equations take under the steady-state
condition pk(t) = pk(t— St), where к is zero or any positive integer, or by any other
means, obtain pn and the mean queue size (in terms if λ and μ) in the steady
state. (C.S.)
19. X is a continuous random variable with mean μ and variance σ2. λ is a
positive constant. Prove Chebyshev's inequality
Ψΐ(\Χ-μ\ > λσ) < 1/λ2.
Show how this inequality may be sharpened, in the case λ = 2, if the
distribution of X is assumed to be (i) normal, (ii) uniform.
20. Spacecraft land on a spherical planet of centre O. Each is able to transmit
messages to, and receive messages from, any spacecraft on the half of the surface
of the planet nearest to it.
(i) It is known that spacecraft have landed at points A and В on the surface
of the planet. Show that the probability that a space craft, landing at random on
the planet, will be able to communicate directly with the spacecraft at A and В is
(π-0)/2π
where θ is the angle AOB.
(ii) What is the probability that three spacecraft, all landing at random on the
planet, will be in direct contact with each other? (C.S.)
21. Engine crankshafts are manufactured so that the diameters, in centimetres,
form a normal distribution with mean 5 and standard deviation 0-03.
Crankshafts with diameters less than 4-94 or greater than 5-06 are rejected. The
accepted product is classified into three grades of size 4-940-4-988, 4-988-5-012,
5-012-5-060.
Show that:
(i) fxe-*a/2dx = -e-*!'2,
(ii) Γ x2 e-^'2 dx = - χ е-*2'2 + Г е-*2'2 dx.
Hence find the average diameter in each of the three grades and the ratio of the
standard deviation in the middle grade to the standard deviation of the
unclassified product before any rejection of under- and oversize shafts. (M.E.I.)
22. The police force in a certain district carries out tests on the brakes of
automobiles chosen at random on the road. Each man is required to test 20 cars.
Calculate the distribution of the number of cars with defective brakes in sets of
20 cars if the probability that a single car has defective brakes is 10 %.
Show the distribution graphically together with a plot of the normal
distribution with the same mean and variance. Comment on the relation between the
two distributions. (M.E.I.)
23. A population contains two strata I and II in proportions }, f respectively.
Both strata are exponentially distributed the first with mean 1 and probability
density function ДлтО = λιβ-Λ'^ (χ1>0);
the second with mean 3 and probability density function
Αχύ = Л2е~л^ (х2 > 0).
550

MISCELLANEOUS EXERCISE 25
Find the mean and standard deviation of the population and the probability
that a randomly chosen member of the population is greater than 2. (M.E.I.)
24. Initially a machine is in good running order but is subsequently liable to
breakdown. As soon as a breakdown occurs repairs begin. If the machine is in
good order at time t then the probability that a breakdown occurs in a small
interval it, t + d t) is α di, and if it is under repair at time t the probability that the
repair is completed in time(i, ί + di) is/?di. Let pit) be the probability that the
machine is under repair at time t.
Write down an equation relating pit + di) to pit) and hence show that pit) is
^{l-exp[-(a+/?)i]}. (C.S.)
551

26. Numerical solution of equations
1. ACCURACY
When equations arise in practical problems, their solutions are generally
required only to some given order of accuracy. General algebraic methods
of solution (for example, for the cubic equation), even if they are available,
are often less suitable than approximate methods for deriving numerical
solutions. Thus, to find the real root of the equation
х3-2-7л;-5-3 = 0
by Cardan's method would require burdensome calculations and no such
method exists to solve, for example, the equation
χ6-2·7λ;-5·3 = 0.
Because of its great practical importance, the estimation of roots of
numerical equations has been extensively studied; in this chapter we shall
confine ourselves to some of the most elementary methods available: the
reader who wishes to study the subject further should consult one of the
standard texts on numerical methods (for example, Henrici, Elements of
Numerical Analysis). It may be mentioned in passing that this subject allows
more scope than most others in elementary mathematics for students to
devise their own methods; although these may often prove less efficient
than the standard techniques, much profit will be gained by developing
them as far as possible—apart from the obvious enjoyment of producing
something original.
Before considering the solution of equations it is worth pausing to
consider some problems connected with the accuracy of calculations. In
numerical calculations, errors occur through rounding-offto a given number
of significant figures. If a number is to be rounded-off to N digits and the
discarded digits form exactly half a unit, round-off to the nearest even digit.
For example, to 3 significant figures,
3-864 к 3-86; 21990 m 2-20 χ 104;
0-03765 at 3-76 χ 10~2; 21-55 в 2-16 χ 10.
*Εχ. 1. Suggest any advantage you see in adopting the convention above. Can
you suggest any reason why even rather than odd digits have been selected ?
552

1] ACCURACY
Once numerical data have been rounded-off, their subsequent use in
calculations introduces further errors which will be cumulative. It would
be pleasant to be able to calculate exactly the error in any numerical answer.
This is usually impossible, but we may be able to produce a positive
'acceptance' for the answer; that is, a quantity which we are sure is more
than the absolute error in the answer. For example, suppose we round-off
a given angle Θ, measured in degrees, to 3 significant figures to give 37-2°.
The actual angle lies between 37-2°-0-05° and 37-2° +0-05°. From four-
figure tables, tan 37-2° = 0-7590 and the difference for +0-05° is
1(0-7618 -0-7563) к 0-0014.
Thus tan0 = 0-7590 + 0-0014;
in other words, tan θ is 0-7590 with an acceptance 0-0014.
*Ex. 2. If possible inaccuracies in the tangent tables are taken into account, show
that tan θ = 0-7590 ± 0-0015.
Now suppose у is a rational approximation to the number x; thus
x = y+e,
where e is the error involved in writing у for x. The absolute error is
|e| while the relative error is \e\jy (more correctly \e\jx, but the two forms
are nearly equal for small relative errors and the form \e\jy has the
advantage of possessing a known denominator).
Usually we do not know the exact error, only an upper limit to the
absolute error—the maximum absolute error. For example, if a number is
rounded to 4 decimal places, the maximum absolute error is 5 χ Ю-5.
Similarly, we have a maximum relative error. In a calculation in which we
know the maximum absolute errors in the given numbers it is important
to be able to estimate the maximum absolute error in the answer.
Suppose that x1 is written as yx with maximum absolute error \ex\
(we use modulus signs to emphasize that maximum absolute errors are
positive) and that x2 is written as y2 with maximum absolute error |e2|.
Then the maximum absolute error in writing yx+y2 for x1+ x2 is \ex\ +\e2\,
while the maximum relative error is
(Ν + Ν)/0ί+λ).
Ex. 3. Show that the maximum absolute and relative errors in writing у^-Уг for
xi-x2 are respectively |et| + |e2| and fle^ + \ег\)/(у1-у2).
*Ex. 4. Neglecting terms such as |e1e2|, show that the maximum absolute and
relative errors in writing у^уг for хгхг are respectively
ΐΛβ,Ι + ΐΛ^Ι and Μκ\ + Μγ%\.
Show also that, writing у^уг for xtlx2 they are respectively
ЫУг\ + МА\ and \еЫ + \чЫ-
553

NUMERICAL SOLUTION OF EQUATIONS [26
Ex. 5. If you have a calculating machine available, calculate:
(i) Х1 + Х2,У1 + Уг\ (ii)*i-*a, Уу-Уг, (Ш) ХуХг, У\Уг,
(iv) x-ijx^, Уг1уг (6 significant figures) where
Xl = 2-914, x2 = 0-3472, yi = 2-9, y2 = 0-35.
Ex. 6. What are the maximum absolute and relative errors in writing
(i) y\ for xl; (ii) Vtt for V*i5 ("О VCfi+^a) f°r Vfe+^2)?
Before leaving the subject of errors, it is worth noting one particular
case in which significant figures may be lost. Consider 3-144-3-097: both
of these numbers are given to 4 significant figures, but their difference,
0-047, is correct to only 2 significant figures. It is sometimes possible to
modify a calculation in such a way to minimize this loss of accuracy.
For example, if it is required to calculate the difference between the
values of a function of χ for two given values of χ that are close together,
it may help to use a Taylor expansion. Thus, using four figure tables,
sin3r-sin30° = 0-0150,
but, by Taylor's expansion,
sin 31° = sin 30°+(я/180) cos 30°-KW180)2 sin 30°...
= 0-5000+0-01512-0-00076
and sin ЗГ-sin 30° к 0-01504; again using four figure tables and
obtaining a result correct to 4 significant figures.
Again, suppose we have to solve the quadratic equation
х*-\6х-\ = 0.
16 + V260 . 16-V260
The roots are ^ and ~ .
Direct evaluation of these two quantities gives
16-06 and -0-06
(using four figure tables). However, in obtaining the negative root, we
lost significant figures by subtraction. A better approximation may be had
by calculating the negative root using the more accurate positive root 16-06
and the known relation that the product of the roots is — 1:
««-las*- °·0627·
Before leaving the subject of significant figures, it is worth noticing that
a calculation can be carried out to a high degree of accuracy and yet not
furnish a correct result even when corrected to 1 significant figure. For
554

1] ACCURACY
example, if the correct answer is 0-649 999 9 and our numerical answer is
0-6500001 then, to 1 significant figure our answer is 0-7 but the correct
answer is 0-6 although our calculation was only 0-0000002 in error.
Example 1. Given that J31 =5-568 and J 30= 5-477, find J 31-J 30.
Direct computation gives ^31—^30 = 0-091; but using
(V31 - V30) (V31 + V30)=ι
we have ^31-^30= 1/(5-568+ 5-477)
= 0-090539 (from 5 figure reciprocal tables)
and this is in fact correct to 5 s.F.
2. LOCATING THE ROOTS OF THE
EQUATION f(x) = 0
Before embarking upon the accurate estimation of a root of the equation
f(x) = 0 it is usually necessary first to locate the root roughly.
One method of doing this is to draw the graph of у = f(x) and find
where it cuts the χ axis. Indeed, it is possible, by accurately plotting a
succession of graphs of the relevant regions of f(x) on increasing scales,
to obtain quite good approximations to roots of given equations. When
using a graphical technique it is sometimes convenient to write the equation
in a form other than f(x) = 0, e.g. g(x) = x. See, for example, Ex. 7.
*Ex. 7. Verify from a freehand sketch that the equation
χ = 2 sin л:
has a root lying between \π and Jw. Draw the graphs of у = χ and у = 2 sin л:
accurately on graph paper, taking values of χ from \π to \n and hence obtain
a better estimate of this root. Suggest how you could continue the process. If
you have a table of sines of angles in radians, show how these may be used to
obtain rapidly a good approximation.
The graphical method just mentioned suggests a simple analytic method
for locating the presence of a root: if /is continuous and f(a), f(b) have
opposite signs, then the equation f(x) = 0 has at least one root lying
between χ = a and χ = b.
Ex. 8. Show that the equation _
has just one real root and that this root lies between χ = 1 and χ = 2.
Ex. 9. Show graphically that it is possible for roots of the polynomial equation
f(x) = 0 to occur between χ = a and χ = k without f(a), /φ) being of different
signs.
555

NUMERICAL SOLUTION OF EQUATIONS
[26
When a root has been located between χ = x0 and χ = хъ an
approximate value for the root may be obtained by linear interpolation; that is,
by approximating to the graph of у =f(x) between χ = x0 and χ = хг
by a straight line (see Figure 26.1, in which О A = x0, OB = xx; we take
ОС as our approximate root, the exact root being OD). Since the equation
of the linePg, in the notation of Chapter 18 with x1—x0 = h, is
Ky-fo) = A/o(*-*o)
we have, putting у = 0,
as an approximation to the required root.
_hfo
0 Δ/ο
*Ex. 10. Show from graphical considerations how the computation of Δ2/0
enables one to tell whether the approximation by linear interpolation is likely to
be too big or too small.
Example 2. Show that the equation
x3 + 2x-4 = 0
has only one real root, and find its value, correct to 1 decimal place, by linear
interpolation.
WriteДх) = x*+2x-A; then/'(x) = 3x2+2, and/"(x) = 6x.
Thus we see immediately that, since the graph of/has no maximum or
minimum, it can cut the χ axis only once, giving just one real root (Figure
26.2).
556

2] LOCATING ROOTS
By direct computation,/(1) = —1, /(2) = 8 and the real root lies
between χ = 1 and χ = 2, probably nearer to χ = 1.
By linear interpolation, a second approximation is given by
x= l+h
Since f\x) > 0, f"(x) > 0 between χ = 1 and л- = 2 (see Figure 26.2)
the root χ = Ц will be an underestimate.
Fig. 26.2
By direct computation again we have
Λ1·1)= -0-469, /(1-2) =+0-128,
showing that the root lies between χ = 1-1 and χ = 1-2, and our next
approximation to the root is л л , 469 1
1·1+Τ9ΤχΤο·
As before, this is an underestimate and we can be confident, without
further working, that χ = 1-2 is the value of the required root correct to
1 decimal place.
Another valuable method of locating roots of a polynomial equation is
based on a consideration of the number of sign changes among the
coefficients in the expression „ , „ , ,
Ex. 11. Prove that, if all of the coefficients are positive, then the equation
a0xn + a1x"-1+...+an = 0
can have no positive roots.
Prove further that, if the coefficients of all odd powers of χ are zero and
remaining coefficients are positive, then the equation can have no real roots

NUMERICAL SOLUTION OF EQUATIONS [26
The result of Ex. 11 may be extended to the following result, known as
Descartes's Rule of Signs: the number of positive roots of a polynomial
equation cannot exceed, and has the same parity as, the number of sign
changes among the coefficients, reading from left to right. (Two numbers
have the same parity if they are both even or both odd.)
For example, the equation
2χ»-ΊχΊ-4χ* + 1χ· + 2χζ + χ*-5χ-4 = 0
+ -- + + +--
ft t
has three sign changes and thus either one or three positive roots. Again,
theequation л.-бх.-4л» + Зх«-^+2 = 0
has four sign changes and thus has zero, two or four positive roots.
By writing — χ for χ one may similarly find an upper limit to the number
of negative roots.
Ex. 12. Show that the equation
χ7-3χ*-χ2-ί = 0
has just one real root, by considering separately the possible positive and negative
roots.
The proof of Descartes's Rule is not difficult, but is complicated by the
number of special cases that have to be considered. The reader who wishes
to follow the proof through should try Exs. 13-17; in all these equations
we adopt the notation
P(x) = aoxn + a^-i+.-.+a^x + an,,
where an > 0 and we suppose that P(x) has к sign changes.
Ex. 13. Prove that P\x) has к sign changes if a„_t > 0 and (k-1) sign changes
if e„-i < 0.
Ex. 14. Prove graphically that, if P'(0) > 0, no roots of the equation P(x) = 0
lie between χ = 0 and the least positive root of P'(x) = 0 and that if P'(0) < 0,
at most one such root of P(x) = 0 exists.
Ex. 15. What happens in Exs. 13, 14 if α„_ι = 0?
Ex. 16. Complete the proof of Descartes's Rule of Signs using mathematical
induction.

2] LOCATING ROOTS
Exercise 26(a)
1. If χ is rounded-off to 3 significant figures to give the numerical value 48-7,
give the value and range of acceptance for (i) sin x°; (ii) tan x°, using 4-figure tables.
2. If x, у are rounded-off to 3 significant figures, their values are χ = 17-2,
у = 5-16; give the range of acceptance for (i) x+y; (ii) xy; (iii) x/y.
3. Compare the relative accuracy obtained from your square root tables for
x = V3-V2 with that obtained for χ = (V3 + V2)-1·
If V2 = 1-41421... and V3 = 1-73205... find as accurately as you can a value for
V3-V2-
4. Given that V130 = 11-4018 and V132 = 11-4891 find as accurately as you
can a value for V132-V130.
5. The triangle ABC is right angled at В. AC, АВ are measured to the nearest
millimetre, their lengths being found to be 7-4 and 4-4 cm respectively. Use the
theorem of Pythagoras to calculate the length of ВС, stating what confidence
you have in the reliability of your answer.
6. If χ = y + e, use the Taylor expansion to find the absolute and relative error
in taking tan χ to be equal to tan y.
7. With the notation of Question 6, find the absolute and relative error in
taking (1+ЛГ2)-1'2 to be (1+У)-1'2.
8. If χ = 4-6 ± 005 and f(x) = x3 - 2x +1, and if χ is taken to be 4-6, find the
relative errors in (i) x; (ii) f(x). Comment upon your result.
9. Find graphically, using the method of enlarging scales, the least positive root
of the equation , χ
correct to 2 decimal places.
10. Solve graphically the equation 2x = 1 + In x.
11. Solve graphically x3 = 10.
12. Find 21'5, correct to 2 decimal places, by a graphical method.
13. Locate the three roots of the equation
2х3-6х-Ъ = 0
and find their approximate values, using linear interpolation.
14. Locate the two real roots of the equation
*4 + x410x-24 = 0
and find their approximate values, using linear interpolation.
15. Use Descartes's Rule of Signs to prove that the equations
(О^-глг'-З = 0; (11)^-^ + 5^ + 2 = 0; (iii) χ6 + **-4*3 + 5л: + 2 = 0
each have at least two complex roots.
559

NUMERICAL SOLUTION OF EQUATIONS [26
16. Use Descartes's Rule of Signs to show that the equation
x* + x+l = 0
has only one real root, and that this root is negative.
Show that this result can also be obtained from simple graphical considerations.
17. What can you say about the reality of the roots of the equation
xn + x2 + 2 = 0
where и is an integer greater than 2?
3. ITERATIVE PROCESSES FOR
SOLVING EQUATIONS
In this section we consider the application of iterative methods for finding
numerical solutions of equations; that is, methods which develop
successive approximations to a root of a given equation by a simple repetitive
process depending upon a recurrence relation.
As a first example, consider the equation
x2-5*-5 = 0.
(A quadratic equation is chosen for simplicity, but the method to be
outlined below is applicable to any polynomial equation.)
The given equation has roots lying between — 1 and 0 and between 5 and
6. We shall denote successive approximations to one of the roots by
X0, Χι, X2, ....
We first seek the negative root: take x0 = — 1. Since the equation may
be rewritten , , .
χ = jx2— 1
we try, as a plausible attempt at developing successive approximations,
the recurrence relation , , .
Xr+i = yx?-l.
Starting with x0 = — 1, this gives successive approximations
-1, -0-8, -0-872, -0-847, -0-857.
Thus xt = —0-857, and the process is seen to be converging (albeit rather
slowly) towards the negative root —0-8541 ....
Now suppose we adopt the same recurrence relation to find the root lying
between 5 and 6: , . , „ , „ A
x0 = 6 gives 6-2, 6-69, etc.
diverging.
x0 = 5 gives 4, 2-2, -0-32, -0-980, -0-808, etc.
and we are clearly converging towards the wrong root!
560

3] ITERATIVE PROCESSES
To see what has gone wrong let us write
where α is the exact root of the given equation and er is the error in the
estimate xT. _ j. _ 1
gives α - er+1 = |(a - erf -1
which reduces to er+1 = fae„
on neglecting e2. and recalling that α = ^-α2—1. Thus |eT+1| < |eT| only if
\oc\ < 2-5; in other words, the absolute errors diminish in this iterative
process only if the root to which we are approximating is less than 2-5
a magnitude.
Graphically, the solution о
= ΐχ2-1
means finding the χ coordinate of the intersection of the straight line
with the parabola у = jx2 — 1.
Our iterative process consists in starting with a given value of x,
Oo = — 1) finding the corresponding point on the parabola, moving from
there to the line у = χ (χχ = jxl—l), and thence to the parabola, then
on to the line у = x, again (x2 = \x\-1), and so on, spiralling in to the
root χ = — 0-854... in the cobweb pattern shown in Figure 26.3.
Fig. 26.3
However, if we start at x0 = 5, a quite different pattern emerges. The
reader should draw the graph and illustrate the various stages of the
iterative process: he will see that the path taken moves rapidly across the
graph to the negative root.
561

NUMERICAL SOLUTION OF EQUATIONS
*Ex. 17. Write the equation 2 _
in the form χ = 5 + 5/л:
and obtain the recurrence relation
Show that, with x0 = 6, this gives xb = 5-8541 which is correct to four decimal
places.
Show further that, in this case,
*Ex. 18. Illustrate graphically the iterative process of Ex. 17 by sketching the
graphs
у = χ and у = 5 + 5/x.
Ex. 19. Explain geometrically why convergence is more rapid for finding the
positive root from xr+i = 5 + 5/xr than for finding the negative root from
Ex. 20. Explain why, in both the iterative processes considered so far, the
successive approximations oscillate from side to side of the exact root.
Now consider a polynomial equation rewritten in the form
(We have already seen that there are various ways of doing this: our
object now is to choose the best one for a given root.)
Suppose we take „ ч
Xr+l = f(Xr),
where xr, xr+1 are successive approximations to the exact root a, with
errors er, er+1:
oc = xr + er, cc = xT+1 +eT+1.
Since a is a root of the given equation,
α =/(α).
But xr+i=f(xr),
giYing a-er+1=/(a-er)
» /(a) - erf'{a), on using Taylor's expansion.
Thus er+1 « erf'(oc).
Successive applications of this result give
er+1 « erf{cc) « e^JfW *... * ejf (*)]'*;
thus, provided \f'(oc)\ < 1, |er+1| < |e0|
562

3] ITERATIVE PROCESSES
and er decreases in magnitude as r increases. If, however, |/'(a)| > 1,
the error increases as r increases. The reader should now reconsider the
two iterative processes for the quadratic equation considered previously,
in the light of this analysis.
Ex. 21. Explain graphically what happens to the iterative process if /'(a) = 1.
*Ex. 22. Illustrate the significance of the condition |/'(a)| < 1 for the
convergence of the iterative process
by drawing four graphs with:
(i) fix) < l,/"(a) < 0; (ii) /'(a) < l,/"(a) > 0;
(iii) Γ (μ) > 1,/"(α) < 0; (iv) /'(a) > l,/"(a) > 0.
As we have already seen, the iterative procedure outlined above
converges to the required root rather slowly. A more powerful iterative
procedure is provided by the Newton-Raphson process. Consider the equation
f{x) = 0
and suppose, as before, that α is an exact root, with x0 the first
approximation and error e0; that is _
We then have Да) = О,
and thus, on using a Taylor expansion and regarding e0 as being sufficiently
small for us to be able to ignore e£ and higher powers,
f(x0)+ej'(x0)*0.
This yields an expression for the error term
F ~ /(*o)
0 ~ ГЫ
and we may take as our next approximation
r _ r Я*о)
10 /w
Repetition of this process leads us to the recurrence relation
f{Xr)
*+1 - Xr f'(xrY
Example 3. Show that the cubic equation
л:3-5л:-8 =0
has just one real root, and find its value, correct to 2 decimal plac

NUMERICAL SOLUTION OF EQUATIONS
we have
and the α
f'(x) = 3^-5,
= f(x) is seen to have two stationary points, a maximum
at χ κ —■s/l-l, and a minimum at χ « +^\·Ί. But —^/1-7 » —1-3 and
/(—1-3) < 0 therefore the curve cuts the χ axis just once and the equation
thus has only one real root (which is clearly positive). (See Figure 26.4.)
Fig. 26.4
The next step in the solution is to locate the root: to do this we substitute
integral values for χ until we discover a sign change. In order to write
down the differences Д/, Д2/,... we tabulate our working, writing the values
of/in a vertical line:
/
-8
-12
-10
Δ/
Δ2/
6
12
Δ3/
The root we seek lies between 2 and 3; using linear interpolation we
take as our first approximation
x0 = 2 + ifxl t
2-7.
Since ДД2) > 0, the curve у = f(x) is increasing between χ = 2 and
χ = 3 and since Δ2/(2) > 0, it is increasing at an increasing rate: we
564

3]
ITERATIVE PROCESSES
deduce that 2-7 is an underestimate of the exact root (see Figure 26.5,
which has not been drawn to scale).
SinceA*) ξ 3χ2-5,
Xl = 2-7-
= 2-7+
a 2-
Л2-7)
/'(2-7)
1-817
16-87
У
0
1 2 /
/Ax =
1
.1
ι ·
4Λ2) =
Fig. 26.5
The process is now repeated, taking, xx = 2-808 as our second
approximation. (The process is much facilitated by the use of a hand-calculating
machine: if one is available, recall the method of nested-multiplication for
the evaluation of polynomials—see Chapter 18.)
= 2-8(
Л2-8(
/'(2-8
Since we may regard this as a more accurate approximation to the
required root than 2-808, we take
χ = 2-80
as the root to 2 decimal places.
Ex. 23. Use a calculating machine to obtain the root of the equation
л:3-5л:-8 = О
correct to 4 decimal places.
A feature of the Newton-Raphson process is that, since at each stage the
value off(xr) is calculated, a running check may be kept on the residuals;
7 ppmii 565

NUMERICAL SOLUTION OF EQUATIONS [26
that is, the values obtained by substituting our successive approximations
into the polynomial. (We want the residuals to be zero eventually.)
We now investigate, as before, how rapidly the process converges.
Suppose that the exact root of the equation J{x) = 0 that we seek is α
and that we obtain a sequence of approximations x0, хъ x2, ..., where
From the recurrence relation
we thus have
giving
Now write
then
and also, since
we have
Xr+1-*r f(Xry
a_e x c Л«-*г)
Λ«-«γ)
8(x)=Mr for /,(*) + 0;
g(a) = 0, provided/'(a) + 0
1&)f(x)=№,
SW'M+£(*)/"(*) =/'(*)
and g"(x)f(x) + 2g'(x)f'(x) + g(x)f"(x) =f"(x).
Thus
From the equation
g'(pi)=l, g"(cc)=-f(cc)lf'(cc).
er+1 = er + g(cc-er)
we now have, using Taylor's theorem and ignoring powers of er higher
than the second, . N ,. N , 2 „, N
- 2/'(a)e-
The analysis above shows us that each error is proportional to the square
of the error in the preceding term: we deduce that convergence is more
rapid in the Newton-Raphson process than in the first iterative method
discussed in this Section. (The Newton-Raphson method is a second-order
process.)
Geometrically, the Newton-Raphson process is equivalent to drawing
a sequence of tangents to the curve у = f(x).
Let y0 =f(x0)', then the tangent at the point (xq, y0) on the curve
У = Λχ) has equation „. ч . ч
566

and this meets the χ axis
ITERATIVE PROCESSES
(see Figure 26.6).
Ex. 24. Show by graphical considerations, that, once the process starts to
converge to a root, it always does so from one side. Try to produce an analytical
argument to support this assertion.
Ex. 25. Explain by a graphical argument how in accuracies may arise in the
neighbourhood of two nearly equal roots.
The Newton-Raphson process involves division by the awkward
number/'^,.). A simplification is effected by using the von Mises iteration:
Xr+1 Xr ПхоУ
where the variable denominator f'(xr) is replaced by the constant f'(x0)·
The convergence is less rapid but nevertheless fairly good.
Ex. 26 Find the negative root of the equations
by Newton-Raphson and von Mises's iteration, taking x0 = - 1.
Ex. 27. Interpret the von Mises process graphically. Explain why a very efficient
procedure is to use the Newton method strictly for a suitable number of stages
and then stick with a constant value of/', e.g.f'(x^) after two stages.
We conclude this chapter with an example of how an iterative process
of a required order may be developed—in this case determining the
reciprocal of a number.
7-3 567

NUMERICAL SOLUTION OF EQUATIONS [26
Example A.Iflja is calculated from the recurrence relation
xr+1 = 2xr — ax\
prove that er+1 = αή..
Find also the connection between eT, eT+1 if I/a is calculated from the
recurrence relation „ „ „ „ „
xr+1 = Зхг-3ах}-а2х1
and suggest a fourth-order iterative process for finding \\a.
If xr+1 = 2xr-a4
we have xr+er = -
a
1
4
.. l_^x
a4=1-(2xr~a4)
=> er+1 = ее*.
Similarly, if xr+1 = 3χτ-3αχ} + α24,
-*--1,
. ι 3χτ 34 ,
=> а*4=--(3хт-3а4 + а24)
=> er+1 = а2$.
Now suppose we seek a recurrence relation such that erH
xr+er =-
n«4-
=> a34 = - - (4xr - 6ax2r + 4a24 - aa4).

3] ITERATIVE PROCESSES
Thus, if we set xr+1 = 4χτ-6α4+4α24~α3χ%
it follows that er+1 = a3e*
and we have a fourth-order iterative process for finding \\a.
Exercise 26(b)
Find the real roots of the equations 1-8, using any suitable iterative process. Use
linear interpolation to find the first approximation and give your final answer to 3
significant figures.
I. x3-3 = 0. 2. л:3-100 = 0.
3. x*-5 = 0. 4. хг-Ъх-П = 0.
5. χ3- 6x2 + \0x- 9 = 0 (1 root). 6. х3-Ъх2-Ъх-1 = 0 (1 root).
7. jc*-7x-12 = 0(2 roots). 8. х3 + Ъх2-9х-\6 = 0(3roots).
9. Find to 4 significant figures the least positive root of the equation
л4- 13л:2- 18л:- 5 = 0.
10. Show that the equation
2лл-10л:3+10л:-1 = 0
has two roots between 1 and 2 and find their numerical values, correct to 4
significant figures.
II. Use the Newton-Raphson method to find an approximate value for the least
positive root of the equation
3 tan χ = Ax (4 decimal places).
12. Find an approximate value of χ such that л: + е* = 3.
13. Solve, correct to 2 decimal places, the equation
sin^=3*-l.
14. Show that the equation
(2k+l)x3-k(x+l) = 0,
where к is large and positive, has a root near to 1.
Find the equation of the tangent to the curve
y= (2k+l)x3-k(x+l)
at the point χ = 1. From the equation of this tangent find a better
approximation to the indicated root of the original equation. (O & C)
15. Find the root of the equation
sin χ = хг
other than χ = 0, to 3 decimal places.
569

NUMERICAL SOLUTION OF EQUATIONS [26
16. Establish Newton's formula for obtaining a closer approximation to a real
root of the equation f(x) = 0.
Use this method to find, correct to 3 significant figures, the positive root of the
^иай°П 4cos*-2*-l = 0. (L.)
17. By using Newton's method of approximation, or any other method, find
the value of χ correct to 3 decimal places for which the expression
x+l
\nx
has a stationary value. (L.)
Miscellaneous Exercise 26
1. Find the greatest root of the equation
х3-Ъх+\ = О,
correct to 3 decimal places.
2. The roots of the quadratic equation ax2 + bx— 1 = 0 are calculated from the
recurrence relation .
Χτ+1 = ^Го-
Interpret this process geometrically, and prove that
eT+1 = -aX4T,
where Xis the exact value of the root being calculated and X = xT + eT.
With the help of reciprocal tables, use this method to solve the equation
3. Give a sketch showing the general shape of the graph of у = sec л: for values of
χ from χ = 0 to χ = %π.
Deduce from the graph that large roots of the equation
are approximately equal to (и+ i) π, where и is a large integer; and prove that
closer approximations are given by
where the positive sign is taken when η is odd, and the negative sign when η is
even. (O & C)
4. A root of the equation
sin3 ix0 + cosx0-f£§ = 0
is close to 60. Find the value of the root, correct to 0· 1. (O & Q
5. Draw an accurate graph of у = \π sin χ between χ = 0 and χ = π. Draw in
the same diagram the lines у = mx for m = f, f, f, f, 1. Determine the values
of χ where these lines cut the graph of у = \π sin χ, giving your answers in the
form ктт, where к is correct to 2 decimal places.
570

MISCELLANEOUS EXERCISE 26
Use the values obtained to draw a separate graph of у = in sin x/x between
χ = in and χ = η. (Ο & С)
6. Verify that χ = § η is an approximate solution of the equation cos χ = ix,
and show that a better approximation is 1-03. (O & Q
7. If Or + er)2 = a, prove that
Ι/β \ er2
and deduce the recurrence relation
—1Ы
for finding Vft
Suggest an intuitive argument leading to this recurrence relation and show
that the same relation is obtained by applying the Newton-Raphson process to
the equation x2-a = 0.
8. Vй is calculated from the recurrence relation
_ α2 + 6αχΙ + χξ.
Xr+1~ AxM+xl) '
""""""" ^~ 4xr(a+xl)
and find Vll correct to 6 decimal places.
9. Develop a recurrence relation for finding av* in which er+1 = ke\ (where к
depends upon r) and hence find $6 to 5 decimal places.
10. Prove graphically, or otherwise, that the equation
cos χ = mx (иФО)
has one and only one root in the interval — in < χ < \π.
The angle α is denned as that root of the equation
cot α = -α
which lies between %n and n. Prove that, when m lies in the range
the equation cos χ = mx has three and only three roots in the interval
-in <x<\n.
By means of careful graphs of the functions cot χ and — x, or otherwise, obtain
the value of a, giving your answer in radians to 2 significant figures.
11. If a root of the equation f(x) = 0 is obtained by an iterative process in which
er+1 « keT (k constant) show that the expression
(Xr-Xr-d*
xT — 2лгг_! + л:г_2
may be taken as an approximation to the required root.

