Pi and the AGM
A Study in Analytic Number Theory
and Computational Complexity
JONATHAN M. BORWEIN
PETER B. BORWEIN
Department of Mathematics
Dalhousie University
Halifax, Nova Scotia
A Wiley-Interscience Publication
JOHN WILEY & SONS
New York • Chichester • Brisbane • Toronto • Singapore

Copyright © 1987 by John Wiley & Sons, Inc.
All rights reserved. Published simultaneously in Canada.
Reproduction or translation of any part of this work
beyond that permitted by Section 107 or 108 of the
1976 United States Copyright Act without the permission
of the copyright owner is unlawful. Requests for
permission or further information should be addressed to
the Permissions Department, John Wiley & Sons, Inc.
Library of Congress Cataloging-in-Publication Data:
Borwein, Jonathan M.
Pi and the AGM.
(Canadian Mathematical Society series of monographs
and advanced texts = Monographies et etudes de la
Societe mathematique du Canada)
"A Wiley-Interscience publication."
Includes index.
1. Numbers, Theory of. 2. Computational complexity,
3. Functions, Elliptic. 4. Pi. I. Borwein, Peter B.
II. Title. III. Series: Canadian Mathematical Society
series of monographs and advanced texts.
QA241.B774 1986 512'.7 86-15811
ISBN 0-471-83138-7
Printed in the United States of America
10 9876543

To our mathematician father, David Borwein

Five thousand digits of pi.
1415926535 8979323846
8214808651 32S2306647
4428810975 6659334461
7245870066 0631558817
3305727036 5759591953
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5024459455 3469083026
5982534904 2873546873
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8583616035 6370766010
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6782354781 6360093417
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4547762416 8625189835
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0674427862 2039194943
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4962524517 4939965143
6868386894 2774155991
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5869269956 9092721079 7509302955
5181841757 4672890977 7727938000
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4668049886 2723279178 6085784383
2694560424 1965285022 2106611863
0865832645 9958133904 7802759009
9909026401 3639443745 5305068203
5977297549 8930161753 9284681382
3836736222 6260991246 0805124388
4220722258 2848864815 8456028506
7745649803 5593634568 1743241125
8617928680 9208747609 1782493858
4775551323 7964145152 3746234364
8493718711 0145765403 5902799344
3191048481 0058706146 8067491927
1426912397 4894090716 6494231961
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2708266830
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5961458901
5403321571
6639379003
1900042296
0386743513
4132604721
From the eleventh iteration of
x„., := (xj'2 + x;"2)/a
yo«:=(y„xJ'2 + xS"2)/(y» + 1)
xo := a"2
y, := a"4

Preface
When I was a student, abelian functions were, as an effect of the Jacobian
tradition, considered the uncontested summit of mathematics and each of us
was ambitious to make progress in this field. And now? The younger
generation hardly knows abelian functions.
How did this happen? In mathematics, as in other sciences, the same
processes can be observed again and again. First, new questions arise, for
internal or external reasons, and draw researchers away from the old
questions. And the old questions, just because they have been worked on so
much, need ever more comprehensive study for their mastery. This is
unpleasant, and so one is glad to turn to problems that have been less
developed and therefore require less foreknowledge—even if it is only a
matter of axiomatics, or set theory, or some such thing.
And so there is nothing for it but to collect together the old subjects in
good references... so that later developments may continue them, if fate
should so decree.
Felix Klein (1849-1925) [79 p. 294]
A central thread of this book is the arithmetic-geometric mean iteration
of Gauss, Lagrange, and Legendre. A second thread is the calculation of v.
The two threads are intimately interwoven and provide a remarkable
example of the application to twentieth-century computational concerns of
the type of nineteenth-century analysis whose neglect Klein so deplores. The
calculation of digits of v has had a fascination that has far exceeded
utilitarian concerns—a fascination that has driven some to dedicate their
lives to calculations we may now electronically effect in seconds. The
methods that make the computation of hundreds of millions of digits of v or
any elementary function within our grasp are rooted in the AGM and this is
where our interest in the material began. Making sense of this material took
us in three directions and motivated our writing this book.
The first direction leads to nineteenth-century analysis and in particular
the transformation theory of elliptic integrals. This necessitates at least a
vii

Preface
ix
tions helped produce a better book. The cheerful technical assistance of P.
Flemming, of our departmental office, and of the staff at Wiley is also
gratefully acknowledged, as is the assistance of the Canadian Mathematical
Society and the support of the Natural Sciences and Engineering Research
Council of Canada.
Jonathan M. Borwein
Peter B. Borwein
Halifax, Nova Scotia
November 1986
1
Complete Elliptic
Integrals and the
Arith metic-Geometric
Mean Iteration
\
2
Theta Functions,
TheAGM
and
Algorithms for Pi
\
3
Jacobi's Triple-
Product,
Theta Functions and
Number Theoretic
Applications
1
4
Modular Equations
and the
Transformation
Theory of
Elliptic Integrals
1
5
Modular Equations
and Algebraic
Approximations
to Pi
^--,
<.
6
The Complexity
of Calculating
Algebraic Functions
♦
7
The Complexity
of Pi and the
Elementary Functions
1
8
General Means
and
Mean Iterations
9
Further
Applications of
Theta Functions
->
^
^-^
->
10
Other Approaches
to the
Elementary Functions
I
1
11
The Story of Pi
Computation
and
Transcendence
- Necessary Reading Order
- Possible Reading Order ,

V1U
Preface
brief discussion of a number of topics including elliptic integrals and
functions, theta functions, and modular functions. These attractive and once
central concerns of analysis have been dropped from the standard
curriculum—and much that is beautiful has become relatively inaccessible
except to the expert or the archivist. In presenting this material we have not
striven for generality. This is available in the specialty literature. At times
we have settled for giving only a taste of the material and a few pointers on
where it can be pursued.
We have found this excuse to consult the nineteenth-century masters a
pleasurable and rewarding bonus—as Hermann points out in his
introduction to Klein [79], "We are so used to thinking in terms of the 'progress' of
science that it is hard for us to remember that certain matters were better
understood one hundred years ago."
The second direction takes us into the domain of analytic complexity.
How intrinsically difficult is it to calculate algebraic functions, elementary
functions and constants, and the familiar functions of mathematical physics?
Here part of the attraction is the surprising answers—the familiar methods
are often far from optimal.
Finally, an honest treatment invites exploration of applications and
ancillary material, particularly the rich and beautiful interconnections
between the function theory and the number theory. Included, for example, are
the Rogers-Ramanujan identities; algebraic series for v; results on sums of
two and four squares; the transcendence of v and e; and a discussion of
Madelung's constant, lattice sums, and elliptic invariants.
Our primary concern throughout has been the interplay of analysis and
mathematical application. We hope we have elucidated a variety of useful
and attractive analytic techniques. This book should be accessible to any
graduate student. Only rarely does it assume more than the content of
undergraduate courses in real and complex analysis. It is, however, at times
terse, at times computational, and some of the exercises are difficult. A fair
amount of the material, particularly on the approximation and computation
of v and the elementary functions, is new and only partially available in
research papers.
The accompanying flow-chart gives possible routes through the material.
Chapters 8 and 11 are largely self-contained. Chapter 8 is a treatment of
general mean iterations, while Chapter 11 sketches some of the history of pi,
its calculation and its transcendence. There are numerous exercises
(frequently with hints). The exercises often develop substantial additional
examples and bodies of theory, and even the casual reader is encouraged to
look at them.
The contributions of family, friends, students, and colleagues have been
many and varied and have greatly facilitated the production of this book. To
all these people we offer our thanks. A particular debt of gratitude is owed
to Professors R. Askey, B. Berndt, R. Brent, K. Dilcher, W. Gosper, Y.
Kanada, D. Shanks, and J. Zucker. Their thoughtful comments and sugges-

Biographical Information
Jonathan M. Borwein was born in St. Andrews Scotland in 1951. In 1971 he
obtained an Honours B.Sc. in mathematics from the University of Western
Ontario, where his father is still head of the mathematics department. He
was awarded an Ontario Rhodes Scholarship that year which he held at
Jesus College Oxford, where he was awarded a mathematics D.Phil, in 1974
under the supervision of Michael Dempster. Since then he has been on the
faculty of Dalhousie University and has been Professor of Mathematics since
1984. He has also been on faculty at Carnegie-Mellon University (1980-
1982) and has spent visiting research periods at Cambridge, Limoges, and
the University of Montreal. His other research interests are in classical
analysis, functional analysis, and optimization theory.
Peter B. Borwein was born in St. Andrews Scotland in 1953. He obtained an
Honours B.Sc. in mathematics from the University of Western Ontario in
1974, and a Ph.D, under the supervision of David Boyd, from the University
of British Columbia in 1979. He spent 1979-1980 as a NATO research
fellow in Oxford. Since then he has been a faculty member at Dalhousie
University and is now Associate Professor of Mathematics. He has spent a
sabbatical year at the University of Toronto. His other research interests are
in approximation theory, classical analysis, and geometry.
xi

Contents
Chapter 1 Complete Elliptic Integrals and the Arithmetic-Geometric
Mean Iteration 1
1.1 The Arithmetic-Geometric Mean Iteration 1
1.2 Gauss's Derivation of the Fundamental Limit Formula 5
1.3 Basic Properties of Complete Elliptic Integrals 7
1.4 Quadratic Transformations and Iterations and a Third
Proof of the Fundamental Limit Formula 11
1.5 Jaeobi's Differential Equation and a Fourth Proof of the
Fundamental Limit Theorem 18
1.6 Legendre's Relation 23
1.7 Elliptic Functions 28
Chapter 2 Theta Functions and the Arithmetic-Geometric Mean
Iteration 33
2.1 A Theta Series Solution to the AGM 33
2.2 Poisson Summation 36
2.3 Poisson Summation and the AGM 40
2.4 The Derived Iteration and Some Convergence
Results 44
2.5 Two Algorithms for it 46
2.6 General Theta Functions 52
2.7 The Landen Transformation 57
Chapter 3 Jaeobi's Triple-Product and Some Number Theoretic
Applications 62
3.1 Jaeobi's Triple-Product Identity 62
3.2 Some Further Theta Function Identities 67
3.3 A Combinatorial Approach to the Triple-Product Identity 76
3.4 Bressoud's "Easy Proof of the Rogers-Ramanujan
Identities 78
xiii

XIV
Contents
3.5 Some Number Theoretic Applications 81
3.6 The Mellin Transform and the Zeta Function 87
3.7 Evaluation of Sums of Reciprocals of Fibonacci
Sequences 91
Chapter 4 Higher Order Transformations 102
4.1 A First Approach to Higher Order Transformations 102
4.2 An Elementary Transcendental Approach to Higher
Order Transformations 109
4.3 Elliptic Modular Functions 112
4.4 The Modular Equations for A and j 119
4.5 The Modular Equation in u-v Form 126
4.6 The Multiplier 136
4.7 Cubic Modular Identities 142
Chapter 5 Modular Equations and Algebraic Approximations to it 152
5.1 Singular Values of the Second Kind 152
5.2 Calculation of a 155
5.3 Further Formulae for a 164
5.4 Recursive Approximation to it 169
5.5 Generalized Elliptic Integrals and Rational and Algebraic
Series for 1/tt and \IK 111
5.6 Other Approximations 191
Chapter 6 The Complexity of Algebraic Functions 200
6.1 Complexity Concerns 200
6.2 The Fast Fourier Transform (FFT) 204
6.3 Fast Multiplication 209
6.4 Newton's Method and the Complexity of Algebraic
Functions 212
Chapter 7 Algorithms for the Elementary Functions 219
7.1 77 and Log 219
7.2 Theta Function Algorithms for Log 224
7.3 The Complexity of Elementary and Elliptic
Functions 226
Chapter 8 General Means and Iterations 230
8.1 Abstract Means 230
8.2 Equivalence of Means 239
8.3 Compound Means 243
8.4 Convergence Rates and Some Examples 249

Contents
xv
8.5 Carlson's Integrals and More Examples 256
8.6 Series Expansions of Certain Means 263
8.7 Multidimensional Means and Iterations 266
8.8 Algebraic Iterations and Functional Relations 273
Chapter 9 Some Additional Applications 281
9.1 Sums of Two Squares 281
9.2 (Chemical) Lattice Sums 288
9.3 Odd-Dimensional Sums and Benson's Formula 301
9.4 The Quintuple-Product Identity 306
9.5 Quintic and Septic Multipliers and Iterations 309
Chapter 10 Other Approaches to the Elementary Functions 316
10.1 Classical Approximations 316
10.2 Reduced Complexity Methods 326
Chapter 11 Pi 337
11.1 On the History of the Calculation of ir 337
11.2 On the Transcendence of v 347
11.3 Irrationality Measures 362
Bibliography 387
Symbol List 396
Index
405

Chapter One
Complete Elliptic Integrals and
the Arithmetic-Geometric
Mean Iteration
Abstract. The focus of this chapter is the arithmetic-geometric mean (AGM)
iteration of Gauss, Lagrange, and Legendre and its relationship to elliptic
integrals. The iteration converges quadratically to a nonelementary
transcendental function simply expressible in terms of complete elliptic integrals. This
result, which is fundamental to this monograph, is established in a variety of
ways. Some of the basic properties of elliptic integrals and functions are
discussed.
1.1 THE ARITHMETIC-GEOMETRIC MEAN ITERATION
One of the jewels of classical analysis is the arithmetic-geometric mean
(AGM) iteration of Gauss. It is the following two-term recursion:
C1 1 1\ „ ■ an + bn
(1-1-1) «»+!•= n
(1.1.2) *W=VoA.
It is customary and useful to introduce an auxiliary variable,
(1-1-3) W=i(a„-6„).
If we assume that 0<60^a0, then from the arithmetic-geometric mean
inequality we have
1

2
The Arithmetic-Geometric Mean Iteration
and
(1.1.4) 0^«„+1-6„+1--(v_; + v^)2.
Whence we observe that an and bn converge to a common limit determined
uniquely by a0 and b0. This common limit will be denoted (on letting a'= a0
and b '• — b0) by
(1.1.5) M(a, b) := lim an = Urn bn .
We will refer to this limiting process as the AGM and usually reserve the
symbols an, bn, and c„ for variables bound by the AGM relations. We also
reserve the symbol AG(a, b) for the common limit.
We say that a„ —> a ,with pth-order convergence if
(1.1.6)
(a„ - a)p
= 0(1)
where, as usual, an = 0(j8„) means that, for some constant c and for all n,
an :£ cj8„ . If the an are functions defined for all x in a set jRT, and if the
implicit constant concealed by the O symbol in (1.1.6) is independent of x,
then we say that the convergence is uniformly pth order. Roughly speaking,
quadratic (second-order) convergence doubles the number-of-digits
agreement between successive iterates and the limit, cubic (third-order)
convergence triples the agreement, and so on.
It is an easy observation from (1.1.4) that the AGM converges uniformly
quadratically for a0, b0 restricted to compact subsets of (0, °°). Very precise
estimates for the rate of convergence will be established later.
We also observe that M(a, b) is homogeneous, that is, for A > 0
(1.1.7) AM(a, b) = M(Xa, Xb)
and thus there is little loss of generality, though often some loss of
symmetry, to setting a = 1.
The function M satisfies
(1.1.8) M{a,b) = M(~^-,VaT))
or
(1.1.9) M(l,6) = i^M(l,|^).

1.1 The Arithmetic-Geometric Mean Iteration
3
The analysis of the limit of the AGM rests on finding a two-variable function
M invariant under the transformation (1.1.8) or, equivalently, on finding a
function / satisfying the functional relation (1.1.9), namely,
(1.1.10) /W-HMlS)-
If we set k0 '■= x G (0,1) and
(1.1.11) kn+1: =
*/K
l + kn
then since
"+1 (i + VU2(i + U
we have that kn-*l quadratically. In fact, the function g defined by
(1.1.12) g(x):=af[^r^ k0:=x
n=0 z
is the unique solution of (1.1.10) analytic in a neighbourhood of 1 that
satisfies g(l) = a. (See Exercise 4.) This form of the AGM, as a single
variable iteration, is usually called the Legendre form.
It is convenient and standard to define the complement x' of x by
x . Differentiation will be denoted by /.
Comments and Exercises
The early history of transformations of elliptic integrals, of which, as we
shall see in the next section, the AGM is an example, is laid out in an
entertaining article by G. N. Watson [33] entitled "The Marquis and the
Land-Agent: A Tale of the Eighteenth Century." The marquis is Fagnano
and the land-agent is Landen. Landen's transformation, published in 1775,
will be discussed later. The names of Euler and Lagrange should also be
associated with the early transformation theory. Lagrange uncovered the
AGM iteration sometime before 1785. Gauss rediscovered it independently
in the 1790s. He apparently first considered the iteration in 1791 at the age
of 14 (Almquist and Berndt [Pr]). It is, however, Gauss and Legendre who
develop the theory fully. As Watson [33] points out, "in the hands of
Legendre, the transformation became a most powerful method for
computing elliptic integrals." Gauss is unique in having deduced the invariant
function from the functional equation rather than proceeding in the opposite
(and easier) direction.

The Arithmetic-Geometric Mean Iteration
Deduce that, for the AGM,
an = an+1 + cn+1 bn = an+1-cn+1 c2n = a2n-b2n.
Hence the AGM is well defined for negative n. Show that
a_„ = 2"a„* 6_B=2"c; c_n = 2nb*n,
where a*, b* and c* are generated from the AGM commencing with
a% := a0, b% '■= c0, and cj ".= b0. Show that
c»-i 1
c = =S — c
"■ 4a„ 2 "-1-
Observe that this formula avoids the subtractive cancellation problems
inherent in calculating the very small number c„ from cn+1 = \{an - bn).
Show that
«n + V«n-i(2«n-«n-i)
a„^ = r
Consider the harmonic-geometric mean iteration
._ 2a„)3„
«» + A,
Show, for a0, /30E(0, °°), that the above iteration converges quadrati-
cally to H(a0, )80), where
H(a0, &)'=
M(l/a0,l/P0) ■
Consider the arithmetic-harmonic mean iteration
.= <*„ + /?,,
«»+i • 2
2«„)8„
A,+i: =
«» + A,
Show, for a0, )80 G (0, °°), that the above iteration converges quadrati-
cally and that
lim an = lim /3„ = V^A ■

1.2 Gauss's Derivation of the Fundamental Limit Formula
5
Show that the AGM is a well-defined quadratically convergent iteration
for starting values a0 '■= 1, b0- = z, where re(z) > 0. Likewise the
function g of (1.1.12) is a single-valued analytic function on re(z)>0. In
both cases the convergence is uniformly quadratic on compact subsets.
[The root must always be chosen to lie in re(z) > 0.] ■ Show that
g(z) = M(l, z) for re(z)>0.
1.2 GAUSS'S DERIVATION OF THE FUNDAMENTAL
LIMIT FORMULA
By May 30th, 1799, Gauss had observed, purely computationally, that
2 f1 dt
(1-2.1) ^77-^ -d " J0 ^^
M(l, V2)
agreed to at least eleven decimal places. He commented in his diary that this
result "will surely open up a whole new field of analysis"—a claim
vindicated by the subsequent directions of nineteenth-century mathematics. The
inverse of the above (indefinite) integral is the lemniscate sine, a function
Gauss studied in some detail. He had recognized it as a doubly periodic
function (see Section 1.7) by the year 1800 and hence had anticipated one of
the most important developments of Abel and Jacobi: the inversion of
algebraic integrals.
We now outline Gauss's derivation of the limit of the AGM (Gauss
[1866]). This is not the easiest development but it may be the most
motivated.
Theorem 1.1
2 r12 dd
Jo
M(l, x) it Jo yi - (1 - x2) sin2 8 '
First proof. Observe that [M(l + x, 1 -x)]~\ defined by (1.1.5), is
analytic and even in some neighbourhood of zero. (See Exercise 4, Section
1.) Thus we may suppose that
(1.2.2) 777— : r = 1 + d,x2 + d2x4 + d,x6 + ■■•.
v ' M(l + x, 1-x) x 2 3
Upon application of the AGM transformation we have
1 1
(1.2.3)
M(l + 2Vx/(l + x), 1 - 2V*/(1 + x)) M(1> yi-4x/(l + xf)
- 1 + x
~ M(l + x, 1 - x) "

6 The Arithmetic-Geometric Mean Iteration
Comparing (1.2.2) and (1.2.3) gives
(1 +x)(l + dlX2 + d2x4 + ---) = 1 +d^—j +d2(TT^) +■■-.
(1.2.4)
We leave it as an exercise to the reader to follow Gauss's footsteps by
solving the above equation for dt. In fact,
t2
42'-1
(1-2.5) -J'-L «0^1)1
If we observe, as in (1.2.3), that
(1,2'6) M(l,Vl^x~2) = M(l + x,l~x)
we see by (1.2.2) that we are finished if we show that
/1 o ^ 2 r'2 de V
(1.2.7) - . =Z I .,,. 1X,
irJ° Vl-*2sin20 #_0 L*!(» — 1)!
2 I"*'* <*» _y f(2f-l)I12 -a
-1)!
2 _■ 2 /i\-l/2
4«-»
This final equation requires expanding (1 - x sin 8)~ and integrating
term by term.
It is perhaps possible to be guided to the limit of the AGM by the above
method. If, however, one has correctly guessed the limit, then proving it
correct is much more straightforward and only involves establishing the
invariance of the limit under the transformation.
Second proof. Let
(1.2.8) na,b):=lfQ
r/2
de
Va2 cos2 e + b2 sin2 0
Then, as the substitution f- = b tan 6 shows,
dt
(1.2.9) T{a, b) = -\
IT J -
V J— V(«2 + t2)(b2 + t2) '
Now the substitution u- = \(t — abli) (and some care) yields
(1.2.10) T(a,b)=T(~~-,V^b).
It follows that T(an, bn) is independent of n and hence, since (1.2.9)

1.3 Basic Properties of Complete Elliptic Integrals 7
evaluates as an arctan when a = b, we have on interchanging limit and
integration
(1.2.11) T(a0, b0) = T(M(a0, b0), M(a0, b0)) = ^ . □
Comments and Exercises
An excellent account of the development and the importance of elliptic
function theory in the nineteenth century is to be found in Felix Klein's
classical work, "Development of Mathematics in the 19th Century" (Klein
[79]). Gauss's works are, of course, available in collected form (Gauss
[1866]). Before establishing the limit formula, Gauss produces partial
expansions of M(l, x) and M(l + x, 1 - x) plus a number of AGM
calculations carried to as many as 20 decimals. It is apparent that both the
observation of the limit and the route to a proof were indicated by
prodigious numerical experimentation.
The second proof may be found in Carlson [71], Newman [82, 85], Todd
[79], or Wimp [84]. Carlson [71] offers some interesting generalizations.
These are discussed in Section 8.5.
1. Fill in the details in the above proofs. In particular prove (1.2.5),
(1.2.7), and (1.2.10).
2. Show that [M(l + x, 1 - x)]'1 and [M(l, x)]"1 both solve the second-
order differential equation
(*3 - *) 71 + (3*2 - 1) T~ + xy = 0 .
dx dx
Hint: For the second solution consider equation (1.2.6).
1.3 BASIC PROPERTIES OF COMPLETE ELLIPTIC INTEGRALS
The two basic integrals we will encounter are the complete elliptic integral of
the first kind,
dd
rn/2 j.
2 • 2 n
sin 8
- f1 dt
~J° V(l - t2)(l - k2
V(l - t2)(l - k2t2)
and the complete elliptic integral of the second kind,

8 The Arithmetic-Geometric Mean Iteration
ir/2 /
de
ctt/2
(1.3.2) EW:=)o Vl - k2 sin2 8
-I
VT^tV
o VT1
<fc.
r
The complementary integrals E' and £' are the integrals in the
complementary variable k' = VI - k2,
(1.3.3) K'(k):=K0/l-k2) = K(k')
(1.3.4) E'(k):=E0/l-k2) = E(k').
As is traditional, we willuse the notation f'(k)'-=f(k') to indicate any
function in the complementary variable. The variable k is often called the
modulus, and k' is the complementary modulus.
The second integral arises in the rectification of ellipses. The arc-
length A of an ellipse with semiaxes a and b is given by
A = 4jo W cos2 8 + b2 sin2 8 d8 = 4aEM-) .
The first integral has the following physical interpretation. If p is the
period of a pendulum with amplitude a and length L, then
K-WfiMD)
8
where g is the gravitational constant. Note as a—>0, K(sm(a/2))—> v/2,
and we are left with a simple harmonic approximation.
The Gaussian hypergeometric series, discussed by Gauss in 1812 in what
is one of the first rigorous discussions of convergent series, is defined by
,..<,« ., ^ , a-b a(a + \)b(b + V) 2
(1.3.5) F(a,b;c;z):=l+ — z+ \.2.^+1)^2
a(a + l)(a + 2)b(b + !)(& + 2) 3
l-2-3-c(c+l)(c + 2) Z •
For the complete elliptic integrals we have the series expansions, for
1*1 <1,

1.3 Basic Properties of Complete Elliptic Integrals 9
where (2/ - 1)!! := 1 • 3 • 5 • • • (2/ — 1). The derivations are left as Exercise 1.
We have, as in Exercise 2 of Section 1.2, that K and K' are both solutions
of
(1.3.8) (k3-k)^+(3k2-l) ^~ + ky = 0.
This equation has a regular singular point at zero (the roots of the indicial
equation are both 0) which, as the reader familiar with the elementary
theory of second-order differential equations knows, says that
(1.3.9) K'(k) = alogk K(k) + f(k)
where /is analytic in a neighbourhood of zero. (See, for example, Birkhoff
and Rota [69].)
In fact,
(1.3.10) K'(k)=llog(f)K(k)
This logarithmic asymptote at zero will be of considerable interest later in
the discussion of the complexity of log. Also,
(1.3.11) ^) = 1+1(^)-^
^M-iVsV^K---
Once again the above verifications are left as exercises.
The first and second integrals are linked by the equations
/.o,^ dE E~K
<L3-12> ik = -nr
/--,.,^ dK E-k'2K
(L3-13) * = "W~-
(See Exercises 2 and 3.)

10
The Arithmetic-Geometric Mean Iteration
Comments and Exercises
The functions K and E are nonelementary transcendental functions. This is
a result of Liouville's. In general an elliptic integral is an integral of the form
J R(x, y)dx,
where R is a rational function of x and y and where y2 is a quartic
polynomial in x. Except in special cases, such as repeated factors, this is
always a nonelementary function of u. Incomplete elliptic integrals of the
third kind are of the form
f" dx
•>° (1 - t,*2)V(1 - x2)(l - k2x2) '
The integral is complete when u = 1. (The complete third integral can be
expressed in terms of K and E.) Analogously we may define incomplete
elliptic integrals of the first and second kind. The basic result due to
Legendre is that any elliptic integral may be algebraically reduced to a linear
combination of elliptic integrals of the first, second, and third kind. (See
also Exercise 5 of Section 1.4 and Exercise 6 of Section 1.6.)
The formulae in the section may be found, in tabular form, in Ab-
ramowitz and Stegun [64], Gradshteyn and Ryzhik [80], and the Bateman
project (Erdelyi et al. [53]), so may transformation formulae for the
hyper geometric functions. Of course the indispensible companion volume is
Whittaker and Watson [27].
A number of the seminal nineteenth-century papers, including Gauss's on
the hypergeometric series and Jacobi's on theta functions, are available in
translation in Birkhoff [73].
1. Establish the series expansions (1.3.6) and (1.3.7) for K and E by
expanding the radical by the binomial theorem and integrating term by
term.
2. Establish, by differentiating the integral (1.3.2), that
dE E-K
dk~ k '
3. Verify, from the series expansions (1.3.6) and (1.3.7), that
dK _E- k'2K
dk kk'2 "
4. From the integral representations show that

1.4 Third Proof of the Fundamental Limit Formula
11
a) £(0) = f £(0) = f £(1) = 1.
b) K'(k) = log(|) + 0(A:2|log k\) k I 0 .
Hint:
r12 de = r
J° Vl - (1 - it2) sin2 e ->c
A:' sin 8 d8
Vl - (1 - A:2) sin2 8 Jo Vit2 + (it')2 cos2 8
+
r/2 /r=~F
Jo V l + if
sin 0
sin 8
dd .
The second integral evaluates to log [(1 + k')/k\. (See Borwein and
Borwein [84a].)
c)
*'(*)- log (I)
;4A:2(8 + |logA:|) A: G (0,1].
5. Verify the expansions (1.3.10) and (1.3.11). (See also Exercise 1,
Section 2.3.)
6. Establish the relation, for re(c) > re(6) > 0,
F(a, b; c; z) = C:jQ t"-\l - t)c-b~\l - tz)~
dt
where C is a constant independent of z. [In fact C = T(c) /T(b)T(c - b)
where, as usual, T is the gamma function. (See Section 1.6.)] Note that
this provides an analytic continuation of F to C — [1, °°).
7. Show that F(a, b;c;z) satisfies
z(l - z) ^4 + [c - (a + 6 + l)z] ~^-aby = 0.
This is the hypergeometric differential equation.
1.4 QUADRATIC TRANSFORMATIONS AND ITERATIONS AND A
THIRD PROOF OF THE FUNDAMENTAL LIMIT FORMULA
The complete elliptic integrals satisfy the following functional relations
which we collect together as

12 The Arithmetic-Geometric Mean Iteration
Theorem 1.2
For A: £(0,1),
(«) m = ^-kK{^) (upward)
(b) TO = T^7*(yTf) (downward)
(c) E(k) = ^E(^) + ^K(k) (upward)
(d) E(k) = (l + k')E(~j^j-k'K(k) (downward).
Proof. Parts (a) and (6) are equivalent and follow from the previous
discussion. We give a direct proof of (b) based on Ivory [1796]. This is also a
third proof of Theorem 1.1. Let 1:=(1-k')/(1 +k'). Then we have
V7 = kl(l + k') and 1 + /2 = 2(1 + k'2) /(1 + k'f. Thus
d6
m^-u'^?
V(l + /2) + 2/cos20
on replacing sin2 0 by (1 - cos 20) 12. Then
(~r")*w=\ Jo*(1+/e~2l'flr1/2a+/e2l'fl)~1/2 d0
2m,»=o \ m /\ n /Jo
Here we have used the binomial theorem twice. Since only the terms with
m = n are nonzero, we have
(^■fjO-f'd.!*')-'"
on using (1.3.6). This completes (b). If g(k): = 2Vk/(l + k), then g'1 =
(l-k')/(l + k') and [1 +g'(k)]/2= 1/(1 +k). Now (a) follows by
substituting g(k) for /: in (6). To derive part (c), we differentiate (a) to get, for
K = dK/dk,
(1.4.1) (1 + /:)¾ + K(k) = K(g(k))g(k) .
This is coupled with the differential equation (1.3.13) in the forms

1.4 Third Proof of the Fundamental Limit Formula 13
(1.4.2) E(k) = kk'2K(k) + k'2K(k)
and
(1.4.3) E(g(k)) = g(k)[g'(k)]2k(g(k)) + [g'(k)fK(g(k)) .
We now use (1.4.1) to eliminate K(g(k)) from (1.4.3) and then employ
(1.4.2) and (a) to solve for E(g(k)) in terms of K(k) and E(k). The
algebraical details are left to the reader. Part (d) may be derived
analogously from (b) or by substituting g~1(k) for k in (c). □
The transformations are termed upward and downward because, for
k G (0,1), iterating (a) leads to a sequence of k values increasing to 1 and
iterating (b) forms a sequence of k values decreasing to 0.
It is convenient to introduce homogeneous forms of E and K and to
recast Theorem 1.2 in AGM terms.
Let
(1.4.4) Ha,*):-!"* ., ,« , , .ilffl
Jo Va2 cos2 8 + b2 sin2 8 a Vfl/
(1.4.5) J(a,b):=JQ ■ Va2 cos2 8 + b2 sin2 8 d0 = «£'(-) •
Theorem 1.3
If a„+1 := (a„ + 6„)/2, 6„+1 : = V«A. and 0 < 6„ < a„, then
(«) '(«n+i,fen+i) = /(«n>&n)
(6) 2/(a„+1, 6„+1) - /(a„, 6„) = anbnI(an, bn).
Proof. Part (a) has been observed in the second proof of the
fundamental limit theorem (Theorem 1.1).
To see part (b), notice that if kn ■= cnlan, then k'n= bjan,
2/(«B+i, bn+1) = 2an+1E(yjl - ^) = 2an+1E(kn+1)
and
J(aH,bn) = aHE(^l- ^)= <*„£(*„).
Recall that c2 = a2 — b\. The relationship between kn+1 and A:„ is given by
. =Qn-bH = l-k'n
Kn+1 a„ + b„ 1 + k''

14 The Arithmetic-Geometric Mean Iteration
Thus, establishing (b) is equivalent to establishing
2an+1E(kn+1)-anE(K) = bnK(kn)
which on setting an+1 = (an + bn)l2 and dividing by an may be seen to be
equivalent to part (d) of Theorem 1.2. □
These transformations may be iterated to produce quadratically
convergent algorithms for K and E. (See also Exercise 1.)
x I! (1 + k'„) (upward iteration)
2 „=i
*oe(o,i].
CO
x ll (1 + kn) (downward iteration)
2 „=i
where
*"+i:=TT*^ and ^o e [0,1).
The first product is the unique solution of Exercise la) analytic in a
neighbourhood of 1 that takes the value irl2 at 1, while the second product
is the unique solution of functional relation (b) of Theorem 1.2 that takes
the value v/2 at 0 and is analytic in a neighbourhood of 0. This observation
verifies that the above products are analytic in neighbourhoods of 0 and 1,
respectively, since they are uniformly convergent products of analytic
functions. Also, specifying the value of an analytic solution of either functional
relation at any point in (0,1) in fact specifies the function at an infinite set of
points with limit point within the domain of analyticity and hence uniquely
defines the function up to a constant multiple of the value at the limit point.
(See Exercise 4.)
The algorithms for E and K in AGM form are particularly attractive.
Algorithm 1.2
For a0:= 1, b0 ■= k'E.(0,1], and c0:= k,
Algorithm 1.1
(a) *'(*„)= fn^nr—
z „=0 l ^ Kn
where
(6) ^)-111^ =

1.4 Third Proof of the Fundamental Limit Formula
15
(b) ^"fi-irV.W
Proof. Part (a) is Theorem 1.1. For part (6) we use Theorem 1.3,
and since 4a2+1 - 2a2 - 2a„6„ = -c2,
2"+1[/(a„+1, 6„+1) - a2+1/(a0, &„)] - 2"[/(a„, 6„) - «^(«0, 60)]
= 2-^/(^,6,,).
Thus on summing
(1.4.6) J(a0, b0) = (a2 - 2 2^)/(^, 60)
N n=0 '
which specializes to (6). Here we must observe that
An:=2n[a2J(an,bn)-J(an,bn)]
^ («2 - 62) sin2 8
y/a2n cos2 0 + 62 sin2 0
dd
sin2 0
Va2, cos2 0 + 62 sin2 0
Thus Os A„ <2"c2/(an, 6„), and A„ tends to zero as n—>°° since c2 tends to
zero quadratically. D
Comments and Exercises
The algorithms of the section provide remarkably efficient methods for the
calculation of E and K and related functions. The analysis of these
algorithms will be reserved for later chapters. We have restricted our attention
in this section primarily to a real variable k. This is simplifying but entirely
unnecessary. All of the algorithms and functional relations extend naturally
to the complex domain. In fact, all the algorithms and functional equations
of this section hold at least for kE.{re(z)>0} — [1,00). The interested
reader may readily establish the exact domains of validity for the various
relations. The analysis of the AGM iteration for complex starting values is
reasonably complicated. (See Cox [85].) The problem is to decide which

16
The Arithmetic-Geometric Mean Iteration
root is appropriate in the computation of 6„+1 = ^Janbn. The right choice is
made to ensure \an+1 - bn+1\ < \an+1 + bn+1\ [with im (6„+1/a„+1) >0 in the
case of equality]. The surprise is that no matter how the roots are chosen,
the AGM iteration converges (provided a0^—bQ) though unless the right
choice is made all but finitely often, the limit will be zero.
Exercise 5 on the Landen transform provides an algorithm for calculating
incomplete elliptic integrals. We will revisit Landen's transform in Chapter
2. A wealth of formulae on the calculation of elliptic integrals may be found
in King [24].
1. Show that
a) K'(k)
L_ r.( *rt\
K
l + k"\l + k/
b) kW^^TTf)
c) E'(k) = (l + k)E'(j^)-kK'(k)
d) ^)-(^)^)4^)
and observe that K'lK satisfies a multiplication theorem, namely,
Show that for every integer n there is a unique algebraic x G (0,1) so
that
Show that if kn+1 :=(1- k'n)/(1 + k'n) and )n : = yfk^H, then
and hence
b) K(x) =
i+^w:
(1 + VF)2*VVL1 + ^T7
K
)■
Also

c) *(/&=? n
1.4 Third Proof of the Fundamental Limit Formula
77 TT 4
17
f3 = f ri(i + ;„)2
2-0(1 + ^1^)2 2^
Show that /„ converges quartically to zero.
(77ie quartic AGM) Let a„, 6„, and c„ satisfy the AGM relations. Set
an := a2n and Pn'-=b2„2- ShoW that
a) «n+l =
and
<*n + Pn
b) ft + l = (^^)
1/4
Show also that
c) s 2-¾=s 4-«:- (^ .
n=0 n = 0 L \ Z, / J
Show that the convergence is governed by
4 _ (<*„ ~ ft,)4
d) a»+i-jSll+i*
16
Derive the following quartic algorithms. For a0:=l and j80:=V^fe7(
(0,1],
e) K(k) =
21im«:
f)
K(k)
-14-(^)1
Consider a functional relation
F(z) = s(z)F(g(z)) F(0) = a*0
where s and g are analytic in some complex neighbourhood U of zero.
Suppose that, for some p>\, limn_>eo g(n)(z)—>0 with />th order
convergence uniformly on U. [g(n) denotes g composed with itself n times.]
Show that the above relation has a unique nonzero analytic solution on
U if and only if s(0) = 1.
(The Landen transform) Consider the incomplete elliptic integral of
the first kind
P(4>:
■'*>:-j«
de
0 Vl - k2 sin2 6
kE[0,l), ^0.

18 The Arithmetic-Geometric Mean Iteration
Show, as in the second proof of Theorem 1.1, that if we set kn+1 •'=
(1 - k'n)l{\ + k'n) and tan (<£„+1 - <£„) = /^ tan <f>n, then
a) F(<t>n+1,kn+1) = (l + k'n)F(<l>n,kn) kn+1<kn, 4>„+1^„
and
b) ^.*.)=(noT^)um(^).
Show that if kn+1 : = 2yfTj(l + kn) and sin(2<£„+1 - <f>n) : = /:„ sin <£„,
then
c) F(<£„+1,/;„+1) = |(l + *„W„,/;„) kn+1>kn, <£„+1^„
and
d) J**,, k0) = ( fi -^) log tan (f + I lim *„) .
e) Establish the quadratic convergence.
Similar methods for calculating incomplete second and third integrals
may be found in King [24]. (See also Section 2.7.)
6. Prove Euler's addition theorem. Let g(x) : = (1 - x2)(l — k2x2). Then
Jo VgW+Jo V^W J° Vg(x)
where c : = [by/g(a) + ay/g(b)] /V1 - k2a2b2. This result, which dates
from 1753, is, according to Birkhoff [73], the "first notable theorem
about elliptic integrals."
1.5 JACOBVS DIFFERENTIAL EQUATION AND A FOURTH PROOF
OF THE FUNDAMENTAL LIMIT THEOREM
The second-order linear differential equation (1.3.8) satisfied by j&Tand K' is
(*3 - k)^2 +(3k2 -1) ^ + ky = 0 .
Equivalently
G(k):=kV2k'K(k) and G*(k):= kll2k'K'(k)
satisfy

1.5 Fourth Proof of the Fundamental Limit Theorem 19
d*y 1 /i + £2\2
Also, the functional relations [(Theorem 1.2(a) and Exercise la) of Section
1.4] become
and ,
rTTTT /2Vk\
(1.5.3) G*(k) := V2(l + Qyjj^ G*[TTk)-
If we set g(fc) : = 2Vk/(l + k), these become
(1.5.4) <Kk) = £^<K&))
and
(1.5.5) G*(k) = {j^G*(g(k))
2_
8(k)
Theorem 1.4
Suppose that /, g, and a are all in C2(0,1) and that g maps [0,1] into [0,1],
If, for x G (0,1),
. d2f(x) .
(«) -^2 = «(*)/(*)
and
(fe) /(*) = V5a /(#(*)) c a constant
then
(c) a(x) = (g(x)Ta(g(x))-
1
*(*) 2 \g(x)
\g(x)J J
Proo/ Use (b) to change variables in (a). Then, on suppressing x,
2£
(1.5.6) «/ = /= (2pg +P8) ^f+pf(8) + 82P ^fi
where p '- = y/clg (remember that a is the derivative with respect to x). Also
from (a) and (b),
(1.5.7) f=af=apf{g).
Substituting (1.5.7) into (1.5.6) to eliminate / yields

20 The Arithmetic-Geometric Mean Iteration
dg2 g2p dg \ g2p IJK6)
which on comparison with (a) gives
ap-p ,
•2„ =<*(g)
8 P
since Ipg + pg = 0. Finally we compute p, p and p in terms of g to get
(c). D
The bracketed quantity on the right of (c) is often called the Schwarz
derivative of g.
Corollary 1.1 (Jacobi's Differential Equation)
Suppose G(x)'- = xll2x'K(x). Up and g are algebraic functions that map
[0,1] into [0,1] and
G(x)=p(x)G(g(x))
then g satisfies an algebraic differential equation
1 T a(r\ ^ f S(y\^2
r(x) = [g(x)fr(g(x))
where
M-1( MX
g(x) 2 \g(x)I
... 1 (l + x2\2
r{x) -=-^1^72)
Proof. We may assume that G(x) does not satisfy any equation of the
form
G(x) = p(x) G(x)
with )8 algebraic since by (1.3.13), (1.3.10), and (1.3.11) GIG has a
logarithmic singularity at 1. Thus since G satisfies equation (1.5.1), as in
(1.5.6) we must have
2pg+pg = 0
or
[c
P <S
The result now follows from Theorem 1.4 applied to (1.5.1). □

1.5 Fourth Proof of the Fundamental Limit Theorem 21
This corollary has a converse which provides an algorithmic check on
whether g is an algebraic transform of K. (See Exercise 1.) Theorem 1.4 has
a partial converse.
Theorem 1.5
ippose that j
Suppose that
Suppose that/, g, and a are all in C2(0,1) and that g maps [0,1] into [0,1].
2 . ... 1
(a) a(x) = (g(x)ya(g(x))-
8(x)
M)2]
g(x)J J
2 Lg(*) 2 \g(*)
and
(b) ^ = «(*)/(*).
Then
(<0 \/?/U(*))
is also a solution of (b).
Proof. This is essentially the computation of (1.5.8). Replacing / by
yfcig /(g) leaves (6) invariant provided (a) holds. □
We now offer a proof of Theorem 1.1 based on Theorem 1.5.
Fourth proof of the fundamental limit theorem. For
r \. l (1±jl\2 a t \- 2V*
aW-=~i?il^7i and ^-=1^
condition (a) of Theorem 1.5 holds. This is a tedious though (eventually)
entirely rational calculation. Thus if G is a solution of
d2G(x) ,
v2 =a(s)G(s)
a*
then so is
>SS <***»■
f2gW
However, the above second-order linear differential equation has a regular
singular point at zero and has fundamental solutions of the form

22
The Arithmetic-Geometric Mean Iteration
Vxgt(x)
and
Vxg^x) log x + Vxg2(x)
where gt and g2 are analytic and nonzero. Now observe that
is analytic and nonzero in a neighbourhood of zero, and hence G(x) must be
a constant multiple of the first fundamental solution. Moreover,
_L /ZL
Vx V 26(x
2g(x)
G(g(x))
cannot have a logarithmic singularity at zero, and hence (2g(x)y1/2G(g(x))
must also be a constant multiple of the first fundamental solution. In
particular for some c,
G(x) = cy[^~G(g(x)).
If we multiply by yg(x)7x and take the limit as x—>0, we deduce that c = 1.
We have thus proven that G satisfies equation (1.5.4). Equation (1.5.4)
transforms into
and we have shown that K, defined as an analytic solution of the differential
equation (1.3.8), satisfies the above functional relation, and hence we have
found an analytic invariant for the Legendre form of the AGM. [See
(1.1.10).] □
Comments and Exercises
A form of Jacobi's highly nonlinear differential equation may be found in
Cayley [1895], which is an excellent account of nineteenth-century elliptic
function theory with particular emphasis on the transformation theory.
The general question of when a functional relation
f(x) = p(x)f(g(x))
has a closed-form solution in terms of familiar functions is difficult. We shall
consider it again later. Theorems 1.4 and 1.5 do, however, suggest how one

1.6 Legeudre's Relation
23
might proceed to check whether a solution of the above functional relation
satisfies a second-order differential equation with rational coefficients. (See
Exercises 2 and 3.)
1. Suppose that g(x) is an algebraic function, g maps [0,1] into [0,1], and
g(0) = 0. Suppose that g satisfies the algebraic differential equation of
Corollary 1.1. Show, as the fourth proof of Theorem 1.1, that there
exists a constant c so that
G(x) = 4^-G{g{x))
2. Suppose that
F« = ^F^*»
and that
d2F(x) , , ^ .
where F{x)IVx is analytic in a neighbourhood of zero, r is a rational
function, and g(x) '■ = 2Vxl{\ + x). Without identifying F explicitly,
show that r has double poles at 0 and 1. (Considerations of this nature
can turn the fourth proof into less of a verification.)
Suppose that a, p, and g mapping [0,1] into [0,1] are algebraic, that
4-£$- = a(x)f(x) x G (0,1)
and that
dx>
f(x) = p(x)f(g(x)) * £(0,1)
Suppose also that / does not satisfy a linear differential equation with
algebraic coefficients of order less than 3. Show that
g(x) = ~^2 and P(x) = <cx + df ■
1.6 LEGENDRE'S RELATION
The four quantities K, E, K', and E' are related by a remarkable relation.

24 The Arithmetic-Geometric Mean Iteration
Theorem 1.6 (Legendre's Relation)
For 0<fc<l,
E(k)K'(k) + E'(k)K(k) - K(k)K'(k) = | .
Proof. Let G : = A^'ViC and G* := kU2k'K', as in Section 1.5. Then,
by (1.5.1),
G = G^
G G*
and hence there is a constant c so that
(1.6.1) GG*-G*G = c.
In other words, the Wronskian of G and G* is constant. On writing G and
G* in terms of K and £' this becomes
(1.6.2) k{\ - k2)(KK' - KK') = c .
We now use equation (1.3.13),
dK = E-(l-k2)K
dk k(l - k2)
to eliminate the derivatives in (1.6.2) and deduce that
(1.6.3) EK' + E'K-KK' = c.
It remains to evaluate the constant c. This may be done directly from the
series expansion at zero (see Section 1.3, Exercise 5), or by using Exercise 4
of Section 1.3. □
We define the gamma and beta functions by
/* CO
(1.6.4) rW:=j0 e~'f~ldt re(*)>0
(1.6.5) p(x, y) := jQ f~\\ - 0'"1 dt re(«), re(y) > 0 .
The function T satisfies the functional relation
(1.6.6) r(*)r(i - *) = "
sin TTX

1.6 Legendre's Relation 25
The relationship between T and /3 is
These standard results will be discussed further in Section 3.6. Our present
objective is to evaluate K(l/V2) and £(1/a/2).
Theorem 1.7
«-■ T'(i)
<*)
and
V2/ 4V^?
. , 4r2(|) + r2(i)
<£)-
a/2/ 8V¥
Proo/.
^(V2/ Jo a/7T"Z
dt
V(i-*2)(i-i*2)
V(l-^2)(2-r2)
The change of variables x2 '■= t2/(2— t ) gives
*
a/2/ Jo a/J~
t
The above arclemniscate (giving the arclength of a lemniscate) can be
evaluated in terms of /3. We set u ■ = t4 and see that
I 1 \ V2 f 1/4-1/! sl/2-1 ..
V2 /i i\ a/2 r(j)r(i)
a/2 /i i\
= T Ki« 2)
4 r(3)
Finally, by (1.6.6),
r(|)^ ana r(|) = ^
The evaluation of £(1/a/2) is left as Exercise 1. □

26
The Arithmetic-Geometric Mean Iteration
Since £(1 A/2) = £'(1 A/2) and £(1 A/2) = £'(1 A/2), the evaluation of
K(l A/2) and £(1 A/2) gives an alternate route to evaluating the constant in
Legendre's identity that avoids developing the asymptotic expansion of K'
at zero.
Comments and Exercises
We observe that at 1A/2 Legendre's identity reduces to
*(£)M£M£)
2L
2
or
2£(lA/2~) p£(lA/2) £(1 A/2)
IT \- IT IT
Now KIit and £/77 can be calculated quadratically by Algorithm 1.2 using
only the operations of addition, multiplication, division, and square-root
extraction and commencing with a rational starting value. This provides an
excellent approach to calculating v—an approach we will pursue in some
detail in Chapters 2 and 5.
Legendre derives the constant in (1.6.3) by evaluating K((V3± l)A/8).
(See Whittaker and Watson [27] and Exercise 6.) Note that (V3-l)A/8
and (V3 + l)A/8 are complementary. It transpires that
(4^.
EL ^-
K \ V
Values of k for which K'IK is a rational surd are of considerable interest and
importance. (See Chapter 4.) Such k are called singular values, the simplest
of which is 1A/2, where
-(—) =
K \V2/
1. Show that
Start by writing
«&)-
2fl\,r2fV
1 \ 4^(3) +r(i)
V2/ 8V7F
VV2/ Jo vrr^
Then set u •'= Vl — f2 and show that

1.6 Legendre's Relation 27
^)=711 vrb^"4"^ vrb^"
Show that
KvlMvl)-^
Show that
* 1 •■"
2 , V?T
Show first that
f{X):={\oe~'2dt) and ^):=^-^
V+i)
satisfy /' + g' = 0 and f + g = it I A.
b) Use a) to show that Y{\) = Vtt .
Use Theorem 1.2 to deduce that
^(V2-1) = V2.
[In fact K(V2- 1) = (V2> 1)^(1)1^(1)/213^172.] Thus V2- 1 is a
second singular value. We return to both singular values and T function
evaluations of K subsequently.
Establish Legendre's relation directly from (1.3.12) and (1.3.13) applied
to E, E', K, and K'. Use Exercise 4 of Section 1.3 to determine the
constant.
From the general theory of the Weierstrass function developed in the
next section, or more directly, one can show (Whittaker and Watson
[27, p. 516]) that for a, b >0,
J-" \l(n2 - Y2\(h2 4- y2\ Je
dt
yj{a2-x2){b2 + x2) •>' y4t3-g2t-g3
where e is the real root of the above cubic and
g2:=h(a2-b2)2-(abf
g3-=-ie (a2- b2)[(a2 - b2)2 + 36(afe)2] .
a) If g2 = 0, then
a2 + b2 = 2V3|2g3|1/3 a2 - b2 = -3(2g3)113

28 The Arithmetic-Geometric Mean Iteration
b) If g2 - 0 and g3 > 0, then
•^ f" dt - 2 K( a )
■^ r dt = 2 K( b )
J-V^+ft V7TV2 ^VT+T2''
c) Let e: = g3 ■ = \. Then a + b2 = 2V5, and bi) gives
Also, bii) gives
2X3-^'= f _*_= p-^L.+ r *
J-i VFTT J° Vi- *3 ^°
VT+
r
Now substitution of s : = 1 /f and s-=(1 + t ) , respectively, leads
to
2X3-1/4K' = !
>(|.i)-"(|.§)]-^"(i.i)--
d) Thus when k := sin(ir/12) = (V5 - 1)/2V2
K'(k) = V3K(k)
and
r,/V3-i\_,_1/4 r(|)r(|) 31/4r(i)3
*\ 2V2 I 4V¥ 27/3tt '
This evaluates K in terms of T functions and shows that sin(7j712) is
a third singular value.
1.7 ELLIPTIC FUNCTIONS
The study of elliptic functions began in the 1820s when Abel and Jacobi
independently discovered that the inverses of elliptic integrals are doubly
periodic functions. As noted earlier, they had been anticipated by Gauss
who at the turn of the century had studied a particular elliptic function, the
lemniscate sine. This work, however, had not been published. This was one
of the critical discoveries of the era and was vital to the concomitant
development of complex analysis and the later development of modular and
automorphic functions.

1.7 Elliptic Functions
29
Except for minimal application in Sections 2.6 and 2.7, we have no great
need for this body of material and offer only a brief sketch of this attractive
theory, primarily in the exercises at the end of this section. Details may be
found in Whittaker and Watson [27] and, more recently, in Bowman [53],
Du Val [73], Eagle [58], and Lang [73].
An elliptic function is a meromorphic function with two periods wt and
w2, with imiwjwj^0. That is, for all z, f(z)=f(z + w^^ftz + w2). We
assume that wt and w2 are minimal in the sense that they are not multiples
of any smaller period. The function / is completely determined by its
behaviour on any parallelogram that is a translate of the parallelogram with
vertices 0, wx, w2, and wt + w2. Any such parallelogram is called
fundamental. Associated with the periods wx and w2 there is the lattice L := {nw1 +
mw2\m, n integers}. It transpires that given any lattice L there exist elliptic
functions with the appropriate associated periods. However, there are not
many such functions. As we shall indicate in the exercises, any two such
functions are connected by an algebraic equation. The circular functions
may be thought of as degenerate elliptic functions (one of the periods is
infinite), as may rational functions (both periods infinite).
A function / is said to have an algebraic addition theorem if there exists a
polynomial il in three variables with complex coefficients so that for
x,yEC,
il(f(x),f(y),f(x + y)) = 0.
One of the many remarkable facts, due in part to Weierstrass, is that a
meromorphic function has an algebraic addition theorem if and only if it is
elliptic or degenerate elliptic. (See Exercises 11, 12 and 13 or Hancock [09].)
The Jacobian elliptic function sn is defined by
(1.7.1) u = [
sn(u,k) dt
V(l - t2)(l - k2t2)
The value k is thought of as a parameter and is often omitted. It is true,
though not obvious, that sn(w) is a meromorphic function with periods AK
and 2iK' and, hence, is elliptic. (See Exercise 2 of Section 2.7.) The two
other basic Jacobian elliptic functions are en and dn, defined by
(1.7.2) „=/;
cn(u,k) ^
V(i - t2)(k'2 + k2t2)
(1.7.3) « = Ji
and
•dn(u,k) df
y/(i-t2)(t2-k'2)'
Note that sin and tanh are limiting cases

30
The Arithmetic-Geometric Mean Iteration
sn(w, 0) = sin u
and
sn(w, 1) = tanh u
A "half-angle" formula for sn is
.2/ i
(1.7.4) snzl - = -====== .
■ ' \o) . + yi - A:2 sn2(w)
Again this is not obvious. It is worth observing that a half-angle formula is a
specialization (x = v) of an algebraic addition theorem. These matters are
pursued further in Sections 2.6 and 2.7. In particular, Exercise 6 of Section
2.7 presents an addition formula for sn.
Comments and Exercises
The following exercises develop some of the elementary theory of elliptic
functions. The approach is via the Weierstrass p function and is standard.
See, for example, Erdelyi et al. [53], Du Val [73], or Lang [73].
1. Show that the elliptic functions (with respect to a fixed lattice L) form
a field that is closed under differentiation.
2. Show that an entire elliptic function is constant.
3. Let/be elliptic with respect to L. Assume/is nonconstant. Let P be a
fundamental parallelogram whose boundary B contains no zeros or
poles of/.
a) Show that the sum of the residues of / in P is 0 by using the
periodicity of/to show that jB /=0. Thus any nontrivial elliptic
function has at least two poles in P. The number of poles in P,
counted according to multiplicity, is called the order of /
b) Show that / has the same number of zeros as poles in P counted
according to multiplicity. Hint: Consider $B flf and observe that
///is elliptic.
c) Show that / assumes every complex value exactly order of / times
in P.
4. Show that the Weierstrass function
p(z) :=-2+ 2
Z weL'
1 1
(z - wf w2
is meromorphic, even, and has double poles at each lattice point
[L' := L -(0,0)].
5. a) Show that

1.7 Elliptic Functions
31
p(z) = -2 2 7 ~3 ■
Show that p is an odd elliptic function of order 3.
b) Use a) and the fact that p is even to deduce that p is elliptic.
Observe that p has order 2.
Show that every even elliptic function / is a rational function of p.
Hint: Observe that if z is a zero of /in P (a fundamental
parallelogram) then so is — z mod L. (By — z mod L we mean the unique point in
P, that is equivalent to — z with respect to L; that is, there exist m and
n so that mw^ + nw2 — z = —z mod L and - z mod L G P.) Observe
that there are exactly four points in P where z =
— z mod L (0, wxl2, w2/2, (vvt + w2)/2) and that if/has a zero at one
of these four points, it must have even multiplicity. Consider
g(z):=U[p(z)-p(zi)f
where the product is extended over the zeros of /, choosing only one
representative from each pair (z,-zmodL) and where 8i is the
multiplicity of the zero (except in the case zt = -z,modL, in which
case 8i is half the multiplicity). Now shqw that / and g have the same
zeros with the same multiplicities. Treat the poles similarly.
Show that any elliptic function / is a rational function of p and p.
Hint: Consider
.ffr,-_Az) + A-z) , ([/(*)-/(-*)]/2)/K*)
A) 2 + p(z)
which decomposes / into an even elliptic function and an even elliptic
function divided by p.
Show that there exist constants g2 and g3 so that
[p(z)f = 4p\z)-g2p(z)-g3.
Hint: Consider the order of the pole of p2 — 4p3 at zero. Observe that
g2 and g3 depend on L. Different lattices lead to uniformizations
(parametrizations) of different cubics.
Show that any two nonconstant elliptic functions / and g (with respect
to the same lattice) satisfy an algebraic equation
ft(/,*) = 0
where il is a polynomial in two variables with complex coefficients.
Suppose that f is meromorphic and singly periodic with period w.

32
The Arithmetic-Geometric Mean Iteration
11.
Suppose that / assumes no value infinitely many times in any period
strip. Show that/is a rational function of el7rlzlw. (One can use this as
a definition of the class of trigonometric or circular functions.)
Prove that the function p satisfies the algebraic addition theorem
tK " 4lp(x)-p(y)
Hint: Show that, as a function of x,
-p(x)-p(y)
12.
13.
'Kx)-p(y)
p(x)-p(y)
-p(x)-p(x + y)
has no singularities at the points 0, ±y and, hence, is independent of x.
With Exercise 8, this is an addition theorem.
Show that every elliptic function/ has an algebraic addition theorem.
Hint: Connect f to p algebraically and use the addition theorem for p.
Show that if / meromorphic and has an algebraic addition theorem SI,
then either
a) / is a rational function,
b) / is a trigonometric function, or
c) / is an elliptic function.
Hint: Case a) occurs if / has a pole at infinity. So suppose to the
contrary that infinity is an essential singularity. Now for some w,
f(z) = w has infinitely many solutions z1; z2,. ... Choose z* so that
z*, z* + zx, z* + z2,.. . are not poles of/. Show that there exist m and
n, both less than a constant depending only on SI, so that/(z* + zn) =
f(z* + zm). Show that in any neighbourhood there exist infinitely many
points and some n and m so that f(z + zn)=f(z + zm). Thus / has a
period zn — zm. Case b) occurs if/has only one period, in which case/
can assume any value only finitely many times within any period strip.
In that case Exercise 10 applies. Otherwise we have case c).

Chapter Two
Theta Functions and the
Arithmetic-Geometric Mean
Iteration
Abstract. In this chapter we solve the AGM iteration in theta function terms
and derive a variety of useful properties of theta functions. The central tool is
the Poisson summation formula. These results are then applied to produce
quadratically convergent products for it and e* and quadratically convergent
sums for v.
We finish the chapter by discussing the Landen transform and the
relationship between theta and elliptic functions.
2.1 A THETA SERIES SOLUTION TO THE AGM
The basic theta functions are defined for | q\ < 1 by
CO
(2.1.1) 62(q):= 2 q(n+U2)2 8,(0) = 0.
n = — co
CO
(2.1.2) fl,(s):= 2 q"2 <%(0) = 1.
n=— co
CO
(2.1.3) 84(q):= 2 (-1)"/ 04(0) = 1.
These series are more properly viewed as functions of two complex
variables one of which is presently set to zero. (See Section 2.6.) Theta
functions have various number theoretic connections. Note that 03 is a
33

34 The Arithmetic-Geometric Mean Iteration
generating function for squares, while as 64(q) = 0j(-?), % also considers
the parity of the square. Thus
(2.1.4) 0,(4) + 0,(4) = 2 2 qn=283(q4).
n even
Also
CO CO
(2.1.5) e23(q) = 2 r2(n)q" 824(q) = 2 (-l)nr2(ri)q»
n=0 n=0
where r2(n) counts the number of ways of writing n = j + k . Here we
distinguish sign and permutation [so that, for example, r2(5) = 8 since
(±2)2 + (±1)2 = (±1)2 + (±2)2] and set r2(0) := 1. Now it is elementary
that r2(2ri) = r2(n). (See Exercise 2.) It follows that
CO
(2.1.6) 823(q) + 824{q) = 2 2 r2{2n)q2n = 2823{q2) .
n = 0
Also, (2.1.4) and (2.1.6) allow us to solve for 83(q)84(q). We have
%{q)Mq) = \ [<%(*) + o4(q)]2 - \ [e23(q) + e24(q)]
= 2823(q4)-823(q2) = 82(q2).
Thus
(2-1-70 2 = *3(« )
(2.1.7b) V02(?)04(?) = 04(?2)
which bears an obvious resemblance to the AGM. Similarly,
CO CO
0l(q) ~ 023(q2) = 2 r2(n)q" - 2 r2(2n)q2n
CO
= 2 r2(2n + l)q2n+1 .
n = 0
This last term may be rewritten as
2 r2(2n + l)q2"+l = 2 ^
n = 0 k,m——<*>
k+m odd
which, on setting k = i — j and w = i + / + 1, gives

2.1 A Theta Series Solution to the AGM 35
2 (q2f+1'2)2+U+1'2)2 = e22(q2).
Hence
(2.1.8) e\{q2) + e22{q2) = e\{q).
This combines with (2.1.7/) to produce
(2.1.9) el(q2)-el(q2)=e24(q)
and these last two and (2.1.7//) yield Jacobi's identity
(2.1.10) e\{q) = d\(q) + 8\{q).
Now set k:=k(q):=822(q)/el(q). Then (2.1.10) shows that k' =
#4(9)/03(9)- K we return to (2.1.7) and set an : = 823(q2") and bn := 824(q2")
we observe that an and bn satisfy the AGM iteration. Moreover, since
03(O) = 1, the limit is 1. Thus
(2.1.11) M(8l(q),824(q)) = l.
We recast these last observations in:
Theorem 2.1
Let 0 < k < 1 be given. The AGM satisfies
(2.1.12) M(l,k') = 8;2(q) for k'= 024(q)/823(q)
where q is the unique solution in (0,1) to /: = 8\{q)l8\{q). In particular,
(2.1.13) K{k)=\e\{q).
Proof. This follows from the previous discussion and Theorem 1.1. The
uniqueness of q will be obvious from the results of Section 3.1, which will
show that 84 decreases and 03 increases on (0,1). □
The results of Theorem 2.1 remain true more generally in the complex
plane. This is discussed in Exercise 4.
Comments and Exercises
The solution of the AGM in theta function terms can be found in Gauss

36 The Arithmetic-Geometric Mean Iteration
[1866]. The systematic investigation of theta functions in the context of
elliptic function theory originates in Jacobi's masterpiece Fundamenta Nova
Theoriae [1829].
While we have given a number-theoretically motivated development, it is
possible to give a very elegant formal verification. (See Exercise 1.) This
technique, used frequently by Liouville, is discussed in detail in Bell [27].
1. a) Let /: Z X Z-> R. Establish the formal identity
CO CO CO
2 f(m,n)= T f(l + k,l-k)+ 2 f(l + k,l-k-l)
(2.1.14)
valid whenever both terms on the right-hand side converge.
b) Apply (2.1.14) to qm2+"\ (-l)m+nqm2+n\ and (-l)mqm2+"2 to
derive (2.1.8), (2.1.9), and (2.1.7«) respectively.
c) Hence rederive (2.1.7) and (2.1.10).
2. Prove that r2(n) = r2(2n). Hint: 2a2 + 2b2 = (a + b)2 + (a - b)2.
3. a) Show that the downward transformation [Theorem 1.2(b)] sends
k(q) to \(q) := k(q2). Precisely,
,„,« 1-k' _82(q)-82(q) _022(q2)
(2-L15) 1 + *' el(q) + el(q) 023(q2)-k-
b) Show that the corresponding transformation for K [Theorem
1.2(6)] is
(2.1.16) 83{q )- -
0l(q)
2
4. Show that (2.1.12) is valid for complex q with \q\ <1. Precisely,
for such q. [That 83(q) does not vanish will be apparent from Section
3.1.]
2.2 POISSON SUMMATION
A most analytically accessible route to the behaviour of the AGM lies in the
Poisson summation formula, which we now describe. We then give some
examples of its use before returning in the next section to its relationship
with the AGM. The formula we need is:

2.2 Poisson Summation 37
Theorem 2.2
Let / be a nonnegative function, increasing on (-oo, 0] and decreasing on
[0,0°). Assume that JT„ f(x) dx exists as an improper Riemann integral.
Then, for each x in 05,
-j.mKt j,
e at
(2.2.!) ± f(n + x+)+f(n + x-) = £ ^ P ^
n=—oo 2 &=-°o J —oo
each series being absolutely convergent.
Proof. A complete proof may be found in Apostol [74, pp. 332-333]. In
essence one considers the function F(x) •'= S^=_eo /(n + x), which under the
given hypotheses is of bounded variation on compact intervals, and which is,
by construction, periodic. The left-hand side of (2.2.1) is merely the average
[F(x+) + F(x-)]/2, while the right-hand side is obtained by computing the
Fourier coefficients for F and regrouping. □
example 2.1. We apply the formula to / given by
The right-hand side becomes
- ei«ikx ^ 1 . ycos(27rfcc) + (2irk)*ui(2irkx)
fc~. y + 277//: ~y + tx y2 + (2irk)2
and the left-hand side becomes
2 e-*"+*> + fJ
„>_* w otherwise
If x'■=(), we derive
2V-1 y+2yiiy2 + (2^k)2
which yields
1 °° 1
(2.2.2) 77 coth(77*) = - + 2x 2 ^-
„=1 ;*;' + n2
on replacing y by 277*.
If x'-= I, we derive
1 °° (—IV
(2.2.3) 77 cosech (77*) = - + 2x 2 2 :
^ , 2

38 The Arithmetic-Geometric Mean Iteration
By elementary analyticity considerations, (2.2.2) and (2.2.3) remain valid in
the complex plane and produce the classical formulae for cot, cosec, and so
on. (See Exercise 1.)
example 2.2. This time apply the formula with
f{x):=e-sx2" s>0.
Then (2.2.1) becomes
CO 00 /* CO
n=— eo k — ~ <»' J —oo
Now the integral on the right is
2 j\-'2 cos (2Trkt)dt=^F(^k)
where
f 2 A/77" 2
(2.2.4) F(y) : = I e~x cos (2xy) rfx = — e~y .
(See Exercise 2.) Thus we deduce
(2.2.5) i ^-4-^ = ,-172 2 e2'***"
lirikx — irh2/s
r
Again, analyticity considerations show that (2.2.5) holds for re(s) >0. This
is a general form of the theta transformation formula. For future reference
we make the notational agreement that Bt{s) ■= O^e'™) and observe that for
x '■= 0, (2.2.5) can be written as
(2.2.6) ^8,(1) = 8,(1-1).
Note that for large s the sum Vsfl,(s) converges much more rapidly than
03^-1). For example, if $:=100, 03(O.O1) and 10 + 20e~100,r coincide
through more than 500 digits.
Comments and Exercises
The result of (2.2.1) was known to Poisson by 1827. With x :=0 he had
obtained the formula in 1823. Equation (2.2.5) was first obtained by Jacobi
using elliptic function theory in 1828.

2.2 Poisson Summation 39
Establish the formulae
' sin'--^ 2 2
b) * = 2 (-l)"+i(2n + l)
sin(7rz) z „=1 z2-n
\n + l
cos(ttz) ^o z2-(n+§)2
j. / s 1 v1 2z
C) 77 COt (TTZ) =- + 2, "2 2
z „=i z - n
d) sin (77 z) =7rzll(l 2 ) •
«=i ^ n I
Establish (2.2.4) by showing that
F(y) + 2yF(y) = 0
and using F(0) = Vtt/2. (See Exercise 3 of Section 1.6.)
Let I(s, y) := jr. e-*"2-2*" dt. Evaluate I(s, y) for real s, y by
completing the square. By analytic continuation I(s, ik) = s~112 e"""*/s.
Recall that the Laplace transform is defined by
~y'f{t)dt.
F(y):=\0
Provided that f(t)= 0(ebt) as f->°°, F will be analytic for re(;y)>6.
Show that the Laplace transform of (2.2.5) with * := 0 produces (2.2.2).
This entails evaluating integrals of the form
/* CO
F(a,b):=\o
-[a2t+b2lt] .-1/2
r1'2 dt,
This is a special case of a Bessel function transform which can be
evaluated explicitly as F(a, b) = {VJfla) e~2ba by substituting s '■= t and
v'= bis — as. (Various extensions and related matters are discussed in
Bellman [61].) In principle, therefore, one can derive (2.2.5) as an
inverse Laplace transform of the derivative of the product form of sin
given in Exercise Id).
Let / be nonnegative, continuous, decreasing, and Riemann integrable
on [0, oo). Let
S(y) := V- J0 /(*) c°s (xy) dx .
Show that

40
Va
The Arithmetic-Geometric Mean Iteration
1 °° t r 1 °°
~ /(0) + 2 /(ma) = Vfi ~ g(0) + 2 g(n/?)
»1 = 1 J ^-^ n = l
whenever a, /3 >0 satisfy a/3 = 2tt. Deduce that
Va
\+i e-a2-2/2'
•2 OT=i
(Exercises 2 and 5 follow Apostol [74].)
2.3 POISSON SUMMATION AND THE AGM
We commence by specializing (2.2.5). Setting x'-=0 produces
(2.3.1) ya%(e-") = «,(«-")
while a; : = § gives
(2.3.2) V5%(e~") = 04(e-,r")
and dually on setting s = s_1,
(2.3.3) Vs84(e'^) = ^(e-^)
where re(s)>0. On dividing (2.3.2) by (2.3.1) we have
(2.3.4) k(e~") = k'(e~'u).
From Theorem 2.1 we see that
M(l,k) = e;\e~'u) q:=e~™.
Thus
M(l,k') _ eKe-*")
(2.3.5)
IT
M(l,k) «&"")
tts= -log q
on using Theorem 2.1 and (2.3.1) again. This produces the following
fundamental theorem.
Theorem 2.3
For all k in (0,1),

(«)
and so
2.3 Poisson Summation and the AGM
M(l,k') , e\{q)
" M(l,k)=-l°^ k=0l(q) k =
el(q)
41
K'(k)
Proof. We have established (a) and (b) follows from our identification
of M(l, k') and ir/(2.Kik)). This second equation is often written as q =
e-*K ik an(j ^ jg caj^e(j tjje nome associated with k. In principle it solves the
inversion problem for q in terms of k. □
We know that k= 0\iq) 10\iq) = ^Vq~ + Oiq) as q tends to zero from
above. Theorem 2.3(a) and this information show that since M(i, k') tends
to 1 as k tends to zero,
(2.3.6) lim
k-*0+
2M(1
jtrlogQ)
:0
which reproduces the asymptotic of Exercise 4 of Section 1.3. As we shall
see (Exercise 1), we can derive the exact asymptotic (1.3.10) from these
considerations.
Now consider the AGM iteration commencing with a0: = 1 and b0 '■ = k'.
Then as we saw in Section 1, bjan = k'iq2 ) and cjan = kiq2"). Hence
(,3.7, ^-^)-1¾.
In Exercise 2 one establishes that the differential identity
(2.3.8) 2-"b;2dlog(^) = b-0>dlog(^)
holds. It follows from (2.3.5) and (2.3.7) that
and from (2.3.8),
''n"0 ...
while ir0=-T-r-,.
bn^0 . -, 2 dfc
~bj WMe ff«=-* A"
Since 6„ tends to Mil, k'), we have established that

42
(2.3.9)
The Arithmetic-Geometric Mean Iteration
dk
ds
tt kk'2
2 M(l,k'y if
2 kk'2K2
and since , = — it log q,
(2.3.10)
dk
kk'2
2kk'2K2
dq lq M(l,k')2 qir2
Rewriting k, k', and K in (2.3.9) in theta terms produces
03 e2 4 4"
(2.3.11) f - -t = T 01. (w.r.t.,)
Differentiation of (2.3.1) and (2.3.3) yields
(2.3.12)
s%(s) + - 6,(8) = -s-l'%(s-1) (w.r.t.,)
(2.3.13)
,26>4(,) + - 04(,) = -1-11%^1). (w.r.t.,)
On using (2.3.1) and (2.3.2) again we deduce that
(2.3.14) , f (,) + ,- | (,-1) = - \ = , £ (,) + ,-1 £ (,-) .
Now, since s20l(s) = 6\(s-1), this shows that
(2.3.15)
fy _ ^ = 5 «4
»4 03 4 °2
(w.r.t.,)
is equivalent to (2.3.11). Finally, adding (2.3.11) and (2.3.15) gives
(2.3.16)
^ _ A - E a*
e4 e2 4°*
(w.r.t.,)
on using (2.1.10).
We can now also express E in terms of theta functions. We have from
(1.3.13),
E- K=kk"
dK
dk
kK=~K
1 dK . k2K2
2K ds
+
because of (2.3.9). Hence
*Differentiation with respect to s.

2.3 Poisson Summation and the AGM
43
and
(2.3.17)
Similarly
and
(2.3.18)
E-K-l
'83 77
U+4
E=K~^i-
77 TTK' e2
E K ~ 2K + k2 e2
E' = — +
ttK' e4
V2 ft
+ \ »2J (w.r.t.s)
(w.r.t.s)
(w.r.t.s)
(w.r.t.s)
Comments and Exercises
1. a) Observe that k = ^Vqfiq) with / analytic for |q\ < 1 and /(0) = 1.
Standard real reversion arguments show that for 0<k<l,
Vq=\g{k2)
where g is (real) analytic with g(0) = 1. Show that
(2.3.19) K'(k) = - K(k)
log (£) + **)]
where h(0) = 0 and h is analytic,
b) Observe that the right-hand side of (1.3.10) and K' both solve
(1.3.8). Moreover, both can be expressed in the form of the
right-hand side of (2.3.19). Deduce that (1.3.10) holds.
2. a) Show that 2~"b~2d log (an/cn) is independent of n.
b) From a) deduce that 2~nc~2dlog(bJan) is independent of n.
c) Use a) and b) to show that 2~na~2d log (bn/cn) is independent of n.
d) Show that all three coincide with b$2d log (a0/c0). This again is due
to Gauss.
3. a) Establish (2.3.18) by using (2.3.16).
b) Observe that (2.3.18) and (2.3.17) immediately give another proof
of Legendre's relation.
4. a) Show that 02(e-,r) = 04(e-,r).

44 The Arithmetic-Geometric Mean Iteration
.2.
b) Show that 477 = (S:__. e~" ')/(£:._. «V" ').
c) Let r^(n) give the number of distinct representations of n as a sum
of k squares. Show that
477 = ( i *rt(«) «—)/( i "/•*(«) «—) .
Xn=0 '/ \„-0 '
M/tf: Take the kth power of (2.3.1) and differentiate.
All of results in this section can be derived without Poisson summation as
the following exercises show.
5. Derive (2.3.7) by combining Exercise le) of Section 1.4 and Exercise
4b) of Section 1.3. This can also be found in Gauss [1866], but not
apparently in later nineteenth-century authors. (See Borwein and Bor-
wein [84a] and King [24].)
6. Use k~Ayfq to rederive Theorem 2.3 from Theorem 1.1.
7. Observe that Theorem 2.3(a) and Theorem 2.1 show (2.3.1) and
(2.3.4). Hence deduce (2.3.2) and (2.3.3). Thus we have established the
0 transformation formulae [(2.3.1) to (2.3.3)] directly from the AGM.
8. Suppose that in (2.3.7) the 4 is omitted. Show that the convergence,
which was quadratic, is now linear. Of course the limit is unchanged.
2.4 THE DERIVED ITERATION AND SOME CONVERGENCE
RESULTS
It is generally the case with AGM related approximations that it is relatively
easy to establish the convergence rate and much harder to determine to
what the given iteration converges. We now consider various preparatory
convergence results. In the next section these will be used to produce two
surprising algorithms for v.
Motivated by the fact that V2K(l/V2)K(l/V2) = tt (Exercise 2 of
Section 1.6), we commence by computing K. If we consider the AGM
sequence with a0 := 1 and b0:= k, it is apparent that a„ and bn viewed as
functions of k converge uniformly and analytically to M(l, k). It follows that
the derived iterations d„ and bn converge to M(l, k). Since M(l, k) =
irl2K'{k), we see that
(2.4.!) Up,*)-! *|gl.
on noting that (dK'/dk)(k) = -(k/k')(dK/dk)(k'). Equivalent^,
(2-4.2) *(*')- o ,, „2.
77 k[ M(l, k)
2 k M\l,k)
Now the derived iteration is

2.4 The Derived Iteration and Some Convergence Results 45
(2.4.3) an+1 - —2— 6»+i 2 '
Since bjan converges quadratically to 1, it is easy to show directly that dn
and bn converge quadratically to M(l, k) for 0<k<l. Moreover bn
decreases and dn increases, at least eventually (Exercise 1). For our purposes
we will generally consider the Legendre forms xn ■= ajbn and yn '■= bnldn.
Then
^.t.tj xn+1 — 2 yn+\ v +1
where x0 ■= k~ , yx •'= VTQ, and y0 is undefined. Moreover,
(2.4.5) 1 < xn+1 <yn+1<Vx;<xn.
(See Exercise 2.) We also have
2V^(1 + V^)2~8^" 1}
and
(y„ - !)(*„ - 1) 2(*n+1 - 1)
y"+1 {yn + l)^x-n + l) yn + l
^Hyn-i)2 + Uxn-if^Uyn-i)2
so that for n s 1
(2.4.6/) *„+1-l<§(*„-l)2
(2.4.6b) ^1-1=^1(^-1)2.
This establishes the quadratic convergence of xn and ynto 1. From (2.4.1),
(2.4.2), and the previous discussion it is now apparent that
,„ ,^ „ ^M3(l,l/V2)
(2.4.7) 77 = 2V2 -T-H 7=1
V ' M(l,l/V2)
and since both M and M are quadratically computable, so is v. In the next
section we turn this identity into an explicit algorithm.
Comments and Exercises
1. Show that dn and bn defined by (2.4.3) converge quadratically to
M(l, k), by showing that dn increases and bn decreases. Then show
bn - dn < (Vx~l - 1) /2. For this to hold for all n, assume x0 ^ 3.

46 The Arithmetic-Geometric Mean Iteration
2. Establish (2.4.4) and (2.4.5).
3. a) Use *„ = e\{ q") I0\( q2") to show that *„ ~ 1 + 8e '^"'H'o1) as
n—»oo.
b) Show that cn^Aane~2'"i<KIK')(-x°i) :=8n, and that cn~8n as
n—>°°.
4. Show that with a0 := 1 and b0'-= k,
£(*') = £ £lim%.
v ' 2 fc «— ft2
Hence deduce that with x„ and yn, as in (2.4.4),
In particular with x0 := V2 and ^j := 21/4,
CO
n
" r2(l) •
5. Show that
«(l, £)- .-»r>(|) and 44) = 2^-^(¾.
2.5 TWO ALGORITHMS FOR it
The systematic use of the derived AGM leads directly to quadratic
algorithms for 77.
Algorithm 2.1
Let x0 := V2, tt0 := 2 + V2, and y1 := 21/4. Define
l + yn_4r2(|)
„ = 1 it„ 77
(0 W-j(v*; + ^;) "a0
<«) W-^+//^ «l
*« + !
(ui) ^:=^^-^ n>l
Then 7r„ decreases monotonically to 7r. Moreover, for n>0,

2.5 Two Algorithms for n 47
(2-5.1) Uyn+i~xn+1)^TTn-7T^l(yn+1-xn+1)
1 2
(2.5.2) 77„ + 1 - 77 =S — (77„ - 77)
and, for n > 2,
(2.5.3) 77„ - 77 < 1(T2"+1 .
Proof. We first establish the limit. This follows from (2.3.7) (see
Exercise 1) or from (2.4.7) as we now indicate. Let 77„ : = 2V262+1an+1/dn+1
where a0:=l and 60:=1/V2. Then by (2.4.7) of the previous section,
77„—> 77. From (2.4.4) we have
^n _ (t>n+i/bn)\an+1/an)
*■„-! an+1/dn
l + x„
l + y„
Moreover, 770 = 2 + V2 and the algorithm converges to 77 as claimed. Since
y„sx„> 1, it is obvious that 77„ decreases. Let us observe that
(2.5.4) y.+r^+i-i^TTTrr^B^""*")
provided that
Next
2V^(l + ;y„) 8
*" 1<2-V3.
y„-l
(2-5-5) T» - «■» + ! - —j ■
/n + 1 T x
Hence
77 77
~ ~TJ (yn + l ~ Xn + l) =^n~ ^n + l ~ ~f (^„+1 ~ *„ + l)
/n+1 ' x ^
and

48 The Arithmetic-Geometric Mean Iteration
Here we have used (2.5.4) and a geometric estimate. From the above
77
7T„ — 77 S ■
(.^n+l Xn+l) ■
yn+i +1
Thus, for n >0, upon checking the early cases,
(2.5.6) | (yn+1 - xn+1) < tt„ - 77 < | (yn+1 - xn+1)
and with (2.5.4),
77„ + 1 - 77 < I (yn+2 ~ Xn+2) S= & (yn + 1 - Xn + 1f < ^ (77„ - 77)2 .
Finally (2.5.3) follows from (2.5.2). □
The first nine iterations give 1, 3, 8, 19, 41, 83, 170, 345 and 694 digits of
77. The 24th will produce more than 45 million digits at the expense of only a
few hundred arithmetic operations. A more exact asymptotic will be derived
in Chapter 5. (See Exercise 5.) Note also that the number of leading zeros
+1 gives the number of digits of agreement between 77„ and 77 to within
1.
The second algorithm, based on an identity of Gauss [1866], was
discovered by Brent [76a] and Salamin [76] independently.
Algorithm 2.2
Let a0 := 1 and b0 := 1/V2. Define
2a»+i
77„ .=
1 _ y" 2kr2
where c„ :=V^ - b2n = c2n_1/4an and a,„ and bn are computed by the AGM
iteration. Then 77n increases monotonically to 77 and satisfies
(2.5.7) 7T-77„<
77 2 e
M2(l,l/V2)
and
2-(n+l)
(2.5.8) 77 - 77„ + 1 ^ — (77 - 77„)
77
2
Proof. This algorithm is based on the use of the second integral E
rather than K. With k:= 1/V2 Legendre's relation is (2E- K)K= tt/2.
Combine this with Algorithm 1.2 of Section 1.4 to derive

2.5 Two Algorithms for *r 49
2M2(1,1/V2)
(2.5.9) tt = ,
which on truncation shows that irn converges to it. We leave the
convergence estimates as Exercises 3 and 4. □
The first eight iterations produce 0, 3, 8, 19, 41, 84, 171 and 344 digits,
which agrees extraordinarily well with the asymptotic.
Both of these algorithms generalize in many ways. (See Chapter 5.) At
the moment we only exhibit two additional identities.
If we use the differential equation for K equation (1.3.13), we may
rewrite Legendre's identity as
(2.5.10) | = kk'\KK' - KK') .
[See equation (1.6.2).] With (2.4.2) we now derive
r?5in 2 ,,,2 K'(k) M(l,k') 2 K'(k') M(l,k)
( ' ir K(k) M\l,k') KK K(k') M\\,k)
and, as in Algorithm 2.1, we can produce
2
77 :
<-'>!In(i^)+(1-*,|§n(i±§)
(2.5.12)
Here x0'= k'~\ x%'=k~1, yx'-=VTQ, yX'-=^fx*Q, and the iterations are
given by (2.4.4). When /r-=l/V2, (2.5.12) reduces exactly to Algorithm
2.1. Also whenever K'{k)IK{k) = Vr for rational r (a singular value of k),
we can in principle find k algebraically, as we will see in Chapter 4. We
already know of four such values of k. (Exercise 7.)
In a similar fashion we may use Algorithm 1.2 to substitute for both E
and E' in Legendre's relation. This as observed in Salamin [76] produces
/,,,« 2M(1, k)M(l, k')
(2-5-13) ™~~1 ^ o«-l/„2 , „*2x
^:..2-^+0
Again, c„ and c* are computed from complementary AGM iterations. When
k:= 1/V2, this identity reduces to (2.5.9).
From (2.5.2) and (2.5.8) we observe explicitly that the corresponding
algorithms converge quadratically. We are primarily interested in the

50
The Arithmetic-Geometric Mean Iteration
cumulative error given by (2.5.3) and (2.5.7). We will say, informally, that
any such algorithm allows for quadratic or fast computation. In similar
fashion, we talk about mth-order computation if the cumulative error (i.e.,
digits correct) is of order m" after n steps.
Comments and Exercises
Algorithm 2.1 is derived in Borwein and Borwein [84a] by the method of
Exercise 1.
1. a) Differentiate (2.3.7) and apply (2.3.8) to establish that with
a0'-= 1 and &„'•= k',
2 (a*a )2
IT = T—T hill
kk'2 «-~ d*an - dna*n
Here a* as usual denotes the AGM iteration commencing with
b0 '■ = k, and all derivatives are with respect to k.
b) Let k ■= k' '■= 1/V2 and observe that the previous formula
reduces to (2.4.7).
a) Fill in the details in (2.5.3) to (2.5.6).
b) Show that
Note that while Algorithm 2.1 relies on evaluating K, Algorithm
2.2 in fact relies on evaluating E. Indeed (1.3.12) and Algorithm
1.2(6) combine to show that
E{k)-~\ (2 2-^(/:)
a) Use (2.3.7) to show that (c„/4a„) increases monotonically
and quadratically to q. Thisgives a quadratic algorithm for e7' on
using a0 := 1 and b0 ■= 1/Vl
b) Show explicitly that, with a0 := 1 and bQ := k',
(2.5.14) e^W^)=16n(^±i)21".
As in Algorithm 2.2, show that
(2" -2-1) ii
2 J- 20n„2
1T„,,C„, , IT 2 C„
+ 1^+1 _ _ _ _ " ^ '-n+l
=S 7T„ + 1 - 7T„ S
1+2
M2(l,l/V2)
Then deduce that

2.5 Two Algorithms for *r 51
r"+1(^~)C"+1 * * ~ ^- * M\l,llV2) 2
n + l 2
Cn + 1
Now use Exercise 3a) to show (2.5.7) and (2.5.8).
5. As in Exercise 4), show that in Algorithm 2.1 irn — v is of order
2" e_,r2" . Convergence proofs of this type can be found in detail in
Salamin [76] and Borwein and Borwein [84a] with discussion of the
asymptotics.
6. Prove (2.5.12).
7. Recall from Section 1.6 (see also Chapter 4) that
ii) K'(V2 - 1) = V2K(V2 - 1)
iv) K'(3 - 2V2) = V4K(3 - 2V2).
Observe that for these values of k and k' (2.5.12) reduces to an
algebraic combination of two infinite products. Equally (2.5.13)
simplifies.
8. Establish the general identity of Gauss, Brent, and Salamin given as
(2.5.13).
9. Use the quartic transformation to produce a quartically convergent
infinite product for tt. (See Exercise 3e) of Section 1.4.)
10. Show that with xn and yn as in Algorithm 2.1,
(V5+2)H*.(^)=r,(!)^
and
(V2 + 2>Sfix-(^)S
V32
11. There are actually eight natural products implicit in (2.4.7). One can
select either bn or an (bn or d„) for each of the means. While
Algorithm 2.1 is the cleanest of these, it is not the best approximation.
a) Show that there are four decreasing and four increasing products.
b) Show that the two best approximations are given by 2V26^/d„ and
2V2V„/6„.

52 The Arithmetic-Geometric Mean Iteration
c) Show that with xn and yn as in Algorithm 2.1,
2xm .WArt.nfr.^-^+W.
m = l l ^ ym m = l *-
+ ymx„
with both products converging monotonically to v.
d) Similarly, produce a decreasing analogue to Algorithm 2.2.
12. The identity (2.5.14) can be found in Gauss [1866]. Show similarly the
following identity due to Jacobi. (See King [24].)
K(k)
(2.5.15)
= log(^) + 3i2-"log(^) a0:=l b0:= k'
2.6 GENERAL THETA FUNCTIONS
Theta functions are more properly considered as a function of two
variables—a parameter q and an analytic variable z. So far we have considered
only special theta functions (z = 0). In this section we sketch some relevant
parts of the general theory. We write
^(z, q) := ^(z, t) := 2q114 sin z - 2q9'4 sin 3z
+ 2q25'4 sin 5z
02(z, ¢) := 02(z, 0 := 2#1/4 cos z + 2#9/4 cos 3z
+ 2?25/4cos5z+---
03(z, ¢) := 03(z, 0 := 1 + 2q cos 2z + 2#4 cos 4z
+ 2q9 cos 6z + • • •
04(z, ¢) := 04(z, t)'-=l-2q cos 2z + 2¾4 cos 4z
— 2q9 cos 6z + • • •
where q = emt and im (t) > 0.
When z = 0, then 0^0, ¢) = 0 and q(0, q) = 8j(q) for j = 2, 3, 4. When
the precise value of q is unimportant, one writes 0;( q) = 0;. and 0^(z, ¢) =
0;(z). When / is unimportant, one writes 0(z). It is straightforward to
establish the following functional identities (Exercise 1):
(2.6.1)

2.6 General Theta Functions 53
e1(z) = -02[z + |) = -iM8,(z + f + y) = -iM84(z + y)
82(z) = M83(z + y) = M84[z + | + y) = ^(z + |)
03(z) = 84{z + |) = M^z + § + y) = M02(z + y)
04(z) = -/M0! (z + y) = /M02(z + | + y) = 03(z + |)
where the multiplier M is given by
(2.6.3) M:=q1,4eiz.
From (2.6.2) or directly one has
81 (z) = - 81 (z + it) 8x (z) = - qe2% (z + Trf)
^(z) = -02(z + 7t) 02(z) = qe2i\{z + 7r0
(2.6.4)
fl,(z) = fl,(z + tt) J,(z)=?e2'z03(z + 7rO
04(z) = 04(z + it) 04(z) = -qe2i\{z + irt) .
These identities show us that any theta function is entirely determined by its
values on any fundamental parallelogram
P(Z0) ■■= {z\z = Z0 +r1TT+ ^77^,02^,^^1} .
We assume as we may that the given function has no zeros on the boundary
of P(z0). It is obvious that z = 0 is a zero of 0t(z) so that v/2, tt/2 + irtl2,
and TTt/2 are zeros of 02(z), 0j(z) and 84{z), respectively. Moreover, (2.6.4)
shows that z0 + mir + nirt (m, n integral) is a zero of a theta function
whenever z0 is. We now show that each theta function has exactly one zero
in each fundamental parallelogram. Thus we will have specified all the zeros
above.
Consider the integral
2iri JpOo) 8{z)
which gives the number of zeros of 8 inside P(z0). Explicitly,

54
The Arithmetic-Geometric Mean Iteration
0(z) 0(z + TTt)
0(z) 0(z + 77-r)
z0+irr
liriN
-n
-/:
dz
Lfl(z)
0(z + y)
8(Z + 77-)
dz
Now logarithmic differentiation of (2.6.4) shows that
0(z)
(2.6.5)
8(z)
0(z+tt) 8(Z+7Tt)
8{z + 77-) 8{z + ttO '
Thus 27r/iV = 277-f and N = 1 as required.
The identities in (2.6.4) show that each ft/ft, i¥=j, is doubly periodic. In
combination with our knowledge of the zeros of each ft we can painlessly
apply Liouville's principle (bounded entire functions are constant) to
establish many identities. We illustrate this with the following.
Proposition 2.1
(2.6.6)
Proof. Consider
«fc)«4 :
M-=
8\{z)8\
8\{z)8\
8\{z)8\ - 8\{z)8\
By (2.6.4) /(z) is doubly periodic with periods v and 7rf. Moreover 82A{z)
has a double zero at 7rf/2 in P(0). Also (2.6.2) shows that 8\{TrtlT)8\ =
8\{jrtl7)8\. Thus/is elliptic with at most one simple pole in P(0), and by
Exercise 3a) of Section 1.7,/can have no poles. Hence/is constant, being
bounded and analytic. Since 82{irl2) = 0 and 0,(77-/2) = 04, 04(7t72) = 83, the
constant must be 8\ and (2.6.6) follows. □
Note that on letting z = 0 in (2.6.6), we recover (2.1.10): 03 = 02 + 04-
It is a simple matter to recast Example 2.2 as the classical theta
transformation formula.
Theorem 2.4
For z an arbitrary complex number and im (t) >0,
(2.6.7) 03(^0 = (-^)-^^^^3(7,-7)-
Here one takes the principal square root. The proof is left as Exercise 3.
It is equally simple to use the Jacobi triple-product of Chapter 3 to

2.6 General Theta Functions
55
produce product expressions for 0;(z) (Exercise 4). Various other
relationships are indicated in the exercises.
Comments and Exercises
There is a proliferation of notations for theta functions. We follow the most
usual notations (used in Bellman [61]) and note that both Dickson
[71, vol. 3, p. 93] and Whittaker and Watson [27] give tables of alternate
notations. The abuse of functional notation, in particular the distinctions
among 8}, 0;(#), 8j(t), and 0-(z), necessitates some caution. Both Bellman
[61] and Whittaker and Watson [27] provide an accessible introduction to a
considerable amount of material on theta functions.
1. Verify the identities of (2.6.2) and (2.6.4). Note that (2.6.2) shows that
we can restrict attention to one theta function (say, 03) and lose no
information.
2. Show using Liouville's principle that,
a) e\{z)el = e\{z)e\-e\{z)e\
b) e\{z)e\ = e\{z)e\-e\{z)e\
c) e3(z + w)e3(z - w)e23 = el(w)e23(z) + e\{w)e\{z)
d) 84(z + w)e4(z - w)e24 = e24(w)e24(z) - e\{w)e\{z)
e) 8x{z + w)84{z - w)8283 = el(z)e4(z)6l(w)6i(w)
+ 82(z)83(z)81(w)84(w) .
Results of this kind are discussed in detail by Whittaker and Watson.
Many are given in tabular form in Erdelyi et al. [53].
3. Establish Theorem 2.4.
CO
4. Show that with Q0 := FE (1 — q2n),
n = \
CO
i) fl,(z) = Q0 11 (1 + 2q2n-1 cos (2z) + tf4""2)
n = l
CO
ii) B^z) = 2q1,4Q0 sin z U (1 - 2q2n cos (2z) + q4n) .
Establish similar identities for 82(z) and 84(z). Many variations are listed
in Erdelyi et al. [53]. (This relies on Section 3.1.)
5. Use Exercise 4 to show that
a M-4Y (-l)Vsin(2/iz)
1} fl,(z) 4^ \-q2n

56
The Arithmetic-Geometric Mean Iteration
u) logrMgi=4i(zi^!iBiM
'L 6,(0)
i 1-q'
for |im(z)| < \ log \q\.
6. One of the easy but important properties of theta functions is that they
solve the one-dimensional heat or diffusion equation
d26(z, t) _ 4f d8(z, t)
dz2 ~ it dt '
This provides a significant connection between the analytic properties of
theta functions as functions of z and their number theoretic properties
which revolve around the variable q. Various of these matters are
pursued in Bellman [61] and Rademacher [73].
7. Bellman [61] gives an interesting functional equation approach to
Theorem 2.4. Consider entire solutions /to
i) /(Z+T7)=/(Z)
ii) f(z+7Tt) = be-2,zf(z)
where b is an unspecified function of t independent of z. Suppose,
temporarily, that / possesses an absolutely convergent Fourier
expansion valid for all z:
m-= 2 c„e
2niz
a) Show that c„+1 = b xq2ncn. Hence deduce that for \q\ = |e"'| < 1,
f(z) = c0\ 2 ?«(«-D6-V"'z
*-n = ~ eo
is an analytic solution to i) and ii) and is unique up to choice of
constant. When q = b~ , we recover fl,(z).
b) Show that any entire solution to i) and ii) has an absolutely
convergent Fourier series. Hint: Use contour integration to show
that for any integral k,
I/.
it/2
/(z) e~ ,nz dz
—nil
<e-2knTr max \f(w-ik)\.
c) Use the uniqueness of solutions of i) and ii) to rederive Theorem
2.4. Hint: Show that both sides satisfy i) and ii). Then use Exercise
6 to normalize the equation.
There is also a considerable literature on multidimensional theta
functions. (See Bellman [61].) Since they do not impinge on our main
considerations, we say no more.

2.7 The Landen Transformation 57
2.7 THE LANDEN TRANSFORMATION
We finish this chapter by deriving an expression for sn in terms of theta
quotients and by relating this expression to the transformation of incomplete
elliptic integrals. We begin with the classical Landen transformation in theta
form.
Theorem 2.5
For all z and im (t) > 0,
83(z, t)84(z, t) fl,(0, t)84(0, 0 _ 82(z, M(z, 0
(2.7.1)
e4(2z,2t) e4(0,2t) e^izyit)
Proof. As a function of z, 84(2z, 2i) has zeros when 2z = irt + rmr +
2nirt or when z = mv/2 + (2n + l)vt/2. This is exactly where 03(z, t) or
64(z, t) is zero. Again (2.6.4) shows that/,(z):= 63(z, t)84(z, t)/84(2z,2t) is
doubly periodic with periods it and vt. Thus Liouville's theorem shows that
ft{z) is a constant with respect to z. The second equality follows on
substituting z + irtl2 for z and using (2.6.2). □
To establish the existence and analyticity of sn in theta terms we begin
with
( ' dz 64(z) °* 8\{z) ■
(See Exercise 1.) Now let p := 8^)/8^) and observe that
(This relies on Exercises 2a) and 2b) of the previous section.) Then
replacing z by 11822 and p by y = p83/82, we observe that, since k = 82/83,
(2.7.4) (|)'-(1-/X1-*V).
This is solved by
(2.7.5) Bn(ll)ik):=y=J4^Q
or
rr / ix «i(uvl2K)
^ksn^k)=W^I2K)

58 The Arithmetic-Geometric Mean Iteration
and then, in agreement with (1.7.1),
rsn(u,k) dy
r
'» V(l -y2){l-k2y2)
which solves the inversion problem (at least for real k) for the given
integral. We wish to have en + sn2 = 1 and k2 sn +dn =1, and it is
appropriate to define
(2.7.6) cn(«, k) .= - -^-^ = yjT- ^^
(See Exercise 2.)
Finally we wish to recast Theorem 2.5 in elliptic function terms.
Theorem 2.6 (The Descending Landen Transform)
Let 0 < i/f and 0 < k < 1 be given. If
^1 ' 1 + *'
and 0 < i/r < i/rj is given by
. , (1 + fc')sin i/rcos iA
sm i/fj: = T—
Vl - k2 sin2 ^
then
(2j-8) (i+t,)rvrfei=rviT|
2 sin2 0
or
(l + k')F(ilf,k) = F(ilf1,k1)
where
,, f* de
wt):=l-'- *^.
is the incomplete elliptic integral of the first kind.

2.7 The Landen Transformation
59
Proof. From Theorem 2.5 we deduce
^(z,Qfli(z,0_^(2z,20
03(z, 0«4(«. 0 «4(2z, 20 "
In. terms of sn, en, and dn this becomes
where u = zd\{q) and ux = 2zd\(q2) since kx = 02(q2)/83(q2).
Thus ^=(1-/:0/(1 + ^0 and u1 = 2ue23(q2)/823(q) = (l +k')u.
Collecting information, we have
(2.7.9) Bl = (l + *> t,-^ 811(^,^) = (1 + ^)-^-
since kk\xl2 = 1 + k'. The change of variables sin 6 = sn(w, /:) and sin 0t =
sn(«!, kx) now produces (2.7.8). D
Comments and Exercises
A profusion of information on the numerical use and derivation of various
Landen transforms is given in King [24]. More bibliographic information is
available in Watson [33] and in Whittaker and Watson [27].
1. Prove the differential equation (2.7.2) by showing that
^(z)«4(z)-«4(z)^(z)
*(2): =
02(z)03(z)
is doubly periodic with periods ir and irtll and that <f>(z) has simple
poles possibly only at v/2, v/2 + vt/2, and translated points. Then
show that (f>(z + vt/2) = <f>(z), and hence relative to the periods it and
■ntl2, <f> is doubly periodic with a single pole. Thus <£ is constant.
a) Establish (2.7.5).
b) With en and dn defined as in (2.7.6) and (2.7.7) show that
en2 + sn2 = 1, dn2 + k2 sn2 = 1, and that cn(0) = dn(0) = 1.
c) Show that (d/du) sn(«, k) = cn(w, k) dn(«, k).
d) Show that sn is strictly increasing on (0, k) with sn(0, /:) = 0 and
sn(jRT, k) = 1. [You may find it convenient to use the product
formulae for 0j(z) and 04(z).]
e) Show that dn(K) = k'.
f) Establish the double periodicity of sn(«, k), cn(«, k), and dn(«, k)
with respect to u and verify the following table:

60 The Arithmetic-Geometric Mean Iteration
Periods Zeros Poles Residues
sn
en
dn
4K, 2iK'
4K,2(K + iK')
2K, 4iK'
2mK + 2niK'
(2m + 1)*: + 2niK'
(2m + \)K + (2n + Y)iK'
2mK+(2n + \)iK'
2mK+(2n + \)iK'
2mK+(2n + \)iK'
(-
("
(-
-\)mlk
-l)m+1i/k
3. The ascending Landen transform is given by reversing the roles of k and
ku if/ and i/^. Thus if
,. 2Vk
and
then
l + k
sin (2^ - i/f) = k sin i/r fa > if/
(l + k)
W, k)=Fth, k,).
These transforms and their analogues for incomplete second and third
integrals clearly lead to quadratic iterations which are studied in detail
in King [24] without reference to theta functions. Note that if iff = it 12,
we recover the quadratic transformation for K.
a) Show that, in the notation of Theorem 2.6,
cos2 if/ — k' sin2 tp
COS fa = —■ =r- .
Vl - k2 sin2 if,
b) Hence show that
1 — k'
sin (2i/r - i/O = l + k, sin fa = K sin fa .
c) Thus show that, if k = 2\fk[l(\ + kx) and sin (2i/r - fa) = kx sin i/^,
then
^,^) = -^-^^,^).
Compare Exercise 5 of Section 1.4.
a) Show that if u : = 2Kx/tt, then
sn(«, k)=2q1,4k-ln sin (a;) fl f T^
n = l L 1 —
- 2qz" cos (2s) + q™
2n-l /0n , 4n-2
■ 2^-1 cos (2*) + ^
b) Find similar expressions for en and dn

2.7 The Landen Transformation 61
5. Combine Exercise 4a) and Theorem 2.6 to produce a quadratically
converging approximation to sn(w, k) given one for sin x.
6. Use (2.7.5) and Exercise 2 of Section 2.6 to show "the addition
theorem"
sn(w) cn(y) dn(y) + sn(y) cn(w) dn(w)
sn(w + v) = -J—27^—27T •
\-k sn2(«)sn (v)
Obtain similar expressions for cn(« + v) and dn(« + v). (See Whittaker
and Watson [27] and Exercise 11 of Section 1.7.)

Chapter Three
JacobVs Triple Product and Some
Number Theoretic Applications
Abstract. We establish JacobVs triple-product identity and apply it quite
variously. We first use it to derive the fundamental product identities for the
theta functions. We then rederive the triple product via Cauchy's q-binomial
theorem and present Bressoud's beautiful elementary proof of the celebrated
Rogers-Ramanujan identities. After this we derive JacobVs formula for r4(k)
(the representation of k as a sum of four squares) and two partition results
due to Ramanujan. We also establish the Gaussian sum formula and indicate
another proof of the theta transformation formula. Then we briefly discuss the
Mellin transform and use it to give the classical reflection formula for the
Riemann zeta function. Finally, we show how certain reciprocal series can be
evaluated in terms of theta functions. In particular, we give a result due to
Landau on the Fibonacci numbers. We also sum the squares of the reciprocals
of the Fibonacci numbers.
3.1 JACOBVS TRIPLE-PRODUCT IDENTITY
Our first proof of the triple-product identity is:
Theorem 3.1
For each pair of complex numbers x and q, with x ¥=0 and \q\ < 1,
co eo
(3.1.1) n (l + V-1)(i + x~V"_1)(i - i2n) = 2 A"2 •
n = l n--«>
Proof. Let F(x, ¢) : = Id (1 + V-1)(l + *"V""1). Now F(-, q) is
62

3.1 Jacobi's Triple-Product Identity
analytic except at zero and has a Laurent expansion at zero. Observe
F(x, q) = F(x~\ q) and F(xq2, q) = (*?)"¥(*, q). Thus if
CO
F(x,q)= 2 cn(q)xn
then cn(q) = C-n(q) and cn(q) = q2n~1cn_1(q) for n3=0. It follows
cn(q) = q"2c0(q) and
(3.1.2)
F(x,q) = c0(q) 2 A"
"t
I
It remains to evaluate c0(q). Letting x'-=l in (3.1.2) gives c0(#)fl,(
iCi (1 + q2"-1)2 and letting x := -1 gives co(?)04(?) = 11^ (1 - q2n
Since V03(?)04(?) = S4(q2), we deduce that
n = l
c0(q)84(q2)=ll(l-q4n-2)
But on replacing # by q2,
c0(q2Mq2) = U(l-q4"'2f
Hence
n = l
^-mi-c*2)2"-1].
Since c0(0) = 1 and c0 is analytic at zero,
CoiqT1 = 111^1) = 111111-^2^]
k = lLCQ(q )J t=l«-l
= fi(i-?2m).
m = l
This establishes (3.1.1). □
It is convenient to make the following notational abbreviations.
(3.1.3)
Qo-=U(l-q2n) Q1-=U(l + q2")
n=l n-1
22--=11(1 + ^-1) Q3-=fl(l-q2n-1)
n=l n»l

64 Some Number Theoretic Applications
From these definitions one easily verifies Euler's identity QiQ2Q3 = 1,
which may also be written
CO CO
(3.1.4) U(l + q")U(l-q2"~1) = l.
n=l n-1
Also
(3.1.5/) Q0Q1 = Q0(q2)
(3.i.5«) Q0Q3 = Qo(qin)
(3.1.5m) Q2Q3 = Q3(q2)
(3.1.5/v) QiQ2 = Qiiq1'2) ■
We gather the first three specializations of the triple-product identity into:
Corollary 3.1
For \q\ <1, one has
CO
(3.1.6) <%(*) = Q0Q22 = II (1" q2")(l + q2n"f
n = l
CO
(3.1.7) 84(q) = Q0Q23 = II (1 - q2")(l ~ q2n"f
n = l
CO
(3.1.8) e2(q)=2qV4Q0Q\=2q1UY{(l- q2n)(l^ q2nf .
n = l
Proof. These follow on using x-=l, —1, and ¢, respectively, in (3.1.1)
(Exercise 2). □
More generally, let k and I be real numbers, and let q'= qk and * := ±ql
in (3.1.1). Then
(3.1.9) I! (1 ± q2kn+k~')(l ± q2kn+k+')(l - q2kn+2k)
= 2 (±i)nqkn2+ln:
n = —co
When k'.= l and /: = \, this gives
CO CO
(3.1.10) 11(1-?")= 2 (-l)V3n+1)n/
This is Euler's pentagonal number theorem, which he found empirically and
which affords a combinatorial interpretation [Exercise 3b)], as most of these
identities do.
When /::=/:=3, we have

3.1 Jacobi's Triple-Product Identity 65
co co co _ 2n
(3.1.11) I[(l + q")(l-q2")=~ 2 ^^ = 11:^-4=1.
„ = 1 Z n = -«> „_i l — q
Finally, k'-=\ and /:=5,5give two formulae which play a central role in
the Rogers-Ramanujan identities (Section 3.4):
(3.1.12a) U(l-q5n+1)(l~q5n+4)(l-q5"+5)= 2 (-l)V5"+3>"/2
00 00
(3.1.126) U(l-q5"+2)(l-q5n+3)(l-q5n+5)= 2 (-l)V5n+1)n/2 .
n = 0
We finish the section with a slightly less immediate corollary of the triple
product. If q '■= q1'2 and x '■= q1 w, then (3.1.1) becomes
(3.1.13) 11 (1 - qn)(l + q"W)(l + q"^1) = (-^-) f W"q«»^>2 .
The right-hand side is
7 (W +W )i7m(m+l)l2 V -«/* + 1\ m(m+l)/2
If we now let w tend to — 1 from above this gives
co co
(3.1.14) EI (1 - q"f = 2 (-l)m(2m + i)qm(-m+^'2 .
n=l m=0
Thus with (3.1.10),
(3.1.15) [ 2 (-i)-y3»+1>»/2l =2 (-l)m(2m + l)?m(m+1)/2.
Lm = -co J OT=0
Comments and Exercises
The proof of (3.1.2) is due to Gauss [1866]. The AGM identity (2.1.7) was
also in his possession, but Jacobi, by elliptic function techniques, was the
first to publish a proof of the triple-product identity. There are many proofs
in the literature. Given the AGM, none perhaps is as simple as the one
given here.
1. Verify the identities (3.1.4) and (3.1.5).
2. a) Prove Corollary 3.1.
b) Show that
(3.1.16) ft (1 + q2"'1)8 = 11 (1 " q2"'1? + 16? 11(1 + q2"f .

66 Some Number Theoretic Applications
3. a) Establish (3.1.9), (3.1.10), and (3.1.11).
b) A pentagonal number is a number of the form n(3n ± 1)/2. Show
that (3.1.10) implies that every nonpentagonal number can be
partitioned into an even number of distinct parts as often as into an
odd number of distinct parts. Show that for pentagonal numbers
there is a surplus or deficit of 1, depending on whether n is odd or
even. This was first observed by Legendre in 1830. (See Dickson
[71, vol. 2].)
4. Establish (3.1.12) and give an interpretation in terms of partitions.
Ewell [81] observes that Euler's pentagonal number formula (3.1.10)
allows one to establish Jacobi's triple product, given (3.1.2). Thus one
can base the identity on a combinatorial proof of (3.1.10) such as is
given in Hardy and Wright [60].
5. Use (3.1.10) to show that c0(q3) n^=1 (1 - q6n) = 1.
6. A complex analytic approach to Jacobi's triple-product is as follows.
a) Show that (3.1.1) is equivalent to
CO CO
(3.1.17) 04(z, q) = EI [1 - 2q2"~1 cos 2z + q*n~2] Y\{1- q2n)
n=l n=l
where q '■ = emt.
b) From the discussion of Section 2.6 show that both sides of (3.1.17)
are analytic with zeros at z = (n + \ )irt + mir (n, m integral). By
Liouville's theorem they differ only by a multiplicative constant.
c) Use Exercise 5 to show that this constant is 1.
7. a) Justify taking the limit in (3.1.14).
b) Prove that (3.1.15) is equivalent to
(3.1.18) [ 2 (-l)Y6n+1)2l =2 (-ir(2m + l)?3(2m+1)2.
c) Observe that (3.1.18) has the following number theoretic
interpretation due to Catalan (Dickson [71, vol. 2]). The excess in the
number of even values of x + y + z in
(6x ± 1)2 + (6y ± 1)2 + (6z ± 1)2 = 3(2n + 1)2
over the number of odd values of x + y + z is (2n + 1)(-1)" In
particular, any number of the form 3(2n + 1)2 must have at least
(2n + 1)/6 decompositions as a sum of three squares.
Show that 03 and 04 never vanish (\q\ < 1).
Show that 82 and fl, increase monotonically on (0,1) and 04
decreases monotonically on (0,1).
8. a)
b)

3.2 Some Further Theta Function Identities 67
9. Let p(n) denote the number of partitions of a natural number n into
positive integral parts (the order being irrelevant). Thus p{A) = 5 and
,,(5) = 7.
a) Show that
(3.1.19) 1 + 2 p(n)qn = f[(l- q")^ = 1+2 7^-%
k\2
"iXlUil-q*)
For the second equality see also Proposition 3.4 of Section 3.3.
b) Use
\l+l P(n)qn] 2 (-1)V(3"+1)/2 = 1
L „ = 1 J n=-»
to write (and implement) a recursive formula for p (n). This is how
MacMahon (1918) computed />(n), l<n<200; /»(200) =
3972999029388. For more information on partition theory the
reader is referred to Hardy and Wright [60] or Andrews [76].
c) (Euler) Establish that
2 p(n-a(m))= 2 p(n-a(m))
m even m odd
- where a(m):= m(3m + 1)/2 is the mth pentagonal number.
3.2 SOME FURTHER THETA FUNCTION IDENTITIES
In this section we collect a number of definitions and relations, some of
intrinsic interest and some for future reference.
Let r G (0, oo) and define A*(r) := A:(e_W7). As in Chapter 2, we consider
k as a function of q (=e-irV7). Then
(3.2.1) 0£4e-W7/2 -A*(r) = 16e""3W7/2 + 0(e""SW7/2)
SUIV^C /I ^/^ — (72V
(3.2.2)
From Corollary
(3.2.3i) .
e ;/
3.1 we
e2{q)
o3ve
K'
£
lave
= Vfc =
^ vOCC JL
[A*(r)] =
2?i/4 ri
n = l
^ACltlSC J..
Vr.
( W"
\\ + q2n-
)■ "
*)'

68 Some Number Theoretic Applications
(323«) Mi)-vF=nf^^y
when 9 := e^™^.
We next derive another beautiful identity due to Jacobi, in which 8^
denotes the derivative of 8X with respect to z at zero. [See equation (2.6.1).]
Proposition 3.1
For \q\<\,
CO
(3.2.4) 82(q)83(q)84(q) = 2 2 (-l)"(2n + l)?(n+1/2>2 = O^q) .
Proof. Identity (3.1.14) can be rewritten as
OUq) = 2qV4Ql = 2qXI*Q\{Q,Q&,f
= Q-qllAQoQ\)(QoQ2i)(QoQl) = <%<%«4. □
For various applications to lattice sums it is natural to augment our theta
definitions by
CO
(3.2.5) 65{q): = 2 2 (-l)Y2"-1/2)2
and
(3.2.6) 06+(?): = 2i (-l)"("-1)/2(2n + l)?(n+1/2)2.
By similar arguments [see (3.1.14)] one can show that
(3.2.7) 85(q) = Iq^QvQJQziq2) = V2l# VXV)
and
(3.2.8) e+6{q) = 85823(q2) = 0,8,(q2)84(q4) .
We leave these identities as Exercise 6 and note that (3.2.4) and (3.2.8)
have number theoretic interpretations like that of (3.1.15). These three
identities and their remanipulations are among the very few known
reductions of three theta terms to one known theta expression. (See Glasser and
Zucker [80].)
Following Weber [08] and others, it is usual to identify the following
quantities. With q and r as above and t : = iVr,

3.2 Some Further Theta Function Identities
(3.2.9/) i} = i,(V=r) = t,(t) : = ?1/12eo
(3.2.9b) A =/x(V=r) =/x(t) := ?-1/24Q3 = (4*'2/*)1/12
(3.2.9m) /2 =/2(V=r) =/2(r) : = 21/V/12Q1 = (4*2/*')m2
(3.2.9/v) / = /(V=7) =/(r) : = <f 1/24Q2 = (4/kk'f12 .
Then
(3.2.100 /1/2/= V2
(3.2.108) /8 = /?+/2
and jRT satisfies
(3.2.11) K=-r)f -^VJi-^jvfi-
As will be discussed in Chapter 4, whenever r is rational, /1} /2, and /
satisfy algebraic equations (whose degree is determined by the number of
quadratic forms with determinant -4r). Thus once tj and one other of flff2,
f, k, k' is known, all six are determined,
A significant identity which follows from (3.2.9iv) is
(3.2.12) 2. 2 ' 2 = ~vf log [/(V=7)] = — log I —- I .
Here and hereafter, the prime over a summation indicates that the term
m = n = 0 is omitted and summation is over expanding rectangles. One
establishes (3.2.12) by writing
-^iog(Ga)=£i£ tf2q^k
v' v '■ t=ln=0 K
= ir2 —7=r- cosech (irVrk) = 2 2
00 00 / 1\*+«
(-ir_„„w„^_v v tiy
t^\ Vrk ~" "v/ ^xm^1-™ m2 + rk2
[using (2.2.3)], and so
V' ("I) 47T , 77
2, 2, 2 =~ v?log(<22)- —
since S^=1 (-l)m+1/m2 = 7r2/12. This is the desired result. One may
observe that whenever one can evaluate the double sum, one can also evaluate
kk'.
In some future work we will be following Ramanujan rather than Weber.
Ramanujan studied

70 Some Number Theoretic Applications
G„ := (2kk'yin2 = 2~1/4/(V^n)
, ,,2\ 1/12
..-(If) -2-'V,<V=*>.
For some purposes these give slightly cleaner results. For example,
_V5+1 _ V5+1
O25- 2 fto-\ 2 '
For the moment we denote k{ q") by 8 and jK(S) by A. We have
,3
(3.2.14) kk'^f) =4V?Go
so that
(3215) mV6ji- qU12Q°
^■2A5) \88'J VA _ qnn2Q0(q")
We also have
(3.2,6) (M)'(1-2*»)-l-24Ze5±iJ£
\ 7T / „=0 1+ q
(3.2.17) (^)V - k2k>2) = 1 + 240 £ -¾
v t / „=1 1 — 0
and
(3.2.18) (—) (1 - 2*2)(l + \ k2k'2) = 1 - 504 2 ^
\ IT I \ I I n = l 1 "
" -y
12n
These formulae lie considerably deeper, either in elliptic function theory, or
as direct computations. Indeed each formula entails formulae for powers of
theta functions. We prove only (3.2.16) and leave (3.2.17) as an exercise.
We need an identity whose proof we also leave as an exercise.
Lemma 3.1
For \q\ <l,
nqn _ ^ {In + l)q2n+x
(3.2.19) 2^ = 2^
2n+l
9
Proposition 3.2
For \q\<\,
(3.2.20) 6t(q) = 1 + 8 £ ^C + 8 £ ^¾^
„=0 1+9 n = 0 1 — ^

3.2 Some Further Theta Function Identities 71
,„n ,4,, of (2n + l)g2"+1 f (2n + l)g2"+1
(3.2.21) 02(g)-8Z ., 2«+i +8Z—;— 2«+i
„ = 0 1 + ^ n = 0 1 ~ 9
(3.2.22) ^) = 1 + 82^-82^^1^ ■
„=0 1+ ^ n=0 1 + ^
Proof From (2.3.16), on differentiating with respect to q = e~ns, and
Corollary 3.1 we have
*«>~««<H)-*£*(i)+>
which yields (3.2.20). The other two follow similarly. □
Theorem 3.2
For \q\ <1,
(3.2.23) «J(«) = l + 8 2 7¾
n^0(mod4) X 9
nal
and
2n + l
(3.2.24) flj(*) - e\{q) = 1 - 24 2 ^¾
„=o 1 + 9
Proo/. Apply Lemma 3.1 to (3.2.20) to establish (3.2.23). To prove
(3.2.24) one uses the lemma with q and —q. □
Now (3.2.16) is immediate from (3.2.24). Formula (3.2.17) is similarly
derived, if one knows in addition (see Rademacher [73]) that
(3.2.25) 8l(-q) = 8\{q) = 1 + 16 2 ^"^ .
Finally, (3.2.18) follows from (see Rademacher [73])
(3.2.26) 0«(g) = i + 8f JtiL-oS ^^+(0+)4
„=i 1- q „=i 1~9
Of course, implicit in (3.2.23) is a formula for the number of
representations of n as a sum of four squares. We discuss this more fully later.

72
Some Number Theoretic Applications
Comments and Exercises
1. Establish the asymptotic of (3.2.1).
Proposition 3.1 is often established by analytic arguments like those of
Exercise 6 of Section 3.1. Proposition 3.1 may then be used, as it is in
Whittaker and Watson, to establish the triple-product identity. (See
Exercise 6.)
2. a) Show that
n[l-e-(2-+1)'] = 21/8e-
n = 0
b) Show that
2 e-(2"+1>2- = (21/4-i)77-3/42-"/4r(i).
n=0 ^4/
(This was set in Trinity College, Cambridge, in 1881.)
c) Recall that if *:= (V3-1)/V8, then kk'= \ and K'(k) =
y/3K(k). Deduce that
n (l+e-ini"^)=2~i3/24(V3+1)" vi2V3.
n = l
d) Similarly, if k:= V2(3 - V7)/8, then kk' = £ and K'{k) =
VlK(k). Deduce that
CO
11 (1 + e-2nWV7) = 2-5/8(3 + V7)1M e"lnV~7 .
n = \
3. Establish the identities in (3.2.9), (3.2.10), (3.2.11), (3.2.14), and
(3.2.15).
4. a) Justify the derivation and convergence of (3.2.12).
b) Evaluate the left-hand sum when r '■= 1, 2, 3, 4, 7.
c) Prove that as r tends to infinity, the sum converges to —tt2/6. It is
already "close" by the time r = 7.
d) Prove that
Y V' (-1)" 477, r,, „ ir. (4k'2\
m=—n=- n + rm Vr °LJ^ 3Vr \ k /

3.2 Some Further Theta Function Identities 73
5. Show that
a) g*n = 2l"gnGn
b) Gn=GVnwAgnl = gun
c) {gnGn)\Gl~gl) = \.
6. a) Prove Lemma 3.1 by expanding both sides.
b) Prove (3.2.7) and (3.2.8).
c) Establish Theorem 3.2.
d) Prove (3.2.16) and (3.2.17). Try to prove (3.2.18). Exercise 9 will
help.
The next two exercises sketch out some facts about Jacobi's imaginary
quadratic transformations.
7. a) The transformation kx •'= k1 is given in theta terms by
(t)V">=(f)V-"-v
b) K'(kx) = kK' and K(kx) = k(K + iK').
c) The transformation kx:=k'~l gives K'{kx) = k'K and K{kx) =
k'(K' + iK).
d) The transformation kx'-=iklk' gives k[ = -k'~\ K(kx) = k'K,
and K'(kx) = k'{K' - iK), since re(s) > 0.
e) The transformation kx = k'lik gives k'x = k~l, K{kx) = kK', and
K'(kx) = k(K + iK').
8. a) Use sx ■'= i + s~x to derive from (2.3.14) that
_i= Aw. -.to
2 *<%(,) * 04(Sl)
and
Mill = _ • AM + AM + I
S>04(*i) l04(Sl) * 6,(5) + 2-
b) Now use Exercise 7a) and (2.3.17) and (2.3.18) to show that when
kx := k~\ kE{kx) = E(k) - (k')2K(k).
c) Derive similar formulae for E(kx) for the other three imaginary
transformations.
d) Evaluate K'lK when k'-= VZ + V2.
9. a) For r := 1, 2, 3, 4, show that ^(fq) = 83(q4) + 1¾¾4).
b) trrie"3(irq) = 4 i (?k-V)«2(*4)-
&=y(mod 4)
c) Evaluate 03~*0* for A: := 1, 2, 3, 4.