NUMERICAL SOLUTION OF EQUATIONS [26
12. Find approximately the root of the equation
χ* + 3χ3-7χ* + 18χ-1& = 0
which lies between 1 and 2.
13. Prove, graphically or otherwise, that the equation
/(*)= l+|2-tanx=0
has a root near χ = ^π + kn, where к is any large integer.
Denoting (&+£) π by K, prove that a better approximation to the root is given
by K+A, where /да + Л/'Ш = 0,
and that λ « —.
Find an approximation to the root correct to the term of order K~*. (O & Q
14. The equation хг — ax + b = 0 has roots α, β. Form the equation with roots
a2,/?2 and, if the equation with roots a2",/?2" is x2-a„x+b„ = 0, write down
the equation with roots a2"+1, /?2"+1.
If α = λ/? where |λ| < 1, show how the formation of a sequence of such
equations may be used to give successive approximations to β and hence a.
Under what circumstances does this method provide a very satisfactory
method for solving quadratic equations ? Discuss the magnitude of the errors
involved and consider the cases of (i) equal, (ii) complex, roots.
Discuss the application of this method of solution to equations of degree
higher than the second.
15. Prove that the equation e* = λχ has a real solution if and only if λ > e,
but that xex = μ has a real solution for all real values of μ.
Solve the equation xsx = 1 to 5 decimal places.
16. xT, xT+u xT+2 are successive approximations to a root of the equation f(x) = 0
obtained by the Newton-Raphson process. Writing CT for the correction
-АхЖ(Хт), so that CV+1 = -fixT + Cr)/f'(Xr + CT) show that
CV+1 lffe)
Cr2 ~ If'ixr)
by using Taylor's result,
Ax+h) *Ax)+hfXx) + m"(x),
where h is small.
Deducethat _ иытгы
χ»* χτ+1« 2[Πχτ)Υ ·
17. It is required to find the nth root of the number a by the Newton-Raphson
process. Show that this can be done by solving the equation xn+k—axk = 0.
Using the analysis of Question 16, show that most rapid convergence is
obtained by choosing к = £(1 - η). Use this method to obtain a value for $2,
carrying out two iterations.
572

27. The ellipse and hyperbola
1. CONICS: FOCI AND DIRECTRICES
In Chapter 22 we considered the geometry of the parabola and rectangular
hyperbola; both these curves are examples of a type of plane curve called
a conic, which we shall now define. Given a point S and a line / not
containing S, a conic is the set of all points Ρ in the plane of / and S such that
the ratio of the distance of Ρ from S to the distance of Ρ from / is a fixed
number.
If we call the foot of the perpendicular from Ρ to /, M, and the fixed
ratio e (not to be confused with the base (e) of the natural logarithms) then
a conic is the set of points Ρ in the plane containing S and /, such that
\~\ = e (Figure 27.1).
The point S is called a. focus of the conic; /is a directrix. The number e
is called the eccentricity of the conic. The type of conic is determined by
Fig. 27.1 Fig. 27.2
the value of e. The reader will no doubt already have noticed that, with
e = 1, the conic defined is a parabola. More generally, we define the
following types of conic:
(i) if 0 < e < 1, the conic is called an ellipse;
(ii) if e = 1, the conic is called a. parabola;
(iii) if e > 1 the conic is called a hyperbola.
The shape of the three types of conic are shown in Figures 27.2 (ellipse),
27.3 (parabola) and 27.4 (hyperbola). (The dotted lines in Figure 27.4 are
573

ELLIPSE AND HYPERBOLA [27
not part of the hyperbola: they are, in fact, asymptotes and are included
as an aid towards drawing the curve.)
Ex. 1. Explain why the set of points
{(х,у):* = 41(х-1У + <у-1П
represents an ellipse.
Fig. 27.3 Fig. 27.4
2. THE ELLIPSE
The ellipse has been defined in Section 1 as a conic with eccentricity e < 1.
Before attempting to obtain the Cartesian equation of an ellipse, it is
worthwhile to consider a little of the geometry of the curve, in order that
we may be able to choose the most suitable coordinate axes. Suppose that
S is the given focus and / the corresponding directrix and let К be the foot
of the perpendicular from S on to /.
If e is the given eccentricity, then there are two points A, A' on the line
SK which belong to the ellipse—namely the points dividing SK internally
and externally in the ratio e : 1 (see Figure 27.5 and notice that A' lies on
KSproduced, since e < 1). Let О be the mid-point of AA' and set OS = s,
OK= k. Then, if A A' = 2a,
SA = a—s, A'S = a+s,
AK=k-a, A'K= k+a,
and thus, by the definition of A, A' as points of the ellipse,
a—s = e(k — a)
a+s = e(k + a).
Solving these two equations we have
s = ae, к = ale.
574

2]
ELLIPSE
Let us now take О as the origin, OK as the л: axis and the perpendicular
to this line through О as the у axis. Then, by what we have just shown, S
is the point (ae, 0) and /is the line x — aje = 0. LetP(x, y) be any point of
the ellipse; by the definition of the ellipse
SP2 = e2PAP,
where Μ is the foot of the perpendicular from Ρ on to / (Figure 27.6).
У
в
л\ о
В'
Ρ
\)
s Ια
μ
Fig. 27.6
Thus, the ellipse is the set of points
{(x,y): {x-aef+y* = e\ale-xf).
The equation may be rewritten
x2 — 2aex + a2e2+y2 = a2 — 2aex + e2x2
=> x\\-e2)+y2 = a2{\-e2)
If we write
a2^a2(l-e2) ' L
b2 = a\\-e2),

ELLIPSE AND HYPERBOLA
the equation becomes 2
-2 + Γ2 = ^
a2 b2
which is called the standard form of equation for the ellipse.
*Ex.2. Show that the ellipse ,, „
b2x2 + a2y2 = a2b2
is symmetrical about the χ and the у axes and that (with the notation of Figure
27.6) В is the point (0, b) and B' is the point (0, - b).
*Ex. 3. Deduce, by an appeal to symmetry, the existence of a second focus S'
(—ae,0) and second corresponding directrix x+a/e = 0.
*Ex. 4. О is called the centre of the ellipse. Show that every chord PP' of the
ellipse which passes through О is bisected at O.
AA', BB' are respectively the major and minor axes of the ellipse; a is
the length of the major semi-axis, b is the length of the minor semi-axis
(b < a, hence the word 'minor'). From the original definition of b, we have
the important equation connecting a and b:
b2 = a\\-e2).
Ex. 5. If a circle is regarded as a special case of an ellipse with equal major and
minor axes, show that it corresponds to a conic with zero eccentricity.
Ex. 6. A chord drawn through a focus perpendicular to the major axis of an
ellipse is called a latus rectum of the ellipse. Show that the length of the latus
rectum is 2b2/a.
Ex. 7. Show that the eccentricity of the ellipse
is у and deduce the coordinates of its foci.
Ex. 8. Show, by translating the coordinate axes, that the equation
(x-1)2 Q4-2)a =
3 + 2
represents an ellipse with centre (1, — 2). Find the coordinate of its foci and show
the eccentricity to be yV3·
The ellipse arises in nature as the typical orbit of objects moving under
the action of an attractive force varying inversely as the square of their
distance from a fixed point. For example, planets describe ellipses (with
slight variations and discrepancies due to the attractions exerted on each
other) with the sun at one focus—hence the use of the letter S for the focus.
Similarly, for the motion of satellites around the Earth, the centre of the
Earth being situated at a focus (again with a slight discrepancy, this time
576

2]
ELLIPSE
because the Earth is neither exactly spherical, nor uniform). The
aesthetically pleasing shape of the ellipse has long been admired: it can be seen, for
example, in the elliptic arches of some bridges which with their reflection
in still water yield complete ellipses.
Ex. 9. The eccentricity of the Earth's orbit is approximately -^ while that of the
planet Pluto is \. Compare the shapes of their orbits.
Ex. 10. An astronomical unit (a.u.) is the mean distance of the Earth from the
Sun. (It is approximately 500 light-seconds.) At its nearest approach to the Sun,
the planet Mercury is distant about 0-308 a.u. from the Sun, while its maximum
distance is about 0-466 a.u. Calculate the approximate eccentricity of the orbit.
*Ex. 11. Show that χ = a cos Θ, у = b sin θ gives a parametric representation for
points of the ellipse b2x2 + a2y2 = a2b2.
A rod has three points P, Q, R marked on it, where PQ = a, QR = b. If
P, Q are constrained to move along two fixed perpendicular lines, show that R
moves along the arc of an ellipse. (This is the engineer's paper trammel method
for drawing ellipses.)
*Ex. 12. Draw in the same diagram the ellipse
b2x2 + a2y2 = a2b2
and the circle x2+y2 = a2
(the auxiliary circle of the ellipse). Given a point Ρ of the ellipse, let the
perpendicular through Ρ to the major axis of the ellipse cut the auxiliary circle at Q.
If Ο β makes an angle θ with the major axis, show that Q is the point (a cos Θ,
a sin Θ) and that Ρ is the point (a cos Θ, b sin θ), θ is called the eccentric angle of
PQ.
The relationship between the ellipse
b2x2 + a2y2 = a2b2
and its auxiliary circle x2 + y2 = a2
outlined in Ex. 11 and Ex. 12 is worth a little further study.
It will be recalled (see Chapter 13) that linear transformations of the
plane into itself map straight lines into straight lines and in particular,
parallel straight lines into parallel lines. They also map ratios of lengths
on a line into the same ratio of corresponding lengths on the image line;
in particular they map the mid-point of a line segment into the mid-point
of the image line segment. We shall now consider the particular linear
transformation, T, defined by the equation
x' = x, y' = by/a.
Ex. 13. Show that perpendicular lines do not map into perpendicular lines
under T, unless a2 = b2. Identify the two linear transformations for which a = b
and a = —b, and explain why these do preserve perpendicularity.
577

ELLIPSE AND HYPERBOLA [27
*Ex. 14. Show that the image of the circle.
{(x,y):x2+y2 = a2}
under Τ is the ellipse , ,.,.„,. „ ....
{(x , у ): b2x2 + a2y2 = a2b2).
It follows from the preceding remarks and the result of Ex. 14 that
properties of parallel chords of circles, mid-points of chords of circles, etc.
are preserved by Τ and thus correspond to identical properties of the
ellipse. For example, since the mid-points of parallel chords of a circle lie
on a straight line through the centre, the same is true for an ellipse. The
line obtained is called a diameter of the ellipse. Furthermore, the two lines
parallel to the given chords passing through the intersections P, Q of the
corresponding diameter with the circle are tangents: the same is true of the
ellipse (see Figure 27.7).
Again, in the circle, the centres of chords parallel to PQ define another
diameter, RS, such that all chords parallel to RS are bisected by PQ. The
same property is thus true of the ellipse: P'Q' and R'S' are called
conjugate diameters; they have the property that each bisects all chords parallel
to the other (see Figure 27.8).
The geometrical interpretation of Τ is as follows. Given a point P, draw
578

2] ELLIPSE
Ρ X perpendicular to the χ axis to meet it at X {PX is the ordinate of X);
then P' is the point on PX such that
P'X = (b/a) PX.
In particular, using the language of Ex. 12, the image of a point Ρ on the
auxiliary circle lies on the ellipse (see Figure 27.9).
Fig. 27.9
*Ex. 15. Show that Γ is a non-singular linear transformation, that is, that T~x
exists.
Example 1. LP is the tangent at Ρ to an ellipse, centre O, and PR is the chord
parallel to LO. Show that, if RQ is a diameter of the ellipse, then LQ is a
tangent.
Suppose that the ellipse is taken in standard form and let us denote image
elements under T'1 (see Ex. 15) by dashes. The corresponding property
of the circle (see Figure 27.10) is immediately obvious, either by an appeal
to symmetry, or, equivalently, by proving the triangles L'P'O, L'Q'O
congruent. Thus, since the properties with which we are concerned remain
invariant under the linear transformation Τ (under which the image of P'
is P, etc.) the result is also true for the ellipse (see Figure 27.11).
Notice that no appeal can be made to symmetry in the figure for the
ellipse, but since the final result is couched in terms of tangents, diameters,
and parallel lines only, the results for ellipse and circle correspond
completely.
In solving problems on the ellipse it is sometimes necessary to obtain the
equations of particular lines, notably the tangent and normal. The quickest
579

ELLIPSE AND HYPERBOLA [27
way to arrive at these equations is to use calculus, although alternative
methods are available.
Fig. 27.10
*Ex. 16. Show that, if
then
6V + sy = aV,
Deduce that the equation of the tangent at the point (лгь y^ on the ellipse is
■-X\_ У-Ух _
а2уг -b*xi
and show that this may be rewritten in the form
xxi УУ1 _ j
*Ex. 17. Show that the equation of the tangent to the ellipse at the point
(a cos Θ, b sin Θ) is
: 1,
χ cos 0 ysma
~V~ + ~b~
and that the equation of the normal is
axsind-bycosd = (α2-ό2)ο
*Ex. 18. Show that the line L, with equation
cuts the ellipse
/ m
Ьгхг+агуг = a2b2

at points with parameters λ1; λ2 which are roots of the quadratic equation
Deduce that the point Piixi, уг) is the mid-point of the chord L if
and that this gives the equation of the chord with mid-point Px as
*-*i = У-У\ =
Show how to deduce from this the equation of the tangent at the point
-P2U2, уг) of the ellipse.
The ellipse, like the parabola, has a rich geometry. We shall now prove
two of its most famous properties. (Since both are focal properties, neither
is derivable from the circle.)
(i) Focal distance property
For any point Ρ on the ellipse with foci S, S' and major axis la,
SP + S'P = 2a.
Take the ellipse in the standard form
b2x2 + a2y2 = a2b2
and let M, M' be the feet of the perpendiculars from Ρ on to the two
directrices (see Figure 27.12).
У
[ S' О
Ρ
s J
Μ
Let P(h, k) be any point on the ellipse; then, by the focus-directrix
definition of the ellipse
SP+S'P = e(PM+M'P)

ELLIPSE AND HYPERBOLA [27
*Ex. 19. Prove that SQ + S'Q < la for all points Q within the ellipse and that
SR + S'R > la for all points R outside the ellipse.
*Ex. 20. Prove a converse of the above result, namely that if S, S' are fixed
points and Ρ is any point such that SP+S'P is constant and greater than SS',
then the locus of Ρ is an ellipse with S, S' as foci. (If the constant distance is
taken as la, show that a unique ellipse exists with S, S' as foci and major axis
of length la; then use Ex. 19.)
*Ex. 21. If a, b are complex numbers such that \a—b\ < c, where с is a real
positive constant, describe the set of points
{zeC: \z-a\ + \z-b\ < c)
in the Argand diagram.
Ex. 11. Explain the theory underlying the following well-known mechanical
construction for an ellipse. Two drawing pins are stuck in a sheet of paper and a
loop of cotton is placed loosely around the two pins. The loop is made taut by
the point of a pencil which is then made to trace out a curve on the paper,
keeping the loop taut at all stages of the construction.
(ii) Reflection property
The tangent and normal at any point Ρ of an ellipse bisect the angle SPS'.
Take the ellipse in standard form and let Ρ be the point
{a cos Θ, b sin Θ).
With the notation of Figure 27.13, the equation of the normal, PG, is
(see Ex. 17)
ax sin Θ — by cos (9 = (a2—b2) cos θ sin Θ,
У
---
is' о
^p
7V
G Si
Μ
Τ >
Fig. 27.13
and thus ΰ is the paint ([(a2_ft2) cos ,]/a> 0).
Using b2 = a\\ — e2), this reduces to
(ae2 cos Θ, 0).
582

Thus S'G = ae + ae2 cos Θ, GS = ae2 cos θ
, GS^ _ ae-ae2 cos θ _ aje-a cos θ _ PA£ _ PS
an S'G ~ ae + ae2 cos θ ~ aje + a cos θ ~ M'P ~ PS'
and so, by the angle bisector theorem for a triangle, PG bisects LSPS'
internally. Since LGPTis a right angle, it follows immediately that PTbisects
£SPS' externally.
Ex. 23. ABC is any triangle and Ρ is any point on the external bisector of the
angle ВАС. Prove that
BP+CP^ BA + CA,
with equality only if Ρ coincides with A.
Example 2. Prove that the chord of contact of tangents from the point
C(h, k) to the ellipse 2 2
is the line -5-+7T = 1·
Deduce that, if the tangents at the extremities of a variable chord through
the fixed point D{<x, β) meet at R, then R lies on the straight line
αχ /S> _
a2 + b2 ~
1.
From Ex. 16, we know that the equation of the tangent to the ellipse
at the point Ρχ{χν jj is
a2 + b2 ~ l-
Now suppose that Ρ^, jj, P2(x2, Уг)are the points of contact of tangents
from C(h, k) to the ellipse. Then the coordinates (h, k) satisfy the equations
and we have ^+^=1 and Щ+^ = 1.
α' οΔ α ο*
Thus PjOj, yx) and P2(x2, у2) lie on the line
hx ky= j
a2 b2 '
which must represent the required chord of contact.
If Ζ»(α, /?) lies on this chord of contact,
Ы k£_
a2 + b2 ~

ELLIPSE AND HYPERBOLA [27
for all positions of the point (h, k). Thus (h, k) lies on the line
Example 3. The normal at the point Ρ to the ellipse Σ with equation
b2x2 + a2y2 = а2Ьг
meets the χ axis at Η and the у axis at K, and Q is the fourth vertex of the
rectangle OHQK. Prove that the locus of Q is a concentric similar ellipse Σ',
but with the minor axis ο/Σ' lying along the major axis ο/Σ.
[Locus questions are usually best attempted using the parametric form
(acos Θ, b sin Θ): the coordinates of the point whose locus is sought are
then obtained in terms of the parameter Θ, which may be eliminated by
using the identity , . , . , .
6 J cos2 Θ+ sin2 θ = 1
to yield the equation of the locus.]
Let Ρ be the point . . , . ..
{a cos Θ, b sin Θ).
r
\ ·
к
1
=».
/
tl
h
•)
Q
Then the normal atP has equation ax sin Θ — by cos θ = (a2—b2) cos θ sin θ
(see Ex. 17). Thus we may write down the coordinates of the points
Я (у = 0) and Κ (χ = 0) and hence of Q (see Figure 27.14):
„ /a*-b* a A
a2-b2 .
b S1
ta2-b2
Κ: (θ, ■

If we call the coordinates of Q(x, y) we have
and the equation of the locus of Q is
a2*8 by
which, by comparison with the equation
A2 + B* L'
is seen to be an ellipse, Σ', with
a ' b
The centre of Σ' is the origin and Α.Β = b:a; the ellipses Σ and Σ' are
thus concentric and similar. Furthermore, if a > b, then A < В and
the minor axis of Σ' lies along the χ axis, that is, the major axis of Σ.
(Strictly speaking, the first part of our solution is incomplete until we
have verified that every point of Σ' is a point of the locus. The reader may
care to supply the details.)
Exercise 27(a)
1. Find the eccentricities of the following ellipses:
(i) |+^2 = 1; (ϋ) ^+Уг = Ι"; («Ο 3^+5^ = 1.
2. Draw a rough sketch of the ellipse
Find its eccentricity, the coordinates of its foci and the equations of the
corresponding directrices.
3. Show that the equation
3(x-2)2+4(y+l)2 = 36
represents an ellipse and find its eccentricity and the coordinates of its foci.
4. Showthattheequation χ2 + 2γ2+6χ_4γ + 9 = Q
represents an ellipse and find its eccentricity and the coordinates of its foci.
585

ELLIPSE AND HYPERBOLA [27
5. Show that the equation
Ах2 + Ъу2-\6х+\2у+\6 = 0
represents an ellipse and find its eccentricity and the coordinates of its foci.
6. The arch of an elliptic bridge is in the form of half of an ellipse. The span is
12-5 m and the maximum height of the arch is 1-75 m. Find the eccentricity of the
ellipse.
7. Find the equation of the tangent to the ellipse
at the point (3, - 2).
8. Find the equation of the tangent to the ellipse
4-^ =
at the point (1, 2).
Prove that the line x-y = 3 also touches the ellipse.
9. Prove that the line χ-Ъу-П = О
touches the ellipse 2л:2 + 3у2 = 14,
and find the point of contact.
Where does the line ^x—y + 7 = 0
meet the ellipse. 3x2 + 2y2 =14?
10. An ellipse has foci S, S', minor axis BB' and major axis of length 2a. Prove
that BSB'S' is a rhombus of side a.
11. Ρ is a point on the ellipse
6¥+аУ = a2b2
and TV is the foot of the perpendicular from Ρ to AA', where A is the point
(a, 0) and A' the point (-a, 0). Prove that
PN2 _ b2
A'N.NA ~ a2*
If the tangent at Ρ meets the directrix corresponding to the focus S at T, prove
that LPST is a right angle.
12. The feet of the perpendiculars from the foci S, S' of the ellipse
b*xs+a2y* = a2b2
to the tangent at the point Ρ are Υ, Υ'. Prove that Υ, Υ' lie on the auxiliary
circle of the ellipse and that , _
13. Ρ is a variable point of an ellipse, focus S. Prove that the locus of the
midpoint of PS is an ellipse and locate its centre.

14. The tangents at the points P, Q of an ellipse, centre O, meet at the point
T. Prove that ОТ bisects PQ.
15. The parallel chords Ρλ Qu P2 Q2 of an ellipse are bisected by the diameter UV.
Prove that Λ Q2 and P2 Qx meet on UV, as also do ΡλΡ2 and βα β2.
16. С/У is a diameter of an ellipse and Ρ is any point on the ellipse. Prove that
the diameters parallel to PU, PVais conjugate.
17. With the notation of Question 16, the tangent at U meets PV at Τ and the
tangent at Ρ meets TU at M. Prove that Μ is the mid-point of TU.
18. The diameter UV of an ellipse bisects the chord PQ. UP and VQ meet at
X, UQ and VP meet at У; prove that XY and ^Q are parallel.
19. What is the locus of the mid-points of chords of an ellipse which pass through
a common point?
20. UVis a diameter of an ellipse and the tangents at U, Fare u, ν respectively;
the tangent at any point Ρ meets u, υ at Xand Y. Prove that Xand У lie on
conjugate diameters of the ellipse.
21. Prove that tangents at the extremities of a focal chord of an ellipse meet on
the corresponding directrix.
22. The tangent to the ellipse 2 2
at the point Ρ meets the axes at Q and R. Find the locus of the mid-point of QR.
23. The perpendicular from the centre of an ellipse with focus S to the tangent
at a point Ρ meets SP produced at Q. Prove that the locus of β is a circle, and
find its centre and radius.
24. Define geometrically the eccentric angle φ of a point Ρ on the ellipse
-У^ =1 (a > b)
a2 b2
and express the coordinates of Ρ in terms of φ.
Prove that the equation of the normal at Ρ is
О is the centre of the ellipse and QP is the ordinate of P, the normal at Ρ cuts
the χ axis at TV. Show that
NQ = — cos φ.
If the normal at Ρ bisects the angle OPQ prove that the eccentricity e satisfies
the equation ^(1 + sin^) = 1,
that OP = ae, and that , „ _
iV2 < e < 1. (O & Q
587

ELLIPSE AND HYPERBOLA [27
25. S, Sf are the foci of the ellipse
аг b2
and Pia cos φ, b sin φ) is a point on the curve. Calculate the lengths PS, PS' in
terms of a, e and φ, where e is the eccentricity of the ellipse and verify that
PS+PS' is constant.
The tangents to the ellipse at Ρ cuts the χ axis at T, and the normal at Ρ cuts
the χ axis at N. Prove that
(i) ОТ. ON = a2e2; (ii) PT/PN = tan ф/(1 - e2)
where О is the centre of the ellipse. (O & Q
26. Show that the coordinates of any point Ρ on the ellipse
can be expressed as (a cos φ, b sin φ). Prove that the equation of the tangent to
Ρ and Q are two points on the ellipse, such that φ has the value & at Ρ and φ
has the value фг at β. If the tangents to the ellipse at Ρ and Q meet on the line
ay = олг, prove that , , , . . /л_ . _.
Ф1 + Ф2 = in or ψ. (O&C)
27. Prove that, ifa2P + b2m2 = n2, then the line
lx+my + n = 0
touches the ellipse —+ — = 1
and find the coordinates of the point of contact.
Find the equations of the common tangents to the two ellipses
28. If Ζ is the foot of the perpendicular from the centre of the ellipse
b2x2 + a2y2 = a2b2
to the tangent at a variable point P, prove that the locus of Ζ is the curve
(x2+y2)2 = b2y* + a2x*.
3. THE HYPERBOLA
A hyperbola is a conic with eccentricity e > 1. The analysis of the
hyperbola follows closely that for the ellipse and there are many striking
similarities between the geometry of the two curves. To obtain the standard
form of the equation of the hyperbola we first derive certain geometrical

3]
HYPERBOLA
results, just as we did for the ellipse in Section 2. Suppose that S is the
given focus and / the corresponding directrix and let К be the foot of the
perpendicular from S on to /. There are two points, A, A', on the line
SK which belong to the hyperbola—namely, the points dividing SK
internally and externally in the ratio e:l. Since, for the hyperbola, e > 1,
/ will lie between A and A' (see Figure 27.15, and compare with Figure
27.5). Again we take О as the mid-point of AA' and set OS = s, OK = k.
Then, if A A' = 2a,
SA --
К А --
a-k,
A'S =
A'K =
a + k,
and thus, by the definition of A, A',
= e(a + k).
Solving these two equations we have
-a/e.
(Compare these results with the corresponding results obtained for the
ellipse in Section 2.)
Following closely the corresponding analysis for the ellipse, we take О
as the origin, OK as the χ axis, and the perpendicular to this line through О
as the у axis. Then, exactly as for the ellipse, Sis the point (ae, 0) and / is
the line x — a/e = 0. If P{x,y) is any point of the hyperbola, by the focus-
directrix definition we have
SP2 = ίΡΡΜ2,
where Μ is the foot of the perpendicular from Ρ on to /.
Thus, the hyperbola is the set of points
{(x,y):(x-aef+f = ^(x-a/ef}.

ELLIPSE AND HYPERBOLA
The equation may be written
x2-2aex+a2e2+y2 = e2x2-2aex + a2
=> x\e2-\)-y2 = a\e2-\)
=> t t_ = l
a2 a2(e2-l) '
Since e2 > 1, a\e2-\) > 0 and we may write
b2 = a2(e2-l).
у
Μ
к
/л/
\s
/
JP
(Notice that this is the first slight point of departure from the
corresponding analysis fortheellipse. Notice also that, forahyperbola, b can be greater
than a.) With this notation the equation may be written as
which is called the standard form of equation for the hyperbola.
*Ex. 24. Show that the hyperbola is symmetrical about both the χ and у axes
and deduce by an appeal to symmetry, the existence of a second focus S'(—ae, 0)
and second corresponding directrix.
*Ex. 25. Prove that all chords through the centre, O, are bisected at О and that
the length of the latus rectum (see Ex. 6) is 2b2/a.
The form of the complete curve is shown in Figure 27.17. Notice that,
if P(x, y) lies on the hyperbola,
590

3] HYPERBOLA
and thus no part of the curve lies within the interval — a < χ < a. It
follows that the points B(0, b) and B'(0, -b) do not lie on the curve in
contrast to the corresponding points of the ellipse. A {a, 0) and A'{ — a, 0)
are called the vertices of the hyperbola; AA' is the transverse axis of the
hyperbola, while BB' is the conjugate axis (see Ex. 26).
Fig. 27.17
The equation of the hyperbola may be written in the form
(H)(->
that is
As χ, j> both tend tooo, (--r) tends to zero and the equation of the curve
approximates to
This line is thus an asymptote to the hyperbola; by a similar argument so
also is the line
*Ex. 26. Show that the equations of the asymptotes may be written in the form
а* Ьг
and prove that the acute angle between them is 2 arctan (b/a).

ELLIPSE AND HYPERBOLA [27
*Ex. 27. Sketch in the same diagram the two hyperbolas
-,-ζ= 1 and -,-ζ= -1
a2 b2 a2 b2
and mark in the points A(a, 0), A'(- a, 0), 5(0, b), B'(0, - b). Two hyperbolas of
this form are said to be conjugate; show that conjugate hyperbolas have the same
asymptotes.
A rectangular hyperbola was defined in Chapter 22 as the curve which,
with a suitable choice of axes, has equation
To justify the use of the word hyperbola in this definition, that is, to
show that such a curve has the required focus-directrix property, consider
the linear transformation, T, of the plane into itself defined by the equations
It is not difficult to see that Γ preserves distance; for, if the matrix of
Τ is A, then
V2 V2
\ V2 V2/
fcos( — $π)
(,sin(-i7r)
-sin(-M|
cos (-in))
which represents a rotation through an angle -\n (see p. 243, vol. 1).
It follows that, under T, the image of a curve С has precisely the same
appearance and geometrical properties as the curve С itself, but is rotated
through an angle of —\v. But, since
y = j2(X'+y%
we see that any point of the set
{(x,y):xy = c2}
maps into a point of the set
{(х,уУ.х'*-у'* = 2с*}.
592

3] HYPERBOLA
This latter set clearly represents a hyperbola, with a2 = b2 = 2c2 and
asymptotes , „
Thus, the equation xy = c2
represents a hyperbola with perpendicular asymptotes: we may therefore
redefine a rectangular hyperbola as a hyperbola with perpendicular
asymptotes.
Ex. 28. Prove that the eccentricity of a rectangular hyperbola is V2 and that the
latus rectum is equal to the distance between the vertices of the curve. (A
rectangular hyperbola is sometimes called an equilateral hyperbola. All rectangular
hyperbolas are similar to one another.)
Ex. 29. Show that the foci of the rectangular hyperbola
are the points (± cJ2, ± сц/2) and find the equation of the corresponding
directrices.
Ex. 30. A parametric form for a point on the curve
xy = i<z2
is (aty/l, a/(tJ2)); by considering the linear transformation T, deduce that the
parametric form for a point on the curve
is (a(l/t+t)/2, a(l/t—t)/2). Examine how this point moves along the curve as t
varies between —oo and +oo; in particular, explain what happens as t approaches
the value zero from below and from above.
Ex. 31. Show that χ = a sec θ, у = btand gives a parametric representation
for all points of the hyperbola
6V-es/ = Μ*
and explain how P(a sec Θ, b tan Θ) moves on the hyperbola as θ varies from 0
to 2π. In particular, what happens as θ approaches the values %π, ftf from below
and from above?
Ex. 32. The circle. x* + y2 = а2
is called the auxiliary circle of the hyperbola
(see Ex. 12). Sketch in the same diagram a hyperbola and its auxiliary circle. Ρ is
a point of the hyperbola with positive coordinates, TV is the foot of the
perpendicular from Ρ to the transverse axis and NQ is a tangent to the auxiliary circle
at the point Q with positive coordinates. The angle NOQ = Θ. Show that Q
is the point (a cos Θ, a sin Θ) and that Ρ is the point (a sec в, b tan Θ).
593

ELLIPSE AND HYPERBOLA
Ex. 33. Show that the equation of the tangent to the hyperbola
at the point P(a sec Θ, b tan Θ) has equation
χ sec θ у tan θ _
Corresponding to the property SP+S'P = 2a for points of an ellipse
we have the following property for the hyperbola.
Focal distance property
If Ρ is any point on a hyperbola, foci S, S', then \SP—S'P\ = la.
Take the hyperbola in the standard form
UV
-azf = aW
and let M, M' be the feet of the perpendiculars from Ρ on to the two
directrices (see Figure 27.18). Let Ρ be the point (h, k); then, by the definition
of the hyperbola sp = ePM, S>P = ePM
and \SP-S'P\ = e\MM'\ = 2a.
It is not difficult to see that SP-S'P = -la if Ρ lies oi
hyperbola enclosing S, and that SP-S'P = +2a if Ρ 1:
branch.
the branch of the
ss on the opposite
*Ex. 34. Prove that, if S, S' are fixed points and Ρ moves in such a way that
SP-S'P is constant, then the locus of Ρ is a branch of a hyperbola.