74 Some Number Theoretic Applications
10. Show that
j. ("ircTnc"(r+i)7r=\ ** w - ^
[Let /-:=9, 2kk' = (2 - V5)2 in (3.2.12). See (4.6.10).]
The formula for sums of two squares is derived from Jacobi's identity
CO „
(3.2.27) »3 = l + 4lrru.
n=\ 1 + q
This will be shown in Chapter 9.
11. A comprehensive list of hyperbolic identities can be derived, as was
(3.2.12). We have, for q : = e~",
1
(3.2.28/) -log Q0 = 2 , 2™ n
i n(e - 1)
_ 1 Y coth(7rns) — 1
~2 Y n
its 1 (2K3kk'\
= - n - 6 lo^ [-J-)
(3.2.28«) -ioget=i (2;ir
l n(e - 1)
_ 1 y (-l)"[CQth(ff«8)-l]
2 j n
= _ tts _ J_ / A:2 \
12 12 g \ 16/V I
„ 1 v (-l)"cosech(7rns)
(3.2.28m) -log Q2 = ■= L ^-2 * L
L J ft
ITS
24
- Tilog (*f)
/-^ no' \ i st, ! V cosech (7ms)
(3.2.28*v) -log 03 = -Z -*

3.2 Some Further Theta Function Identities
(3.2.28v) log Q0Q\ = log ^¾ == S
1 °°
= -2
2 ,
2# i n(e +1)
1 y 1 — tanh (jrtis)
ITS
T
1 , (kK\
(3.2.28vi) | log Q0q2 =3 log fl,
y 1 =1, /¾
""f (2n-l)K(2"-1)s + l] 410gW/
(3.2.28v//) - i log e0e2 = - | log 04
i (2n - i)[e<2"-^ - 1]
-i^h-.-.-ib,^)
(3.,28W,.01og[|^] = |_I-)l_
= ly (-!)"[!-tanh(ff/u)]
2 j n
T + 2log VIFV
/o^^o-\ , ^ ^2 v cosech [7r(2n - l)s] 1
(3.2.28**) log ete22 = Z ^^ ° = - 4 log k' .
These are taken from Zucker [84]. Many more identities follow
differentiation or as below.
12. a) Show that
Hmf: Use (3.2.28k). Then replace q by q1'2 so that k' is replac
by (1 — /:)/(1 + k). Then replace # by —q so that /: is replaced
iklk' (as in Exercise 7).
b) Show that

76
Some Number Theoretic Applications
+ 1)tts/p
V (-1)" sech
„t'02n + lSeCtl
f"(2n
tt/8
tt/24
isin_1[(3-V7)/4V2l
= 1
= 3
= 7
3.3 A COMBINATORIAL APPROACH TO THE TRIPLE-PRODUi
IDENTITY
The Gaussian or q-binomial coefficients are polynomials defined by
0/q \n/c
and by
(3.3.1)
(") =-
\m/ q
(?)„ n"=„-m + l(l - 91)
* («)»(«)-» nr=1(i-<7s)
when 0<m<n. Here (^)„ is defined by
(3.3.2)
(*).:= n
In
-q
l-qs+
This allows for any complex value of s, but we will consider only integral
values. Thus for n in N, (q)n = U"=1 (1 - q') and (q)Z\ = 0. For m<0 or
m > n we either define ( ) to be zero, or observe that it is implicit in our
definition of (q)m (for integral m). The next proposition gathers up some
easy facts.
Proposition 3.3
For \q\ <1,
(fl) (m + l)t = (m + l)t + «"'"(m)t
(6) Hm(n) =(")
(c) ( I is a polynomial in q.
\m/ q
Cauchy's binomial theorem is, for n = 1, 2, 3, . . . ,

3.3 A Combinatorial Approach to the Triple-Product Identity 77
(3.3.3) 2 y^-O-iW n) = fl (i + M*}.
m=0 \m/g fc=1
This is easily established inductively by showing that both sides of (3.3.3)
agree for n = 1 and satisfy
(3.3.4) Fn(y, q) = *„_,(>», q) + yqnFn_x{y, q) .
One can also argue, combinatorially, that the coefficient of ysq' on each side
gives the number of partitions of t into s distinct parts not exceeding n.
If we now let y ■ = xq~N and n '■ = 2N, we obtain, after some manipulation,
(3.3.5) £ *v(m+i)/2L2*J =n(i+x^)(i+*-v-1)'
m-~N Xly m/ 1 1
a result also due to Cauchy.
This bears a striking resemblance to the triple product in the form of
( 2iV \_1
(3.1.13). Indeed on letting N tend to oo one has \N_ ) tending to
Q(q) '■= n^=1 (1 - qm), and we are left with (3.1.13) if we can justify the
exchange of limit and summation (Exercise 3). Thus (3.3.5) deserves to be
considered as a finite form of the triple product.
The following Eulerian result will also be used in the next section. For
notational simplicity the empty product equals 1.
Proposition 3.4
For k = 1, ,2,.. . ,
rr 1 y *V2 (k\
fA l-v /-o(i-**)---(i-V) WV
Proof. Inductively one establishes that each side satisfies
xqk
Gk(x, q) = Gk„x{x, q) + y^—• Gk_x{xq, q) .
Obviously the result is true for k = l. Alternatively one can argue that the
coefficient of x'qs on each side counts partitions of s into t parts not
exceeding k. □
Comments and Exercises
Here and in the next section we follow Bressoud [83].
1. Prove Proposition 3.3 and (3.3.3).

78 Some Number Theoretic Applications
2. Verify (3.3.5).
3. a) Prove that for \q\ < 1,
limf 2" ) =Q-\q).
b) Justify the exchange of limit and summation in (3.3.5) by observing
that the terms possess uniform majorants that are summable.
4. a) Prove Proposition 3.4.
b) Deduce that
CO CO
(3.3.6) I! j = 2
,=i 1 - xq' ,=o
*v
(l-xq)(l-q)---(l-xq')(l-q>)
and compare with (3.1.19).
3.4 BRESSOUD'S 'EASY PROOF' OF THE
ROGERS-RAMANUJAN IDENTITIES
The Rogers-Ramanujan identities are
„2
n 1
(3.4.1) 1 + Zj ~7\ vTl ZJ\ /1 „m\ ~ 1 ■*■ 7l „5m+l\/1 5m + 4\
m = l (1-?)(1-? )•••(!-? ) m = o(l-? )(1-9 )
m = l (1-9)(1-9 )---(1-9 ) m = 0(l-9 )(1-9 )
They afford a remarkable combinatorial interpretation (Exercise 1). We
use equations (3.1.12a) and (3.1.126) and the notation of the last section to
rewrite the identities as
(3.4.3) 2 7^r = G_1(9) 2 (-iyy5m2+m)/2
m = 0 \H)m m = -»
00 mz + m <*>
(3.4.4) 2 fT-==G"1(9) 2 (-iyy5m2+3m)/2.
»"=0 K^m m=—<*>
It is this form that we establish. The key observation is the following.
Lemma 3.2
For n positive and integral and any complex a,

3.4 Rogers-Ramanujan Identities 79
m am2 s2 m (a-l)m2
(3.4.5) S, '« -2-^-2"%, ■
Proof. Note that the above sums are all finite. Set k'.— n — m and
x:= #2m in Proposition 3.4. Then multiply each side by (¢)^. We get
(?)n + m=2 T-
'^^ (*)-
Make this substitution for (q)~lm in the left-hand side of (3.4.5). We have
m am2 m am2 j2+2mj, \
y x q _ y x q y q K1)n-m
m KQjn-mKQjn + m m n — m—j
x qK
j +2m
We now sum over m and s-=m + j and exhibit the right-hand side of
(3.4.5). □
m2
The effect of Lemma 3.2 is to reduce the power of q by 1. Repeated
use of the lemma results in an expression which can be handled by the finite
triple-product. As we will see, the k = 2 case of the next result contains the
desired identities.
Theorem 3.3
Given positive integers k and iV, we have
_s\ + s\+---+32. Sk
x —^— — nd+x^xi+x-v1)
m \Jy — ml q
Proof. Commence with applying the lemma to the right-hand side of
(3.4.6). We have
(^2*v(2*+i)m2+m]/2L ) =s(7x}*, x—
- \N-m/q m (q)N-m(q)N+m
N+m
2 - l/2yn <2k-l)m2/2
= V 1 v w ) g
(q)s-m(l)s +

80 Some Number Theoretic Applications
We continue by applying the lemma /:-1 more times and arrive at
/ \sl+s%+---+si / l/2\m m2/2
y (?) v (*g ) g
^,¾.....¾ (#)^-^( g^-JjC #)^^-¾ m \1)sk-m\4)sk+m
We use the finite triple-product (3.3.5) to write
, l/2yn m2/2 s*
2 ffi >« = (¾ n (i+^m)(i+*-v-1) •
m \H)sk-m\<l)sk+m m = l
This completes the proof. □
The theorem is applied by specifying k and/or x. When x:=-1, we
observe that unless sk = 0, the products on the left-hand side are zero. Thus
(3.4.7)
=(^2(-1)-,---(^),.
If we let N tend to », we arrive at
(3.4.8) 2 2
*1>*2 "k-l (¢)^-¾ " ' " (¢)^-2-^-1( ?)»*_!
n = l m = -»
For /::=1, the left-hand side is 1 and the formula recaptures Euler's
pentagonal number theorem. For /::=2, (3.4.8) coincides with the first
Rogers-Ramanujan identity (3.4.3). The second identity is derived similarly
by specifying x '■= —q and /::=2 (Exercise 3).
Comments and Exercises
The Rogers-Ramanujan identities were discovered by Rogers in 1894 and
rediscovered by Ramanujan in 1913 (a letter to Hardy) and Schur in 1917.
They have continued to receive a great deal of interest. As recently as 1979
Basxter rediscovered them in a physical context.
Hardy and Wright say "no proof is really easy (and it would perhaps be
unreasonable to expect an easy proof)." Bressoud's proof certainly comes
close to contradicting this. Much of the early history can be found in
Hardy's footnotes to Ramanujan [62].

3.5 Some Number Theoretic Applications
81
1. Show that (3.4.1) says that the number of partitions of n into parts with
minimal difference 2 is the number of partitions into parts congruent to
1 or 4 modulo 5. Equally, (3.4.2) says that the number of partitions of n
into parts with minimal difference 2 and minimal part 2 is the number of
partitions into parts congruent to 2 or 3 modulo 5.
2. Verify the equivalence of (3.4.1) to (3.4.3) and (3.4.2) to (3.4.4).
3. Derive the second identity (3.4.4) from Theorem 3.3. There is an
interesting continued fraction associated with the identities.
n
(l-x5n+2)(i-x5n+3) _ K=-A-i)nxi5n+1)nn
„io (l - x5n+1)(i - x5n+4) e:__. (-i)Vs"+3),,/2
(3.4.9)
2 3
= 1+ 1+ 1+ 1+"" '
When x'-=e~,r/" (n rational), this is in principle evaluable in closed
form. Ramanujan gives
(3-0, 1+£^-[.-(^.v?ii)]-.
In a recent paper, Bhargava and Chandrashekar Adiga [84] detail
(3.4.9) and other continued fraction identities. The closed form (3.4.10)
and its extensions are elaborated on in Ramanathan [84].
4. Investigate the analogue of Theorem 3.3 in which 2k replaces 2/: + 1.
3.5 SOME NUMBER THEORETIC APPLICATIONS
Our first application is a proof of Jacobi's formula for r4(ri), the number of
representations of n as a sum of four squares [including sign and
permutation so r4(2) = 24 and r4(l) = 8]
Theorem 3.4
For each positive integer n,
(3.5.1) r4(n) = 8 2 d.
d\n
d>^0(mod 4)
In particular every positive integer is the sum of four or fewer squares.
Proof. Expand the right-hand side of formula (3.2.23) for 6\{q) to
obtain

82 Some Number Theoretic Applications
CO CO CO y ,
4-Tm 4fd
Now compare coefficients with 0\(q) written as
+ 00 00
n-=-oo
The subsidiary conclusion is Lagrange's famous result. Our second
application is an analytic proof of Fermat's theorem that any prime of the form
4/: + 1 is the sum of two squares. It is convenient to define
o-j(n) := Zi d
d\n
and
(3.5.2) w(n) := <rx{ri) + o-j(odd n)
where odd(n) is the odd part of n (that is, the largest odd divisor of n). We
set w(n):=0 for n=sO. The proof of the following lemma is left as
Exercise 3.
Lemma 3.3
The value w(n) is divisible by 4 unless n is an odd square.
Theorem 3.S
An odd prime p is the sum of two integral squares if and only if it is
congruent to 1 modulo 4.
Proof. The 'only if is immediate on consideration of residues mod 4.
Now argue as in Proposition 3.2 and Theorem 3.4 to show that
\ri (2n + 1)# \ri n#
2j \ „2n+l -^ 1 _ _"
n=0 I" ? « = 1 A 9
(since Q0G3 = G)- Hence, on expanding these expressions (as in Exercise 12
of Section 3.7)
On multiplying by 04 and comparing terms we have

3.5 Some Number Theoretic Applications 83
(3.5.3) »v(n) + 22(-l)Mn-y2) = {2(01)r V ^^
Suppose now that n = p = 4m + l with p prime. Then w(p) = 2<rx(p) =
8m + 4 and we see that
Am + 2 + 2 (-l)Mf - ft = 0 •
Thus some w(p — j2) is not divisible by 4 and Lemma 3.3 implies that
p = j2 + /:2 for some integers j and /:. □
Our third application is to establish the following reciprocity result for
Gaussian sums. Let
S(p,q):=2e-^'"q
where p and q are nonzero integers.
Theorem 3.6
For positive integers p and q with pq even,
«(/>, ¢) = ^^5(<?>/>)•
(Here the bar represents complex conjugation.)
We will prove the result from the transformation formula for 6^ (2.3.1).
This needs the following:
Lemma 3.4
For pq even and q positive,
liraiVa,(e + ^) = ½ S(P>1)-
e-»o+ \ q I q
Proof. Write 03(e + i/>/?) := £"_ e-™2(«+*/?) j^
fl,(e+ ^) = 1 + 2^ «-**"[2 e-^'"^']
on taking the periodicity of e~niTiplq into account. (Here we have used
e~niq2p,q = 1, which entails pq even.) Thus

84 Some Number Theoretic Applications
q
ds + 0(Ve)
Vee^e + &) = 2 e-"*«[2Ve [ <T(r
q/ r-l
on using the integral test. We now take the limit as e—> 0+ and calculate that
/•CO /*CO
lim Ve e-^sqf™ ds = lim Ve e~£,r?V ds
e^0+ JO e^0+ Jr/g
= lim Ve f e-"**2 <b = -4=- f e~'2 * = T~ •
e^o+ Jo y/Trq Jo 2q
Since S(/>, q) = S*=1 e^"2''7', this gives the desired result. □
Proof of theorem. We now use the theta transform (2.3.1) to write
Thus the lemma shows that
7f S<* *> = '""* W ,¾ ^(6 / - 'j + °(e2))
= e~"'*Vp lim VF03(e' -^ + 0(e'2)) .
By Lemma 3.4 applied to -^ and p we deduce that
(3.5.4) ■^S(p,q) = e-M4~S(-q,p). □
Our final application is a partition result of Ramanujan's which relies on
the triple-product identity. (See Exercise 9 of Section 3.1 for the definition
of P.)
Theorem 3.7
(a) p(5n + 4) is divisible by 5.
(b) p(7n + 5) is divisible by 7.
Proof. With Q(q) = n^=1 (1 - q") as before, we write
CO CO
qQ\q) = qQ(q)Q\q)=Z 2 (-l)"+"(2m + 1)**
m=0 « = — eo
where /::=1 + (3n + l)n/2 + m(m + 1)/2. This uses the triple-product
identities (3.1.10) and (3.1.14) multiplied together. One now considers when k is

3.5 Some Number Theoretic Applications
85
divisible by 5. Since 2(n + 1)2 + (2m + 1)2 = 8fc(mod5), we must have
2(n + 1)2 + (2m + 1)2 divisible by 5. An inspection of residues shows that
this can only happen if 2(n + 1)2 and (2m + 1)2 are both divisible by 5.
Hence, 2m + 1 is divisible by 5 and so is the coefficient of qim 3 in qQ (q).
From the binomial theorem one deduces that
(3.5.5) (l-qy5 = (l-q5y\mod5)
in the sense that all coefficients are congruent. It then follows that Q{q)l
Q5(q) = l(mod5), and hence the coefficient of q5m+5 in
gQ(q5) = qQ\q)Q(q5)
Q(q) Q\q)
is divisible by 5. But
qQ-\q) = qQ^Q-\q5)
_ qQ(q5) rr (v sm«\
so that the coefficient of q5m+5 in qQ~x(q) is divisible by 5. However, by
Exercise 9 of Section 3.1,
qQ'\q) = q+2 p(n-l)q".
n=2
Case (b) is similar, but uses the square of (3.1.14) instead of the product
with Euler's series. (See Exercise 5.) □
Comments and Exercises
The identity (3.5.1) was discovered by Jacobi on April 24th, 1828. He also
subsequently observed similar number theoretic interpretations of the
formulae for 02 and 8\, and he gave an arithmetic proof of his theorem. The
identities can also be found in Gauss's unpublished work. A wealth of this
and similar information can be found in Dickson [71, vol. 2].
An analysis of the components of our proof of (3.5.1) shows that it is
rather simpler than that in Hardy and Wright [60], which proceeds from
rz(n), or in Rademacher [73], which uses elliptic function arguments. We
use only the triple-product identify and the AGM.
1. Show that r4(n) is 8 times the sum of the odd divisors when n is odd,
and r4(n) is 24 times the sum of the odd divisors when n is even.
2. Show that

86
Some Number Theoretic Applications
n=0 J n=0 1 —
1)?
4(2n + l)
8(2n + l)
and deduce that the number of representations of n as a sum of 4 odd
squares is 16 T.4dd,=n d.
d,d' odd
The argument of Theorem 3.5 is due to Ewell [83]. We give the general
formula for r2(n) in Chapter 9.
3. Prove Lemma 3.3. Consider odd and even cases and use the multip-
licativity of crx.
Gaussian sums arise naturally in the study of cyclotomic polynomials and
hence occurred to Gauss while studying constructible polygons. Apparently
this led Gauss to the lemniscate and thence to elliptic functions. The
formula (3.5.4) (with p'-=2) plays a key role in establishing the class
number formula for binary forms. Landau [58] is sufficiently taken by the
result that he presents three proofs. The result we give is due to Dirichlet.
The case with p := 2 is due to Gauss save for the "detail" of determining the
sign of the complex square root. (See Landau for the significance of the
sign.) Our proof follows Bellman [61].
4. Show Gauss's result. For ?>1,
i + '
This, generalized, leads quickly to a proof of quadratic reciprocity. (See
Apostol [76a].)
There is a host of more recondite modular results on partitions. (See
Andrews [76] or Hardy and Wright [60].) The rule of thumb that additive
number theory is generally harder than multiplicative theory is born out by
the relative paucity of partition information.
5. a) Show that {\- q)"p = {\- qpyx{mod p) holds for any prime p.
b) Prove Theorem 3.7(6).
There is a combinatorial proof of the theta transformation formula which
is suggestive of the arguments used in the sections on the Rogers-Ramanu-
jan identities, in that it produces a 'finite theta transform' and moves to the
limit. The proof is due to Polya. Again we follow Bellman [61].
6. a) (zin + z~ll2)2m = Zj ( ,)z for any z and integral m.

3.6 The Mellin Transform and the Zeta Function 87
2mll
b) Let w ■= e m , I a' positive integer. Then
-//2=3fc=3//2 *:=-|m//J v"tT»"'/
(Here [jcJ is the greatest integer less than or equal to *.)
c) Fix s and f with t real and positive. Let 1-= [(mt)ll2\ and z = es".
Then
■ <s+2irtk)l2l , -(i+2ir/fc)/2/-|2in
-ins,ks,in "- ^
2
2=3*
= 2 {
-//2=3*:=3//2 *■
/ i 1 .-/,\2 1 8/2(m/4/2)
//2=3*:=3//2 "- O*
|m//J
= 2
fc=-|m//J
d) Now let / go to infinity and use
i) lim (1 + -) =ex if lim ;c„ = x
. n ( In \ e
u) lim —nr I = —y= if
H^n=x-
e) Deduce from c) and d) the following form of the general theta
transformation:
co t~~— CO
V (s+2irik)2/4t _ /_£. V -/A:2+JA:
3.6 THE MELLIN TRANSFORM AND THE ZETA FUNCTION
We continue our tour through theta function theory with a discussion of the
Riemann zeta function. This also allows us to catalogue a few useful
properties of the Mellin transform for future use.
The Mellin transform is a specialized Laplace transform defined by
/•CO
(3.6.1) M(f):=Ms(f)-\0 f{x)xs-ldx.
For integrable functions with suitable behaviour at zero and infinity, Ms is
analytic in a strip a<re(s)<6. For example, the gamma function,
r(s) '■= $o e~xxs~x dx, is analytic in re(s) >0. A most useful identity is
/•CO /*CO
(3.6.2) jo /fry)**"1 dx = y"s Jq fW1 dx y>0.

88
Some Number Theoretic Applications
Under mild conditions, the Mellin transform is invertible and one can
identify two functions whose transforms agree for re(s) > 0. (Certainly this is
true if the functions are transformable and continuous.) This is an
appropriate place to state the following functional characterization of the gamma
function, whose proof is left as a guided exercise.
Theorem 3.8
The gamma function is the unique function /: (0, oo)—>[0, °°) such that
(1) /(1) = 1
(2) f(x + 1) = xf(x) for x > 0, and
(3 ) log f(x) is convex.
Many otherwise tedious facts are easy consequence of this functional
characterization of the gamma function. We list three whose proofs are left
as exercises. These are the functional relation T(s)r(l — s) = 7;7sin its
(1.6.6), the beta function formula /3(s, t) = T(s)T(t)/T(s + t) (1.6.7), and
the duplication formula
(3.6.3) T(2s) = 77--17¾2^1^)^ + i) .
We now derive the functional equation for the Riemann zeta function
£(s) '•= S^=1 n~s, re(s) > 1. We observe in passing that £ has an immediate
analytic continuation to re(s) > 0 simply by writing
CO CO CO
n=l n=l n=l
so that
(3.6.4) £(s) = -A^S 2 (-1)"+V* re(s) > 0 .
More interestingly, consider g{i) •"= [#,(*) - l]/2. For re(s) > \ we have
Thus
r(§) *«»-'" "JV2-1^)*
=[t^g(t)dt+for^y^dt
+ \fo(t^-iy>^dt.

3.6 The Mellin Transform and the Zeta Function
89
Here we have used the theta transform for 03 (2.3.1) to substitute for g on
[l,oo). This leads to
if we evaluate the third integral and replace t by lit in the second integral.
Since \g(t)\ = 0(e~*') as t—>°°, we see that the integral is an entire
function of s. Thus r(s/2)£(s) is analytic except for simple poles at s = 0, 1.
Since T(s) has a simple pole at 0, we see that £(s) is analytic except for a
simple pole at s = 1. In particular (3.6.5) gives an analytic continuation of
£(s) to the entire complex plane. Moreover, as Riemann discovered, the
right-hand side of (3.6.5) is invariant under the change of variable s '•= 1 — s.
Thus we have
(3.6.6) T(±)„-'nto) = T(±^)v-»-*nC(l-').
This is the celebrated functional equation for the zeta function.
Comments and Exercises
It is also possible to deduce the Poisson summation formula from the
functional equation for the zeta function. (See Bellman [61].)
1. a) Use Holder's inequality to show that T satisfies theorem 3.8.
b) Conversely, g(x) ■= log f(x) satisfies g(n + 1) = log (n!) and
x log n =s g(n + 1 + jc) — g(n + l)^x log (n + 1) .
Thus
0^(*)-log x{x + 1ynX.{x + n^*log(l+l)
and
(3.6.7) r(je) = lim , , "!" , , , = /(*) .
y v ' »-» x(x + 1) • • • (x + n) J v '
This argument is due to Bohr and Mollerup. Interesting extensions
can be found in Askey [80] and in Denninger [84].
2. a) Prove formulae (1.6.6), (1.6.7), and (3.6.3). In the latter case write
the ostensible identity in the form T(x) = f(x) and verify that /
satisfies Theorem 3.8.
b) Establish that T has an analytic continuation to the entire plane
with simple poles at the negative integers and zero, and with no
zeros.

90 Some Number Theoretic Applications
3. Let
e-<'+*2/4'> r1'2 dt.
Verify that f(x) and Vtt e~x have the same Mellin transforms by using
the duplication formula. Hence reprove the result of Exercise 4 of
Section 2.2. Alternatively, establish the duplication formula from that
exercise.
4. a) Use (3.6.4) and the fact that £{s) has a pole with residue 1 at 1 to
show that
b) Show that £(-2n) = 0 for positive integral n. The reflection formula
(3.6.6) at least hints of the centrality of the line re(s) = | in the
behaviour of the zeta function. The factorization, due to Euler,
&)= 11 (l-p'T1 ">(*)> 1
p prime
shows the connection between prime distribution and the zeta
function. The Riemann hypothesis is that all the nontrivial zeros of
£(s) lie on re(s) = \. The asymptotic distribution of the primes is
inextricably tied up with this famous conjecture. (See Rademacher
[73].)
5. Use Theorem 3.8 to establish Gauss's multiplication formula,
T(nx) = (2fl-)(1-"),2«"-1'2 it r(* + -) .
6. (Stirling's formula) The formula is
-^=4= = 1 + 0(1) s-*c°.
Outline: Substitute x = s(\ + u) in Ms(ex) and write
T(s + 1) = /+1 e~s J [(1 + «) e-"]s du .
Now replace u by tVTIs and obtain
T(s + l) = ss e"sVTs J__ ij,s(i) dt
for a (dominated) kernel ifrs(t) which approaches e~' uniformly for
bounded t.

3.7 Reciprocals of Fibonacci Sequences 91
3.7 EVALUATION OF SUMS OF RECIPROCALS OF FIBONACCI
SEQUENCES
Since the theta functions provide quadratic analogues for the geometric
series, it is natural to ask about their relationship to sums of the form
(3.7.1) 2 «„ ' where a„+1 := Man + Na
n-l
with a0 and at given (M, iV^O). The one-term (N = 0) recursion leads to
the geometric series. The two-term recursion leads to theta series and their
relatives. If a ¥^ /3 are the roots of x = Mx + N, then
,~ - ~ («i ~ P«o)"" ~ («i ~ aao)P"
(3.7.2) an = —
(Exercise 1) and
a + /3 = M a/3 = -N .
We will consider only the case in which N = ±1 and M is real. We may write
an+l = (2c)a„ + ea„_j, |e| = 1, and to assure that a, /3 are real, we assume
c> max {0, -e}. Then ap" = ±1 and
(3.7.3) <*:=c + Vc2 + e /3 := c-Vc2 + e.
We must consider summing series of the form
The following proposition provides the key. We define the Lambert series
(3.7.5) L(/3) :-2-7^ IJ31 <1.
Proposition 3.S
For 0 < /3 < a with a/3 = 1,
SO
(») ^^ = ^ = 7^)-1]
~, a2" + /32" „" 1 + /34" 4
and

92 Some Number Theoretic Applications
00 i <» o2n + l i
(»0 2 2„+l4,fl2n+1 = 2 ,; -4n+2 = 7[^3()8)-823(p2)]
„=o a + /3 „=o 1 + /3 4
' =^()62).
Similarly,
(iv) i-7r^ = i-^r = L()8)-L(J82)
so
(v) 2 2n ft2n=21J1^r = L(j82)-L(j84)
„=! a -/3 „=1 1- /3
and
(W) 2 2n+1 * a2n+x = 2 , P "4n+2 = ms) - 2L()82) + L(/34) .
„=o a -jo „=o 1- jo
Proof, (i) The first equality follows on multiplying top and bottom by
/3" and the second now follows from (3.2.27), the formula for 0\. (ii) Now
observe that a2 and /32 satisfy the hypotheses of (i). Then (Hi) follows on
subtraction. We leave (iv), (v), and (vi) as Exercise 2. □
Proposition 3.5 will allow us to sum E"=1 a~l for certain starting values.
Specifically, we can handle e = ±1 and A = ±B in (3.7.3) and (3.7.4).
case 1 (A = -B) Let a0:=0 and a1'=l. [This normalization comes by
setting A := (2Vc2 + e)-1.]
(i) (e = -1) Then a/3 = 1 and c > 1. Now
(3.7.6) 2 a;1 = 2V7^1[L(/3) - L(/32)]
n = l
as follows from (3.7.2) and Proposition 3.5 (iv). Similarly, one can
evaluate
2a-1=2V7^1[L(J82)-L(J84)]
n-l
and
2 a^, =2V7^1[L(/3) -2L(/32) + L(/34)]

3.7 Reciprocals of Fibonacci Sequences 93
(ii) (e = 1) Then a/3 = -1. Now /3 < 0 and
CO CO ^
2 ail =2V7TT 2 2n 2n =2V7TT[L()82) - l(/34)]
n-l n = l a "JO
while
(3.7.7) 2 a^+1 = 2Vc2+l 2 2n+i \ aim+i
n=0 n = l a + |j8|
= ^4^[03(|j8|)-02(j82)].
case 2 (A = B) Let a0 := 1 and at-= c. [This normalization comes from
setting A = \.]
(i) (e = -1) Then a/3 = 1 and c> 1. Now
2j a„ = 2
(3.7.8)
- i2/„2>,
f -1 03(/n-i
2, «2» = o
2
2/o2x /i2/o2x
'i-' fl2n+l 9 O
n-0 z z
f r iv -1 2fl2(/32)-fl2(/3) + l
2 (-1) a„ = = .
(ii) (e = 1) Then )8 < 0 and a | /31 = 1. It follows that
^, _, 02(j82)-l
(3.7.9) 2 a2xn = -^
and
CO
2 a2-„1+1=2[L(|)8|)-2L()82) + L()84)].
In certain cases the theta series involved above are particularly simple to
evaluate. For example, in Case 2(i), we have )8 = c — Vc2 — 1. Thus c =
()8 + j8-1)/2. If )8 := l(Tm, the series 03(/3) and 0,(/32) in (3.7.8) can be
evaluated entirely by writing down sequences of l's and O's. Thus for

94 Some Number Theoretic Applications
1c = 10m + 10-m, the sums in (3.7.8) can be evaluated at the same speed as
one can multiply n-digit numbers. (See Chapter 6.) Moreover, since theta
functions can be fast computed for any algebraic )8, the series are always fast
computable. (See Chapter 7 and Exercise 6.) Hence the series of Exercises
3, 4, and 5 are quadratically computable.
Finally, consider (3.7.7) for c: = sinh (tts) and with s>0. Then
an+1 :=2smh(irs)an + an_1, a0- = 0, ax '■= 1, and
(3.7.10) 5(,) := S a£+1 = ^^- [*&-') - *&-*»)].
If we use Theorem 2.1, we have
S_, cosh (tts) ( 1 - k' \ „,..
n = 0 ?> N z '
where
on using
For singular values of fc (see Section 4.6), this formula becomes particularly
pretty. [See Exercise 6b).]
Comments and Exercises
1. a) Show that when x = Mx + N has distinct roots a and )8, a +1 =
Man + Nan_1 is solved by (3.7.2).
b) If a = )8, show that
an = naxan~ — (n — l)a0a" .
2. Establish (iv), (v), and (vi) of Proposition 3.5.
3. a) Use (3.7.7) to show that for the Fibonacci numbers
Here F0 := 0, Fx := 1, and F„+1 := F„ + F„_j. This result is due to
Landau [1899], as is

b)
3.7 Reciprocals of Fibonacci Sequences
95
This is also discussed in Ribenboim [85].
c) The Lucas numbers are defined by the same recurrence,
Ln+1 := Ln + L„_j but with L0 :=2 and L1 := 1. (See Hardy and
Wright [60].) Show that
s^1-
n=0
3-V5^
e\[ ^-^) + 1
'4.
d) Note that xn '■= F2n+1 satisfies xn+1'-=3xn- xn_1, x0'=l, and
*! := 2. Similarly, xn ■= L2n satisfies xn+1 ■= 3xn - xn_x, x0'=2,
and *! :=3.
a) Verify (3.7.8) and (3.7.9).
b) Show that for an+1 :=4a„ - an_1, a„: = 1, a1 :=2, one has
2j an =
«=i z
and
V ( iv -i 2^(7-4V3)-^(2-V3) + l
2, (-1) a„ = 5
Let a0 := 0, ax := 1, and a„+1 := 9.9a„ + a„_!. Show that
2 2 ^ = 5.05
2f-
3Vio
)-3(^):
a) Recall that d\(q) = (2/ir)K(k), where ^ = e--*'/*(*)_ As shown in
Chapter 7, it is possible to quadratically compute k given q. Since
K is quadratically computable given k, we can fast compute 63(q)
given q.
b) Show that with S(s) given by (3.7.10), we have
5(1) =
cosh(Tr) /V2-1
2V2
M£)
^.g*fe^('-^^),^-¾
^-==^ l^^)K(^i>).
cosh (W3) /2-V2 + V3\ r^V2(V3-l)N
7T

Some Number Theoretic Applications
c) Show that S(s) = cosh (tts) E~=0 sech [(2n + l)irs], so that
V ur/o a.i\ 1 gs(e ) ~ g3(e )
2j sech [(2n + 1)-7T5,J =
and
d) Show that
2 sech (mrs) =
ei(e-") + i
n=0
Zj tt sech [(2n + l)ir] =
(2-V2)r2(|)
16Vw
Equations (3.2.28) and (3.2.29) can be used to derive closed forms for
a host of other reciprocal sums. (F„ and L„ are as in Exercise 3.)
a) Use (3.2.29) to show that
VI
„f0(2n + l)F2n+1 4
(-1)"
arcsin
03()8)
where 0:=(3-VI)/2.
b) Use (3.2.28/v) to show that
h 2^:=12-log
02()82)03()82)
2V)802()82)
where /3 := (VI -1)/2.
Use (3.2.28/) and transformation formulae to show that
i) X cosech2 (n™) = - + — (—-—)K2-KE
„ = 1 0 7T L\ 3 /
k-
11
ii) 2 cosech2 (2mrs) = — -\ 5
„ = 1 O 7T
l + k
/2
K£
~ 1 r/i + &'2\
iii) Z cosech2 [(2n + 1) its] = — (—-— }K2 - KE
„=0 7T L\ 2 I
d) Similarly, use (3.2.28v) to show that
i) Z/ sech2 (rnrs) ■■
n = l
7T

3.7 Reciprocals of Fibonacci Sequences 97
ii) 2j sech2 (2mrs) =
EK + k'K2 1
„ = 1 TT2 2
.... X1 ■ 2 r,~ *s ^ EK — k'K
in) 2j sech [(2n + l)irs\ = 5
e) Show that
i) z, cosech2 (mr) = -: — -r—
n = l 6 2tt
(2n + l)7r
2
J_
2tt"
ii) 2 sech2
n=0
f) Combine results of c) and d) to show that, with /3 := (3 - VI)/2,
CO
(3.7.12) 2 f;2 = & [e*(/8) - e*4(p) +1]
n = l
and
(3.7.13) El;2 = | [e*(/3)-i]
«=1
and deduce similar formulae for E~_x (—1)"F2„2 and for E"=1
(-1)¾.
g) Show that the Lucas numbers satisfy
co / °° \ 1 °°
El;2 = 2 El- h-El-
and show that a similar formula holds for all recursions of the form
an+i = (2c)an + an-i> aQ:=2c, and ^:=1.
h) Show that
co co , co ^2
3Ef„-2 + 5 2l;2 = 4(Ef2~„1+1 .
a) Show that
i (-ir+i #*=i f „-2.
/Zmf: Differentiate (3.2.28m) and compare the result to Exercise
7f), equation (3.7.12).
b) Show that

98 Some Number Theoretic Applications
CO CO
-2 2 (-l)mm cosech (2rmrs) = X sech2 [(2m + 1)tts]
and hence that
vs 2 (-1)-^ = 2/^.
^2n n=0
9. Show that
^ <?" _034(?)-l
«l2
«-i[g" + (-l)"]
y (-l)V _et(q)-l
and
ei(q)-et(q) + i
„-![9" + (-l)"+1]2 24
10. a) Show (by expanding both sides) that
L(,):=E /^ = 2^^-
n = l x ■* n = l x ■*
b) Hence show that all the series considered in this section can be
computed with at most 0(Vn) operations for n digits. (See
Chapter 6.)
A remarkable elementary result is
V 1 =kV5
«=o F2n+1 + Fk 2Fk
k = l,3,5...
This is a specialization of identities established in Backstrom [81]. (Related
results can be found in Carlitz [71].) Thus
^ 1 V5
A related formula is
«=o F2n+1 + 1 2
V 1 =7V5
«=o F2n+1 + 13 . 58

3.7 Reciprocals of Fibonacci Sequences 99
11. Show that
v 1 KE \
where q := (VI - 1)/2. Now k is close to 1, so that KE/2v2 is close to
K = e23(q) = e^e^ioric^-D/gj
2tt2 4tt 4 log [(V5 + 1)/2]
where the last equality follows from the theta transform. One sees that
KEHir1 + g is close to
1 1
+
4 log [(V5 + 1)/2] 8
which explains, in some part, Backstrom's formal manipulation in
Backstrom [81]; and which evaluates (3.7.14) as requested therein.
Almqvist [Pr] treats this sum and some relatives. Note that
gw /iog[( 5-1)/2] _ jq-9 sQ t^at^ as |n ajj ^ese Fibonacci series,
transformation considerably speeds convergence.
Lambert series occur naturally in multiplicative number theory, as the
following exercise shows.
12. a) Show formally that for any real valued function /,
CO fi CO
n = l -1 ■* n = l
where
F(n):=2f(d).
This is due to Laguerre. Hence show that
i) 2 r—r = 2 *»)*"
„=i i X „_j
where r(n) is the number of divisors of n.
ii) E ^Ar = E ^(n)^" fc = 1,2,. . .
n = l L x n-1
where <rfc(n) is the sum of the fcth powers of divisors of n.
b) Let

100 Some Number Theoretic Applications
'. f 1 if n is square
F(n):=\n ., . M
10 if n is nonsqu
; nonsquare.
Show that/(n) = (-I)26'where
m
n = n P;' (in prime decomposition) .
i=i
Hint: Use Mobius inversion. Since F is multiplicative, so is /. Thus
using Liouville's function e(n) : = (-l)s'»ie<
2 e(n) „
c) Show that
«,(*) - 1
n = l ^2« z
where 0:=(3-VI)/2.
Zucker [79] gives general formulae for sums of powers of hyperbolic
functions (in which the coefficients are defined recursively and have been
computed extensively by Ramanujan and Zucker). Using these one can
evaluate X~ml (-l)"+1F;4fc and 2~ml F;4k+2 in terms of 6 when
k := 1,2,3,. . . . There are similar Lucas number results, and if K and E are
used, many more sums are expressible. We give two examples:
i (-i)"+1l;4 = ^ {3 + [e*(/8) -1]2 - [eJ(/3) +1]2}
and
96
2^i = ^<&/8)[1-044(/8)].
Here, as before, j8 : = (3 — VI)/2.
13. Let a0:=0, u1 := 1, and a„+1 := aun + un_1. Let u0: = 2, vx := a, and
u„+i '-=avn + »„_!•
a) Establish that
2 ^¾ = «V +4) 2 ^-^- = | (VTTi- a) .
/fim: The second sum in a) can be made to telescope.

3.7 Reciprocals of Fibonacci Sequences
b) In particular,
°° / i\«+i
I
c) Compare
i) 2^7^ = ^-3)
and
ii) 2^-72=6(3-41082).
«-»1 "jt=i f
d) Show that
„=1 z,^.! a*. * . . . .
Witxi - a
Jt=l "-tit + l
= UV7T4-a) + 2a2 .
^ fc=l M2*a2*: + 1

Chapter Four
Higher Order Transformations
Abstract. We develop algebraic transformations of prime order for the
elliptic integrals. For small numbers this can be managed purely algebraically.
However, the development of modular equations for arbitrary primes is
most comfortably effected via transcendental methods. This requires some
rudimentary modular function theory. The cubic equation is studied in
particular detail.
4.1 A FIRST APPROACH TO HIGHER ORDER TRANSFORMATIONS
The fundamental relation from Theorem 1.2
(4.1.1) KTO = TA^(f^)
is remarkable for a number of reasons. One notable consequence is the ab
initio unlikely observation that when k is algebraic and
(4.1,) «:-?£
the values of the transcendental function K at I and k are algebraically
connected. Equation (4.1.2) is one form of the quadratic modular equation.
It can be rewritten as
(4.1.3) /2(1 + A:)2-4A: = 0.
We will develop a class of algebraic equations (modular equations) that
induce algebraic transformations on K in a similar fashion.
We commence by sketching, a la Cayley [1895], a purely algebraic
102

4.1 A First Approach to Higher Order Transformations
103
approach to the modular equation. It transpires that this approach becomes
unduly complicated for all but a few simple cases and is hard to use as a
rigorous basis for the general theory. Thus in Sections 4.3 and 4.4 we will
derive the general theory using function theoretic techniques.
We are looking for a relation of the form
(4l4) M(l, k) dy = dx
V(i-/)(i-/V) V(i-*2)(i-fc2*2)
where I and k satisfy a polynomial equation O in two variables. Such an
algebraic connection between the two moduli (/ and k) is called a modular
equation or modular transformation for k. Specific but equivalent modular
equations for algebraically related functions are defined in (4.4.2), (4.4.6),
and (4.5.1). The function M will be an algebraic function of k and I and is
called the multiplier. With this in mind, let P and Q be polynomials in x2 so
that deg(P ±xQ)\\±x) = n (n odd) and write
(4.1.5) W_(P-*G)21-*
l + y (P + xQy l + x
The important condition to impose on P, Q, and l-=l(k) is that (4.1.5)
must be invariant when (x, y) is replaced by (1/kx, l/ly). (See Exercise 1.)
We now try to solve for P and Q. Set
U := x(P2 + 2PQ + x2Q2) V : = P2 + 2x2PQ + x2Q2
(4.1.6)
A := P ~xQ B- = P + xQ
and observe that
y = UIV
(4.1.7a) 1 - y = (1 - x)A2IV 1 + y = (1 + x)B2IV.
Also, from the invariance of (4.1.5),
(4.1.76) l-fy = (l-fcc)C2/y 1 + ly = (1 + kx)D2IV
where C and D are polynomials of the same form as P and Q. Thus we
deduce that
(4.1.8) "y - ^-^ *
V(i-/)(i-/V) ABCD V(i -*2)(i-*V)
It can now be computed (Exercise 1) that

104 Higher Order Transformations
1 UV-VU = Q(0)
(4 9) M(/, k) ~ ABCD ~ P(0)
is independent of x. Let us return to (4.1.5). For n = 4p + 1 we must have
(4.1.10) degP = 2p degQ = 2(p-l).
For n = 4p + 3,
(4.1.11) degP = 2p degQ = 2p.
Since P and Q are even functions, we have in either case \{n + 1) degrees of
freedom. It remains to use the invariance in (4.1.5) as (x, y)-*(l/kx, l/ly).
This leads immediately to the equation
(4112) ViMk*) uvl
since y = U/V. Thus we are reduced to solving
(4.1.13) (P2 + 2x2PQ + x2Q2)* = yjj k(n~1)l2{P2 + 2PQ + x2Q2)
where the * operation is defined as follows. If
S'-=a0 + axx + • • • + amxm
then
5* := a0(kx)m + a^kx)"1"1 + ■ ■ ■ + am .
Equating coefficients in (4.1.13) leads to a system of \(n + l) nonlinear
equations. There are \{n + 1) degrees of freedom in the coefficients of P
and Q. Since the equation is homogeneous in P and Q, one of the
coefficients may be assumed to be 1. This leaves one additional condition to
be satisfied by k and I and leads to the desired unique algebraic relation
between k and I. The pitfalls of this approach are now, of course, apparent.
We end up with a large system of nonlinear equations that are virtually
impossible to solve or analyse directly.
We illustrate with the cases 3 and 5.
cubic transformation (n = 3). We have P — l and Q = a. Equation
(4.1.13) becomes
k2x2 + 2a + a2 = ^ k{\ + 2a + a2x2) .

4.1 A First Approach to Higher Order Transformations 105
This leads to two equations,
*« yjl a2
(4.1.14)
2a + a2=yJ- k(l + 2a).
This is solved parametrically by
(4.1.15a) k2-^^- and /»- «(2+«>'
2a + 1 (2a + if
with similar expressions for (k')2 and (/')2, namely,
,2 (!-«)(! +a)3 2 _(! + «)(!-a)3
" 2^n and l (2«+i)3
From this one can deduce that
(4.1.16) Vkl + VkT^l
or, equivalently,
(4.1.17) (k2 - I2)4 = I28k2l\l - k2)(\ - l2){2 - k2 - I2 + 2k2l2).
In the a:
namely,
In the associated variables u : = k114 and i; := I1'4 this has a simpler form,
(4.1.18) u - v4 + 2u»(l - wV) = 0.
(See Section 4.5.) The multiplier M has any of the following forms:
(4.1.19) M =
1 i; 2u3 —w
2a + 1 u + 2w3 3«
(See Exercise 2 of this section and Exercise 2 of Section 4.6.)
Quintic transformation (n = 5). We have P = 1 + j8*2 and Q = a.
Equation (4.1.13) leads to the three equations (A2 := k5ll)
0a-A
(4.1.20) k2(2ap + 2f3 + a2) = A(2a + 2)8 + a2)
A:4(2a + l) = A(j82+2aj8).
This eventually solves, with u := k1'4 and v'—l11*, as

106 Higher Order Transformations
(4.1.21) u - v6 + 5wV(w2 - v2) + 4wv(l - wV) = 0 .
The multiplier M has any of the forms
1 v(l - uv3) u + v5
(4.1.22) M=__»__»__^.
The transformations are called cubic, quintic, and so on, because of the
underlying order of the transformation (and so of convergence).
Comments and Exercises
The very classical approach of the section to cubic and quintic modular
equations is due to Jacobi [1829]. This algebraic approach was extended to
the septic (n = 7) case by Cayley [1874], who also treated the endecadic
(n = 11) case partially. The calculations are formidable. We have followed
Cayley closely in this discussion. The associated variables u'=k114 and
V-—1 /4 of Jacobi's considerably simplify the calculations, as we will see in
Section 4.5.
In order to find the relationship between K(l) and K(k) implicit in (4.1.4)
we must show that the underlying transformation (4.1.5) is one to one and
onto on the interval [0,1]. This can be done directly for n = 3 (Exercise 3)
and other small n. However, as with most of the details, it is easier to use
the general transcendental approach of Sections 4.3, 4.4, and 4.5.
1. a) Show that an equation of the form (4.1.5) that is invariant under
the change of variables (x, y)-*(l/kx, 1/ly) exists. Hint: Set
_ x n * - x2/a2
Ml i 1 - k a,x
Replacing (x, y) by (1 Ikx, 1 lly) gives
n^
,2 2 2
k a,x
ly xMk"UtayiX \-x2la2
which holds, provided that
/ = M2r(n«,)4.
[See also (4.1.13).]
b) Establish (4.1.7b) and exhibit C and D.
c) Establish (4.1.9).
Hint: Let U and V be as in (4.1.6). Let y-=UIV and let
Y(a, b) := (a2 - b2)(a2 - l2b2). Then

4.1 A First Approach to Higher Order Transformations 107
Y(V, U) = (V2 - U2)(V2 - l2Tl2)
and
dy _ UV- UV dx
^/Y(Xy)~ VWTU) '
Now observe that any square factor (x — a)2 of Y(V, U) is a linear
factor of VU - UV Since
and
V2 - U2 = (1- x2)A2B2
V2 - l2U2 = (1- k2x2)C2D2
the fact that (UV- VU)IABCD is independent of x now follows.
The explicit form of the multiplier M(l, k) is derived by setting
x = 0 in 0V - VU) IABCD.
a) Establish the four forms of the cubic modular equations (4.1.15),
(4.1.16), (4.1.17), and (4.1.18).
b) Complete the calculation of the quintic modular equation (4.1.21).
Show that the n = 3 relation between x and y underlying the cubic
transformation is one to one and onto on [0,1], and hence (4.1.4) can
be integrated over [0,1].
(An explicit cubic algorithm for K) The cubic modular equation
(4.1.18) is of degree 4 and, hence, can be solved explicitly for u in terms
of v.
a) Show, for i; e (0,1), that u E. (0, v) is a solution of (4.1.18), where
v3 D-i?
y +
and
S :=^(1-08)
R:=V7+S
,„ 6 „ 4»-2d9
R
b) Show, for u G (0,1), that i; G (u, 1) is a solution of (4.1.18), where

108 Higher Order Transformations
u D + R
„ —y +
and
S:=V4w2(l-w8)
i?:=Vw6 + S
n ■ /o« „ , 4w - 2w9
D •= \/2w - 5 +
c) Show, for uG(0,1), that there is a unique wG(0,u) so that
(4.1.18) is solved by (w, v). We can define an iteration as follows.
For u, G (0,1), let vi+1 G (0, u,) be such that u '= uJ+1 and 1; := vt
satisfy the modular equation (4.1.18). Show, using a) and b), that
(4.1.23) vn+1 = v3n- ]/v6n + y/4v2n(l-v*n) + vn_1
where v0 G (0,1) and v1 G (0, v0) is computed from v0 by Exercise
4a).
d) Show that, for v0 G (0,1),
(4.1.24) *(„;)-|n(1 + M)
(Use Exercise 3.)
e) Show that vt tends to zero cubically. (For further details see
Borwein and Borwein [846].)
5. a) Let «6(0,1) and uG(w, 1) satisfy the cubic modular equation
(4.1.18). Show that
(4.1.25) ^=3"F5V-
K(u ) K(v )
Hint:
b) Let u G (0,1) and 1; G (u, 1) satisfy the quintic modular equation
(4.1.21). Show that

4.2 An Elementary Transcendental Approach 109
(4-L26) K(u*) ~5 K(v*) ■
Hint: Show that if (u, v) : = (k1H, I1'4) satisfies (4.1.21), then so
does (u, v) = (I'1'4, k' /4). Now Exercise 1 of Section 1.5 shows that
the two sides of the above equation differ by a constant. Use the
logarithmic asymptotic at 0 and the relationship between u and i; as
u-»0 to evaluate this constant.
These important identities will be revisited in Section 4.4.
6. a) Verify Schlafii's form of the modular equation of degree 5,
<«"> (f)'+(i)'-^-^)
where u := 2~1/4/(t) and v := 2~1m/(5t), / as in (3.2.9).
b) Compute the corresponding equation for /j.
4.2 AN ELEMENTARY TRANSCENDENTAL APPROACH TO
HIGHER ORDER TRANSFORMATIONS
In terms of the nome q we have, by Theorem 2.3, the identification
(4.2.1) KqY^k^^fi-
(4.2.2)
k'(q)- =
K(k) =
¢ =
k'
■ e~
el(q)
e\(q)
■r,K'(k)IK(k)
(4.2.3)
and
(4.2.4)
From Exercise le) of Section 1.4 and Exercise 5 of Section 4.1 we see that
the quadratic modular equation (4.1.3) is satisfied by l'-—k(q112) and
k:= k(q), while the cubic equation is solved by I := k(q113) and k'-~ k(q)
and the quintic equation is solved by I := k(q /5) and k'-— k{q). [To see this
just observe that (4.2.4) uniquely determines q.] In general we will see in
the next sections that the pth-order modular equation for k is a polynomial
in two variables with integer coefficients that is satisfied by k(qp) and k(q).
We observe from (4.2.4) that for these algebraically connected moduli

110 Higher Order Transformations
(4-2'5) P K(k(q)) K(k(q")) '
Before turning to the general theory we wish to give an elementary
derivation of the cubic transformation in theta function terms. From
(4.1.16) the cubic modular equation for k is
(4.2.6) Vkl + VFF=l.
From the preceding discussion this is seen to be equivalent to:
Theorem 4.1
(4.2.7) d4(q)d4(q3) + 62{q)92{q3) = e3(q)63(q3) .
Proof. From the definitions and Exercise 1 of Section 21 applied to
n2 + 3m2 „_j / 1\n+m n2+3m2 u„.,„
q and (—1) q we have
and
Now
(ft + / + l)2 + 3(ft-/)2 = (2ft-/+i)2 + 3(/+i)2.
Thus subtraction of the two theta identities produces
^)^)-04(4)04(^) = 2 i ^+^3(-^
m,« = —co
m+n even
= 62{q)62{q3)
(as replacing m by 1 — m shows). □
Comments and Exercises
This is as far as we wish to pursue the transformation theory on an ad hoc
basis. The next section introduces enough of the theory of modular
functions to provide a general framework for the development of modular
equations. We have only considered modular equations for p a prime. If we
view the modular equation as the algebraic relation between k(q") and k(q)
and can find this relation for p a prime, then for composite n a relationship

4.2 An Elementary Transcendental Approach 111
can be constructed out of the modular equations corresponding to the prime
factors of n. Rational n are treated similarly.
1. Construct modular equations of order 4, 6, and 8 [that is, construct
algebraic relationships that are satisfied by k(q) and k(q2), k(q6), and
k(q8), respectively].
2. Show, for
that
and that
c:=f Y(k(q))
0<k(qp")<4e-cpn ¢£(0,1)
k(qp )~4e~cp as n-*°°.
Many modular identities follow from:
3. (Schroter's formula) Consider a general theta function written as
CO
T{x, <?):= X x"q"
where x ¥- 0, | q\ < 1 (as in Section 3.1). Let a and b be positive integers,
a) Show that
(4.2.8) T{x,qa)T{y,qb) =
a + b-l
Sk bk2rr, 2bk a + b\rr, a -b 2abk ab(a+b)\
y q T(xyq , q )T(y x q ,qK >).
Hint: Write
x,q )T(y,q )= Z x y q
m,n
Let s be chosen so that n = m + (a + b)s + /:(0 <k<a + b) and let
u'-~ m + bs. Then w and 5 range over Z as m and n do. Also
m n / \u/ -~b a\s k
x y ={xy)(x y)y
and
am2 + bn =(a + b)u2 + 2bku + ab(a + b)s2 + labks + bk2 .
Now rearrange. (See Tannery and Molk [1893].)

112 Higher Order Transformations
b) A form of the seventh-order modular equation is
V03(<7M(<77) - y04(q)64(q7) = \J 62{q)62{q1) .
(See (4.5.4).)
Use part a) with a '■= 7 and b'-=\ to establish this formula.
Hint: Set x-=y-=±l to find a formula for 63(q)63(q1) +
04(q)64(q1). Now set x'= y1''■— ±q7 to similarly write
^.lq)^(q) + 62(-q)S2(-q1)'=02{q)e2(q1). On making simple
rearrangements this yields
0,0,(47) + OM q7) = 2T(l,q8)T(l,q56)
+ 2quT{q8,q8)T{q56,q56)
+ 4q*T(q\q8)T(q28,q56)
and
d2d2{q1) = 2q2T{q8, q8)T(l, q56) + 2quT(l, q8)T(q56, q56)
+ 4q4T(q\q8)T(q28,q56).
Since q2T(q8, q8) = ^(q8) and T(l, q8) = 63(q8), we may write
030,(<77) + 0,6^)-^0^) =
2[03(q8) - 02(q8)][63(q56) - 62(q56)] = 284(q2)d4(qu) .
Thus an application of equation (2.1.7ii) gives
(V0303(<77) - V0404(47))2 = 0202(47)
as required,
c) Establish the cubic modular equation (4.2.7) as above.
4.3 ELLIPTIC MODULAR FUNCTIONS
The theory of elliptic modular functions and more general automorphic
functions is, in part, a natural extension of the theory of elliptic functions.
The basic defining property of elliptic functions is their invariance under a
group of linear transformations. Automorphic functions are functions
meromorphic in the upper half-plane $f: = {im(t) > 0} that are invariant
under a group of linear fractional transformations. We will, of necessity,
explore only the rudiments of this remarkable and difficult theory.

4.3 Elliptic Modular Functions
113
Definition 4.1
(a) The (inhomogeneous) modular group T (J-group) is the set of all
transformations of the form
w ■■
at+ b
ct+ d
a, b, c, d integers, ad — be = 1
(b) The A-group is the subgroup A of T with a, d odd and b, c even.
That both of the above are groups (under composition) is straightforward.
The transformation
at+ b
w --
ct + d
can be represented as either of the two matrices
=(::)
and the group product becomes matrix multiplication (See Exercise 1.). The
(homogeneous) modular group SL(2, Z) distinguishes these matrices. Note
that any element of the modular group fixes the real axis and maps $f onto
itself.
Definition 4.2
(a) The set Fr of t e X* := {im(t) > 0} U {»«>} U {0} is denned by
Fr := {|re(0| < \ and |f| > 1} U {re(t) = - \ and |f| > 1}
U{|f| = land
; re(f) < 0}
(b) The set FA of t E "M* is defined by
*"a := (lre(0l < 1 and \2t ± 1| > 1} U {re(0 = -1} U {\2t + 1| = 1} .
These two sets described in a) and b) are fundamental sets for the T- and
A-groups. The interiors of these two sets (F° and F°x) are fundamental
regions in the following sense.

114 Higher Order Transformations
Theorem 4.2
(a) Every point in "M is the image under some element of T of some point
of Fr.
If A e T is not the identity, then A(F°r) D F°r = 0.
(b) Every point in $f is the image under some element of the A-group of
exactly one point of FA.
If a e A is not the identity, then A(F°X) D F°x = 0.
The proof of this theorem is elementary though not entirely
straightforward. (See Exercise 2.) Any set F (with interior F°) which satisfies (a) or (b)
of the above theorem is also a fundamental set.
Definition 4.3
(a) A T-modular function is a function / which satisfies:
(/) / is meromorphic in "M.
(ii) f(A(t)) = f(t) for all t e $f* and A e T.
(Hi) f(t) tends to a limit [possibly infinite in the sense that 1//(£)-» 0]
as £ tends to the vertices of the fundamental region Fr where the
approach is from within the fundamental region F°. [In the case
of i°° the convergence is uniform in re(x: + iy) as .y-*00.] The
vertices of the fundamental region are (0, 1), (-1/2, V5/2) and
i°°. Since / is meromorphic in "M, this condition is automatically
satisfied at (0, 1) and (—1/2, V5/2) and need only be checked at
(b) A k-modular function is a function / which satisfies (/), (ii), and (Hi)
above with the T-group replaced by the A-group. For condition (Hi)
the vertices of the fundamental region F° are (—1, 0), (0, 0), and /°°.
Our notation is not entirely standard. What we have termed T-modular is
often just called modular or automorphic with respect to the T-group, while
what we have labelled as A-modular is often referred to as automorphic or
modular with respect to the A-group.
The existence of a A-modular function is the content of the following
theorem.
Theorem 4.3
The function
A(0:=*2(0 =
W1 Q-e
is a A-modular function.

4.3 Elliptic Modular Functions 115
Proof. From Corollary 3.1 we have
(4.3.1) A(0 = l6gn(1 , f„-i)
and it is clear that A is meromorphic in $f. For the invariance of A under the
A-group it suffices, by Exercise le), to show that
(4.3.2) A(* + 2) = A(0
and
(4.3.3) <27TT) = ^-
The first equation follows since emt = em(f . The second equation is a
consequence of (2.3.1), (2.3.3), and (2.1.10), which combine to yield
(4.3.4) A(-i) = l-A(0.
(Note that t = is.) Hence
A(27TT) = A(2TT7i) =l~ A(-2- ?) =l~ A(- ?) = A« ■
Finally we observe that, in a limiting sense,
(4.3.5) A(ioo) = 0 A(0) = 1 A(±l) = oo.
The first value is immediate from (4.3.1), while the value at zero can be
calculated from (4.3.4). (Observe that as f-»0 in Fx, l/f-»°° in FA.) The
value at 1 is computed from Jacobi's imaginary transformation
(4.3.6) A(f + 1)= A(f)
A(0 - 1 •
(See Exercise 4.) D
Some additional properties of A are established in Exercises 4, 5, and 10.
From (4.3.4) and (4.3.6) one can prove, as in Exercise 6, the following
theorem.
Theorem 4.4
The function
A_ [1-A(Q + A2(Q]3 = ± {l-[k(t)k'(t)]2}3
U' 27 A2(0[1-A(0]2 27 [k(t)k'(t)]*
is T-modular. J is called Klein's absolute invariant.

116 Higher Order Transformations
The basic result we need is a version of Liouville's theorem.
Theorem 4.5
A T-modular function that is bounded on FT is constant. Similarly, a
A-modular function that is bounded on Fk is constant.
Proof. Suppose/is T-modular and is bounded and nonconstant on FT.
By Theorem 4.2 this implies that / is analytic on $f. Consider f(t) -/(foo).
This function has no poles interior to Fr, and so achieves its maximum
modulus at some finite point on the boundary of Fr. By the invariance of
/— /(i°°) under T this is a global maximum at an interior point of $f, which is
impossible.
For the second part consider the A-modular function
[/(0 - /(0)][/(0 - /(-i)][/(0 - /(*-)] ■ □
This is sufficient theory for our discussion of modular equations in the
next section.
Comments and Exercises
This is only the very tip of the iceberg. We have restricted our attention to
two particular groups where we can directly establish the existence of
modular functions. In general this restriction is unnecessary. Only slightly
further into the theory are results such as: any modular function takes each
complex value the same number of times in the fundamental region. An
important consequence of this is that A takes every value exactly once in Fx
and A has a well-defined inverse that has branch points only at 0, 1, and °°
(Exercise 10). J has similar properties on Fr. One can now prove, much as
for elliptic functions, that two nonconstant functions which are modular with
respect to the same group are algebraically connected. Furthermore, if one
of these functions is univalent on the fundamental region, then the other is a
rational function of it.
This wide-ranging and difficult body of theory that is intimately tied in to
many questions in number theory and algebraic geometry may be pursued in
any number of texts, such as Apostol [76b], Chandrasekharan [85], Lang
[73], Lehner [66], Rankin [77], or Schoeneberg [76].
Two of the seminal papers of the subject, both dating from 1882, are due
to Klein and Poincare (selections of which may be found in Birkhoff [73]).
Poincare was interested in studying linear differential equations with
algebraic coefficients. Klein, who considered this his main field of work, in
keeping with his Erlanger Programme had more algebraic and geometric
interests (Klein and Fricke [1892]).