3] HYPERBOLA
*Ex. 35. Interpret the conditions \SP-S'P\ < 2a and \SP-S'P\ > 2a
geometrically.
Ex. 36. Devise a mechanical construction for a branch of a hyperbola with given
foci and length of transverse axis, based upon the relation SP— S'P = 2a.
*Ex. 37. By adapting the method used for the ellipse, prove that the tangent
and normal at the point Ρ οι a hyperbola with foci S, S' bisect the angle SPS'.
Hyperbolic orbits arise under the action of forces of repulsion varying
inversely as the square of the distance from the centre of force, such as arise
in the case of electrically charged particles. They also arise under
attractive forces, such as gravity, when the energy content of the orbit is too
great for it to be an ellipse, such as, for example, comets which orbit the
sun only once before retreating into outer space. (Comets such as Halley's
comet, which reappear, obviously follow elliptic orbits: the orbits are
usually highly eccentric, that is, e χ 1.) The property \SP-S'P\ = 2a is
used in range-finding. If a gun at Ρ is fired and the times taken for the sound
to reach two listening posts, S and S', recorded, \SP—S'P\ = 2a may be
determined and Ρ lies on the unique hyperbola with foci S, S' and
transverse axis 2a. If the same experiment is conducted from S and a further
listening post S", a second hyperbola is obtained and Ρ lies at the
intersection of the two hyperbolas. A more sophisticated application of the
same idea is employed in navigation. [See the article 'Sound Ranging' in
No. 195 of the Mathematical Gazette (July 1928) by W. Hope-Jones.]
The ellipse and hyperbola have very similar geometrical properties—
most of the properties peculiar to the hyperbola are associated with its
asymptotes. Both the ellipse and hyperbola possess a centre, that is, a
point at which all chords are bisected and, for this reason, they are called
central conies; the parabola is not a central conic. If we are concerned with
a property common to all central conies, we may take the equation of a
typical conic in the form 2 „ 2 _ 1
(where α, β are not both negative).
Example 4. Prove that the line
у = mx+c
touches the central conic
xx2 + £y2=1 (a,/? + 0)
if and only if c2 = (a+/?m2)/(a/?).
Prove that perpendicular tangents to a central conic meet, in general, on a
circle, the director circle of the conic, and point out what exceptional cases
may arise.
595

ELLIPSE AND HYPERBOLA [27
The given line and conic meet in points with χ coordinates given by
*χ2+β(τηχ + ό)2 = 1,
that is, by (a+/3m2) x2 + Ifimcx + (fie2 -1) = 0.
The line у = mx + c touches the conic αχ2+βγ2 = 1
о fi2m2c2 = (a+/?m2) (/?c2-l) (the condition for double roots of the
quadratic equation above)
о a/?c2 = χ+βτη2
о с2 = (χ+βηι2)Ι(ΰφ), since α,/? + 0.
Now let у = тх+сЪеа tangent through the point (h, k). Then we have
к = mh+c
and thus, using the condition proved above for the line to be a tangent,
^(к-mh)2 = *+βτη2,
which reduces to m2(ocfih2-/3)-2oLfihkm+(oLfik2-oL) = 0. This is a
quadratic equation in m (showing that two tangents can, in general, be
drawn through the point (h, k)). Let the roots be тъ т2: then
mim2 = Wk2-cc)IWh2-p).
Thus, (h, k) lies on perpendicular tangents if
afiV-jl = -Wk*-a),
that is, if h2 + k2 = -+\,
and the locus of (h, k) is thus
χ2 + ^ = ϊ+-β·
This is a circle provided (l/a) + (l//?) > 0. This is certainly always true
for an ellipse. For a rectangular hyperbola, (l/a) + (l//?) = 0 and the
director circle reduces to a point (the origin); for any other hyperbola, the circle
exists provided (1/a) > —(1//?), that is, in the usual notation, provided
a2 > b2: a hyperbola in which the conjugate axis is greater than the
transverse axis has no director circle. Since the angle between the
asymptotes is 2 arctan (b/a), a hyperbola has no director circle if the angle
between those parts of the asymptotes containing the curve is greater
than \ti.
Ex. 38. An elliptic lamina moves in a plane in such a way that it touches each
of two fixed perpendicular lines. What is the locus of its centre?
596

HYPERBOLA
Exercise 27(b)
1. Find the eccentricities of the following hyperbolas:
(i) ix*-b* = 1; (ii) 2x2-y2 = 1; (Ш) (x-l)2-20>-2)2 = 1.
2. Sketch roughly the hyperbola
i*2-ib2=l
and find its eccentricity, the coordinates of its foci and the equations of the
corresponding directrices.
3. Sketch roughly the hyperbola
find its eccentricity, the coordinates of its foci and the equations of the
corresponding directrices.
4. Show that the equation , . „ r
x2-4y2 + 6x + 8y + l = 0
represents a hyperbola, and find its eccentricity, the coordinates of its foci and
the equations of the corresponding directrices.
5. Find the equations of the asymptotes of the hyperbolas
(i) 9x2-4y2 = 12; (ii) 3x2-y* = 1.
6. Find the equations of the asymptotes of the hyperbolas
(i) x2-4y2 + 2x+8y-5 = 0; (ii) 2x2-4y2-8y-5 = 0.
7. Find the equation of the hyperbola with asymptotes
2x-y = 0, 2x+y = 0,
which passes through the point (2, 3).
8. Find the equation of the hyperbola with asymptotes
3x-2y-9 = 0, 3x + 2y+3 = 0,
which passes through the point (2, —4).
9. Find the equation of the tangent to the hyperbola 2л:2- Ъу2 = 6 at the point
(3, —2) and prove that the line x + y+1 = 0 also touches the curve.
10. Find the equation of the normal to the hyperbola 3x2—y2 = 2 at the point
(1,1) and find the χ coordinate of the point where it meets the curve again.
11. Find the equation of the tangent to the rectangular hyperbola
x2-y2 = 3
which passes through the point (3, — 3).
12. Prove that a hyperbola intersects a line parallel to one of its asymptotes at
just one point.
13. Find the equation of the rectangular hyperbola which has the points (4, 0),
(-4,0) as foci.
14. If the eccentricity of a hyperbola is e, prove that the eccentricity of the
conjugate hyperbola is e/V(e2-l).

ELLIPSE AND HYPERBOLA [27
15. Prove the results of Exercise 27(a) 11 for the hyperbola
b2x2-a2y2 = alb\
16. Prove the result of Exercise 27(a) 12 for the hyperbola
Ъ2х2-Фу2 = a2b2.
17. If Ρ is a variable point on a hyperbola with vertex A, show that the locus of
the mid-point of PA is another hyperbola and find its centre and eccentricity.
18. Find the equation of the hyperbola (x2/a2) - (y2jb2) = 1 when the origin is
changed to the point whose coordinates are (h, k), and the new axes are parallel
to the original axes.
The coordinates of a point are given by
χ = 2 cos2 0/(2 cos2 Θ-1), у = 2 tan 20.
Show that the locus of the point is a hyperbola, and find the coordinates of the
centre.
Find the equation of the tangent to the curve at the point θ = \v. (O & Q
19. Obtain the equations of the tangent and normal to the hyperbola
4хг-у2 = 36
at the point Д5, 8).
The tangent at Ρ meets the у axis at Q and the normal at Ρ meets the χ axis
at R. Prove that the area of the triangle PQR is 145 square units.
Obtain the coordinates of the point S such that PQRS is a rectangle. (O & Q
20. Prove that the normal to the hyperbola
α2"^2 = l
at the point (a sec φ, b tan φ) is given by the equation
ax sin φ + by = (α2 + ό2) tan φ.
The normal at a variable point Ρ on the hyperbola meets the axes at Хгпа Y;
the mid-point of XY is Z. Prove that, if О is the centre of the hyperbola:
(i) the length of OZ is not less than (a2 + b2)/2a;
(ii) OZ is inversely proportional to the perpendicular distance from О to the
tangent at P;
(iii) if a = b, then Ζ always coincides with P. (O & Q
21. The tangent at the point Ρ of the rectangular hyperbola
xy = c2
meets the asymptotes at Q, R. Prove that PQ = RP.
By considering a linear transformation Τ with matrix of the form
show that the result above is true also for the hyperbola
Ь2х*-а2у2 = aV.
598

3] HYPERBOLA
Repeat the above process to prove the following more general result: if a chord
PQ of a hyperbola meets the asymptotes at P', Q', then PP' = QQ'.
Suggest a further generalization of this result by considering a second
rectangular hyperbola xy = d2 in place of the asymptotes.
4. POLAR EQUATIONS OF CONICS
Given coordinate axes Ox, Oy, the position of a point is uniquely
determined by its Cartesian coordinates (x, y). However, other systems of
coordinates are available; in Figure 27.19 the position of the point Ρ is
determined if, given the point О and the line OX, we know the length
r = OP and the angle θ = ХОР. {r, θ) are called the polar coordinates
of Ρ relative to the origin or pole О and the initial line OX. We adopt the
usual convention that positive values of θ are measured in the
anticlockwise sense.
The polar coordinates of a point are not uniquely determined. For
example, consider the point Ρ with Cartesian coordinates (1, — ^3) (see
Figure 27.20). Taking О as origin and Ox as initial line, the polar
coordinates of Ρ can be taken in any of the alternative forms
(2,- fr), (%br), (-2, in), etc.
If unique polar coordinates are desired, we define the principal polar
coordinates of a point to be the ones for which
r > 0, -π < θ < тт.
Generally, however, we allow r to take both positive and negative values.
In the same way that a set of points in a plane could be defined by an
equation or inequality involving Cartesian coordinates, sets may be
defined using polar coordinates. For example, the set
{ir, в): г
-a)

ELLIPSE AND HYPERBOLA [27
represents a circle, centre О and radius a; the set
{(r, Θ): r cos θ = a)
represents a straight line (traversed twice) perpendicular to the initial line
and at a distance a from O; the set
{{r,0):a<r<b}
represents the annulus defined by two concentric circles of radii a and b.
Ex. 39. Describe in words and draw sketches of the following sets of points:
(i) {(r, ff):rsme = a}; (ii) {(r, θ):θ=ο};
(iii) {(r, Θ): r = a sin α cosec (β-a)}.
{Hint: rewrite the defining equation and think of the Sine Rule.)
(iv) {(r, 0):a<r < b, 0 < θ < Щ; (ν) {(r, Θ): r sec θ < 2a}.
Ex. 40. Find the polar equation of:
(i) the line passing through the points with polar coordinates (1, 0) and
(1, Ϊ");
(ii) the circle through the points with polar coordinates (2, in) and touching
the initial line at the origin;
(iii) the circle of radius 1 with centre at the point with polar coordinates
(V2, iw).
Using the relation „ . a
χ = r cos О, у = r sin a
it is a simple matter to rewrite the defining equation of a set of points in
the plane in terms of polar coordinates. For example, the ellipse Г may
be written in the alternative forms
Г = {(x,y):bV+aY = azb2}
= {(r, Θ): r\b* cos2 θ+α2 sin2 Θ) = azb2}.
However, it may well happen that a curve is best expressed in polar form
without recourse to Cartesian coordinates. Suppose, for example, that
we are given a conic of eccentricity e and semi-latus rectum /. The focus-
directrix definition suggests that it would be reasonable to take the origin,
O, at a focus and the initial line along the major axis.
In Figure 27.21, О is the focus of the conic, MN is the corresponding
directrix and OL = /is the semi-latus rectum.
Then, since Qp = ePM>
we have, r = ePM.
But PM = LN-r cos θ
= Ije—r cos Θ,
giving us r{\ + e cos Θ) = I.
600

4] POLAR EQUATIONS OF CONICS
This equation is the standard form for the equation of a conic in polar
coordinates with the origin at a focus and the initial line along the major
axis.
Ex. 41. Sketch the parabola r{\ +cos Θ) = 2.
Ex. 42. Show that the equation r(2 + cos Θ) = 2 represents an ellipse and sketch
the curve.
^^
L
\p
/v\
0 J
N
Μ
X
Fig. 27.21
The polar form for a conic is of great value in dealing with properties
of focal chords. We conclude this section with an example illustrating its
use in this context.
Example 5. PQ, UV are perpendicular focal chords of a rectangular
hyperbola, focus S. Prove that:
(i) \PS.SQ\ = \ US.SV\, (ii) PQ = UV.
Taking S as origin and the major axis SX as initial line, the polar
equation of the rectangular hyperbola is
<1+V2cos0)= /.
Let P, Q, U, V have polar coordinates
(гъв), (г„ <?+*), (r„ <?+**), (rtoe+\ri).
Then, from the equation of the conic we have
r1= /(1+V2COS0)"1,
r2= /(1-V2 cos θ)~\
r3 = /(l-V2sin0)-\
rx = /(1 + 42 sin θ)~\
601

ELLIPSE AND HYPERBOLA [27
(Notice that θ cannot be an odd multiple of \π if the chords PQ, UV are
to exist.) Taking sense along each chord into account, this gives
PS =/(l+V2 cos ff)~\
SQ = /(1-V2COS0)-1,
US = /(l-^sintf)-1,
SV= /(l+^sintf)-1.
(i) PS. SQ = /2(1 - 2 cos2 Θ)-1 = - /2 sec 2Θ,
US. SV = /2(1 - 2 sin2 Θ)-1 = I2 sec 2Θ.
Λ PS.SQ+US. SV=0.
(ϋ)
and sir
nilarly
PQ = PS+SQ
I
1+V2cos0' 1
2/
l-2cos2<9'
\PQ\ = \2l sec 2Θ\,
\UV\ = |2/sec2(9|.
/
-V2cos^
Ex. 43. .Ρβ is a focal chord through the focus S of a conic of semi-latus rectum
/. Prove that

4] POLAR EQUATIONS OF CONICS
Ex. 44. PP', QQ' are perpendicular focal chords of a conic; prove that
PP' QQ'
is constant.
Ex. 45. Show that the polar equation of the directrix corresponding to the focus
Show also that a polar equation of the form
l/r = acos0+osin<?
represents a straight line.
Deduce (with the help of Exercise 27(a), question 11 (second part)), that
the equation
-= cos(0-a) + ecos6>
is the equation of the tangent to the standard conic at the point with vectorial
angle a.
Prove that the tangents at the ends of a focal chord of a conic meet on the
corresponding directrix.
5. SECTIONS OF A CONE
This section is included for the attention of readers interested in the
historical development of the geometry of the conies.
The conies, or conic sections, were first studied, as their name suggests,
as sections of a right circular cone. Their history extends back to the time
of Ancient Greece: they were first extensively studied by Apollonius
of Perga (247-205 B.C.) who wrote a treatise on their properties. The focus-
directrix property, which we have made the basis of our definition, was
not discovered until later.
The various types of conic arise as sections of a cone made by planes
making various angles with the axis of the cone. For the purposes of the
following definitions, the cone is taken to extend infinitely in both
directions from the vertex. (See Figure 27.23.)
First observe that any plane perpendicular to the axis of the cone (and
not through the vertex) cuts the cone in a circle, which may therefore be
regarded as a particular type of conic.
If the plane is oblique, not parallel to one of the generating lines of the
cone and cuts only one half of the cone, the resulting section is an ellipse
(Figure 27.24).
If the plane is parallel to one of the generating lines of the cone, the
resulting section is a parabola (Figure 27.25).
If the plane is oblique and cuts both halves of the cone, but does not
603

ELLIPSE AND HYPERBOLA [27
pass through the vertex of the cone, the section is a hyperbola (Figure
27.26).
Ex. 46. Show how a pair of straight lines arises as a conic section.
Fig. 27.25
Fig. 27.26
The connection between the conic section and the focus-directrix
definitions is exhibited in Exs. 47-52. The notation refers to Figure 27.27
(for convenience of drawing, we take an elliptic section and show only one
half of the cone).
604

5] SECTIONS OF A CONE
V is the vertex of a right circular cone with its axis vertical and we
consider the section of the cone by the plane Π. A sphere may be drawn to
touch the plane Π at S and also to touch the cone in a circle lying in a
horizontal plane Π'. The vertical plane containing V and S cuts Π and
the cone at A and A' and Π' and the cone at С and C". /is the line of
intersection of Π and Π' and A'A meets latK.P is any point on the cone lying
in the plane Π and VP touches the sphere atP'. Μ is the point of / such that
Ρ Μ and AK are parallel, NPQ is a horizontal section, N lying on A A' and
Q lying on VA.
Fig. 27.27
Ex. 47. Show that PNKM is a rectangle.
Ex. 48. Show that SP = QC and PM = NK.
Ex. 49. Show that QC/NK = AC/AK.
Ex. 50. Deduce that SP/PM is fixed for a given plane Π, and locate a focus and
its corresponding directrix for the resulting conic section.
Ex. 51. Show how the parabolic cross-section arises.
Ex. 52. Show that, both for elliptic and for hyperbolic sections, a second sphere
may be drawn to touch the cone and the plane Π, giving rise to a second focus
and a second directrix.
Miscellaneous Exercise 27
1. Show that the equation
хг + 2/ -2x +12^+8 = 0
represents an ellipse, and find its eccentricity and the coordinates of its centre.
Prove that „
3x-2y+2 = 0
is a tangent to the ellipse and find the point of contact.
605

ELLIPSE AND HYPERBOLA [27
2. The tangents drawn from a point P(xu yd to the circle x2+y2 = r2 touch the
circle at L, M. Show that the equation of LM is ххг+ууг = r2.
If Ρ moves on a hyperbola of eccentricity e with its centre at the origin, show
that LM touches a hyperbola of eccentricity e, where
e2 e2
Draw a figure for the case when e < J2 and the circle touches the given
hyperbola. (London)
3. Show that the equation of the tangent at P(a cos a, b sin a) to the ellipse
ftV+flV = fl№
is bx cos a+ ay sin a = a6.
The tangent at Ρ cuts the χ and .y axes at A and 5 respectively and the normal
at Ρ cuts the χ and д' axes at С and £> respectively. Find the ratio PC.PD.
If AD and ЯС meet at E, prove that Я£ is perpendicular to AD and hence, or
otherwise, find the equation of the circle through А, В, Е. (London)
. Prove that the conies
X*l_
a^b* «""" cP+λ b* + X
-- 1
have the same foci, whatever value A takes (provided Α Φ - φ or - b").
Find the equation of the rectangular hyperbola whose foci coincide with those
of the ellipse b2x2 + d>y2 = <fb2.
Show further that two ellipses which have the same foci cannot intersect in
real points but that if an ellipse and a hyperbola have the same foci, they
intersect orthogonally at four real points.
5. Ρ is any point on a rectangular hyperbola, centre О and foci S, S'. Prove that
OP2 = SP.S'P.
6. Ρ is the point (a cos Θ, b sin Θ) of the ellipse
and Q(a cos Θ, a sin Θ) is the corresponding point of the auxiliary circle. Prove
that the perpendicular distance from S to the tangent at Q to the circle is equal to
SP.
7. Write down the equation of
(i) an ellipse which has its minor axis along the у axis and touches the χ axis;
(ii) an ellipse which has its minor axis along the χ axis and touches the у axis.
Two such ellipses are given. Assuming that they meet in four (real) points,
write down the equation of any conic through these four points, and prove that
its centre lies on a certain rectangular hyperbola. (O & C)
8. A tangent to the ellipse , „ „ , , , ., „
b2x2 + a2y2 = <fb2
(a, b > 0) meets the ellipse
at the points P, Q. Prove that the tangents to the second ellipse at Ρ and β
meet on its director circle.
ab(a+b)

MISCELLANEOUS EXERCISE 27
9. Two hyperbolas, S and S', have the equations
a· w
-- A,
where A < 1. Prove that the tangent to S at any point Ρ meets S' in two points
Q and R which are equidistant from P.
If A = -1, prove that the tangents to S' at Q and Λ meet on S at i*, the
reflection of Ρ in the origin. (O & C)
10. The asymptotes of the hyperbola
cfi b*
(ii) PX.PY =
(Ш) py.py' = ""Г7™· (°&Q
are / and /', and Ρ is a point on the hyperbola. The perpendicular from Ρ to I
meets / and /' at X and Y, and the perpendicular from Ρ to /' meets /' and / at
X' and Y'. By expressing the coordinates of Ρ in parametric form, or otherwise,
prove that, for all positions of P,
®PX-PX' = ^->
" a*-b*;
11. Show that, if the point (хг + r cos Θ, yx + r sin в) lies on the central conic
ax2 + by2 = 1, then r satisfies the quadratic equation
r\a cos2 θ+b sin2 Θ) + 2^-*! cos (9+6л sin ff) + ax\ + by\ - 1 = 0.
Deduce that, if (xj, >ί) is the mid-point of the chord PQ, then PQ has gradient
-axjbyt.
Prove that the locus of mid-points of parallel chords of a central conic is a
diameter of the conic
12. Use the analysis of Question 11 to prove Newton's Theorem for a central
conic: if PQ, RS are two chords intersecting at X, then the ratio
PX.XQ
RX.XS
depends only upon the directions of the chords PQ and RS, and not on their
positions.
13. Given the outline of an ellipse, show how you would construct the centre,
the axes, the foci and the directrices.
14. Prove that, if tangents are drawn to a hyperbola from any point of the
conjugate hyperbola, their chord of contact touches the opposite branch of
the conjugate hyperbola and is bisected by it.

ELLIPSE AND HYPERBOLA [27
15. Find the equation of the perpendicular bisector of the line joining the points
(xu yu, (*2, yd-
A fixed circle has centre С and radius 2a. A is a fixed point inside the circle and
Ρ is a variable point on the circumference. Prove that the perpendicular bisector
of AP touches the ellipse whose foci are at С and A, and whose major axis is of
length 2a. (C.S.)
16. A point Ρ is taken at random inside an ellipse of eccentricity e. Calculate
the probability (in terms of e) that the sum of the focal distances of Ρ should be
not greater than the distance from a focus to the opposite end of the major axis.
(C.S.)
(Note: the area of an ellipse with major and minor semi-axes of lengths a and b
respectively is nab—a result easily deduced by the usual calculus methods, or
alternatively by considering the effect on areas of the linear transformation Τ of
Section 2 of this Chapter.)
17. Σ! is a circle, centre O, radius b, Σ2 is a circle, centre A radius a(a<b) which
touches Σ! internally. Describe the locus of the centre, P, of a variable circle, Σ,
which touches Σ! internally and Σ2 externally.
If Σ! is a straight line, Σ2 is a circle, centre A and radius a which touches Σ!
and Ρ the centre of a variable circle Σ which has Σ! as a tangent and touches Σ2
externally, describe the locus of P.
18. Prove that at most four normals may be drawn from a point A to a central
conic with centre O.
Prove further that, if the normals at the points Pu P2, P3, P4 on a central conic
intersect at A, then P^PaPi lie on a rectangular hyperbola which passes through
О and A and has its asymptotes parallel to the axes of the given conic. (The
rectangular hyperbola of this question is known as the hyperbola of Apollonius.)
19. Prove that the vector equation (referred to the vertex as origin) of the tangent
to the parabola уг = Aax at the point ρ = at4+2at] is г = p+Au where u is
the unit vector (ii+j)/(l+i2)*.
Prove that the locus of the meets of tangents to the parabola уг = Aax which
cut at a fixed angle α is a hyperbola of eccentricity sec a.
20. Prove that, if the chord PQ of any conic with focus S, when produced, cuts
the corresponding directrix at R, then SR is a bisector of the angle PSQ. Deduce
that, if the tangent at Ρ cuts the directrix at T, then the angle PST is a right angle.
21. If X is any point on the tangent at the point Ρ of any conic with focus S,
and if Η, Κ are respectively the feet of the perpendiculars from X to SP and the
directrix corresponding to S, prove that SH = eXK, where e is the eccentricity.
608

28. Further matrices
Throughout this chapter it is to be assumed, unless explicitly stated
otherwise, that the matrix A of a linear transformation Τ is referred to the base
vectors i, j {or i, j, к in three dimensions).
1. EIGENVALUES AND EIGENVECTORS
FOR 2x2 MATRICES
Let us write the matrix of the linear transformation, T: R2 -» R2 as A
where
Μ ьл
U ν*
Whatever form T, and therefore A, has, the origin О is mapped into itself.
In general, no other point remains fixed under Τ but, if there is a point P,
distinct from O, such that T(P) = P, then all points of the line OP are
mapped into themselves.
Ex. 1. Prove the assertion that if Τ maps a point Ρ (other than the origin) into
itself, then it maps every point of the line OP into itself.
Ex. 2. Prove that, if Τ maps two points, Ρ and Q, into themselves, where О, Р, Q
are distinct and not collinear, then Τ is the identity transformation, that is, the
transformation with matrix
We may now investigate the answer to the question: Even if Τ does not
map any point other than the origin into itself, is there a line / through the
origin such that every point of / is mapped into a point of /? If such a line
exists, we say that / is an invariant line and that / maps into itself under T.
Ex. 3. If S is a point such that T(S) e OS, prove that every point of the line OS
is mapped into a point of OS.
Suppose that S is a point other than the origin, with the property that
T(S) e OS; write OS = s. Then As = As and thus
(A-AI)s = 0.
Since s Φ 0, we must have det (A —AI) = 0 (see p. 294). This is a quadratic
equation in A, the characteristic equation of the matrix A, with, in general,
609

FURTHER MATRICES [28
two distinct roots, λ1 and A2, called the eigenvalues of the matrix A. Any
non-zero solution, s^, of the homogeneous equation
(A-A1I)s = 0
is called an eigenvector corresponding to the eigenvalue λχ; similarly, we
can find eigenvectors corresponding to the eigenvalue A2.
Example 1. Find the eigenvalues of the matrix
Ά
■С
and determine their corresponding unit eigenvectors.
The characteristic equation
2 \_n
*(Y
о А2-9А + 8=0
о A = 1 or A = 8.
Consider first the eigenvalue A = 1: if Sj is any corresponding
eigenvector then, writing . ,
we have (A-I)s! = 0,
6
that is
Э©-©·
On solving these equations we obtain χ = 2μ, у = — 3μ, giving as an
eigenvector
In particular, a unit eigenvector is given by
2/V13\
-3/V13J
All the points of the line г = к&,
that is, all points of the line 3x+1y = 0,
map into points of the same line (in fact, in this case, into themselves
since A = 1): 3x + 2y = 0 is an invariant line under the transformation.
Now consider the eigenvalue A = 8: if s2 is any corresponding eigen-
vector we have (A-8I)s2 = 0,
610

1]
(1 -W-Q-
giving s2 = (^ .
In particular, a unit eigenvector is given by
е2 = (ад-
All points of the line г = к2е2,
that is, of the line 2x-y = 0,
2x2 MATRICES
map into points of the same line (but, apart from O, not this time into
themselves): 2x—у = 0 is a second invariant line under the
transformation.
We shall now analyse the linear transformation represented by the
matrix A of Example 1 in greater detail. The matrix
formed by taking as its columns the eigenvectors
s1 = (_2) and s2=Q
(these eigenvectors being chosen for their simplicity, although any
nonzero multiples of S! and s2 would do just as well) has an important property.
Since . , . 0
Α% = Si and As2 = 8s2
it follows that the product AP is a 2 χ 2 matrix with s^ as its first column
and 8s2 as its second column. Thus
/1 0\
>(l I
But, since det Ρ = 7, Ρ is non-singular and we may multiply both sides
of the above equation on the left by P_1 to obtain
The matrix A has been reduced to diagonal-form. Notice that the
elements of the diagonal are precisely the eigenvalues of A.

FURTHER MATRICES
[28
The linear transformation T, with matrix A relative to i, j as base vectors,
maps the point with position vector as!+/?s2 where s^, s2 are defined above,
into the point with position vector as!+8/?s2; it follows that, if we take
Si, s2 as base vectors, the matrix of Τ is
as required.
Thus, when considering the transformation T, it is simpler to express
vectors in terms of the eigenvectors sb s2 rather than in terms of the more
usual base vectors i, j. For example, the point Q, where
q = 2^ + ^,
maps into the point Q', where
q' = 2s!+4s2.
Figure 28.1 shows the effect of the transformation Τ upon Q:
OQ = 2^ + ie, = OP + PQ,
OQ' = 2s!+4s2 = OP'+P'Q',
where OP'Q'P is a parallelogram.
Fig. 28.1
612

1] 2x2 MATRICES
Ex. 4. Calculate the eigenvalues and corresponding unit eigenvectors of the
matrix
■UD·
Hence describe geometrically the linear transformation Τ which has matrix В
relative to the base vectors i, j.
Reverting to Example 1, the reduction to diagonal form enables us to
calculate powers of A. For example,
/1 0\
\0 83)
= (P-iAP) (P"1AP) (P-1AP)
= P-1A(PP"1)A(PP-1)AP
= p-iA3P
■*GJ·)-
More generally we have
7\-3 2Д0 512ДЗ
_ 1 / 2 512\ /2 -1\
71-3 1024J \3 2J
_ 1/1540 1022\
7^3066 2051 j
_ /220 146\
Ц38 293 j'
*-H 2 l)(l °W2
7^-3 2J l0 8"j ^3
-и.
4+3.8й -2+2.8η
-6+6.8" 3+4.8"J
Ex. 5. Verify by direct calculation the form given for A™ above and also prove
the result by mathematical induction.
*Ex. 6. If st and s2 are eigenvectors corresponding to the distinct real
eigenvalues Яь λ2 of a matrix A and if Ρ is the matrix with first column Sj and second
column s2, and assuming that Ρ is non-singular, show that
■Gi)·
Eigenvalues need not be real. For example, if
■U !)

FURTHER MATRICES [28
the characteristic equation of A is
(A-l)2+l =0
giving eigenvalues λ1 = 1 + j, A2 = 1 — j, with corresponding eigenvectors
(9· (-1
In this case, the invariant lines are imaginary.
Of the linear transformations of the plane into itself, those which
preserve distance are of particular importance—that is, linear transformations
Τ with the property that, if T(P) = PT,f then OP = OPT for all points Ρ
of the plane. In Theorem 28.1 we shall show that the matrix A of such a
transformation is of a special type; to this end we make the following
definition (which holds for η χ η matrices, although we are considering here
only 2x2 matrices): A is called an orthogonal matrix if AA' = I.
*Ex. 7. Show that, if A is an orthogonal matrix then det A = ± 1, A is non-
singular, and A-1 = A'.
Ex. 8. In the course of the proof of Theorem 28.1 we shall require the fact that
the transpose of a product AB of two matrices is the product of the transposes of
A and В in reverse order; that is
(AB)' = B'A'.
Prove this result
(i) when A is 2x2 and В is 2x1;
(ii) when A is 2 χ 2 and В is 2 χ 2.
Show how your proofs may be generalized to cover all cases of conformable
matrices А, В up to order 3 χ 3 (or m χ η if you can manage it).
Theorem 28.1. Let Τ be a linear transformation of the plane into itself
with matrix A. Then Τ is distance preserving {that is, |OP| = ΙΟΡ^Ι for all
positions of the point P) if and only if A is orthogonal.
First observe that, if OP = г = xi+yj, then
r'r = (*J0Q =(х2+У2У,
that is, the 1 χ 1 matrix r'r represents OP2.
Now suppose that Ρ is any point of the plane, that OP = r, and that
OPy = rT.
Then ГуГг = (Ar)' (Ar) = r'A'Ar (by Ex. 8)
and thus r' r _ r'r
о r'(A'A) r = r'r
■^ r'(A'A) r = rlr
or'(A'A-I)r = 0.
t We use the notation Рт rather than P' here to avoid possible confusion with the
notation for the transposed matrix.
614

1] 2x2 MATRICES
(i) Suppose A is orthgonal.
Then A'A = I and thus ГуГг = r'r for all points Ρ of the plane and Τ is
distance preserving.
(ii) Suppose Τ is distance preserving
Then r'(A'A-I)r = 0 for all position vectors г and, in particular, for
Г! = i, r2 = j and r3 = i+j. Writing
we thus have
Thus bx =
But (A'A)'
<">G80-©^-*
<"«(!) =©-'+™·=°·
bt = 0 and b2 = —b3 = b say, and we
B-(_J ») .nd A'A-(_
= A'A and therefore
and 6 = 0. This shows that A'A = I and thus A ii
have
1 Μ
b l)'
i orthogonal.
*Ex. 9. Prove that an orthogonal matrix Ρ preserves separations, in the sense
that , „ . . .
|Px-Py| = |x-y|.
*Ex. 10. Prove that, if Ρ is a 2 χ 2 orthogonal matrix, then either
о a rotation through an angle Θ, or
sin*\
-COS0/
(COS
sin
corresponding to a rotation through — θ followed by a reflection in the χ axis
(that is, a reflection in the axis у = χ tan $θ).
*Ex. 11. Prove that the eigenvalues of a 2 χ 2 orthogonal matrix Ρ are
(i) conjugate complex numbers of modulus 1 if det Ρ = +1, or
(ii) the numbers ± 1 if det Ρ = - 1.