4.3 Elliptic Modular Functions
117
a) Verify that T and A are groups.
b) Verify that composition of transformations is equivalent to
multiplication of the associated matrices.
c) Show that two transformations represent the same function if and
only if they have the same coefficients (associated matrix up to
sign).
d) Show that T is generated by
Sr:-(J j) and rr:=(J _J).
e) Show that A is generated by
Prove Theorem 4.2.
a) Hint: To prove part (a), fix z E. ~Mand let A = ( j e T. Show
that
im(z)
im A(z) =
cz + d\
Pick any element of Y that minimizes \cz + d\ and let w be the
image of z under this transformation. By considering Tr(w) show
that |h>|>1 and by considering 5^(^) show that for some k,
|re[5w(iv)]|<|. For this k, |5w(iv)|>l. Thus every element of
X is the image of some element of FT.
b) Show that no two elements of F° map to each other under an
element of T. Examine the image of Fr under 5r and Tr.
c) Deduce part (b) of Theorem 4.2 from part (a).
The upper half-plane can be tesselated by images of the fundamental
region under the generating transformations. Sketch pictures of the
tesselations associated with the T-group and the A-group.
a) Show that X(A) is transformed into one of A, 1 - A, 1 /A, 1 /(1 - A),
A/(A-1), 1-1/A by any AST. Hint: Examine X(t+l) and
A(— l/t) using similar arguments to those of Theorem 4.3. Use
Exercise Id).
b) Show that X(m/n) = 0, gcd(m, n) = 1, if and only if m is odd and n
is even.
Show that with q : = e'
16 l 4- Y hn"

118 Higher Order Transformations
where the bn are integers. Show that 1/A(0 is finite at every point of
6. a) Prove Theorem 4.4 from (4.3.4) and (4.3.6).
b) Let q:= elir". Show that
/(0 - 1728/(0= \ + i.cnqn
where the c„ are integers. [In fact, qj{t)= Qf{256q + Q^f.\
c) Show that f\t), -fT(t), and -f2\t) are the roots of
(x - 16)3 - *;(0 = 0 .
[See (3.2.9) for definitions.]
7. a) Show that if / is T-modular then for some integer k and nonzero
constant c
/(0 ~ ce
as f-» foo.
tfinf: Let /(^):=/(0 where q := e2vil for f in Fr. Show, by
modularity of /, that / is meromorphic in q in a neighbourhood of
zero. Show also that/has a pole at zero. Thus/has a convergent
^-expansion at i°° with finite principal part,
b) Show that the T-modular functions form a field. Likewise the
A-modular functions.
8. Establish that /f, f\\ and /24 [see (3.2.9) for definitions] are A-
modular.
9. The Schwarz derivative of any function / is
sm-=2//~3(/)2
Show that if / is modular, then so is 5(/)/(/)2. Show that/is not in
general modular.
10. {On the inverse of A)
a) Show that A maps the set A := {re(z) = -1, 0<im z) one to one
onto (-oo, 0). Hint: X(t ± 1) = A(0/[A(0 - 1]. Now consider A on
the imaginary axis.
b) Show that A maps the semicircle B '■= {\z + \\= \, im(z)>0} one
to one onto (1, oo). Hint: \(-\lt) = 1 - A(0-
c) Show that A maps the interior of Fx one to one onto C-

4.4 The Modular Equations for A and j
119
d)
{(—oo} 0] U [1,0°)}. Hint: The number of zeros of A — c is
Itri h
A'(0
t A(f) - c
dz
where 7 is a contour of the form seen in the accompanying figure.
Use the invariance X(t) = X(t + 2) to estimate the integral on the
sides of the contour. Use the relation A(— 1H) = 1 — X(t) to relate
the integral on BC to the integral on ED. Use the relation
k(t ± 1) = A(f)/[A(f) - 1] to estimate the integral on CD and DC
in terms of BAB and then (by t-*—llt) in terms of EE. Finally
take limits.
Thus with respect to Fr, A has a well-defined analytic inverse with
branch points at 0,1 and 00.
11. (Picard's theorem) Show that a nonconstant entire function assumes
every complex value except possibly one.
Hint: Suppose F does not assume either a or )8, then G(z)-=
[F(z) — a]/(j8 — a) never assumes 0 or 1. If co '■= A-1, then co(G(z))
is entire (by analytic continuation). Show that (i)(G(z)) is constant
since co(G(z)) G {im(z) 5:0} and hence ea(G(z)) is a bounded entire
function.
It is worth observing that the apparently special case analysis of the
function A leads directly to the celebrated general theorem of Picard.
4.4 THE MODULAR EQUATIONS FOR A AND j
A transformation of order p is a matrix
(4.4.1) ( , J a, b, c, d integers, ad-bc = p
or the associated linear fractional transformation. We assume throughout

120
Higher Order Transformations
that p is an odd prime, though for much of the development this is
unnecessary. We will denote the set of all such transformations by Tp. We
say that M is equivalent to N mod G (M = N mod G) for a group of
transformations G if there is an S E.G so that M = SN. We need the
following purely algebraic result.
Lemma 4.1
(a) Every M G Tp is equivalent mod T to one of the p + 1 transformations
of the set si, where
"■-W-Co ?)■ *■■■< ;).<-<u.....,-i}.
(b) The p + 1 elements of si are pairwise inequivalent mod I\
(c) Every B of the form B '■= BtC, where C is in the A-group and Bt G 08,
is equivalent mod A to some element of S3, where
•■■-{>A',, ;)■ *■■< 2;)' '-<m.-.»-i}.
(d) The p + 1 elements of S3 are pairwise inequivalent mod A.
The proof is left as Exercise 1.
Theorem 4.6
(a) The p + 1 functions
/(A,.(0) i = 0,..., p
are permuted by any element of the T-group.
(b) The p + 1 functions
AOB,(0) i = 0,..., p
are permuted by any element of the A-group.
Proof. For part (a) we must show that {/° At ° S}?=1 = {J° A,.}f=1
for any 5 G T. We first observe by the lemma that if S G T, then since
AjS^-^modr for some;
Thus by the modularity of J,
JoA^S^JoAj

4.4 The Modular Equations for A and j
111
which with part (b) of the lemma finishes the proof. The second part is
identical via parts (c) and (d) of the lemma. □
The modular equation for A of order p is the polynomial
p
(4.4.2) W,(*, A): = IT (* - A,) A,: = A ° B, ■
i=0
This is obviously of degree p + 1 in x and has a root at each A,. Note that
Xp(t):= X(pt) and Xt(t) = X((t + 2i)/p), i<p. Thus as functions of q,
Xp(q) = X(qp) and Xt(q) = X(a'qllp), where a" = 1. We now show that
(independent of t) Wp is also a polynomial in A. This relies on two basic
facts. First, any symmetric polynomial in the A, is A-modular and second,
any A-modular function is a rational function of A.
Theorem 4.7
Wp(x, A) is a polynomial of degree p + 1 in x and A with integer coefficients.
The coefficients of xp+1 and Ap+1 are both 1.
Proof. Consider Wp (x, A): = Of=0 ( y - Xt), where A;: = 16 /X, and
y •= 16/*. This is a convenient form to work with. Observe that Wp and Wp
are connected by
(4.4.3) l6p+1Wp(x, A) = \xp+1 J! x]wp(x, A) .
L i'=0 J
Now by Theorem 4.6 any symmetric polynomial in Aq , Aj ,..., A is
left invariant by any element of the A-group. (See Exercise 6 of Section
11.2.) It follows that any such polynomial is A-modular, and by Exercise 4b)
of the last section A, is finite valued in the fundamental set except possibly at
t := ioo. In particular if st is the coefficient of y' in Wp (viewed as a
polynomial in y), then st is A-modular. From Corollary 3.1,
(4.4.4) X(q) = 16qTl{.+ 1-J
„=i \1 + q I
In \ 8
I
<r
and we have integers an such that
- 16 1 v
A:= — = - + 2, a„<5
A 1 n = 0
and
A,. = -7iT7F + Ea„«V/p i<P
a q „=o
1
r „=o

122
Higher Order Transformations
where a is a primitive pth root of unity. It can now be established that there
are integers ci so that
CO
(4.4.5) 5,.= 2 cAl.
/=-(P+i)
It is a consequence of the symmetry that the nonintegral powers of q vanish.
(See Exercise 2.) Next there is a polynomial P of degree at most p + 1 with
integer coefficients so that
CO
5;-P(A) = 2<W
«=0
and 5, — P(A) has zero principal part. This is easily proved. First remove the
^-(p+i) term ^ considering
Tp + l
and then proceed inductively. Obs_erve from (4.4.4) and (4.3.5) that the
only candidate for a pole of st — P( A) is q = 0 but P has been chosen so that
5; —P(A) is finite at q = 0. Thus we see that s, — P(A) is a bounded
A-modular function and is hence, by Theorem 4.5, constant. Since
we have that 5,. is a polynomial of degree at most p + 1 in A: = 16 /A with
integer coefficients. Hence Wp(x, A) is a polynomial of degree p + 1 in 16/*
and 16/A with integer coefficients. We can prove directly (see Exercise 3)
that
ru=A'+i
i=0
Exercise 7 shows that 16p+1 divides every coefficient of Wp(x, A) when
viewed as a polynomial in l/x and 1/A. Thus with (4.4.3),
xp+1xp + 1
Wp(x, A) = w+l Wp{x, A)
is of the required form. □
Analogously we have a modular equation for ;':= 1728 J. (See Exercise
4-)

4.4 The Modular Equations for A and j
123
Theorem 4.8
The modular equation for j of order p
p
(4.4.6) Fp(x, j) :=U(x~ jt) jt : = ;»At
is a polynomial with integer coefficients of degree p + 1 in x and /. The
coefficients of xp+1 and/p+1 are both 1.
The modular equation (4.4.2) is irreducible over C(A) (the rational
functions in A) since any root can be transformed into any other by an
appropriate transformation. (See Exercise 5.) Likewise (4.4.6) is irreducible
over the rational functions in j. The Galois group of Fp over ®p(j) is a
group of order p(p2 — 1), which is nonsolvable for p^5. (See Exercise 10.)
Here Qp is Q adjoin the pth roots of unity. For nonprime n, modular
equations can be constructed from the modular equations corresponding to
the prime factors of n. (See Exercise 8.)
Comments and Exercises
Further properties of modular equations are chronicled in Lang [73],
Schoeneberg [76], and particularly in Weber [08]. We have chosen a path of
limited generality focusing on the modular equations for A and j. It should
however be fairly clear that analogous equations hold for other modular
functions.
1. Prove Lemma 4.1.
Hint: For (a) prove that M is equivalent to an upper triangular matrix
mod T. Then write out the system of equations required for two
triangular matrices to be equivalent. For part (c) show that BtC is
equivalent to a triangular matrix mod A. Note that BtC has
determinant p, and hence the diagonal entries of this equivalent triangular
matrix are ±1, ±p.
2. a) Suppose that v
CO
/(?)••= 2 cnq" c„real.
n--h
Show, for a := e2mlp and p prime, that
p
2 [/(«V/P)f m integer
n = l
has no fractional powers of q in its expansion. (See Exercise 4 of
Section 6.2.)
b) Prove that (4.4.5) holds by applying Newton's formulae to express
st in terms of powers of the roots. (See Exercise 6 of Section 11.2.)

124
Higher Order Transformations
3. Show directly from (4.4.4) that
fU = A'+1.
4. Prove Theorem 4.8 by modifying the proof of Theorem 4.7. Use
Exercise 6b) of section 4.3. The proof is somewhat easier since we can
consider IIf=0 (x — /,) directly.
5. Show that the modular equations (4.4.2) and (4.4.6) are irreducible in
x, over C(A) and C(;'), respectively, by elaborating on the comments
following Theorem 4.8. This requires a minimal knowledge of Galois
theory. Note that the transformations of Lemma 4.1 act transitively on
the roots.
6. Show that
a) W„(*,l) = (*-1)'+1
b) Wp(x,0)~xp+\
Hint: Consider the orbits of 0 and i°° under the A-group.
7. a) Show that
Wp(x,X) = Wp(X,x)
and
Wp(^,A) = Wp(^iy+V+1
and if c(.; is the coefficient of xlx' in Wp, then
Ci,j ~ Cp + l-i,p + l-j ~ Cj,i ~ Cp+l-j,p + l-i •
Thus there is a fourfold symmetry in the coefficients of Wp.
Hint: Wp(X(qp), X{q)) = 0 and Wp(X(q), X(qp)) = 0 and by
Exercise 5, Wp(x, A) is irreducible in x over C(A). Also, Wp(x, A) and
Wp(X,x) have a common root x = X(qp) and are of the same
degree in x. Hence,
Wp(x,X) = R(X)Wp(X,x)
where R is a rational function of A. Show that this implies that
R = l. For the second symmetry use X(t/(t — 1)) = l/X(t).
b) Consider Wp(x, A), as in the proof of Theorem 4.7, as a
polynomial in l/x and 1/A. Let c,_; be the coefficient of x~'X~!. Show
directly that
16''+/'|c,;/ ;+;>p + l

4.4 The Modular Equations for A and j 125
and, by part a), that
162(P + l)-0+;)|-^ i+jsp.
c) Show that the coefficient of x'x' in WJx, A) is divisible bv
16Ip+i-(<-+/)I p }
8. In general, for n not necessarily prime, the modular equation (4.4.2)
has degree i/f(n) := n Up\n(l + 1 lp). Let n '= px • • • pk be a product of
distinct primes. Prove that there exists a two-variable polynomial
Wn(x, y) of degree at most ip(n) that satisfies
Wn(X(q),X(q")) = 0.
Hint: Let n = plp2. Since
Wpi(A( q"), A( q)) = 0
and
WP2(A(^^),A(^)) = 0
we deduce that \(qPlP2) is algebraic over Q(A(q)) and is of degree
ty{PiP2)- Now proceed inductively.
9. Show that the coefficients of W3(x, A) are
1
-256
384
-132
384
-762
384
-132
384
-256
1
Note that by Exercises 6 and 7 is suffices to determine cl l and cx 2.
The cM coefficient is always -16p_1. (See Exercise 6 of the next
section.)
10. {On the Galois group of Fp ) If A G Y then A induces an automorphism
on the ;\ that fixes /. Hence A is an element of the splitting field for F
over 0(/). Any two elements 5 and T of T induce the same
automorphism exactly when
for some a. Equivalently MST~lM G V for all M G Tp. One can show

126
Higher Order Transformations
that T, equivalenced as above, is the Galois group for Fp over Qp(j)
and that it is a group of order p(p2 — 1) which is nonsolvable for p > 5.
(See Klein and Fricke [1892], Schoeneberg [76], and Exercise 8 of
Section 4.5.)
4.5 THE MODULAR EQUATION IN u-v FORM
The actual calculation of the modular relation for A is most readily carried
out in the associated variables u '■= k114 '■= A1/8 and v '■= I114 '■= x118.
Analogously to Theorem 4.7, though with a few substantial additional details (see
Exercises 1 and 2), we have the following theorem.
Theorem 4.9
If p is an odd prime, then the modular equation in u~v form is given by
(4.5.1) ap(v, u):=(v- u0)(v - mx)• • • (v - up)
where
Up := (-1)^-1>/8[A(^)]1/8 := (-l)(p2-1)/8M(¢0 q := e^
uk := [A(«V ^)]178 := u(a8kqVp) k = 0, 2,. .. , p - 1
and a is a primitive pth root of unity. This modular equation is a polynomial
in u and v of degree p + 1 (independent of t) with integer coefficients.
For p = ±l (mod 8),
np(v, u) = ap(u, v) = ap(-v, -a).
The coefficients of vp+1 and up+1 are 1.
For p = ±3 (mod 8),
np(v,i) = (v + iy(v-i)
ap(v,u) = -ap(-u,v) = ap(rv,-u).
The coefficient of up+1 is —1 and the coefficient of vp+1 is 1.
The most striking property of the modular equation in u and v is the
"octicity." Because ui(qp) is of the form a'qll8f(a8'q), where/is analytic in
q, only every eighth coefficient of the modular equation is nonzero. (See
Exercise 3.) We illustrate with some examples which give the nonzero
coefficients and the column sums.

4.5 The Modular Equation in u — v Form
4 3 2 *
V V V V 1
u
3
u
n3 u2
u
1
u
5
u
4
u
u2
u
1
+1
+2
-2
-1
1+2 0-2 -1 = (v + lf(v - 1)
+1
+4
+5
-5
-4
-1
1+4+5 0 -5 -4 -1 =(v + lf(v-l)
u
7
U
u5
3
U
2
U
U
1
0
+1
-8
+28
-56
+70
-56
+28
-8
+1
0
1 -8 +28 -56 +70 -56 +28
+ 1 =(v-

Higher Order Transformations
T
o
1
+
1
+
oo
oo
+
1
o
+
f
5
+
T
o
+
oo
oo
1
1
5
+
1
+
o
+
~3 ~3 ~3 °3 3 3 ^3 "a **S *S ^3 3

4.5 The Modular Equation in u - v Form
%
t
T
o
•
1
-3-
+
1
o
oo
o
+
-3-
1
o
+
o
+
°5
-3-
1
o
oo
o
ts
1
oo
o
+
o
-3-
+
o
1
1
o
-3-
+
00
o
.1
o
+
-3-
1
+
o
-

w
vm
0
0
+1
v"
-256
+272
-34
,16
0
-272
+425
,1S
+ 1632
-2418
v»
-4080
+7140
-
*13
+4896
-13464
*12
-4080
+22644
V11
+ 1632
-33456
v10
-272
+44030
0
V*
+272
-49164
+272
Vs
0
+44030
-272
v7
-33456
+1632
,6
+22644
-4080
v5
-13464
+4896
V*
+7140
-4080
v3
-2448
+uh
v2
425
-272
0
V
-34
+272
-256
1
+ 1
0
0
-18 +153 -816 +3060 -8568 +18564 -31824 +43758 -48520 +43758 -31824 +18564 -8568 +3060 -816 +153 -18 +1 =(i>-1)1!

-^^(0^^3^3^^3^^^^9^^^^^^
„2° „19 „B .." „16 „15
V» v"
v6 v5 v" V3 v2 V 1
0
1
+512
-608
+114
-2432
+2584
-2432
+3344
-114
0
-3952
+6859
+5472
+2280
+2432
+ 10488
+2584
+25536
-2280
+3952
+21242
0
-608
"""
+20748
-3344
-10488
+ 10488
+3344
-20748
+608
0
-21242
-3952
+2280
-25536
-2584
-10488
-2432
-2280
-5472
-6859
+3952
0
+ 114
-3344
+2432
-2584
+2432
-512
-114
*
+608
-1
0
0
1 +18 +152 +798 +2907 +7752 +15504 +23256 +25194 +16796 0 -16796 -25194 -23256 -15504 -7752 -2907 -798 -152 -18. -1 = (v + l)"(v -

132
Higher Order Transformations
u24
ua
u22
un
u2a
u"
u18
u"
H16
«15
u"
u"
«23 „12
u"
«10
9
U
«8
«7
u6
u5
u4
«3
u2
u
0
v24
0
0
0
+1
v°
-2048
+2944
-920
v22
0
-13248
+13524
v2i
-23552
+75072
-53544
v2a
+52992
-124752
+82386
19
V
-138368
+149408
-53544
u"
+334144
-213072
+ 13524
v"
-712448
+367264
-920
v"
1159200
-423729
0
1 -24 +276 -2024 +10626 -42504 +134596 -346104 +735471
The exhibited portion of il23 is sufficient to easily calculate the remainder
of the coefficients because of the symmetries. For p = ±1 mod 8 the table is
symmetric through both diagonals. For p = ±3 mod 8 the reflections through
the diagonals change the sign of the entries according to ct j = (-l)'+1c;, and
c,j = (-i)'cp+1_hp+1_t, where clyj is the coefficient of v'u' in O. (See
Exercise 4.)
The numerical calculation of these modular equations is fairly
straightforward. From the q expansions for the 6 functions we can compute
u0, w1;.. . , up for a variety of q values (for p : = 23 one must use three
values of q). We then use (4.5.1) to calculate the coefficient of v' at these
values. However, we know the form of the coefficient (for p : = 23 and
/:==23 one has, for example, that the coefficient of v' is of the form
au + bu + cu1), and we can easily calculate u at the same q values we

4.5 The Modular Equation in u - v Form
133
used to calculate the coefficient of v\ This leads to a system of linear
equations to solve. We know that the system has an integral solution. We
also know what the column sums are in each case. One can use this
information to reduce the size of the system by 1 or, perhaps more
reasonably, as a check on the solution.
The main limitation is that the size of the entries grows exponentially
with p. The pth modular equation will have entries of size roughly 2P, so for
large examples one must work to a high degree of precision. Note that the
size of the linear system only grows as p/8.
Comments and Exercises
A fairly complete account of modular equations up to 1928 is given in
Hanna [28]. This includes equations of degree 103, 107, 127, 167, 191, and
239.
The u — v modular equations up to degree 20 are presented in Cayley
[1874] as we have presented them. We easily computed il23 by the method
outlined in the section. The others were originally calculated by Sohnke
(1836), whose method roughly parallels the one we have described, except,
that he computed an expansion for u(q)/q118 and computed sufficient
coefficients of um, to calculate the elementary symmetric functions directly.
Then instead of solving a linear system, he compares coefficients. As Cayley
[1874] points out, "The process is a laborious one (although less so than
perhaps might beforehand have been imagined)."
A particularly simple form of the modular equation for p =23, due to
Schroter, is
(4.5.2) (A;/)1'4 + (k'l')1H + 22'\klk'l')1'12 = 1.
For many theoretical purposes modular equations for j are preferable.
However, for calculations the modular equation for u is usually simpler. The
extent of the numerical simplification is quite remarkable. Du Val [73]
exhibits low-order modular equations for the T-modular function /'.= //
(/— 1). The cubic modular equation for I tabulates as a 5 x 5 matrix where
all but one of the 25 entries are either 15- or 16-digit integers. Modular
equations for j up to order 11 have been calculated. (See Kaltofen and Yui
[84].) For the llth-order modular equation the coefficients are enormous.
For example, the coefficient of f is
27090964785531389931563200281035226311929052227303
x292319520ll2-53.
Various of the A and j modular equations are presented in Greenhill [1892].
In particular, clean forms for p =29, 31, 47, and 71 are given for Wp.

134
Higher Order Transformations
Exercises 1 and 2 outline the proof of Theorem 4.9. Some of the details
are rather complicated. The flavour should come through.
1. a) Observe using (3.2.3) that
„=i\l + <7 I
b) Analyze the action of A-group on u. Show that
u°SA8) = u and u°Tk = u .
Observe that u°Sk = j8w where )8 is an eight root of unity.
c) Analyze the action that SA induces on the functions A0,. .. , Ap and
on the functions u0,. . . , up. What is the permutation SA8) induces
on u0,. . . , up1
d) Analyze the permutation that Tx induces on the functions
A0,. . . , Xp and on the functions «„,..., up. Identify Ap(rA) and
A (rj1) in particular.
2. Prove Theorem 4.9.
Hint: Use Theorem 4.7 and Exercise 1. Consider how Wp(v8, u8) splits
over Q(w). (Note that u is invariant with respect to group generated by
Sf and TA.)
3. Prove the "octicity" of the u-v modular equation. That is, for p = ±1
mod 8 the nonzero entries of the table associated with Op lie only on the
main diagonal and every eighth sub and super diagonal, while for
p = ±3 mod 8 the nonzero entries are only in every second entry of
every fourth diagonal.
4. a) Prove the symmetries (or antisymmetries) of the u-v modular
equation with respect to reflection through both diagonals.
b) Evaluate the row sums explicitly from Theorem 4.9.
c) Observe that with the aid of the "octicity" one can read off the
modular equations of degrees 3, 5, and 7. Verify the modular
equations of degrees 11 and 13. This requires either calculating the
ui at a single value of q, or using Exercise 6.
5. Show that il7 can be written as
(4.5.3) (1 - w8)(l - v8) - (1 - uv)8
or as
(4.5.4)
(kl)1'4 + (JfcT)1'4 = 1.

4.5 The Modular Equation in u — v Form 135
6. From our analysis we know that
a.(v,u) = vp+1+ E ct/u' + (-l)(p2-1),8«p+1.
a) Show that Cl, = -(-1)(^-^2^-1^2
Hint: ilp(uQ,'u) = 0 and u = V2q118 + 0(u2) while w0 = V2q1,8p +
O(u20). Thus
0 = (-l)(p2~1);V + clAu0 + O(u0u)
and
Cu = -(-D(p2-1)/8lim,^.
b) Show that if 0 < u < v < 1 and Q,p(v, u) = 0, then
V<2("-1)/2M and lim-=2(p-1)/2.
c) Hence if W (y2, y2+1) = 0 with 1 > yn > yn+l > 0, one has
7„+1^t4 and lim-^-=4
7. In 1858, Hermite and Kronecker separately gave solutions of a general
quintic using quintic modular equations. Hermite's method is
outlined below.
a) Let
¢,. := («5 - ul)(ui+l - m,_1)(m,+2 - w,_2) i = 0,1,. . . , 4
where i + / is chosen mod 5. Then
/ 1 \1/4 1
v 5 ' w(l - u )
are the five roots of the quintic
2 1 + w8
*5 — x ■
r5/4 2/1 8\l/2
C5/4 Z/1 8\1
5 w (1 - u )
b) The quintic modular equation (4.1.21)
6 6 , c 2 2/ 2 2\ 1 ,1 /! 4 4\ «
w — v +5u v (u — v ) + 4uv(l — u v ) = 0

136 Higher Order Transformations
can equivalently be transformed into
5 o4c3 4/1 8\2 /-,6^5/2 3/i 8\2/1 , 8\ rv
x -25u{l-u)x — 25 w(l — w)(l + w) = 0.
The details are formidable,
c) Any quintic can be algebraically reduced (via solution of a quartic
equation) to the Bring form
x5 — x — a
and hence a) and c) provide a solution of the quintic in terms of the
roots of the modular equation. This requires solving
2 1 + w8
55M u\l - u8)in
and leads to a quartic equation in u .
The amount of calculation required above is prohibitive.
However, that some combination of the uf solves a quintic is not overly
surprising since the Galois group of the quintic modular equation
(4.1.21) is 55. The reduction to Bring form in c) is effected via the
Tschirnhaus substitution.
(On the Galois groups for W5 and W1 over <Q>( A))
a) For p = 5, SA and Tx induce the following permutations of the roots
(*' = *,)
5A: (0,1,2,3,4,5)-^(1,2,3,4,0,5)
TA: (0,1, 2, 3, 4, 5)-» (0,5, 3,1,2, 4).
b) For p = 7
5A: (0,1, 2, 3, 4, 5, 6, 7)-* (1, 2, 3, 4, 5, 6, 0, 7)
TA: (0,1,2,3,4,5,6,7)-^(0,3,1,4,6,7,5,2).
c) Show, using a), that the Galois group of W5 contains A5 and is not
solvable.
d) Show, using b), that the Galois group for W7 is not solvable.
In both cases these permutations actually generate the Galois
group. (See Exercise 10 of 4.4, and Exercise 1 of 4.5.)
4.6 THE MULTIPLIER
As we saw in Section 4.4, thepth-order transformation can be considered as
determined by k := k(q) and 1:= k(qllp). We define Mp, the multiplier of
order p, by

4.6 The Multiplier
l2/
137
1(4-6.1)
MM,k):
6i(q) _K(k)
e\(qVp) K(l)
|> and in the future will denote K(l) : = L.
Theorem 4.10
|If k := k(q) and /:= k(qVp), then Wp(/2, A;2) = 0 and
f2_ ll'2 dk v(l-v8) du
1(4.6.2)
pMt
kk'2 dl u(l-u8) dv
In particular, Mp is an algebraic function of k and I.
Proof. Since 7rK'/K= —log q [by Theorem 2.3(6)], we have
1 7T
f (4.6.3a)
A(£.
dk\K
2 A*'2*:
21^2
I on using (2.3.10). Similarly,
f (4-6-36)
1
dl
\LJ 2 U'2L
Now as pL'/L = K'/K [again by Theorem 2.3(6)], we can write the I, k
form of (4.6.2) on dividing (4.6.3a) by (4.6.36). We now verify the u-v
form directly. □
When Mp is given by v(l — v8)/[u(l — w8)] (du/dv) as a function of u and
v, we write Mp(v, u). In this form we see that M2 is rational. In fact
Mp(u, «) is rational. (See also Exercise 1.)
Let us use (4.6.2) to compute M2 and M3. When p'-=2, we have
/ = 2V£/(1 + k) and A: = (1 - /')/(! + /'), by equation (2.1.15). Thus
(4.6.4)
2M2 =
dk ll'2 Vk(l + k)2
kk'2 dl kk'2 1-k
(1 + kY
and
M2(l, k) :
1 + /'
1 + A:
- which corresponds with Theorem 2.6.

138
Higher Order Transformations
When p'-=3, we have Vlk + VVk7 = 1, by equation (4.2.6). This
produces, on differentiating implicitly and using (4.6.2),
3M, = -
l'2^M-l2VFF
VE-l2
1
TO
13 k'WJk-kWTF y/Tk-k2 \-
Let a4 := k3/l. Then I3Ik = [(2 + a) /(1 + 2a)]4 and
(46-5) 3M'=(dn? - m>=2Stt-
(See Exercise 2 of Section 4.1.) But a = ulv so that
M3 =
lai
2v -u
v + 2u
3w
because
3 J2+a
1
2a + 1 \2a + l-
For p : = 5 or 7, similar, but more elaborate, calculation produces
(4.6.6) Ms =
(4.6.7) M7
u + v
v(l - uv3)
5 5m(1 + u3v) v — us
- ^(1 ~ uv)[l - uv + (mi;)2] _ _
w — i;
v — u
7w(l — mu)[1 — uv + (uv) ]
(See Cayley [1895].) Many other multipliers have been calculated and can
be found in Ramanujan's collected works, Cayley [1874], Tannery and Molk
[1893], in Weber [08] and elsewhere. The main technique for larger p is via
manipulation of theta series. Thus one has Ramanujan's form of the
multiplier for 13:
(4.6.8)
»*..M>-ap.(£r-(&r-«(&)
1/3
1
Ml3(k, I)
We list also
l i\in /i'V2 i w V2
(4.6.9) 17M„ft«H*) +(f) +(W)

4.6 The Multiplier
139
We conclude this section by touching on the matter of singular values, kp,
which for us are defined to be the solutions in (0,1) of Wp(k' , k2) = 0.
These are often called singular moduli for the function A. Corresponding
values for J are discussed in Exercise 6.
Then, since K'(k)/K(k) is isotone, Theorem 2.3(b) shows that this is the
unique solution to
Y<*p) = Vp
0 < kp < 1.
In the notation of equation (3.2.1), k = A*(p) and k' = A*(l/p), so that
kp = k(e~7ry/p) and lp : = k'p = k(e~^p). Sophisticated number-theoretic
techniques are available for computing kp for large p, without knowledge of
Wp. This is discussed briefly in Exercise 5. For small p one can solve directly
for kp. Thus
'*! =
_1_
V2
fc, = V2-l
^3 =
V2(V3-1)
4
3-2V2
^7 =
VV5-1-V3-V5
V2(3 - V7)
, _ (V2-31/4)(V3-1)
/c9 _
_1_
V2
= V2V2-2
V2(V3 + 1)
4
= 21/4(2V2-2)
VV5-1 + V3-V5
2
V2(3 + V7)
8
(V2 + 31/4)(V3-1)
2
2^:^ = 1
^3^3 0
2A:5/5 = V5 - 2
ZiKjlj
2A:9/9 = (2 - V3)2
(4.6.10)
A more comprehensive list is given in the next chapter. A profusion of
modular equations of degrees 3, 5, and 7 are given in Chapter 19 of
Ramanujan's Second Notebook.
Comments and Exercises
1. From the results of Section 1.5 (in particular Theorem 1.5 and Exercise
1) we know that G(k) := kV2k'K(k) satisfies

140
Higher Order Transformations
where k and I are solutions of the pth-order modular equation Wp. Use
this to show that
M2(l,k): =
ll'z dk
= c
K(l)
K(k)i " kk'2 dl
where c is a constant. This provides an easy alternate derivation of
Theorem 4.10 up to the evaluation of the constant c.
2. a) Verify the computation of M3 in (4.6.5).
b) Compute Ms in (4.6.6).
Cayley [1874] discusses algebraic methods for computing Wp and Mp at
length. These seem only to be entirely reasonable for 2, 3, 5, 7, and in part,
11. The discussion therein also illuminates the rational nature of Mp(u, v).
3. a) Show that pMp(l, k)Mp{k', V) = 1.
b) Use Theorem 4.10 to show that for all u and v,
[PMp(v,u)Mp((-l)^-l)l8u,v)]2 = l.
Hint: Consider the similar symmetry of Op.
4. a) Cayley observes that given any polynomial identity F(u, v) = 0
which satisfies F(u, v) = F(-u, -v), one can produce a similar
identity G(u2, v2) = 0, with G of the same degree. One uses
(4.6.11) [F(u, v)F(u, -v)F(~u, v)F(-u, -v)]1'2 = 0 .
b) Use this technique to develop modular equations and multipliers in
terms of the u, v2n (n := 1,2, 4) for p : = 2, 3, 5, 7. (See Cayley
[1874].)
5. Verify the singular values in (4.6.10). In each odd case one verifies
t„ -Ikk' first and uses
kn = (VT+7n-VT=tn)/2
ln = (VT+7n + y/T=Tn)/2.
The invariants of (3.2.9) to (3.2.13) lie at the heart of calculating
singular values. Armed with these and either Ramanujan's insight or
some knowledge of group theory, singular values can be calculated in
profusion. Watson, in a long series of papers commencing with Watson
[32], has recreated what he believes to be Ramanujan's procedure,
while Weber [08] explains the classical theory and lists many examples.
Zucker [77] indicates an attractive way of calculating many large
singular values such as Ramanujan's celebrated

4.6 The Multiplier
141
(4.6.12) k2W = (V2 - 1)2(2 - V3)(V7 - V6)2(8 - 3V7)
x (VlO - 3)2(4 - Vl5)2(Vl5 - Vl4)(6 - V35) .
Zucker's methods are described in Section 9.2.
6. In this exercise we sketch the relationship between binary quadratic
forms and singular invariants or values for Fp [solutions of Fp(j, j) = 0].
a) Let
M: $)
where |5| = ad — be = p > 0 and a, b, c, d are integral, and assume
with no loss of generality that c > 0, b^O. We say t with im(t) > 0
satisfies complex multiplication by jit if (/jut, jit) = S(t, 1) so that
t = (at + b) /(ct + d). In particular, f is the solution with im(t) > 0 to
an integral equation Ax + Bx + C = 0 with -D-=B2 - 4AC <0.
[This corresponds to asking for which \x, one can solve
p(lJ,z, w1, w2) = R(p(z, wl, w2)) for some rational function jR. Here
p{z) : = p(z, h>1; w2) is the Weierstrass p function of Section 1.7. Or,
in other words, for which /jl does the lattice L generated by wt and
w2 contain the lattice jitL? Hence, the multiplication.]
b) Suppose that
at + b , „ a*t+b*
ct+d c*t+d*
(4.6.13) ad-bc = p, a*d* - b*c* = 1.
Then ;(f*)= j(t). Also t* solves a quadratic with the same
discriminant as that for t, and t* possesses a multiplication by \l if and
only if t does. Now (4.4.6) implies Fp(x, ;') = 0 is solved by j(t),
because 5 = A,modr for some i and hence j(t) =/,(0- Thus
Fp(i>J)~Q 1S solved by j(t) exactly when t possesses complex
multiplication. Note that Fp{j, j) will have lower degree than
Fp(x, j).
c) Since / is one to one on the fundamental region, f(t) =/(**) if and
only if the associated primitive binary quadratic forms are properly
equivalent. A binary form ax2 + bxy + cy = 0 is primitive if
gcd(a, b, c) = 1. Two forms are properly equivalent if there is an
integral linear transformation of determinant 1 converting the one
into the other. (See Dickson [71, vol. 3].)
d) Thus the study of the degree of Fp(j, j) becomes the study of
h(—D): the class number of primitive forms of negative
discriminant — D. More of this is sketched in Hardy [40, chap. 10], and in
Tannery and Molk [1893].

142
Higher Order Transformations
7. For fixed m and p let y0: = km be the mth singular value and let yn be
the sequence of solutions of Wp(yn, yn+l) = 0 as defined in Exercise 6 of
Section 4.5. Use Theorem 2.3 and (3.2.3) to show that
n—*<» \y /
lip"
Wm/2
e
■TV
This provides an algorithm of order p for computing e~in/m. When
p "- = 2, this reduces to the algorithm in Exercise 3 of Section 2.5.
By the same process as in Section 4.5 one can produce polynomials in
x := M~l{l, k) and k. Cayley [1874], following Joubert, gives
(4.6.14) *4 - 6*2 + 8(1 - 2fc2)* - 3 = 0 p:=3
x6 - 10*5 + 35*4 - 6(k3 + 55*2 + [64(2kk'f - 26]* + 5 = 0 p := 5
(4.6.15)
and
(4.6.16) *8 - 28*6 + 112(1 - 2*2)*5 - 210¾4
+ 224(1 - 2A:2)*3 - [140 + 1344(2A*')2];c2
+ [48 + 512(2A£')2](1 - 2A:2)* + 7 = 0 p := 7 .
He gives a similar expression when p := 11.
8. a) Verify (4.6.14) and observe that 3A/3(1/V2, k9) = V3 + 2V3.
b) Explicitly solve (4.6.14) to produce M^1 as a function of k.
c) Use (4.6.15) to determine ks.
4.7 CUBIC MODULAR IDENTITIES
Ramanujan in his notebooks gives the following remarkable identity:
03( q3)
$5-11
04^9)
1
[See (24.28) and (24.29) of Chapter 18 of his second notebook in Berndt
[Pr].]
An equivalent form and some variants of this formula are established in
the next theorem.
Theorem 4.11
ith
k:
el(q)
0l(q)
. e22(q3)
y'~el(q3)

4.7 Cubic Modular Identities
143
one has
(a)
(b)
(c)
e*(q)
3^5-1
^( ^)
= 4(^-) =9^^-!
V 77' 1 6t(q)
4/ _3>
e:(?)
02(<z)
024(jLl
1(9)
1.
Proof. We establish only (a). Then (6) and (c) follow (Exercise 2). It is
most convenient to use the quintuple-product extension of Jacobi's triple
product, which is developed in Section 9.4. Equation (9.4.3) with/:= 6 and
k: = 1 may be remanipulated to show that
CO
303(<79) -8,(9) = 211(1- g")(l + <72")(1 + q6"'3) ■
n=l
Hence, with (3.1.6),
CO
0s(q)pe3(q9) - e3(q)f = 8 IT tt + q6n+3f
n = 0
00
X I! (1 - q2n)\l - q2"'1)3^ + q2")\l + q2"~l?
and Euler's identity (3.1.4) reduces this to
s fi (l+q6-+y n (l+q2-+y n (-^4^) ■
Now (3.2.9iv) shows
and
EI (1+ q2n + 1f =VZql'\kk'rUA
n(i+?6n+3)3=V2^/8(77'r1M
n = 0
while (3.1.4), (3.1.7), and (3.1.8) combine to yield
In \ 4
/ 1 ~z" \ 4 1
(?)
These identities result in

144 Higher Order Transformations
(4.7.2) e3(q)[303(q9)-e3(q)f=4
606,
(kkTXyy'Y
■40
(kk')3'4
3 / ,\l/4
(77 )
This establishes the first equality for part (a). The second is most easily
seen by using equation (4.1.15), with y'■= k and k'-=l, to write k3/y =
[(2 + a)/(2a + l)]4, k'3/y' = [(1 - a)/(2a + l)]4, and using (4.1.19) or
(4.6.5) to write 043(q3)/643(q) = 1/(2« + 1)2. Then both sides of the second
equality become 4(2 + a)(l - a) /(2a + 1)2. □
These identities have many remarkable consequences some of which we
leave as exercises. Two, however, are worthy of more explicit analysis.
an iteration for the cubic multiplier. In decreasing form, as above, we
consider the multiplier Mn : = Kn+l/Kn where Kn : = K(q3"). Then part a)
of Theorem 4.11 shows that, with mn := 3A/„, we have
(4.7.3)
where
(4.7.4)
m„+, =
mn =
1 + 2V2
(2kk')
(277 01
3/4
Alternatively, part (b) shows that if we set rn '■= 3624(q3
mn = (/■„ + 3)/(rn - 1) (see Exercise 8). We have
)/04(<73")>sothat
(4.7.5)
where
(4.7.6)
[(^-1)^ + 1]
rn-
1+2V2) '
(27/7 )
2^3/4
4
1/2
This latter is more suitable when we begin with an even singular value.
We have written (4.7.4) and (4.7.5) in a form consistent with Ramanujan's
invariants Gn and gn. [See (3.2.13) and Exercise 5, Section 3.2.] In these
terms, the iterations are initialized by
(4.7.7)
and
(4.7.8)
m0:=m(n):=[l + (V2G9„/G^)3]1/2
r0:=r(n):=[l + (V2g9n/g3n)3]U2.
q:=e
-7rVn
We also write M(n) := m(n)/3.

4.7 Cubic Modular Identities
cubic recursions for G„ and gn. We have
9 g'(g3) =1 + 2V2^
03(<7) G«
and, using the theta inversion formula in part (a),
03( 1) _, ,../. G
3
= 1 + 2V2 —£- .
Ot(ql G
Thus
(4.7.9) 9 = (l + 2V2%)(l + 2V2^)
V G„/\ G9„/
Similarly
(4.7.10) 9 = (l + 2V2%)(l-2V2-%-).
on 69«
Thus given G„ or gn, we have a simple equation to solve for G9n or g9n.
example, with n"= 5 we have G3 = G1/3, and with *:= G3 we know
(1 + 2V2V6)2 =9 or G3=21/12. Similarly, g2~/3 = g6 so that (4.7.10) y
of -g6" = 4V2 and g* = V2+T.
A more tractable recursion can be attained by observing that
3^-1 = ^%
and
Thus
^)-1 = V2G'
03(<?y) G
3
81n
(4.7.11) 3 = (V2^ + l)(V2^ + l)
G81«
and, rearranging,
(4.7..2) ei.^^ig

146 Higher Order Transformations
Correspondingly
3 gl + V2g9„
(4.7.13) 881"~89" g9n-^2gl'
From these two formulae, and other singular values given by Ramanujan,
one can give explicit equations for some very large invariants (G\025, G\991,
#4698> f°r example). We illustrate with g4698. Ramanujan gives
2 _ 5 + V29
8s8- 2
'5
6*522
±^)"2(5V5+11V6r(^ + ^M)
Then manipulation of (4.7.13) yields
3 _
64698
/V29-5\1/2(V29 + 5) + V2(llV6+5V29)1/6(V9 + 3V6 + V5 + 3V6)
\ 2./ (V29 -5)- V2(ll V6 - 5V29)1/6(V9 + 3V6 - V5 + 3V6) "
(See also Exercise 13.) Analogous results for quintic and septic multipliers
are treated in Section 9.5.
Comments and Exercises
The quintuple-product identity was certainly known in essence to
Ramanujan so that our derivation of Theorem 4.11 is in all likelihood similar to that
which he had in mind. The proof we give in Section 9.4 is self-contained and
can be read with ease now. Biagioli [Pr] has given a modular function proof
of (4.7.1).
1. a) Show that (4.7.1) is equivalent to
3
03(<f)
03(q)
4/ _3>,
9g;(gj)
ot(q)
l
1/3
(either by modular considerations or by direct inversion of the
underlying quartic polynomial),
b) Show that the AGM iterations in theta form can be written as
e3(q) x rn2/~2
e3(q*) ' I-03(94)
el(q2) ^1
1

4.7 Cubic Modular Identities
147
or as
03(<74)
03(q)
1 =
Ol(q)
1
(Note that in this case the replacement of q by —q does not give a
formula in 04.)
a) Show that Theorem 4.11, parts (b) and (c) are equivalent to part
(a).
b) Derive part (c) directly from (9.4.3).
Use (4.7.9) and (4.7.10) to verify that
i) Gp = (V3 + l)/V2
ii) g?8 = V3 + V2
iii) G^7 = 21/4/(21/3-l).
a) Use Exercise 3 and (4.7.12) and (4.7.13) to verify that
.. 3 (2V3+2)1/3 + l
(2V3-2)1/3-l
.., 3 1 + (2V6 + 4)1;
u) giffl" " —
3
1/3
1-(2V6-4)J
b) Calculate G243 and g54.
Show that, in the notation of Proposition 3.1,
a) 11^(9)-30^)1 = 40^9)
/=2
b) 11^)-0/(9)1 = 40^(99).
/-2
Show that
9\l3
02{ql[e2{q) - Wtf + 0A(qv)[e4(q) - 6A(q*)]
= 0s(q9M(q)-d3(q9)}3.
Use Schlafli's equation (Exercise 6 of Section 4.1) to establish that
i) G5- 2
n) gio = —2—
iii) G25 =
V5+1

Higher Order Transformations
In the notation of (4.7.7) and (4.7.8) show using (4.1.15) that
i) [m(n) - l][r(n) - 1] = 4
and
ii) r(4n) = '\/m(n) + r(ri) + 3 = y/r(ri)m(ri) .
Also (4.7.9) gives
«0 ^)-^(1)-3-.-(1)-
a) Given that
^-(^OW
verify that
__6 /3-V7\/V7 + V3\3/2
G^=\-vr)\~t~ ) and
M-(|) = V3(^1)V^^V3.
b) Use the modular equations of degrees 3 and 7 to verify the value
of G21.
In each case, given the first invariant verify the following ones.
a) ^ = (-^^)(^^)^^3 = (-^-)(^1) and
Ki r~* /3V19-13\/V3-1\3 6 /3V19-13\/V3 + 1\3
b) G57 - ^ ^ )\-^) > Gl9l3 - {—^—)\~Vr)
and
Af^y) = V3V2V19 + 5^(^1)3.
N _6 /39-7V3T\/V3T-3V3\3/2
c) G93=( ^ )( ) ,
'39 - 7V3T\IV31 + 3V3\3/2
'31/3
G:L = t= = and
vr-A—2—J
M_1(|)=^V3T_3V3)3^y^l)3

4.7 Cubic Modular Identities 149
d) g- = (v^-5)(^^)3,g2t/3 = (V^+5)(^1^-)3 and
r(f) = V3-(^|^-)3/2V6 + 34V2 + 15Vl3.
e) g-06 = (VlO - 3)(V5 - 2), g-% = (VlO - 3)(V5 + 2) and
A"(y)=V3(V2-l)(V5 + 2).
11. a) Verify that m($) = V3 and r{§) = V6 + V3.
b) Hence verify that m(§) = 3V2 + 2V3 - V6-3 and that K§) =
V3(V3 + V2). =====
c) Verify that m(l) = V3 + Vl2 and r(4) = V3 + V3 + V9 + 6V3.
d) Verify that a-(2) = V2+V3 and m{\) = V6- V2 + 1. Hence
m(^) = 3(3V6 + 7 - 4V3 - 5V2).
e) Verify that m(|) = (3 - V6)(V2+1) and m(^) = (6V3+ 9)-
(4V6+6V2).
12. a) Show that (4.7.9) can be written as
(j)2 + (^)2 = 2V5[(^)_(^rl]
where x'= G3N and y '■= G9N. Hence compute G3 and Gg.
b) Find the parallel expression for g3N, g39N. [Compare (4.1.27).]
13. One can explicitly solve (4.7.9) and (4.7.10) to obtain the following
formulae.
a) Show that G39N and G3NI9 are the two solutions to
x2 - V2GN[G8N + ^G^+Gl + ljx + G2N[G8N + 1 + ^/gY+G% +
= 0.
b) Show that g39N and ~g3NI9 are the two solutions to
*2 - ^2gN[g8N+y/gN-g8N+i]x - gtigl -1+V^6-^+i]
= 0.
c) Given that G2S = (V5 + 1)/2, show that
G225 = (2 + V^)1/3(^i)[V4TVl5 + 151/4]
and
G25/9 = (2 + V3)1/3(^i)[V4TVl5 - 151/4].

150 Higher Order Transformations
d) Given that g58 = (5 + V29)/2, show that
^ = 1^^)^(5^+11^)1
9 + 3V6 /5 + 3V6
and
#58/
^~^y/2(5V29 + llV6)
1/6
4 + 3V6 9 + 3V6
14. a) Entry 23 in Chapter 18 of Ramanujan's second notebook (Berndt
[Pr]) gives
i) V2 E exp( 2W7r^)cos( ""y2)
M / 2 . 2 . \l/2 V -n2TTX / 2 -,
= (yx + y + x) Zj e cos (n Try)
w=-co
CO
. /« I 2 . 2 \l/2 V -n2!! -/2 \
+ (V* + y - x) Zj e sin (n Try)
W = —00
••x /~ V (-n\x\ . ( n-ny \
u) V2 2, expl 2 2 ) sin t 2 2)
«=-co \* + y / \* +y /
= (yx + y - x) Zj e cos (n 7ry)
/I=—CO
CO
/.. / 2 . 2 . \l/2 V -nVr -/2 \
- (V* + y + x) Zj e sin (n 777) .
/! = — co
Use Exercise 5 of Section 2.2 to prove these when x and y are
complex with re(* ± iy) > 0.
b) Deduce that if re(s) > 0,
S e~"2-cos (nW) =^^-^ S <r"2™sin(n2™')-
/1 = -00 VI S n = -co
c) Use ai) to deduce that
03(e~") = V5V5 - 10 fl,(<r5*) .
d) Use b) with s : = V5/3 to deduce that
(V5 + 3)030-7^3) = (3 + V3)03(e-"3V~5) .

4.7 Cubic Modular Identities 151
e) Use d) to conclude that
M(5) = £V 1 + 2V3 + 2 V5
and
G« = KV5 + 2)3(V5 + V3)4.
f) i) Use b) with s := V3/2 to obtain
kl2 = (V3 - V2)2(V2 - 1)2 and g% = V2(V3 + 1)3.
ii) Use b) with s := Vl5/4 to obtain
jT = (4+ Vl5)(2 + V3)2 and G" = 4(7 + 3V5)
iii) Use b) with s := V7/4 to obtain
^- = 8 + 3V7 and G" = ;
'7
g) Use c) to show that
vnr ^ (^ + 2] r2(')
15. a) As in Ewell [86] show that
04(g)3 1irJy g3""1
3n-l n3"~2
+ q3-2
and so develop a formula for r,(n).
Hint: Use the quintuple-product of Section 9.4 and mimic the
derivation of equation (9.1.14).
b) Evaluate
CO
El/[l + F«,-3]-

Chapter Five
Modular Equations and
Algebraic Approximations
to 71
Abstract. In this chapter we study the algebraic relationships between elliptic
integrals of the first and second kind. This study is applied to produce nth-
order iteratons for ir, rapid series for ir~ , and assorted other algebraic
approximations to ir.
5.1 SINGULAR VALUES OF THE SECOND KIND
In Section 4.6 singular values were introduced. We will call A* the singular
value function (of the first kind) where A*(r) := k(e~ r) as in Section 3.2.
We introduce the singular value function (of the second kind) a which we
define for positive r by
(5.1.1) «W:=f-~2 * := *(*-*"') = A*(r).
Since \*(r) tends to 0 as r tends to °°, we have a(r) converging to tr"1 as r
increases. Indeed, as we shall see, the convergence is exponential which
allows us to use a(r) to approximate 1/tt effectively.
Using Legendre's identity and the fact that K'(h*(r)) = VrK(\*(r)), we
have
(5.1.2) aW_ - m_,)
152

5.1 Singular Values of the Second Kind 153
and k:=X*(r)
(5.1.3) «W = V7f-^-
If in (5.1.2) we use the differential equation for K, equation (1.3.13), we
may also write
(5.1.4) «(r) = ± (^)2 - V?(kk* § - ¢) k:^ A*W ■
Now since A*(l/r) = A*'^), we discover that
(5.1.5) a(,-1) = ^W.
In particular a(l) = \. We may rearrange (5.1.2) and (5.1.3) as follows:
(5.1.6) j = K[V?E - (V7 - a(r))K]
and fc:=A*(r)
7¾ - ^ X'"
(5.1.7) l = K'
which may be viewed as one-sided forms of Legendre's identity. In the next
sections we will give these identities substance by showing that a(r) is
algebraic for rational r and by computing many values. Another useful
equivalent form is
(5.1.8) E'=V?E-8(r)K
where k-=X*(r)
(5.1.9) 5(r):=Vr-2a(r).
Theorem 5.1
The function a is montonically decreasing for rsl.
Proof. Since A* is decreasing, it suffices to show for k<l/V2 that
f'-=(E'K-ir/4)/K2 is an increasing function of k. This we establish by
computing / and using the differential equations for E' and K. (See

154 Modular Equations and Algebraic Approximations to jt
Exercises 2 and 3 of Section 1.3.) We deduce from Legendre's identity that
, = irlljEIK -1) + k2E(K' - E') + (k')2E'(K - E)
i{) ~ k(k')2K2
= k2[(K - E)(E - irl(2K)) + E(K' -E')-E(K- E)] + (k'f(K - E)(E' - irl{2K))
k(k')2K2
Since, for k<l/V2,
K>E>E'>l>irl2K
we finish by observing that, for k < 1/V2,
K'-E'>K-E
and so substitution into the last equality completes the proof. □
We next provide a theta function expression for a. We combine (2.1.13)
and (2.3.17) with (5.1.2) to write
,^ „ „™ , x tt"1- VrAqe.ld,
(5.1.10) a{r) = fl4 (w.r.t.q)
where q := e~ r. Expanding this gives
(5.1.11) 0 < a(r) - tt"1 = 8(V7 - tt"1) e'7^ + 0(re~27r/~r)
^leVfe"^ r>l
and
(5.1.12) 0<«(a-)-7t"1<V7A*2(a-).
One should compare (3.2.1).
Comments and Exercises
The function a is implicit in Ramanujan's work. In the next section we
indicate how. Zucker [79] computes a(n) for n '• = 1, 2, 3, 4, 5, 7, while a(3)
was known to Legendre. Formula (5.1.10) allows one to numerically
compute a very easily.
a) Show that
tt"1 = [a(r) - V?\*Xr)]el(q) + AVfq
-1-r.v.A .^,,2,.^4,^ ,„.,= _ ¢,(9)
03(q)
where q'- = e~'*v~T.

5.2 Calculation of a
155
b) Use a) to reprove Exercise 4b) of Section 2.3:
4v=
K=-~ne-^
2. By Exercise 4 of Section 1.6, A*(2) = V2 - 1. Use Theorem 1.2 to show
that a(2)=V2-l. Hence a(|) = |.
3. Prove (5.1.11) and (5.1.12).
5.2 CALCULATION OF a
We begin with an appropriate generalization of the quadratic transformation
formula for E given in Theorem 1.2.
Propostion 5.1
Letp>0, k:=k(q), and l:=k(qllp) be given. Then
U>2 dMp(l, k)
(5.2.1) pM2Mk)
| (*) - *'2
MM, k) dl
+
Lf <'>-<"
Note: Herein (dMp(l, k)ldl) is the full derivative of Mp{l, k) with respect to
Proof. We have Mp(l, k)K(l) = K(k). We differentiate both sides with
respect to k and use (4.6.2) to write
(5.2.2) pMp{l, k)~^^ (k) = Mp(l, k) ^ (/) + K(/) ^ (/, /:) .
Next we use the differential equation for K to write
dK _ E{k) - k'2K(k) dK E(l) - l'2K(l)
dkW~ kk'2 dl{l)~ ll'2
and we substitute these two identities into (5.2.2). We then have
2 dMM, k)
PM2p(l, k)[E(k) - k'2K(k)] = Mp(l, k)[E(l) - l'2K(l)] + K(l)ll'2 "^
On dividing each side by K(k) we have (5.2.1). □
We now derive the key identity for a.

156 Modular Equations and Algebraic Approximations to n
Theorem 5.2
Forp and r>0 let /: = \*(r) and k: = A*(pV). Then
a(p2r) = Mp2(l, k)a(r) - Vr
(5.2.3.)
W-*+^^
Proof. We suppress variables in the multiplier when convenient. From
(5.1.2) and (5.2.1) we have
a(p2r) = ^(k~)-pV?
Ml
AK\l)
' E
-K
V?
(*)-!_
[!«
-i
-V7
\,2+»'2dMP
I Mp dl
pMlk2
This gives
(5.2.4) a(p2r) = M'2\a(r)~V?
l2 +
ll'2 dMD
Mp dl
-pMlk2
on using (5.1.2) again. Another application of (4.6.2) produces the desired
formula. □
In particular, if we let r := lip above, then / = k', and with (5.1.5) we
derive that
(5.2.5)
a(p) = Vpk2 - | k'k2 j{ Mp(k', k)
or
(5.2.6)
a(p) = Vpk2-Plkk'2^Mp(k',k)
where k-~ A*(p).
Let us observe that, since A*(p) and Mp are algebraic for rational p, this
shows that a(p) is algebraic for rational p.
example 5.1. When p : = 2, Af (/, k) = 1/(1 + fc) and fc = A*(2)=V2-1.
Then dMp(k', k) ldk = -\ and a(2) = V2(V2 - 1)2 + (V2 - 1) = V2 - 1.
(Compare Exercise 2 of Section 1.) Similarly a(3) = (V3 - 1)/2 and also
a(7) = (V7-2)/2. (See Exercise 3.)

5.2 Calculation of a
157
Before continuing to evaluate a it is necessary to connect it to Ramanu-
jan's multiplier (of the second kind).
(5.2.7)
where
(5.2.8)
nfl .y-PPiqi-W)
P(q) :-1-242 nq
1£.
- ~ q
Proposition 5.2
Let p, k, and I be as in Theorem 5.2. Then
Rp(l, k) =p(l ~2k2)Mp(l, *) - (1 - 2l2)M;\l, k) + 3pkk'2 dMjlkk) •
(5.2.9)
Proof. We start with (3.2.15), which gives t}(<7)/tj(<71/p) in the form
l/12p/
(1-^)(1-^)(1-^ P)"-
(5.2.10)
We differentiate logarithmically and obtain
P[,)-1*(,»)-2,S
1 rf(fcfc') 1 rf(fl') rff 3 dMp
kk' dk IV dl dk Mn dk J
Now we use (4.6.2) for dlldk and (2.3.10) for qdkldq and obtain
PP(q)-P(qllp) =
4LK
TT
p(l - 2k2)Mp -(1- 2l2)M;1 + 3pkk'
dMp
dk J'
This gives (5.2.9). □
It is convenient to introduce two additional quantities:
dk
(5.2.11) ^)-.= -^
and
(5.2.12) <r(p):=Rp(k',k)
k-.^e'7^ ,

158 Modular Equations and Algebraic Approximations to jt
In these terms we have:
Theorem 5.3
(a) With r, p, k, and I as above, we have
spV, k) = 3
(5.2.13)
(5.2.14) a(p2r) = Mp"2(/, *:)«(>■) - Vrep{l, k) .
(b) With fc(<TWp) = A*(p), we have
1 + A*(p)2 cr(p)
(5.2.150 «(/>) = VJ> 3 6
(5.2.15,) ,(p),V?[l-2Ay]+a<fl
Prao/. We deduce (5.2.13) and (5.2.14) by comparing (5.2.9) and
(5.2.3). We obtain (5.2.15) on comparing (5.2.9) (with 1 = k') to
(5.2.6). □
Ramanujan has computed Rp for p := 2, 3, 4, 5, 7, 11, 15, 17, 19, 23, 31,
35. From these, many values of a are obtainable. The following table is
taken from Ramanujan [14] with the entry for i?4 corrected.
The verification that Rp has the given form is tedious but straightforward
for small p. [See Exercise lc).] For larger p we rely on Ramanujan.
example 5.2. For p: = 7, the modular equation in the form (kl)1,A +
{k'l')VA = 1 shows that 2k7k'7 = £, k7 := A*(7). Then k7 = (3 - V7)/4V2,
and so o-(7) = R7(k'7, k7) = 3(1 + 2k7k'7) =27/8, a(7) = (V7-2)/2, and
6(7) = 2.
We now make (5.2.14) explicit for p : = 2, 3, 4.
Proposition 5.3
If /:= A*(r) and A: := A*(4r), then
1-/'
(0 k-j^jr
and
(5.2.16) («) a(4r) = (1 + kfa(r) - 2V7ft .