FURTHER MATRICES [28
The reduction of a matrix A to diagonal form P_1AP is of particular
importance if A is a symmetric matrix (A' = A). We shall show (Theorem
28.5) that, in this case, the reducing matrix Ρ may be taken to be orthogonal.
Before proving this, it is necessary to obtain some simple preliminary
results. In the following four theorems, A is a 2 χ 2 matrix with eigenvalues
Ab A2 and corresponding eigenvectors sl5 s2 and Ρ is the matrix with first
column Si and second column s2.
Theorem 28.2. If the eigenvalues of the 2x2 matrix A are distinct then the
matrix Ρ is non-singular.
Proof det Ρ = 0
=> Si = &s2, where к is a non-zero number,
=> A^-Msa = 0, on multiplying by A,
=> Ад —A&s2 = 0
=> (At — Ag)^ = 0, since ks2 = s^,
=> λ1 = A2, since Si φ 0.
Thus, since det Ρ = 0 => At = A2, we have At Φ λ2 => det Ρ φ 0.
Theorem 28.3. //^ and s^ are unit eigenvectors of A then s^ and s2 are
perpendicular if and only ifP is orthogonal.
Proof
write * = (;;), S2 = (;:);
" Vi У J
and pt = (*1 *) fa *») = ( *+* *х;+у?>).
\X2 у J \л yj ил+лл χΙ+Α ι
(i) Suppose Ρ is orthogonal then
/ x\+y\ Xlx2+yiy2\ = /1 0\
и*2+лл 4+A 1 \o ι/
giving Xi+yt = xl+yl = 1, ΧχΧι+УхУг = 0 and thus ^ and s2 are
perpendicular unit vectors.
(ii) If Si and s2 are perpendicular unit vectors then
P'P = / *?+■>* *А+ЙЛ]
-Π
and Ρ is orthogonal.
616
then Ρ =

1] 2x2 MATRICES
Theorem 28.4. If к is a 2x2 real symmetric matrix Φ kl then its
eigenvalues are real and unequal.
Proof. Write . ,4
Then det(A-AI) = 0,
о (a-A)(c-A)-Z>2 = 0
oA2-(a+c)A + (ac-62) = 0.
But the discriminant of this quadratic equation is
(a + cf-Щас-Ъ*)
= (a-c)2+4b2
> 0, since, if b — 0, а Φ с.
Thus, At and A2 are real and distinct.
Theorem 28.5. 77ге reduction to diagonal form P^AP/oj- a symmetric matrix
Афй may always be effected, and Ρ may be taken to be orthogonal.
Proof Notice first that, since st and s2 are non-zero,
st perpendicular to s2osis2 = 0.
Now suppose that st and s2 are unit eigenvectors of A.
ASi = A^
=> SgASi = A1S2S1, on premultiplying by s2,
<*- (s^Asl)' = A^sisg), on taking the transpose of each side,
о s[As2 = A1sis2, by Ex. 8 and noting that A' = A,
о A2sis2 = A1sis2, since As2 = A2s2,
o(A1-A1)sis2 = 0
<*- sis2 = 0, since At Φ λ2 by Theorem 28.4.
Thus, Si and s2 are perpendicular and, by Theorem 28.3, Ρ is orthogonal.
Furthermore, by Theorem 28.2, P_1 exists and the theorem is complete.
(Notice that orthogonal matrices are always non-singular.)
■G3-
find an orthogonal matrix Ρ such that
and interpret Ρ, Ρ' as matrices of a rotation transformation.

FURTHER MATRICES [28
Ex. 13. A is a symmetric matrix corresponding to the linear transformation T.
ВУ writing A = PDP',
where Ρ is orthogonal and D is diagonal, give a geometrical interpretation of T.
Illustrate your answer with the particular matrix A of Ex. 12.
Exercise 28(a)
1. Find the eigenvalues of the matrix
■еэ
and the corresponding unit eigenvectors. What lines in the plane map into
themselves under the linear transformation Τ with matrix A ?
and the corresponding unit eigenvectors. What lines in the plane map into
themselves under the linear transformation Τ with matrix A ?
3. Find the eigenvalues of the symmetric matrix
«
and hence find an orthogonal matrix Ρ such that P_1AP is diagonal.
4. Find an orthogonal matrix Ρ such that the matrix P_1AP is diagonal, where A
is the symmetric matrix . _ .
(-1 5Γ
Give a geometrical interpretation of the linear transformation Γ with matrix A.
5. Give a geometrical description of the linear transformation Τ with matrix A,
where /2 2\
6. Give a geometrical description of the linear transformation Γ with matrix A,
where
■e -:)·
7. The 2x2 matrix A has the property that
A2 + I = 0.
Prove that the eigenvalues of A are ±j.
8. A matrix A which has the property
A2 = A
is said to be idempotent.
618

1] 2x2 MATRICES
If A is a 2 χ 2 idempotent matrix, prove that its eigenvalues are either 0 or 1
and that, if Α Φ I, then A is singular.
Show that the matrix
G3
is idempotent and give a geometric interpretation of the linear transformation
for which A is the matrix.
9. Find An in the following cases:
10. A is a 2 χ 2 matrix and the eigenvalues of the matrix A-1 are ± j. Prove
that det A = 2.
Is the converse of this result true?
11. a, b, c, d are unequal non-negative real numbers such that
a+b = c+d = 1.
Prove that the eigenvalues of the matrix
la b\
are 1 and λ, where 0 < |λ| < 1.
Find λ for the matrix
СЭ
and describe geometrically the linear transformation with matrix An where η
is a large positive integer.
12. Find the eigenvalues of the matrix
A ~ ll4 -20J
/11.2n+1-21 -33.2"+3:
\ 7.2n+1-14 -21.2- + 2:
and deduce that /(, ,„„ „, „ r+33\
;n+22J·
= An + An-1+A"-2
find an explicit form for B.
What are the eigenvalues of В ?
2. EIGENVALUES AND EIGENVECTORS
FOR 3x3 MATRICES
The results we have proved for eigenvalues and eigenvectors of 2x2
matrices hold in large part for 3 χ 3 matrices too. As before, we define the
619

FURTHER MATRICES [28
eigenvalues of the 3 χ 3 matrix A to be the roots A1; A2, A3 of the character-
isticequation det (A - AI) = 0,
and, corresponding to each Аг-, any non-zero vector sf such that
Авг = Агвг
is called an eigenvector of A.
Geometrically, each real A,· gives rise to a real яг which defines a line
through the origin, the points of which all map into points of the same line
under the linear transformation for which A is the matrix. In contrast to
the two-dimensional case, since a real cubic must possess at least one real
root, there must be at least one such line.
Ex. 14. Show that, if T(S) = S where S is not the origin, then Г maps each point
of the line OS into itself.
Ex. 15. U, V, W are distinct non-collinear points in space with T(U) = U,
T(V)= V, T(W) = W. Prove that, if the plane UVW does not contain the origin,
then Τ is the identity transformation.
As in two dimensions, if the 3 χ 3 matrix A has three distinct eigenvalues
\, λ2> λ3 with corresponding eigenvectors Sj, s2, s3 and if the matrix P,
with first column slt second column s2, and third column s3 is non-singular,
then P_1AP is a diagonal matrix. For
Asi = ^ι18!» As2 = A2s2, As3 = A3s3
AP =
Mi 0 0\
Ρ I 0 A2 0 I
\0 0 Aj
Mi 0 0\
= I 0 A2 0 J, i
\0 0 A3/
г Р is non-singular.
If such a diagonalizing procedure exists, A is said to be reducible
to a diagonal matrix. Not all 3 x 3 matrices are reducible to a diagonal
matrix; the discussion of necessary conditions upon A for it to be so and
the possibility of reducing a matrix A which does not satisfy these
conditions to a form (the Jordan canonical form) which approximates to a
diagonal matrix is beyond the scope of this book: the interested reader
should consult one of the more advanced algebra texts mentioned in the
bibliography in Volume 1. (See, however, Exercise 28(6), Question 25.)
*Ex. 16. Given two 3x3 matrices A and B, if a non-singular matrix Ρ exists
such that в = Р-АР,
В is said to be similar to A.
620

2] 3x3 MATRICES
Prove that:
(i) A is similar to A;
(ii) if В is similar to A, then A is similar to B;
(iii) if В is similar to A and С is similar to B, then С is similar to A.
Example 2. Find the eigenvalues and corresponding unit eigenvectors of the
matrix
/10 OX
= (10 -7 10
\ 7 -5 8/
and interpret the results geometrically.
Find a matrix Ρ such that P"1AP is diagonal.
The characteristic equation of A is
Ι 1-λ 0 0 I
10 -7-Λ 10 =
I 7 -5 8-Λ|
that is (1-λ)(λ2-λ-6) = 0,
giving the three eigenvalues λ = 1, λ = — 2, λ =
(i) λ = 1
The equations
10x-8j>+10z = 0,
7x-5y+7z = 0,
have solutions χ = 1, у = 0, ζ = —1 and the corresponding unit
eigenvector is
V-1/V2/
All points of the line
map into points of the same line (in fact, in this case, into themselves, s
λ=1).
(ϋ)λ=-2
The equations „ _ _
Kk-5j>+10z = 0,
7x-5y+10z = 0,

FURTHER MATRICES [28
have solution χ = 0,y = 2,z = 1 and the corresponding unit eigenvector
= (2/V5).
\1/V5/
All points of the line - = У = -
0 2 1
map into points of the same line (but not, apart from the origin, into
themselves).
(iii) λ = 3
The equations — 2x = 0,
10x-10j>+10z = 0,
7x- 5y+ 5z = 0,
have solution x = 0,y=l,z=l and the corresponding unit eigenvector
All points of the line
map into points of the same lim
The matrix
P =
has the required property, that
P1AP =
3=(l/V2).
\1/V2/
c =l = z-
) 1 1
/1 0 0\
0 -2 0 .
\0 0 3/
Exactly as in the two-dimensional case (see Example 1) the geometrical
interpretation of the linear transformation T, which has matrix A, relative
to i, j, к as base vectors, is facilitated by expressing the position vector of
a point in terms of s^ s2 and s3. For the point Q, with position vector
q = а^ + ^г + ГЯз,
maps into the point Q' with position vector
q' = a&i-2fis2 + 3ysz.
622

2] 3x3 MATRICES
Since the coefficient of s^ is unchanged, the displacement QQ' is seen to
be parallel to the plane determined by s2 and s3.
Ex. 17. Verify that, in the notation of Example 2, P_1AP is a diagonal matrix.
Pursuing the analogy with 2x2 matrices, we define a 3 χ 3 orthogonal
matrix Ρ as a matrix which has the property that PP' = I.
*Ex. 18. If Ρ is a 3x3 orthogonal matrix, prove that Ρ is non-singular,
det Ρ = ± 1 and P"1 = P'.
*Ex. 19. Follow through the steps of the proof of Theorem 28.1 to show that a
3x3 matrix is distance preserving if and only if it is orthogonal.
Theorem 28.6. If Ρ is a 3x3 orthogonal matrix with detV = + 1, then 1 is
an eigenvalue of P.
Proof
SinCe (P-I)P' = I-P'=-(P-I)',
we have det(P-I)detP' = det (-(P-I)').
But det P' = det Ρ = 1, and also
det (-(P-I)') = det (-(P-I)) = - det (P-I)
(since P —Iisa3x3 matrix)
and thus det (P-1) = - det (P-I)
=> det (P-I) = 0
=> 1 is an eigenvalue of P.
Corollary. If J? is a 3x3 orthogonal matrix with det Ρ = — 1, then — 1 is an
eigenvalue ofP.
This follows immediately on writing —Ρ for P, since det (—P) = +1.
*Ex. 20. In the case det Ρ = +1 (Ρ orthogonal), let s be an eigenvector
corresponding to the eigenvalue λ = 1. The linear transformation Γ of which Ρ is the
matrix:
(i) preserves distances between all pairs of points;
(ii) leaves all points of the line OS fixed.
Show that Γ may be interpreted geometrically as a rotation about the axis OS.
*Ex. 21. Show that, if det Ρ = — 1 (Ρ orthogonal), then the corresponding linear
transformation may be interpreted as a rotation about the axis denned by the
eigenvalue λ = — 1 followed by a reflection in the origin.

FURTHER MATRICES
Ex. 22. Show that the matrix
(cos θ -sin θ 0\
sin θ cos θ 0)
0 0 1/
is orthogonal and represents a rotation about the ζ axis through a positive
angle Θ.
Ex. 23. Show that the columns of a 3 χ 3 orthogonal matrix Ρ represent three
mutually perpendicular unit vectors and interpret this result geometrically.
In summary, if Ρ is a 3 χ 3 orthogonal matrix representing the linear
transformation T, then the columns of Ρ are the components of three
mutually perpendicular unit vectors and Τ represents either a pure
rotation, or a rotation followed by a reflection in the origin.
Ex. 24. If
α ι :i)·
prove that Ρ is orthogonal and interpret Ρ as the matrix of a linear transformation.
*Ex. 25. If u is a unit vector, show that it is possible to construct an orthogonal
matrix with u as its first column.
*Ex. 26. If P, Q are two 3x3 orthogonal matrices, prove that:
(i) P_1 is orthogonal; (ii) PQ is orthogonal.
Interpret both these results geometrically.
Ex. 27. Find a result corresponding to Theorem 28.6 for 4 x 4 orthogonal
matrices. (Be careful when you take determinants!)
We saw earlier (Theorems 28.4 and 28.5) that the eigenvalues of a 2 χ 2
symmetric matrix Афй were always real and distinct and this enabled us
to construct an orthogonal matrix Ρ such that P_1AP was diagonal. 3x3
symmetric matrices are not quite so amenable: although it can be shown
that their eigenvalues are always real, they are not necessarily distinct.
Nevertheless, the diagonalization process is still always possible as we shall
now show.
Theorem 28.7. Given any 3x3 symmetric matrix A it is always possible to
find an orthogonal matrix Ρ such that P_1AP is diagonal.
Proof. We reduce the problem to the 2 χ 2 case. First, if λι is a real
eigenvalue (such an eigenvalue must exist, since the characteristic equation is a
cubic) with Si the corresponding unit eigenvector, then Asx = A1s1 and thus
AP =
Ρ ο Α βλ,
\0 7i yj

2] 3x3 MATRICES
where Ρ is an orthogonal matrix with sx as its first column and at... γ2 are
constants (see Ex. 25). Premultiplying by P"1 we then have
IK -ι «Λ
P-!AP= 0 β1 βΛ.
\o η yJ
But (P-iAP)' = P'A'CP"1)' = P'A(P')' = Р~1АР (see Ex. 8), since
A' = A and Ρ' = Ρ-1. It follows that P_1AP is a symmetric matrix:
Mi 0 0\
P1AP = (θ Α βλ.
\0 βг УJ
Now, by Theorem 28.5, there exists an orthogonal matrix
e: *;)
which reduces the symmetric matrix
U yJ
to diagonal form. Thus the matrix
/1 0 0\
Q= о 9l 9J
\0 rx rj
reduces P_1AP to diagonal form:
Mi 0 0\
Q-ip-lAPQ = / 0 μ% Ο .
\0 0 fj
Finally, we may write PQ = R where, by Ex. 26 (ii), R is orthogonal.
Ex. 28. Is the converse result true that, if Ρ is orthogonal and P"1AP is diagonal,
then A is a symmetric matrix ?
The determination of an inverse matrix is generally a burdensome
operation, but, if Ρ is orthogonal, we have
p-i = F
and thus the diagonalization process for a symmetric matrix is particularly
simple to effect.
625

FURTHER MATRICES [28
3. THE CAYLEY-HAMILTON THEOREM AND
POWERS OF MATRICES
A matrix A is said to satisfy the equation
α0λη + α1λη~1 + α2λη~2+... + αη-1λ+αη = 0,
if a0An + % A""1 + a2An~2 +... + ап_г\ + аи1 = 0.
Theorem 28.8. {The Cayley-Hamilton Theorem.) Every square matrix A
satisfies its own characteristic equation.
Proof. (We prove the result for a 3 χ 3 matrix A, but the proof clearly
generalizes very readily to the и х и case.)
If det (A-AI) = α0+α1λ + α2λ2-λ3, we have to show that
αοΙ + ύ^Α + α^-Α3 = 0.
Now each element of the matrix adj (A-AI), being a cofactor of
(A —AI), is a polynomial of degree at most two in λ. Thus
adj (Α-ΛΙ) = C0+Ad + A2C2,
where C0, d, C2 are 3 χ 3 matrices whose elements do not contain λ. But
(A-AI) adj (A-AI) = I det (A-AI) (Theorem 14.2)
and thus (A-AI) (Co+Ad + A2C2) s Ι(α0+αιλ + α2λ2-λ3).
Comparing coefficients of powers of A in this identity
AC0 = a0I, (1)
Ad-C0 = aj, (2)
AC2-d = αΛ (3)
-C,=-I. (4)
Multiplying (2) by A, (3) by A2, and (4) by A3 and adding this gives
0 = αοΙ + ^Α + α^-Α3
and the proof is complete.
*Ex. 29. Verify the Cayley-Hamilton theorem for 2x2 matrices by direct
substitution.
Ex. 30. If
\ 0 0 -3/
find A3 by applying the Cayley-Hamilton theorem.
626

3] CAYLEY-HAMILTON THEOREM
Example 3. If
A-(2 ~з)·
./ЫА8.
The characteristic equation of A is
А2-4Л + 5 = 0
and thus by the Cayley-Hamilton theorem
A2-4A+5I = 0.
But λ8 ξ (Λ2 - 4λ + 5)/(λ) + αλ + β,
where α, β are integers. Since Λ = 2+j => λ2-4λ + 5 = 0 we have
(2+j)8 = a(2+j)+/?.
Using the Binomial Theorem and equating real and imaginary parts,
|28-28.26 + 70.24-28.22+l = 2a+/?
l8.27-56.2B+56.23-8.2 =a
Π6[16-112 + 70-7]+1 = 2a+/?
* ll6[64-112 + 28-1] =a
(a =-336
* I/? = 145.
Thus λ8 = (λ2 - 4λ + 5)/(λ) - 336λ +145
and A8 = -336A+145I, since A2-4A+5I = 0,
_ /-336 336\ /145 0\
1-672 -1008J + 1 0 145 j
/ — 191 33(
1-672 -863;
Alternatively, one can 'build up' A8:
A2 = 4A-5I, as before;
A4 = 16A2 —40A+25I, using the commutative rule,
= 16(4A-5I)-40A+25I
= 24A-55I;
A8 = 242A2-48.55A+552I
= 242(4A-5I)-48.55A+552I
= -48.7A-5(576-605)I
= -336A+145I,
and proceed as before.
627
!36\
S63J'

FURTHER MATRICES [28
Notice that, by multiplying the Cayley-Hamilton result by Ar we
establish a recurrence relation between successive powers of A :
Ar+2 = 4A-+1-5A7.
Exercise 28(b)
1. Find the eigenvalues of the matrix
and determine corresponding unit eigenvectors.
2. Find the eigenvalues of the matrix
/1 -2 IX
A= 3 -4 1 .
\3 -7 4/
What does the fact that zero is one of the eigenvalues tell us about A?
Find a matrix Ρ such that P_1AP = D, where D is a diagonal matrix.
3. Find the eigenvalues of the matrix
/3 1 -2V
A = 14 0 -21
\4 -1 -1/
and hence write down the value of det A.
Find A-1 and determine the eigenvalues of A-1.
Suggest and prove a general result about the eigenvalues of an inverse matrix.
4. Find the eigenvalues of the matrix
/1 2 -2V
A = 6 4 -6
\6 5 -7/
and corresponding unit eigenvectors. Find a matrix Ρ such that Ρ"*ΑΡ = D
where D is a diagonal matrix.
Show that, if A is the matrix of the linear transformation T, then Γ maps all
points of the plane 2x+y-2z = 0 into points of the same plane. Find the
equations of the other two planes through the origin which have this property.
5. Find the eigenvalues of the symmetric matrix
/-1 -6 -4\
A = H-6 -2 -2
\-4 -2 3/
and hence find an orthogonal matrix Ρ such that P_1AP = D, where D is
diagonal.
628

EXERCISE 28
6. Find the equations of the planes through the origin, which map into
themselves under the linear transformation Τ with matrix
*e ■ э-
7. Ρ is a 2 χ 2 orthogonal matrix with det Ρ = 1. Show that the eigenvalues of
Ρ are eje and e->°, and interpret Θ.
Show that, although the geometrical property of a 2x2 orthogonal matrix Ρ
is characterized by its eigenvalues if det Ρ = +1, this is no longer true if
det Ρ =-1.
8. The trace of the matrix A, written tr A, is defined to be the sum of elements of
A in the leading diagonal (top left to bottom right). If A is a 3 χ 3 matrix with
eigenvalues Ab λ2, λ3, prove that
tr A = λ^λ, + λ^
Prove also that tr (A + B) = tr A + tr В and that tr (M) = /* tr (A).
9. A and В are both 2x2 matrices. Prove that A and В have the same eigenvalues
if and only if both det A = det В and tr A = tr B. (See Question 8 for definition
of trace.)
10. Prove that, if A is a non-singular 3x3 matrix, then
det(A-AI) = detA-AdetAtr(A-1) + A2tr Α-λ3.
11. If А, В have a common eigenvector s, with corresponding eigenvalues λ, /t,
prove that s is an eigenvector of (i) A + B; (ii) AB. What are the corresponding
12. A is a 3 χ 3 skew-symmetric matrix (A = -A'). Prove that
det(A-AI) =- det(A + AI)
and deduce that if A possesses a non-zero eigenvalue a, then - a is also an
eigenvalue.
If ДА) = det (A2 - λΐ), prove that - /(λ)/λ is the square of a linear
polynomial in λ.
13. If В = Р_1АР and s is an eigenvector of A prove that p-"s is an eigenvector
ofB.
14. If λ is an eigenvalue of the 3 χ 3 matrix A prove that λη is an eigenvalue of
An. Deduce that, if ж
Ar = a^A' + a-iA + aol,
15. A matrix A is said to be nilpotent if there exists a positive integer η such that
An = 0. Prove that, if A is nilpotent, then all the eigenvalues are zero.
16. Find A4 + A2 +1 if
■и -Э-

FURTHER MATRICES
-(ΰ :?)·
find An, where и is a positive integer.
18. If
■U
findA3 + 3A2+12A.
19. Show that the characteristic equation of the matrix
is λ3-11λ2+15λ-1 = 0.
Deduce that A is non-singular and that
A"1 = A2-11A+15I.
L2+ 15A-1 =
that
A2-11A+15
show that A3 = (5A -1) (A -1)
ind A"1 = (A -31) (A -21).
Deduce explicit forms for A3 and A-1.
21. The matrix A is denned by
с; *
Show that A3 - ЗА2 -16A - 161 = 21 - A
and express (21-A)-1 as a quadratic polynomial in A.
22. Show that 1, 2, 3 are the eigenvalues of the matrix
(17 -16 8\
10 -8 61.
-10 11 -3/
Hence, or otherwise, find an explicit form for An.
Write down the inverse matrix of A.
23. The 2x2 matrix A has equal eigenvalues λ. Prove that, for и > 2
A" = иА»-1А-(и-1)А"1.
Prove further that, if λ Φ 0, this formula is true for all integral n.
630

EXERCISE 28
24. If S is a 3 χ 3 skew-symmetric matrix, prove that its eigenvalues are 0 and
±ja, α real.
If Ρ is a 3 χ 3 orthogonal matrix and Ρ +1 is non-singular, prove that
S = (Ρ-ΙΧΡ + Ι)"1
is skew-symmetric, and that, if λ is an eigenvalue of S, then
A = Ρ-Ι-λ(Ρ+Ι)
is singular.
The eigenvalues of Ρ are 1, eie, s~ie; what are the corresponding eigenvalues
ofS?
25. The 3x3 matrix A has characteristic equation
(Л-ЛОЧА-Я.) = 0,
where Ab λ2 are real and At Φ λ2.
Show geometrically that, if the equations
Ax = Atx
have a plane of solutions, then it is possible to form a matrix Ρ such that
(At 0 0\
0 At 0 1.
0 0 V
If A is the matrix of the linear transformation T, and if / is the invariant line
through the origin corresponding to the eigenvalue λ2, show that every plane
containing / is mapped into itself by T.
If
(-4 -10 30\
-3 -5 18
-2 -4 13/
show that rieaves all the points of a certain plane containing the origin fixed and
find the equation of this plane. Show also that every plane containing the line
- =y- = -
5 3 2
is mapped into itself by T.
Find a diagonalizing matrix Ρ for this particular matrix A.
26. The linear transformation Τ has matrix A where A is non-singular with an
eigenvector d = OD. Ρ is a given point not on OD and Ί\Ρ) = Q. Show that
Τ maps PQ into itself if PQ is parallel to OD. Show further that, if this is the
case, then the linear transformation S with matrix A-I maps the line PQ into
the line OD.
9-2 631

FURTHER MATRICES [28
27. Ρ is a non-singular 3x3 matrix, ω is a cube root of unity and r is an integer.
Prove that, if
,.r 0 OX
1 0 ω^1 Ο Ρ
\0 0 ω-"/
Prove conversely, that if Ω is a 3 χ 3 matrix, with real determinant, having the
property
(ω' Ο 0 \
0 W+1 0 J Ϊ
0 0 ω'+2/
for some matrix Ρ and integer r.
Prove further that, if Ω Φ I, then the matrix
Ι+Ω + Ω2
is singular.
28. Show that the linear transformation Τ with matrix
■(:! i Э
leaves just two lines through the origin invariant and find their equations.
Explain geometrically why it is impossible to find a matrix Ρ such that the
matrix p_1Ap
is diagonal.

29. Further coordinate geometry
1. PARAMETRIC FORMS FOR
PLANE CURVES
We have seen in Chapters 22 and 27 that the coordinates of points on
certain plane curves may be given in terms of parameters. Indeed, we could
define, for example, the parabola as the set of all points Ρ with position
veCtOTS r = at4+2ati
relative to some origin O, where t is a scalar parameter. More generally a
plane curve is completely specified if we are given a parametric representa-
tion r=f(t)i+gm
where/, g are two (continuous) functions. We shall also demand that there
is a 1-1 correspondence between points of the curve and values of the
parameter - with the possible exception of a limited number of multiple
points, for which several values of the parameter may yield the same point
(see Ex. 5). If f{t), g(t) happen to be algebraic expressions, the resulting
analysis is much simplified and we are often able to employ the theory of
polynomial equations in problem solving.
Ex. 1. Show that an algebraic parametric form for the ellipse
Wx* + a*y* = а2Ьг
α(1-ί2). 2bt .
r = -TT^1+Τι?1·
What point of the ellipse is excluded by this parametric representation?
Find a similar algebraic parametric form for the hyperbola.
i¥-flY = a*b*.
In the following two examples we illustrate some of the methods available
for the solution of geometrical problems employing algebraic parameters.
In the first example, we define a curve parametrically and deduce a
geometric property; in the second, we obtain a geometrical property of the
parabola using the standard parametric form.
Example 1. The semi-cubical parabola is defined parametrically by
г = at2\ + ataj.
633

FURTHER COORDINATE GEOMETRY [29
Prove that a straight line not parallel to the у axis cuts the curve in either
one or three real points and that, if a straight line not through the origin cuts
the curve at P, Q, R and the tangents to the curve at these three points cut the
curve again at P', Q', R', then P'Q'R' is a straight line.
Any line not parallel to the у axis has an equation of the form
This cuts the curve in points with parameters given by the roots of the
equation .4 .2
at3 = mat2+c,
that is, by at3-mat2-c = 0.
Since this is a cubic equation in t, it has either one or three real roots,
due regard being paid to their multiplicity. This proves the first result
Now suppose that the parameters of the points P, Q, R are respectively
p, q, r. Then, if the equation of the line PQR is у = mx + с we see that the
roots of the cubic equation
at3-mat2 -с = 0
are/>, q, r; thus qr+rp+pq = 0.
But the line does not contain the origin; thus p,q,r + 0 and we have
l/p+l/?+l/r = 0.
Again, since the tangent atPcuts the curve atP', parameterp', we have,
as above, ,,, ч ,,. ч ,,, ,ч .
(i/p)+(i/p)+(i/p) = Q,
giving, lip' = —2jp, similarly \\q' = —1\q, and \\r' = —1\r and we have
α/ρΟ+α/ϊΟ+α/ο = -2[(ΐ//0+(ΐ/?)+(ΐ/ι-)] = o.
It follows that^', q', r' are the roots of a cubic equation of the form
at3-m'at2-c = 0
(where m' = p' + q' + r' and c'ja = p'q'r') and thus the points P\ Q', R'
lie on the line , ,
у = mx + c .
The shape of the semi-cubical parabola is shown in Figure 29.1. О is
called the pole and Ox is the axis of the curve. The Cartesian equation is
x3 = ay2. A connection between the parabola and the semi-cubical
parabola is obtained in Example 2.
Example 2. Given a parabola and a point C, prove that, in general, either one
or three normals may be drawn to the parabola to pass through С but that,
if just two such normals may be drawn, then С lies on a certain semi-cubical
parabola.
634

1] PARAMETRIC FORMS
Fig. 29.1
If the normals at three points, P, Q, R on a parabola are concurrent,
show that the circumcircle of the triangle Ρ QR passes through the vertex of
the parabola.
Take the parabola in the standard form y2 = Aax and let С be the point
(h, k). The equation of the normal at the point P(at2, 2at) is
tx+y = at3 + 2at
and this passes through С if
at3 + (2a-h)t-k = 0.
This is a cubic equation in t, with, in general, either one or three real roots,
and the first part of the question is proved.
If just two normals pass through C, then the cubic equation above has
a repeated root and the condition for this (see Chapter 23) is
4 /. h\3 к2 п
showing that С in this case lies on the curve
4(x-2a)3 = Π ay2.
By Example 1, this is a semi-cubical parabola with pole at (2a, 0) and axis
coinciding with the axis of the parabola.
Now suppose that the normals at P(ap2, lap), Q(aq2, 2aq), R(ar2, 2ar)
meet at C; then p, q, r are the roots of the equation
at3 + (2a-h)t-k = 0
and thus p+q + r = 0.
635

FURTHER COORDINATE GEOMETRY [29
Suppose that the circumcircle of triangle PQR has equation
This cuts the parabola at points with parameters given by the quartic
equation
(at*-x)4(2at-py = y\
But the coefficient of t3 in this equation is zero and thus the sum of the
roots is zero. Since three of the roots are, by definition p, q and r, and since
p + q+r = 0, the fourth root must be zero and we have proved the final
result.
Ex. 2. Prove that the normals to the parabola у2 = Лах touch the semi-cubical
parabola of Example 2. Draw in the same sketch the parabola y2 = 4ax and the
semi-cubical parabola 4(x-2a)3 = Пау2. (It is worth drawing the parabola
accurately and then constructing a large number of normals—which may be
done very rapidly using the geometrical property that, in the notation of Exercise
22 (ft), NG = 2a.)
Ex. 3. Draw the parabola y2 = Aax and shade in the set of all points from
which three real normals may be drawn to the parabola.
Ex. 4. The Folium of Descartes has Cartesian equation
By considering the intersection of the line у = tx with the folium, show that the
curve may be represented parametrically by
_ 3at _ 3at2
X ~ 1 + f3' У ~ 1 + f3'
Taking t = 0 it is seen that the origin lies on the curve and that the χ axis is a
tangent there; by writing и = Г1, show that the у axis is also a tangent at the
origin (which is thus a double point of the curve, that is, a point through which
pass two separate branches of the curve).
Show that the line χ - у = 0 is an axis of symmetry and that the line χ+у = a
is an asymptote. Sketch the curve.
Ex. 5. Using the method of Ex. 4, find a parametric form for the curve
y2 = x\l + x),
and deduce that no point of the curve has an χ coordinate less than -1. Prove
also that the χ axis is an axis of symmetry for the curve.
Show that the origin lies on the curve and that this point arises from two
distinct values of the parameter. Deduce that the origin is a double point of the
curve and determine the equations of the two tangents there.