5.2 Calculation of a
159
TABLE 5.1
RP(l, k)
1
11
15
17
19
23
31
35
I' + k
l + kl + k'l'
3(1 + /')(! + k)
2
l + kl + k'l'
(3 + kl + k'l')
3(1 + kl + k'l')
2[2(1 + kl + k'l') + VE + VkT - VkkW]
[1 + (kl)1'4 + (k'l')1'4]4 - (1 + /c/ + /c7')
[44(1 + /c72 + k'T2) + 168(/(:/ + k'l' - kk'll')
-102(1 - kl - k'l')(4kk'll')U3 - 192(4/c/c'//')2/3]1/2
6[(l + /c/ + /c7') + V&/ + VFF - VitFF]
11(1 + kl + /c7') - 16(4/c/c7/')1/6[l + VFl + VTT]
-20(4/c/c7/')1/3
3{3(1 + /c/ + /c7') + 4(V&/ + VkT + VkkW)
- 4(/c/c7/')1/4[l + (W)1M + (/c7')1/4]}
2[Vkl + VkT - VkkW]
+ (4/c/c7/')"V6[l - V&/ - VF77]3
Proo/. Since R2(l,k) = l' + k and Af^/, A:) = 2/(1 + /') = 1 + k, we
have
3e2(/, k) = (1 + fc)z(l + O - 2(1 + k:) + (1 + *:)(/' + A:)
and
v*\
3e2(/, /:) :
4(1 + O + 2(1+ /'")-4(1 + I")
>\2
(! + /'>
since/'+ A: = (1 + /'2)/(! + /')■ Thus
For example,
A*(4) = (V2-1)2 and a(4) = 2(V2 -1)2

160 Modular Equations and Algebraic Approximations to n
Proposition 5.4
If /:= A*(r) and fc:= A*(9r), then
(i) M;\l,k) = yjl + 4^p-=:s(r)
and
(5.2.17) (if) a(9,) = /W«W-^^±^^
where m(r) : = 3 /s(r) satisfies
r~\ fa ^ ([^W ~ if3 + I)2
(ui) m(9r) = ————7^ — .
m(r)
Proof, (i) and («7) were established in Section 4.7. For (h) write
3e3(/, fc) = /(/-)(1 + I2) + 5(r)(l + kl + k'V) - 3(1 + k2) .
Substitution in terms of a, as in Section 4.6, and some easy algebra yield
s(r) = 2a + 1 and
e3(/, fc) = 2a(a +2) = [s(r) - l][s(r) + 3] /2
as claimed. □
A convenient variant of (5.2.17) is
(5.2.18) 8(9r) = s2(r)8(r) + 2VTs(r)
where, as before, 8(r) = V7 - 2a(r).
example 5.3. If /-:=3, then as in Section 4.6, m(r) = s(r) = V3. Then
(5.2.17) gives a(3) = 3a( £ ) - 1. But (5.1.5) shows that a(3) = V3 - 3a( 5).
Thus a(|) = (V3+ 1)/6 and a(3) = (V3-1)/2. Now we have m(3) =
(21/3 + l)2/V3 so that s(3) = (41/3-l)V3 and 6(27) = 3(24/3 - 1). Thus
a(27) = 3[(V3 + 1)/2-2173].
Proposition 5.5
If /:= A*(r) and fc := A*(16r), then
(1) Vk = 77==
1 + ^1-/2

5.2 Calculation of a
161
and
(5.2.19) (ii) a(16r) = (1 + y)*a(r) - 4Vry(l + y + y2)
where y ■ = Vk.
Proof. This may be derived similarly but is easily deduced from
Proposition 5.2 and the quartic iteration of Exercise 3 of Section 1.4. □
example 5.4. For r: = 4 we have a(4) = 2( V2 - 1) and A*(4) = (V2 - 1)2
Thus A/A*(64) = (1 - 25/8VV2-l) /(1 + 25/8VV2 + 1) = (VVl+T - 25/8) I
(VV2 + 1 + 25/8) and a(64) = 8[2(V8 -1)- (21M - 1)4] /(VV2 + 1 + 25/8)4
which gives eight digits of tt~\ Similarly, VA*(16) = (21/4 - 1)/(21/4 + 1)
and a(16) = 4(V8-l)/(21/4 + l)4.
Combining some of these calculations with (5.1.6) we have established
that
i-<**-(^
= k( Vie
k: =
k: =
V3-1
2V2
3-V7
4V2
and
J = K\3V3 E -
V3-1
+ 21I3)]k)
A::=A*(27)
with a similar identity whenever a(r) and \*(r) are known.
Comments and Exercises
Computation of A*(r) will be discussed further in Section 9.2. In many of the
following numerical exercises the algebra is not entirely straightforward.
1. a) Verify that the quadratic case of (5.2.1) coincides with the formula
given in Theorem 1.2.
b) Verify (5.2.6).
c) Verify that R2, R3, and i?4 are as claimed.
2. a) Show that in terms of Ramanujan's Gn and gn of (3.2.13) one can
write
i) A*(„)=iG„-3[VOG„-6-V^-G„-6]
ii) A*(n)=HVl + G„-12-Vl-G;12]
iii) A*(n) = gUVs" + ^12-s£]-

162 Modular Equations and Algebraic Approximations to n
b) Find similar identities for A*'(").
c) Verify the singular values given in (4.6.10).
d) In each following case, given G„ or gn verify kn. Then verify Gn or
8n-
i) G35 = 2~1/4(V5 + 1),
kl5 = (3- V5)(V5 - V3)(2 - V3) /(8V2)
ii) G35 = V5 + 2, k25 = (V5 - 2)(3 - 2 ■ 51/4) /V2
iii) g66 = V2 + 1, fc6 = (V2 + l)(V6-V2-l)
= (2-V3)(V3-V2)
iv) g6l0 = V5 + 2, A:10 = (V5 + 2)(3V2-V5-2)
= (VlO-3)(V2-l)2
v) gi8 = 5 + 2V6, A:18 = (5 + 2V6)(7V2-5-2V6).
3. Use (5.2.6) with p '■ = 3 to compute a(3).
4. Use (5.2.15) to obtain the following values of a (or 6).
i) 6(5) = V2(V5-1)
ii) 6(11) = [2x2 -(*-§) + VlTVl-(*-§)2] /3
Note: G^12 = * - § where *4 - x3 = 2.
iii) a(15) = (VB-V5-1)/2.
5. Use (4.6.8) and (5.2.6) to compute 6(13) = (7 +3VT3)G1"36 where
G1"34 = (Vl3 -3)/2.
6. Show that Rp(l, k) = /?,(*', /'), that Rp-i(k, I) = -p^R^l, k), and
that d(p'l) = -a(p)lp.
7. Use (5.2.14) to show that
i) 6(25) = 10-51/4(7-3V5)
ii) 2A*(49)A*'(49) = (1863 + 704V7) - (810 + 306V7)V273/4
'71/4-V4 + V7y
and
«(49) = I - VV[V273/4(33011 + 12477V7) - 21(9567 + 3616V7)].
8. Use Proposition 5.3 to calculate
i) A*(8) = (V2 + 1)2(1 - VV8-2)2,
a(8) = (20 + 14V2)(1 -VV8-2)2
ii) A*(12) = (V3-V2)2(V2-1)2,
a(12) = 264 + 154V3 - 188V2 - 108V6.

5.2 Calculation of a 163
9. Use Proposition 5.4 to show that
i) a(18) = (21-6V6)^8A*(18)
ii) 6(9) = 33/4(V6 - V2), as m(l) = V3 + V12
iii) 6(81) = 9V2 31/4(V3 + 1)(3 + a)a'\ where
a:=[(2 + Vl2)1/3 + l]2.
10. Prove Proposition 5.5.
11. a) Use (5.1.4) to write
-i- ^,,,2 AG(l,fc') , a(p)-Vpk2
' -^^ X5(l^)5+ AG(1,*')2
where k'-= \*(p) and the derivative is with respect to k.
b) Then (as in Section 2.5) show that
tt = lim
1
Vpk'k2Pi + [a(p)-Vpk2]qi
where x0 : = 1/k', qx '- = x0, yt :=y/~x~0, and pl := x2al{xQ + 1) while
. VxZ + 1 /Vx~
y n+i ■
2
ynYx~ + l/Yx~
P"^-\lTx~JP"
In
Qn + l'-— v
An
c) Show that convergence is quadratic.
d) When p :=1, this is Algorithm 2.1.
The next exercises develop results in J.M. Borwein [85] and Ramanujan
[14]
12. a) Show that with P as in (5.2.8),
pP(e~'vli)+P(e-"vp)
-Wp\ , p/\,-w/VP\ _ ^Vp
7T
Hint: Let q := e-w? in (5.2.10) before differentiating.
Hence show that
7T

164 Modular Equations and Algebraic Approximations to jt
b) Show that
(5.2.20) P(e~^) = ^ AG-2(1, A*'(p)) + ^~=
Thus P(e~m/^) is quadratically computable.
c) Evaluate P(e-,r3).
13. Show that
(5.2.21) - = 1-2 p'
^r^Rpi^k^-ip-l)
where Kt:= K(kt) and A:,. := A*(p2'). In J. M. Borwein [85] these
identities are studied in detail.
14. Show that when p'- = 2, (5.2.21) is equivalent to Gauss's identity
(2.5.9).
15. a) Show that
P(q) =
K(k)
3 | (k) -(1 + k'2)
This identity, which follows from Exercise 7c) of Section 3.7, is
entry 2 in Chapter 18 of Ramanujan's second notebook,
b) Prove that
CO
P(q) = 1-24^ cr^q2".
Here cr^ri) is the sum of the divisors of n, as in Exercise 12 of
Section 3.7.
5.3 FURTHER FORMULAE FOR a
We commence by establishing a multiplication result for o\
Theorem 5.4
Let p, r>0 be given. Let /:= A*(l/pr), y-=\*(plr), and k'-= \*{pr).
Then
(5.3.1) tr(pr) = M~r\y, *)[*„(/, y) + Vf ^(% *)
and
(5.3.2) *(£) = Af,(% *)[*„(/, 7) - Vf *,(?, *)

5.3 Further Formulae for a 165
Proof. Let s : = pr and ql = q1/r. Then y = k(ql) and 1= k(q\/p). We
have
i/p\
flfol)^") J 03(9)
PP(^)-P(^i'P)
+ P
rP(g)-P(g1/r)1 0¾^)
L 03(9)0'3(9l/r)
03(91/S)
so that with r: = K(y),
(5.3.3) /?,(', *) = RPV, y) j; + pK(y, k) j-
For k' = I we have L = Vjp^ and (5.3.3) becomes
<*& = £
RPV,y) + y%Rr(y,k)
This is (5.3.1). Then (5.3.2) follows from (5.3.1) and Exercise 6 of Section
5.2. (See Exercise 1.) □
For p-r, (5.3.1) reduces to
2Rp(l/V2,k)
(5.3.4)
-(P2)=^P
Mp(l/V2,k)
k:=k(e~"p).
Corollary 5.1
Let p > 0 and k ■= A*(2p). Then
¢.3.5) S(2p):=Vf^ + ^«,(^|,S).
Ptoo/. Observe that M~\y, k) = 1 + k and i?2(7, fc) = * + 7', while
y' = (1 -/:)/(1 + *). Then ' 7'
<r(2p) = (1+ fc)[^(/, y)+yfl(k+yl)
and since Rp(l,y) = Rp(y',k),
(5.3.6) .(2/0 = (1 + k)Rp(i^|, /c) + Vf (1 + fc2)
Now (5.3.5) follows from (5.2.15). □

166 Modular Equations and Algebraic Approximations to n
example 5.5. For p: = 3, k = (2 - V3)(V5 - V2) while /?3(y,A:) = l +
7& + 7'fc'. Now
(1 + k)R3(j^j ,kj=2k + k'2 + 2Vkk'
which we rewrite as 2k(l + V2g66 + gf). Similarly \l\k'2 = Vlgfk, and we
can write 6(6) = [V6(V2 + 1)2 + 2V2(V2 + l)]k. Thus
8(6) = ggA*(6)(2V2 + 2V3 + V6)
and
«(6) = g6A*(6)(3-V2) = 5V6 + 6V3-8V2-ll. D
Corollary 5.2
Letp>0and/:=A*(l/3p), y:= \*(p/3), and k-= A*(3p). Then
(5.3.7) cr(3p) = Af 3^(7, *:)[/?,(7', fc) + Vf (1 + 7^ + y'k')]
and
(5.3.8) cr(|) = M3(7, A:)[k„(7', *) - Vf (1 + 7* + ?'*')_
where
3p
"1(|) = Af 3^(7,^) = )/1 + 2^
Pwo/. This follows from Theorem 5.4 and (4.7.9). (See Exercise
3.) □
example 5.6. By piecing together various formulae many more values of a
can be obtained. We illustrate this as follows. Given a(6) and A*(6) from
the previous example, we may use Proposition 5.3 to compute a(|). Then
the functional relation for a, equation (5.1.5), yields a(§) = (5V6 — 6V3 -
8V2 + 11)/3 = 0.5138118 Similarly, given a(3) and A*(3) from
Example 5.3, we can use Proposition 5.3 to find that a( f)= 66 + 47V2 - 38V3 -
27V6 = 0.5138837 ....
Indeed numerical computation of the maximum of a [using (5.1.10) and
Newton's method] shows that it occurs around 0.709 with a value of
approximately 0.514275. [Note that (| + f )/2 is very close to this point.] In
addition, one may observe graphically that a increases up to this value and
then decreases.

5.3 Further Formulae for a 167
Comments and Exercises
From the given formulae and appropriate singular values, many values of a
can be found in closed form. Some of the cleanest are given below.
1. Establish (5.3.2) of Theorem 5.4.
2. Use Corollary 5.1 and the given value of gn to compute a(ri) or S(ri).
i) g?o=V5 + 2, a(10) = g?0A*(10)(3V5-4)
ii) g?8 = (V3 + V2)2, «(18) = g?8A*(18)(21-6V6)
iii) g222=(V2 + l), a(22) = g«2A*(22)(33-17V2)
iv) g28 = (V29 + 5)/2, a(58) = g568A*(58)(99V29-444).
(This requires having a tractable form of W29, which we have not
given but which may be found in Greenhill [1892].)
v) gl42gl4 =2 + V2, 6(14) = gt4A*(14)[(8 + 6V2) + Vl4>n
vi) g3° g3° =6 + 5V2, 6(30) = g*0A*(30)[(56 + 38V2)
6.-6
vii) g46 g46 = 18 + 13V2, 6(46) = g46A*(46)[(200+142V2)
g646A*(46)[i
+ V46g646].
3. Establish Corollary 5.2.
4. Use the values of G3p, Gpl3, and M(p/3) given in Exercises 8 and 9 of
Section 4.7 to prove that:
i) o-(21) = 3G2-16(11 + 6V3 + 2V7 + V21)
ii) o-(|) = G7/63(11-6V3 + 2V7-V2T)
,_6 /2V3 + V11
iii) 0-(33) = 3 G;36tJ (11 + 13V3 + 5VII+V33)
) ^^r.W^^ai + ^-svn-vsj)
,_6 /2VI9 + 5V3
v) o-(57) = 3 G576y (5VB7 + 13V19 + 49V3 + 19)
) .(f)-G^?™^(SV57-13Vl9 + 49V3-19)
and

168 Modular Equations and Algebraic Approximations to n
IV3 +1 \3
vii) o-(93) = 6G~36^——J (15V93 + 13V31 + 201V3 + 217)
viii) o-(y)=2GJ16/3( 32-1) (15V93-13V31 + 201V3-217).
Hint: In each case express the right-hand-side of (5.3.7) or (5.3.8) as a
function of (yy'kk')1'4 by using the cubic modular equation (4.1.16).
5. The values of a given in Exercise 2 are all expressed in the form
a(p):= gpA*(p)ap for a quadratic number ap. This is also true for
some other even p.
a) Since g6p\*{p)~l/(2g6p), deduce that
«,:=-£-«
and estimate the accuracy of the approximation,
b) Show that
_ 14 + 10V2 140 + 26V29
77-22 ~ 33 - 17V2 and ^58 ~ 99V29 - 444
which give four and eight digits of ir, respectively.
6. a) Weber gives G27 + G~2 = (1 + VT7)/2 and gj4 + ^m = (3 +
VT7)/2. Hence evaluate o-(17) and o-(34) as cleanly as possible,
b) Similarly, /(V11!^) =: x solves *3 = 2x + 2 and £0/=38) =: V2x
solves x3 = 2*2 + (2x + 1)(V2 + 1). Attempt to evaluate a(19) and
a(38).
7. Use (5.3.4) to evaluate 6(25). (Compare Exercise 7 of Section 5.2.)
8. Given that G% = V37 + 6, one can show that
6(37) = (101 +21 V37)G
-6
37
Again we have not given a tractable form of W31.
Show that the perimeter of an ellipse with major axis a and eccentricity
k is given by
2a
V?
V +[V?+ 8(r)]K(k)
L2K(k)
where k'-= X*(r). In particular, if k '■ = tan (71-/8)
flV?
r(l) + r(|)
r(i) r(|)J

5.4 Recursive Approximation to tt 169
(See Exercise 4 of Section 1.6.) If k ■ = 1 /V2 = sin (ir/4)
fir
p = a-
r(!) r(|)J
Further such evaluations follow from Table 9.1, which gives K(\*(n))
in T terms for 1 < n < 16.
5.4 RECURSIVE APPROXIMATION TO tt
Theorem 5.3 is easily recast as a pth-order iterative method to compute tr.
A General Iteration 5.1
Let p^l be a positive integer. Let r > 0 and set
aQ:=a(r) and kQ '■= A*(r) .
For n in N compute kn+l by solving Wp(k2n, k2n+l) = 0 and let
mn:==M~\kn,kn+l) rn:=Rp(kn,kn+l)
sn:=[mnrn + m2n(l + k2n)-p(l + k2n+l)]/3
and
(5.4.2) o„+1:=mX-p"Vre„.
Then, for rp2">l,
(5.4.3) 0 < «„ - tt"1 = 8(pnVr - tt"1) e^"^ + 0{p2nr e'""2^)
sl6p"Vrg"i',VF*.
Proof. This is a straightforward consequence of Theorem 5.3 and
(5.1.11) because an : = a(p2"r). D
We may also write this as an identity for v.
Theorem 5.5
Let p ^ 1 be a positive integer and let r > 0. Let m, and e„ be as in (5.4.1).
Let aQ := 1 and an+1 := m~1an. Then
(5.4.4) ,._J^ki^_
a(r) - Vr E„_0 p a„+1e„

170 Modular Equations and Algebraic Approximations to n
Proof. Rewrite (5.4.2) as
(5.4.5) a2n+lan+l = »««« - ^rp"a2n+isn ■
Then summation yields
m
«0)-"i+i«m+i = V7:2 p"a2n+le„
n-0
and, as am converges to it 1 while am converges to AG(1, k'Q) [since
am = K(km)tK(ka)], we obtain (5.4.4). D
EXAMPLE 5.7.
a) When p' = 2, the proof of Proposition 5.3 shows that sn = 2kn+l. In
terms of the AGM: a2n+lsn =4a„+1c„+1/2 = c2nl2, and (5.4.4) gives a
family of identities extending (2.5.9), or Algorithm 2.2 (the case r : = 1).
b) When p : = 3, we similarly use Proposition 5.4 to deduce that a2n+lsn~
(an - fl„+1)(fl„ + 3an+1)/2, where an+l = m~1an can be computed from
the cubic recursion in Proposition 5.4. (See also the cubic iteration
given below.)
We now specialize Iteration 5.1, changing notation as appropriate.
A Quadratic Iteration 5.2
Let r>0. Let a0 := a(r) and k0 ■= A*(r). For n in N let
(5-4.60 k"^:=TTK
and
(5.4.6b) «„+1 := (1 + kn+lfan -2"+lVrkn+l.
Then, for a-22">1,
0< a - rr~l ^16-2"Vr e
-2"v7ir
A Quartic Iteration 5.3
Let r>0. Let a0 := a(r) and y0 : = VA *(/■). For n in N let
l-</\
y
(5-4-70 y„+i:~ 4/l :
and
(5.4.7b) «„+1 : = (1 + yn+l)\ -4"+1Vry„+1(l + j„+1 + y2n+l)

5.4 Recursive Approximation to jt 171
Then, for M2";
-!_..,; All ,— -4"V?1T
0<a„ -ir <16-4"Vre
Iteration 5.3 just peforms two steps of Iteration 5.2 as one, with some
computational saving. The error bound given in each case is very accurate.
A Cubic Iteration 5.4
Let r >0. Let a0:=a(r) and m0: = m(r) = \/l + 2V2G3gr/G9r. For n in N let
(5.4.8j) ,^.^-ir+if ±
and
(5.4.8«) an+l:=s2nan-3"Vr^
Then for a-32">1,
si + 25-3
-1--1^ 1« .r- -3"V7ir
0<a„-7r"1<16-3"V7e
One can also provide a cubic iteration using kn instead of sn. This is
somewhat more inelegant and no easier to initialize. [See Exercise 3b).]
Large numbers of initializations for Iterations 5.1, 5.2, and 5.3 are
available in the previous sections. We collect some of the cleanest in Tables
5.2a) and b). The cubic information is given in Table 5.3. Recall that
[m(N)-l][r(N)-l]=4.
All the information necessary to verify these values lies in Section 4.7 and
the previous sections of this chapter.
When p : = 7 we can write the iteration cleanly in terms of w„ := klJ*.
Indeed
(5.4.9) (l-«„«„+i)8 = (l-«*)(l-«*+i)
while
7"A+i(l - «„«„+i)[l - «„«„+i + ("„"„+i)2]
m„ .=
4
and
r„ : = 3[1 + (uBun+1)4 + (1- «„uB+1)4]
. ml(l+u8n) + mnrn-l(l+u8n+l)
*-•" 3

TABLE 5.2a. Values of G~12 and a(N) for N odd
N 2A*(JV)A*'(iV) = G~12 a(N)
1
2
V3-1
1
3
5
7
9
13
15
25
27
37
1
1
2
V5-2
1
8
(2-V3)2
5V13-18
sm
(V5-2)4
(2l/3 _ 1}4
2
(V37 - 6)3
V5-V2V5-2
2
V7-2
3 - 33/4V2(V3 - 1)
2
Vi3 - V74VI3 - 258
2
VB-V5-1
5(l-2-51/4(7-3V5))
2
3(V3 + l-24/3)
2
37-(171- 25V37)(V37 - 6)1/2
2
TABLE 5.2b. Values of g~12 and a(N) for JV even
N 2\*(N)/\*'2(N) = g-12 a(N)
2 1 V2-1
2(V2 -1)2
6 (V2-1)2 (V5-V2)(2-V3)(3-V2)(V2 + 1)
10 (V5-2)2 (7 + 2V5)(VT0-3)(V2-l)2
18 (V3-V2)4 3(V3+V2)4(V6-l)2(7V2-5-2V6)
22 (V2-1)6 (V2 + l)6(33~17V2)(3V22-7-5V2)
58
1
2V2
(V2-
(V5-
;V5-
(V2-
'V29
-I)2
-2)2
V2)'
-I)6
-^
/V29 +5\6
( = ) (99V29 - 444)(99V2 - 70 - 13V29)
V
172

TABLE 5.3. Selected cubic invariants
m(N) r{N) a(N)
1019 + 416VS - 720V2 - 588V3
6
47V2 + 27V6 - 38V3 - 66
3
8V2-4V6-6V3 + 11
6
V3 + 1
6
1
2
5V6-6V3-8V2 + 11
3
1
2
2[54V6 + 77V3 - 94V2 - 132]
3
V2-1
16(3V6 + 4V2-2)
1/18
1/12
1/6
1/3
1/2
2/3
1
4/3
2
8/3
3(3V6-4V3-5V2 + 7)
6V3+9-4V6-6V2
(3-V6)(V2+l)
V3
V6-V2 + 1
V3(V2-i)(V3 + V2)
V3 + 2V3
V6 + V2-1
— ■
-
(3 + V6)(V2 + l)
3 + 2V3
V5+V2 + 1
V6 + V3
—
V3(2 + V3)
V3 + V2
\/3(V3 + V2)
3[V3 + 1 + V2(V3-V2)]4
[1+21/3]2 _ V3-1
V3 2
4
5
6
7
8
9
18
—
V1+2V3 + 2V5
-
/ 6 + V21 + V27 + 6V2T \ "2
[(2V3 + 2)"3 + i]2(V3-i)
V2-31'4
-
V3 + V3 + V9 + 6V3
[V2(V2 + l)2/3 + l]2(V2-
V3
V(V2 + 1)(2 + V3)
-
1)
(V3-V2)[l + (4 + 2V6)"3]2
6-4V2
V5-V2V5-2
2
5V6 + 6V3-8V2-11
V7-2
2
(2 + V2)3(l-V2V2-2)2
3-33"V2(V3-l)
2
3[721 V2 - 1019 + 588V3 - 416V5]
173

174 Modular Equations and Algebraic Approximations to jt
We present a few illustrations of the behaviour of the algorithms. We
note that an in the quartic algorithm is just a2n in the quadratic algorithm.
r = 2
r = l
Digits Correct in Quadratic Algorithms
n=\ 2 3 4 5 6 7
0
2
5
3
5
12
8
13
26
19
28
55
41
56
112
84
120
227
171
242
458
344
489
919
Digits Correct in Cubic Algorithms
n=\ 2 3 4 5
r=l
r = 7
2
8
10
30
34
93
107
288
327
873
Digits Correct in Septic Algorithms
n = \ 2 3
r = l
7
22
63
173
464
>1000
The updates for a(25r) and a(49r) are studied again in Section 9.5, where
solvable versions of the quintic and septic iterations are given.
Comments and Exercises
Additional information on these iterations can be found in various of our
papers. In particular, Borwein and Borwein [86] indicates the genesis of the
quadratic iterations and [84b] that of the cubic iterations.
1. Verify the claims in Example 5.7.
2. a) Prove that, in the notation of Theorem 5.5,
E
-(/:) = 1-2 p"a2n+lsn k=:kQ.
b) Observe that this extends Algorithm 1.2.
3. a) Verify Iterations 5.2, 5.3, and 5.4.
b) Use (4.1.24) to produce another form of the cubic iteration.

5.4 Recursive Approximation to -n
175
a) Determine the initial values for the cubic iteration with r '■— Nl 3
and N: = l, 11, 19, 31.
b) Compute 6(45), 6(63), 6(243), and 6(54).
Derive the following version of the septic (seventh-order) iteration. Let
a0 := a(r) and u0 '■= A*(r)1/4, and generate un+l decreasingly from
Let
(l-u8n)(l-u8n+l) = (l-unun+l)8
u„u„
ft ••=
' ™n n + 1
8
u — u u ,,
n ™n n + l
and
«„+i : = *n«n +7"V7(7 -sn - tn).
In .
Then for rT" > 1
0<a-w x<\(s-TVre
-7"v7ir
Derive the following version of the quintic (fifth-order) iteration (given
in Hughes [84]). Let a0:=a(r) and u0 = A*(r)1/4, and generate un+l
from u6n+l -un- 5(w„w„+1)2(w„ - u2n+l) - 4w„w„+1[l - (w„w„+1)4] = 0.
Let
yn
an
bm
o 5
™n™n+l
^u2n + 5u2n+l + 2xn
= 5u2n + u2n+l-2yn
6-
dm: =
._ (1 ~ M«+i)[5(w'+i + *„) + c„(y„ - w*+1)]
4a„
+
(1 - u*n)[u2n + xn + 5cn(yn - u2n)}
4b„

176 Modular Equations and Algebraic Approximations to n
Then
satisfies
for r52">l.
«« + 1 := 5cnan + 5" + W^„ + Mn + 1 "~ CnUl)
< a„ — 7r :£ 16 • 5 V7 e
7. Show that, in Iteration 5.1, convergence is indeed pth order.
Observe that for a variety of other values of p we have the information to
make Iteration 5.1 entirely explicit. For example, p '•= 17 is satisfactory since
we have i?17, and M17 is given in (4.6.9). Moreover, (4.6.9) also gives a
form of the modular equation of order 17. Using G\7 + G^12 =40 + 10VT7
and
^(17)= (llV?+45>
(42 + iiVn)V13+25VI7-(W7+277VI7>
we have an algorithm for ir which gives more than VT7 • 17" digits at step n.
8. Let pal, let /:=A*(p) and fc:=A*(49p). Set zp : = (GpG49p)~V
V2.
a) Show that
M7(/, fc) =
Vi-
Vi-
2zp(l
.-,-24
_Cj49p
-4z -
-¾)
-Vi
Vi-
-4z
p
y-,-24
GP
UzJl - z.)
b) Ramanujan [14] gives
and
,_i V4 + V7-71
'49
G147 -2
\2 + V3
Verify that
V7
(28)1
vt^
where zl = G493/V2, and

5.5 Generalized Elliptic Integrals 177
«) M7-(,.(3).A.(14,))-vi^:^/2
where
_ [V7 + V3-2(28)176]3
H 48V3
c) Compute that
i?7(/,fc) = 6(l-zp)2.
9. Verify, from Table 5.2, that
i) a(22) = (V2 + 1)3(33 - 17V2)(3VII - 7V2)(10 - 3VTT)
and
/V29 + 5\3
ii) a(58) = (V2-l)6(13V58-99)(99V29-444)(^^ J .
10. Ramanujan gives the following form of the modular equation of degree
7:
(5.4.10) (04 + (J)' + 7 = 2V2{M3 + (xyy3}
where x : = G49N and y := GN. This is Entry 19(ix) in Chapter 19 of the
Second Notebook (Berndt [Pr]). One should compare Exercise 6 of
Section 4.1 and Exercise 12 of Section 4.7.
a) Verify G7, G49, and G147.
b) Show that
(5.4.11) (j)4 + (^)4-7 = 2V2{(^)3 + (^r3}
where x-= g49N and y := gN.
c) Verify that g2u + gu=V2 + l and that
+ gss1 = 2 {V2 + V14 + 4V14}
5.5 GENERALIZED ELLIPTIC INTEGRALS AND RATIONAL
AND ALGEBRAIC SERIES FOR 1/ir and 1/K
We begin with some results on hypergeometric functions [see (1.3.5)].
Changing notation slightly we write

178 Modular Equations and Algebraic Approximations to n
V (a)n(b)n *"
(5.5.1) JP^a, b; c; x) ■= Z, —tr— -7
n-0 \c)n m
for appropriate values of the variables. Here we use the rising factorial or
Pochhammer symbol (a)„ := T(a + n)/T(a) = a{a + l)(a + 2) • • • (a + n - 1).
Similarly the generalized hyper geometric function 3F2 is defined by
(5.5.2) 3F2(«, 6, c; d, e; *) .= 2 ^^ ^
again where appropriate. (See Slater [66].) We define the generalized
complete elliptic integrals of the first and second kind by .
(5.5.3) K,{k) := f • 2Fl(\ -s,\+s;l; k2)
and
(5.5.4) E,(k) := I • 2FX(- \ - s, \ + s; 1; fc2)
for |s| < I and 0<k^l. We still denote the complement k' ■= Vl — fc2 and
write K;(fc) := £,(*') and E;(fc) := £,(*)• Now K "•= ^o and E := £0 are the
classical elliptic integrals, and each Ks and Es admits many integral
continuations. (See Erdelyi et al. [53, Section 2.12].) Moreover one has
kk'2
(5.5.5) Es = k'2Ks + — Ks.
This may be verified directly or by using Erdelyi et al. [53, Section 2.8].
[Here K, = {dldk)Ks.} Similarly, using Erdelyi et al. [53, vol. 1(13), p. 85]
we have
(5.5.6) E.K> + Kn-K,K>=l?gg.
When s = 0, this is Legendre's relation (Section 1.6). The following
relationships will be helpful.
Proposition 5.6
For 0< h < 1/V2 we have
(") ~Ks{h) = 2Fx{\-^\+S-;l;{2hh'f)
(b)
TT
= ,F2[\- s,\ + s,\;l,l;{2hh'f).

5.5 Generalized Elliptic Integrals
179
Proof, (a) is a special case of Kummer's identity given in Rainville
[60, p. 67] or in Erdelyi et al. [53, Section 2.11]. It may be verified by
showing that both sides satisfy the appropriate hypergeometric differential
equation (given in Exercise 7 of Section 1.3), are analytic, and agree at
zero, (b) is a special case of Clausen's product for hypergeometric functions
given in Slater [66, p. 75] and Exercise 13. □
In the sequel we will again use Ramanujan's invariants of (3.2.13)
(5.5.7) G:=(2kk')~in2 g-={2klk,2yin2
and
2U4gG = (k2/2k'yUl2 ■
We also need Klein's absolute invariant J, which was introduced in
Theorem 4.4. This is
Ysstt (4G24-1)3 (4g24 + l)3
*" j 27G24 27g24 '
Ramanujan [14] talks about "corresponding theories" for Ks, s •"= 3,4, g,
to that for K. For 5:=5,3 this is explained by the next result.
Proposition 5.7
(a) Kll4(h) = (l + k2)U2K(k)
if 2M' = [(g12 + g~12)/2]_1 and 0<fc<l/V2, 0<A:<V2-1.
(b) KU3(h) = [l-(kk'f]U4K(k)
ii2hh' = J~V2 and 0<fc<l/V2, 0<A:==1/V2.
Proof. These may be discovered by piecing together the quadratic and
cubic transformations given in Erdelyi et al. [53, Section 2.11]. They may be
verified by establishing that both sides satisfy the same differential equation
(derived from the appropriate hypergeometric differential equation), and
both functions involved have the same finite value at zero. D
There is a corresponding relation for Kll6. Since it is a little less concise,
we consider it at the end of the section. Combining these last two
propositions leads to a variety of alternate hypergeometric expressions for K and
K2.
Theorem 5.6
2K /1 1 \ 1
(fl) (0 —{k) = 2F\-A,-A;l;{2kk')2) °^^^vf

180 Modular Equations and Algebraic Approximations to jt
(ii) fw^-^H;!;-!)2) 0^V2-1
(Ui) ^^) = k^%Fi}A,\-,l-,-{-^)2) 0^k2^2V2-2
(b) (iv) ^(^) = (1 + ^)-^(^^^(^-^)-2
(") vw=(^2-^2)"1/2^(iI;1;-/Gl2 G"2x"2
0<A:<V2-1
)1
0<A:<
21/4-V2^V2
(c) (,i) ^(k) = [l-(kk')2]-1'\Fl(^,-^;l;r1) O^fcs^
V2
Proo/.
(a) We let 5 := 0 above to deduce (i). Then («) follows on replacing q by
—q in the theta function representations of K and (2kk')2. This is
Jacobi's imaginary transformation of Exercise 7d) in Section 3.2. We
derive (Hi) from (ii) by replacing /c: by kl-= (1- k')/(l + k') and
using the quadratic transformation K(kr) = [(l +k')/2]K(k) of
Theorem 1.2.
(b) (iv) comes from letting s'-=\ above. Then (v) again follows from
Jacobi's imaginary transformation.
(c) (vi) comes from letting 5:=3 above. □
Similarly,
Theorem 5.7
For k restricted as in Theorem 5.6
(a) (i)
(ii)
7T
2K
(k)
= ,F2(\,\,\;lA;(2kk'f)
= Z,'~2

5.5 Generalized Elliptic Integrals
181
(b)
(c)
(iv)
(v)
(vi)
2K
2K
«]-(1^)-^4 f.lu.iKS41-)")
Proof. We combine Theorem 5.6 and Proposition 5.6. □
Thus we have provided series for K and K2 in terms of each of the six
invariants. One can produce other such formulae by further use of
transformation identities. For example, Bailey's formula in Erdelyi et al.
[53, (2), Section 4.5] with a'.= \ and b-= 1 gives
— (k)\ =[l-4(2kk )] 3¾ 6'2^^-4(2/^13J
(5.5.9)
= {k>* + l6ey^F2[\,\,\-,l,l;-21g
3X2\6'6'2'
(r + 4)
+ 4?)
Note also that we may use (5.5.5) with $:=0 and Theorem 5.6 to
produce similar series for E. We are now ready to build our series. Recall
(5.1.4), which we write as
1 AKK AK2
- = VNkNk'N2 — + [«(A0 - VNk2N] — kN := A*(AT)
(5.5.10)
or
1 AK AK
- = VNkNk'N2 — + [«(Af) - VNk2N] — kN := A*(iV) .
(5.5.11)
Thus given a(N) and A*(AT), we can combine (5.5.10) with Theorem 5.7 to
produce series for 1/tt. In like fashion we derive series for 1/K or for the
Gaussian AGM, M(l,k')= ir/2K. In each case we have [(2KI>n)(k)]2 =
m(k)F(<f>(k)) for algebraic m and <f>, while F(<f>) has a hypergeometric-type
power series expansion E~_0 an<f>". Then AKK/tt2 = |mF + jm.(j>F(<f>).
Substitution in (5.5.10) leads to
- = 2 an -~ kk'2m + [a(N) - VNk2]m + -^- m J kk'2
<f>
<t>".
(5.5.12)

182
Modular Equations and Algebraic Approximations to jt
Thus for rational N, the bracketed term is of the form a + nb with a and b
algebraic. We now specialize this for our invariants.
series mGjy: For N>1,
1
(5.5.13) - = 2
where
TT
n-0
(*)■
n\ J
~12\2n
an(N)(G-Nli)
an(N) := [a(N) - VNk2N] + nVN(k'N2 - k%) .
series in gN: For AT s= 2,
(5.5.14)
where
1
^ = 2 (-i)"
T „=0
r/-1
(_a
n! J
12N2n
bn(N)(g-N")
bn(N) := «(A0^~2 + "Viv(^|f) .
series in gm = 2 gNGN: For N > \,
(5.5.15)
^2(-1)-
T „ = o
G)„
n! J
c„(iV)(g4^2)2"
where
c„m-
a(N)~^fk2N
k'N-' + nVN(k'N+ klT1)
series in xN := (g» V*")"1 = (T+W ForAf>2'
(5.5.16)
1_ _ Y (■*)n(2)n(4)n , />A 2n + l
_ _ ^ ,- .^.3 an\ly)XN
TT „_0 (n!)
where
W:=
<*(#)*/ _ ViV _12'
1 + 4 4 8n
+ nVN
8%-g-12
Z)

SERIES in yN:
(5.5.17)
where
5.5 Generalized Elliptic Integrals
■ 4(kNkN)
183
- = 2j (-1) 7-7^ en(N)yN
(my
en(N):
a^y-1 VN
k'J-kl 2
'AT*-7 AT
!(^)
+ nsmiG" + G:"'
series in JNl. For N > 1,
(5.5.18)
where
A_Y (6)n(l)n(6)n ^ /An/ r-l/2y2n + l
H" «=0
w
w)(j-Ni,r
fn(N) := 3^ [ViVVl-G-24 + 2(«(AT) - VNk2N)(4G2N4 - 1)]
+ nVN 3Vf [(8G" + ^Vl-G^24] •
There are many rearrangements of these formulae. In similar fashion we
may deduce that for all N,
(5.5.19) M(l, k},) =
where
n = 0
mn(N) := [«(iV) - VM^] + nlVNk'J
and for N>1,
(5.5.20) M(l, ^) =
where
CO
7T 2 (-l)X(Af)
n-0
L n! J
7**
In
/-1
nn(N):=a(N)k^ l + n2VNk'N
These use the hypergeometric definition of K and (5.5.11). Also using
(5.5.11) and Theorem 5.6(ai) and (aii) leads to, for N>1,

184
Modular Equations and Algebraic Approximations to jt
(5.5.21)
where
and for JV>2,
M(1,^) = tt2 o„(N)
n=0
(U,
n! J
(G^12)2"
on(N) :- [«(AT) - VlM2,] + n2VN(k'N2 - ^)
(5.5.22) M(l,^) = 7r2(-irp„(JV)
where
n! J
-\2\2n
urn
pn(N) = a(N)k'N-1 + n2VN[l + fc^"1 .
Similar formulae exist in the other invariants.
From our formulae for ir-1 and the values of a(N) and X*(N) = kN
previously derived we have explicitly computed all but two of the 14 series
which Ramanujan gives without justification in [14, Section 14]. Ramanujan
gives series of the form (5.5:13) for N '•= 3, 7, 15, of the form (5.5.16) for
N := 6, 10, 18, 22, 58, and of the form (5.5.17) for N := 5, 9, 13, 25, 37. He
gives series of the form (5.5.18) for N:=3 and 7 and two in terms of Kll6
which we derive below. In each case manipulation of the formulae produces
the desired result. Indeed a(37) and a(58) were obtained by calculating
e0(37) and d0(58) to high precision. In fact, with N ■= 58, using (5.5.16) and
Exercise 2 of Section 5.3 produces
(5.5.23) - = 2V2 2 (3)"(2)"(^" (1103 + 26390n)(-^-)
ir „„0 (n!) ^99/
2n + l
irVlV
which adds eight digits a term! Since kN behaves like 16e~"v"', it is very
easy to estimate the convergence rate in each series. For N at all large, the
rate while linear is most impressive. In the exercises we give various other
examples. Bailey [35, p. 96] gives (5.5.14) with AT : = 2 [equivalently (5.5.15)
with N'■= f] and ascribes this to Ramanujan. The series is
(5.5.24)
\ = 2 (-D-
ti)n
n\ J
{An + 1) ,
Correspondingly, (5.5.22) with N'- = 2 yields
(i)J2
(5.5.25) M(l,l/^) = ^-E(-l)"
n\ J
(8« + 1) ,
while for N '•= 1, the series in (5.5.20) diverges.

5.5 Generalized Elliptic Integrals 185
We now return to Kll6. From Goursat's exhaustive list of transformations
(Goursat [1881]) we obtain:
Proposition 5.8
For k < 1 A/2 and h the smaller of the two real solutions of
.2x3
(9-8Ji')J (4G"-1)J
64h6h'2 K ' 21G2
(5-5-26) ^71677^ = ^)-^^4
one has
Proof. Formula (126) in Goursat [1881] and Propositions 5.6 and 5.7
combine to produce (5.5.26) and (5.5.27). □
There is a corresponding formula for the larger solution. This implicit
formula for I in terms of k can be solved explicitly as follows (Exercise 19).
For ft<(V3 + l)/2V2,
(5.5.280 ^:=(MT2 = (^+21/V")2
where
(5.5.28b) x := (^^) + § V3 + 2(71/3 - 1)A - A2
and
(5.5.28m) A: = Vl + J1'3 + J2'3 .
In each case (s '■ = j, \, g ) the transformation from k to h can be
described very simply analytically. Consider the generalized singular value
function A* defined by
(5 5 291 K's(K(N)) rj-
Then it transpires that in all three cases,
/rr^x r7TK's(K(N)) K'( A*(C,JV)) „ „ 2/ x
(5-5-3°) ^ iMm = KindN)) Cs := 4 cos2 (.,).

186
Modular Equations and Algebraic Approximations to jt
(This can be verified from formulae in Goursat [1881].) In other words, the
Mh singular value of K is sent to the Mh, (A/72)th and (A/73)th singular
values of Kll3, Kll4, and Kll6, respectively. Thus
12 _, ~12N -1
(5.5.310 (^2^) =2A?/4(Af)A?;4(A0
and
-1/2 .
(5.5.31«) J-^ = 2X*ll3(N)XX'l3(N),
while various singular values for KV6 are given in Exercise 19b). If we now
define as by
(5.5.32) ^.^2^-^1-0 *:-*:(*>
we may use (5.5.5) and (5.5.6) to write
(5.5.33) ^ = l[as(N) + nbs(N)]
where
(n!)3
G;24"(iV)
(5.5.340 as(N):= [as(N) - VN\f(N)] * + 2"
cos (its)
(5.5.34/0 5!W: = VWl-G;>)^
while
G;12(iV): = 2A:(iV)Ar(iV)
The details are left as Exercise 20. Now, with some perseverance, we can
derive series including Ramanujan's missing formulae, which come with
s:= | and N: = 4 and 5 in (5.5.33) [See Exercise 20b).]
Finally, we observe that the result of Exercise 22 in combination with the
discussion of T values in Section 9.2 shows that the formulae for Kll3, Kll4,
and Kll6 are in essence the only such formulae.
Comments and Exercises
In Section 13 of Ramanujan [14] one finds an explanation of seriesof the
form (5.5.13) without many details. Then in Section 14, with essentially no

5.5 Generalized Elliptic Integrals
187
explanation, he gives his other 14 series. Hardy quoting Mordell (in
Ramanujan [62]) observes that "it is unfortunate that Ramanujan has not
developed in detail the corresponding theories." The explanation as
provided by this section is a bit disappointing, since for all these theories, all we
have are well-concealed versions of the original theory for K. Nonetheless
we can explain all of the beautiful and mysterious series.
1. Prove the generalized Legendre identity of (5.5.6).
2. Prove Proposition 5.7.
3. Verify formulae (5.5.13) to (5.5.23).
4. Show that
-^-iriJ}§f^(im+2imnim)
2n + l
5. Show that
V2 _y
,1/4 ^-1
6. Show that
2
= 2 (-i)"
G)„
OX,
[(3 - V3) + 12n](2 - V3)
4n + l
3tt „=o
Show that in (5.5.18)
/„(2) = (28n + 3)V3/9
[(7 - 2V6) + 28n] (V3 - V2)
8n+2
3\3
while
and
(!)
•C = (£)3
/„(4) = (63n + 5)V6/3
/„(7) = (133n + 8)9V3/4 /^ = (^)3.
8. Show that
9. Show that
4'7iHf0"
n\ J
4'w) = f„!/12"+1>
ti)n
n\ J

188
and
Ml,
Modular Equations and Algebraic Approximations to jt
(ai2,
10. Show that
CO
2V2M(21/4, 2~1/4) = 7T E (-l)"(12n + 1)
n = 0
11. a) Show, using (5.5.13) with n := 7, that
_1 = y (2n\3 42n + 5
* n%\n) 212n+4
This series of Ramanujan's has the property that, as J. Holloway
has observed, it can be used to compute the millionth (binary)
digit of IItr without computing the first half million digits. Note
that the terms are exact binary fractions whose numerators grow
roughly like 26" while the denominators are 16-212^
b) Observe that formula (5.5.23) can be recast as
1 2V2
Siw11103"6390"1^)"-
7T 9801 „=0 44"(n!)4
12. Use Exercise 6 of Section 1.3 to show that when re(c — a — b) > 0,
T(c)T(c -a-b)
2F1(a, b; c;l) =
r(c - a)T(c - b)
[This is easy for re(c)>re(6)>0.]
13. Clausen's product formula is
[2Fx(a, b; a + b + \; z)]2 = 3F2(2a, a + b, 2b; a + b + \, la + 2b; z) .
Prove this by showing that both sides satisfy the same generalized
hypergeometric equation and are analytic at zero, with value 1 there.
14. Verify that
r K(k)dk = r12 r12
J° Vl - k2 J° J° Vl
da d6
(sin2 a sin2 0)
b) ^,>imV(T^)£4(2;)4-]V.

5.5 Generalized Elliptic Integrals 189
Use Exercise 12 and Theorem 5.6(ai) to compute K(l/V2). Similarly
use Theorem 5.6(biv) to compute K(V2 — 1). (Compare Section 1.6.)
a) Verify that
i) sin (tx) = (t sin x)2Fl y -^-, -^-; -; sin xJ
ii) arcsin* = x-2Fl(j, \; \;x )
iii) log (x + Vl + x2) = x ■ 2Ft(k, h \; -x2).
b) Use Clausen's product to deduce that
c) Similarly deduce Euler's formula (Bromwich [26])
. 2 If (2*)2
arcsin x= - 2j
"'»f;)
[This is also the limiting case of b). See also Exercise 16 of Section
11.3.]
d) Find similar formulae for sinh [t log (x + V1 + x2)] and for
log (x + Vl + x2).
e) Establish that
2l0g>(iW!)=i(_-ir'
»f;)
and that
f) Prove that
2 °° 1
7T _ Vi 1
18 «-i 2/2n\
"I J
sin (ttQ _ y (0,,(-0,,
irt «=o (n!)2
a) Prove that
y "'»f;)

190 Modular Equations and Algebraic Approximations to •n
b) Show that the previous integral is £(3)/10. (See Exercises 12d)
and 12e) of Section 11.3.)
18. Let Ga(x):=2Fl(a,(2a + l)/2; 2a + l;x).
a) From Clausen's product, establish that
b) Show that G_in(x) = (1 + Vl - x) 12. Thus
(Note that G_1/2 must be evaluated as a limit.)
19. a) Establish the solution of (5.5.28). Hint: x:=h'2/h2 satisfies a
simpler cubic than h2.
b) Now verify that the following solutions obtain:
i) Gf = 2 gives H24 = l
ii) g12 = (V2 + l)2 gives H? = 2
iii) Gl2 = (2 + V3)2 gives H? = 2(2^f
iv) g^=V2(V3 + l)3 gives HT=Y
N ^12 o/^5+1\4 • rT24 125
v) G is =*{—£—) gives ^" = -4-
12 . /14V3 + 13V2\2
vi) g^ = (V3 + V2)4 gives Hf = [ ^= J .
Hint: For iii) and vi) use the increasing form of (5.5.28), which
gives H3N in terms of GN. This entails changing the central sign in
(5.5.28«).
20. a) Establish the general formula for ir~l given by (5.5.33) and
(5.3.34).
b) Verify the following values of aU6(N) + nbll6(N):
i) N •■= 2 gives (6n + l)/3V3
ii) N : = 4 gives (60n + 8)/27
iii) N := 5 gives (66n + 8)/15V3.
c) Compute the values of aU6(N) and bll6(N) for N'-=3 andN' = 6.

5.6 Other Approximations
191
21. By comparing (5.5.33) with (5.5.16) or (5.5.18), verify the assertions
of (5.5.31) for various N.
22. a) Use Exercise 12 to establish that
1 \ T(\ + s/2)T(i-s/2)
K
(£)-
-V2/ 4Vtt
b) As in Exercise 15, compute K(X*(3)).
cos (tts) .
5.6 OTHER APPROXIMATIONS
We begin with equation (5.2.20), which we rewrite as
(5.6.1) VP
~. v> n 1 3 o-(
MM]2
IT' J
where /::= A*(p). Then this shows that
(5.6.2)
--vp- —
2K(k)
+ 0(k4Vp).
Thus on approximating 2K/ir by an algebraic quantity we produce various
approximations for tr. The simplest [which also follows from (5.2.15/)] is
(5.6.3)
-=Vp-^ + 0(k*Vp)
We do better, however, by using Theorem 5.7(a) to write
2K(k)
TT
= 1 + | (kk'f + 0(k4)
so that
(5.6.4)
-=Vp- ^r [2 + (kk'f] + 0(k4Vp) ■
Ramanujan [14] uses different estimates of 2Klir. Motivated perhaps by
symmetry considerations, he uses (3.2.16) to expand [2K(k)/-n-]2 as
(1 -2k2)f1 + 0(k2) and then (3.2.17) and (3.2.18) to obtain
2K(k)
>\2
1 - (kk')
(l-2k2)[l+\{kk'f]
7757 + 0(k«).

192 Modular Equations and Algebraic Approximations to n
Then his approximations to tr are
(5-6>5) ^):=VF-(p)/[2(l-2*2)]
with error O^xTpe"^), and
3
(5.6.6) 7r2(p) := ^_ _ Kp)(4G24 _ 1)]/[(1 _2k2)(8G2* + 1}]
with error 0(\^pe~2'nV^). (See Exercise 1.) Moreover, (5.6.5) and (5.6.6)
produce very simple approximations. Thus
,e,* Mrs 3(3+ V5) , /OCN 63(17+15 V5)
(5.6.7) ffl(25) = -S—-i and .,(25) = ^^ •
The latter gives 11 digits of tr. Similarly,
84 j rt„ <„„ 145V37 + 1134
ffl(37)=2lV^7-101 ^ Wa(37) = 14722399V^7- 41916"
(5.6.8)
(The values of a and a are not quadratic surds.) For even p it is better to
use
(5.6.9) tt3(p) =
Vp ~ [<r(p)(4g" + 1)]/[(1 + k*)(8g" - 1)]
which is again an 0(Vpe_2,rV?) approximation. Thus we derive
(5.6.10) ,(22).^¾^.
v y 3V y 887 +1045 V2
An even more classical approximation to ir is obtained through taking
logarithms of GN or another invariant. Thus in the notation of the previous
section we may write that ir is approximately equal to
(5.6.110 ^log(8G^)
(5.6.11«) ^=log(8^2)
or
(5.6.120 ^,eg (£

5.6 Other Approximations 193
(5.6.12,7) 3s**{B
or
(5.6.13) ^108(1728/^).
We leave it to the reader to estimate the error in each expression. (See
Exercise 4.) For example, when N'■= 58, (5.6.11«) produces
/V29 + 5V
12 , /V29 + 5\
V58l0gl^72-j
which gives 10 digits of tr. Not surprisingly, Ramanujan [14] gives a host of
examples of this kind. When the invariant is large these give very good
algebraic approximations for e17. Following Shanks [82] one can take this
analysis considerably further. In (3.2.9) we gave ^-product formulae for
various invariants. Thus (3.1.4) and (3.2.9«) yield
nx24
(5.6.14) ^:=/i(V^V)-24 = (^) =qn(l + q")
and there is a similar expression for (kk'/4)2. We may expand this product as
a power series and compute as many terms as we wish of its reversion. This
will produce a series of the form
(5.6.15) q = x - 2Ax2 + 852*3 - 35744*4 + ■ ■ • .
We may also take logarithms in (5.6.14). Then we can write
log x + VNtt = 24 2 log (1 + q")= 24 2 ^- qk(l - q")"1
n=\ k=l &
(5.6.16)
which may be expanded as a power series in q. When we substitute (5.6.15)
into (5.6.16) we will recursively compute
CO
(5.6.17) log(\x\) + VNTT=:^ (-1)""1 — x"
for a fixed sequence {an}. Indeed ax '•= 1, a2 : = 47, a3 '. = 2488, a4 := 138799,
a5 : = 7976456, and a6 :=467232200. In fact one can show (Newman and
Shanks [84]) that 24a„ is the coefficient of q" in 11^ (1 + 02*"1)24".
Moreover, 24a„ <64". A now standard trick of replacing q by —q shows that
(5.6.17) still holds for x := -{kk'IA)2. In Shanks [82] several large invariants
are computed (including g3502 and G2737) to which (5.6.17) may be applied.

194 Modular Equations and Algebraic Approximations to n
Each series adds roughly 7rViV log10e digits a term. Hence when we use g469g
given in (4.7.14), we gain more than 90 digits a term! But of course we have
to compute the logarithm as well. Thus in our context the formula should be
viewed as a rapid series for log (|*|) + VNtt, not as a computation of tr.
Comments and Exercises
Approximations (5.6.5) and (5.6.6) were the reason why Ramanujan
computed Rp. He did not give (5.6.9). Many very large invariants, including
G14155, G19947, and G20i55 are derived in Shanks [82].
1. a) Establish that (5.6.5) and (5^6.6) have the claimed errors without
using the rather deep formulae (3.2.16) to (3.2.18).
b) Show, using (3.2.16) to (3.2.18), that
>rrx(p) - ir ~8ire_w3?(ir\/p - 3)
and
7r2(p) - 7T ~247re~2W?(107rVp - 31) .
2. a) Use (5.6.5) and (5.6.6.) to deduce that
.... 3(3Vl3 + 7) 103V13 + 125
^1 (13) = yj and ir2(13) = ^ •
b) Verify (5.6.7) and (5.6.8).
c) Verify that (5.6.9) has the claimed error.
d) Show that
180 + 52V3
^ ^ ~ 45V93 + 39V3T - 201V3 - 217 "
3. a) For even p, the estimate
-¾
Vp- <r(p) /2(1 + kl)
will often be cleaner than irx{p). It also gives 0(Vpe w?) error
in estimation of tr.
b) Obtain
c) Compute 7r3(58) and tt2(22).

5.6 Other Approximations
195
a) Estimate the error in each of (5.6.11), (5.6.12), and (5.6.13).
[Compare (2.5.15).]
b) Use (5.6.12) to estimate tr by
~= log (396) and -~ log (84V2)
a) Show that
and
g-24 = 64q + 1536<?2 + 19200,7 + • ■
64g = q - 24 + 276<? - 2048^2 +
with similar expressions for G24.
b) Thus
nViV
^(8n + 8n24) + 24 = e^ + 4372c-^ +
and
Vn
-Wn
64(G£ + G~Z4) - 24 = e"" + 4372*-™ +
c) When gN or GN is a quadratic surd, this gives an expression for the
integer part of e17 N (and the proximate 0's or 9's). Thus the
integer part of e17^ is 2,508,951. That of e17^ is 199,148,647, and
that of e"^ is 24,591,257,751.
a) Show as discovered by Beukers that, with {an} as in (5.6.17),
i)
ii)
TT
2K(k)A
2K(k)i
Vl - (2kk'y
1 + 24
= (1 +A:2)
2At)
1+24I(-Wip)
b) Combine ai) and Theorem 5.7(ai) to produce a recursion for {fl„}.
c) Show that an is an integer.
A variety of other approximations to tr and to p, the perimeter of an
ellipse, can be found in Chapter 18 of Ramanujan's second notebook
and in Ramanujan [14]. For example, given an ellipse of major axis a,
minor axis b, and eccentricity k:=(b/a)', he gives
i) p = 21^0, - hi; k1) = 7r(a + b)2Ft(- \, - b i; 0
where t := [(a - b) /(a + b)f. Then, as t-*0,

Modular Equations and Algebraic Approximations to n
ii) p ~ 7r[3(a + b)- V(3a + b)(3b + a)]
and
iii) p ~ <rr(a + b)[l + 3/(10 + V4-37)~1/2]
where the error in ii) is about 2~9t3 and the error in iii) is about
3 • 2-1V. This is pleasantly developed in Almqvist and Berndt [Pr]
and in Berndt [Pr]. Truncating the approximation in iii) leads to
iv) p ~ tt(o + b)[l + yf
-83
with an error about 2 t. This is due to Nyvoll [78].
a) Prove i).
b) Justify the error estimates in ii), iii), and iv).
One may avoid the Landen transform in a) by following Ivory
[1796]. We write
k2
l-y(l-COS20)
1/2
d0
12 dd
p = 4aE(k) = 2a J
= (a + b) Jo (1 + tine2iy\l + tV2e-2W)in
= (a + b) 2 (-lY"+"(~s)";(~2)"f<m+">/2
m,n-o mini
r e2i{m-")e dd .
Jo
a) Combine (5.1.2) and (5.2.14) to derive that
p=4aE(k)~2rra[M;\k,f)-Mn(k,f)en(k,f)] as k -*0
where Wn(k2, f) = 0.
b) Show that the error is roughly of order ak2n.
c) Deduce that to order ak"1,
o (a2 + b2\
and to order ak8,
o ( a + b \2
v~2ir\—- 7=1 n: = 4.

5.6 Other Approximations 197
d) Use the cubic identities to establish that
p ~ ira[3m~l(r) + m(r) - 2]
when k = X*(r). Thus, with order ak6,
k: = V2-l gives 27ra(5V6-6V3-7V2 +9)
and 2kk': = VI - 2 gives 7ra[3(2 + V3)V(2V5 - 2V3 - 1)
+ V(2V5 + 2V3 + l)-2].
9. In the notation of the AGM iteration let k'-= c0/a0 and let x2n '■ =
Cn+Van + l-
a) Establish that
^ 1 (l + il-x*\ , (2\ K'(k)
b) Let x : = {xJ2)A so that x = [\ (1 - VF) /(1 + VF)]4 and
K'(k) 1 , „ 26 2 368 3
"W)=~4logx~2x~Tx -Tx +----
c) Thus with k ■= A*(r) one has another estimate for ttVt with jc of
order <T4,n/~r. When r := 1, x = ^[(21/4 - 1)4/(21M + l)]4, and the
given terms yield 19 digits of ir.
From our point of view possibly the most remarkable result in Chapter 18
of Ramanujan's second notebook is the following continued fraction identity
given in entry 12. Let n>0 be fixed. Define
w flV- a P2 (2«)2 (3^)2 (4«)2
for a, /3 > 0. Then
(5.6.18) a.fefi.Vg)-*■(*"> ;*■<*■>
whenever )8 > a > 0 and
AG(j8,Vj82-«2) = 1

198 Modular Equations and Algebraic Approximations to n
[or, equivalently, whenever K(alfi) = (irl2){i\. Even more surprisingly, a
slight adjustment of the proof given in Berndt [Pr] shows that (5.6.18) holds
for all a, /3 > 0.
10. (The moments of K and E) Let
Kn-=\o knK(k)dk
and
£„:=/„ k"E(k)dk.
a) Use the differential equations for K and E to establish that
nK + E„
and
K"+2 n + 2
„ *C + 1
E =
n + 2
b) Show that
ir/2 n °° 1 rir/2
rir/2 n °° 1 f'2
u Jo sin0 „_0 2n +1 Jo
4"
«=o /2n\ ,
( J(2n + 1)2
c) Use contour integration of 0/sin 6 (on the infinite rectangle above
[0, 7r/2]) to deduce that
*° „4-0(2« + 1)2
which is twice Catalan's constant, j8(2) or G.
d) Establish that Kt = 1, El = §, and E0 = \ + /3(2).
e) Observe that all the odd moments are rational while each even
moment is of the form a + bj8(2) with a and b rational. Thus all
moments lie in <Q>(/3(2)).
f) Analogously define K'n and E'n. Determine recursions for these
conjugate moments. Show that K'0 = tt2IA and E'0= v2/8.
Compute K[ and E[.

5.6 Other Approximations 199
g) Use the quadratic transformations to show that
)o 1 + k dk 2K° 8 •
h) Show that

Chapter Six
The Complexity of
Algebraic Functions
Abstract. The aim of this chapter is to analyze the complexity of algebraic
functions in general; and of multiplication, division, and root extraction in
particular. There are two primary tools, Newton's method and the fast
Fourier transform.
6.1 COMPLEXITY CONCERNS
It is obviously inappropriate to consider the multiplication of two many-
thousand-digit numbers to be of equal difficulty to the multiplication of two
single-digit numbers. A reasonable measure of complexity that takes this
into account is the bit complexity. The bit complexity of an algorithm is the
number of single-digit operations required to terminate the algorithm.
Single-digit operations include addition, multiplication, logical comparison,
and storage and retrieval of single-digit numbers. We are exclusively
interested in how the complexity increases with the size of the problem. For
example, addition of two n-digit integers by the usual algorithm has bit
complexity 0B(n)—the subscript B on the order symbol is for emphasis.
This is a serial notion of complexity in the sense that on a serial machine it is
an appropriate asymptotic measure of the time required for the calculation.
We use the slightly nonstandard notation
an=n(bn)
if
an = 0(bn) and bn = 0(an).
200

6.1 Complexity Concerns 201
If an = 0,(bn), we say that {an} and {bn} are equivalent. Since accessing an
n-digit number requires O(n) bit operations, it is apparent that "usual"
addition of two n-digit integers is in fact O(n). These trivial lower bounds
are a consequence of uniqueness considerations—if we change any digit of
one of the numbers being added, we change the answer. Thus any algorithm
for addition must at least "inspect" every digit. So, up to a constant, usual
additon is asymptotically optimal. As we shall see later, one of the
interesting consequences of this body of theory is that "usual" multiplication is far
from asymptotically optimal.
A detailed approach to complexity requires a model of computation and
is perhaps most readily made rigorous in an analysis of Turing machines.
(See, for example, Aho, Hopcroft, and Ullman [74].) This much detail is
unnecessary for our purposes.
We will usually content ourselves with merely counting single-digit
additions and multiplications. In all the algorithms we present the comparisons
[note that the comparison of two n-digit numbers is ftB(n)] and the storage
concerns will be bounded by the arithmetic operations—provided the
algorithms are sensibly implemented. This is almost always transparent and
will rarely even elicit comment.
Operational complexity counts the number of operations (addition,
multiplication, division, and extraction of kth roots performed to a precision
bounded by the precision of the output). When all the operations in an
algorithm are performed to roughly the same precision, this is a useful
measure. The reasons for the particular choice of operations will be made
apparent in Section 6.4. Thus the algorithms of Chapter 5 compute n digits
of ir with operational complexity 0op(log n)\ once again the subscript on the
order symbol is for emphasis.
Comments and Exercises
One of the primary tools for the analysis of algorithms is the use of recursive
functions. The idea is to divide a problem into smaller subproblems that can
be solved by essentially the same technique and then recurse, a strategy
often called "divide and conquer." The reader unfamiliar with this approach
might like to examine the exercises. One of the lessons of complexity theory
is that many of the usual algorithms of mathematics are far from optimal.
Multiplication, taking Fourier transforms, and matrix multiplication are
but three examples. (See Exercise 3 and the next section.) A second lesson
is that good lower bounds are very difficult to obtain. For the analytic
algorithms we are considering, the only lower bounds we can establish are
the trivial ones. Thus unless the algorithm is of the same order, as is the case
for addition, we cannot achieve exact results.
We shall not discuss combinatorial complexity except to mention that an
introduction to this well-developed and important field may be found in
Aho, Hopcroft, and Ullman [74].