И
PARAMETRIC FORMS
Exercise 29(a)
1. Show that, if OP = г = at'\ + at3], then Ρ lies on the curve Γ: x3 = агу.
Prove that any line in the plane cuts Г in either one or three real points.
The tangents at three points P, Q, R of the curve meet Г again at P', Q', R'
respectively. Prove that:
(i) if the centroid of the triangle PQR lies on the у axis, so also does the cen-
troid of the triangle P'Q'R';
(ii) if P, Q, R are collinear, so also are P', £>', R'.
2. The cissoid is defined parametrically by
at2 . at3 .
Find the Cartesian equation of the cissoid and give a rough sketch of the curve.
Prove that, if the chord PQ subtends a right angle at O, then the mid-point
MoiPQ lies on a fixed straight line.
3. A curve Г is given parametrically by
r^i + '-^J HO).
Prove that, if the points with parameters tu i2, t3 are collinear, then
1 + ίι+ί2 + ί3 + ίιί2ί3 = 0,
and conversely.
Show that, given a point Ρ lying on Γ, two lines may be drawn through Ρ to
touch the curve at Q and R. If QR cuts Γ again at S find the parameter of the
point S in terms of the parameter of P.
Show that the roots of the equation i2 +1-1 = 0 give the double point of the
curve.
4. A curve Γ is given parametrically by
r = (i2+l/i)i + (i2-l/OJ·
Obtain a necessary and sufficient condition for the three points P, Q, R of the
curve to be collinear.
The tangent at Д meets Γ again at P2, the tangent at P2 meets Γ again at P3
and the tangent at P3 meets Γ again at Pt. РгР3 meets the curve at Q while PaP4
meets the curve at R. Prove that QR is a tangent to Г at Q.
5. A curve is given parametrically by the equations
x = a(\-t\ y=a(t-t3).
Prove:
(i) that an arbitrary line Ix+my+na = 0 meets the curve in three points;
(ii) that if three points t = tu t = i2, t = t3 are collinear then
hh + tsti + htz - -1; (over)
637

FURTHER COORDINATE GEOMETRY 29
(iii) that if tu t2 and t3 satisfy the above equation then the points
t = tj, t = t2, t = t3
are collinear.
A chord through the point (a, 0) meets the curve again at the points Ρ and Q.
Prove that the locus of the middle point of PQ is a curve with parametric equa-
ti°nS x = -iaT\ у=-ОХ7* + 2). (О&С)
6. Prove that, in general, three tangents may be drawn from a point С to the
cubic curve x3 = a2y and that, if these tangents cut the curve again at P, Q, R
then the tangents at P, Q, R are concurrent.
7. A rectangular hyperbola is given parametrically by the equations
χ = ct, y= c/t.
If the four points of the curve with parameters h, t2, t3, i4 lie on a circle, show
that ίιί2Μ4 = 1· Show conversely that, if ht2tzti = 1, then the four points lie
on a circle.
A variable circle passes through the fixed points Α, Β οία rectangular
hyperbola, and meets the hyperbola again at P, Q. Show that the direction of PQ is
fixed.
8. P, Q are variable points on the parabola y2 = Aax such that PQ is parallel to
the fixed line x + ky = 0. The normals to the parabola at Ρ and Q meet at R.
Prove that the locus of R is the normal to the parabola at a fixed point on the
parabola, and find the coordinates of this point. (O & C)
9. Find the equation of the normal to the rectangular hyperbola xy = c2 at the
point (ct, c/t).
The normals to the rectangular hyperbola at the points P, Q, R, S are
concurrent; prove that each of these points is the orthocentre of the triangle formed by
the other three.
10. The normals at three points P, Q, R of the parabola y2 = Aax meet at a
point N. Prove that the centroid of the triangle PQR lies on the axis of the
parabola.
If N coincides with P, prove that QR passes through a fixed point (that is, a
point whose position is independent of P, Q, R). (O & C)
11. Find the coordinates of the point P, other than the origin in which the line
у — tx meets the curve x3+y3 = axy. (t is called the parameter oiP.)
If a line meets the curve in three points whose parameters are h, t2, t3, prove
that h t2t3=-1.
If Q is any point on the curve with negative parameter, not equal to -1, prove
that there are two points РъРг of the curve (other than Q) such that the tangents
at Д and P2 pass through Q. Prove that ОД and OP2 make equal angles (apart
from the sense) with either of the coordinate axes. (O & C)
12. Prove that the equation of the normal to the curve у = χ3 at the point

1] PARAMETRIC FORMS
By considering the maximum and minimum values of a certain function of t
and drawing a rough graph of the function, or otherwise, prove that three normals
can be drawn to curve у = χ3 from a point (0, b) when b > 4/3V3.
How many real normals can be drawn to the curve from a point (0, c) when
0< с < 4/3V3? (O&Q
13. A circle has a diameter О A of length a, and the tangent at A is /. A variable
line through О meets the circle again at Q, and / at R; Ρ is the point on OR
such that OP = QR. If О is the origin and OA is the χ axis, show that Ρ has
coordinates (at2/(1 + t2), at3/(1 + i3)), where t is a suitable parameter.
Prove that, if a line meets the locus of Ρ in three points with parameters
tu t2, tz then t2t3+ t3t!+ ht2 = 0.
Hence, or otherwise, prove that the tangent to the locus at Ρ meets the locus
again at the point with parameter — %t. (O & C)
14. A point on the curve ay2 = χ3 is given parametrically in the form (at2, at3).
If the points on the curve with parameters ii, t2, t3 are collinear, prove that
(l//i) + (l//i) + (l//s) = 0.
Hence show that, if the tangent at the point with parameter it meets the curve
again at the point with parameter i4, then it+2i4 = 0.
Perpendicular lines through О meet the curve at P, Q; PQ meets the curve
again at R and S is the point of the curve such that the tangent at S passes through
R. Prove that OP, OQ are the bisectors of the angles between OS and Ox.
(O&C)
15. The rectangular hyperbola xy = k2 is met by a circle, passing through its
centre O, in four points Au A2, Bu B2. The lengths of the perpendiculars from
О to АгА2 and ВгВ2 are a and b. Prove that ab = k2. (C. S.)
16. Show that there are three values of t, not necessarily real, for which the point
(t2, t3) lies on a given straight line.
P, Q and R are distinct points (p2, p3), (q2, q3) and (r2, r3) on the curve j>2 = x3.
Show that:
(i) if these points are collinear then ~Zpq = 0;
(ii) if the tangents at these points are concurrent, then Σρ = 0;
(iii) there are no real points on the curve for which these two conditions
co-exist.
The tangent at the point Ρ on this curve meets the curve again at P'. Find
the ratio in which PP' is divided by the χ axis. (London)
2. SURFACES AND CURVES IN
IN THREE DIMENSIONS
The simplest surface in three-dimensional space is the plane. Given three
points А, В, С, the vector equation of the plane through ABC is
r = a + A(b-a) + /t(c-a),
where λ, μ are scalar parameters. Notice that we need two parameters to
define the surface.
639

FURTHER COORDINATE GEOMETRY [29
Two surfaces intersect in a curve (which need not necessarily be a
plane curve). Thus, two planes intersect in a line; the vector equation of
the line through А, В is ч
6 г = а + Л(Ь-а),
where we have the single scalar parameter λ.
Another familiar surface is the sphere, which is defined as the set of
points in three dimensions lying at a fixed distance from a fixed point.
If the fixed distance {radius) is с and the fixed point {centre) is A, then the
vector equation of the sphere is
(r-a).(r-a) = c2.
Ex. 6. Prove that
г = a + cos θ cos 0i + sin Θ cos iij + sin фк
gives a parametric representation of the sphere. (Notice now that we have two
parameters; suggest names for them.)
*Ex. 7. Prove that the vector equation (r-a).(r-b) = 0 represents a sphere,
and locate the points A and В as points on the sphere.
The Cartesian equation of a sphere, centre А{аъ а2, a3) is thus
{x-a1)2 + {y-a2)2 + {z-a3)2 = c2,
which may be rewritten
x2+f+z2+2ux + 2vy+2wz + d = 0.
Conversely, an equation of the form
x2+f + z2+2ux+2vy+2wz + d= 0
may be rewritten as
{x+uf + {y+vf + {z+wf = и2+1?+п*-а,
and thus represents a sphere, centre {—u, —v, —w), provided
ιΡ + ΰ+ν? > d.
Ex. 8. Show that the equation
хг+уг + гг-2х-4у+2г-\0 = О
represents a sphere, and find its centre and radius.
*Ex. 9. Show that four points in space in general define a unique sphere. What
exceptional cases may arise?
The intersection of the sphere
{x-a1)2+{y-a2f + {z-a3f = c*
640

2] SURFACES IN THREE DIMENSIONS
with the plane ζ = 0
is the curve in the Oxy plane with equation
{x-a^f + iy-a^ = сг-а\,
which is seen to be a circle provided c2 > c%. If c2 = a$, the equation
represents a single point and, if c2 < of, the sphere and plane do not intersect.
Thus, a sphere and a plane, if they intersect at all, intersect in a circle,
which may be of zero radius if the plane is tangential to the sphere. The
centre of the circle lies on the perpendicular from the centre of the sphere
on to the plane. If the sphere and the plane have vector equations
(r-a).(r-a) = c\ r.n=p (where |n| = 1),
this perpendicular will be r = a + An.
To find the value of λ which gives the centre of the circle determined by
the sphere and the plane we solve
r = a + An and r.n=^,
giving (α + λή).η = p.
Thus, λ =p — a. n, since η. η = 1 and the position vector of the centre of the
r = a + (p-a.n)n.
circle ii
Ex. 10. Find the centre and radius of the circle determined by the sphere
(x-l)2 + (y-3)2 + (z-2)2=20
and the plane x—y-3z=3.
Now consider two spheres, with equations
(r-a).(r-a) = c2, (r-b).(r-b) = d\
These equations may be rewritten
r.r-2r.a = c2-a.a and r.r-2r.b = uP-b.b
which on subtraction, give
2r.(b-a) = c2-uP-a.a+b.b.
But this is the equation of a plane, perpendicular to the vector b—a,
that is, to the line of centres AB and, since a plane and a sphere determine
a circle, we see that two spheres, provided they intersect at all, intersect in
a circle.
641

FURTHER COORDINATE GEOMETRY [29
The line г = a + λη (|n| = 1)
cuts the sphere (r-b).(r—b) = c2
in points with parameters given by
(An + a-b).(An + a-b) = c2;
that is, by the quadratic equation
A2 + 2An.(a-b) + (a-b).(a-b)-c2 = 0.
Thus, a line cuts a sphere in two points (which may be coincident or
imaginary).
If η is given and we take A to be the mid-point of any chord in the
direction defined by n, the two roots λ1 and λ2 of the above quadratic equation
have the same magnitude but opposite sign; that is, λι + λ2 = 0. It follows
that n.(a —b) = 0 and A lies on the plane n.(r—b) = 0 which passes
through the centre, B, of the sphere and is perpendicular to n; thus the locus
of the mid-points of parallel chords of a sphere is a plane through the
centre of the sphere (a diametral plane).
Example 3. Find the equation of the tangent plane to the sphere
x2+y2 + z2-Sx-6z-2 = 0
at the point (1, 3, 0).
Prove that the plane . , , „„ _
F x-5y+z + 20 = 0
is a tangent to the sphere and find its point of contact.
Rewriting the equation of the sphere in the form
(*-4)2+/ + (ζ-3)2 = 27
we see that the centre of the sphere is the point (4, 0, 3).
(i) The vector (4i + 3k)- (i + 3j) = 3i- 3j + 3k is normal to the required
tangent plane whose equation is thus
l(*-l)-l(y-3)+l(z-0) = 0,
that is x-y+z + 2 = 0.
(ii) The radius of the sphere is ^27; but the perpendicular distance from
(4, 0, 3) to the plane x-5y+z + 20 = 0 is
4+3 + 20 _
V(l2+52+l2)- VZ/·
Thus the plane x— 5y+z+20 = 0 is tangential to the sphere.
A normal to the given plane is i —5j + k and thus any point on the
642

2] SURFACES IN THREE DIMENSIONS
radius to the point of contact has coordinates (4 + λ, — 5λ, 3 + λ). This lies
on the given plane if
(4 + Λ)-5(-5λ) + (3 + λ) + 20 = 0,
that is, if λ = — 1. This gives the coordinates of the point of contact as
(3, 5, 2).
The vector equation of the tangent plane at a point Τ of the sphere,
centre A and radius c, is easily derived by the same method as that used
in the last example. For a unit vector in the direction AT is (t —a)/c and
thus the equation of the tangent plane is
(r-a).(t-a) = c\
Notice that the equation of the tangent plane is obtained from the
equation of the sphere by writing t for г in one of the brackets; this simple rule
enables us to write down the equation of the tangent plane at any point of
the sphere. For example, the equation of the tangent to the sphere
x*+y2 + z*-2x + 4y+12z + 4 = 0
at the point (1, -3,0) is
*.l+j>.(-3) + z.0-(*+l) + 20>-3) + 6(z+0) + 4 = 0
or y-6z+3=0.
Example 4. Determine whether or not the circles
Si:xz+f + z2-lSx-6y-4z+14 = 0, x+3j>+5z = 0,
S2: x2+y2 + z2-6x+l4y-12z + 22 = 0, x+y+z = 2,
are linked (as in the links of a chain).
The planes of the two circles meet in the line, /, with equations
x + 3j>+5z = 0, x+y+z = 2
,•1 j j. ·* —3 У+1 z
which reduce to —j— = —у = у
and any point on this line has coordinates (3 + λ, — 1—2λ, λ). Thus /
meets the given sphere through 5Ί at points given by
(3 + λ)2+(-1-2λ)2+Λ2-18(3 + λ) + 6(1+2λ)-4λ+14 = 0,
or 6λ2-24 = 0,
giving λ = +2 or —2.
643

FURTHER COORDINATE GEOMETRY [29
Again, / meets the given sphere through S2 at points given by
(3 + λ)2+(-1-2λ)2+Λ2-6(3 + λ)-14(1+2λ)-12λ + 22 = 0,
or 6λ2-36λ = 0
giving λ = 0 or 6.
Thus, if Si. meets / at A1 (λ = -2) and B1 (λ = +2) and S2 meets / at
Α2(λ = 0) and52(A = 6), the order of the points on the line is A1A2B1B2
and the circles must therefore be linked.
Another familiar surface in three dimensions is the circular cylinder.
If the axis of the cylinder passes through the point A(alt a2, a3) and is in
the direction of the unit vector u = /i + mj + ик, any point P(x, y, z) lies
on the cylinder if its perpendicular distance from the axis is a constant, b.
Thus, if OA = a, OP = r, we have (see Figure 29.2) |(r-a) л u| = b,
or in Cartesian form,
Z[n(y-a2)-m(z-a3)]* = b\
Fig. 29.2
If we take the χ axis as the axis of the cylinder
(and thus 1=1, m = η = 0 and аг = a2 = a3 = 0)
this equation reduces to the much simpler form
y2+z* = b\
*Ex. 11. Show that a parametric form for a cylinder, radius b, with its axis along
the ζ axis, is , a. , , . a. ,,
г = b cos θι+b sin 0j + Ak.
Notice once again that we have two parameters, θ and λ for this surface.
Ex. 12. What does the surface ,
-A = '
a2 b2
represent in three dimensions?
A cylindrical spring is a three-dimensional curve lying on the surface
644

2] SURFACES IN THREE DIMENSIONS
of a circular cylinder. The curve is called a helix and is defined parametric-
ally by θ
r = 6cos0i + 6sin0j+|-k
where;> (thspitch of the helix) and b (the radius of the cylinder on which the
helix lies) are constants. Note that the parametric representation of the
helix involves only one parameter.
*Ex. 13. Interpret geometrically the constant p, the pitch of the helix.
Ex. 14. Show that the helix defined above is one of the two curves of intersection
of the cylinder x2+y2 = b2 and the corrugated surface χ = a cos . Show also
Ρ
that the other curve formed is an oppositely twined helix.
The right-circular cone is the surface traced out by a variable line through
a fixed point О (the vertex of the cone) and making a constant angle with
a fixed line though О (the axis of the cone). If we take О as the origin and
the unit vector u = li + m\ + rik as defining the direction of the axis we have,
for any point Ρ of the cone,
r.u = |r| cos a,
where r = OP and α is the angle of the cone (Figure 29.3). In Cartesian
form, this gives ., N, , „ , , , „4 ,
6 (lx + my+nz)2 = (x*+y2 + z2) cos2 α
Fig. 29.3
and we see that the equation of a cone with its vertex at the origin is
homogeneous in x, y, z. We saw in Chapter 26 that the section of a plane
with a cone is a conic.
Example 5. Obtain the equation of the cone with vertex V (2, —1, —2)
which touches the sphere x^+yP+z2 = 1 at the points of a circle on the
surface of the sphere.
Since the centre of the sphere is the origin, O, and the cone touches the
sphere, VO is the axis of the cone. Let FT be a line on the surface of the
645

FURTHER COORDINATE GEOMETRY [29
cone (a generator of the cone) (see Figure 29.4). Then, if LOVT = a, since
OV = 3 and the radius of the sphere is 1,
_2V2
cosa 3 .
Now consider any point P(x, y, z) on the cone. A unit vector along
OF is u = f i — Jj — f к and thus, since
VP.u= |VP|cosa
we obtain
[f(x-2)-Ky+l)-f(z + 2)]2 = l[(x-2y + (y + lf + (z + 2n
which may be rewritten as
4(*-2)2 + 7(y+l)2+4(z + 2)2-4(>+l)(z + 2) + 8(z + 2)(;c-2)
+ 4(*-2)O+l) = 0.
3. CHANGING THE COORDINATE SYSTEM
IN A PLANE: ROTATION OF AXES
In Chapter 13, a linear transformation of the plane into itself of the form
x' = ax + by, y' = cx + dy
was regarded as a mapping of the plane into itself, in which the point
P(h,k) is mapped into the point P'(ah + bk, ch + dk), coordinates being
referred to the same axes. In particular, the transformation Τ whose
matrix P, is orthogonal, . „ . _4
_ /cos0 -sin0\
Г - Ism θ cos θ)
has the effect of rotating any line О A through the angle Θ, measured
in a positive sense, into the position OA', where О A = О A'.
646

3] CHANGES OF COORDINATE SYSTEM
There is, however, an alternative way of looking at an orthogonal linear
transformation. Suppose we leave each point of the plane where it is but
re-name the coordinates by taking a new pair of perpendicular axes, with
the origin still at O, but with the new axes Ox', Oy' making an angle — θ
with the old axes Ox, Oy (Figure 29.5).
Fig. 29.5
Let OP = r, and suppose OP makes an angle α with Ox. Then
Thus,
*' = г cos (« + <?),
x' = r cos α cos 6
= л; cos θ—ysin Θ,
y' = r cos α sin θ + r sin α cos
= л; sin θ+ycos Θ.
In matrix notation, this means
sin(a+0).
;sin0
(Й-
\sin0
cos θ
and we have precisely the orthogonal linear transformation discussed ii
the previous paragraph.
In summary, the linear transformation T, with orthogonal matrix
may be regarded either (i) as mapping each point A into another point A',
where О A = О A' and АО A' = + θ or (ii) leaving each point in its original

FURTHER COORDINATE GEOMETRY [29
position but rotating the axes of coordinates through an angle — θ (i.e.
through θ clockwise).
We shall now show how this new interpretation of an orthogonal matrix
may be employed to find the major and minor axes of a central conic.
Consider first the equation
ax2 + by2 = 1 (a, b not both negative),
which represents a conic with its centre at the origin and its major and
minor axes along the axes of coordinates. If we rotate the axes through
an angle - Θ, that is, if we rename each point by the rule
χ' = χ cos Θ—у sin Θ,
у' = л; sin θ+y cos Θ,
then χ = x' cos θ+y' sin Θ,
у = — χ' sin θ+y' cos Θ,
and the new equation of the conic takes the form
a(x' cos θ+y' sin 0)2 + b(—x' sin θ+y' cos Θ)2 = 1,
or
(acos2e + bsin20)x'2 + '2(a-b)x'y' sin θ cos θ + (a sin2 θ + b cos2 Θ) y* = 1.
We may rewrite this as
a'x'2+2h'x'y' + b'y'2 = 1,
which is the typical form of equation of a central conic with its centre at the
origin.
Now consider the converse problem: given the equation of a central
conic in the form ^ + lhxy +by2 = \,
how can we rotate the axes to obtain the equation in the form
a'x'2 + b'y'2 = 1?
The equation a^ + lhxy+by2 = 1 may be rewritten in matrix form as
or u'Au = 1,
where o=(J) and Α- β J).
The problem thus reduces to that of diagonalizing the matrix A for
then the equation would become, in the new coordinate system,
648

3] CHANGES OF COORDINATE SYSTEM
or λ1Λ;'2 + λ2>''2 = 1.
But A is a symmetric matrix and, by Theorems 28.3 and 28.4, we can find
an orthogonal matrix Ρ such that PAP' = D where
Ai and λ2 being the eigenvalues of A. (Recall that, for an orthogonal matrix,
P"1 = P'. Notice too that, to obtain PAP' in diagonal form (rather than
P'AP), the matrix Ρ is obtained by transposing the matrix with columns
which are the eigenvectors of the matrix A; that is, the eigenvectors appear
as the rows) The fact that Ρ is orthogonal tells us that the new axes, Ox',
Oy' will be perpendicular. The details of the transformation are given
below.
Consider the linear transformation defined by
ν = Pu,
where ν = I , I, the new position vector of the point u = I I. Then
P'v = Р-Ч = u
and the equation u'Au = 1
transforms into the equation
(P'v)' A(P'v) = 1
or, since (P')' = P, and using the results of Ex. 8, Chapter 28,
v'PAP'v = 1.
But PAP' = D, a diagonal matrix and the equation has been transformed
into
<*»(*£)(?)■
or λιχ'*+λ?' = \.
If we define the trace of the matrix A, tr A, as
trA = a+b
(see Exercise 28 (b), Questions 8-10), the characteristic equation of A,
Ι α-λ h I
\ h b-X\ '
may be written in the form
A2-AtrA + detA = 0.

FURTHER COORDINATE GEOMETRY
Thus we have
but
tr A = λ1 + λ2,
detA = AiA2;
trD = Ai + A^
detD = λ^,
and thus both the trace and the determinant of the matrix denning the
conic remain invariant under the given linear transformation.
*Ex. 15. Show how the invariance of the trace and determinant enables us to
deduce immediately the nature (ellipse, hyperbola) of a central conic
ax2+2hxy + by2 = 1.
Ex. 16. Prove that, if Q is any non-singular 2x2 matrix, then
tr (Q^AQ) = tr A and det (Q^AQ) = det A.
Can these results be generalized for higher-order matrices (the trace still being
denned as the sum of the elements in the leading diagonal)?
Example 6. Determine the nature of the central conic
5хг + %ху+\\уг = 1
and find the lengths of its axes.
The matrix A of the conic is given by
Let the eigenvalues of A be Λ1( λ2; then
Λ1+λ2= 5+11 = 16,
λ1λ2 = 55-16 = 39
giving λι = 3, λ2 = 13.
The equation of the conic may thus be reduced to the form
3x'2+13/2= 1.
This is seen to be an ellipse, with major semi-axis 1/^/3 and minor semi-
axis 1/V13.
To draw a sketch of the ellipse, showing it in relation to the original
axes, we need to find the inclination of the axes of the conic to the axes
Ox, Oy. This may be done by obtaining an explicit form for the reducing
matrix P.
Corresponding to the eigenvalue λ1 = 3 we have the unit eigenvector
(_ ι/ /5) > wnile corresponding to λ2 = 13 we have U, /5) ·
650

CHANGES OF COORDINATE SYSTEM
Thus
>-4 D-
and
The transformation ν = Pu
may be interpreted as a rotation of axes clockwise through a
θ = arctan i
y,
xtCx
(Figure 29.6).
/
1 ^4
^__J/^^*
Fig. 29.6
In other words, in terms of the original coordinate system, the major and
minor axes of the conic lie respectively along
x + 2y = 0 and Ix-y = 0.
*Ex. 17. Show that the equation
f+ihxy+by* ■■
(i) represents an ellipse if h2-ab < 0, and a > 0;
(ii) represents a hyperbola if h2-ab > 0;
(iii) represents a rectangular hyperbola oi a+b = 0;
(iv) represents a pair of parallel straight lines if h2 = a
Ex. 18. Let us write the conic of Ex. 17 in the form
x'Ax = 1,
И·»©'
u is a unit direction vector and Ρ is a point with position vector p. The point
with position vector p + λιι lies on the conic if
(р + Ли)'А(р + Ли) = 1;

FURTHER COORDINATE GEOMETRY [29
show that this reduces to
A2u'Au + 2Au'Ap + p'Ap = 1.
If Ρ is the mid-point of a chord parallel to u, prove that
u'Ap = 0;
what does this tell us about the vector Au ?
The axes of a central conic (other than a circle) may be defined as the (unique)
pair οι perpendicular conjugate diameters: use the results proved above to deduce
that u is in the direction of an axis if there exists a number к such that
Au = ku.
Interpret your results in terms of eigenvectors.
Exercise 29(b)
1. Obtain the equation of the sphere:
(i) with centre (0, - 1, 0) and radius 1;
(ii) with centre (2, - 3, - 1) and radius 5;
(iii) with centre (a, b, c) and radius J(a2 + b2 + c2).
2. Find the centre and radius of the sphere:
(i) withequationx!+y! + z2-6;c-2>' + 2z-5 = 0;
(ii) with equation x2 + y2 + z2 + Ax -Sy-2z- 60 = 0;
(iii) with equation x2+y2 + z2-2ax-2by + 2cz + 2bc-2ca+2ab = 0.
3. Find the equation of the sphere:
(i) through the points (4, 4, 4), (5, 6, 1), (0, -4, 2), (7, 3, 2);
(ii) through the points (2, 4, 4), (5, 1, 4), (3, 4, 3), (-3, - 1, 0);
(iii) through the points (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c).
4. Find the centre and radius of the circle:
(i) with equations x2+y2 + z2-12x + 2y-10z+A4 = 0, x-y + z = 6;
(ii) with equations x2+y2 + z2-lSx- 10y+ 12z- 55 = 0, 2x+3y-Az = -1.
5. The position vectors of points A and В relative to some origin О are
respectively a and b. Show that the locus of a point Ρ which moves in space in such
a way that PA2 + PB2 = d2, where d is a constant, is a sphere with its centre at
the mid-point of AB. Find the radius of the sphere in terms of a, b and d. What
can you say about the sphere if d = |a- b| ?
6. With the notation of Question 5, show that the locus of a point Ρ which moves
in space in such a way that РАг+РВ2 +РС2 = d2, where С is a point with
position vector c, is a sphere and find its centre.
Can you generalize this result in any way?
7. Show that the line г = Au, where u is a unit vector, cuts the sphere
(r-3i-2j-2k).(r-3i-2j-2k) = 6
at two points with parameters given by the quadratic equation
A2-2Au.(3i+2j + 2k)+ll = 0.
652

EXERCISE 29
Deduce that the line through the origin in the direction of the vector i + j + 3k
touches the given sphere.
Prove more generally that the line through the origin in the direction of the
vector /i + wj + ик is tangential to the sphere if and only if
(3/+2»г + 2и)2 = \\{P + m2 + n2),
and hence show that the equation of the cone with vertex the origin which
circumscribes the sphere is
(3x+2y + 2z)2 = lHx'+y + z2).
8. Find the equation of the cone obtained by rotating the line
x/p = y/g = z\r
about the line x/a = y/b = z/c.
9. A right-circular cone of semi-angle α has its vertex at the origin and contains
lines in the directions i, i + j and i +j + k. Show that
sec2 α = 9-2V2-2V6.
10. Show that the equation
yx+zx+xy = 0
represents a cone, vertex the origin and semi-angle arctan V2· What is the
curve of intersection of this cone:
(i) with the plane x+y+z = 1;
(ii) with the plane x-2y + z = 0;
(iii) with the plane x- 2y + ζ = 1 ?
11. A circle, S, is defined by the vector equation
(r-a).(r-a)-c2 = 0, r.n = p.
Show that any sphere passing through the circle S has vector equation
(г-а).(г-а)-с2 + Л[г.п-р] = О
and deduce that, if В is a point not on the plane r.n = ρ and if OB = b, then
the vector equation of the sphere through S and containing the point В is
(Ь.п-р)[(г-а).(г-а)-с2]-[(Ь-а).(Ь-а)-с2](г.п-р) = О.
Find the centre and radius of the sphere which passes through the point
(1, -1, 1) and contains the circle.
x2+y2 + z2= 4, x+2y+z=l.
12. A is the fixed point with position vector ak, and L, Μ are variable points
with position vectors λΐ, μ) respectively, where λ, μ are scalar parameters. If
Ρ is a point such that the angles LPM, MPA, APL are all right angles show that,
whatever the values of λ and μ, Ρ lies on a sphere, centre A and radius a.
13. Find the equation of the tangent to the sphere
x2+y2 + z2- 10x-2y-12z+ 35 = 0
at the point (4, 2, 1).
Prove that the plane x+y + z = 3 is also a tangent to the sphere and find the
point of contact.
653

FURTHER COORDINATE GEOMETRY [29
14. Find the values of к for which the plane
4x+2y+3z = к
touches the sphere „ „ „ ,„
x2+y*+z*-10x-6y-10z+30 = 0.
15. The point В lies outside the sphere (r-a).(r-a)-c2 = 0. If OB = b,
prove that the length of the tangent from В to the sphere is (b - a). (b - a) - c2.
Prove that the set of points from which the tangents to two non-intersecting
spheres are of equal length is a plane which is perpendicular to the line of centres
of the two spheres.
Deduce that the set of such points for three non-intersecting spheres is a line
and that, given four non-intersecting spheres with non-coplanar centres, there is a
point В from which the tangents to the four spheres are of equal lengths.
Explain how the last result can be used to show that a sphere may be
circumscribed about any tetrahedron.
16. Describe the curve of intersection of the surface with equation
хг + уг-гг = О;
(i) with the plane ζ = 1;
(ii) with the plane χ = 1.
What surface does the equation represent?
17. Show that the parametric equation of the helix of pitch 8 which lies on the
surface of the cylinder x2 + z2 = 1 is
40
r= cos0i+— j + sin#k.
Find the coordinates of the point in which the helix cuts the plane
J2x-y+2j2z = 0.
Show also that the helix cuts the plane x+z = 0 at an infinite number of
points, and explain this geometrically.
18. Prove that the mid-points P, Q, R, S, T, U of the edges ВС, CA, AB, AD,
BD, CD of a tetrahedron ABCD, in which the opposite edges (ВС, AD),
(CA, BD), (AB, CD) are perpendicular pairs, lie on a sphere. Is the converse
result, that if the points P, Q, R, S, T, U lie on a sphere then the opposite pairs
of edges are perpendicular, true or false?
If the sphere cuts ВС again at P', prove that ВС is perpendicular to the plane
P'AD.
19. Two spheres, centres Аг and Аг and radii cx and c2 intersect in a circle of
radius d. If АгАг = /, where Ρ = c\ + <%, prove that d = c^/Zand find the ratio
in which the centre of the circle divides the line АгАг.
20. Prove that there is one and only one sphere which contains a given circle
and passes through a given point not in the plane of the circle. Two circles, not
in the same plane, meet at points A and B. A plane meets one of these circles
at С and D and the other circle at Ε and F. Prove that:
(i) the points C, D, E, F lie on a circle;
(ii) the lines AB, CD, EF are either concurrent or parallel. (O & C)
654

3 + Г, + й=1·
EXERCISE 29
21. Show that, if the ellipse (x2/a2) + (y2/b2) = 1 lying on the plane ζ = 0 is
rotated through four right angles about the χ axis, the equation of the surface
obtained is „
x2 y2 z2
α2 ό2 ό2
The surface thus obtained is called an oblate spheroid if a < b and a prolate
spheroid if a > b. Both surfaces are particular cases of the ellipsoid.
2 y2 Z2
By considering sections of this ellipsoid with planes χ = ρ, у = q, ζ - r,
describe the shape of the general ellipsoid and in particular, show that it is a closed
surface.
22. Describe the nature of the hyperboloid of one sheet
a2 + b2 с2
and of the hyperboloid of two sheets
й2 о2 С2 '
23. Describe the nature of the elliptic paraboloid
cP Ь* с
and οι the hyperbolic paraboloid
α2 ό2 _ с '
24. Prove that the hyperboloid of one sheet (see Question 22)
xl+yl-t = ι
a2 b2 c2
contains the line determined by the two planes
?+5= λ(ΐ+^) and λ(^-ή= 1-
Call this line a λ generator. Prove also that the surface contains μ generators of
the form
-ί-Ή-'ίΝ)-1-
pair of λ generators and any ρε
lut that each λ generator meets
of the form
/α(λ + μ) Κ1-λμ) <*λ-μ)\
\ 1 + λμ ' 1+λμ ' 1 + λμ ) '
Prove also that any pair of λ generators and any pair of μ generators are
necessarily skew lines but that each λ generator meets every μ generator at a
point with coordinates of the form

FURTHER COORDINATE GEOMETRY [29
(Notice that we have obtained a parametric representation of the hyperboloid
of one sheet in terms of the two parameters λ, μ.)
25. (See Questions 23, 24.) Prove that the hyperbolic paraboloid
a2 b2 с
contains a system of λ generators and a system of μ generators with the
properties that any pair of λ generators are skew, any pair of μ generators are skew, but
each λ generator meets every μ generator in a point of the form
(α(λ + μ), Κλ-μ), Ιολμ).
26. Determine the nature of the central conic
7x*-8xy+13y* = 150
and find its eccentricity and the equations of its axes.
27. Determine the nature of the central conic
6x* + 5xy-6y2 = 169
and find its foci and the equations of its axes.
28. Show that the equation
x* + 6xy+y* + 6 = 0
represents a hyperbola and find its eccentricity and the equations of its
asymptotes.
29. Two lines, h and /2, intersect at an angle a. Find the locus of the centre of а
sphere of given radius which touches both lines.
656

Revision exercise С
1. The polynomials ^(л:), h(x) both leave the same remainder on division by
(x—a). Prove that the polynomial
xgQc)-ah(x)
is divisible by (x—a).
The polynomial/(л:) (degree > 4) leaves a remainder of rx+s on division by
(х-ЬУ. Prove that f"(x) is divisible by (x-b).
2. If ζ Φ 1 and -
show that \z\ = 1 if and only if Re (w) = 0.
3. If a < b => a" < bn for all α, ό e Λ where и е Z, what further conditions does η
satisfy ? Are these further conditions still necessary if a, b are restricted to positive
real numbers? (S.M.P.)
4. The function/: R -» R is defined by
№ = χ*-2\χ\.
Sketch the graph of/and find what values of л: are invariant under/.
5. Solve the equation 3 sin (9 + 2 sin (60° - Θ) = 2
for values of θ lying between 0° and 360°.
6. A random number table consists of a succession of digits each chosen at
random independently from the set 0, 1, 2,..., 9. Two successive digits are taken
from such a table; show that the probability that their sum is 9 is 1/10.
Four successive digits are taken from the table. Calculate the probability, p,
that the sum of the first two equals the sum of the third and fourth. (M.E.I.)
7. Prove that ,- _ n,
a + ar + ar2 + ...+arn-1 = ^~-^,
where гФ1,
Prove that in general
(1 + x+x2 + ... + x*") (1 - x+x2 -... +x2n) = l+x2 + xi+...+xi\
State any values of χ for which your proof does not apply and obtain the
appropriate results in each of these cases. (O & C)
8. Prove that, if the points О, А, С are non-collinear, then the position vector
with respect to О of any point in the plane О AC may be expressed uniquely in
terms of a = OA and с = ОС. (over)
657

REVISION EXERCISE С
О ABC is a parallelogram, L is the mid-point of AB and Τ is the point which
divides OB in the ratio 3:2. CT meets О A at X and LT meets ОС at У. XY
meets 05 at Z. Find the ratio OZ.ZB in which Ζ divides ОЯ.
■π·
prove by induction that
An = i(6n-2") (A-21)+ 2*1.
form the products MQ and NQ. What law, which is true for products in the
algebra of real numbers but which is not true for products of matrices, is
exemplified by this? (S.M.P.)
11. Express ^
* = гЩ and q = Y^b
in the form x+jy, where χ and у are real.
Find also the modulus and the argument of ρ and of q. Hence write down the
argument of ρ +q. (M.E.I.)
12. Express in partial fractions:
6-Зл: 2х-3 Зл:2-7л: + 3
W (x-D(x + 2)' W х(1+х2У (Ш) (1-х)2 (1-2*)'
13. Two concentric circles have radii 10 and 15 cm. If a point is taken at random
in the annulus formed by them (all points of the annulus being equally likely) find
the expected distance of the point from the centre of the circles. Explain why
you would expect your answer to be greater than 12-5 cm.
14. Ρ is the point (at2,2at) on the parabola^2 = 4ax and OQ is the chord passing
through the origin О and parallel to the tangent at P. Find the coordinates of
the point of intersection, R, of the tangents at Ρ and Q. Give a reason to show
that the locus of R is another parabola.
If S is the mid-point of О Q, prove that PSR is a right-angled triangle, and that
the area of triangle PQR is \аНг. (О & С)
15. The sets А, В, С, D are denned as follows:
A = {xeR: -2< x^l}, В = {xe R: -1 ^ χ < 2},
C= {xeR:x> 1}, D = {xe R: χ *ζ -2}.
Express in the form {:}, the sets:
&A(]B; (ЩЛ'ПС; (iii) А П C; (iv) Β' η С П D';
(ν) (Α η Β) υ (Β η С')·
Express the sets
Е= {xeR-.x <-l} and F={xeR:x<2}
in terms of unions and intersections of the sets А, В, С, D and their complements.

REVISION EXERCISE С
16. OAB is a triangle, right-angled at О; О A = a, OB = ό. Points X, У are
taken on OA, OB (or on these lines produced) in such a way that
OX+OY= a+b.
A У and BX meet at P. Find the locus of P.
17. Assuming a population in which equal numbers of births occur in each of
the twelve months of the year calculate the probabilities that, of four persons
taken at random:
(i) no two will be found to have birthdays in the same month;
(ii) exactly two will have birthdays in the same month and the others in
different months. (M.E.I.)
18. The polynomial ax2 + bx+c can be represented by the vector
Θ-
Find a 3 χ 3 matrix Μ which premultiplies this vector to give the vector
representing the derivative of the polynomial. Check that M3 = 0 and comment on
the significance of this result. (S.M.P).
19. Prove that
1*_2*+3*-4*+... + (-1)п-1и* = (-1)"-Чи(и3 + 2и2-1).
20. Find all angles in the interval 0 =ξ θ =ξ 2π satisfying the equation
sin 20 + cos (0-;|π) = 0.
21. If к is very small, show that χ = 1 - \k is an approximation to a root of the
equation **4-*·-ι = ο,
and find approximate values for the other two roots.
22. Solve the inequality , _..., ..
(x-7)(*+3)<1-
23. Express Αχ_χ
"(2x-l)(3x-l)
in partial fractions and deduce a quadratic approximation for^ in the
neighbourhood of the point (0, -1).
Deduce that the line x+y = -1
is the tangent to the graph of у at the point (0, -1) and that, near this point, the
curve lies in the region „ ^
{(x,y):x+y > -1}.
24. a,p,q, r are four non-zero complex numbers such that
p+q + r = a, p2+q2 + r2 - a2, pqr = a*.
Find, in terms of a,
(i) \{a-p){a-q){a-r)\; (Щ l/p+l/q + l/r.

REVISION EXERCISE С
25. Three points P,Q,R vary in position on the surface of a fixed sphere in such
a way that PA>+QA> + RA>
is constant, where A is a fixed point. Find the locus of the centroid of the triangle
PQR.
26. If θ is real and ζ = cos Θ+j sin Θ, find
1 , 1
zn-\—- and zn—-.
zn zn
Determine a, b, c, d such that
27 cos3 θ sin3 θ = a sin 80 + b sin 6Θ + с sin Λθ + d sin 20. (M.E.I.)
27. Show that the cubic equation
2x?+3x*-12x+k = 0 (ureal)
has three distinct real roots if and only if
- 20 < к < 7.
Solve the equation completely if к = -13.
28. Two real numbers are taken at random from the set of real numbers
{xeR:0^ x^ 1}.
Find the probability:
(i) that the square of their sum is greater than 1;
(ii) that the sum of their squares is greater than 1.
29. Show that the point
(β(1-ί2)/(1+ί2), 26ί/(1 + ί2))
χ* ν2
lies on the ellipse — + -^ = 1
for all values of t.
Deduce that, if the line lx+ my — 1
is a tangent to the ellipse, then
IW + mW = 1.
30. Prove that Р + 22+... + и2 = £и(и+1) (2и+1).
If Sl= 12 + 42 + 72 + ... + (Зи-2)2,
and j2 = 22 + 52 + 82 + ... + (Зи-1)2
show that j2 — Ji = Зи2.
and Ji+j2 = 6n3-n.
Hence, or otherwise, find Ji and s2. (O & C)
31. Prove that the medians of a triangle ABC are concurrent at a point G.
О is an arbitrary point in the plane of triangle ABC and L, Μ, Ν are the
midpoints of ВС, CA, AB. Prove that the lines through L, Μ, Ν drawn parallel to
О A, OB, ОС respectively are concurrent at a point H.
Prove also that the line GH passes through О and determine the ratio OG: GH.
660

REVISION EXERCISE С
32. It is observed that
1 + 2 = 3
4+5+6 = 7+8
9 + 10+11 + 12 = 13+14+15
16+17+18+19 + 20 = 21+22 + 23 + 24.
Enunciate a general result of which the above are particular cases and prove
your conjecture.
33. For what values of θ in the interval 0° < θ < 360° is the expression
6 cos2 0-cos#
less than one ? Give your answers to the nearest tenth of a degree.
Answer the same question for the expression
6 sin2 θ — ύτιθ.
34. Express
(x-3)(x-2)
in partial fractions.
If л: is so large that (l/л:)3 and higher powers of 1/x may be neglected, show that
the given expression is approximately equal to (_хг + х- 1)1 χ3.
35. Show that there is just one real value of the constant λ for which the equations
λχ+y + z = 2x + Xy+3z = 4x—2y + Xz = 0 have a common solution other than
χ = у = ζ = 0. Find, for this value of λ, the solution of these equations which
satisfies also the equation x+y + z = 3. (M.E.I.)
36. A and В play a match of five games, each of which must be won or lost. In
each of the first three games the probability that A will win is 2/3 and in the
remaining two games the probability is 3/4. Find:
(i) the generating function for the probability that A wins r games;
(ii) the probability that A will win the match;
(iii) the average number of games that A will wia (O and С modified)
37. Prove that, if A, B are two 3x3 matrices, then
(BA)' = A'B'.
What is the transpose of a 1 χ 1 matrix ?
Let Τ, Τ be two linear transformations of three-dimensional space into itself,
with matrices A, A', where A' is the transpose of A. Prove that, if
T(P) = Q and T'(P) = R,
then OP is perpendicular to QR.
38. Show that the points O(0, 0, 0), /1(0, 2, 2), 5(2, 0, 2), C(2, 2, 0) are the
vertices of a regular tetrahedron. (over)
10 PPMll 661

REVISION EXERCISE С
L, Μ are the mid-points of the edges О A, OB respectively. The tetrahedron
О ABC is now cut through by a plane CML. Find the cosines of the acute angles
between the plane CML and
(i) the plane OML; (ii) the plane OCM.
39. Prove that, for any complex number z, zz* = \z\2.
The reflection of the point zt in the line θ = α through the origin in the Ar-
gand diagram is z2 and the reflection of z2 in the line θ = β through the origin is
z3. Prove that , „ . . „ .
z2 = z1{ooi2a+ism2a)
and that z3 = zjcos (2/?- 2a)+j sin (2/?- 2a)].
40. Prove, by induction, or otherwise, that
Σ r(r+l) (iy-1 = 16-(и2+5и + 8) (i)«-i. (M.E.I.)
41. If OA = a, OB = b, prove that the position vector, relative to O, of the
mid-point of AB is K» + b). Deduce that the lines joining the mid-points of
opposite edges of any tetrahedron are concurrent.
If ABCD, A'B'C'D' are any two parallelograms, not necessarily coplanar,
and if the mid-points of AA', BB', CC, DD' are respectively W, Χ, Υ, Ζ, prove
that WXYZ is a parallelogram.
42. Give rough sketches of the curves:
(i) у = sinx; (ii) у = sin(x+|77); (iii) у = l + 2sin(x+3-7r).
How many real roots has the equation
x= l + 2sin(x + -37r)?
43. Sketch the graph of
* x2-5x-W
For what values ofxis^ < — 1 ?
44. Find the coordinates of the point of intersection, P, of the plane
х+Ъу+z = 0,
and the line joining ДО, 1, 15) and 5(3, - 5, 3).
Find the ratio АР: РВ.
45. The line / is a tangent to a parabola with focus F and N is the foot of the
perpendicular from Fto /. Prove that N lies on the tangent at the vertex and that
FN produced meets the directrix in a point Μ such that FN = NM.
The triangle ABC is right-angled at B. It is reflected in a line m in such a way
that the image of С lies in the original line of AB. Prove that m is a tangent to a
parabola with focus at С and directrix AB. (M.E.I.)
46. Interpret geometrically the transformations performed by premultiplying
the position vectors of points in a plane by each of the matrices
. № 0\
■С'^Э-'ЧЗ·
State in the form of a single matrix each of the expressions R", S", T", where
ι is a positive integer, and prove one of these results by induction. (M.E.I.)

REVISION EXERCISE С
47. Show that there are two complex numbers ζ such that
|z-2-j| = 1
where a, b and с are non-zero constants, are α, β and the roots of the equation
ca? + 2bx + c = 0
are γ, δ. Without solving either equation, show that the equation whose roots
are ουγ+βδ and οώ+βγ is
a3x2-2ab*x + c(5b2-4ac) = 0.
Show also that if this last equation has equal roots then one and only one of the
original equations has equal roots. (M.E.I.)
49. Show that:
w K L -*· + **+!'
(ii) (,l+x)*<l+x/3 (x>0),
and deduce that, when χ > 0,
(1+XJ ^27 + 9^ + ^·
Hence show that the error in taking (1 + x)* as 1 + x/3 is 1
9л:2 + χ3
81+27л: + Зл:а"
ess than
(M.E.I.)
50. A machine produces components to any required length specification with a
standard deviation of 1-40 mm. At a certain setting it produces to a mean length
of 102-30 mm. Assuming the distribution of lengths to be normal, calculate:
(i) what percentage would be rejected as less than 100 mm long;
(ii) to what value, to the nearest 001 mm, the mean should be adjusted if this
rejection rate is to be 1 %;
(iii) whether at the new setting more than 1 % of components would exceed
107 mm in length. (M.E.I.)
51. Prove that, if z, w are complex numbers such that
|z-2| = \w + 2\ = 1,
then -%π =ξ arg (z-w) < %π.
Find the limits between which \z—w\ must lie.

REVISION EXERCISE С
52. If A = ί! 3X u = P\, A«u = П prove that, for large n, xn « yny/3
and interpret your result geometrically.
53. Prove that, for any positive integer r,
tan (2Г-Щ = cot (2τ~1θ)-2 cot (2rff).
Hence evaluate Σ 2r_1 tan (2r_10).
54. Two players stake £1 each in a game in which eight counters are tossed on
the ground. The counters are painted black on one side and white on the other.
If the number of blacks showing is odd the thrower wins the other stake, if all
counters are of the same colour he wins a double stake, and otherwise he loses
his stake. Calculate the expected gain of the thrower and state what assumptions
you make. (M.E.I.)
55. Prove that „
Zr! = \n(n+\)(2n+\).
The χ axis from χ = 1 to χ = 2 is divided into η equal intervals by the points
(1,0), (l + ^.o), (l + ^.o), ..·, il+^,oj, (2,0).
Rectangles with sides parallel to the axes are drawn on these intervals with one
vertex on the curve у = χ2 and the rest of each rectangle between the curve and
the χ axis. Show that the sum of the areas of these rectangles is
1+1
-9(Ю-
Show further that as η increases this sum increases but remains less than J.
56. Show that, if the roots α and β of the quadratic
ax* + (b + c)x + d = 0
are unequal, the equation
aun г^+1 + bun + cun+1 + d = 0
can be put in the form -^^—- = λ ——-,
where λ is given by the equation
57. If a/b - 1 is a small positive quantity not greater than 10"^, show that
/д\* За+ 6
W a + 3b
is not greater than 10_3iv_1.
664

REVISION EXERCISE С
58. The linear transformation, T, has matrix (relative to i, j, k)
.π -: -:v
\ 0 0 1/
Show that Τ leaves every point of the plane
2x+3y + z = 0
fixed and that, if Π is any plane containing the line of intersection of the two planes
x+y = 0, ζ = 0
then Π is mapped into itself by T.
59. OAB is a fixed triangle and points X, У are taken on О A, OB (or these sides
produced) in such a way that AB and XY are parallel. A Y and BX meet at P.
Show that, for varying positions of X, Y, the locus of Ρ is ОС, where С is a vertex
of the parallelogram OACB.
60. Explain what is meant by the set Rx R.
The function /; R χ R -» R is denned by the equation
Ax,y) = x+y.
Find the image, under /, of:
(i) {(x,y)eRxR:2x+y= 1}; (ii) {(x,y) e Rχ R: x*+y* = 1}.
61. Find the modulus and argument of 1 +cos θ+j sin Θ, where —π<θ<π,
and hence or otherwise express (1 +cos θ+j sin Θ)" in the form a+jb, where η is
a positive integer.
By writing cos Θ+j sin θ = ζ, and using the binomial expansion, prove that
*0—Θ
1+ COS0+ COS20+...+COS«0 Ξ (2 COS 45)" COS in0,
and obtain a similar identity with sines on the left-hand side. (O & Q
62. If the equation x3 + ax2 + bx + c = 0
has the property that one root is equal to the product of the other two, prove
that (b-cy + cia-iy = 0.
If a = f- and b — \, find the two values of с which satisfy this condition, and
solve the resulting cubic equations. (O & C)
63. Prove that the ellipse „ „
and the hyperbola xy = c2
touch each other if 2c2 = ab. Is the converse true?
Let D be one of the points of contact in the case 2c2 = ab. Show that if the
common normal at D meets the hyperbola again at A and the coordinate axes аХР,
Q, then the mid-point of PQ coincides with the mid-point of AD. (M.E.I.)
665

REVISION EXERCISE С
64. A probability density p(x) is given by
t N fC*(4-x) (0^x^4),
(О (л: < 0, 4 < χ),
where С is a positive constant. Find the value of С and sketch the distribution.
Explain why the standard deviation of this distribution is the same as the
standard deviation of the distribution with probability density
Pix)
jC(4-x2) (|x| <2),
" to (|x| > 2).
For the first distribution, calculate the mean, the standard deviation and the
mode. Calculate also the probability that a value of the variable χ taken at
random will lie further than one standard deviation away from the mean. (M.E.I.)
65. If a circle passes through the origin, show that its equation, written in
complex form, is zz*+g*z+gz* = 0, where ζ = x+jy, z* = x—\y, and g is a
complex constant.
If a variable point Ζ on the circle is transformed into a point W by means of
the equation w = a2/z, where α is a real constant, find the equation satisfied by w.
What geometrical locus does this equation represent? (O & C)
66. By solving the set of equations
-x1+x2 = b,
Злг!+2л:2 + x3 = с
or otherwise, find the ii
\0 0 1/
find B_1 and also, by any other method, the inverse of AB. State a set of
equations which this inverse (AB)-1 would help you to solve. (M.E.I.)
67. An unbiased six-sided die is thrown. If a six is thrown the die is thrown a
second time and the scores for the two throws are added; find
(i) the chance of a score of more than eight;
(ii) the average score.
If the die is thrown again whenever a six has been thrown, find the average
number of throws and the average total score. (O & C)
68. The position vectors relative to the origin О of the points Α,Β,Ρ are
respectively a, b, p. X is the point of trisection of AP nearer A, and Υ is the mid-point
of BP. Ζ is a vertex of the parallelogram PXZY. Write down, in terms of a,
b, p, the position vectors of Χ, Υ, Ζ.
666

REVISION EXERCISE С
If А, В are fixed points on a unit sphere, centre O, and Ρ is a variable point
on the sphere, deduce that the locus of Ζ is a sphere and find its radius and the
position vector of its centre.
). Prove that the lines
K*- 3) =
x+2 =
K*+2) =
y + A
Ky-3)
-b
= z-3,
= z+3,
= iz
are skew and find the equations of the line through the origin which intersects
all three lines.
70. UV is a diameter of an ellipse and P, Q are points on the ellipse. PU, QV
meet at X and PV, QU meet at Y; prove that the chord defined by XY is
bisected by UV.
71. Prove that, if ξ, is any root of the equation
z"=(z-2j)",
thenlm© = 1.
Draw a sketch to show the position of the roots in the case η = 3 and show
how they are related to the equations
(i) z3 = (z + 2j)*; (ii) z*=(z + 2)3.
72. Express the vector i-4j- 5k as
(i) the sum of three vectors in the directions i, i+j, i + j + k;
(ii) the sum of two vectors, one in the plane 2x + 4y + 3z = 1 and one
perpendicular to this plane.
73. A single die is thrown six times. Find the probability that:
(i) exactly one six is thrown;
(ii) at least two sixes are thrown;
(iii) each face of the die turns up once.
Find the probability that, when a die is thrown repeatedly, a six appears for
the first time at the rth throw. Find the least value of r such that there is more
than a 50 per cent chance of obtaining a six on the rth throw or earlier.
Write down expressions, in terms of ρ = f and q = f, for the expectations
Sir), £(r*) and calculate the numerical values of these expressions. (M.E.I.)
74. A transformation is defined by
: Ш
Show that the set of planes perpendicular to the у axis maps into a set of parallel
lines. What is the direction of these lines ? Which of these planes is mapped onto
a line through the origin? (S.M.P.)
75. In the binomial expansion of (l-л:)*, where χ is positive and less than 1,
show that the ratio of the term in x"+1 to that in xn is

REVISION EXERCISE С
By comparing the sum of all the terms after the first three with a suitable
infinite geometric series, prove that the magnitude of the error involved in
neglecting these terms is less than (5л:3/81) (1 -л:)-1.
Show that, if χ =ί 0-1, this error is, in magnitude, less than 0-00007 and hence,
using the binomial expansion, evaluate (0-9)* correct to three decimal places.
76. Prove that the points with coordinates (0, -2, -\\ (1,1,1) (5, 3, 4) and
(8,2, 5£) are the vertices of a trapezium in which the non-parallel sides are equal
in length. (S.M.P.)
77. The fixed point С and a variable point Ρ have position vectors с, г
respectively in three-dimensional Euclidean space. Describe in geometrical terms the
locus Г denned by г. с = Irl |c| cos α
where α is a given acute angle. If С is the point (a, b, c) and Ρ the point (x, y, z)
express the equation of Γ in cartesian form and show that, for all real numbers λ,
{χ, y,z)eV^. (λχ, Ay, λζ) e Г.
Taking the special case with С the point (4, 0, 3), cos α = 4/5, show that Γ
meets the plane ζ = 6 in a parabola and that the sphere |r- c| = 3 touches the
plane at the focus of this parabola. (M.E.I.)
78. Obtain a quadratic approximation for the expression
Vd+x)/(l-2x)
in the neighbourhood of χ = 0.
Write down the equation of the tangent to the curve
y= V(l+*)/0-2x)
at the point (0,1) and draw a sketch showing the relative position of the curve
and the tangent at this point.
79. ABCDA'B'C'D' is a parallelepiped, with faces ABCD, A'B'C'D' and
parallel edges AA', BB', CC, DD', etc. The mid-points of CC, AB', B'C', CD' are
respectively W, X, Y, Z. Prove that the plane XYZ cuts A W externally in the
ratio 1:3.
80. Find the five roots of the equation
z5+l = 0,
and plot their positions on an Argand diagram.
A regular pentagon ABCDE is inscribed in a circle of unit radius. Prove that,
if Ρ is any point of the circle, then PA .PB.PC.PD.PE ^ \j2.
81. If A is a 2 χ 2 singular matrix, prove that there is a number a such that, for
any integer к > 0, д* _ at-i\
Find α if A is the matrix
В is the matrix
prove that
for any positive integer n.
668

REVISION EXERCISE С
By considering the product (1 - x)" (1 + x)n,
prove that Д,(-1)'[ОТ = C«)"
83. A boy and girl play a game of' spotting' approaching red cars when travelling
along a road on which 25 % of the cars are red. If a car is red the first one to
'claim' it gains one point (a claim is always made by either the boy or girl for a
red car, but never by both simultaneously). If a car is not red it is found that on
half the occasions either the boy or the girl (but not both) makes a claim, for which
there is a penalty of two points. The first to be five or more points ahead is the
winner.
The events 'a claim by the boy' and 'the car is red' are independent and the
probability that he makes a claim on a car is 2/5.
(i) Draw a tree diagram in which the first two branches lead to the events
'red' and 'not red', and mark on the secondary branches the probabilities of
claims by the children.
(ii) If it is known that a claim has been made for a car which is not red, what
is the probability it was made by the girl ?
(iii) If the score is 8-5 in favour of the girl, show that the probability that she
will have won by the time two more cars have passed is 0-48. (S.M.P.)
84. Find the solution set of the equations
in the following cases :
(i) a = 9, b = - 1; (ii) a = 8, b =
For what value of a does the equation
7 J not have a unique solution;
Θ-ί
J 7 j not have a unique solution?
w
85. Use de Moivres's Theorem to prove that
cos ηθ = Σ(-1)' Γ" ) cos"-21· θ sin2' Θ,
and deduce that cos ηθ may be expressed as a polynomial in cos Θ.
Write д 2, n
cos ηθ - Σ ar cosr Θ.
By taking 0=0 prove that an = 2"-1. What is a„ if η is even ?

REVISION EXERCISE С
86. Find and illustrate graphically the set of points (x, y) that satisfy
simultaneously the inequalities
(x- l)3 < x- 1,
87.
M =
\\
By considering the matrix product M'M, or otherwise, prove that
I 2 a+a2 b+b2 I
a+a2 a2 + cfi ab + a2b2 = 0.
| b + b2 ab+cPb2 b2 + b*
Evaluate the determinant
x+y
x2+y2
2xy
x+y
2xy
x2+y2
88. Describe geometrically the sets of points in the Argand diagram defined by
(i) |z-l| = |z+j|; (ii) |z-l| = 2|z+j|;
(iii) |z-l| + |z+j| = 2; (iv) Re(z+1) = |z+j|.
89. The points P(ca,c/a) and Q (cfi.c/β) lie on the rectangular hyperbola
xy = c2, and (X, Y) is the mid-point of PQ. Find Xand Yin terms of c, α and
β. Write down a quadratic equation in t whose coefficients involve only c, X
and Υ and whose roots are α and β.
Determine what points (X, Y) can arise as mid-points of real chords of the
hyperbola. On a rough sketch shade in the regions they occupy. (O & C)
90. Show how the matrix
/cos 2
-cos 20/
can be associated with reflection in the line у = χ tan 0.
If В is the matrix associated similarly with reflection in the line у = х^З,
find the values of θ for which AB = BA.
Describe how these values could be predicted by geometrical argument
(M.E.I.)
91. A rectangular box has faces OABC,0'A'B'C and OO', etc. vertical. О А = 2
units, ОС = 1 unit, OO' = 1 unit. Taking О as origin and axes Ox, Oy, Oz along
О А, ОС, OO', find the equations of the planes OA'C, ACM, where Μ is the
mid-point of B'C Hence show that
(i) the line of intersection of these planes is parallel to the face О ABC;
(ii) the cosine of the acute angle between planes is 7V6/18.
92. А, В are two fixed points with position vectors a, b relative to some origin O.
Χ, Υ are two variable points with position vectors x, y. Show that, if
x+y = a and χ л у = b,
then Ο, Χ, Υ, Α are coplanar and OB is perpendicular to this plane. Hence find
670

REVISION EXERCISE С
the most general solution, in terms of a, b and a scalar parameter, of the vector
equations
x + y = а, хлу = b.
93. Prove that the matrix
is orthogonal and interpret Ρ as the matrix of a linear transformation. Show that,
if χ is very small, the matrix
Q=L 1 0
\0 0 1/
represents a small rotation, if we ignore powers of χ higher than the first. If SS
is a matrix all of whose elements are very small, and if
1+SS
is orthogonal, prove that the matrix <SS is skew-symmetric.
94. Two men, A and B, play a match consisting of separate games, the probability
of A's winning a game being ρ and the probability of B's winning being q.
(p+q is not necessarily equal to one.) They start the match with η counters each
and the winner of each game receives a counter from the loser; the first player to
win all In counters wins the match. If the probability that, when A has к counters
he will eventually win the match, is denoted by uk, prove that
(p+q) "к = puk+1+quk-!.
What is the initial probability that A will win the match?
95. Prove that, if и is a positive integer, then
0< (2-V2)"< 1.
Use the Binomial Theorem to deduce the existence of an integer m such that
(2m- 1) (2-V2)" < 2" < 2m(2-J2)n.
96. In the complex equation
zn+an_1z»-1+a„_„z«-a+...+a1z+l = 0
it is given that the complex numbers α„_1; α„_2,..., аг satisfy
K_!| < 1, |a„_.| < 1, ..., k| < 1.
Show that any root of the equation in the complex plane must lie in the annular
regi°n }<И<2. (C.S.)
97. The expressions Еъ Е2, E3 are defined as follows:
£Ί = x+y-2z-3,
E, = 2x-3y-4z + 4,
E3 = 3x + 5y + 3z - 5. (over)

REVISION EXERCISE С
If λ, μ are two numbers, explain the geometrical significance of the equation
Ε!+λΕ2 + μΕ3 = 0.
Find the equation of the plane which contains the line
(x+l) = -7(y-l) = -7(z-l)
and passes through the point of intersection of the planes £Ί = 0, Ег = 0
Ε, = 0.
E3 = 0.
98. Show that the equation
х*-Ъх2-Ъх+\ =
has real roots between 0 and 1 and between 2 and 3. Find the larger of these roots
correct to three significant figures and check that three figure accuracy has been
achieved. (O & C)
99. The plane Π is said to be invariant under the linear transformation Τ of three-
dimensional space into itself if every point of Π maps into another (possibly the
same) point of Π.
Find the equations of the planes through the origin which are invariant under
the linear transformation Γ with matrix A (relative to the usual base vectors) given
by
-1
100. The complex numbers z, z' are represented in the complex plane by the
points P(x, y) and P\x', y'). State in geometrical terms the transformation of the
plane Ρ -> Ρ' given by z' = z(cos Θ+j sin Θ). Verify your statement by deriving
from z' = z(cos 0+j sin0) the matrix form
β-Θ·
What geometrical transformation is given by
z' — c = (z—c) (cos 0+j sin Θ),
where с is a complex constant? If the transformation ζ ->z' corresponds to a
turn of 120° about the point (1,0) and the transformation z' -» z" corresponds to
a turn of 60° about the point (-3,0), show that z" — -z + 2^3j, and hence
express in geometrical terms the single transformation equivalent to the two
transformations in the order given. (S.M.P.)
101. The points Flt F2 are the foci of hyperbola and Pisa point of the branch
defined by F^B- F2B = к > 0. Indicate on a sketch the position of points P, Q
such that PlP-FtP>k, FlQ-F>Q<k,
and justify your answer.
By reflecting the figure in the line / which bisects the angle F2PFi internally, or
otherwise, show that В is the unique point of the hyperbola which lies on /.
Deduce a theorem relating the tangent at a point of the curve and the lines joining
that point to the foci. (M.E.I.)
672

REVISION EXERCISE С
102. If α is one of the complex fifth roots of unity, and χ = α- α4, show that
χί + 5χ2 + 5 = 0.
Express the other roots of this equation in terms of a. (O & Q
103. The tangent to the cubic curve у = ax3 at Д(лг1; y^ meets the curve again
at Рг(х2, Уг)- (Л is not at the origin.) The tangent at P2 meets the curve again at
Рз(*з, Уз) and so on. Show that xu x2, x3, ■·■ are in geometric progression.
The tangent at Pt meets the perpendicular from P3 to the χ axis at P. Show that
Pj, is the mid-point of P2P and that the locus of.P is another cubic curve. (M.E.I.)
104. S stands for three-dimensional space in which points are specified in the
usual way by coordinates (x, y, z), and Ρ stands for the plane ζ = 0 in which
points are specified by just two coordinates (x, y). A transformation
'(")
Л1 °
\0 1
maps points of S onto points of P; and a transformation
4Vg :!)©
maps points of Ρ onto points of S. Show that for 7Ί any given image point in
Ρ is associated with a set of points of S, and describe this set; and that for Тг
the range is a proper subset of Slt to be specified.
Show that for both transformations the line joining a point and its image is in
a fixed direction. Hence describe in geometrical terms the transformations
rtra and ГаГ!. (S.M.P.)
105. Show that the number of solutions of the equation
x+2y+3z = n,
where x, y, ζ, η are positive integers, is equal to the coefficient of tn in
1/{(1 _i)(l_,.)(!_,»)}.
106. Show that the roots of the equation
where А, В, С are positive real numbers and a > b > с lie one between a and b
and one between b and с
If A is small compared with В and С show that one of the roots is
approximately
a +
A(b-a)(c-a)
3(с-аНСф-аУ