202 The Complexity of Algebraic Functions
1. Prove the following. Let a, b > 0 and c > 1. Suppose that/is monotone
on (0, °°).
a) If f(n) < af(n/c) + bn and /(1) = d, then
f(n) = 0(n) if a < c
f(n) = 0(n log n) if a = c
f(n) = 0(nlogO if a > c .
b) If f(n) < af(n/a) + bn(log n)c~l and /(1) = d, then
/(«) = 0(«(log«r).
Hint: Analyze a) with equality. Then establish the general principle
that the equality solution is the maximal solution.
2. Show that the usual algorithms for multiplying and adding two nx n
matrices have complexity Oop(n3) and Oop(n2), respectively. (We are
counting the number of multiplications and additions of the entries of
the matrices.)
3. (Fast matrix multiplication (Strassen 1969))
a) Show that
An
- -^21
"*M2
-^22-
"fill
- fi21
fil2
fi22-
"Cu
- V21
£-12
^-22-
can be computed from
Mi = (A12-A22)(B21 + B22)
M2 = (An + A22)(Bn + B22)
M3 = (An-A21)(Bn + B12)
M4 = (An + A12)B22
M5 = An(B12-B22)
M6 = A22(B21 - Bn)
M7 = (A2l + A22)Bn
and
Cn = M1+M2-M4 + M6
C12 = M4 + M5
. C21 = M6 + M7
C22 = M2- M3 + M5-M7.
b) Observe that the above method reduces the multiplication of
2m x 2m matrices to 7 multiplications and 18 additions of n x n

6.1 Complexity Concerns 203
matrices. Thus if W(n) is the operational complexity of multiplying
two n x n matrices, then by iterating the procedure in a)
W(2n)<7W(n) + an2 .
The final term comes from using usual matrix addition for the 18
additions. The constant a can be chosen independent of n and can
be used to include the "overhead" of actually breaking the problem
up. Use the above inequality to show that
W(n) = Oop(nl°^).
Note that log27 < 2.81, so the above method is asymptotically faster
than the usual method. [Extensions of this method can reduce the
bound for multiplication down at least to 0(n2'5~), much as in
Knuth [81]. The best lower bound known is the trivial one, en2.]
Suppose A is a nonsingular 2m x In triangular matrix. Write
A = (B °
A \C D
where B, C, and D are n x n matrices,
a) Show that B-1 and D~x exist, and that
B'1 0
-D~lCB~l D~l
A~l =
b) Show that a) iterates to produce an algorithm for inverting 2m x 2m
triangular matrices. Let I(n) be the operational complexity of this
algorithm. Show with W(n) as in Exercise 3 that
7(2«) < 2W(n) + 2I(n) + en2
and hence that
I(n) = 0(W(n)) = 0(nlog27) .
(One can show that in general matrix inversion and matrix
multiplication are asymptotically equivalent. See Aho, Hopcroft, and Ul-
lman [74].)
Show that the bit complexity of calculating n! by multiplying 1 x 2 x 3 x
• • • using usual multiplication is 0((n log nf).
Hint: Analyze the complexity of multiplying an M-digit number by an
m-digit number. Use Stirling's formula to estimate the number of digits
in k\. Exercise 10 of Section 6.4 explores this further.

204
The Complexity of Algebraic Functions
6.2 THE FAST FOURIER TRANSFORM (FFT)
Let w be a primitive (n + l)th root of unity either in C or in a finite field Fm,
that is, w"+1 = 1 and wk ¥= 1 for k < n + 1. In the complex case we may take
w:=;e2^'/("+1). Consider the following two problems.
interpolation problem. Given n + 1 numbers a0,. . . , an, find the
coefficients of the unique polynomial pn{z) '■= a0 + axz H h anzn of degree n
that satisfies
pn(wi) = ai 0<i*<n + l.
evaluation problem. Given the coefficients of a polynomial pn of degree
n, calculate the n + 1 values
pn(wl) 0<i'<n + l.
These are the two directions of the finite or discrete Fourier transform.
The classical approaches to either part of the Fourier transform problem
have operational complexity at least en2. This is the operational complexity,
for example, of evaluating pn at n + 1 points using Horner's rule [writing
pn(x) = (((anx + an-i)x + an_2)x + ■ ■ •)]• We wish to prove that, in fact,
both directions can be solved with complexity Oop(n log n). Actually, we
only treat the case n + 1 := c2m, which is sufficient for our purposes and
somewhat simpler.
Theorem 6.1 (Fast Fourier Transform)
If n + 1 = c2m with c an integral constant, then both the interpolation and
the evaluation problem have operational complexity Oop(nlogn).
Proof. We assume c = 1, that is, n + 1 = 2m, the case for general c is
entirely analogous. We treat evaluation first. Suppose
p(x):=a0 + --- + anxn .
Let
q(x2) '■= a0 + a2x2 + a4x + h an_1xn~1
and
xr(x2) ■= x(al + a3x2 H V anxn~x) .
Then with y'=x2,

6.2 The Fast Fourier Transform (FFT)
205
(6.2.1) P(x) = xr(y) + q(y)
where r and q are both polynomials of degree 2m-1 -1. The observation
that makes the proof work is that for w an (n + l)th root of unity,
(w')2 = (w(n+1)/2+;)2.
Hence, evaluating p(x) at the n + 1 roots of unity in (6.2.1) reduces to
evaluating r and q each at the (n + 1)/2 points (w2)1, (w2)2,..., (w2)(n+1)/2
and amalgamating the results. Observe that w2 is a primitive (2m-1)th root
of unity and we can iterate this process. Let F(2m) be the number of
additions and multiplications required to evaluate a polynomial of degree
2m - 1 at the 2m points wk, k = 1,. . . , 2m, where w is a primitive (2m)th root
of unity. As above,
(6.2.2) F(2m) = 2F(2m~1) + 2-2m F(1) = 0.
The final term comes from the single addition and multiplication required to
calculate eachp(w') from r{w2') and q(w2'). The recursion (6.2.2) solves as
(6.2.3) F(2m) = 2m+1 • m
and the bound for the evaluation problem follows.
The interpolation problem is equivalent to evaluation. This can be seen
as follows. Let w be a primitive (n + l)th root of unity and let
1 1
w w2 ...
(6.2.4) W:=\ 1 w2 w4
Then
.„n In
W w
1
-1 -2 -n
w w ... w
(6.2.5) ^1 = ^Tl( 1 vv-2 w-4 ... w-2"
w~n w-2n ... w-»2
and w-1 is also a primitive {n + l)th root of unity. (See Exercise 4.) The
interpolation problem can be written as: Find (a0,..., an) so that
W(a0,..., an) = (a0,..., an).

206
The Complexity of Algebraic Functions
However, this can be solved by
W~\a0,... ,an) = (a0,. ..,an)
which is exactly the evaluation problem. □
While restricting the FFT to powers of 2 poses no problem for us over C,
it is a nuisance over finite fields. The problem is that Fm has primitive kth
roots if and only if k divides m — 1. Hence, our approach is restricted in the
finite case to considering primes of the form m = c2k + 1, which are not
particularly abundant. There are many ways around this difficulty. This is
discussed in Winograd [80].
Comments and Exercises
The FFT is an enormously useful and widely used algorithm. Depending on
exact form and implementation, it can outperform traditional methods for
values of n well below 100. For a history of the FFT consult Cooley, Lewis,
and Welch [67]. While antecedents for FFT methods are plentiful, Cooley
and Tukey [65] are primarily responsible for introducing the FFT in its
modern form as a complexity-reduced method. More extended discussion of
the FFT and related matters may be found in Aho, Hopcroft, and Ullman
[74] and Winograd [80]. As a theoretical tool the FFT and related methods
are central. They form the basis for the next section's discussion of fast
multiplication. In the exercises we show how these ideas can be used to
construct asymptotically fast polynomial multiplication, division, and
interpolation algorithms. It is even possible to accelerate integer factoring
algorithms by FFT methods. In Chapter 10 an application to the estimation
of certain transcendental functions is provided. Our proof of Theorem 6.1
and some of the exercises follow Borodin and Munro [75].
1. a) (Fast polynomial multiplication) Consider the following algorithm
for multiplying polynomials. Given the coefficients of p and q (both
of degree <« - 1), compute the coefficients of pq as follows:
Step 1: Evaluate p and q at 2m points w1,. . . , w2", where w is a
primitive (2n)th root of unity.*
Step 2: Form the 2m products
p(w')q(w') i = 1,..., In.
Step 3: Solve the interpolation problem for pq to find the
coefficients.
Show, using an FFT, that the above algorithm has operational
complexity
Oop(n log n) .
* Strictly speaking, we should be using (2m)th roots of unity where we have established an
FFT. This can be arranged by padding with leading zero terms, if necessary, without changing
the order of complexity.

6.2 The Fast Fourier Transform (FFT)
207
[The usual convolution product algorithm has operational complexi-
ty 0(«2).]
b) Given
n
p(x)-=U(x-xi)
1=1
show that the coefficients of p can be calculated in Oop(n(log nf).
Hint: Treat the problem recursively and recombine the pieces using
a fast polynomial multiplication.
(Fast polynomial division) Given p of degree n and q 6f degree m < n,
both with integer coefficients, it is possible to find u and r with
deg r < deg q so that
p(x) = u(x)q(x) + r(x)
in Oop(n log n).
Outline: Simplify by observing that it suffices to calculate u since r may
then be computed by Exercise 1. Set x ■= 1/x. Then
p(l/x)= (1\ r(llx)
q(llx) \x) q(llx)
and so
^- = u(x) + xn-m+h^\ v(x):=x^vv(-)
q(x) q(x) v ' \x)
where /i > 1. To calculate « (and hence u) it suffices to calculate the first
n-m (= deg u) Taylor coefficients of II q. This can be done by
Newton's method as follows. Suppose that deg st = j- 1 and that
Establish that
-=^ - [2st(x) - s2t(x)q(x)] = -^ [1 - st(x)q(x)]2 = 0(x2J) .
[Note that we may assume q(0) ¥^ 0.] Now the computation of si+x '■ = 2s,. -
s2q can be performed using an FFT-based polynomial multiplication and
need only be performed using the first 2/ — 1 coefficients of q and 5,. By
starting with an appropriate first estimate of s0 [say, s0(x) '■= l/q(0)] and
proceeding inductively as above (doubling the number of coefficients
utilized at each stage), show that the required number of terms of the
expansion can be calculated in Oop(n log n). (See Section 6.4.)

208
The Complexity of Algebraic Functions
3. (Fast polynomial evaluation) Given p of degree n and n + 1 distinct
points xn,. .. ,xn, show that p(x0),..., p(xn) can all be evaluated in
Oop(«(log«)2).
Hint: Let qx(x) = n"„o * (x - xt) and let rx be the remainder on dividing
p by qv Note that ^1(^) = ^(^,) for i<n/2. Similarly use
q2(x):=Kni2(,x~xi)- Thus two divisions reduces the problem to two
problems of half the size. Use Exercise 2 and evaluate the recursion.
[The other direction of this problem, namely, constructing Lagrange
interpolating polynomials, is also Oop(n(log n) ). See, for example,
Borodin and Munro [75]. In fact, both directions are Oop(n log n).]
4. Show that if w is a primitive (n + l)th root of unity, then
y (, = {" + l / = 0mod(n + l)
,=0 I 0 otherwise.
Show that (6.2.4) and (6.2.5) are inverse to each other.
5. (Reversion of power series) Let f(x) '■= E^„0 akx be a formal power
series, with known coefficients.
a) Show, as in the proof of Exercise 2, that the first n coefficients of
the formal series expansion of 1//00 can be computed in
Oop(n log n).
b) Discuss the complexity of computing the coefficients of the formal
inverse of / by Newton's method.
6. (On calculating xn) The 5-and-X binary method for calculating x" is
the following rule. Suppose n has binary representation 8,,8^2---8^
with S0 = 1. Given symbols 5 and X, define
rsx if s, = i
''IS if 8t = 0
and construct the sequence
SXS2 ■ ■ ■ Sk.
Now let 5 be the operation of squaring and let X be the operation of
multiplying by x. Let the sequence of operations SXS2••• Sk operate
from left to right beginning with x. For example, for n = 27,
8,,8^8384 = 11011
and
5!525354 = (SX)(S)(SX)(SX) .

6.3 Fast Multiplication 209
The sequence of calculations for x21 is then
2 3 6 12 13 26 27
a) Prove that the above method computes x" and observe that it only
requires storing x, n, and one partial product.
b) Show that the number of multiplications is less than 2 [log2 n\.
c) Show that the above method is optimal for the computation of x2™
(considering only multiplications).
d) Show that the 5-and-X method is not optimal for computing x15.
An extended discussion of this interesting and old problem is presented
in Knuth [81].
6.3 FAST MULTIPLICATION
We wish to present a strategy for multiplying very large numbers that is
considerably faster than the usual 0B(n2) method. The idea is to exploit the
FFT. Let a and /3 be two n-digit integers and write
(6.3.1) a(x) := anxn~l + an_xxn~2 + ■ ■ ■ + ax
and
(6.3.2) (3(x) := by1 + bn_xxn~2 + --- + b1.
Then if at and bt are the decimal digits of a and /3, respectively, we have
(6.3.3) a = a(10) and /3 = /3(10).
We can now calculate a • /3 by using the FFT, as in Exercise 1 of the last
section, to compute the coefficients of
(6.3.4) y(x):=a(x)p(x).
Finally we evaluate y(10).
Let T{n) denote the bit complexity of multiplying two n-digit integers (or
equivalently, floating point numbers) by the above method.
The analysis of the complexity of the algorithm that follows assumes that
we are working with complex roots of unity (and that we are working with a
problem of size 2m, as can always be arranged by padding with leading
zeros). This introduces rounding error problems into an intrinsically integer
algorithm; however, this is the setting which we have established an
abundance of values for which the FFT works. Since in practice (and in

210
The Complexity of Algebraic Functions
theory) there are plenty of FFT analogues in a finite setting, one may, if one
prefers, consider the entire algorithm performed mod (O(n)).
Step 1:. Evaluate a(x) and fj(x) to precision 0(log n) at the In points
w1, w ,. .. , w2n with w a primitive (2n)th root of unity. Using the FFT, this
has operational complexity
Oop(n log n)
and bit complexity
(6.3.5) 0B{n{\ogn)T{\ogn)).
[It suffices to use w to precision 0(log n) and thus every multiplication in
Theorem 6.1 is of complexity 0B(r(log «)). Observe that the coefficients of
a and /3 are single-digit numbers and that the coefficients of y are hence
0(log n)-digit numbers. See Exercise 3.]
Step 2: Form the 2n products
y(w') = a(w')/3(w')
computed to 0(log ri) digit precision. This is of bit complexity
(6.3.6) 0B(nT(logn)).
Step 3: Interpolate the coefficients of y to precision 0(log n) using the
FFT. This has operational complexity
Oop(n log n)
and bit complexity, as before,
(6.3.7) 0B{n(\ogn)T(\ogn)).
Step 4: Evaluate y(10). This has bit complexity
(6.3.8) 0B{n).
This step essentially requires only addition. The coefficients of y are closely
related to the digits of a/3 except that they may be too large and a "carry"
must be performed.
The total complexity thus satisfies

6.3 Fast Multiplication
211
(6.3.9) T(n) = 0B(n(log n) T{\og n))
or
(6.3.10) T(n) = 0B(n(log «)2(log(log n)f ■■■)
with the product terminating when the iterated log is less than 1.
This is not optimal. (See Exercise 2.) The same analysis as above using a
base n representation of a and /3 reduces the time for multiplication to
Ob(«(log «)(log log «)•••) .
This is still not quite the asymptotically fastest known algorithm. The best
bound is due to Schonhage and Strassen [71] and is
0B(w(log«)(loglog«)).
We shall call any multiplication that performs with this speed a Schonhage-
Strassen multiplication. It should be emphasized that in all the reduced
complexity multiplications we present, the order estimates include the
overhead additions (and storage concerns).
Comments and Exercises
The observation that multiplication is not intrinsically 0B(n2) was made by
Karatsuba in 1962. He proposed an Ofl(nloS23) algorithm. (See Exercise 1.)
Subsequent refinements and improvements are due to Toom, Cook,
Schonhage, Strassen, and others, culminating in the Schonhage-Strassen
multiplication of 1971. This algorithm is of the same flavour as the one
presented above using size 2^ representations and performing FFT
operations mod(22V" +1). A presentation may be found in Aho, Hopcroft, and
Ullman [74]. Knuth [81] has an extended discussion of multiplication
strategies which includes a discussion of the precision concerns of
performing a fast multiplication over C. Cook and Aanderaa [69] conjecture that
multiplication is not 0B(n). Here one must be careful about the model of
computation allowed. Under some more powerful than usual models 0B(n)
multiplication is possible (see Knuth [81]), while under other more
restrictive than usual models it can be shown to be not possible.
Once again it is possible to implement a fast multiplication that will
outperform traditional methods for n in the several hundred-digit range. For
the many million-digit calculations of it discussed in Chapter 11, use of a fast
multiplication is imperative. For a discussion of the multiplication used in
this setting see Tamura and Kanada [Pr] and Bailey [88].

212 The Complexity of Algebraic Functions
1. (An 0B(n'og23) multiplication) Observe that
(a + 610" )(c + rflO") = ac + [(a - b)(d - c) + ac + bd]10n + bdl02n.
Use this to reduce multiplication of 2/j-digit numbers to three
multiplications of n-digit numbers and some additions. Show that this can be
used to produce a multiplication of
0B(n10^) .
[This method can be refined to produce an 0B(n1+s) algorithm; details
are in Knuth [81].]
2. Construct a multiplication of complexity
0BO(log n)(log log «)•••)•
Hint: Instead of using base 10 representations, use base n
representations so that the polynomials (6.3.1) and (6.3.2) are polynomials of
degree n with coefficients of length log n. Now proceed as in the
algorithm of this section.
3. Discuss the bit complexity of the FFT. Show that if the input is given to
precision 0(m) and the output is required to precision 0(m), the bit
complexity is
0B((n log n)M(m))
where M(m) is the complexity of whatever multiplication is employed.
6.4 NEWTON'S METHOD AND THE COMPLEXITY OF
ALGEBRAIC FUNCTIONS
We wish to show the equivalence, from a complexity point of view, of
multiplication, division, and root extraction. The primary tool is Newton's
method.
Theorem 6.2
Suppose that/is analytic in a complex neighbourhood of z. Suppose/(z) = 0
and /(z) ¥= 0. Then the iteratipn
(6-4.1) xn+1 -=^-^),
converges uniformly quadratically to z for initial values x0 in some
neighbourhood of z.

6.4 Newton's Method
213
The reader unfamiliar with the proof is directed to Exercise 1. Newton's
method is useful for calculating inverse functions. Observe that g~1(y) is a
zero of f(x) = g(x) - y. This is the content of the corollary. The uniform
nature of the convergence is discussed in the exercises.
Corollary 6.1
Suppose that/is analytic and one to one in a neighbourhood of z0. Then
there is a neighbourhood of /(z0) where /-1 can be computed uniformly
quadratically by Newton's method.
The quadratic nature of the convergence is only half the story. The other
half is the "self-correcting" nature of Newton's method. Suppose that
/(z) = 0 and that we are computing z by Newton's method. If the nth iterate
xn is perturbed by an amount 0{\xn - z|), then provided we stay in the
domain of uniform quadratic convergence, computing xn+1 from the
perturbed value of xn will preserve quadratic convergence. In other words, if
xn - z agree through M digits, then the calculation of xn+1 need only be
performed to precision 2M.
Let M(n) denote the bit complexity of multiplication of two n-digit
numbers by some method. We make the following regularity assumptions:
2M(n) ^ M(2ri) < AM{n) and M(n) is nondecreasing .
(6.4.2)
Since two multiplications of length n can be viewed as subproducts of a
single multiplication of length 2n, while four multiplications of length n
comprise one of length 2n, the first part of the assumption is reasonable.
Since multiplications of length n can be padded with leading zeros to
multiplications of length n + k, the second part is also reasonable. Of course
it is easy to imagine a perversely designed multiplication for which (6.4.2)
does not hold.
Let D(ti) and R(n), respectively, denote the bit complexity of division
and extraction of square roots, where the input and output are to precision
n. We say that two operations are equivalent if the complexity of one is
bounded by the complexity of the other and conversely. For example, we
say multiplication and division are equivalent if, given a multiplication with
bit complexity M(n), we can construct a division with bit complexity
D(ri) = 0(M(n)); and conversely, given a division we can so construct a
multiplication. The following remarkable theorem, the first part of which is
due to Cook, may be found in Brent [76c].
Theorem 6.3
Multiplication, division, and root extraction are all equivalent.

214
The Complexity of Algebraic Functions
Proof.
(a) We first construct a division. Applying Newton's method to the
function f(x) '• = 1 Ix — y leads to the iteration
(6.4.3) xk+1: = 2xk-x\y
which employs only multiplication and addition. Note that
l ( V2
(6-4.4) ^+1-- = -3^-
and the quadratic nature of the convergence is manifest. We assume
|_y| < 1 and that y lies in a neighbourhood V bounded away from zero
(which if we are working in floating point, is no restriction). We may
also assume that, by using a usual Os(n ) division performed to a
fixed low precision, we have already computed x0-=x0(y), so that
\x0 — l/v| <1/10. Assume n is a power of 2. Then log2 n
iterations of (6.4.3) will by (6.4.4) produce an error bounded by
_ 1
X0 y
HT"
and hence provide n digits of IIy. Furthermore, by the self-correcting
nature of Newton's method, the fcth step of (6.4.3) requires two
multiplications and two additions of precision only 2k. Thus the total
complexity of the iteration is given by
log2n
(6.4.5) E [2M(2*) + 2-2*]<8M(n)
k~l
since 2M(2*)<M(2*+1). We have shown that lly can be calculated
with complexity 0B{M{n)). Hence since alb = a ■ (lib),
D{n) = 0B{M(tij) .
(b) The equivalence of division and multiplication is now obvious since
u a
ab = TFb-
(c) Square roots can be extracted by Newton's method applied to x — y,
which yields the classical iteration
(6.4.6) **+*:=!(** + i~

6.4 Newton's Method 215
This satisfies
(6.4.7) xk+i-Vy = ^(xk~Vyf
and the quadratic convergence is again apparent. We can proceed
exactly as in (a) to show that
R(n)=0B(D{n)) = 0B(M{n)).
(d) The proof that
M(n)=0B(R(n))
is Exercise 7. □
It is apparent that any time a function may be quadratically computed by
Newton's method from an iteration involving only addition, multiplication,
and division, that function will be of complexity 0B(M(n)). This applies to
any algebraic function over Q(x), that is, any function / satisfying an
equation
(6.4.8) 4>(x,/(x)) = 0
where $ is a polynomial in two variables with rational coefficients. More
precisely:
Theorem 6.4
If/is algebraic over Q(x), then the complexity of calculating n digits of f(x)
is 0B(M(n)).
The preceding results, of course, assume that we are avoiding the branch
points of the function in question.
Comments and Exercises
Newton's method has a host of refinements and variants. See, for example,
Householder [70] and Exercise 3. The iteration for square roots can in some
form be traced back to the Babylonians, who used one or two steps of the
method.
It should be observed that not only is D(n) — 0B(M(n)), but the constant
concealed by the order sign is fairly small. [From (6.4.5) we see that a
constant 8 works. Indeed, for all known multiplications, the additions term
is negligible and a constant of 4 is appropriate.] This is also the case for root

216
The Complexity of Algebraic Functions
extraction. See Brent [76c], where a number of constants for these and
various other equivalences, as in Exercise 7, are established.
Further discussion of the calculation of algebraic functions may be found
in Kung and Traub [78]. Related matters may also be pursued in Lipson
[81].
1. a) Prove Theorem 6.1 by observing that
fan) ="fa) + (*» - *)fo) + 0(Xn ~ zf .
Substitute this into (6.4.1) to get
'fan) ~fo)
* = (*»- z)
fan)
+ 0(xn - zf .
Use explicit estimates to prove uniformity.
b) Show that for real / and real xn, the tangent to / at xn intersects the
x axis at xn+1.
c) Suppose that / is convex and strictly increasing on [a, b] and that
f(x) = 0 for some x in (a, b). Show that if b > x0 > x, then {xn}
decreases to x and convergence is guaranteed.
2. Construct the Newton iteration for y1/p by inverting xp. Write xn+1 —
yllp in terms of (xn — y1/p)2, thus exhibiting explicitly the quadratic
convergence. For which real starting values does the method converge?
3. a) Consider the iteration
(l//)(n)
yk
where the notation indicates that the derivatives are evaluated at
yk. For sufficiently well-behaved (for example, analytic) /, this
method will find a zero of / with (n + 2)th-order convergence
provided various derivatives are nonvanishing. Prove these
assertions. For n'-=0 this is just Newton's method, for n'-=l it is
Halley's method. (See Householder [70].)
b) Let
xk+i :=xk(l + (1- yxk) + (1- yxkf) .
Show that
1\ ..2( 1X3
, :y \xk~
yl ' \ y
and that for x0 sufficiently close to 17y, xk converges cubically to
1/y.

6.4 Newton's Method 217
c) Let
xk+1-=lxk(15-10x2ky + 3y2xk)-
Show that
yx2k+1 -l = h (9/4 - 33y*2 + 64)(y*2 -1)3
and that for x0 sufficiently close to 1/y/y, xk converges cubically to
l/Vy.
d) Show that b) and c) require fewer multiplications than (6.4.3) or
(6.4.6) for computation of reciprocals and square roots. These
are, in practice, very good high precision algorithms.
Let x0 '■ = 1 and let
Kk + l ' *-Xk XfrX
be the iteration (6.4.3) for computing 1/x. Show that xk+l is the
(2 +1 - l)th Taylor polynomial of f(x) '■= 1/x expanded around the
point 1. Thus for division we may think of Newton's method as a
means of accelerating the computation of the Taylor series.
Let x0 : = 1 and let
_ 1 ( x_
xk+i — 2{xk + Xk
be the iteration (6.4.6) for computing Vx. Show that xk+l is a rational
function r(x) with numerator of degree 2 and denominator of degree
2—1 that satisfies
|V3e -r{x)\ = 0{x -1)2
[This implies that r(x) is the (2k, 2k - 1) Pade approximant to Vx at 1.
See Section 10.1.]
6. Invert 1/x — y to calculate Vy without using any divisions.
7. (Other equivalences; Brent [76c])
a) Show that squaring is equivalent to multiplication by considering
(a + bf - (a - bf.
b) Complete the proof of Theorem 6.3 by showing that square root
extraction is equivalent to multiplication.
Hint:
S2
(1 + 2Sx)112 = l + 8x- — x2 + 0((8xf) .
Use 8:=10 m for appropriate m to reduce the computation of x
to root extraction and 0B(n) operations.

218
The Complexity of Algebraic Functions
c) Show that inversion is equivalent to multiplication.
d) Show that pth-root extraction is equivalent to multiplication.
8. a) Suppose that Cf(ri) is the bit complexity of calculating n digits of/.
Assume that Cf(n) is increasing and that Cf(2n)^2Cf(n).
Suppose that / satisfies the conditions of Corollary 6.1. Show that/has
at worst the same bit complexity as /, and that
Cri(n)=0B(Cf(n) + M(n)).
b) Write down Newton's method for computing exp from log and log
from exp. These particularly simple iterations combine with a) to
show that the problems of calculating exp and log are effectively
equivalent, from a bit complexity viewpoint.
9. (Fast base conversion; Schonhage) Let k and / be fixed integers. Show
that an n-"digit" base k number can be converted into base / with bit
complexity
0B(lognM(n)).
Hint: Break the number to be converted in half (base k), convert each
half, and recombine. (See Knuth [81] for a lengthy discussion of this
problem.)
10. a) Show that n\ can be calculated with bit complexity
0B(log n M(n log n)) .
Hint: First calculate 1 x 2, 3 x 4, Then calculate 1x2x3x4,
5 x 6 x 7 x 8, .. . etc.
b) Compare this to calculating n! as in Exercise 3 of Section 6.1.
Show that no multiplication can reduce this method below 0(n2).
[The best known bound for n\ is 0B((log log n)M(n log «)); see
P. B. Borwein [85].]

Chapter Seven
Algorithms for the
Elementary Functions
Abstract. We analyze algorithms for the transcendental elementary functions
based on the transformation theory for elliptic integrals and in particular on
the AGM.
7.1 w AND LOG
All the elementary transcendental functions can be calculated with bit
complexity 0B(lognM(n)). This is a consequence of the fact that log has
operational complexity Oop(logn), and hence has bit complexity
0B(logn M(n)). The approach to log rests most easily on the logarithmic
asymptotic of K at 1. Before proceeding with this analysis it is convenient to
record the complexity of the algorithms for ir, based either on iterating the
modular equation, Wp, or more specially on the AGM.
Theorem 7.1
The initial n digits of tt can be calculated with operational complexity
Oop(log«)
and with bit complexity
Ofl (log «Af(n)).
Proof. Both Algorithms 2.1 and 2.2 as well as most of those of Chapter
5 perform with the above complexity. □
219

220
Algorithms for the Elementary Functions
In all of the above algorithms some of the calculation may be done to
reduced precision. For example, in Algorithm 2.1 the computation of xn and
y„, which both tend to 1 quadratically, can be performed at successively
lower precision. The saving, however, is only in the constant term, reducing
it by a factor of less than 2.
We record the following estimates:
Theorem 7.2
(7.1.1)
(7.1.2)
(7.1.3)
(7.1.4)
*'(*)-log I £
E'{k) -1| =
*'(*)-log
= 0(\k2 log k\)
0(\k2 log k\)
4l ;10|fc2logyt|
\E'(k)-l\<lO\k2logk\
re(Jfc)>0,Jfc-»0
re(fc)>0, Jfc-»0
ike (0,10-3)
yte (0,10-3).
Proof. The relationships for K are in Exercise 4 of Section 1.3. (See
also Exercise 1 of Section 2.3.) For (7.1.2) observe that by (1.3.2), for
0<yt<l
E'(k) -1
1 Vl - (1 - *V - Vl-f2 dt
Vv^T2
k2t2 dt
< k2K'(k).
,0 y[l-(l-k2)t2](l-t2)
The constant in (7.1.4) requires a little additional scrutiny. □
The following approach to calculating log is essentially due to Salamin (in
Beeler et al. [72]).
Algorithm 7.1
For x e(s,l) andn>3,
(7.1.5)
|log x - K'(10-) + K'(1(T"*)| < -~^x
where
£'(10"") =
IT
2AG(1,10-")
and

7.1 it and Log
221
K'(10 nx) 2AG(1; 10-Bje)
are computed from the AGM iteration.
This algorithm has operational complexity 0op(log n) and bit complexity
0B(M(n) log n).
Proof. The estimate (7.1.5) is immediate from Theorem 7.2. The
computation of the two elliptic integrals requires precomputing ir, which has
complexity Oop(logn). The final detail is that AG(1,10~nx) and
AG(1,10~") can be calculated to precision 2m using 0(log n) iterations of
the AGM iteration. (See Exercise 1.) □
A related algorithm that avoids precomputing ir can be established from
Algorithm 1.2 (which provides a direct calculation of K'IE').
Algorithm 7.2
Let R'(k) := K'(k)/E'(k). For* £(1,1) andn>3,
(7.1.6) |log x - R'(10~n) + R'(10~nx)\ < ^rsy
where
(7-1.7) R'(k)= *-n-12
1-E„.02 c„
and c„ is associated with the AGM process commencing with a0: = 1 and
b0 := k. [See equation (1.1.3).]
This algorithm has operational complexity 0op(log ri) and bit complexity
0B(M(n) log n).
The proof is straightforward and is left as an exercise. Both of the above
algorithms are based on an underlying quadratic method for calculating K'
and E'. Algorithms based on pth-order methods can be constructed as in
Chapter 5. A quartic version, for example, can be derived from Exercise 3
of Section 1.4. Instead of calculating R' from the AGM, we use
(7.1.8) R'(k) =
l-E:=04>^[(a2„ + /32„)/2]2]
where
«M:-H^ and ^-J"^±^'"

222
Algorithms for the Elementary Functions
and
a0:=l and po:=kin.
In this, as in other quartic versions of quadratic algorithms, there is a
substantial computational saving (as much as 35%).
Comments and Exercises
The algorithms of this section for log all suffer from the drawback that they
are not truly iterative. Increasing the precision requires choosing new
starting values. This is more of an aesthetic than a computational problem;
even with iterative methods one only computes to a fixed precision, and
increasing the accuracy usually entails starting at least one of the calculations
all over again. While two different AGM or related processes must be
calculated for the initial value of log, subsequent values require computing
only a single AGM since one of the terms can be reused. These methods are
quite stable, requiring only 0(loglog n)-guard digits. They will outcompete
traditional methods—depending enormously on implementation—in the
several-hundred-digit range.
The algorithms of this section are the asymptotically fastest known
algorithms for log (see Section 10.2) and are faster than any known
algorithms based on other methods, although 0B((log n)2M(n)) can be
achieved by techniques of Chapter 10. These types of algorithms were first
examined by Salamin (Beeler et al. [72]) and independently by Brent
[76a, b, and c]. Newman [82] gives a self-contained account, as do Borwein
and Borwein [84a]. The second algorithm is in Borwein and Borwein [84d].
Finally, while we have only presented the algorithms for real k, they
extend naturally into the complex plane; only the error estimates become
slightly more complicated. (See Exercise 1.) Matrix versions due to Stickel
[85] are discussed in Exercise 6.
1. Show that Algorithm 7.1 can be used to calculate log uniformly for
{z£C| |z - l| < j} with operational complexity 0op(log n) and bit
complexity 0B(logn M(n)).
2. Examine the convergence of the AGM for a0:=l and b0-=10~n.
Specifically, estimate the number of iterations required to produce an
answer within 10"" of the limit. For Algorithms 7.1 and 7.2, find a
reasonable bound on the number of iterations of the AGM required to
produce a 1000-digit precision algorithm for log x, xE.( \, 1).
3. (An asymptotic algorithm for ir) Show that, for n >3,
1 1 1
AG(1,10~") AG(1,10~" + 10~2n) J
< «102~2n ,
log (1 + 10-)--
and hence

7.1 it and Log
223
10" 10"
AG(1,10~") AG(1,10~" + 10-Z")J
< «102~
This provides an 0op(log n) algorithm for -rr. (See Newman [82] or
Borwein and Borwein [84a].)
4. Given thepth-order modular equation 0,p (as in Section 4.5), show how
to construct asymptotic algorithms for log analogous to Algorithms 7.1
and 7.2, but with an underlying pth-order iteration.
5. Show how the series expansion for K' and E' of Section 1.3 can be
combined with Algorithms 7.1 and 7.2 to provide 0op(log n) algorithms
for log that provide n digits of log using starting values lQ~nlkx and
10-n/k
6. (The matrix AGM) Let &N denote the N X-N self-adjoint positive
definite matrices and let I denote the N X N identity matrix. Let
Pn,B0:=I,
A0
(7.
(7.
:=A6
1.9/)
1.9«)
a) Show that if Ae.&N, then there exists a unique C e SPN so that
C2 = A. Hint: The iteration C„+1 '•= C„ + \(A - C2n), C0 : = 0
converges to C.
b) Suppose that X0 £ 3PN commutes with AE.SPN. Show that, for X0
sufficiently close to VA, Newton's method Xn+1 ■= \(Xn + AX~l)
converges to VA £ 3PN quadratically.
c) Show that the matrix AGM (7.1.9) converges to a matrix
AG(A, I) £ &N and show that An - Bn converges quadratically to
zero.
d) Show that
J AG(A, By1 = [(x2I+ A2)(x2I+ B2)]~112 dx .
Hint: Imitate the second proof of Theorem 1.1 and use the fact that
A„ and B„ commute.
n n
e) Let
•tt/2
r it
K(Ay.=\o
dd
yi-A2sin2d
and let K'(A) = K(\/l- A2). Suppose that AE. @N and also that
0<a<m||00<6. Show for large n that (7.1.5) holds, namely,
there exists c so that
(7.1.10) \\logA-K'(10-nI) + K'(10-nA)\\ ' C"
102"

224
Algorithms for the Elementary Functions
f) Show that this provides an 0 (log n) algorithm for the matrix
2\
logarithm of A £ &N and, by inversion, an 0op((log n) ) iteration
for the matrix exponential.
For further details, and extensions beyond the positive definite case see
Stickel [85], where computational experience is also indicated.
7.2 THETA FUNCTION ALGORITHMS FOR LOG
We start with the fundamental identity of Chapter 2
and the series expansion
We recall that
(7.2.3) ^ = -¾^
«;(<?)
e:.*c
-» q
The algorithm for log (IIq) is now:
Algorithm 7.3
Fix ^ £ (a, 6) with 0 < a < b < 1.
Step i
Step 2
Step 3
Step 4
Calculate K(k)/ir from (7.2.2).
Calculate k from (7.2.3).
Calculate K(k') using the AGM commencing with 1 and k.
Calculate log (IIq) = Tr[K(k')/K(k)].
This algorithm has operational complexity Oap(Vn) and bit complexity
0B(VnM(n)).
The algorithm's complexity is determined by the series expansions
employed in steps 1 and 2. These "sparse" series yield n-digit accuracy after
Vn nonzero terms. While the asymptotic complexity is far from optimal, the
algorithm has the advantage of not requiring very small starting values for
the AGM iteration (Step 3). Also, only a single AGM iteration is required.
Sasaki and Kanada [82], who proposed and analyzed the above algorithm,
show that it out performs the methods of Section 7.1 for numbers in the

7.2 Theta Function Algorithms for Log
225
3000-digit range. Note that it must be precomputed. As Sasaki and Kanada
observe, the algorithm may be accelerated by using
(7.2.4) mlQg (1) = log (-L)
for various m. This speeds up the series calculation at the expense of the
AGM. For n-digit precision, using m = n effectively reduces this algorithm
to Algorithm 7.1. (See Exercise 1.).
For certain choices of q we get reduction in complexity. If q is any small
integer, then the series expansions are particularly easy to evaluate. (See
Exercise 2.) This leads to a very fast algorithm for ir/log 10 (using base 10
arithmetic) as follows. From (7.2.1), (7.2.2), and (7.2.3) we have
(7.2.5) log (I) = „£!(*)
ql K" AG(dt(q),dt(q))
,2
IT . „/ I V „Z\ I V („ + l/2);
or
Now for q := 1/104 both of the above series are just sequences of 0's and l's
and the starting values for the mean iteration can be calculated very quickly
[0B(M(n))]. The remainder of the work involves calculating a single AGM.
Similarly ir/log p is amenable to very fast computation in base p.
Properly interpreted, (7.2.6) remains valid for matrices and provides a
theta-based computation of the matrix logarithm of a positive definite
matrix. (See Exercise 3.)
Comments and Exercises
Further discussion of material in this section may be found in Sasaki and
Kanada [82] and in Borwein and Borwein [84rf]. Sasaki and Kanada
compare Algorithm 7.3 to algorithms for log based on Taylor series for
log (1 + x) and log [(1 + x) /(1 — x)] and conclude that for more than
(roughly) 100 decimal digits the Taylor series methods are slower.
1. Show that (7.2.4) for various m can be combined with Algorithm 7.3 to
provide algorithms for log of any complexity between Oop(Vn) and
Oop(log n).
2. Discuss the bit complexity of calculating d2(q), 03(#), and d4(q), where
q is the reciprocal of a fixed integer. Show, forp integral, that (7.2.6)
can be used to calculate ir/log p with bit complexity OB(log n M(n)).
(See Exercise 9 of Section 6.4.)

226
Algorithms for the Elementary Functions
3. Consider the matrix AGM of Exercise 6 of the previous section.
Establish a matrix version of (7.2.1) to (7.2.6). Then construct an
algorithm for the matrix log based on (7.2.6).
7.3 THE COMPLEXITY OF ELEMENTARY AND ELLIPTIC
FUNCTIONS
The algorithms of Section 7.1 can be inverted by Newton's method to
provide algorithms for exp with bit complexity Ofl(log n M{n)) and
operational complexity 0op((log nf). From a bit complexity point of view, this is
the best known bound. Since we can invert log in {|z-l|^3}, we can
produce an algorithm for exp in a complex neighbourhood of zero and
hence have 0B(log n M(n)) algorithms for all the trigonometric functions.
Exercises 1 and 2 give some variations. In fact, for any elementary function
we have the following:
Theorem 7.3
Any elementary function / over Q(x) can be calculated uniformly (in
bounded regions where / is single valued and analytic) with bit complexity
Ofl(log n M(n))
and with operational complexity
Oop((log«)')
where s is a constant depending only on /.
Proof. For our purposes the elementary functions are the rational
functions, log and exp, and any function that can be formed from these
functions by a finite number of compositions, multiplications, additions, and
solutions of algebraic equations. (See Davenport [81] or Ritt [48].) The
point of the proof is that such an/is constructed from Q(x) by taking a finite
number of exponentials, logarithms, and solutions of algebraic equations in
these quantities. The number of algebraic equations to be solved determines
the constant s. As in Chapter 6, solution of the algebraic equation in
question can be effected in a time proportional to the complexity of
evaluating the equation. □
A number of comments are in order. First, we can formulate the above
theorem for / algebraic over U(x) if we assume that the requisite real
numbers are given. Second, while multiple solutions of an algebraic
equation pose no theoretical problem, in practice, determining the "correct root"

7.3 Elementary and Elliptic Functions
227
can be a major nuisance. For a multiple-valued function the theorem must
be interpreted as guaranteeing some value of the function. This is inevitable.
It is not even clear what it should mean to compute an infinite-valued
function.
That the operational complexity behaves like (log n)s instead of (log n)
reflects in part that operational complexity is an inappropriate measure
when Newton's method is involved. While the preceding algorithms for log
require most of the operations to be done to full precision, this is no longer
the case for this approach to other transcendental elementary functions.
Since we can calculate elliptic integrals with bit complexity
0B(log n M(n)) (Exercise 5 of Section 1.4 and Exercise 2 of this section), we
can calculate the Jacobian elliptic functions with similar dispatch.
It is not clear which other nonelementary transcendental functions have
bit complexity 0B(lognM(n)). Does, for example, the gamma function?
Nor is it clear whether the bit complexity 0B(log n M(n)) is best possible for
the nonalgebraic elementary functions. The best known lower bound for log
and exp is the virtually trivial bound of 0B(M(n)). (See Exercise 3.)
Comments and Exercises
The algorithms for exp (see also Exercises 1 and 2) require inversion and are
much less satisfactory than the algorithms for log. A direct 0B((log n)2M(n))
algorithm is presented in Chapter 10. There are a number of issues that
remain unresolved in this discussion. The most obvious is a discussion of
lower bounds. Observe that, by Theorem 6.4, showing that exp does not
have complexity 0B(M(n)) would show that exp is a transcendental
function. Likewise, showing that ir.does not have bit complexity 0B(M(n))
would imply the transcendence of it. The gap between the known
operational complexities for log [Oop(log n)] and exp [Oop((log n)2)] is probably
specious, but this also is not known. We will show in Section 8.8 that direct
algorithms for exp and log of the type that we derived for K in Chapter 1
cannot exist. There are no quadratically convergent fixed iterations for these
functions.
1. From (2.3.7) and Exercise 3 of Section 2.5,
lim(^r = exp[flim(%Y
where an and c„ are generated from the AGM commencing with a0 '■ = 1
and b0 '■ = k', while a* is generated from the AGM commencing with
ao:=l and b%'-=k. Show that this leads to an 0B(lognM(n))
algorithm for ex, which begins by solving for anla* = irx/2.
2. (A more direct approach to tan) As in Exercise 5 of Section 1.4 and
Theorem 2.6 of Section 2.6, we have the Landen transform for

228
n&
Algorithms for the Elementary Functions
* de
,k):=l
0 Vl - k2 sin2 d
ike[0,1), ^0.
if
l-K
kn+i:= Y^n? and tan^n+i ~ ^=k'ntan ^"
then, as before,
(7.3.1) F(<pn+1, kn+1) = (1 + k'n)F(<Pn, kn)
and
2
(7.3.2)
^>o> ^o) :
n
l + k'
lim^.
n-»co 2
a) Show that
F(<j)0, k) = <j>0 + 0{k2) as A;-* 0 .
b) Show that
F(0O, A:) = log tan (^+ ^) + 0(1-A:) as k->l
c) Show that if w„ : = tan </>n, then
(! + *>„
*>„_,., =
K"+1 1-^vv2
d) Thus
<£„ = F(<£„, A:„) + 0(k2n)
"frn + kL
or
m=0
tan x w„ =
logtan(^ + |>) + O(A:2„ + (l-A:0))
n
2
it tan w,
log tan ( - + 2
+ 0(^ + (!-*„))
^1 /1 + ^:
m = 0
logS + O(A:2 + (l-A:0))

7.3 Elementary and Elliptic Functions
229
where S •"= yw\ + 1 + w0. Show, by inverting c), that tan-1 can be
calculated from log with complexity 0B(log n M(n)). Show that, by
inverting the above, tan can be calculated from log with complexity
0B(lognM(n)).
This approach, which avoids using complex arithmetic to access
the trigonometric functions, is essentially due to Brent [76a].
3. Let L(n) denote the bit complexity for evaluating n digits of log. Show
that
D(n)=0B(L(n))
and hence, log and exp are at least as complex as multiplication.
Hint: Consider computing the derivative of log.

Chapter Eight
General Means and Iterations
Abstract. In Section 8.1 we define abstract means and discuss their behavior.
In Section 8.2 we discuss equivalent means. In the next three sections we
consider general mean iterations and examine their convergence properties.
Later sections concern Taylor expansions of means, multidimensional means,
and related questions. The final section considers algebraic mean iterations
and the possibility of extracting elementary limits from such iterations.
8.1 ABSTRACT MEANS
There is a large literature on means but little agreement as to what
exactly constitutes a mean. For our purposes we have
Definition 8.1
(a) A mean is a continuous real-valued function M of two strictly positive
real variables a and b such that
(8.1.1) aAb<M(a, b)<avb
for all a>0 and b>0. We denote a a b : = min(a, b) and
a.v b : = max (a, b). (Continuity is not always essential. On occasion
we will refer to possibly discontinuous functions satisfying the above
definition as discontinuous means.)
(b) A mean is strict if, in addition, it is diagonal:
(8.1.2) M(a,b) = a or M(a,b) = b
if and only if a = b.
230

8.1 Abstract Means 231
(c) A mean is homogeneous if
(8.1.3) M(Xa, Xb) = XM(a, b)
for a, b, A>0.
(d) A mean is symmetric if
(8.1.4) M(a, b) = M(b, a)
for a, b>0.
(e) A mean is (strictly) isotone if, for a, b > 0,
(8.1.5) M(a,) and M{-,b) are (strictly) increasing.
We will find it convenient to consider the trace tM of a mean M given by
*m(X): =Af(jc, 1). We gather up some useful properties of means whose
proofs are left as Exercise 1.
Proposition 8.1
(a) Every diagonal continuous (strictly), isotone mapping is a (strict) mean.
(b) Suppose that M is symmetric and homogeneous. Then M is isotone if
and only if its trace tM is isotone.
(c) The isotone, (symmetric), (strict), (homogeneous) means form a
convex set.
(d) The symmetric, (homogeneous) means form a uniformly closed convex
set.
(e) Let M and JV be (strict) means. Then any continuous mapping P such
that
M(a, b) > P(a, b) ^ N(a, b) a,b>0
is a (strict) mean.
Corresponding to each mean M we associate another mean Mp, defined
by
(8.1.6) Mp(a, b):=[M(ap, bp)]llp p^O.
Then M = (M_1)_1 and M_x is strict, symmetric, homogeneous, or
isotone whenever M is, and tM_ (x) = t^(\lx). This is a special way of
building an equivalent mean as we now discuss.
For any strictly monotone (increasing or decreasing) continuous function
/: IR+ -> IR+ we define the mean
Mf(a, b) := f~\M(f(a), /(&))).

232
General Means and Iterations
When f(x) '■ = xp (we write / := lp) we denote Mf by Mp [consistently with
(8.1.6)]. It is easy to check that Mf is a strict, symmetric, or isotone mean
whenever M is. (See Exercise 2.) To give our discussion some flesh, we
introduce four of the most useful classes of means. In this chapter we
reserve the letters M and N for general means.
The Holder Means
ForpeUx let
/a" + bp\Vp
(8.1.7) Hp(a,b):=[——) a,b>0.
Then Hx is just the arithmetic mean A, and thus Hp is a strict,
homogeneous, symmetric, isotone mean. Moreover, limp_0 Hp(a, b) = Vab is the
geometric mean G and may reasonably be denoted by H0. (See Exercise
14.) Since H_x < H0 ^ Hx, Proposition 8.1(e) shows H0 to be a strict mean.
For allp one can unambiguously define Hpfora,b^0. Then the trace of H
satisfies
(8.1.8) VR ^{[0,2-^) p<0-
Note also that (Hp)_1 = H_p and that Hp = Ap.
Another useful way of building means is based on the next proposition.
Proposition 8.2
Let M be an isotone, homogeneous mean. Then, for p £ IR,
(8.1.9) pM(a,b): =
M(ap,bp) _ Mpp{a,b)
M(ap~\ b"'1) MppZ\(a, b)
defines another homogeneous mean, pM, which is strict or symmetric
whenever M is strictly isotone or symmetric.
Proof. This is left as Exercise 4a). □
The Lehmer Means
For p £ IR we let
(8.1.10) V»*):=flP-x + frP-i-
Since Lp = p{Hx), each Lp is a symmetric, homogeneous, strict mean. Since
tL(x) = (xp + l)/(xp~1 + 1), Lp is isotone only for 0<p<l. Moreover,
Li = Hx, Lll2 = H0, L0 = H_x, and (by Exercise 5 of Section 8.6) these are

8.1 Abstract Means 233
the only means common to the Lehmer and Holder classes. Again there is
no difficulty extending Lp to a, b s 0 and
II p>l foo p>0
\ p = l and tLp(oo)= J 2 p = 0
0 p<\ " U p<0
where, here and below, we write /(0°) for lim^^ f(x). To see this, use
(8.1.12) (Lp)_x(a, b) = "^**f = (1_rt(Lx)(a, 6) = L(1_p)(a, 6).
Indeed, generally
(8.1.13) (pM)_x = (1_p)M.
Tfee Gim Means
Let /■ and 5 be given. Consider f=iq where q'= s — r. Then
(8.1.14) r%l(f-Ma),Kb)) =
ar + br
= -G,Aa,b)
defines the Gini mean Gsr(a> b). (See Gini [38].) Moreover (8.1.14) shows
that Gs r is indeed a strict, homogeneous mean.
Proposition 8.3
Let / be a continuous strictly monotone function of a nonnegative variable.
Then
(8.1.15) MSf(a, 6):=/-1
b — a
a¥=b
extends to a symmetric, (generally nonhomogeneous), strict, continuous
mean.
Proof. The integral mean value theorem gives the conclusion, bar the
continuity. This follows from the continuity of the definite integral. □
We immediately apply this to Stolarsky's power means (Stolarsky
[75, 80]).
Stolarsky's Means
Let p £ IR and denote Mjlf-i by Sp. Then Sp is a homogeneous, symmetric,
strict mean given by

234 General Means and Iterations
(8.1.16/) Sp(a,b) =
with
ap-bp
p(a - b) J
i/(P-i)
P^O.l
(8.1.16b) S0(a, b) = lim Sp(a, b) = ^1^. =■ *fi, b)
and
(8.1.16uz) Sx(a, b) = lim S(a, b) = e~\aab~b)ll(a~b) =■ $(a, b) .
p—*l *
The mean iE is the logarithmic mean and 3 is the identric mean. These
means have
p-i'(p-D p>0,p*l
3 p=sO
4PW _ 1 (p-l)xp-pxp~1 + l
p
*s (°) = \ e P = 1 and ts (oo) = oo.
lo
Also
'*,(*) (p-i) (*p-i)(*-i)
so that Sp(a, b) is increasing in (a, 6) for all p in U. This, in part, follows
from the inequality (1 -p)xp + pxp~l si for 0<p ^ 1.
All of the means in these classes are piecewise monotone in the sense that
iM{x) has only finitely many sign changes. Thus there is no difficulty in
defining M(0, b) - bM(0,1) = b limxl0 tM{x) in all these homogeneous
cases, and we can freely consider M defined on IR+ := {(x, y)\x >0, y>0}
and at oo. We will do so from now on. Note also that when M is continuous,
'm([°> °°]) = [°> °°] is equivalent to tM(0) = 0 and tM(*>) = oo.
An elementary but very useful proposition is next. The proof is left for
Exercise 11.
Proposition 8.4 {Composition)
If M is defined by
M(a,b):=M0(M1(a,b),M2(a,b))
where Mt, i = 0,1,2, are means, then M is a mean. If two of M0, Mx, M2 are
strict, so is M. If all three are homogeneous, symmetric, or isotone, then so
is M.

8.1 Abstract Means
235
For example, M v N and M a N are strict means whenever M and N are.
So is any mean between them, in the sense of Proposition 8.1 (e).
There are many highly pathological means, as Exercise 12 shows. This is
particularly so in the absence of continuity. For future reference we will say
that a homogeneous mean is ultimately monotone if M(x, 1) and M(l, x) are
monotone in some neighbourhood of zero and °o. We write M < N if
M(a, b) < N(a, b) for all a, b > 0.
Comments and Exercises
There is a great literature on particular means and very little on means in
general. Much classical information can be found in Hardy, Littlewood, and
Polya [59] and in the other references scattered throughout the chapter.
1. Prove Proposition 8.1.
2. Establish that for any mean M and any strictly monotone /: IR+ —> U+,
Mf{a, b) := /_1M(/(a), f(b)) defines a mean which is strict,
symmetric, or isotone whenever M is. Moreover, if/is ip, p £ IRX, then Mf is
homogeneous whenever M is.
3. a) Show that Hp(a, b) is a continuous increasing function of p with
liny^ Hp = v and limp^_0O Hp = a .
b) Establish the assertions about the Holder means.
c) If M is homogeneous and symmetric, then M0 = G, whenever M0
exists as a limit.
4. a) Prove Proposition 8.2.
b) Establish the assertions about the Lehmer means.
5. (Isotonicity of Mp) Let 4>(a)-'=aloga for a>0, and let M be a
differentiable mean.
a) Show that Mp(a, b) is isotone in p > 0 if and only if
(8.1.17) ¢(,) ^¾^ + ¢(6) ™^> ^ <D(M(a, b)) .
b) Suppose that M is also homogeneous. Then
dM(a,b) , dM(a,b) ... ,.
a—^ + b^r^^M(a,b).
c) Use a), b), and the convexity of $ to show that Hp is increasing for
p rnU.
d) If M is homogeneous and symmetric and if Mp is increasing for
p >0, then Mp is increasing for p in 03, whenever M0 exists.

236 General Means and Iterations
6. Suppose Mp is isotone in p. Then
i) pM>Mp pal
ii) pM<Mp p<l
iii) pM<Mp_x p<0.
7. a) Suppose M is symmetric and homogeneous with tM(0) > 0 or with
'm(°°)<00> then
lim Mn(a, b) = a a b and lim M(a, b) = av b.
b) In particular, this holds for Hp.
c) If, in addition, Mp is isotone in p, then
lim „M(a, b) = a a b and lim „M(a, 6) = a v 6 .
d) In particular, this holds for Lp.
8. a) Show that pAf is isotone in p if g(p): = log M(ap, bp) is always
convex, since then g(p + 1)- g(p) increases withp. (See Becken-
bach [50].)
b) Use .Cauchy's inequality to show that this holds for
pA(a, b) = Lp(a, b) = flP-x + &P-x •
9. a) Establish the assertions about Stolarsky's means,
b) Show that
^Vfl + VPj <<£(fl)6)<#i/2(a6)
c) Show that
log*
1
1 + X
and deduce that
2~* . ,2-*
Jfc-1 Z
10. a) Show that i£p{a, b) is isotone in p by applying Exercise 5a) and
observing that (8.1.17) becomes $>(a, b)^iE{a, b). [This in turn
follows by calculus from (t - 1)2 2= dog t, t > 1.]

8.1 Abstract Means 237
b) Show that 3>p{a, b) is isotone in p This can be done by showing
that (8.1.17) becomes
<B(a) log (a)-¢(6) log (&) | 1:
a - b
which again reduces to (t- 1) s/log t.
c) The generalized mean Er is defined by
(8.1.18) Er>s(a, b) :
¢^)-^6)12
s(ar - br)
r(as - bs) J
V(r-s)
and extended appropriately on the boundary (as in Leach and
Scholander [78]). Show that each Ers - (5 ) for some p and q.
Let A(r) : = log \{d - l)lr\ so that A(r) = log 3>T(a, 1)- BY b) A(r)
is convex. Thus for 5 > /-,
log £^,1)=^:^^108^,1)
and £^ r increases in r and 5 because Err = $r.
d) The mean
„ , ,x „ / ,x a + 6 + Vofc 2 , 1
He(a, 6) := E3l2>ll2(a, b) = ^ = -A+-G
is very classical and is called Heronian mean.
e) Show that if -1 < p < \ or p > 2, then
(8.1.19) 5p(a,6)^i/(p+1)/3(fl,6)
with the inequality reversed whenp < — 1 or ^ ^p ^2. Moreover
(8.1.19) fails if (p + 1)/3 is replaced by any smaller number
(larger in the reversed case). (See Stolarsky [80] for details.)
f) Show that SB and 3>p tend to v as p tends to infinity.
g) Use the condition of Exercise 8a) to show that pSB is isotone in p.
h) Show that Ep p_y= pSB> SBp forp>l and so Ers> SBr for r> s>
0.
i) Show that
i) (£r„)-i = £-
") Er,s = Es,
iii) Er-' = Er,:'E'-'.
Prove Proposition 8.4 on the composition of means.
Let q be a nonnegative function satisfying
i) ¢(1) = 1
ii) 1 a x < q(x) < x v 1 x^l.
~r,~s

238 General Means and Iterations
a) Then
M(x,y): = xq(£)vyq(Z)
M{x,y):=xqy-j/\yq(^
and
are (possibly discontinuous), homogeneous, symmetric, strict
means,
b) Suppose in addition that
Hi) q(x) = xq[-J.
Then
M(l, x) = q(x) = **(*) .
Hence if we take an arbitrary (even analytic) function satisfying iii)
with l<q(x)<x for x>l, there is a strict, homogeneous mean
with tM = q.
c) Let q(x) '■ =
1 + 3x
x > 1, x rational
4
1 + x
otherwise
Then M is a densely discontinuous mean which is not ultimately
monotone. Moreover M lies between two continuous means.
(n \ 1/n n
2ckakbn-k) c,>0, 2ck = l.
1/n n
I c,>o, 2
Then P„ is a homogeneous strict mean which is symmetric if and only if
14. If M is a continuously differentiable mean then M0 : = limp^0 Mp exists
and is given by
M0(a, b) = aqb
«i,i-«
where q:= (dM/da)(l,l).
Hint: Use L'Hopital's rule.

8.2 Equivalence of Means 239
8.2 EQUIVALENCE OF MEANS
Let ¢: IR+ —> IR+ be continuous. We say that a mean M dominates a mean N
if
M(<D(a), ¢(6)) = <$>(N(a, b)) a,b>0
and we write M>^ N or M> N. If M and N dominate each other, we call
M and N equivalent. Since domination is transitive, this is an equivalence
relation which we write ~. Unfortunately most of our results demand that
we consider more restrictive notions of equivalence, so we require that $ be
one to one from now on. Hence 4> is monotone.
Theorem 8.1
Suppose that M and N are two means with M>$N.
(a) Suppose that N is homogeneous. For each t>0, consider
gt{x) ■= <b{t$?~x{x)). Then gt is isotone and
(8.2.1) M(a, b) = g;xM(gt(a), gt(b))
for each a, b in rng (¢).
(b) Suppose that M:=A. Then
¢ = (2(/ + /3 a>0,p^0
and N=Hp, p ^0.
(c) Suppose that M■= G. Then either
or
<& = aefl'P a>0, p^O
and N=Hp, pBU.
(d) It follows that the only homogeneous means equivalent to A are the
Holder p means (Hp, p t^O) and the only homogeneous mean
equivalent to G is G.
Proof
(a) By homogeneity of N we have, for t > 0,
<&_1Af(«&(ta), ¢(/6)) = rfTxAf(<&(«), ¢(6))

240 General Means and Iterations
and (8.2.1) follows. Moreover gt is isotone as a composition of two
co-monotone functions.
(b) Now (8.2.1), with M-= A, becomes
(8.2.2) 2g/(^p) = g,(a) + g,(b) a, b e rng (¢) .
This is solved [Exercise la)] by
(8.2.3) gt(x) = a(t)x + 6(0 xerng(4>)
for some a(i) and b(t) in IR. Thus
4>(fcc) = a(t)$(x) + b(t) x,t>0.
Now Exercise lb) shows that this is solved by
Since 4> is positive, we must have ¢ = atp.+ /3, a >0, and N is as
claimed.
(c) In this case we have
(8.2.5) Vft(«)ft(&) = ft(VS) fl,6erng(4)).
Let ht: = log ( gt ° exp). Then
h(a + b\ht(a) + ht(b)
As above ^,(^) = a(t)x + b(t) and
(8.2.6) <D(fcc) = 2?(0<&(jc)fl(,)
for B(t) positive. This is solved by
(8.2.7) ¢= aP p*0
1 ae
[as in Exercise lc)]. If 4> is of the first type, N= G; while if $ is of the
second type, N = Hp, p¥=0.
(d) Thus all Hp, p ^0, are equivalent and G dominates them all. D
A more general program may be undertaken, based on Theorem 8.1(a).
Unfortunately, without extra hypotheses, solutions of (8.2.1) are very hard
to characterize. The following simple result is accessible.