REVISION EXERCISE С
107. Under what circumstances is it true that a.b =0?
Prove that, for any angle θ and any positive integer n,
Σ cos #+
r=o \
Prove also that Σ cos2 [Θ +
2m\
!>*K)·
108. A square matrix is called 'magic' if there exists a number к such that the
sum of the elements of each row is к and the sum of the elements of each column
is k. Prove that, if A is a non-singular magic matrix, so also is adj A.
109. Χ, Υ and (и- 2) other players play a game in which each player has an
equal chance of winning each round. The game is won by winning 8 consecutive
rounds of the game. Let
кк = 1,2, .
Show that
(и-
qk = Pr (Xwins\Yhas wt
rk= Pr (X wins\Xhas w,
..,7.
-Di* + r*= 1,
on the last к times),
on the last к times)
r* = ->Vn+11--I?i, for & = 1,2,...,6,
and find the corresponding result for к = 1.
Hence show that the chance that X will win the game is (и7- Щи8-!)
if Υ has just won one round but did not win the round before that. (M.E.I.)
110. The remainder when a polynomial Ρ in χ is divided by (л:-a)2 is the linear
polynomial rx+s. Prove that
r = P'(«) and s = P(*)-aP'(a).
The expression - s/r is of significance in the solution of the equation P(x) = 0
by a certain approximate method. Explain this, and show how the division
process for polynomials could be used to approximate to the roots of this equation.
Illustrate your answer by finding in this way closer approximation than -1,
3, 5, to the roots of , „ , „
x3 - 7л:2 + Ίχ +14 = 0.
Find the sum and product of your three approximations and explain how these
serve to check the answers. (S.M.P.)
111. Let Η be the set of matrices of the form
_ (a + bj -c + d}\
X-[c + di a-bi)'

REVISION EXERCISE С
where j = V(~ 1) and a> b, c, d are real numbers not all 0. Show that Η
contains a subset representing the complex numbers and prove that each element of
Η has an inverse in H.
The matrix X is expressed in the form
X = al + cJ + bK + dL,
""--Π-'-ΠΠ-β-^-Π·
Prove that J2 = K2 = L2 = -1 and that JK = -KJ = L.
Given that KL = -LK = J and LJ = - JL = K, evaluate XY, where
(a' + b'j -c' + d'h
\c' + d'i a'-b'i)'
by multiplying the expressions for X and Υ in terms of I, J, K, L. (М.ЕД.)
112. X and Υ are independent Poisson variates from distributions with
parameters μ and λ respectively. Find
Pr(X= r\X+Y= n)
forr= 0,1, ...,η.
The telephone exchange in a large firm handles both internal and external
calls. It is desired to investigate the distribution of internal calls and equipment
is available for recording the number of internal calls in each set of 10 calls.
Calculate the probabilities of 0, 1,2 and 3 internal calls in sets of 10 if the average
number of calls per set is 7. What is the most likely number of internal calls ?
(M.E.I.)
113. Show that if tabular values are given for χ = 0, 1, 2, 3 of the function
Rx) = a + bx+ cxix - 1) + dx(x- D (*- 2),
then the values of b, c, d can be obtained from the differences derived from the
table, e.g. b from/(l)-/(0).
Given the following table for a cubic polynomial g(x), find and check the
values of a, b, c, d required to express it in the form given above; and hence or
otherwise evaluate #(5), #(6), g(J) and #(8),
л: 0 1 2 3 4
g(x) 87-25 89-37 91-49 96-85 108-69 (M.E.I.)
114. Three bags each contain N tickets numbered 1 to N. Three tickets are
drawn, one from each bag; find the probability that the sum of the numbers
obtained is 2N.
115. Two circles, centres A and В and radii one unit, are drawn respectively in the
planes r.n = pu r.n2 = p2- Prove that they lie on the surface of a sphere if and
ОПу (a-b).(nt л n2) = 0 and Oi-^.b)2 = 02-n2-a)2.
675

REVISION EXERCISE С
116. ABCDEF is a hexagon inscribed in a circle centre O. If
ΔΑΟΒ = ACOD = Z£OF = я/3,
prove that the mid-points of ВС, DE, FA are the vertices of an equilateral
triangle.
117. The region E„ is denned by
шР + Ьу1 < 1,
where a > 0 and b > 0. Without assuming any properties of conies, prove that,
if Pi, Рг are the position vectors of points of Ea, then
(i) -ъеЕо, (η) ¥Pi+P,)eE0,
(iii) 2pt e £, where Ε is the region ал:2 + by2 < 4.
A system of elliptical discs congruent to E0 is obtained by translating the centre
О of E0 to all points g whose coordinates are integers. Prove that, if the discs with
centres at gb g2 overlap, then
(iv) for a common t, g1—teE0 and t- g2 e E0,
(v) i(gi-g2)e£o, (vi)gl-g2e£.
Deduce that if Ε contains no point, other than the origin, with integral
coefficients, then no members of the system S overlap. (M.E.I.)
118. The log-normal distribution is a probability distribution given by
for 0 < χ < oo. By using the transformation χ = /эе" and comparing with the
normal distribution show that:
(i) f'jtx) ax = 1, (ii) JJ0 xf(x) dx = /9e*a, (iii) j" xV(x) ax = p*d>«.
Deduce the values of the mean and variance of the distribution. (O & C)
119. If л: is a close rational approximation to the cube root of 2, and
x' = (x2 + x + 2)/(x2 + x + l),
show, by expressing л:'-л: in the form (2*-х)Дл:) and verify ingt hat 1 < f(x) < 2,
or otherwise, that x' is a better approximation than x, and that if one is below the
correct value the other is above. Use this method to find the cube root of 2
correct to two places of decimals. (O & C)
120. The surface of a sphere is divided into regions by r circles; prove that, if a
further circle is drawn on the surface, the number of regions is not increased by
more than 2r.
Prove that η circles on the surface of a sphere divide the surface into at most
n2-n + 2 regions.
Hence, or otherwise, prove that η spheres divide space into at most
\{пг-Ъп2 + Щ
regions. (O & Q
676

REVISION EXERCISE С
121. Prove that Σ ———-—— = -- — — —.
r=i r(r+l)(r+2) 4 2(и+1)(и+2)
It is desired to estimate 1000 л
Σ
,=i r(r+ 101) (r + 2001)
to within 0-001 by summing only the first N terms. Find a suitable value for N.
(M.E.I.)
122. Six integers a,- (i = 1 to 6) are independently and uniformly distributed in
the interval - Ν < α,- < N Find the probability that the simultaneous equations
агХ+а2у = a3, atx+aby = ae
have a unique solution.
Check your answer by enumerating the various possibilities for N = 1.
Show that the probability of obtaining a unique solution tends to 1 as N tends
to oo.
123. Show that, if a, b, c, d,p and q are positive integers such that
ape
b<q<d'
then p>(a + c)/A and q > (b + d)/A,
where Δ = be—ad.
Construct the rational numbers of denominator 20 or less between γ§ and j.
(CS.)
124. If in the expression
g(x) = a0xn + atx"-1 +... + an (a4 all real),
α> = α.-+ι = 0, for some ί in the interval 1 < ι < n—2, prove that the equation
g(x) = 0 has at least two complex roots.
The polynomial f(x) is given by
fix) = b0xn + b1xn-1+ ... + b„ (t>i all real).
Show that, by a suitable choice of α, β, the equation
audt.b.x-bt¥dAxi + №-1x-buf&) =0 (1 < j < и-1)
has at least two complex roots.
Deduce that, if b\ < ЬВ^Ь^Ъ then the equation f(x) = 0 has at least two
complex roots.
125. If ζ, ζ* are conjugate complex numbers, give a geometric description of
those numbers z for which . „. . „..
\ζ-ζ\ < \ζ-ζ*\.
Let z1; ...z„ be η complex numbers, the imaginary parts of which are strictly
positive, and put
Π (z-zd = z" + (a1+jb1)z''-1+... + (an+}bn),
where the аъ ..., an, Ьъ ..., bn are real. Show that the roots of
xn+a1xn~1+... + an = 0

REVISION EXERCISE С
126. From a population of N butterflies which are restricted to a certain habitat
a sample of и is taken. Each butterfly is marked and released again in the habitat.
A second sample of η is taken on the following day. Show that the probability
Ρ that exactly m butterflies in the second sample will be found to be marked is
of the form AuN, where A is independent of N and
uN = (N-n)\j{N\ (N-2n + m)y.
Show that the value of N which makes Ρ greatest is approximately equal to
nVm. (C.S.)
127. Let 5ГО denote the set of all 3x3 real matrices and define an operation *
between members of 5ГО as follows:
X*Y = X+Y-XY.
We define a matrix X to be quasi-regular if we can find a matrix Μ e $t such that
X * Μ = Μ * X = 0.
Prove the following results:
(i) 0 is quasi-regular.
(ii) If X is non-singular and quasi-regular, so also is X-1.
(iii) If Χ, Υ are both quasi-regular, then X * Υ is quasi-regular,
(iv) If X, Y, Z, are quasi-regular, then (X * Y) * Ζ = X * (Υ * Ζ),
(ν) If Xn = 0 for some positive integer n, then X is quasi-regular.
128. The polynomial P(x) is denned by
Длг+1) = P(x) + (l+x)n,
where л is a positive integer. If P(l) = 1, find P(0) and P(-1).
Prove that:
(i) P(-x) ξ Ρ(-χ-1) + (-1)ηχη;
(ύ)Ρ(χ)Ξ(-1)"-ιΡ(-χ-1).
Deduce that, if л is even, then the sum of the nth powers of the first к natural
numbers is a polynomial in k, with integral coefficients, which is divisible by
Щ+1)(2&+1).
What can you say about the sum of the nth powers of the first к natural
numbers in the case when η is odd ?
129. In the tetrahedron ABCD, the perpendiculars from A, B to the opposite
faces intersect; prove that the edges AB, CD are perpendicular.
Prove also that the perpendiculars from C, D to the opposite faces intersect.
130. The parametric vector equation of a line / through the origin in
three-dimensional Euclidean space is
г = tk,
where к is a constant unit vector and t denotes distance measured along / from
the origin. A point Ρ has position vector s. Find the position vector of the
reflection of Ρ in /, i.e. of the point Q such that PQ is bisected at right angles by Λ
If г = tki (i = 1, 2,) are two distinct lines through the origin, and
Si (i = 1, 2) are the operations of reflection with respect to these lines, prove
that 525Ί = SlS2 if and only if the two lines are perpendicular. (C.S.)
678

REVISION EXERCISE С
131. It is required to find the matrix A (relative to i, j, k) of the linear
transformation representing a positive rotation θ about the axis
Find the explicit form for A by proving the following results,
(i) Ifu = /i + wj + лк, ν = l'i + m') + n"k, w = ri + 7n"j + 7i*k
are three mutually perpendicular unit vectors, then
II V
Q=l
is orthogonal.
(ii) If the point Ρ has position vector χ referred to i, j, к as base vectors, then
it has position vector X = Q'x
referred to u, v, w as base vectors.
(iii) The orthogonal matrix
\0 sin θ с
represents a positive rotation of θ about the χ axis,
(iv) A = QRQ'.
(v) Г* + П=\-Р and m'n"-m"n' = l.
(vi)
(/2 + (l-/2)cos6> w/(l-cos0) -nsinff nl(l-cosff) + ,
lm(l-cos0) + nsme m2 + (l-m*) cos Θ nm(l-cos0)-
ln(l-cosff)-msme mn{\ - cos Θ) +1 sin θ л2 + (1-л2) cos θ
132. If a, b are real and a > 0, and if the equation
z3-az+b = 0
has a pair of complex roots Я ± j/t, prove that the point (λ, μ) is one of the points
of intersection of the curves
Зл:2-^2 = a, 2x(y2 + a) = 3b.
Draw rough sketches of these curves and deduce the condition for the given
cubic to have complex roots.
133. The coefficients a, bin the quadratic polynomial
p(z) = z* + az + b
are complex numbers and α2 Φ 4b. Prove that the locus of the point ζ which
varies in the Argand diagram in such a way thatp(z) takes real values is a
rectangular hyperbola with asymptotes parallel to the real and imaginary axes.
By a change of origin, or otherwise, determine the asymptotes.
Determine also the locus of ζ if it varies in such a way that p(z) is purely
imaginary. Sketch both loci on the Argand diagram and interpret algebraically
their points of intersection.
Prove that there are two real values of ζ and two purely imaginary values of ζ
for all four of which p(z) is purely imaginary if
{Re(a)}2 > 4Retf>) > -{Im(a)}2. (О &С)
n-a 679

Bibliography
The following books may usefully be added to the more comprehensive
bibliography of Volume 1.
Armitage, J. V. and Griffiths, Η. Β. A Companion to Advanced Mathematics
(Cambridge). Surveys the underlying concepts of modern algebra and analysis:
an exacting book for sixth-formers but could be read with profit by a student
embarking upon a university course. Contains an excellent selection of
questions.
Brand, T. and Sherlock, A. Matrices: Pure and Applied (Arnold). Illustrates
many of the interesting applications of matrix theory.
Budden, F. J. Complex Numbers and their Applications (Longmans). An
entertaining book illustrating applications of complex numbers in both pure and
applied mathematics.
Cundy, Η. Μ. and Rollett, A. P. Mathematical Models (Oxford). Gives
instructions for making many models. Full of stimulating ideas: very strongly
recommended.
Durran, J. H. Statistics and Probability (Cambridge). One of the best
introductory texts available; contains an outstanding selection of examples: very
strongly recommended.
Hardy, G. H. A Mathematician's Apology (Cambridge). A leading mathematician
writes about what the subject means to him.
Hardy, G. H. and Wright, Ε. Μ. An Introduction to the Theory of Numbers
(Oxford). A famous work on one of the most fascinating of all mathematical
topics—the properties of the integers and their generalizations. Although the
book goes well beyond school level it is charmingly written and contains much
that will appeal to an intelligent sixth-former.
Henrici, P. Elements of Numerical Analysis (Wiley). A very good book on
numerical methods, although rather advanced for sixth-formers.
Hunter, J. Number Theory (Oliver and Boyd). A more modern approach to the
subject than in Hardy and Wright: the text is somewhat exacting as it is
written for university students. Contains some good examples.
Kemeny, J. G. and Snell, J. L. Finite Markov Chains (van Nostrand).
Applications of matrix theory to probability. An interesting but not always easy book.
Kline, M. and others. Mathematics in The Modern World (Freeman). Readings
from 'Scientific American'. Contains a series of fascinating essays on a wide
variety of subjects.
Maynard Smith, J. Mathematical Ideas in Biology (Cambridge). Among much
other interesting material there is an account of the applications of probability
to genetics. The same subject is dealt with in greater detail in D. S. Falconer's
Quantitative Genetics (Oliver and Boyd).
Mendelson, B. Introduction to Topology (Blackie). Perhaps the most readable
introduction available to topology, an important branch of modern
mathematics.

BIBLIOGRAPHY
Midonick, Η. The Treasury of Mathematics (Pelican, 2 volumes). Selected
writings of famous mathematicians—all within the understanding of the sixth-
former.
Ogilvy, C. S. and Anderson, J. T. Excursions in Number Theory (Oxford). A very
elementary introduction to number theory. Contains an interesting chapter
on Fibonacci numbers.
Ore, O. Number Theory and its History (McGraw-Hill). Another very good
introductory work: more advanced than Ogilvy and Anderson, less so than
Hardy and Wright.
Rahman, N. A. Exercises in Probability and Statistics (Griffin). A formidable
collection of problems, going far beyond school level but containing much
challenging material suitable for a sixth-former. Also A Course in Theoretical
Statistics, which takes the subject well beyond school level.
Ribbans, J. Basic Numerical Analysis (Intertext). A good elementary account of
numerical methods.
Uspensky, J. V. Introduction to Mathematical Probability (McGraw-Hill). A
classic: advanced but not too difficult.
Watson, W. Α., Philipson, T. and Oates, P. J. Numerical Analysis (Arnold,
2 volumes). An elementary introduction to numerical methods: contains many
worked examples.
681

^

Answers
CHAPTER 17
PAGE
381 Ex.3, (i) -l+2j; (ii)l-4j; (iii) 7+ 4j; (iv) -8 + 6j;
(v) -26-18J.
382 Ex. 5. (i) - 5 +18j; (ii) - 99 + 20j; (iii) 27 + 36j.
Ex. 6. (i) (2-j)/5; (ii) (5 +j)/2; (iii) (4-3j)/5.
383 Ex. 8. (i) 3,-2; (ii) 5,-12; (iii) Д; ^.
Ex. 9. b, 0.
Ex.10. (iv)8-j; (v) (2-j)/5; (vi)ll-5j; (vii) 8-8j.
Ex. 11. R.
Ex. 12. (i) 5; (ii) 2; (iii) 1Д/2; (iv) 1; (v) 2 sin i<9.
Exercise 17(a)
1. (i)7 + 6j; (ii)9 + 17j; (iii)4 + 3j; (iv)ll+6j; (v)ll+22j;
(vi) 1 + 32j; (vii) 0; (viii) (3 - j)/5; . (ix) - 72 + 368j;
(x)-6-12j; (xi) (7+j)/50; (xii) - (11 + 29j)/37.
2. cos (θ1 - θ2) +j sin (0j - θ2); cos 2вг + j sin 2θλ;
cos 2(θ1 + ej + j sin 2(θ1 + θ2).
3. (i) 2±j; (ii) (1 ±j)/2; (iii) (5±j73)/2; (iv) ± 1 -j.
384 4. 3-2j; (i) (1 +j)A/2; (ii) 2+j. 5. 3 + 2j, 1-j.
6.(i)l, -I; (ii)2-j,2+j.
7. 1 - cos θ + j sin Θ, (1 + j cot Щ/2. 10. 2, pure imaginary.
12. (i)*2-4* + 5 = 0; (ii)*2-4* + 13 = 0; (iii) *2- 4x + 7 = 0;
(iv) *2 - 3(1 + j> + 5j = 0; (v) *2 - (5 + j>+4(1 + j) = 0.
13. (i) (x- 1-j) (x-1 +j); (ii) (x + 2)y) (x-2jy);
(iii) (j*;+jV3^) (^-jV3^); (iv)(j*-l)(* + l-j);
(w)(jx-2y)(x + 2}y); (vi) (x + 1) (2*-1-jV3) (2*-l +W3)/4;
(vii)IT(x±l/V2±j/V2).
14.2,3; (i)2; (ii)* + l.
15. l,j, -1, -j; 1, l+j,j, 0 according as η = 4m,4m+l,4m + 2 or
4m+3.
385 17. -22.
388 Ex. 21. (i) 2, ^π; (ii) 2л/2, £я; (iii) 1, π; (iv) 1, in; (v) 1, -*я;
(vi) 2, -|π; (vii) 2, -\n; (viii) |sec α|, α;
683

ANSWERS
PAGE
388 (ix) |sec «|, (π-2α)/2; (χ) 2 sin \a, (α- π)/2]
(in the last three parts the principal argument depends on a).
390 Ex. 25. (V3-1) + j(V3 +1), cos 75° = (д/3- 1)/2V2, etc.
Exercise 17(b)
396 1. (i) 5, -53°; (ii) Vl3, 56°; (iii) V5, -117°; (iv) V5/2, 63°; (v) V29,112°;
(vi) v'5, -153°; (vii) 13, -23°; (viii) 41, -103°.
2. (i) cos 30 + j sin 3(9; (ii) cos (Θ + φ) +j sin (Θ + φ); (iii) j; (iv) (1 + j) V2;
(v)j; (vi)(l-JV3)/2; (vii)j;
(viii) 4 cos \6 sin i$5(sin (Θ + $5)/2-j cos (θ + φ)/2).
3. (i) (π + 2φ-20)/2; (ii) (-3π + 2φ-20)/2.
7. Circle, centre 2+j, radius 1. 12. — j.
397 13. (i) (1 +j)/2; (ii) (1 ±jV3)/2; (iii) j, 1 -j; (iv) -1 +ij.
14. 9 + 6j. 16. (i) — f, f; (ii) semicircle, radius 1.
17. (1 -j tan Щ/2, (1 + j cot Щ/2. 18. Square.
Miscellaneous Exercise 17
1. (V3 - j) (1 + j), 2V2, ^π. 2. 8, 2.
3. l±V3+j(5±V3).
398 5. -l + 2j, (x-iy + (y-iY = i
399 14. (i) \z\ = V7; (ii)2+j, 3 + 2j.
400 15. #· = 4, 0 = 147-5°.
CHAPTER 18
402 Ex. 1. *> - 9*2 - Зл: + 4, 2.
Ex. 2. 25086278.
403 Ex. 3. 2600.
Ex.4. y= 1957л:-7373.
Ex.5. 2412, 1957, 1190.
405 Ex.7. -26, -8.
Ex.9. -0099,0-357.
407 Ex. 13. 2*2- 5* + 3.
Exercise 18(a)
1. (i)*2-3*-5;6; (ii) *3-9* + 4; 2;
(iii) 3*3-5*2-2*-7; -10; (iv) x3 + 4x* +16x + 59; 231;
(v) 7*4 - 35л3 +175л:2 - 875л: + 4372; - 21865.
684

ANSWERS
408 2. (ϊ)χ*-3χ-7; 7; (ii) x*-x+3; 0; (iii) л:4-4л:3-л:2+Зл:+5; 2;
(iv) (16л:3-44л:2-14л:+11)/8; f;
(v) (9xi - 63л:2 + 6л: - 40)/9; ^.
3.(i)159-l; C") 31-4; (iii) 283-3.
4. (i) (л-- l)3 + 4(л-1)2 + 7(x-1) + 5;
(ii) 2(л:-2)3-5(л:-2)2-14(л:-2) + 8;
(iii) (л-+3)3-2(л-+3)2-21(л:+3) + 50;
(iv) (л: + 2)4 - 8(л: + 2)3 + 24(л: + 2)2 - 32(л: + 2) + 15.
5. (ϊ)λ:3-^2 + 4^ + 2; 1-j;
(ii) *3 + (l+j)*2-(2-j)* + (l + 2j); -j.
6. (i) 2-6; (ii) 3-0. 7. (л: + 4)4 + 4(л: + 4)2; minimum; -4, -4, -4±2j.
8. л:5-Зл:3 + 2л:+1; 1. 9. A = 0, В = I. 10. A = Л, В = 4Л-17.
11.-6,139. 12.86,68.
409 13. л:2-8л:-8. 14. 2л:3-4л:2-х + 6.
15. /г+ 15/г(«- 1)/2+ 25/г(/г-1) (/г-2)/3 + 5/г(/г-1) (п-2) (п- 3)/2
+ /г(/г-1)(/г-2)(/г-3)(/г-4)/5.
Ex. 14. e.g. a = 5, Ъ = -2. Ex. 15. β = 3, й = 3, с = -1.
411 Ex. 17. (i) 4/(л- 1)-3/(л- + 2); (ii) 3/2(л-3)-1/2(л-+3);
(iii) 1/(λ-α)-1/(λ-+α); (iv) 1/(λ-α)+ 1/(лЧ-а).
412 Ex. 18. (i) In л/1 ·5; (ii) In (f).
Ex. 19. (i) 4(л--2)-2-6(л—3)-2;
(п){(л-+1)-2-36(л--6)-2}/7.
415 Ex. 21. (i) 2/(л—2)-3/(л-2)2-2/(л:+1);
(п)1/(л-+1)-4/(л-+1)2-1/(2л-+1).
Ex.22. (ϊ)1/(λ—1)-λ:/(λ:2 + 2);
(п)2/3(л--1) + (л--1)/3(л-2 + л-+1).
Ex.24. (ϊ)1/(λ-1)-1/(λ-2-2);
(π)1/(Λ-1)-{1/(λ-ν2)-1/(Λ-+ν2)}/2ν2.
Exercise 18(b)
1. (i) 2/(3 - χ) + 3/(1 - 2л-); (ii) 2/χ - 3/(χ + 2);
(iii) 7/(л- + 3) - 5/(2л- - 1); (iv) 1 /2(1 - χ) - 3/2(1 + л");
(ν) 2 + 3/л-5/(л-+1); (vi) 1/(л-1)-3/(лЧ-2) + 2/(л:-3);
(vii) 11 χ - 3/(1 - л") - 5/(1 + л"); (viii) - χ + 3/(л- 2) - 7/(* + 2);
(ix) л:2 +5л:+19+81/(л:-3)-16/(л:-2);
(х)1/л:2+1/л:-2/(1 + 2л:);
(xi) 1 /(1 - 2л:)2-1 /(1 -2л") +11(2 +х); (xii) л:/(2 + л:2)-1 /(1 + л");
(xiii) (2л-+1)/3(1-2л:+4л:2)-1/3(1 + 2л:); (xiv) 4/(л—3)2+1/(л—2);
(xv) 1-1/3(л-+1) + (л-4)/3(л-2 + 2);
(xvi) 1/(л-2)-1/л-4-2/л-3-2/л-2-1/л-.
2. А = -13, В =13, С = -11, D= 27.
3. 2/(л- Ι)* + 6/(х-1)3 + 4(л-1)2 + 1/(л- 1)- 1Цх + 4).
685

ANSWERS
PAGE
416 4. }-1/(л + 2); ^-(3/г2+18я + 26)/3(/г + 2) (л+3) (л+ 4).
5. (О 1β(1+χ)-(χ-2)ΐ3(1-χ+χ*);
(ii) 1 /3(1 + χ) - (1 + V3j)/3(2x -1 - V3j) - (1 - V3J)/3(2* -1 + V3J).
6. 1/(jc-2)-1/(jc-1); i-ll2(2y-l) + ll(y-l);
*-1/8(4*-l)+l/(2x-l).
Ex. 25. 0-9802.
Exercise 18(c)
420 1. (i) 1 +3*+9*2 + 27*3; (ii) y+l*+2V*2+sT*3!
(iii) 1 + 4л: +12*2 + 32л3; (iv) 1 + 2л:2;
2. (i) Ι+λ+Ι^ + Ιλ3; (u)1-2a:-2a:2-4^; (iii) 2-|*-^*2-rh*3;
(iv) ΗΑ^ + ϊΙβ^ + ϊΑβ*3; (ν) (i+^+^2 + 2T6^V2.
3. (i) 1 +jx-}x2 + riX3; (ii) 1 -8л:+40л:2-160л3;
(iii) 1+!*+!** +1?*»; (iv) 3V +Α^ + Ί^ + ίίήί*3;
(у)2-1^л-^л:2-То1зб^.
4. (i) 1 + 3* + 7*2 +15л3; (ii) \ + ^х + -2ψέχ* + уЦ^л3;
(iii) i-U+Ч-х2-^; (iv) Зл:-5л:2 + 7л3;
(у)1+8л:+26л:2 + 64л3.
5. ®1-|л: + ^л:2;
(ii)l+* + i*2; (iii) l+fjc+^jc2; (iv) 1-|jc+V*2;
(\)1-3χ+ψχ2.
6.(i)i-K^-i)+K^-D2; (ηΗ+Κ^+υ+^+Ό2;
(iii) 1 + (л-+1)-Кл-+1)2; (iv) l + ^x-D-Ux-^f;
(v) -2-K*+l)-m*+l)2·
8. (i) 0-99933; (ii) 1-0100; (iii) 2-0025; (iv) 0-92793.
421 9.9-997. \2.l+\x + f^x\
13. 2/(2*-1) + (1 -*)/(l + *2). 14. 2-009926.
Miscellaneous Exercise 18
1.Α = λ,Β= arbitrary, С = -Л-1.
3. 1, -(Л-+1), l/(*-2)-(* + l)/(*2 + *+l).
422 4. l/7(*-2)-(* + 3)/7(*2 + *+l); l/9(*-2)-l/9(*+l)-l/3(* + l)*.
5. i, i; -^ - (4л + 5)/8(2и + 1) (2я + 3).
7.(i)V3; GO (4)*; (Ш) 2/^/7; (iv)(3)i.
8. 2л3-13л:2 + 26л:-13, 4(ял:-л:2)/я2.
9. (7-*)/4; tangent at (-1, 2).
10. 249, /·4-8/·3 + 24/·2-34/·+21.

CHAPTER 19
PAGE
423 Ex.2. -29j.
424 Ex.4.j/2».
Ex. 6. 2/(3- 10i2 + 3i4) (1 - 15ί2 + 15ί*- ί6).
425 Ex. 8. (10- 15 cos 2(9+ 6 cos 4(9-cos 60)/32.
Ex. 9. 5π/32.
426 Ex. 10. (1 - χ cos 0 - xn cos ηθ + *n+1 cos (и - 1)0) (1 - 2x cos (9 + л:2)"1.
Ex. 11. χ sin (9(1 -2x cos (9 +л:2)-1.
Exercise 19(a)
1.(0-1; (ii)-j; (iii)j; (iv) (V3+j)/2; (v)l;
(vi) 16(-V3+j)/9V3; (vii) -(l+JV3)/32;
(viii) sec4 0(cos 46·—j sin 40)/16.
2.(i)-l; 00 16; ("О -28(l+jV3);
(iv)cos(n + mW+}sm(n + m)e.
3.(i) 1-8 sin2 (9+8 sin4 (9;
(ii) 5 sin 0- 20 sin3 0+16 sin5 0.
4.(i) 1-8 cos2 0+8 cos4 0;
(ii) 16 cos5 0-20 cos3 0 + 5 cos 0;
(Hi) 6 cos 0- 32 cos3 0+32 cos5 0.
5. 4f(l-f2)(l-6f2 + f4)-1.
6. cos {(2k+1)/14}, A: = 0 to 6; cos (2λ:π/7), A: = 0 to 6.
7. (i) (cos 50 + 5 cos 30+10 cos 0)/16;
(ii)(cos40-4cos20+3)/8;
(iii) (cos 70 + 7 cos 50 + 21 cos 30 + 35 cos 0)/64.
8. (i) (sin 50- 5 sin 30+10 sin 0)/16;
(ii) (35 sin 0-21 sin 30+7 sin 50-sin 70)/64;
(iii)(2sin20-sin40)/8.
9. (sin 90 - sin 70 - 4 sin 50 + 4 sin 30 + 6 sin 0)/256.
10. (i) 3π/16; (ii) 3π/512.
11. (i) (cos 0+cos /z0-cos (n+1)0-1) (2-2 cos 0)"1;
(ii) (sin 0 + sin «0-sin (n+ 1)0) (2-2 cos 0)"1.
12. (cos 0 + 2n+1 cos «0-2-2" cos (n+ 1)0) (5-4 cos 0)"1.
427 13. 2" cos" (0/2) cos (и0/2).
14. i cosec2 (0/2) {sin 0 + (и+ 1) sin (n- 1)0-/г sin ηθ].
15. (sin2 0-sinn+10 sin (n+1)0+ sinn+2 0 sin ηθ) (1 + sin2 0-sin 20)-1.
16. (Σί! - Σί213 Q (1 + h h h h ~ Σίι h)-\
17. (sin 30 sin3 0 - sin (2/г + 3)0 sin2n+3 0 - sin6 0 + sin (2/г + 1)0 sin2n+5 0)
x (1 - 2 sin2 0 cos 20 + sin4 0)-1.

ANSWERS
427 18. a = 1xs,b=-^E, с = f; kn±\n, rm.
428 Ex.l4.(i) ±1, ±j; (ii) ±2, ±2j; (iii) ±(1 ±j)/V2.
429 Ex. 18. -j, (±V3+j)/2.
Exercise 19(b)
430 l.(i)2,(-l±V3j); (ii)-l,(l±V3j)/2; (iii) j, (±V3-J)/2;
(iv) 1+j, {-(V3 + l) + (V3-l)j}/2, {(V3-l)-(V3 + l)j}/2.
2. (i)2-j, (ii) 1-3 + 0-79J; (iii) -0-28-1-8j; (iv) 1-4-1-Oj.
3. 1 + j, { - (V3 + 1) + (V3 - 1) j}/2, {(л/3 - 1) - (л/3 +1) Л/2;
cos π/12 = (1 +V3)/2V2, sin π/12 = (^/3-l)/ZJ2.
4. 1-1-0-088J.
5. (i) cos 3 — j sin 3;
(ii) cos (20^)/6+j sin (20-π)/6;
(iii) cos 20 - j sin 20; (iv) cosec1 0 cjs (π - 20)/8;
(ν)φ(40-π)/8.
7.(±V3±j)/2. 8. -l,(±l±V3j)/2.
9. ij, — wj, — w2j where ω is a complex cube root of one.
10. tan kn/5, к = 0 to 4.
11. cos 0(1 -sin 0)"1, 0 = 2kn/n.
431 15. (α + ύ + ο)(α + ωί, + ω2ο)(α+ω2ύ + ωο),
(α + ύ + c) (a2 + b2 + c2-bc-ca-ab);
(a + b + c + d)(a-b + c-d)(a+jb-c-}d)(a-jb-c+jd)
433 Ex. 25. (i) .ίπ; (ii) -^π; (iii) e""; (iv) e4
Exercise 19(c)
l.\ar + bs\,e. 2. (i) e^(l+j)/V2; (ii) e-».
3. (i) In 2^π/6; (ii) eb(cos In 2+j sin In 2).
4. Moves along real axis from e" to e~". 7. 2c2 cosh ν (cos к + cosh в)-1.
Miscellaneous Exercise 19
434 1. cos (2A: + l)7r/12+j sin (2k + l)n/12, к = 0,2, 3, 5, 6, 8, 9, 11.
3. 2-" cos 2/г0(1+ cos 0)"n.
4. (i) 2 cjs (π/4), 8 cjs (3π/4); (ii) 2 cjs (11 π/12), 2 cjs ( - 5π/12).
5. ±(-0-61 + l-2j), ±(-0-61-l-2j).
6. \{ac - bd+ (α2 + b2)* (с2 + tf2)*}.
9. (i) cos {a + K«-1)/?} sin \ηβ cosec }/ff;
(ii) sin {ot + iin-1)/?} sin ^ cosec ifi.
10. 35π/256.

ANSWERS
PAGE
435 11. (i) 2 cos \θ cjs \6; (ii) 2 sin \6 cjs (θ-π)/2;
(iii) V(cot Щ cjs \tt.
13. $5 = (2/·+ 1)π/5 14. tan {4/c-3)я}/4и.
15. (V3 + 1V2V2, 1/(л/6 + л/2).
16. - (3* + 3*), - (3*ω + 3V), - Ρ*ω2 + 3&ω).
437 23. in(n + l); A =-C = α(1-α)-2, 5 = (1-α)"1.
24. (*-1)Π (*2-2*cos2for/(2« + l) + l).
CHAPTER 20
439 Ex. 2. (2 + 3j)z + (2 - 3j)z* = 13.
Ex.4, (i) 6; (ii) -3j.
Ex.6. zz*-(l+2j)z-(l-2j)z*-4 = 0.
Ex.7. (i)l + 3j, 2; (ii) i-j, 1.
Exercise 20
447 1. (i) |h--3| = 1; (ii) |и>—j| = 1; (ш) |и>| = 2; (iv) |и>| = 1;
(ν) N = V2.
3. -1, ±j. 4. 1+j; 2+j, -1+j.
448 11. z(l -j) + z*(l +j) = 2. Interior of 'shark's fin' bounded by real axis,
\*-i\ = i, k-i-ij| = V6/4.
12. Circle, 2j/3, f. 13. Line. 14. Line. 15. Point j.
16. Circle, 2, V5. 17. Line. 19. arg i^|] = π-θ.
449 21. и- = 3+j- 1/z. 25. Arcs of three circles through origin, centres
j/2, 1/26, j'/2(a-1) and imaginary axis.
26. (i) 3ww*-2(w + w*) + l = 0; (ii) w + w* = }.
27.w = (z-j)(jz-l)-\
CHAPTER 21
450 Ex.1. 1+j, (-l+7j)/5.
454 Ex.2. (i)-Jf; (ii) -^; (iii) V47/2; (iv) ^.
Ex.3, (i) 3z2-10z+12 = 0; (ii) 3z2 + z+l = 0;
(iii)9z2-7z + 9 = 0.
Exercise 21(a)
455 1.3,-7; (ii) 2, 5*; (iii) (l-j)/2, -j; (iv)4-3j, -1;
(v)(2-6j)/5,-(l+2j)/5.
689

ANSWERS
PAGE
455 2. (i) z2-z-6 = 0; (ii) 6z2-z-2 = 0; (iii) z2-6z + 4 = 0;
(iv) z2- (5 + V2)z+ (2 + V2) = 0; (v) z2 - 6z + 25 = 0;
(vi)z2-z+l = 0; (vii)z2-(2+j)z+(3+j) = 0;
(viii)z2-(V2 + l)(l-j)2-3j = 0.
3. (i) 2-8, -1-3; (ii) 075±l-7j; (iii) 0-42 + 0-86J, -1-42-1-86J.
4. (i) 3, -9, 27, -i -972; (ii) i -f, ^-, - j, -^;
(iii)j, -1+j, l-2j,(l-j)/2, -5+j.
5.(0-1; (ii)W. (iii) л/113/2; (iv) 9^113/4.
6. (i) (3/c2 + 2/c- 7)/3; (ii) ψ^; (iii) - 4.
7. (ϊ)2(α2 + 8βό+8ύ2); (ii) 2(α+2ύ)/α2; (iii) 4|0(fl+u)}*|.
456 8. -a2mc/(62 + a2m2), 62c/(62+aW).
9. (i)2z2 + 3z-6 = 0; (ii) z2-5z-350 = 0; (iii) 7z2 + z-2 = 0;
(iv)4z2-29z + 49 = 0.
10. (i) 2z2-2(l -3j)z-5-9j = 0; (ii) z2-2jz + 5-j = 0;
(iii)z2-(2+14j)z-12 + 5j = 0.
11. (i) 4z2 + 7 = 0; (ii) 4z2-(10+ 10j)z+ 9j = 0;
(iii) 4z2-19z+44 = 0.
12. ±24j.
13. ±3(l+j).
14. (i) (4 + 5j)z2-(3+j)z-(2-j) = 0;
(ii)(2+j)z2 + (3-j)z-4+5j = 0.
19. (ca*)*+(c*a)i + b = 0.
20. ύ2 = lea.
Miscellaneous Exercise 21
459 1. (i) llz2-8z-400 = 0; (ii) z2-5(l+j)z + 47j = 0.
2. α2, β\ 7. a/A /?/a.
CHAPTER 22
461 Ex. 2. 2х-Ъу+\Ъ = 0, (5ί-1, 3i-2).
462 Ex.5. bV-αψ = агЪ\
463 Ex.6. (-1,3), (-4,4), (-2,5), (-3, -4).
Exercise 22(a)
464 1. 4*2+.y2 = 4. 2.x3 = y*. 3.y=x2-3x + 3.
4. *2+.y2 = 2. 5. x>+y* = axy. 6. ХЧ.У2 = x2y.
8. (i) (-1, 4); (ii) (2, 0); (iii) (2, 8); (iv) (4,-1).
9. ±i.
465 12. (-2, 5).
690

ANSWERS
PAGE
466 Ex. 8. PSZ = 90°, SPM bisected by tangent.
467 Ex. 11. 4a, a, 3a.
Ex. 14. (Jail, a), χ = 5a/2; (i) (-Πα/4, 0), x + 5a/4 = 0;
(ii)(-a, a),y + 3a = 0.
468 Ex. 20. я =-р-2/р.
Exercise 22(b)
472 1. (i) 4, (3, 2), (4, 2), * = 2; (ii) 2, (3, -1), (3, -i), 2y = -3;
(iii) 1,(1, -1), (f, -1), 4*= 5;
(iv)3,(-2, -3), (-2, -f),4j/ = 11;
(v) 4, (5, -3), (6, -3), χ = 4, (vi) 8, (2, 2), (2, 4), у = 0;
(vii)2, (0,4), (-i,4), 2*= 1.
2. (i)y_4*-2j/ + 5 = 0; (ii) x3+6x + 2y+4 = 0.
3. (i)y*-12x-6y+21 = 0; (ii) *>+2л:+3.у + 4 = О.
4. (i) x+4y + 9 = 0; (ii) x-6y-3 = 0; (iii) 5x-ly + 25 = 0;
(iv) Sx + 3y+30 = 0.
5.(4,-2). 6. (-4, 6). 7. -i,-i+ij.
473 8.^.
474 18. Same axis, focus; directrix is tangent at vertex of original parabola.
19. Parabola. 21. j>2 = lax. 22. Parabola, latus rectum a.
475 28. у = la, y*-2ax = a2. 30. у = 3*2-4*.
31. у = 2jc2-* + 2.
Exercise 22(c)
478 6.x=-3,y= 1. 7.(1, 1).
479 10. 2x-Sy±15c = 0.
Miscellaneous Exercise 22
2. Y/a, (У—2аХ)\2а*\ (b, 0), (b + ia, 0).
481 15. (3V7)c2. 18. (la, 2a J2).
CHAPTER 23
484 Ex. 1. 2 cjs (2k+ 1)я/10, /с = 0 to 4, i.e. iV0« + 2V5) +Кл/5- l)j, etc.
Ex.2. 1, -2,-2.
Ex.6. -l±j, ω, ω2.
488 Ex. 10. (i) 0, 0, ^(2±2V10); (ii) -1,-1, \(5± V10);
(iii)2,2,K-4±j2V2).
691

ANSWERS
Exercise 23(a)
PAGE
488 1. χ*-5*3-4*2 +16*-8 = 0. 2. x*-3x* + x* + 7x-30 = 0.
3. ;t5+3;c4 + 4*3 + 4*2+3*+l = o. 4. 1, 1, ±JV2.
5. -3,3±j. 6.i-|, -|, -f.
7. (i)4z3 + z2-l = 0; (ii)4z3-llz2+10z-4 = 0.
489 8. (i) z3-6z2-12z-32 = 0; (ii) и>3-15и>2-15и>-16 = 0.
9. (i) 27z3+ 3jz+1 +j = 0; (ii) ^-2г-4 = 0.
10. (i) 2z3-13z2 + 32z-21 = 0; (ii) 2z3+17z2 + 52z+64 = 0.
11. (i) z4 + z-5 = 0; (ii) z4-8z3 + 24z2-33z+ 13 = 0;
(iii)z4-10z2-z+25 = 0.
12. 3,3, f. 13. -l±j, -1±V2.
14. -j, ± V2-j.
17. 3 real roots if and only if -11 < к < 16.
18. 4 real if - 7 < к < 9, etc.
490 19. 2 real if к < -23 or -16 < к < 112;
4 real if -23 < к < -16; none real if к > 112.
20. G2 + 4#3 < 0.
491 Ex. 11. f, -f, ψ.
Ex. 12. 0, 0, 4j, 1.
Exercise 23(b)
494 1. CO 14; (ii) 46; (iii) 162; (iv) 10; (v) 17; (vi) 24; (vii) -};
(viii)22; (ix) -2; (x) 30.
2.(i) 16; (ii) -3; (iii) |; (iv) 67; (v) 32; (vi) 48; (vii) 18.
3.(ΐ)2^2; (ii)(p* + W)/9·; ("0 0.
4. -3, i, 4.
5. -|, -i,i 6. f, i, |. 7.-2,21,5.
8. 1, -2, -3.
495 9. 2, 3, -4. 10. -1, 2, 3. 11. ЪаЪс = аЧ+2Ь\
12. -3, -if.
13. (i) ζ3+2ρζ2+^2ζ-^2 = 0; (ii) z3-pz*-q* = 0;
(iii) z3+.pz-$ = 0; (iv) z3 + 4pz-8tf = 0.
14. (i) zi + 2,qz2 + (iq2+ps)z+qs = 0;
(п)23-2^2+^-^2 = 0.
17. (i) x'+Oq-p^+W-P^x+^-p'r = 0;
(ii) *3 + (9<?-3^2)л:-2^3 + 9^-27/· = 0;
(i)G.P.; (ii)A.P.
496 18. r = pq.
497 Ex. 15. (i) *>-5* + 9 = 0; (ii) 2*3 + 4*-7 = 0.
692

ANSWERS
Exercise 23(c)
PAGE
502 1. -2, l±j. 2. -|,(3±jV7)/4.
3. 2 cos π/9, - 2 cos 4я/9, - 2 cos 2π/9.
4. cos a, cos (§π + α), cos (|π + α) where cos 3a = f.
5.3*+3&, ω3* + ω23&,ω23* + ω3&.
6. 2-2*+ 2*, 2-ω2*+ω2*, 2-ω22* + ω2*.
7. 5+ 6*+6*, 5 + ω6* + ω26*, 5 + ω6* + ω26*.
8. -4, -ω, -ω2.
503 9. -2j,j(l±V2).
И- MV2±JV(2 + 4V2)}, M-V2±JV(2-4V2)}.
Miscellaneous Exercise 23
3. ±8, +288.
4. 1. 5. -2±j, ±j/V2. 6. 84, 708.
7. -i,i- 8. 1, -f;(l, 1, 1),(4, -*, -«.
9.3 roots if -81 =S A: < 44.
505 10. -§, —f, 1±V5.
ll.i20'+l)3+jp30' + 2) = 0.
CHAPTER 24
506 Ex.7. ±3c.
508 Ex.9. i+4j-3k.
Exercise 24(a)
512 4. (i)-2i-j + 5k; (ii) 8i-16j + 8k; (iii)i-j + k;
(iv) -4j+4k; (v) -bci + abk.
5. 0; coplanar with O. 6. -k, -j.
7. (6i + 3j + 2k)/7, f. 8. 3i, 1.
9.i(i+2j + 2k), 9, 9.
513 11. |a|"2. 14. α = r.a/{a.(b Л с)}.
Exercise 24(b)
518 1. (i) 3x-2y+z = 6; (ii) lx+2y-5z = 4; (iii) 4x-4y + z = 5;
(iv) χ - 5y + 3z = 7; (v) bcx -3cay + labz = 0.
2. (i) 3*+4j/-2z = -7; (ii) 2*-3j/+2z = 0; (Щ) x+3y-4z = -5;
(iv) 2*-7j/+4z = 1; (v) 2*-5j/-z = -2.
3. (i)6; (ii)V14; (iii) 2V29.
693

ANSWERS
PAGE
518 4. 0; lines intersect. 5. 2x + 3y-5z-21 = 0.
6. 7x-3y+z-16 = 0. 7.x+y-2z = 4.
519 8.(i)(4,-2,3); (ii)i+2k; (Hi) 6i-5j-3k;
(iv) K*-4) = -iC + 2) = -К2-З).
10. л: = iy = \z.
11. r = c+K[bi Л (а!-с)] Л [b2 Л (а,-с)]}.
12. 2δη$α.βγ). 13. aV6/3.
CHAPTER 25
523 Ex. 2. ^5, f{x) = fh>, 0 < χ < 100.
Ex. 3. 2 χ ΙΟ-4, τ^.
528 Ex. 8. i.
Exercise 25(a)
1. 1, i. 2. 2, 0-117. 3. in, I
4. (i) (*- l)jx for * > 1; (ii) 1 -e-2* for χ > 0;
(iii) Kl-cos π*) for 0 < x < 1.
529 5.1,1,1,1- 6. f, f, 1, тй-
7. 0, 0, 0, (π2-8)/4π2. 8. 1/Λ, (In 2)/Л, 0, 1/Л2.
9. Я*) = 0, *2/4, (2*-1)/4, (6*-2*>-5)/4, 1;
G(y) = 0, \y, K2Jy-D, etc.; ^ = dG/dj/.
10. £θ>) = 2/, 0 < у < 1, 2Х2-У), 1 < у < J2;
1, i;8(2V2-l)/15.
И-t£ т&.
530 12. 1, -4, 6.
13. е-*Ч е-3*'», oe-^Hl-e-*'»)2·
14. 1-5, Ц, 015.
15. 4/я, (2/я) Ь 2, 0-0785, V2-1, 0-854.
535 Ex.12, iz2, -l+2z-iz2.
Ex. 13. Triangular.
Exercise 25(b)
545 1. ■&. 2. l-1/ναθπ). 3. i 4. y£g.
5. 1, i. 6. i 7.4/-/Я.
546 8. 2/·2-16/·2/π2. 9. /2/4, /4/48.
10. α4{4π - π2 - (In 4)2}/4я2. 11. 1, 53, 160.
12. 0-149. 13.2-421, 2-641. 14.0-004.
15. 0-154. 16. 2-19, 78%, σ = 0156.
17. 10-825, 98, 0-271.
547 18. 502 g.
694

Miscellaneous Exercise 25
PAGE
547 1. 8a2/3, 1/V2. 2.18-6. 3. (l+ln4)/4.
4. l/(2Jy), jy, h &.
548 5. 0-301, 0-092. 6. 1 + exp [ - πχ/(σ^3)].
7. 0-683. 8. 01336, 1-630, 4-98.
9. Λ, Λ, 2λ. 10. 183р.
549 12. ^ 13. i. 15. 1-β-λ2,2А-2.
17. «1+jO-*, 2/W(1-z2); -f, 2/V45; 2/π, J{(n*-8)/2π2}.
18. λ/Ο*-λ).
550 19. (i)j> < 1/22-4; (ϋ),ρ = 0. 20. |.
21. 4-471, 5-000, 5-029; 0-2280.
23.1, V65/3, 0-387.
CHAPTER 26
554 Ex. 6. Absolute: 2|Ле,|, |е,/2>/л|, (Ы + Ы)1^(У1 + уг)-
Exercise 26(a)
559 1. 0-7513 ± 0-0006, 1-138310-0020.
2.(i) 22-3610-055; (ii) 88-7510-35; (Ш) 3-333 10-014.
3.0-31784. 4.0-043685(5). 5.5-95 + 0-10.
6. |е| sec2 у. 7. \уе\/(1 +у*)К 8. (i) 0-011; (ii) 0-034.
9.0-39. 10.0-23(1). 11. 2-15(4). 12. 1-15.
13. 1-94, -0-56, -1-38. 14. 1-56, -2-56.
17. η even: no real roots; η odd: one real (negative) root.
569 1. 1-44. Ζ4-64. 3. 1-38. 4. 5-14, -
5. 4-10. 6. 4-13 (5). 7. 2-30, -1-30.
8.2-65,-1-42,-4-23. 9.4-193.
10. 1-092, 1-931. 11.0-8447. 12. 0-7920.
13.0-60. 14.1 - (5k + 3)-1. 15. 0-877.
570 16. 0-838. 17. 3-591.
Miscellaneous Exercise 26
1. 1-532. 4. 60-3.
5. 0-88, 0-78, 0-69, 0-60, 0-50.

PAGE
571 8. 3-316625. 9. 1-81713.
572 12.1-162(28). 13. Κ+1/2Κ3-1/4Κ\
CHAPTER 27
576 Ex. 7. ( ± 3, 0). Ex. 8. (2, - 2), (0, - 2).
577 Ex. 10.0-205.
Exercise 27(a)
585 1. (i) i; (ii) V3/2; (iii) V2/V5. 2. f, (0, ± 2), у = ± §.
3. i, (2±V3, -1). 4. 1/V2, (-2, 1), (-4, 1).
586 5. i, (2, -1), (2, - 3). 6. |f. 7. л— 2j/ =
8. 4x+y=6. 9. (-1,2), (-2,1).
13. Mid-point of OS.
587 22. fly+i¥ = 4*2jA 23. 5, b.
27.*±3j/= ±5V2.
596 Ex. 37. Arc of a circle.
Exercise 27(b)
597 1. (i)|; (ii)V3; (iii) V0/2). 2. 2V3/3, (±2^/2, 0), J2x + 3 = 0.
3.|, (6, -l),(-4, -1), 5л:-21 = 0,5л:+11 = 0.
4.V!,(-3±V5,lX*=-3±4/A
5. (i) Зл:± 2j/ = 0; (ii) J3x±у = 0.
6. (i) л:-2^ = -3, л:+2.у = 1; (ii) x±<J2y = +V2.
7. 4л:2-^2 = 7. 8. 9л:2-4.у2-18л:-24.)/-32 = 0.
9. л:+.у-1=0. 10. л:+3>> = 4, -jf.
И. 2л:+^+3 = 0. 13. л:2-^2 = 8.
598 18. (1, 0), j/-2V2*-+2(V2+ 1) = 0.
19. 5a:-2j/-9 = 0, 2*+5j/-50 = 0, (20, -ψ).
600 Ex. 38. (i) Line parallel to θ = 0. (ii) Circle, (iii) Line at α to θ =
(iv) Quadrant of annulus. (v) Interior of circle, centre (a, 0).
Ex. 39. (i) /-(sin Θ+ cos Θ) = 1; (ii) r = 2 sin 0;
(iii)/-2-2/-(sin0+cos0)+l = 0.
Miscellaneous Exercise 27
605 1. 1/V2, (1, -3), (-2, -2).
606 3. ύ2/α2, л:2+уг — ax sec a — by cosec a = 0.
608 15. 2x(Xl -x2) + 2ΧΛ -j/2) = л-2 + y\ -x\-y%.
16. \ V(l + 4e + 3e2). 17. Ellipse, parabola.

PAGE
613 Ex.4. 3, -2.
617 Ex. 12.-L (J -
CHAPTER 28
ANSWERS
Exercise 28(a)
*»·(-» C)· —-
2x+y = 0, Ax- Ъу =
-.-.mi; ;,. 4.42(; "I).
-3/V13,
-2/V5,
1/V5\ (!),
619 9.(i)
)■ <M
5. x+2y = 0 and jc—у = 0 invariant.
6. x—Ay = 0 and 2x+y = 0 invariant. 8. Maps into χ = y.
3.2n+1-5.3n -5.2n+1 + 10.3n\
З.ги-З'н-1 -5.2n + 2.3n+1 }'
7.2"+1 + 15 З.г^-бХ
35.2" + 35 15.2n-14J'
f4(-l)»-3.2» 6(-l)«+
д_1)п_2п+1 3(-1Г+Ч-2"+
/5.3"-22"+2 -2.3" + 22"+1\
UVJ \l0.3n-5.22n+1 -4.3n + 5.22nj·
10. No. 11. -^; maps everything close to χ =
/11.2"+2-21/г-43 -ЗЗ.г^+Ч-ЗЗ/г + ббХ
• \7.2"+2-14/г-28 -21.2"+1 + 22и + 43/'
(« + l)and(2n+1-l).
mr~1J ~J'" "K~1J --rj-2n+1)·
V' l')^_1M»_')n4-l V_1W+1j_1n+! I»
(1 0 -28^
-7 8 7
0 0 -27У
/-5/V33\ / 1/3 V3\ /1/V3\
2/V33 , 1/3 V3, 1/V3 ·
\ 2/V33/ \-5/3V3/ \l/V3/
2. -2, 3, 0; A singular,

ANSWERS
PAGE
628 3.1, -1,2; detA
— 1, 1, i; eigenvah
4.1,
/1 1
\l/V2/ \f/ \l/V2/'
(ίί4
/1 -2 2\
5.1,2, -3; Ш -1 -2 .
\2 2 1/
629 6. ,y-z = 0, x-y = 0, 15*+13j/-8y = 0.
ΙΙ,λ+μ,λμ. 16. (^ _2°).
630 17. A if «odd, I if «even. 18. (J™ ~^.
(10 9 23\ /4 -2 -3\
5 9 14 J, A-1 = I 1 0 -1 J.
9 5 19/ \-2 1 2/
21. (17I + A-A2)/52.
(2n+4_15 _2n+3_4.3n + 12 2"+4-4.3n-12\
10.2"-10 -5.2n-2.3n + 8 10.2n-2.3n-8 .
-io.2n+io s^+s^-s -Б.г^+з^+я)
631 24. 0, ±jtani(9. 25. x + 2y-6z =
, where аг + 2/?г-6уг = 0, det Ρ φ 0.
2/
632 28. i* = \y =
= ΙΛ A 3 ,w
\7i 72 2/
CHAPTER 29
633 Ex. 1. (-α,Ο).
636 Ex. 5. x±y = 0.
Exercise 29(a)
637 3. (2- t)l(2t + 1). 4. i> +<? + r = 0.
638 8. (4ak\ 4ak). 11. (αί/(1 +13), αί2/(1 + ί3)).

ANSWERS
639 16.8:1.
640 Ex. 7. Ends of diameter.
Ex.8. (1,2, -1), 4.
641 Ex. 10. (2, 2, -1), 3.
644 Ex. 12. Elliptic cylinder.
Exercise 29(b)
652 1. (i) x2+y2 + z2 + 2y = 0; (ii) *2+j/2 + z2-4*+6j/ + 2z-ll = 0;
(iii) x*+y* + z2-2ax-2by-2cz = 0.
2. (i) (3, 1, -1), 4; (ii) (-2, 4, 1), 9; (iii) (a, £>, -c), \a-b + c\.
3. (i) *2+j/2 + z2-4*-2j/ + 2z-32 = 0;
(ii)*2+j/2 + z2-2;c-4z-16 = 0;
(iii) x2+y2 + z2 — ax—by — cz = 0.
4. (i) (4, 1, 3), V6; (ii) (5,-1, 2), 9.
5. i*J{2d2-(a-b)2}. AB is diameter.
6. Centroid of A ABC. ΣΡΑ\ = d2; locus a sphere, centre at centroid of
polygon Ax...An.
653 8. (p2+q2 + r2) (ax + by + czf = (ap + bq + erf (x2 +y2 + z2).
10. (i) Circle; (ii) pair of straight lines; (iii) hyperbola.
11. (i, 1, i), 3V2/2.
13. x-y + 5z= 7,(2, -2,3).
654 14. 12, 70. 16. Circle; rectangular hyperbola.
17. (1Д/2, -1, - 1/V2). 18. True. 19. c\:c\.
656 26. Ellipse; 2^/2/3; x-2y = 0, 2jc+j/ = 0.
27. Rectangular hyperbola. (± 5, ± 1), 5x+y = 0, x-5y = 0.
28. V(|); x + O±2j2)y = 0.
Revision exercise С
657 3. η > 0 and odd; η > 0 only. 4. -1, 0, 3. 5. 8-2°, 90°.
6.0-067. 7.-1,1. 8.1:1.
658 10. Cancellation.
11. (V2, -$π), (V2, Ю; £π.
12. (i) l/(*-l)-4/(*+2); (ii) (2 + 3*)/(l +*2)-3/*;
(iii) 1 /(1 - x) +1 /(1 - x)2 +1 /(1 - 2x).
13. 12-67. 14. (2αί2, 3at).
I5.(i){-1 <*< 1}; (ii){*>l}; (iii) {x = 1};
(iv) {-2 < * < -1}; (v){-l<A;<l);£=i'nC,F=BuC.
659 16. Line * +j/ = α + b. 17. (i) ff; (ii) -&.
699

ANSWERS
/О О 0\
659 18. 12 О 01.
\0 1 О/
20. 7я/12, 5я/4, 23я/12.
21. -Ι+ik, -1/к. 22. л: < -5 or -3 < χ < 7.
23. 1/(1- 3*)- 2/(1 -2л:); -1-х + х\
24. |о|, 0.
660 25. Plane perpendicular to О Α. 26. 1, - 2, - 2, 6.
27. -1, (-1 ± Vl05)/2. 28. \, 0-215.
30. Кби3 - Зи2 - и), ЩгР + Зи2 - и). 31. 2:1.
661 33. (i) 60° < θ < 109-5°, 250-5° < θ < 300°;
(ii) 0° < θ < 30°, 150° < θ < 199-5°, 240-5° < θ < 360°.
35. Λ =-2, (l,f,f).
36. (i) (2ί +1)3 (3ί +1)2/432; (ii) ^; (Hi) 3-5.
38. (i) 1/V33; (ii) 3/V33.
662 42. 3. 43. -2 < χ < О, 2 < χ < 7. 44. (2, -3, 7), 2.
663 47. л/2, 2V2. 50. 5-02, 103-26, no (1 % exceed 106-52).
51. 2, 6.
664 53. cot 0-2" cot (2n0). 54.3/128.
665 60. (i) R; (ii) {x e J?: - V2 < χ < л/2}·
61. 2 cos i<9, £0; (2 cos Wn sin i«0.
62. -1, -};i, -1, -2; -i, -l±Vi
666 64. ^; 2, 0-89, 2; 0-374. 65. ^и>+£-*и>* +α2 = 0.
/1 0 0\ /1 -4 14\ /-73 -32 14\
66. /1 1 01, JO 1 -з), I 16 7 -3J.
\5 -2 l/ \0 0 1/ \ -5 -2 1/
67. iff, f,^-. 68.£;£(4a + 3b).
667 69. * = -j//5 = z/2.
72. (i) 5i + (i+j)-5(i+j + k); (ii) (3i-2k)-(2i+4j + 3k).
73. (f)5, 1 -(f)6-(f)5, (6!) (£)·, (f)-V6; 4; 6, 66.
74. -2i+j + 3k, j/=0. 75.0-966.
668 77. 16j/2-24*z + 7z2 = 0, 4f = 36*- 63.
78. 1 + 5*/2 + 39*2/8, 5х-2у + 2 = 0, curve above tangent.
669 83. i 84. (i) (1, -2, 1); (ii) no solution; (iii) (Л, -ЗЛ, Λ + i); 8; 16.
85. (-1)4
670 87.0. 88. (i) Line; (ii) circle; (iii) ellipse; (iv) parabola.
89. c(a+/?)/2, c(a+/?)/(2a/?); cYt*-2XYt + cX = 0.
90.1(3и-1)я.
700

ANSWERS
PAGE
670 92. χ = Ла+(а л b)/|a|2, у = (1 -Л)а-(а л b)/|a|2.
671 94.pnl(pn+qn). 97. 2x+9y+5z-12 = 0.
672 98.206. 99. 5x + 3y+4z = 0,y-z = 0, x-y = 0.
100. Half-turn about (0, V3).
673 102. a4 - a, a2 - a3, a3 - a2.
104. Line in direction 3i + 2j + k; plane x+y-4z = 0.
674 110. -0 96, 2 88, 5 08.
675 112. (") μ"λ»-η(λ+μ)η; Ο, 00001, 0 0014, 00090; 7.
113. 130-25, 164-77, 215-49, 285-65.
114. (iV-l)(iV+4)/(2iV3).
676 И8.реЧр2(е2а-еа). 119.1-26.
677 121. Л^ > 21.
122. 8Л^(Л^+ l)/(2iV+ l)3.
123.1 ϋ, Ц.
678 128. Integral polynomial in k divisible by k(k+1).
679 133. 2z = - Re (a), 2z = - j Im (a). Rectangular hyperbola. p(z) = 0.

^

Index
Abel, 483
absolute error, 553
affix, 385
ApoIIonius, 603
Argand diagram, 385
argument, 386
auxiliary circle
of ellipse, 577; of hyperbola, 593
Buffon, 549
C, 380
Cardan, 496
cardioid, 447
Cauchy distribution, 549
Cayley-Hamilton theorem, 626
central limit theorem, 543
characteristic equation, 609
chi-squared distribution, 549
complex numbers, 381
addition of, 381; argument of, 386;
conjugate of, 382; division of, 382;
equality of, 381; imaginary part of, 382;
logarithm of, 432; modulus of, 382;
real part of, 382
complex plane, 385
cone, 645
conic, 573
central, 595; conjugate diameters of,
578; diameter of, 578; directrix of, 573;
eccentricity of, 573; focus of, 573
conjugate complex number, 382
с roots of real equation, 484
continuous random variable, 520
cover-up rule, 411
cubic equation, 496
discriminant of, 501; irreducible case,
500
cumulative distribution function, 526
cylinder, 644
de Moivre's theorem, 423
density function, 521
Descartes's rule of signs, 558
diagonal form, 617, 620
diameter, 578
diametral plane, 642
differences, 403
director circle, 595
distribution function, 524
double point, 636
eigenvalue, 610
eigenvector, 610
ellipse, 573
construction of, 577, 582
ellipsoid, 655
equations
of curve in Argand diagram, 438;
polynomial, 483; quadratic, 450; sum
and product of roots, 451, 490
equilateral hyperbola, 593
expectation, 527
exponential distribution, 543
field, 379
focal distance property
of ellipse, 581; of hyperbola, 594
folium of Descartes, 636
Fregier point, 482
fundamental theorem of algebra, 380,483
Galois, 483
generator of cone, 646
Gregory-Newton formula, 406
helix, 645
Horner's method of synthetic division,
402
hyperbola, 573
asymptotes, 591; conjugate axis, 591;
of ApoIIonius, 608; rectangular, 593;
transverse axis, 591
hyperboloid, 655
initial line, 599
invariant line, 609
inversion, 441
iterative methods, 560
linear interpolation, 556
log-normal distribution, 676
major axis, 576
703

INDEX
mean, 527
mean deviation, 528
median, 523
minor axis, 576
mode, 523
modulus, 382
modulus-argument form, 387
multiple roots, 484
nested multiplication, 402
Newton
conic theorem, 480, 607; -Raphson, 563
nilpotent, 629
normal distribution, 535
oblate spheroid, 655
ogive curve, 527
orthogonal matrix, 614
parabola, 465, 573
axis, 465; directrix, 465; family of, 475;
focal chord, 468; focus, 465; Iatus
rectum, 467; vertex, 465
paraboloid, 655
parity, 558
partial fractions, 409
polar
coordinates, 599; equations, 447
probability density function, 521
prolate spheroid, 655
pure imaginary, 385
purely random process, 544
quartic, 503
quartiles, 524
real quadratic function, 457
rectangular distribution, 530
rectangular hyperbola, 475, 593
reflection property
of ellipse, 582; of parabola, 470
relative error, 553
roots, coincident (repeated, multiple), 450,
484
roots of unity, 429
rounding-off, 552
scalar triple product, 510
second-order process, 566
semi-cubical parabola, 633
similar matrices, 620
skew symmetric matrix, 629
sphere, 640
standard normal distribution, 537
stationary process, 544
symmetric matrix, 616
synthetic division, 401
trace, 629
translation of axes, 462
triangle inequalities, 390
triangular distribution, 534
uniform distribution, 530
variance, 528
vector
area, 509; product, 506; projection,
570; triple product, 510
volume
of parallelepiped, 511; of tetrahedron,
511
von Mises's iteration, 567