8.2 Equivalence of Means 241
Proposition 8.5
Suppose M is a homogeneous, strict mean with M>hN for some isotone h,
and suppose ^(0) = 0. Then tM(0) = 0.
Proof. Since h(N(a, 1)) = M(h(a), h{\)), we have A(0) = h(N(0,1)) =
M(h(0), A(l)). This is only possible if A(0) = 0, ft(0) = A(l), or h(0) = °o.
Since & is strictly increasing, only h(0) = 0 can occur. Thus h(0) = 0 =
M(0, /i(l)). Since M is homogeneous, fM(0) = 0 as claimed. □
As an example we see that Hp>h AG, p > 0 (AG is the Gaussian AGM)
is impossible for h isotone, as is Lp >h AG withp > 1. These considerations
suggest that we can say more if h is required to be onto. We will write
M~ N if M>hN for some one-to-one mapping h of IR+ onto IR+ [so that h
must be (strictly) isotone with h(0) = 0, /i(°o) = oo or (strictly) antitone with
h(0) = oo, /i(oo) = 0]. This is an equivalence relation stronger than ~, and we
call it strong equivalence. We will only consider strong equivalence of
homogeneous means. If we let gt '.= h(th~ ), gt is surjective and we observe
that (8.2.1) becomes
gt(M(a, b)) = M(g,(a), g,(b)) a,b>0.
Since M is homogeneous, we may replace gt by ht := gtlgt{\). Then ht{\) =
1 and
(8.2.8) h,(M(a, b)) = M(ht(a), h,(b)) a,b>0.
In the next lemma we give several conditions for (8.2.8) to have only the
trivial solutions (normalized). Under these conditions we have gt = c(t)u and
(8.2.9) h{tx) = c{t)h{x) t,x>0.
Then Exercise lb) shows h = aip, a >0, p ¥0, and it follows that the only
homogeneous means strongly equivalent to M are Mp, p^0. Thus we have
determined the strong equivalence class of M in the cases covered by the
next result.
Lemma 8.1
Suppose M is a homogeneous strict mean.
(a) The following two conditions imply that (8.2.8) only has the trivial
solution when ht{\) = 1.
(i) M is ultimately monotone and
(ii) ht{a) = a for some a¥^\, a >0.

242
General Means and Iterations
(b) (i) (ai) holds if M is isotone (or piecewise monotone), while
" (ii) (aii) holds if M is differentiable and (1) tM has a unique positive
zero, a ^ 1 and h is differentiable; or (2) tM(U+) £ IR+.
Proof.
(a) Given that a and 1 are distinct positive fixed points of ht, it follows
from (8.2.8) and continuity that {c>0\hl(c) = c} is a closed interval
C. We show that s '■ = sup C is °o. If not we argue as follows.
For any c<s in C, M(c,s)<s. Hence. M(c, s + e) .< s for some
e > 0. But M(s, s + e)> s. Thus M(c, s '+ e) = s for .some £.< 5, c in C,
e>0. Then for n = 1,2, 3,..., •
s = h"(s) = M(hnt(c), h"(s + e)) = M(c,hnt(s + e)) .
(Here h" : = hn'1°h denotes iterated composition.)
Let bn■:= c_1/i"(5+ e). Then, as bx ¥=ba (because s+ e0C), bnis
a strictly monotone sequence, since ht is strictly isotone. If {bn} were
bounded above, the limit point would be a member of C larger than s.
Thus bn increases without bound and also tM{bn) = sic. This violates
monotonicity of M at infinity. The proof that inf C = 0 is left as
Exercise 2b).
(b) We consider (ii). If h and M are differentiable, we have from (8.2.8)
'm(^W)^00 = ht(tM(x))iM(x) .
Since ht is strictly increasing, we have iM(ht(a)) = 0 and ht(a) = a. If
on the other hand tM is not surjective, we argue as follows. Suppose
M(0,1)>0. Then
ht(M(0,1)) = Af(A,(0), A,(l)) = M(0,1)
as ^,(0) = 0, and M(0,1) < 1 is a second fixed point of ht. Finally if
M(°o, 1)<°°, we argue with M_x. □
Condition (ai) holds for Hp, Lp, and Sp. Condition (2) of (bii) holds for
Hp, p^O; for Lp, p <0 or p > 1; and for 5p, p >0; while condition (1)
holds for Lp, p > 1. (See Exercise 3.) It is reasonable, at least for
computational purposes, to require that equivalence be defined by differentiable
maps. If this is done, the results are more complete.
Comments and Exercises
Theorem 8.1 can be found in Wimp [84] in slightly different form. This is
partly explained by the fact that exp[(logx + log y)/2] = G(x, y). For
Hardy, Littlewood, and Polya [59] or Wimp [84] this means that A >log G.

8.3 Compound Means
243
In a more formal development this is problematical since means are only
defined on (0$+ ) and log is not always positive. Thus we rule out the second
case in (8.2.4).
1. a) Show, using continuity of gn that (8.2.2) is solved as in (8,2.3).
(Proofs without continuity assumptions can.be found in Wimp [84]
and elsewhere. More general results depend on the measurability of
Sr)
b) Let 4/(x) : = 4>(;c) - ¢(1). Then <// satisfies .
■;*""■ ij/(tx) = a(t)ij/(x) + c(t)
for some c(t) in R. Then (f/(t) = c(t) and thus [a(x) -1] lip(x) =
[tp(tx) - tp(t) - tp(x)] /tp(t)4/(x) is independent of x if t/>(*) ^ 0- Thus
ij/(xy) = aj/(x)il/(y) + 4>(x) + ij/(y). If c = 0, then t^(jt) = Clogx,
while if c^O, A(x) : = c(fi(x) + 1 solves k(xy) = X(x)k(y) whose
solution is tp. Thus (8.2.4) holds.
c) Show that (8.2.6) has solutions only of form (8.2.7). Hint: Consider
A(jc) := log [<&(*)]-log [¢(1)]. Then A(jtf) = a(t)X(x) + k(t). Now
proceed much as in b).
2. a) By considering M_1 show that if M >hN, M strict and
homogeneous, h isotone with fM(°°) < °°, then ^(00) < °°.
b) Complete the proof of Lemma 8.1(a).
c) Complete the proof of Lemma 8.1(6).
3. a) Verify the final claims of this section.
b) Calculate the strong equivalence classes within Grs and Ers.
4. a) If M is a symmetric, homogeneous, strict mean, then either tM(0) =
0 or fM(oo) = oo.
b) Similarly, suppose then that M>hN for some homogeneous N with
h monotone. Show that 0 or oo Hes in mg(h).
8.3 COMPOUND MEANS
We now formalize the notion of a mean iteration. The Gaussian AGM and
Archimedes' method provide the central examples. Let M and ./V be any two
continuous means. Let a > 0 and b >0 be given and consider the iteration
a0'-= a b0'-= b
(8.3.1) an+1:=M(an,bn)
bn+1-=N(an,bn).
Under mild hypotheses, Theorem 8.2 shows that the iterates converge to
a common limit, which we call the compound of M and ./V and denote by

244
General Means and Iterations
M(x)N(a, b). We will also denote the entire iterative process by [M, N]. We
say that M is comparable to N if one of the following holds:
(a) M(a,b)^N(a,b) for a, b>0
(b) M(a,b)<N(a,b) for a, b>0
or
(c) M(a,b)<N(a,b) for 0<a<b
and
N(a,b)<M(a,b) for 0<b<a.
When M and N are symmetric, c) cannot occur, and we say M and N are
comparable. In the nonsymmetric case c) can occur, and then M is
comparable to N but not conversely. In the next result comparability is only needed
to establish monotonicity of the iterates. (See Exercise 1 and Theorem 8.8
of Section 8.7.)
Theorem 8.2
Let M and N be means with M comparable to N.
(a) Suppose that M or N is strict. Then [M, N] converges and M (x) N is a
mean which is strict if both M and N are.
(b) M (x) N is a homogeneous, symmetric, or isotone if each of M and N is.
(c) M®N is continuous, and the convergence is monotone and uniform
on compact subsets of {(a, b)\a, b >0}.
Proof
(a) Suppose that a > b and that M>N. Then ax = M(a, b) > N(a, b) = bv
Inductively suppose that an^bn. Then
(8.3.2) an > M(an, bn) = an+1 > bn+1 = N(an, bn) > bn
and {an} decreases while {bn} increases. Since each bounds the other,
both sequences converge, say, to x and y, respectively. By continuity
we have x = M(x, y) and y = N(x, y). Since M or N is strict, x = y. If
a ^ b, we may have to exchange the roles of an and bn. Thus M®N
exists and satisfies
(8.3.3) M aN<M®N<Mv N.
This finishes (a).

8.3 Compound Means
245
(b) The sequences {an} and {bn} are in fact built by repeated composition
of means. Thus an = Mn(a, b) and bn = Nn(a, b) for means Mn and Nn.
These means are symmetric, homogeneous, or isotone when M and TV
are, and so is the limit mean M®N.
(c) Let a, b > 0 and e > 0 be given. Pick n so that \an - bn\< e/2. Now Mn
and Nn are continuous. So we may find 8 > 0 with \Mn(a, b) -
Mn(a',b')\<e/2 and \Nn(a, b) - Nn(a', b')\ < e/2 if \a'-a\<8 and
\b' - b\ < 8, (a', b'>0). Again assume a > b and M>N. Then
M®N(a', b') < M„(a', 6') < M>, 6) + | < #>, 6) + e
<M®N(a,b) + e
and
M®AT(fl', b')>Nn(a',b')>Nn{a, b)~ | >M„(a, 6)- e
>M®N(a,b)- e .
Thus M (x) A^ is continuous. Finally, Dini's theorem shows that Mn
and Nn must actually converge uniformly on compact subsets, since
convergence is monotone. □
The key observation about M(x) N is the following 'invariance principle'
which we have already used repeatedly in Chapters 1 and 2 to show that
AG(1, k') = irllK.
Theorem 8.3 {Invariance Principle)
Suppose that M®N exists. Then M® N is the unique, (continuous) mean
4> satisfying
(8.3.4) 4>(M(a, b), N(a, b)) = <D(a, b)
for all a, b > 0.
Proof. Iteration of (8.3.4) shows that
Thus
<D(a, b) = <D(M ® N(a, b), M ® JV(a, 6))
and since 4>(c, c) = c, ¢ = M®N. □

246 General Means and Iterations
Observe that we need not verify that 4> is a mean, but only that
4>(;c, x) = x for x > 0 and that 4> is a continuous solution of (8.3.4).
EXAMPLE 8.1
(a) Let M:=HX and N:=H_X. Then HX®H_X = G. Observe that
G(a, b)'-= Vab satisfies
la + b lab , ,
G(Hx(a, b), H_.(a, b)) = yj— —^ = Vol <D(a, b) : = Vdb.
Since Vxx = x, we must have Hx (x) H_x = G.
(b) Let M(a, b) : = 9ab2/(a + 2bf and N(a, b): = (a + 2b) /3. Then
Mv3(a, b)N2l\a, b) = aV3b2'3 <D(a, b) : = aV3b2'3
and again the invariance principle shows that M ®N(a, b) = all3b213.
(c) Let tfx(x)tf0 = AG:=,4(x)G. Then for a > 6,
a ^ ^, . x .,-,/., M air ,, ,. air
A®G(a,b) = aAG{l,-J = ^m ¢(,,6) :=^^^
as Theorem 1.1 (second proof) shows by the invariance principle.
We now distinguish two better structured classes of mean iterations.
Definition 8.2
Let M and N be symmetric means.
(a) Suppose M and N are comparable. Then we write
M®gN:=M®N and [M, N]g:=[M, N] .
We call these Gaussian mean iterations and call (x) the Gaussian
product.
(b) Consider the iteration: a0 : = a > 0, b0 := b > 0, and
an+x:=M(an,bn)
bn+1-=N(an+1,bn).
We denote the iteration by [M, N]a and the limit by M (x)a N. We call
these Archimedean mean iterations and call (x)a the Archimedean
product.
The existence of M®gN is guaranteed by Theorem 8.2. For M®aN
existence comes from the next result.

8.3 Compound Means
247
Proposition 8.6
Let M and N be symmetric means. Suppose that M is strict. Then M(x)aN
exists and
M®aN=M®N*
where N*(a, b) := N(M(a, b), b). So by Theorem 8.3 M®a N is the unique
continuous mapping t/> satisfying
<KM(a,b),N*(a,b))=4,(a,b)
with 4>(x, x) = x for x > 0.
In consequence M (x)a N is a continuous mean which is strict,
homogeneous, or isotone whenever both M and N are.
Proof. This follows from Theorem 8.3 since M is comparable to N*
(See Exercise 4.) □
The theorem makes no use of symmetry of M and N. However, this is
critical to our further analysis of convergence rates. Finally, suppose that we
are given a function M which satisfies (8.1.1) and (8.1.2) or (8.1.3), but only
for 0 < a < b. We may extend M to a symmetric mean M via
(8.3.5) M(a,b):=M(aAb,avb).
Similarly we can extend a function defined only on 0 < b < a by using
M(a v b, a a b). The new mean is homogeneous or strict when M is. Thus
in some of our future iterations we will consider means only defined on the
45° sectors (Exercise 3 among others). Moreover, if we have two comparable
means on 0 < b < a, the extensions are comparable so that our convergence
results apply. Note also that in the definition of comparability we excluded
the possibility that M<./Vfor0<6<a while N ^ M for 0 < a < b. In this
case the iterates oscillate, and combining two steps results in a comparable
iteration. We call such iterates partially comparable.
Comments and Exercises
Our treatment is a synthesis and extension of that in Schoenberg [77,82],
Lehmer [71], and Foster and Phillips [846] among others. The term
compound is due to Lehmer [71]. In general it is very hard to determine
M (x) N, but easy to verify a limit once one has found it. Schoenberg [77]
gives a geometric proof of the limit of the AGM due to Jacobi.
1. a) Let M and JV be strict means. Then [M, JV] converges and M (x) N is
a strict (continuous) mean.

248 General Means and Iterations
Hint: {a„ v bn} and {a„ a 6J are monotone sequences, and so
an v bn decreases to x and an a bn increases to y. One may suppose
x = (M v N)(x, y). Thus x = y by Proposition 8.4 and M®N
exists. Since MaN^M®N-^MvN, the limit is a strict mean,
b) M ® JV is symmetric or homogeneous when M and JV are.
2. a) Let M be a strict mean. Then
M® v = v®M = v .
b) Show that
(M®N)f = Mf®Nf
where as before Mf(a, b) '•— /_1M(/(a), f(b)).
c) In particular a) and b) show that
M® A = A®M= A .
d) If M and JV are symmetric, then
M®N = N®M .
[More generally, see Exercise 6.] Thus
M®gN = N®gM.
3. (Carlson's log) Define means M(a, b): = \ja • (a + b) /2 and
N(a, b): = y/[(a + b)ll\-b = M(b, a). Show that
a — b
Hence [using Exercise 2b)],
■M1I2®N1I2 = 2.
icitly, if
. an + y/anbn
the limit is
and
b — a
bn + l
K + VaJn
log 6 - log a
Note that this is neither a Gaussian nor an Archimedean iteration.

8.4 Convergence Rates and Some Examples
249
4. Prove Proposition 8.6 from Theorem 8.3. Note that in this setting
M < N* for 0 < a < b and M > N* for 0 < b < a.
5. a) Show that given strict symmetric means M and N,
M®a N(b, N(a, b)) = N®a M(a, b).
b) Show that in the special case in which M = N the limit
M ®a M =: 4> is characterized by
<D(a, b) = 0(b, M(a, b)).
c) Show [using b) or otherwise] that
i) A®aA(a,b) = ^~
ii) G®aG(a,b) = aV3b2n
(ap +2bp\Vp
hi) Hp®aHp(a,b) = { 3 j p*0.
6. For any function Q let Q(a, b) '■= Q(b, a). Prove that for any means M
and JV,
M®N=N®M
and
M®aN(b, N(b, a)) = N®aM(a, b).
8.4 CONVERGENCE RATES AND SOME EXAMPLES
In this section we show that Gaussian iterations typically converge quadrati-
cally and Archimedean iterations sublinearly. We also give some more
examples of Gaussian and Archimedean iterations for which we can calculate
the limit.
EXAMPLE 8.2 (ARCHIMEDES' METHOD)
(a) Let an denote the area of a regular m ■ 2"-gon inscribed in a unit circle.
Let bn denote the area of the circumscribed regular m ■ 2"-gon. It is
easily verified that an+1 = ^Janbn and bn+1 = 2a„+16„/(a„+1 + bn) while
a0 = \m sin (2ir/m) and b0 = m tan (rr/m). Thus we geometrically
verify that an and bn tend to it. This gives G®aH_1(a0, b0) = it or, on
using homogeneity and replacing 2ir/m by 0,

250 General Means and Iterations
(8.4.1) G®aH_1{sin 6,2 tan (f )) = « ■
There is no need for m to be integral and (8.4.1) holds for all
0 < 6 < ir.
(b) Consider now the same process but using circumscribed and inscribed
perimeters an and bn. Now we have an+1 = 2anbJ(an + bn) and
fr«+i = Va«+i^„ while a0 = 2m tan (ir/m) and b0 = 2m sin {Trim). Then
a„ and bn tend to 2ir and
(8.4.2) //_x(x)aG(tan0,sin 0) = 0.
This is pursued further in Exercises 1 and 2. □
A complete analysis of the iteration of Schwab, Borchardt, Pfaff, and
Gauss is in Miel [83]. The central observation is:
Theorem 8.4 (Schwab-Borchardt)
(8.4.3) A®aG(a,b)
Vb^V2
0<a<b
arccos(a/6)
a a — b
Va^-b2
arccosh(a/6)
0<b<a
Proof. Schwab established the first case by geometric arguments which
Schoenberg [82] reproduces. Given the formulae, it is simpler to use the
invariance principle. Since a2n+1 - b2+1 = (a2 - b2)/4, this reduces to
showing that
2arccos (—— J = arccos (^-) a„<b„
2 arccosh ( -r^) = arccosh (^-) a„>b„
(8.4.4)
Since an+1/bn+1 = \/(anlbn + 1)/2, this is just the half-angle formula for cos
or cosh. □
We now turn to study rates of convergence for Gaussian and Archimi-
dean means.
Theorem 8.5
Let M and JV both be continuously differentiable symmetric means and
suppose that at least one is strict.

8.4 Convergence Rates and Some Examples 251
(a) Consider the Archimedean iteration [M, N]a. Then, if an ¥= bn,
(8.4.5) lim fl"+1~^"+1 = \ .
»— an-bn 4
(b) Consider the Gaussian iteration [M, N]g for comparable M and N.
(i) Then, if an*bn,
(8.4.6) limfl"+1~^"+1=0.
n n
(ii) Suppose, in addition, that M and N are twice continuously
differentiable. Then, if an¥=bn,
(8.4.7) lim '«■"-»■+'' = 1^(^)-^(^)1
where s '■= M®gN{a, b).
Proof. Since M(c, c) = c for all c, we must have Mx(c, c) + M 2(c, c) =
1 for all c. (Here M,. denotes the partial derivative with respect to the ith
variable.) Since M and N are symmetric, it follows that M^{c, c) =
N/c, c) = \, i = 1, 2.
(a) Let 5 := (M ®a Ar)(a, 6). The mean value theorem gives
(8.4.8) an+t -s=±(an-s)+Hbn-s) + o(an - s) + o(bn - s)
and since bn+1 = N(an+1, bn),
(8.4.9) bn+1 - s = K«n+1 ~s)+Hbn-s) + o(an+1 - s) + o(bn - s)
= *(«„-*)+ !(&„ -^) + o(fl„ -s) + o(bn - s).
Thus
an+1 - bn+1 = \{an - bn) + o(an -s) + o(bn - s)
and (a) follows.
(b) (i) Similarly, (8.4.8) still holds, and also
bn+x ~s=\{an-s)+ \{bn -s) + o(an - s) + o(bn - s).
(8.4.10)

252
General Means and Iterations
Thus
an+i ~ bn+1 = o(an -s) + o(bn -s) = o(an - bn)
(since s lies between an and bn). This gives (b).
(ii) Given one more derivative, we have
«„+i -s=Ha„-s)+ \(bn -s) + \Mu(s, s){an - bnf
+ o{an-bnf
and a similar formula for bn. Subtraction gives (ii). Here we have
used V2M(s, s)(an — s,bn — s)2 = Mn(s, s)(an - bn). [Exercise
7d).] □
Considerably more can be said if M and N are two or three times
continuously differentiable. (See Foster and Phillips [846].) In addition the
convergence in each case is uniform. For our purposes Theorem 8.5 suffices.
In particular, Archimedean iterations characteristically converge linearly
and Gaussian iterations super linearly (quadratically if twice differentiable).
In light of Jacobi-type methods in the theory of equations, this may seem
counterintuitive. Exercise 3c) shows that (8.4.6) may fail if M and N are not
differentiable. The theorem justifies our separating Gaussian and
Archimedean iterations. While symmetry is central to our convergence arguments, it
is not always essential, all that is really needed in (a) and (b)(i) Theorem
8.5 is that the two means have the same gradients on the diagonal. (See
Exercise 11.) Also, (8.4.7) shows that better than quadratic convergence is
possible only if Mn = Nn. (See Exercise 7.) Additional information is given
in Exercise 11.
Finally, let us say that two iterations are equivalent ([M, N] ~h [M', N'])
if M~hM' and N~hN'. Strong equivalence is similarly defined. Clearly
equivalence of iterations implies equivalence of the limits, but not
conversely. Indeed, if the mapping h is continuously bi-differentiable, the rates of
convergence must be the same. (See Exercise 8.)
Comments and Exercises
Some of the Archimedean considerations are discussed again in Chapter 11.
If we take m'-=6 in Example 8.2(6), we have a recursive version of
Archimedes' original method. (See also Edwards [79] and Phillips [81].)
1. a) Show that (8.4.1) and (8.4.2) are consistent with the general
formula of Exercise 5a) of Section 8.3.
b) Show directly that in both (a) and (b) of Example 8.2 we have
an+\ ~ bn+1 ~ \(an - bn). This illustrates why computation of any
large number of digits of it by this method is impractical.

8.4 Convergence Rates and Some Examples 253
c) Show that in both cases (an + 2bn)l3 =: dn satisfies
dn - lim dn 16
2. a) Show that changes of variables cause no problems for Gaussian or
Archimedean iterations, that is,
i) (M®gN)f = Mf®gNf
ii) (M®aN)f = Mf®aNf.
b) Show that (8.4.2) implies that
A®aG(x,l) = - 0<*<1.
a arccosx
Hint: Let x-= cos 0 in (8.4.2) and use aii).
c) Rederive Theorem 8.4 from b).
The next exercise shows a class of means which is closed under the
Gaussian product. Contrast this with the Gaussian product of Holder
means.
3. Let Qt{a, b) := t(a v b) + (1 - t)(a a b) for 0 < t < 1.
a) Show that Qt is a homogeneous, symmetric, isotone mean which is
strict for 0 < t < 1.
b) Show that for t > s,
Qt®gQs = Qs/[s+(i-o] ■
This may easily be done by the invariance principle. Alternatively,
suppose a > b > 0 and observe that the limit must be linear.
c) Show that \an- bn\ = {t- s)n\a-b\.
d) showthate,®aa = e,®gar
4. Let M and JV be symmetric, homogeneous means.
a) Show that M®gN = G if and only if N = M_x.
b) Show that M ®g N = A if and only if N = 2A - M. Note that M is
a mean if and only if 2A - M is.
c) Characterize M®gN= Hp.
d) Show that if M{a, b): = \/(a ~ bf + ab, then
M® G = H2.

Explicitly,
*n+x:=^la2n + b2n-anbn and bn+1: = VaJ~n
converges to V(ao + b0)/2.
e) If M:=(HX + H_X)I2 and N: = (Hx + L2)/2 then M®gJV=A
. Explicitly,
: _a2n + b2n + 6anbn __3a2n + 3b,2 + 2anbn
dn+1[~ 4(a„ + &„) and &-- . 4<«. + &„"). ^
* .-
cbnverges to (a0 + b0)/2.
(A variant of the AGM) Show that
2 /w*\ 2 / //x
Explicitly, the common limit of
vi-—4—+ -T" and *»■» - ^+&;+-r-
is 2AG(a0, 60)2/(a0 + 60). (See P. B. Borwein [Pr].)
a) Show that M(a, b):=(a + b + 2ab)l(a + b + 2) is a strict
symmetric nonhomogeneous mean with M = M_,.
b) Show that M ®a M(a, b) = (a + 2b + 3ab)/(b + 2a + 3).
Hint: If A(f)':= f/(l + 0, then M = Ah.
a) Use Theorem 8.5(6) to show that [Hp,Hq]g, [Hp, Lq]g, and
[Lp, Lq]g converge quadratically, not better, for p¥= q.
b) (The quartic AGM) The iteration [A,N]g, where N(a,b): =
[(a3b + b3a)/2]114, converges with order 4. [See Exercise 3 of
Section 1.4]
c) Reformulate the cubic modular equation as a third-order Gaussian
mean iteration to compute AG.
d) Show that for any symmetric twice differentiable mean,
Mn(c,c) = -M12(c,c).
a) Verify the assertions about equivalent iterations.
b) Show that [Hp, Hq]g~[Hp., Hq,]g if and only if p*q = q*p.
(Suppose that p and q are not both zero.)
Hint: One may suppose p = 0 or 1 and use Theorem 8.1.
c) Thus [Hx, H2]g and [A, G]g are not equivalent.
For 0<yt<l let M(a,b):= (l-l/k)a +(l/k)b and N(a,b): =
fl*~16/[(l-l/ik)fl + (l/ik)6]*~1.
a) Show that M®N(a,b) = a1~Ukbllk.
b) Show directly that convergence is quadratic. (Theorem 8.5 does
not apply, but gradients do coincide on the diagonal.)

8.4 Convergence Rates and Some Examples 255
10. If M and N are homogeneous and L'= M (x) N, show that for x > 0,
L(0,x) = xL[tM(0),tN(0)].
Thus although we do not know an explicit formula for L'.= Hx (x) L2,
we have L(0, x) = xL{\, 1).
11. Suppose that M and N are comparable continuously differentiable,
homogeneous means such that M®N exists.
a) Show that if an ¥^ bn,
lim , + 1~b"t =\M1(s,s)-N ,(s, s)\
n n\
where s: = M (x) N(a, b).
b) What conditions, other than symmetry, in conjunction with part a)
now guarantee linear or superlinear convergence?
12. Suppose that M and JV are continuously differentiable, homogeneous,
symmetric means. Let
M*(a, b) := aa + (1 - a)M(a, b)
0<a<l
N*(a, b):=aa + (l- a)N(a, b).
a) Show that [M*, N*] still converges superlinearly.
b) For M:=A and N:=G [so that M*(a, b) = (3a + b)IA and
N*(a, b) = (a + Vab)/2] show directly that convergence is
quadratic.
13. The ten neo-Pythagorean means are /x := A, f2'-= G, /3 := H, /4 := L2
(also called the contraharmonic mean), and
f5{a,b):=\d+\{d2 + Am2)xn
f6{a,b):=-\d+h{d2 + m2)xn
/7(a, 6):= m + d2/m
—2
f9(a,b):--
m + d
f&{a,b)'-=m- d2lm
f10(a, b):=im + l(m2 + 4md)U2
where m '■= a a b, m '■= a v b, and d'-=m- m.

256
General Means and Iterations
a) Show that /5 to /10 are symmetric, homogeneous means which are
not differentiable on the diagonal. Thus Theorem 8.5(a) does not
apply.
b) Show that/7 ®g /g = A, and that [/7, fs]a converges sublinearly for
a < b but quartically for a > b.
c) Show that /8®g/7(2>l)= 1, and that we have one-step
termination—which cannot happen for strictly comparable means.
These means were originally defined by the Greeks in terms of
proportions. For/:= 1 to 10, respectively, x'-= fj(a, b) solves
CD IF"
(5)
(7)
(9)
m ■
x-
m ■
x -
m -
m -
m
m
- X
- m
- X
- m
- X
- m
— X
- m
m
m
m
m
m
X
m
m
m
m
X
(2)
(4)
(6)
(8)
(10)
x — m m
m - x x
x — m m
rn — x m
x — m m
m - x x
rn — m m
x - m m
m — m x
x - m m
d) Verify that x = fj(a, b) in each case.
e) Verify that these are the only such means.
8.5 CARLSON'S INTEGRALS AND MORE EXAMPLES
This section is largely given to a description of a unified approach to Gauss's
and Borchardt's algorithms due to Carlson [71]. It shows both the
possibilities and the limitations of looking for iterative methods based on a
more general hypergeometric transformation.
Let us consider the integral
ll(«;8,8Vx*/):=]S^/0V,-1(r + ^)-'a + /r'A
(8.5.1)
where a + a' = S + 8'; re(a), re(a') > 0, and /3 is the beta function. (The
prime is not complementation in this context.) Then obviously
(8.5.2) R(a; S, 8'\ x2, y2) = R(a;8', 8; y2, x2)
and R is homogeneous of degree -a'mx2 and v . (See Exercise 1.) We are
interested in

8.5 Carlson's Integrals and More Examples 257
(8.5.3) Cij:=Fi®Fj i,/ = 1,2,3,4
where the means Ft, i = 1,2, 3, 4, are given by
Fx(a,b):=^~ F2{a,b):=Vab
^, ,x la + b „, ,. ja + b ,
F3(a, 6):=y-y-a F4(a,b):= y-y- 6 .
Thus C12 = AG, C34 produces Carlson's log (Exercise 3 of Section 8.3), and
Cu ~ A®aG leads to Borchardt's algorithm (Theorem 8.4).
Theorem 8.6 (Carlson)
Let i, j = 1,2, 3, 4 with j ^/.
(a) [F/; .Fj] converges.
(6) Cjy(a,6) = [i?(a;S,S';a2,62)r1/2a
where (a; S, S') is given by the (i, /)th entry Table 8.1.
(c) Convergence is linear except for C12 = C21, which is AG, the limit of
the Gaussian AGM.
TABLE 8.1
1 /=1 1 = 2 / = 4 / = 3
1 * (2! 2) 2) (2! 2) 1) (4) 3> 2)
2 G;M) * (l; 4,1) (i;M)
3 (i;l,i) (1;1,1) (1;1,1)
4 (i;i,3) (l;i,i) * (1-,1,1)
Since Fx and F2 are symmetric while F3(b, a) = F4(a, b), up to exchange
of 8 and 8' the table is symmetric around the main diagonal because of
(8.5.2). Boxes marked with * correspond to trivial iterations.
Proof.
(a) Since F4 < F2 < Fx < F3 for 0 < b < a, each pair of means is partially
comparable. By Proposition 8.4 on composition, each mean is strict.
Thus Theorem 8.2 establishes (a).
(b) Let us denote Ft(a, b) by ft. Make, in (8.5.1), the substitution
^=5(5+/2)/(5+ /2). Then

258 General Means and Iterations
and
* =s +2sf2l+flfl = (^+/3)(^+/4)
ds (s + flf (s+fl)2
.2 _ (s+fl)2 _t2 {s+flf
t+a=- '-%- t + b =
s + fl s + fl
Thus
(8.5.4)
R(a;S,S';a2,b2)=j^-^[sa'-\s + fir-\s + f2r-'
x (s+fiy-^is+f4y-2s' ds.
If we fix i and /, we can determine values of the parameters for which
fk, k¥= /, /, vanishes in (8.5.4). For example, with /:=3 and7 := 4, we
set a : = a' ■ = 1 and have
(8.5.5) i?(l; 8,8'; a2, b2) = R(l, 28-1,28'- 1; f\, f2)
where 8 + 8'= 2. There is a unique value S:=l so that (8.5.5)
becomes invariant for/3 and/4. By Exercise lb), [R(l; 1,1; a2, b2)]~112
is an invariant for [F3, F4] to which Theorem 8.3 applies. Similarly, for
/:=1 and j: = 2, we set 8 '■ = 8': = \, which gives a + a' = 1 and
R(a; \,\;a2,b2) = R(a;l-a,a; f\, f\).
The only possible invariant has a:=a'= \, and by Exercise lb),
[^(5; 5) \',a2, b2)]~l is an invariant for [Fx, F2]. The rest of the table
is similarly verified. [See Exercise lc).]
(c) The convergence assertions are straightforward. [See Exercise
Id).] □
example 8.3. As observed before, C12, C34, and C14 have been previously
identified. In Exercise 2 we indicate how to recover the previous forms of
the limit. In particular Theorem 8.6 gives integral representations for these
means. Now consider /:=1 and /:=3 (or similarly, /:=1 and /: = 4).
Suppose that a> b. Let
(8.5.6) 1-^:=^.
t + a

8.5 Carlson's Integrals and More Examples 259
Then 54 = (a2 - b2) l(t + a2) and
MA 4-,-1/2 , /2 I.2\l/4/. , 2-,-3/4/. , .2-,-1/2 ,.
4(1 - s ) ds = -(a - b ) (t+ a ) (f + o) af.
Since a := |, S := f, and 8' ■= \, we have a' = 1, /3(a, a') = 4, and
ra-iW4
-(a2-6T1/4J0 d-/rl/2^.
Recall that the arclemniscate sine is
arcslx: = Jo (l-s4)~U2ds x2<l
and gives the arc length of the lemniscate (/-2 = cos 20) from the origin to
the point with radial position x. (See also Theorem 1.7.) Thus
(8-5J) c«(fl'fc)=KiF^fF °^b<a-
Similarly,
(8.5.8) Cx3(a, b)
where the hyperbolic arclemniscate is defined by
arcslhx:= f (1 + s4)~1/2 ds .
[See Exercise 2d).] Observe that V2C^I2(1,0) = K(l/V2), by Theorem
1.7.
example 8.4 (Archimedean means) Let us consider Hp ®a Hq for p, q =
±1, 0. Then the previous exercise and Exercise 5 of Section 8.3 show that it
suffices to consider Hx®aHx, Hx®aHQ, and Hx®aH_x. Exercise 5c) of
Section 8.3 gives Hx ®a Hx(a, b) = (a + 2b) /3 while Hx ®a H0 is Borchardt's
algorithm. It remains to study Hx ®aH_x. With a ■= /3 := \, this is a special
case of the mean iteration a0 ■= a > 0 and b0 : = b > 0,
an+x.= aan + (l-a)bn bn+x - ^^ + (1 _ ^
V^T2 n ,
[arcslh(V/a2-l)l/4]2 U<fl<&

260
General Means and Iterations
where 0 < a < 1 and 0 < /3 < 1. In the notation of Exercise 3 of Section 8.4
this computes G-= Qa®a(Qfi)_1 for a>b. If we let kn'-=bnlan, we
derive
^1-l = (a]8)(C-l) = («i8)"+1(f-
and
Thus
T = i + w(H
= a + (1 - a)kn
and the limit is
G(fl|fc)"fliUi-W(i-«/&)J-
If we let 1 - a/6 =: x and a(3 ='■ q, we have
(8.5.9) g(i,~-) = 11
1- (ax)g"
1-V
n:.0[i-(a*)g"]
which is the ratio of thetalike products occurring in the ^-binomial theorem.
(See Exercise 7 in Section 9.4.) Wimp [84] continues a discussion of similar
calculations, all of which give linear convergence.
If we let a/3 : = 5 and a : = \, we have a product expansion for
Hx ®aH_x. Foster and Phillips [84a] show how this function can be closely
approximated by elementary functions and can be given an elegant
asymptotic expansion.
Comments and Exercises
Many other related algorithms can be found in Wimp [84]. This includes an
extensive discussion of quadratically computable trigonometric integrals (in
Chapter 14).
1. a) Show that R(a; S, 8'; x2, y2) = y~2aF(a; 8, S + 8'; 1 - x2/y2),
where F is the Gaussian hypergeometric series. (See Exercise 6 in
Section 1.3.)
b) Show that R(a; 8,8';x2, y2) is homogeneous of degree -a in x
and y2. Also R{a; 8, 8'; 1,1) = 1 and R(a; 8,8';-,-) is continuous.
c) Verify the entries in Table 8.1.

8.5 Carlson's Integrals and More Examples 261
d) Show that [Fit Fj], i <j, is linearly convergent for all cases except
i '■= 1, j '■= 2. Observe that [F3, F4], which is not Archimedean, has
a convergence rate of 2~" and [F2, F3], which is partially
comparable, has oscillatory iterates.
2. a) Observe that Cn(a, b) = (ir/2)/I(a, b) with
/(«,&) :=/o
r/2 de
Va2 cos2 6 + b2 sin2 0
as before.
2 ,2
a - b
b) Show that C34(a, b) := V2log(fl/fe) (fl^6)-
//mf: Use partial fractions.
c) Show that Cu(a, b) is given by (8.4.3).
Hint: Let (f + a2)l{t + b2)=: cos d or cosh2 0.
d) Show that C13(a, b) is given by (8.5.8) for a < b.
e) Use the invariance principle to verify that
C24(b, a) = C32(a, b) = y/aCu(a, b)
and that
C23(b, a) = C42(a, b) = VbC13(a, b).
f) This completes the analysis of all of Carlson's means in explicit
form. Attempt to verify part e) directly from Theorem 8.6.
Use the invariance of C13 to show that
i) arcsl x = V2 arcslh y
where (1 + /)(1 + Vl-x4) = 2, and
ii) arcsl x = 2 arcsl z
where x =: 2zVl - z4/(l + z4) and z2 < V2 - 1.
This is Jacobi's duplication formula for arcsl (Watson [33]).
Show that with M(a, b) = G(a, Hpl2{a, b))=-N(b, a),
M®N(a,b) = Xp(a,b) = [^^]VP.
(Crrlson [75]) Let a, b > 0 and set
/^ r rrl2
F(a, b):= - log (a sin2 6 + b cos2 6) dd .

General Means and Itei
a) Show that
F
b) Using a) (or directly)
\c:bU]
show that
= 2F(a,b).
F(a, b) = 2 log {^Y^-) = log Hll2(a, b).
6. (Carlson [78]) Let a, b, c, A real be given. Assume a + \, b + X,
c + A> 0, and at most one of these is zero. Consider
/• CO
7/( A) := T(a, b, c; A) := J^ [(a + t)(b + t)(c + t)]~112 dt.
Let
k := A + \/(A + a)(A + b) + \/(A + 6)(A + c) + V(A + a)(A + c).
a) Show that 7/( A) = 2 T(k).
b) Show that T(\0) = limn^cc2n+1k~112, where iteratively k:= kn,
fc„+1 := k, and A = : fc0.
c) Show that for a, b,c, d>0,
/• CO
J0 [(* + a2)(t + 62)(* + c2)(t + rf2)]"1'2 * = T(A, B, C; 0)
where A:=(ab + cdf, B := (ac + 6rf)2 and C:= (arf + be)2.
7. (Tricomi [65]) Let 0< yfc < 1 be given. Set
Rk(a,b):=yk2a2 + k'2b2 k':=Vl-k2
and
M(a,b):=ah + Kfbh\b 0<a<b.
' b + Rk(a, b)
a) Show that M is a strict mean on 0 < a < b.
b) Show that
CO
M®aG(a,b):=bH cn(2-"v0)
where cn(u0) := a0/b0 and 0< v0 <2K.
Hint: Use the half-angle formula for en

8.6 Series Expansions of Certain Means 263
(8.5.10) c„2(iM)=^i
cn(u) + V*:'2 + k2cn\u)
2UJ 1 + \Jk'2 + k2cn\u)
and deduce that cn(u„) := a„/fr„ satisfies vn+1 = vn/2 and bn+1/bn =
c) Recover Borchardt's algorithm by considering A:: = 0.
8. Compute a formula for the MacLaurin series for G(l - x, 1) in (8.5.9).
8.6 SERIES EXPANSIONS OF CERTAIN MEANS
For homogeneous means it is particularly easy to compute Taylor series. It
suffices to expand the trace around 1. In the exercises we list various series
taken from Gould and Mays [84], extending results in Lehmer [71].
//(1 l-^) = l-l^ + ^ll^ + £^l^_(p-l)(p-3)(2p + 5) 4
npyi,i X) i 2x-r g x -r. 16 x 3g4 x
+ ••• .
(8.6.1)
Similarly
-^ = 1- ,
2 4 8 48
Lp(l,l-,) = l^ + ^^ + £^,3_(p-l)^-3);c4 +
(8.6.2)
From these it follows that the only means which are both Holder and
Lehmer means are //_x, A, and G. (See Exercise 5.) Similarly, but slightly
more elaborate, analysis shows that the following theorem holds.
Theorem 8.7 (Lehmer)
(«) Hp®gHq = Hs p*q
if and only if p + q = s = 0.
(b) Lp®gLq = Ls p*q
if and only if p + q = 2s = 0,1,2.
(c) Lp®gLq = Hs p*q
if and only if p + q - 1 = 5 = -1, 0,1.
(A) Hp®gLq = Ls p^q
if and only if p = -q and s= \.

264 General Means and Iterations
Proof. Sufficiency is easy in each case. Necessity is more elaborate and
in some cases somewhat tedious. (See Exercise 6.) In essence it follows from
the invariance principle and the computation of the first few terms of the
Taylor series of the limit. □
Thus only A, G, and H can arise in such compounding. Lehmer continues
by studying in detail the Taylor series of A®gL2, various properties of
which had been previously examined by Stieltjes in a letter to Hermite
(Hermite and Stieltjes [05, letter 323]). (See Exercise 7.)
Comments and Exercises
Theorem 8.7, while very pretty, is of limited ambit, since it cannot diagnose
whether the compounded mean is equivalent to a mean in the class.
1. Show that the Holder means satisfy
CO
H(\,l-x) = JJ A(p,n)xn
where
^=-(-^(^1^1^)
/t=o
for n > 1.
2. (Euler) If g(x): = E^0 anxn and gp(x) =: E:=0 bnxn, then
n
3. Show that the Stolarsky means satisfy
CO
Sp+1(l,l-x) = S D(p,n)xn
n — \
where
4. a) Show that the Gini means satisfy
G,,r(l, 1 - x) = GrtS(l, 1 - x)GStr(h 1 - x)
where
• s\ 1/0-r)
GsAa,b):=(^)

8.6 Series Expansions of Certain Means
b) Show that
CO
6,,,(1,1-*) :=2 G{s,r,n)xn
where
c) Thus Gsr can be computed by convolution. A similar result
for Ers.'
Show that H = Lp if and only if p = 0 and q = -1; p = 1 and q =
p = 5 and q = 0.
Prove Theorem 8.7.
(Lehmer) Let
i?0c):=A®gL2(l,l + 4;c).
a) Show that
i?(;c) = 1 + 2 E
x2"
n-0 °0M'2 '" °b
where the P, are defined recursively by P0(x):=l
//in*: If
P
a0:=l
*n p p . . . p
^.,+2*'
2x2
then
"l°2' '°n
Cq • j^A
,. _ bn+i-an+i
t-M4- 1 «

266 General Means and Iterations
Now observe that
R(x) = a0 + 2 0,+1 - a,)
/=0
and use the invariance principle,
b) Show independently that
R(x) = 1 + 2 gnxn
n = l
where gx := 2 and gm, m >2, satisfies the recursion
|m/2J
._/ 1\m V 0m-2/fc/ m-2 \
Observe that the g, are integers.
//inf: Show that the series for R satisfies
«W = (1 + 2,)^(^).
c) Use b) to compute gm for m < 20. (g20 = 23335660.)
8.7 MULTIDIMENSIONAL MEANS AND ITERATIONS
Most of the results of this chapter have direct analogues for functions of
more than two variables. Often these follow by similar arguments. We
concentrate on mean iterations. Let a '■= (ax,. . . , aN) be any strictly
positive vector.
An iV-dimensional mean is any continuous function M such that
(8.7.1) J\at<M{a)<\/ at
/=1 /=1
for all at > 0. The mean is strict if
(8.7.2) A «,<V «/=> A a,<M(a)<\/ a,.
/=i i=i /=i /=i
Also M is symmetric if M(a) = M(6) for any permutation b of a.
Homogeneity and isotonicity are defined analogously. The Lehmer and
Holder means are defined by
(8.7.3) Lp(a):=-^ x
£f=i<

8.7 Multidimensional Means and Iterations 267
respectively.
Given N iV-dimensional means M\ . . . , MN, we consider the iteration
[M\ ..., MN] defined by a0 : = a > 0 and
(8.7.5) a'n+1:=M'(dn) l^i^N.
We write this vectorially as dn+1'-= M(an) and denote the common limit
when it exists by ®/ii M'. Again, when each Mt is symmetric, we consider
N
this to be a Gaussian iteration with limit (x)g M'. Similarly, if
i=i
(8.7.6) a'n+1 := M\axn+X, a2n+1,..., a'~+\, a'n,. . . , a")
we have an Archimedean iteration (there are other possible generalizations)
N i
with limit (x)a M . Under mild hypotheses the convergence results of Section
8.4 remain valid.
Theorem 8.8
Let M1, M2, ..., MN be strict iV-dimensional means.
(a) Then ®^=i Ml exists and is a strict, continuous mean.
(b) Suppose that the means are symmetric and continuously differentiable.
Let an := V"x a'n and bn := /\?-i <■ Then, if an *bn,
(8.7.7) lim lfl"^~^il = o
and convergence is superlinear in the Gaussian iteration.
(c) If, in fact, the means are twice continuously differentiable,
convergence in the Gaussian iteration is quadratic (uniformly on compact
subsets).
(d) Convergence is linear in the Archimedean iteration for continuously
differentiable symmetric means.
Proof.
(a) Much as before, an > an+1 > 6n+1 > 6n, and we may suppose an
converges to a and bn converges to b. Let c be any cluster point of {an},
which is bounded. Then 6 < A /=i ci — V /=i c,^a, and
a = V M'(c) b = A M'(c).
/=i /=i
and
(8.7.4)

268
General Means and Iterations
Since all the means are strict, we must have ci = a and c, = b for each i.
Thus a = b, and the iteration converges, say, to a. As before, the limit
is a continuous strict mean.
(b) By symmetry we have M'k(ce) = 1/JV for any i, k and any multiple of
the unit vector e. Thus
1 N
a'n + 1 - a = - E [(a* - a) + o(a* -a)].-
iy fc-i
Hence
|aj,+1 - <+1| = o« - a) + o(a'n -a) = o(an - bn)
and (8.7.7) follows.
(c) This is argued as in the two-variable case and relies on the fact that the
Hessian V2M'(a, a,. . . ,a) sums to zero and has all diagonal entries
equal to —(N— \)y\ where each off-diagonal entry equals y'. Thus for
each i and j,
<+1 - <+1 = ^T~ 2 (flj - aknf + o((an - bnf)
z h<k
(and unless all y' coincide, convergence will be at best quadratic). Now
0^ «„+1 - &n+1 ^ 5|a„ - bn\2 + o{\an - bf)
where B is an easily computable constant depending on the means,
initial values, and dimension. This shows (c).
(d) We leave (d) as Exercise 2b). □
example 8.5 (schlomilch) Consider
a„ + b„ + c„
n n
= :M\an,bn,cn)
K+i-=yj~n'n ' V—— =--M\an,bn,cn)
a„b„ + b„c„ + a„c„
3
1/3
cn+i:=OAcJ =-M (an,bn, c„).
This is a quadratically convergent Gaussian iteration, as follows from
Theorem 8.8(c). (Since MX>M> M3, M2 is a strict mean.) Let S(a, b, c)
denote the limit. While we cannot identify 5, we can, following Landau,
observe that if ac = b2, then ancn = b2 and thus
(8.7.8) S(a, Vac, c) = He® G(a, c) =: S(a, c)

8.7 Multidimensional Means and Iterations
269
where He(a, c)-= (a + Vac + c)/3 = E3/21/2(a, c) is the Heronian mean.
While this does not appear to have a closed form, it has an attractive
product expansion,
(8.7.9)
nl + VX~n + Xn
. . . ~f
where x0 '■ = x and
3VX~n
*"+1" l + Vx-n + xn-
While it is difficult to identify two-dimensional compound means, it
appears even harder to identify higher dimensional ones.
There is a satisfactory analogue of Proposition 8.3. Let / be continuous
and strictly monotone as before. Let a0, ay,. . . , aN be distinct positive real
numbers. Let F(N) be any iVth antiderivative for/. We define
(8.7.10) Ms («„,«!,...,«„):=/-
>Nf
AH 2
£ F(N)(ak)
k=o nfc^- {ak - Uj) J
Then Mf is a strict, symmetric A^-dimensional mean. (See Exercise 5.)
We should emphasize that the invariance principle continues to hold:
$'■= ®/Ii M' is the unique continuous diagonal mapping satisfying
(8.7.11) ¢ = ^,^,...,^).
Comments and Exercises
A wealth of information on ./V-dimensional means can be found in Hardy,
Littlewood, and Polya [59]. The Lehmer means are studied in Beckenbach
[50]. The Schlomilch mean and Landau's contribution are discussed in
Schoenberg [77]. Various ./V-dimensional quadratic iterations are exhibited in
Wimp [84] and Arazy et al. [85].
1. a) Show that Hp and Lp are strict, homogeneous, symmetric means.
b) Show that (II?=1 a,.)lw = lim^o Hp(a) =: H0(a).
c) Show that Hp is convex for p>l and that Lp is convex for
l<p<2.
2. a) Show that ®*li M' is a continuous strict mean when each M' is.
b) Show that convergence is linear in the Archimedean iteration.
c) Estimate the rate in the three-variable case.
d) Show that

270
General Means and Iterations
®aHp(a) =
N
Ltf(tf + 1) ~
Zj ka
iip
p^O
and
®a H0(a) =
r N
n.
•Jfc = l
2/N(N+l)
Hint: Use the invariance principle when p : = 1.
3. Show that 5(1, jc) satisfies (8.7.9). Compare this to the product
expansion for SS(l,x).
4. a) Let
M\a,b,c): =
M\a,b,c): =
a + b + c
ab + be + ac
a + b + c
and
M\a,b,c):--
3abc
ac + ab + bc 1/a + l/fe + l/c*
Show that
(x) M'(a, b, c) = (abc)
1/3
and that convergence is quadratic,
b) Let Sk be the kth elementary symmetric mean function of N
variables. That is, 5^ : = fkl( Nk) with fk as in Exercise 6 of Section 11.2.
k ..
Thus, 50 := 1, 5X := Hx, etc. Let MK := Sk/Sk_1 for 1 < A; < w. Show
that Mk > M*+1 and that <g) Ml = H0.
c) In a) replace M by //0. Thus
a„ + &» + c„
*n ~n ~n
B+1* 3
c„
kn+i:=(aAO
1/3
""+1" l/an + l/bn + l/cn '
Show that the limit mean M satisfies
M(a, Vac, c) = He®gHe_x(a, c) = Vac = (abc)1'3
This is not the general limit.

8.7 Multidimensional Means and Iterations
271
d) More generally, whenever M is homogeneous and symmetric,
4> = M (x) G (x) M_x satisfies 4>(a, Vac, c) = Vac. Similar results
hold in N dimensions.
a) Show that Mj of (8.7.10) is uniquely defined and is a strict mean.
Hint: Let
-A Uj7tlc (x - Oj)
PN(x) := Z F(N)(ak) n.^ {Uk _ aj) ■
Then PN is the Lagrange interpolating polynomial for F(N) at
a0, 0,-y, • .. , aN. (See Exercise 1 of Section 10.1.) Thus P^° and /
must agree at some point strictly between A;=i at an(* V;=i a,-
Now show that Ms extends continuously to all strictly positive
variables. The harder part is to show it is strict.
b) Show that the three-dimensional Stolarsky means are appropriately
defined (and are homogeneous) if
S(a,b,c): =
+
Lp(p-l) \(a-b)(a-c) (b-a)(b-c)
+
l/(p-2)
for p ^0, 1, 2. Also
S0(a, b, c): =
(c- a){c - b),
{a - b)(b - c)(c - a)
%b - c) log a + 2(c - a) log b + 2{a - b) log c J
(a - b)(b - c)(a - c)
2(6 - c)a log a + 2(c — a)b log 6 + 2(a — b)c log c
and, using f(x) : = log, that the identric mean may be given as
while
5x(a, b, c): =
^"(a, 6, c) := 52(a, b, c)-= exp
a log a
+
fr2logfr
L (a - b){a -c) (b - a)(6 - c)
+
c log c
(c-a)(c-6) 2J-
c) Show that the iV-dimensional logarithmic mean is given by
°*7VV^l> ^2' • ■ ■ ' ^N) '~~
(-1)N(N-1)2
log a,-
/=i nyV,-(a,-- ay) J
-l/(Af-l)
d) Investigate the isotonicity of 5 (a, b, c) as a function of p.

272 General Means and Iterations
6. There are various multidimensional generalizations of the AGM due to
Borchardt [1888] and others. In four variables one may take
fl„ + b„ + c„ + d„
n n n n
"n + 1
K + l
Cn + 1-
dn + X-
4
y/a„bn + y/cndn
2
._ Va„c„ + VMn
2
y/andn + y/bncn
2
a) Observe that, while the means are not all symmetric, the derivatives
on the diagonal (a = b = c = d) all coincide, and hence establish
quadratic convergence.
b) Show that when a0 ■= b0 and c0-= d0, the iteration reduces to the
AGM. This iteration shares many of AGM's attributes (Arazy et al.
[85]).
7. (An extended convergence result)
a) Observe that the convergence result Theorem 8.8 continues to hold
if the condition that the means are strict is relaxed to
N N N N
M(c) = V c,- and M(c) = A ct implies /\ ct = V ci ■
/-1 i-l /-1 /=1
Here
N __ N
M:=/\ M' and M:=\/M'.
/=i /=i
This clearly holds if N - 1 of the means are strict.
b) More interestingly, use a) to establish the following result. Let A be
an entry-positive N by N matrix. Set A0 : = A and
An+1:=|(*VAJ2
for n > 0. Here *VT" represents entrywise square root. Then A„
converges to a constant matrix with entry e(A).
c) KA:=(" ba) then e(A) = AG(a, b).
d) KA:=(" a. J then e(A) = a(a, b).
e) Similarly consider A0 and B0 entry positive and

8.8 Algebraic Iterations and Functional Relations 273
Show that this iteration converges to a constant matrix. (See Cohen
and Nussbaum [87].)
8.8 ALGEBRAIC ITERATIONS AND FUNCTIONAL RELATIONS
Among the more familiar and fundamental properties of the exponential
function is that it satisfies the algebraic functional relation
(8.8.1) /(z) = [/(z/«)]\
Are there any other algebraic functional relations of the form
(8.8.2) /(z) = /3(/(/,(z),z))
where /3 and /i are algebraic? More precisely, /i is an algebraic function if
there exists a polynomial P in two variables with coefficients in some base
field F (for most of this discussion F '■ = Q or F: = C) such that
(8.8.3) P(/x(z),z) = 0
and O is an algebraic functional relation for / if O is a polynomial in three
variables over F and there exists an algebraic function /jl so that
(8.8.4) O(/(z),/(Mz)),z) = 0.
The function /i is termed an algebraic transformation (for /). We assume
throughout that z is a complex variable and unless otherwise indicated that/
is defined on a region U which contains an open set V so that fjb(V)G U.
[This ensures that (8.8.4) can be continued to any open connected
component of the domain of / that contains U.] The collection of all such
transformations satisfying (8.8.4) for some O we denote by TG(/: F), the
algebraic transformation group of / over F. If
(8.8.5) L(x):=M®N(l,x)
where M and N are homogeneous means, then
(8.8.6) L(x) = M(x)L(ijl(x))
where M(x) := M(l, x) and /jl(x) := iV(l, x)IM{\, x). Thus if both M and N
are algebraic, then L satisfies an algebraic functional relation with algebraic

274
General Means and Iterations
transform (jl. It transpires that the transformation groups of the most
elementary transcendental functions are very simple, too simple in fact, to
support quadratically converging mean iterations. (See Exercise 1.) This, in
part, explains why the algorithms of Chapter 7 for exp and log require the
intermediate use of nonelementary transcendental functions.
We wish now to compute the algebraic transformation groups of some
familiar functions. However, we need first to establish the transcendence
[over C(z)] of the functions exp, log, and exp(exp). This is considerably
easier than the arguments for the transcendence of numbers such as e. The
arguments are roughly the same for all the above functions. For exp it
proceeds as follows. Suppose exp satisfies an equation of the form
k
(8.8.7) 2 at exp (iz) = 0
/=o
where the a{ are rational functions, k is assumed minimal, and a0 = 1. We
differentiate (8.8.7). to get
k
(8.8.8) 2 (dt + iat) exp (iz) = 0 .
;=i
If k > 2 and we divide (8.8.8) by exp (z), we obtain a lower order expression
than (8.8.7) and violate the minimality of k. To dispose of the k = 1 case we
must show that exp is not a rational function. Since exp is entire, if it were
rational it would in fact have to be polynomial, and if it were polynomial it
would have to have a finite Taylor expansion.
Theorem 8.9
(a) TG(exp: Q) = {az: a rational}
(6) TG(log:Q) = {z*: b rational}
(c) TG(exp(exp):Q) = {z}.
Proof. We will prove, by elementary methods, that
(a') TG(exp: Q) = {az + b: a rational, exp (b) and b algebraic}
(b') TG(log: Q) = {azb: log (a) and a algebraic, b rational}
(c') TG(exp (exp): Q) = {z + b: exp (b) and b algebraic}.
To see that (a), (b), (c) and (a'), (b'), (c') are the same is equivalent to
Lindemann's theorem which guarantees that exp (a) is transcendental for
any algebraic a^O. (See Exercise 7 of Section 11.2.)
We first prove part (a1). Suppose that /jl is an algebraic transformation for
exp. Then there exists an expression of the form
k
(8.8.9) 2 ai exp (m^) exp (n,/i.(z)) = 0
i = 0

8.8 Algebraic Iterations and Functional Relations
275
where the at are nonzero algebraic functions in z and the mi and nl are
integers. Equivalently,
k
(8.8.10) 2 ai exp (mtz + n,/x(z)) = 0
1 = 0
where we assume k is minimal, that is, we assume (8.8.10) contains a
minimum number of distinct nonzero exp {mtz + «,/i.(z)) terms. If we divide
(8.8.10) by its last term and differentiate, we get
2 [m, -mk + («,. - nk)ii.][-f-j + (-1 j exp [(m. - mk)z + («,. - nk)fji]
= 0
(8.8.11)
This expression has one less term than (8.8.10) and contradicts the
minimality of (8.8.10) unless (8.8.11) contains no nonzero terms. This
implies that each term of (8.8.10) must be constant. Thus there exist m and
n integral and a nonzero algebraic function a{z) so that
(8.8.12) a{z) exp {mz + nfjb(z)) = constant.
Since exp is transcendental, mz + tifjb(z) must be constant. Call this constant
b; since /i. is algebraic over Q, b must be algebraic. Specializing (8.8.12) at
z = 0 shows that exp (b) must also be algebraic. Part (a') is completed by
observing that /jl(z) = (m/n)z + b is an algebraic transform of exp.
The proof of part (b') is similar. If fju is an algebraic transform for log,
then there exists an expression of the form
(8.8.13) 2 a,[log (z)]m< ■ [log (/i(z))]-' = 0
i = 0
where the ai are algebraic functions and the mt and nt are nonnegative
integers. Let (8.8.13) be minimal in the sense that it has the smallest
maximal degree and contains the fewest distinct terms of maximal degree.
(The degree of a term is mi + nt.) Suppose the term corresonding to i = 0 is
such a maximal term. If we now divide (8.8.13) by a0 and differentiate, we
are left with an expression containing fewer maximal terms, which
contradicts minimality unless all the transcendental terms vanish under
differentiation. In particular (8.8.13) must actually be of the form
(8.8.14) b log (z) + c log (/i(z)) = d
where b, c, and d are algebraic numbers. Thus

276
General Means and Iterations
(8.8.15) ti(z) = exp(d/c)z'blc
and we see that fi is algebraic exactly when exp {die) is algebraic and —blc
is rational. (Otherwise, by Exercise 2, z~blc is transcendental.) This finishes
(&')•
The proof of (c') is again similar to (a'). There are, however, a few
additional wrinkles. Suppose now that /jl(z) is an algebraic transform for
exp (exp (z)). Consider minimal length sums of the form
k
(8.8.16) Z, at exp [mi exp (z) - ni exp (/i.(z))] = 0
;=o
where the at are algebraic functions of z, exp (z), and exp (/a(z)). Dividing
and differentiating (8.8.16) leads, as in (a'), to a contradiciton unless
(8.8.17) mt exp (z) - nt exp (/i.(z)) = constant
and the result follows. One of the wrinkles is that we must establish the
transcendence of exp (exp (z)) over the algebraic functions in z, exp (z), and
exp (/i.(z)). This is necessary for our minimality argument and can be done
analogously to proving that exp is transcendental. (See Exercise 2.) □.
The transformations we are considering are algebraic over Q. This is a
natural choice for computationally related questions. The analogous results
for algebraic transformations over C are similar but easier because the
number theoretic details vanish. We have the following:
Theorem 8.10
(a) TG(exp: C) = {az + b: a 6Q and b eC}.
(6) TG(log: C) = {azb: b e Q and a e C}.
(c) TG(exp (exp): C) = {z + b: exp (b) e Q}.
If/(z) = g(a(z)), where a(z) is an algebraic function, then
(8.8.18) TG(g:F) = aoTG(/:F)oa-1.
example 8.6 Let /jb be an algebraic function and suppose /i.(n) is the
identity. [For example, fju(z) := rz, where r is an nth root of unity.] Then
(8.8.19) /:= exp (/x(1)) + exp (//2)) + ■ • • + exp (/x(n))
is invariant under /jl, where /i,(n) := /i.(/i,(n_1)). Note that TG(/) is not in
general conjugate to TG(exp) since TG(exp) contains no finite elements of
order greater than 2.

8.8. Algebraic Iterations and Functional Relations
277
For qz and log^ (z), where q is an algebraic number distinct from 0 and 1,
we have
(8.8.20) TG(qz: Q) = {az + b: a and b rational}
and
(8.8.21) TG(log^: Q) = {azb: a and b rational} .
To derive these results we first observe that (a') and (b') of Theorem 8.9
hold for the above functions (with exp replaced by qz and log replaced by
log9). The only difficult part is to show that q" and a are simultaneously
algebraic exactly when a is rational. This is the celebrated Gelfond-
Schneider theorem. (See Section 11.2.)
In Exercises 1 and 5 we show that functions like /3(e7(z),z) and
/3(log y(z), z), where /3(-, ■) and y(-) are algebraic, cannot support
functional equations that possess quadratic fixed points in their domain of analytici-
ty. In particular such functions cannot be the limit of quadratically
convergent homogeneous mean iterations. This shows why the elementary
functions are always associated with linearly convergent iterations.
Comments and Exercises
The results of the section underscore the difficulties of analyzing in closed
form the limits of mean iterations. The familiar elementary transcendental
functions can only be limits of fairly trivial iterations. The arguments are
extended in the exercises to cover sin, tan, arccos, and so on. Exercise 6
treats the special case of compounding rational means and shows that the
limit of such a mean iteration is either the kth root of a rational function or
is transcendental.
1. a) Suppose that
L(x) = M ® N(l, x)
as in (8.8.5), where M(l, x) and N(l,x) are algebraic functions
analytic in a neighbourhood of 1. Show that if L can be
quadratically computed by iterating (8.8.6), then fi has a fixed point at 1 of the
form (jl(x) := 1 + 0(1 - xf as x-* 1.
b) Call an algebraic transformation /i. quadratically attractive at c if
fjb(c) = c + 0{c - z)2 as z-> c. Suppose that A is algebraically
conjugate to /i.. That is,
A = a° /jl° a*1
where a is algebraic. Show that A is also quadratically attractive
at c.

278
General Means and Iterations
c) Show that none of exp, log, or exp (exp) nor any function
algebraically conjugate to the above functions is the limit of a quadratically
converging algebraic homogeneous mean iteration.
2. a) Prove that z" (a irrational), log, and sin are transcendental
functions over C(z).
b) Show that exp (exp) is transcendental over C(z, exp (z), exp (/i.(z)))
when /i is algebraic.
3. Prove (8.8.20) and (8.8.21) assuming the Gelfond-Schneider theorem.
4. (On the algebraic transformations of sin, cos, and exponential sums)
Let a(z) be an algebraic function over Fm-=C(z,ez), the field of
rational functions in z and ez. Suppose that a(z) is not algebraic over
C(z). Show that
TG(a : F) C {az + b: a, b £ C}
and calculate TG(cos:C) and TG(sin:C). Outline:
a) Show that if fj, £TG(a :C), then there exists a nontrivial relation
d
Z/ bj(z) exp [kj/Jb(z) + hjz] = 0
/-0
with bj £ C(z) and kj,hjB Z.
b) Consider minimal (in d) expressions of the above type and argue,
as in the proof of the transcendence of exp, that for some /,
exp [kjfjb(z) + hjz] £ C(z).
5. a) Suppose that a(z) is algebraic over C(z, es(z)) where g(z) is
algebraic over C(z), and suppose that a(z) is not algebraic over C(z).
Extend the arguments of Exercises 4 and 1 to show that TG(a, C)
contains no elements that are quadratic at any point where g is
analytic. In particular, such a function cannot be the limit of a
quadratically converging algebraic homogeneous mean iteration,
b) Suppose that a(z) is algebraic over C(z, log (g(z))) where g(z) is
algebraic over C(z) and suppose that a(z) is not algebraic over
C(z). Show that TG(a :C) contains no elements that are quadratic
at any point where g is analytic and nonzero.
6. (On the iteration of rational means)
a) Let M and N be homogeneous rational means. (A rational mean is a
mean that is a rational function.) Show that M®N(l,x) is
convergent in some neighbourhood of 1. Show that either
1) M(x)iV(l, x) is transcendental
or
2) [M(x)iV(l, jc)]' is a rational function for some integer i.

8.8 Algebraic Iterations and Functional Relations 279
Outline: Suppose/(x, y) = M®N(x, y) and suppose that/(1, x) is
algebraic. Then one has a finite sum
2s,(*)[/(i,*)]' = o, SieC(x)
and
(8.8.22) lri(x,y)[f(x,y)]i = 0
where rt(x, y) = x^'s^ylx). Thus, on substituting,
(8.8.23) 2 r,(M, N)[f(M, N)]' = 0
while by inyariance,
(8.8.24) ^ri(x,y)[f(M,N)Y = 0.
Consider a minimal expression of type (8.8.22) and deduce from
(8.8.23) and (8.8.24) that, for some i,
rt(M, N) = r,(x, y)
r0(M, N) r0(x, y) '
Observe that
A'r,(x, y) = rfa, $
and deduce by passing to the limit that
/-,.(1,1) /■„(*, yyUi
f(x,y) =
*-r0(i,i) ri(*>y)
b) Let
and
N(x,y):=(l-l)x+ly
As in Exercise 9 of Section 8.4,
N®M(a, b) = (ak-xb)Vk

280 General Means and Iterations
and the convergence is quadratic. In particular case 2) of a) can
occur,
c) Show that A® L2(l, x) is transcendental. Thus case 1) of a) can
also occur. (See Exercise 7 of Section 8.6.)
7. Calculate
TG(sin ~x: C) and TG(cos ~x: C).
Hint:
sin-1 z = —i log (iz + Vl - z2).

Chapter Nine
Some Additional Applications
Abstract. In Section 9.1 we derive the classical formula for r2(n) and its
theta function equivalent. In Sections 9.2 and 9.3 we consider the summation
of various multidimensional series. Results include an alternating series test in
several dimensions, evaluation of various lattice sums, and related invariants.
Section 9.4 gives Watson's quintuple-product identity, and Ramanufan's 1'VP1
product. Section 9.5 considers quintic and septic multipliers and solvable
iterations.
9.1 SUMS OF TWO SQUARES
We need the following simple lemma on Lambert series whose proof
(Exercise 1) proceeds by expanding both sides of each equation.
Lemma 9.1
If \q\ < 1 and un :- ¢7(1 - q"), then
(9.1.1) 2 um(l + um)=l nun
m = \ n=\
and
CO CO
(9.1.2) 2 (-l)m+1"2,„(l + «2«) = 2 (2» - lK„-2 ■
m—\ n=l
Our development hinges on yet another remarkable identity due to
Ramanujan [62]. (See also Hardy and Wright [60].)
281

282 Some Additional Applications
Proposition 9.1
Let 0 be real with 0 < 0 < <n\ Let
1 / 0\
(9.1.3) T := 7X¢,0):-= 4 cot (^)+ 2 "„sin(«0)
(9.1.4) ^:=7^,0)
1 / 0\1 °°
4COt\27J + ? "„(l + "„)cos(
1 °°
(9.1.5) T2:=T2(q,6)=- 2 >w„[l - cos(n0)]
Then
Proof.
T,+ T2= T2
1 /0
7 COtI -
4 \2
+ 5X + 52
where
and
Now
S\'-= 2 ^ "„ cot (^-j sin (n0)
52: = 2 umun sin (m0) sin («0)
1 I 0\ 1 nl 1
- cot (-) sin (nd) = - + 2 cos (kd) + - cos (nd)
while
Thus
2 sin (md) sin («0) = cos [(m - n)0] - cos [(m + n)6].
t cot ( -) + 2j ak cos (kd)
4 \Z/J k=0

9.1 Sums of Two Squares
for constants ak{q) which we proceed to evaluate. Now
m CO j CO
a0 = -z 2 un + - 2 un
A n = l L n = l
where the \u2n term comes from m = n in S2. Thus
283
(9.1.6)
1 °° 1 °°
fl0 = 9 2 "„(1 + "J = X 2 ""„
by (9.11). For fc>l, S1 contributes to ak,
1
-z uk + A «t
and 5, donates
— Zj u u + — Zj mm X MM
■^ m-
for m, n > 1. Thus
^ m-n=k *• n-m=k ^ m + n=k
k-\
Luckily,
and
1 1
^ 1=1 j-1 L 1=1
"/"*-/= "*(1 + "i + "*-,)
"*+,- + uiuk+i = uk(ui - uk+i) ■
Hence
(9.1.7) ak = uk
= uk(l + uk - - k)
Thfs shows that
*:-!
1 1
~+ 2 (",- - uk+i) --z 2 (l + "i + «**-()
rj-t2
T COt - + 2 K*(l + "J COS (*0) +-2 *«**[! - COS (k6)]
(9.1.8)
which is the desired result. D

284 Some Additional Applications
For 6 '■= it72 this result becomes
1 + 4 2 (-1)'
» 9
2n + l
1-9
2n + l
■i + i6 2 (-i)X,(i + «2»)
n = l
CO
+ 82 (2n + l)(«2„+1 + 4M4„+2)
1 + 8^ (2n + l)M2n+1
n-0
CO
+ 8 2 (4" + 2)m4„+2
on applying (9.1.2). Thus
1 + 4 2 (-1)"
n=0
2n+l
1-9
2n+l
:1 + ;
«-i 1-9
n^0(mod 4)
Now (3.2.23) can be used to show that the right-hand side is d\{q). Thus
(9.1.9)
03 = 1 + 4 2(-1)-:7^-2^
1-q
n = 0
= 1 + 42
n = 0
„4n+l
,4n+3
I"?
4n + l
1-9
4n+3
and this last expression is nonnegative for real q. (Hardy and Wright [60] use
Proposition 9.1 to deduce the formula for d\ from that for d\).
For r = 1 or 3 let dr(k) denote the number of divisors of k congruent to r
modulo 4. Then
_4n + r
CO CO 00
,7=2 2 q(4n+r)d = 2 dr{k)qk
n-0 J- — q n = 0 rf=l Jfc = l
and a comparison of the coefficients in (9.1.9) shows that
(9.1.10) r2(k) = 4[d1(k)-d3(k)].
In other words, the number of representations of a positive integer A: as a
sum of two squares, counting order and sign, is 4 times the surplus of
divisors of k congruent to 1 modulo 4 over those congruent to 3 modulo 4.
This recovers Jacobi's classical result, a result also known to Gauss. (See
Dickson [71].) Since

9.1 Sums of Two Squares 285
°° „m °° I -win 2m + \
£-1 -, , 2m ^-1 * 2m + l
m_l 1 + # m-0 I" #
we also have
CO
(9.1.11) 03(0) = 1 + 4 2 ~
«-1 l + q2m
This is the formula exploited in Section 3.7. Also (9.1.9) shows that
CO
(9.1.12) 02(?) = l + 4 2 {-\fn-x)nqnm
n,m — \
n odd
and
(9.1.13) e2A(q) = l + A S (-1)^^)^-^
n,m = \
t odd
= 1 + 4^ (-l)^"1-1^2"-1).
n.m^l
An alternative recent derivation of (9.1.9), due to Hirschhorn [85], relies
only on the triple-product identity. Begin with (3.1.13) and let a2 : = w. This
gives
CO
(a - a'1) 11 (1 - «V)(1 - «"V)(l - 9")
n = l
00 00
V 4n+l 2n2+n V „4n-l 2n2-n
rt = —00 rt = —CO
Now apply the triple product in form (3.1.1) to each of these sums. This
leads to
CO
(9.1.14) (a - a'1) U(l- a2qn)(l - «_V)(1 - qn)
n = l
=a fi (l+aVn_i)(i+<ry~3)(i - qAn)
n = l
-a-1 n (l + «y-3xi+a-y-'xi - <?4n).
Next differentiate each side with respect to a, at 1. (Use logarithmic
differentiation on each product separately.) This, after rearrangement,
produces

286 Some Additional Applications
1 + 4v (-i)V"'1 n:=1(i-g")3
(9 1 15) "=1 1 + ^ ^=1 (1 + I*"'1*1 + I*"'3*1 ~ ^
The right-hand side of (9.1.15) is
n;=1 (i-gnf n:=1(i-g")2
= el(q)
ni=1 (1 + ^)(1 -<T) n:=1 (l + qnf
where the last equality follows from (3.1.4) and (3.1.7). Thus
0^) = 1 + 4 2¼¾
n=l 1 + ^
which is equivalent to (9.1.9).
In the course of his study of Ramanujan's mock theta functions, Andrews
[86] discovered the following remarkable cubic counterpart of (9.1.11),
which we write as
(9.1.16) 6l(q) = 8 1 1 (l+^-^W-C^'tf.
It is an easy consequence of (9.1.16) that every number is a sum of three
triangular numbers (a fact originally observed by Fermat and proved by
Gauss). This also implies that a number is a sum of three odd squares
exactly when the number is congruent to 3 modulo 8 (Exercise 7]) as
observed by Euler.
Comments and Exercises
A comprehensive account of the development of formulae for sums of 2m
squares, from elliptic considerations and Lambert series, can be found in
Rademacher [73]. Many wonderful related identities are to be found in
Ramanujan [62]. Odd sums are much harder to evaluate. They involve
generating functions of class numbers as shown in Mordell [16] and Watson
[35].
1. Prove Lemma 9.1.
2. Verify (9.1.6) and (9.1.7). Show that these are equivalent to (9.1.8).
3. a) Prove formula (9.1.11) for d\(q).
b) Establish (9.1.12) and (9.1.13).
4. a) Let 6 tend to it in (9.1.8) and evaluate the limit.
b) Evaluate (9.1.8) when 6 := u74, -it/3, 2u73. In particular show that

9.1 Sums of Two Squares
287
°° I „3n+l
3n + 2
•— \ 1 „3n + l 1 „3n+2
= 1+122^
= 1 + 12 2 „ g n,2 - 36 2
00 „3n
5 ^ V ?
5. Two identities due to Lorenz (Dickson [71, vol. 3, p. 29]) are
q"
n=\ i q
where x8(n) is 1 if n = 8k + 1, 8k + 3 and -1 if n = 8k + 5, 8k + 7,
l-<74" 14«
/ 1 — /7n \ / 1
ii) 03(#3(^)=i+2S f4K+4^ t .
„=i \1 - # I „=1 \1- q
Let dai+b(k) be the number of divisors of k of the form ai + 6.
a) Show that the number i?2(fc) of integer solutions of n2 + 2m2 = k is
given by
R2(k) = 2[dw+1(*) + dSi+3(k) - dSi+5(k) - dSi+7(k)].
b) Similarly, the number R3(k) of integer solutions of n2 + 3m2 = k is
given by
R3(k) = 2[d3i+1(k) - d3i+2(k)] + 4[d12i+4(k) - d12i+s(k)].
Note that a) follows from Exercise 12 of Section 3.7.
c) Observe that ii) can be used to show that
q + q
/ 1 — n \
e2(q)e2(q3) = 4 2 (f-^V = 4 2 -^
/i odd
q2n + q~2n
Hence
Fin.x 02(/3)02(/33)
y
£0l + L4n+2 4V5
where /3 := (V5 - 1) /2 and Fn and Ln are the Fibonacci and Lucas
numbers, as in Section 3.7.
6. Use Ramanujan's modular identity of order 3 (Section 5.2) to deduce
that

288 Some Additional Applications
Hence deduce that
V4E?_20?(j8)0?(i83)
«=o L12n+6 3 6v5
where /3 := (3 - V5)/2. Note that | E£.2 0?(?)0?(93) can be written
more compactly as dl(q)dl(q3) + dl(q2)dl(q6). Find a closed form for
the Lambert series above when q'-= e~v/ 3.
7. Use formula (9.1.16) to establish that every positive integer is the sum
of three triangular numbers and that every number of the form 8 k + 3 is
the sum of three odd squares.
9.2 (CHEMICAL) LATTICE SUMS
Sums of the form
(—■\y+i+k
(9.2.1) 63(2,):= 2 ,.2./2, ,2y
arise naturally in chemistry. (Here the prime indicates that we avoid
summing i = j= k- 0.) Indeed, b3(l) can be considered as the potential or
Coulomb sum at the origin of a cubic lattice with alternating unit charges at
all nonzero lattice points. This may be considered as an idealization of a
rocksalt crystal. The quantity 63(1) is called Madelung's constant for NaCl.
Different crystals give rise to different lattice sums. We will also consider its
two-dimensional (laminar) analogue
(9.2.2) b2(2s):= 2 tVAt;
u-» 0 + 7 )
and its four-dimensional form
&4(2*)= 2' 7^-7-4^
i+j + k + l
There are some nontrivial considerations about the sense in which these
sums converge. (See Exercises 1 and 2.) We assume all sums denote limits of
rectangular summations. These will converge for re(s) > 0. The general form
of these rectangular sums is
N n
lim sn := 2 2 (-l)Xm' a(mx, m2,. . . , mN)

9.2 (Chemical) Lattice Sums
289
which, for real-valued a, can be shown to converge by an alternating series
test. (See Exercise 2.) The convergence in bN{\) is 0(n~1/2) so that 10"
terms are needed for 0(n) digits. Obviously direct computation is virtually
impossible.
Fortunately, some beautiful analytic reductions are possible. In this
section we illustrate this for b2 and b4. The idea is to observe that, for
re(s)>0.
(9.2.3) r(s)bN(2s) = Mt(e?-l) q-e"
where AT is the Mellin transform of Section 3.6. Thus
co \ 2 ■
2 (-i)Vj -i
n = —co /
T(s)b2(2s) = Ms(6t-l) = Ms
Now (9.1.13) shows that
CO
b2(2s) = T-\s)4 2 (-l)"+m_1M,tem(2,,~1)]
n,m=*l
= 4 2 (-l)n~1+m[m(2«-l)]-
Thus
(9.2.4)
b2(2s) = 4 2
(-1)-
(-1)"
„f o (2n + 1)' ~x ms
-4/3(s)a(s)
and we have factored b2(2s) into a product of Dirichlet L functions (the
alternating £ function a and /3 := L_4) defined by
^):=1^ = (1-2-
')£(*)
In particular, b2{2) = - ir log 2. Equation (9.2.4) is originally due to
Lorenz. Now any reasonable method of computing 0:(5) and /3(5) will
compute b2(l). Such is possible by various integral or summation
techniques. Correspondingly, from Theorem 3.2,
b4(2s) = T-\s)8Ms
; "*(-<?)m
Ui-(-4)"
-2
4mg4i
1-9

290
Some Additional Applications
and expansion gives
CO co -,
8 2 (-1)-^,(^)-8 2 4mMs(q4mk) .
m,k-l m,k-l
Now we compute these transforms and have
b<(2s) = 8 1 ^y1]%8(l-4'-)i ^
m,*-i (mk) m^=1 (mfc)
on adding and subtracting m(mk)~s in each summation. Thus
CO CO
But E"_0 (2n + 1)~* = (1 - 2~*K(*) and so
(9.2.5) b4(2s) = -8(1 - 22~')(1 - 21^)^)^ - 1) = -8o(*)o(* - 1).
In particular,
64(2) = -41og2=^62(2).
TV
Use of the functional equation for £, equation (3.6.6), allows one to
establish other formulae. In general any identity for any even-dimensional
power of theta functions will convert into a factorization for a matching
lattice sum. (See Exercise 4.) In Exercises 6 through 10 we show how
two-dimensional lattice sums may be used to evaluate elliptic invariants.
Comments and Exercises
An excellent and extensive recent survey of lattice sums can be found in
Glasser and Zucker [80]. This also discusses their chemical origin at some
length. Madelung's constant is analyzed from a mathematical perspective in
Borwein, Borwein, and Taylor [85]. There is ambiguity in the literature as
to whether the constants are positive or negative. We have chosen the sign
as convenient.
1. (Analyticity of lattice sums)
a) Considered as a limit of rectangular sums, show that b2(s) and
b3(s) exist and are analytic for re(s) >0.
Hint: Use the Mellin transform.
b) Show that
62(2S)=i(-l)"^ re(s)>\.
n = l " J
b4(2s) = r~\s)

9.2 (Chemical) Lattice Sums 291
Hint: g2(2s): = E"=1 (-l)nr2(n)nrS converges and is analytic for
re(s) > 5. For re(s) > 1, g2(2s) and b2(2s) coincide,
c) The analogous sum g3(2s) '■= E^=1 (-l)nr3(n)n~s diverges for
(Alternating series test) A mapping a: NN-> U is (N-)monotone if, for
m1, m2,. . . , mw>0,
N
2 2 (-l)*s'a(m1 + s1,m2 + s2,. . . , mN + jn)>0.
i = l i, = 0, 1
Thus 1-monotonicity is just a(m) s: d(m + 1), and 2-monotonicity is
a(m, n) + a(m + 1, n + 1) ^ a(m, n + 1) + a(m + 1, n), while 3-mono-
tonicity demands that the alternating sum over any unit cube be
positive if the bottom corner is. We say that a is fully monotone if a
and all its restrictions are monotone. Consider
N 00
2 2 (-1)^0(1^,1112,...,1^).
i-l m,=0
a) Prove the following alternating series test given in D. Borwein and
J. Borwein [86]. If a is fully monotone and lim^^^ a(m) = 0, then
the rectangular sums converge alternatingly. Hint: (In two
dimensions) show that sn := S"y=0 (-l)l+Ja(i, j) satisfies
V S2n— S2n+2
») 52„+1>52„_x
iii) 52„-52„_1-*0
Draw a picture. (The case of b3 and b2 is spelt out in Borwein,
Borwein, and Taylor [85]. An analogous bounded convergence
test due to Hardy can be found in Bromwich [26].)
b) Suppose that a is N times continuously differentiable on (IR4^.
Show that a is totally monotone if the partial derivatives satisfy
a, s 0, a,, s= 0, a, Ms0 (-1)¾ , , 2= 0
for all partial derivatives with ix < i2 < i3 < • • • < iN.
c) Verify that E„ (-l)'+y+*(i2 + /2 + A:2)"" and E„ (-l)'+y(2i + /)_"
converge for p > 0.
Prove that 1^)6^) = Ms(0< - 1).
a) Show that
00 1
H'7Tr^=^)^) re(5)>l
and -(»+»«)
(-1)"
(n2 + m2X
S't-TT^ =2-'62(2s) re(5)>l.

292 Some Additional Applications
b) Show that
b8(2s) = -16 £(s)a(s - 3) re (s) > 1
is equivalent to the formula (3.2.25)
(9.2.6) gfa) = l + 16S \1)nnq ■
c) Use 020304 = 0X+, equation (3.2.4), to show that
(9.2.7) 2 2 ("ir 2 =2* + 1j3(2^-l).
— [m + n +(p+ f) ]
5. The hexagonal sum is
^(2,): = 2' + g;2'mj. ._.,., re(5)>0
-» [(« + m/2) + 3(m/2) J
where <?(«, m) := f {sin [(n + 1)0] sin [(m + 1)0] - sin («0) sin [(m -
1)0]} and 0: = 2u73. This corresponds to calculating the Coulomb
potential on a regular hexagonal lattice. (See Borwein, Borwein, and
Taylor [85].)
a) For re(s) > 1, show using the cubic modular equation that
h2(2s) =
1-3
l-s
2Zi , 2 . „ 2\s ^-1
(-1)"
(n2 + 3m2Y (n2 + 3m2yj
2
b) A formula of Cauchy (Dickson [71, vol. 3, p. 20]) gives
1 + qn
e4(q)e4(g3) = i + 22 (-l)V
n = l
L 1 + q3n
Use this and a) to deduce that
^(25) = 3(1-3^)^)^3(5)
where
L__3(s) := 1 - 2~' + 4~s - 5~* + 7"* - 8~s + ■ ■ ■ .
c) This provides an analytic continuation of h2(2s). Thus
3(V3-l)(V2 + l)o(i)L_3(i) = A2(l)
is Madelung's constant for the hexagonal lattice. Show that
h2(2) = V3ir log 3 .

9.2 (Chemical) Lattice Sums
293
6. (Evaluating invariants) The factorization of two-dimensional zeta
sums into sums of products of L series can be carried a great deal
further. This is described in Glasser and Zucker [80] and in Zucker and
Robertson [76a,b]. By combining number theoretic and transform
techniques, one can explicitly factor all sums whose discriminants are
disjoint (have one form per genus) and a few others. This leads to
formulae such as
(9.2.8) T (m2 + Pn2)'s = 2W 2 L±ixLXAPI^
and
(9.2.9) T(m2 + 2Pn2ys = 2l-'2 L^L,^ .
n\P
In these two formulae P is an odd square-free number [congruent to
1 modulo 4 in (9.2.8)] with t distinct factors. The right hand sums over
all divisors of P and one has
CO
L±d(s):=2 (±d\n)n-s
n = \
which, are primitive L series modulo d. (These can only exist for d = P,
4P, or 8P, and the sign configuration is in fact uniquely specified.) Here
(k\n) is the Kronecker (generalized Legendre) symbol. L±^ is taken for
/i. = ±1 (mod4). For example, with P'~29, (9.2.8) becomes
Zi (m2 + 29n2)~s = L1L_n6 + L_4L29.
Dickson [29] gives an extensive list of disjoint discriminants. In
particular, there are 18 numbers less than 10,000 to which (9.2.8)
applies and 15 numbers less than 10,000 to which (9.2.9) applies.
Indeed, (9.2.8) holds for type one P: = 5, 13, 21, 33, 37, 57, 85, 93,
105, 133, 165, 177, 253, 273, 345, 357, 385, 1365 and (9.2.9) holds for
type two P:= 1, 3, 5, 11, 15, 21, 29, 35, 39, 51, 65, 95, 105, 165, 231.
There are only finitely many disjoint discriminants. We shall call such
P solvable.
Implicit in (9.2.8) and (9.2.9) are corresponding theta series
identities, and formulae for representations as weighted sums of squares.
a) Show that if q is replaced by -q in (9.2.8) and (9.2.9), we
produce formulae for
i) 1]'(-l)m+n(m2 + Pn2)s
ii) l]'(-l)m(m2 + 2Pn2ys .

Some Additional Applications
Show that the effect of replacing q by -q in (9.2.8), (9.2.9), or
similar formulae is to replace L±d by -[1 - {2\d)2x~s]L±d, unless
the L function was multiplied by some factor involving 2~s, or d is
even, in which cases it is unchanged.
Recall that (3.2.12) and Exercise 4d) of Section 3.2 gave
4ir
2' (-ir+n(m2 + m2)"1 -- ~ log /(V=7)
and
4ir
2' (-l)m(m2 + m2)"1 = - ^ log /x(V=r).
Use these formulae with a) and b) to establish that for the
appropriate P,
i) 2'-1^log/4(V=P) = Li4P(l)log2
+ S[l-(2|/*)]L±M(l)L*4/,/M(l)
Mi'
m#i
ii) 2'"1 ^= log /J(V=2P) = L_8P(l)log2
+ 2 [l-(2U)]L±M(l)L¥8/,/M(l)
The classical Dirichlet class number formulae (Landau [58]) allow
us to write L±d(l) algebraically. One has, for d>0 restricted so
that L±d is primitive,
i) L+d(l) = 2^1oge(<0
... _ ,., 2w h(-d)
u) L-(1)=V5^)"
Here h(d) is the number of (broadly) equivalent primitive classes
of reduced forms with discriminant d — b— 4ac or ideals in
Q(VD) where d is D or AD, depending on whether D = 1 (mod 4)
or D = 2,3 (mod 4); e(d) is the fundamental unit in Q(VD), which
may be computed from the fundamental solution of the
appropriate Pell's equation (see LeVeque [77] and Hua [82]); and w(d) is
a factor which counts the number of automorphs of the form, and
is 2 except that w(3) = 6 and w(4) = 4. From formula (4.12) in
Zucker and Robertson [76a] we have Dirichlet's formulae, for
d>4,

9.2 (Chemical) Lattice Sums
295
-1 d_1
h(-d)=-j- 2 n(-rf|«).
Also for d > 0
A(rf) log e(d) = - ^ t (d | w) log (sin ( ^)) .
Moreover, for disjoint forms, one can observe that the class
number h{—d) must coincide with the number of genera g. The
number of genera is as follows. If d is odd, then g = 2m~x, where d
has m distinct prime factors. If d is even and d/4 has m distinct
prime factors (including 2), then g = 1m when d/4 = 0,1,5 (mod 8)
and g — 2m-1 otherwise.
An excellent brief survey of history and of recent advances
regarding the class number can be found in Goldfeld [85]. Now
observe that, for the appropriate P, both \f and \f\ will be
products of powers of fundamental units from some of the divisors
of 4P or 8P. [For our solvable P, /i(-4P) = 2' when P = 1 (mod 4)
and /i(-8P) = 2'.] In particular, verify that for P:=5, 13, or 37,
since t — 1 in each case
4 - lf*Kf-5\-, „(p\KP)
G>^/t(V-P) = e(P)
Since h(P) = 1 in each case, we see that
Similarly, we may now verify the values of G6P, P-=21, 33,57, 93,
given in Exercise 9 and 10 of Section 4.7.
Establish that for P: = 5 or 29,
and for P:= 3 or 11,
g\P = e(P),
gT = s(2).
(Compare Table 5.2.)
For all the type one numbers P listed above, we can now
explicitly give Gp. Similarly for the type two numbers we can give
g2P. This accounts for most of the square-free and nonprime
invariants given by Weber or Ramanujan.
Show that
82i3o
=mm

Some Additional Applications
(Evaluating singular values) We can proceed further to evaluate
A: := A*(2P) for P of type two. We know that yfc/4 =
f1(V=2P)f;8(VI:8P). Hence
2plog('l) = 2'(-inm2 + 2Pnri
-4S'(-l)m(m2 + 8Pn2)-1.
We already know [Exercise 6c)] the first sum on the right. From some
elementary, but skillful, theta transformations we may deduce that
2' 1' (-l)'n(m-; + 8Pn2ys = 1 {[21^ - 1 + 2-'(2|/i)]L±/1L¥8l./M
On setting 5:=1 and substituting above, we derive
(9-2.10) - -^= logfc-22"' 1 L±V(1)L^(1),
a beautiful simplification.
a) Use (9.2.10) to compute A*(2P) for P:= 3, 5, 11, 29.
b) Use (9.2.10) to compute A*(210) given in (4.6.12).
c) Observe, as Zucker did, that there are computable in this form
two larger singular values: those for 2P : = 330 and 462. Verify that
A*(330) = (2 - V3)3(V2 - 1)2(V33 - 4V2)2(Vl0 - 3)2
x (3V5 - 2VH)2(4 - vl5)(V55 - 3V6)(10 - 3VTI)
(9.2.11)
and that
A*(462) = (V3 - V2)4(2 - V3)2(2V2 - V7)2(8 - 3V7)2
x (3VIT - 7V2)2(V22 - V2l)(10 - 3VTT)(76 - 5V231).
(9.2.12)
(Evaluation of K in terms of T) Selberg and Chowla [67] showed for
all rational numbers r that K(\*(r)) is expressible in closed form using
a finite number of T values. This relied on Kronecker's remarkable
'Grenz-Formel,' which has
lim (2' (m2 + rn2Ys - (5_^)v?j = ^ Py - log (4/-) - 4 log v]
(9.2.13)

9.2 (Chemical) Lattice Sums 297
as a special case. Here 17 : = r](Vzzr) is the eta function of (3.2.9) and
(3.2.11), and y is Euler's constant. Using (9.2.13), Zucker [77] applies
the factorization results described above to explicitly compute K
corresponding to solvable sums in terms of ir, surds, and Y values. This
leads to the following table of evaluations (Table 9.1). Elsewhere
Zucker has actually given £(A*(210)) [which involves T(n/840) for
(n, 840) = 1]. The general formula valid for either r-=P(Pof type
one) or r- = 2P (P of type two) is given by
and
4iog,(v-7) = ^S(-4H«)r(^
-log (877/-)- 2 L±Ai(l)L^(l).
(9.2.14) **< Ar) #
There is a corresponding formula when in Zucker's terms 5(1,1,
(1 + r) /4) is solvable.
a) Verify the contents of Table 9.1, for r = 1,2,. . ., 6.
b) Show that T(m/24) is Oop(log n) computable for integral m.
Hint: Express such T values in terms of K at singular values. It
would be interesting to know if this is possible more generally.
c) Compute £((3 - V7) /4V2). (Compare Exercise 9 of Section 5.2.)
d) Show in general that E(k*(rj) is computable in terms of V values
and algebraic quantities for rational r.
{Conjugate divisors and evaluations of k) Consider P as above and
suppose P-=d1d2 and Q-=d1/d2 for divisors dx and d2. Using
Kronecker's genus character sum formulae one can show that GP and
GQ = GQ-\ must have the same general form for P of type one.
Similarly, g2P and g2Q = g2Q-\ are paired, as are A*(2P) and X*(2Q).
This is best illustrated with examples.
a) Show that
g2190 = (V5 + 2)(VT0 + 3) g23g/5 = (V5 - 2)(VT0 + 3).
The first value may be computed as above. The second is easily
verified from Schlafli's form of the quintic modular equation.

TABLE 9.1. Evaluation of K at the First Sixteen Singular Values
K
[T(i)l
i\i2
9
10
11
12
13
14
15
16
4tt1/2
(V2 + i)1/2r(|)r(|)
013/4 1/2
2 TT
31M[r(j)]3
27/V
(V5 + i)[r(i)]2
07/2 1/2
2 TT
(V5+2)1M[
1Mrr(*)r(*M*M*2
1607T
- (V2 - i)(V3 + V2)(2 + V3)r( h )r( A )r( £ )r( a)
384-tt
iW(f)rO)
(71/4)41
2V2+(1 + 5V2)1/2
4V2
1/2(V2 + i)1/2r(i)r(i)
12tt1/2 L U/J
•(2 + 3V2 + Vs)r(A)r(j,)r(|,)r(^)r(a)r(^)r(a)r(a)
2560tt3
31M(2 + V3)'
[2 + (17 + 3V33)1'3 + (17 - 3V33)"3]2 IlAHp|0HAl
(V2 + 1)(V3 + V2)(2 - V3)1/231/4[r(|)]3
^13/3
(18 + 5V13)1'4
Mrr(A)r(^)r(^)r(M)r(i)r(^)r(ipr(i)r(i)r(M)r(i)r(l)i1/2
L 6656-tt5
[(10 + 6V2)1'2 + (2 + 2V2)1'2 + (3 + V2)1'2]1'2
[r(^)r(^)r(^)r(^)r(i)r(i)r(i)r(j)r(i)r(g)r(i)r(g)]1/4
X 16ttV7
(V5 + i)iU)r( 6)r(A)r(6)i"
240-77
(21/4 + i)2[r(i)]2
298

9.2 (Chemical) Lattice Sums 299
b) Correspondingly
_6 _/V5 + l\3/V3 + l\3/V7 + V3\VV7 + V5
G™-\~2~) \~Vr) \~~2 ) \~^7T~
_6 _/V5-l\3/V3 + l\3/V7+y3\VV7-V5
G^~{~2~) l~V2~/ V 2 ) \ V2
6 _/V5 + l\3/V3-l\3/V7 + V3\3/V7-V5
G21/5 V2/VV2M 2 M.V2
_6 _/V5-l\3/V3-l\3/V7 + V3\3/V7 + V5\
G15/7-^-y-j [-^-) { 2 / I V2 /•
One may verify G35/3 and G21/5 from the corresponding cubic and
quintic equations, and so on.
c) Use the techniques of Section 5.3 to compute 0-(105).
d) Observe that
i) A*(6) = (2 - V3)(V3 - V2)
A*(§)= A*'(|)= (2 - V3)(V3 + V2)
ii) A*(10) = (VT0-3)(V2-1)2
A*(|) = A*'(|) = (VI6- 3)(V2 + 1)2
iii) A*(58) = (13V58-99)(V2-1)6
A*(i)-A*'(f ) = (13V58-99)(V2 + 1)6.
e) Indeed, for all P of type two, A*(2Q) and A*(2P) will be
"conjugate", as we illustrate for A*(210). Let
^••-(V^-i)2
«4:=(4-Vl5)2
m7:=6-V35
M2:=(V7-V6)2
«5:=2-V3
«8: = 8-3V7.
«3:=(VlO-3)2
M6:=Vl5-Vl4
Then A*(210) = Ilm=1 um, and each A*(2Q) is a corresponding
product U8m=1 ue™(Q) where each em is ±1. In compact form one
has em{Q) := 1 -2bm{Q), where bm{Q) is the mth binary digit of
6(2(2) defined by
6(f) = 142 6(f) = 201 6(f) = 45
KH) = 71 &(g) = 163 6(1) = 228
6(^) = 106 6(210) = 0.
These correspond to all the reduced forms with discriminant 840.

300 Some Additional Applications
f) Numerically, find the "conjugate" values for P".= 165 and
P : = 231. [This involves computing 28 products and comparing
them with the theta expansion of k(2Q).]
With these conjugate values and the formulae of Section 5.3 one may
observe that a(2P) is now available in closed form for all P of type two (and
for conjugate divisors Q). Again we illustrate with an exercise.
10. a) Generate a recursive version of formula (5.3.3) and specialize this
expression to produce a formula for a(2d1d2) in terms of Rd ,Rd,
Md , M2 and the appropriate singular values.
b) Apply this formula with dx '■= 35 and d2 '■= 3 to compute a(210).
Observe that this uses only values of A* given in Exercise 9e).
c) Compute a(42). Note: g\2 = (2V2 + V7)'[(V7 + V3)/2]3 is
incorrectly, given in Weber [08].
11. Let {an} and {bn} be given sequences,
a) Show that
if and only if
b) Show that
if and only if
£(s) S ann s = S bnn
2 by.
" 1 - r"
a(s) Sa„" s= S bnn
S bnx".
c) Show that, with the notation of Exercise 12 of Section 3.7,
CO
i) £ <rk(n)n-'= £(s - k)£(s)
and
n) S,W« =^

9.3 Odd-Dimensional Sums and Benson's Formula
301
Also
in
j) f, *<">""'-^r
where ¢ is Euler's totient function, which counts the numbers less
than n and relatively prime to n. Thus ¢(1):=1, ¢(5) = 4, and
¢(6)=2.
9.3 ODD-DIMENSIONAL SUMS AND BENSON'S FORMULA
While even-dimensional sums usually factor, only a few odd-dimensional
ones factor. The reader can, however, produce identities like (9.2.7) from
Jacobi's identity (3.1.15) or from (3.2.8). (See Exercise 1.) There are
nonetheless many theta-based techniques of which we establish one based
on the theta transform.
Theorem 9.1 {Benson (1956))
-h(l) = E ,.2-, -2^,2,112 = 12T- 2 Sech2
-» (i + ] + k )
(9.3.1)
m,n=\
odd
- (m + n )
Proof. By symmetry,
(-2/- -ly' + fc
03(1)-3 2( ..2 , .2 , ,2
,-.2 , .2 , /2-,3/2
0 +7 +* )
and
r(| 63(1) = 3 S (-i)Vm3/2 S (-i)/+V2+y2
where # : = e '. Thus
r( 2 )63(1) = 3M3/2
2 (-dv^jco
The theta transform (2.3.2) leads to
-r^)&3(i) = 3Af3/2
»2 w „?/ it
2,(-1) n q — 02y t

302 Some Additional Applications
and since T( §) = Vtt/2,
00/- 00 /• CO -i
-63(i) = i2V¥E (-i)n+V E [g-^'^^^jr1'2* •
n=l <- y,*:=-o<.JO J
odd
The internal integral I(n, j, k) was evaluated in Exercise 4 of Section 2.2.
We have
and
-63(1) = 48tt E E (_i)»+i„e^nV(2^1)2+(2A:+1)2.
y',Jt=0 n=l
Finally, for a > 0,
4 2(-1^--^^--^(1)
and Benson's formula follows. D
The convergence acceleration is astounding. Summing for 0sj, A: < 3
produces b3(l) '■= -1.74756459 ..., which is correct to eight places. Even
the first approximation, by -12ir sech (ir/V2), gives -1.73 ....
For s¥= \ this manipulation leads to an integral involving Bessel functions
of the second kind, and the closed form is lost.' There is, however, an
extension to N dimensions (among others). One can show that
(9.3.2) -»^-2)-^^2i«h"(f^j).
odd
Thus
(9.3.3) -64(2) = 167r2 S sech2( ^[? + j2 + k2) = 4 log 2
«,M=1 ^
odd
and
00
(9.3.4) -62(0) = 2ir E sech2
n = 0
which coincides with Exercise 7eii) of Section 3.7. (See Exercise 2.) For the
final evaluation it helps to know that for primitive L series
(2n + 1)

9.3 Odd-Dimensional Sums and Benson's Formula 303
V*
L_t(0)= — L_t(l)
as follows from the functional equations for L±k:
L_k(s) = C(s)cos[™)L_k(l-s)
(SIT \
YJL+k(l-s)
where C(s) := 2V-1Jfc~,+1/2r(l - 5).
Comments and Exercises
Our derivation of Benson's formula can be found in Glasser and Zucker
[80]. In that paper, and references therein, one finds much further
discussion of odd-dimensional sums. They also illuminate the relationship between
the multidimensional zeta functions of Epstein and lattice sums.
1.
a)
b)
Show that
00
2S(-l)m
— CO
\2m--J + 2n+2p2
— s
= 2s+1L_8(2s-l)
where L_8(s) :=1 + 3~s-5~s-7~s + • • • .
Show that
CO
g(2s):=2(~l)m+n+p
— CO
= 12s/3
(2s - 1)
[(-^K)2+K)1
and g(l) = Vl2/3(0) = VS.
c) Derive similar identities from Exercise 5 of Section 4.7.
2. Establish the generalization of Benson's formula (9.3.2) and its special
cases (9.3.3) and (9.3.4). Exercise 4b) of Section 9.2 shows that
68(6) = -8f(3).
The remaining exercises examine the n-dimensional Hurwitz zeta
function. Let d>0 and define
(9.3.5) L,.(s,,):.(So)|^L_
where 5,- := sign (a,) and a :— (ax, a2, ■ ■ ■ , aN). Thus

304 Some Additional Applications
».«-i (n + m)
and
t ' ,<u- Y (-1)"
3. a) Show that
"'-'-'(-log*)'-1
^"•^['W-*
//inf: Take a Mellin transform. Then make a logarithmic variable
change and sum the resultant series.
b) Let Ls(s, d)~: AN(s,d) and L_-(s, d)=: A_N(s, d), where e is
the vector (1,1,. . ., 1) in UN. Show that for d > 1
(9.3.6) ^(MH^t^-lM^-^-l)
~A±(N_iy(s- l,d-l)].
Hint: Use integration by parts.
c) Combine integration by partial fractions and the recursion (9.3.6)
to show that every sum of the form (9.3.5) factors into a linear
combination of one-dimensional Hurwitz zeta functions (with
coefficients depending on 5).
4. Let AN(s) := A_N(s, N) and PN(s) := AN(s, N).
a) Show that
1
and
Pn(s) = J^l Pn-x(s -I)" PN-x(s) •
b) Deduce that, for appropriate s,
0.2,1¾-^-««■-«.
ii} 1,(^7 = 2^)-^ = 2^)

) 2
in
Thus
9.3 Odd-Dimensional Sums and Benson's Formula
1
305
„=1 (m + ny
IV
■) r
(-1)"
(\m\ + \n\):
= as -i)- «*).
= 4a(s -1)
o r
= 4^-1).
(H + wr
c) Show that
(#-1)1^)=2^-1 + /0
n = l
where a„ are Stirling numbers of the first kind. Thus
p4(S)=u(s-i)- as -2)+%as-i)~ as)
and
P5(s) = &&-*)- &&-$ + %C(s-2)- $C(s-l) + fa) .
There is a similar formula for ^4^(^).
5. a) Show that
S'(H+H+i^r=2^+4^-2)-
b) Show that
(-1)"
1-2
2 y«(5)+2 /3(5).
n,m=o (2n + m + l)s
c) Show that
°° (—■\\n+m 1
JL fe + < = 3[{2 + 3""><'> - «<' - *> - L~^ Dl
and so L_1>_3(1, 4) =
6. Show that
V ("■*■)
7 7T 1
3L3l0g2 3V3 2.
+ s
n,^-i (« + 2m + 4p)s ~0 (8n + 7)
Hint: (1 + jc)(1 + x2) ■ ■ ■ (1 + j2""1) = (1 - x2")/(l - x)
=8-sas).

306 Some Additional Applications
9.4 THE QUINTUPLE-PRODUCT IDENTITY
Jacobi's triple-product identity has an elegant fivefold analogue due to
Watson [29] and Gordon [61]. This is
CO
(9.4.1) II (1 ~ qn)(l - *?")(! - z"V-1)(l - z2q2n-x)(l - z-2q2n-x)
CO
= 2 (z3m - z-3m-1)gm(3m+1)/2
m = —CO
valid for all complex z and # with \q\ < 1 and z^O. The proof is left as an
exercise. (See Exercise 1.)
If we divide both sides by 1 — z~x and let z tend to 1, we derive
ri(l-g")3(l-g2,,-1)2= E (6m + l)gm(3m+1)/2 .
Now this yields
(?.4.Z) V4{q)— oo m(3m + l)/2
Em=_„(-1) <?
on using Euler's pentagonal formula (3.1.10) and (3.1.7). This can be
used to establish a recurrence formula for r2(n), as in Ewell [82]. (See
Exercise 4.)
Comments and Exercises
The identity, implicit in Ramanujan's work, was discovered by Watson [29]
and rediscovered by Gordon [61]. Further extensions are discussed by
Gordon. Other proofs abound in the literature.
1. Prove (9.4.1). As with the triple-product identity, it is easy to establish
the formula (9.4.1) up to a constant relying on q alone. To evaluate the
constant observe that when z:= -1, (9.4.1) reduces to Euler's
pentagonal identity.
2. Show that (9.4.1) is equivalent to
CO
II (1 - q2")(l - q2"^z)(l - q2n-lz-l){l - q4n-4z2)(l - g4"~4z-2)
= S q3n2~2n[(z3n + z~3n)- (z3n~2 + z-(3n~2))].

9.4 The Quintuple-Product Identity 307
a) Let q'=q' and z := q~k in (9.4.1) and deduce that
(9.4.3) EI (1-<?")= 2 qi^"^-*'"'- qV + W]
neN(j.k) m=-oo
where N(j, k) consists of all integers congruent to 0, ±k, j, j ± k,
j±2k (mod 2/) (repeated as appropriate).
b) Deduce that Euler's identity follows for /:=4 and k ■ = 1. For /:=3
and k : — 1 one gets
CO
II (1 - qn)(l - q6n-5)(l - q6"-1) = g(q) - 3ffi(g9)
n-l
where g(q) : = E:=0 g<"2+">'2. With /(¢) := 11^ (1 - q'y1 this
becomes a formula due to Ramanujan
c) Use g(q) = f(q)/f\q2) to obtain a functional equation for /(¢)
(the partition function).
d) Use (9.4.3) with / := 6 and k := 1 to obtain
2f(q2)f(q3)f(q12) „„„„,_9
/(<7)/(<74)/V)
= 3¾^9) -«,(<,)
Hence obtain a functional equation for 03. Thus note that
3#j(# )- #j(#) never has a solution. (Compare Section 4.7.)
a) Establish equation (9.4.2) and
E;=_0O(6m + l)9(6"1+1)2
el(q24)
£;=_(-iyY6m+1)2
b) Use (9.4.2) to derive a recursion for r2(n).
c) Prove that
CO CO
II (1 - q2")(l - q2n~X)2{l - ?4")2 = E (3m + l)?3m2+2m .
(9.4.4)
//i/tf: Use Exercise 2.
d) Show that
04(<?)_ E:,-.(3/i + l)g(3"+2)"
^(¢) £:„_.„ (-1)-(3/1 + l)g(3"+2)"

308 Some Additional Applications
so that
Observe that this is slightly faster to compute than the original theta
series ratio,
e) Show that
^_M_, 1/4£:=-„(3k + 1)<7(3"+2)"
Vk 63(q) lq E;__„(6,i+ 1)^-
so that
E:=_„(6« + l)^6"+1)2/12'
5. a) Show that
CO
2
/ly+i + k
/,/;*— [24i + 24/ + (6k + 1)T
1 1
= 1
2s-l /£, , r\2s-l
(6/n + ir-1 (6m+ 5)
= (1+21_2*)L_3(2j-1).
b) Replace g by -g in a) to express L_24(2,j - 1) as a lattice sum.
c) Combine (9.4.4) and (3.2.7) to prove that
L.3(2s - 1) = X I£t. ,^2, i,1?
i+j+k
[6(i+ \f + 6(; + |)2 +9(/c+i)2]s •
Observe that at 5 = | this equals 3.
Gordon [61] also gives various congruences, like those given in Chapter 3
for the partition function. For example,
/(<7)/(<74) nJA
has c3n+2 divisible by 3.
6. There is yet another remarkable identity due to Ramanujan, which
includes both the triple-product and the q-binomial theorems. This is
the 1M*1 sum, whose derivation and uses are accessibly described in
Askey [80]. In standard notation one writes

9.5 Quintic and Septic Multipliers and Iterations
309
(9 4 5) y (fl> ?)» x" = (ax'> q)~{qiax; g)~(<?; q)~(b/a; g)„
-„ (b; q)n (x; q)m(b/ax; q)„(b; q)„(q/a; q)„
where (a; q)„ := II (1- aqk) and (a; q)n : = ^-.
t=o 1«? . q)oo
This converges at least for \q\ < 1 and \blq\ < \x\ < 1.
a) Verify that the triple product is contained in (9.4.5).
Hint: Begin by setting a '■= c"1, x •'= ex and b ■ = 0. Now let c := 0.
b) For b'-= q (9.4.5) becomes the q-binominal theorem:
V ia' tin xn= (ax; <?)„
«=o (<?; q)n 0; <?)»
valid for |jc| < 1 and \q\ < 1. Verify that Cauchy's binomial theorem
is a special case. Hint: Begin by setting a '■= q~2".
c) Use the q- binomial theorem to express the limit mean of Example
8.4 as a Taylor series.
9.5. QUINTIC AND SEPTIC MULTIPLIERS AND ITERATIONS
In this section we discuss additional modular and multiplier identities and
give a number of applications. Notations are as in Sections 4.6 and 4.7.
References to entries are all to Chapter 19 in Ramanujan's Second
Notebook (Berndt [Pr]).
Proposition 9.2
(a)
(9.5.1) 5M5 = 1 + 2G25„/G^
(9.5.2) 1/M5 = 1 + 2G„/G^„
(9.5.3) 5M5 + l/M5=2{2 + /c/ + /c7'}.
(b) Let 11 (It + 1): = M5. Then
(9.5.4) 1 - 2/2 = M\{\ - lit - t2)V(l + t2)M5
(9.5.5) 1 - 2k2 = (1 + t - t2)V(l + t2)M5 .
Proof.
(a) These are given in Entry 13. Berndt [Pr] provides proofs which may
also be deduced from (4.1.20).
(b) These may similarly be found in Entry 14. □

310 Some Additional Applications
We will write, as in Section 4.7,
Mp(n):=Mp(kn,kp2n).
Ramanujan also gives the following beautiful counterpart to Theorem 4.11
(Entry 12(iii)).
Theorem 9.2
03(q25)
(9.5.6) ^ = 1 + ^ + ^
where for i = 1 or 2
(9.5.7) ri-=\x{y±yJy2-Ax3)
and
(9.5.8) ^:=5-^^-1, >>:=(*-l)2+7.
This provides a solvable update for M5 despite the nonsolvable nature of Ws
(See Exercise 8 of Section 4.5). Indeed with mn : = 5M5(r52") we have
(9.5.9) mn+1 = (1 + r11'5 + r12'5f/mn
with x: = mn - 1, y and r; as above.
If we combine (9.5.9) with the following formula for e5 we obtain a
remarkably simple solvable 5th-order iteration for it.
Proposition 9.3
If r >0 and s(r) := M^\r) then
(9.5.10) a(25r) = s\r)a{r) - Vr[S (^~5 + Vs(r)(s\r) -2s(r) + 5)} •
Prao/.
We begin with e5, as given by (5.2.13), and the explicit formula for R5-
We use Proposition 9.2(a) to rewrite s(r)R5 and 9.2(b) for the terms
involving /2 and k2. This leads reasonably directly to
e5 =. S-&ll + ys(r)(s\r) - 2s(r) + 5).
Now (5.2.14) becomes (9.5.10). (Exercise 2.) Q

9.5 Quintic and Septic Multipliers and Iterations 311
The identical manipulations to those in Example 5.3 immediately yield
a(5)=i{V5-V2(V5-l)}.
We next list several similar identities for M1.
Proposition
(a)
(9.5.11)
(9.5.12)
(b) Let t := i
9.4
{kl)1/4
, / /' //' ( IV \213
.,,- k k' kk' Jkk'\2'3
. Then
(9.5.13) 1M7 - 1 /M7 = 6 - 16t + 12t2 - 8t3 .
Proof. These are to be found in Entry 19. □
From (9.5.13) and (4.6.7) we may establish that for r>0 and
s{r):=M~\r)
(9.5.14) a(49r) = s\r)a(r) - Vr{^~7 + s(r)(4t2 -4/ + 3)} ,
where tA = kl. (See Exercise 3.)
Ramanujan does not give a septic analogue to Theorem 9.2. He does,
however, give quintic and septic updates for the eta-multiplier. Let 17 be
given by (3.2.9) and (3.2.11). Let
(9.5.15) Np:=JlLti
v\q'P)
so that Np corresponds to Mp. This is the eta-multiplier of order p. Now
(3.2.15) can be written as
(9.5.16) kk'M3p = ll'N3p or Mp = Np{^)
where Wp(l2, k2) = 0.
Theorem 9.3
(a)
(9517) 6=M2(1-M2)
1 -^ ■U) ■ 2 4(2M2-1)
1/3

312
(9.5
.18)
(9.5.19)
(b)
(9.5
.20)
Some Additional Applications
3 M3(l - M\)
3 9M2 -1
3_M5(1-M5)2
5 (5M5-1)2 •
(49M2 - 1)JV7 - (8M7) JV2 + (8M2)iV7 + M7(M27 -1) = 0.
Proof.
(a) In each case one combines (9.5.16) with appropriate multiplier
equations. For p ■ = 5 use (9.5.1) and (9.5.2). For p ■= 3 use the identities
preceding equation (4.7.9). (The details are left as Exercise 6a).)
(b) We use (9.5.11) and (9.5.12) to write
//'M7"2(l + 8JV7) = kl' + Ik' - kk'
and
A*'M2(49 + 8/JV7) = kl' + Ik' - IV .
We now subtract one from the other and divide by IV. □
We finish the section by listing Ramanujan's updates for N5 and N7. Entry
12(i) can be recast as
(9.5.21) 5JV5(<?5) = (M1/5 + v115 - l)2/(5N5(q))
where fi and v are the solutions to
fiv = -l, ix + v = 11 + (5N5(q)f .
Entry 18(ii) becomes
(9.5.22) 7iV7(^7) := (fiin + vin + a>in - l)2/(7iV7(^))
where /jl, v and w are the roots of
x3 - ax - bx + 1 = 0
and a and b are given by
a:=57 + 14[7JV7(<7)]2 + [7JV7(<7)]4
b : = 289 + 126[7JV7(^)]2 + 19[7JV7(^)]4 + [7JV7(^)]6 .

9.5 Quintic and Septic Multipliers and Iterations
313
Theorem 9.2 gives 63(q2S) solvably in terms of 83(q5) and 63{q). Likewise
(9.5.22) gives 63(q49) solvably in terms of 03(<?7) and 63(q). Thus 64(q) =
03(-q) is similarly solvable and since k = 62AI0\ and ;' is solvable in k we see
that for /any of 04, k, g, G or J, f(qp) is solvable over <Q(/(<?), f(ql'")); for
p '■ = 5 and for p ■= 7. In view of the nonsolvability of the quintic or septic
modular equations for A (and hence k and /), this is at first surprising. What
is happening is that the Galois group for Fp, /?>5 and prime, is a
nonsolvable group of order (p - l)p(p + 1). However j(qllp) is a root of Fp
and, since Fp is irreducible, it is of order/? + 1. Thus the splitting field for Fp
over Qp(j(q), j(qVp)) has order dividingp(p - 1). For/? := 3, 5, 7, and 11,
(/? - 1) /2 is prime and the corresponding group is obviously solvable. For
/? : = 7, for example, we expect seventh roots of cube roots to comprise the
solution ((/?-1)/2 = 3). This is consistent with equations (9.5.22) and
(9.5.20).
Comments and Exercises
Knowing that a solution exists and exhibiting it, particularly in simple form,
can be very different matters. The components of the quintic and septic
algorithms for it are far less complicated than one might initially expect.
Both can be packaged very elegantly. (See Exercises 2 and 7, and Borwein
and Borwein [Pr].)
1. a) Combine (9.5.1) and (9.5.2) to obtain Schlafli's form of the quintic
modular equation.
b) Compute that
i) M5"1(1/5) = V5
ii) M5_1(l) = 5(V5-2)
iii) m;\3/5)=£^±
iv) M5_1(2/5) = V5(V5+2)(V2-1)2.
c) Find closed forms for M^l{n) for « := 1, 5, 9, and 25.
d) Use G85 = [(V5 + 1) /2][(9 + V85) /2]1/4 and the conjugate nature
of G17/5 to compute 5M5(-f). Use Theorem 5.4 to compute cr(85),
2. a) Verify the formula (9.5.10) for a(25r).
b) Obtain closed forms for a(n) « = 25, 125, 625, 225, and 1225
(Ramanujan [14] givesG1225).
c) Observe that Exercise lb) (9.5.9) and (9.5.10) combine to give
several explicit iterations for it. For example, we may begin with
r:=l, a(l)=i,*(l) = 5(V5-2).
d) Use Theorem 9.2 to compute G62Sn in terms of Gn and G25n.

314 Some Additional Applications
3. a) Show that (9.5.14) holds and that
,_s2(r) + 2s(r)-7
e7.=
+ 2s(r){VJk + VFF}.
b) Compute the corresponding updates for S(25r) and S(49r). In
particular
8(49) = M~\l){l + 2\fkl + 2VW} .
c) Verify a(J).
d) Establish that R5 and R7 are as given in Table 5.1.
4. Ramanujan also gives (Entry 19)
m- - lV
M
(k'i'y"-(kiy
\7i 1/12
1-41«
- kk'
1Mi=7T7Jn
a) Thus show
(«)1/4-(*:'/')
1M2_(GA9n\1 Gl-2V2G,9n
/Mt~\ Gn I 2V2G -Gl
b) Compute M7( f) and M7( f).
5. Ramanujan in his letters (Hardy [40] p. 353) gives the following
beautiful hybrid identity. Let
(C C \3/2
Q Q5") > P:=(GnG9nG25nG225n)in ■
Then
Ap+1p) = Q+i
Q
a) Verify that G\5 = Qf±^ and G\n = 8^f + *> .
6. a) Establish the eta-multiplier formulae of Theorem 9.3a) and b).
b) Show that Np(k'p,kp) = ^=.

9.5 Quintic and Septic Multipliers and Iterations 315
c) Show that
49M72 = §(/>(>>) + ^p\y) + 196/)
1 IM2 = \ (P(x) + ^P2(x) + 196a;3)
where x : = (kk'/W)113, y :={ll'lkk')113 and
P(x):=l + 8x-8x2-x3.
d) Prove that if x ■ = \/M13 and y : = yN[z
(1 - l?>x2)y3 + (4x)y2 - (4x2)y + x(l -x2) = 0.
Hint: Use (4.6.8).
7. Combine (9.5.20) and (9.5.22) to produce a solvable update for
mn '^IM^rl2"); and so a solvable 7th-order iteration for it.
8. Ramanujan's letters also contain the modular equations of degree 5 for
KllA and K1/6. (See Section 5.5.) For K1/6 one has
(Ik)213 + (I'k'f3 + 3(ll'kk')U3 = 1
and for K1/4 one has
Ik + l'k' + 8(ll'kk')V3{(lk)V3 + (I'k')1'3} = 1.
a) Verify that, in the notation of (5.5.34),
b) Similarly
Gr,?(5) = ^y
G,7.2(5)=5
Hint: The appropriate /?th-order modular transformation for Ks
sends / =: A*(n) to k : = \*(p2n).

Chapter Ten
Other Approaches to the
Elementary Functions
Abstract. We examine some of the standard polynomial and rational
approximations to elementary functions, particularly to exp and log. We discuss
methods for reducing the complexity of calculating these functions based on
accelerating the evaluation of the approximants. While these methods are
usually less than optimal, they are of more general application than those of
Chapters 6 and 7.
10.1 CLASSICAL APPROXIMATIONS
We commence with an analysis of the standard approximations to exp on a
disk Ds := {\z\ s 8}. The notations we will require are as follows. Let Pn
denote the algebraic polynomials of degree at most n with real coefficients.
Let \\f\\A denote the supremum norm of a continuous function/on the set
A, that is,
(10.1.1) H/IL:=sup|/(*)|.
For a continuous / on an infinite compact set A C C, let
(10.1.2) E„(f,A):= rmn\\ f-p\\A
PSPn
and let
(10.1.3) /?„(/, A) :=/?„(/)= min \\f-p/q\\A.
p,q<s.rn
These quantities are, respectively, the error in best uniform polynomial and
316

10.1 Classical Approximations
317
best uniform rational approximations. The existence of the best
approximants is fairly straightforward and is left as Exercise 2.
The most commonly used polynomial approximations to exp are
undoubtedly the partial sums of the Taylor series. This is reasonable since in any
neighbourhood of zero the partial sums are asymptotically optimal. (See
also Exercise 3.)
Theorem 10.1
(a)
d1 (n + 1)!
1 +
n + 1
ion
(b) KpePB, then
\W -/WIU* (^Ti)t(i-^t)
Proof. Part (a) follows from the estimate
1
(n+1)!
1 +
1 +
+
+
n+2-V ' « + 3 (n+ 3)(rc +4)
For part (b) we observe that if
Ik2-p(z)\\D <mmy-sn(z)\
where sn is the nth partial sum to exp at zero, then by Rouche's theorem,
p(z)-s„(z) and ez-sn{z)
have the same number of zeros (counting multiplicity) in Dx. As ez -
s„(z) has a zero of order n + 1 at zero, we deduce the contradiction that
p = sn. To finish the proof we need only observe that
min |e2 - sn(z)\ > , .
|z|=i (n + 1)!
1
n + 1
O
We now turn to the Pade approximants to exp. These are rational
approximations that are a natural extension of the Taylor approximants. We
define
(10.1.4)
j;t\t + z)me-'dt
rm,«W $Z(t-z)ntme-'dt
and observe that rmn is a rational function of z with numerator of degree m
and denominator of degree n. In closed form,

318
Other Approaches to the Elementary Functions
(10.1.5) rmJz)=2
m
yi
vl (-z)'
v=0(n + m\ v u=0//n + «x v-
Theorem 10.2
(a)
(b) Iip,qEPn,then
r (z)|| < 8(^)(^)
rn^)\\Dl- (2/,)!(2/i + l)! •
p(z)\\ ^ (n\)(nl)
qiz)^ 8(2n)!(2n + l)! '
- Q(_Z)U r
(10.1.6) q„(z) := (2n)! X = I 0 - *)"<"«-'
Proof. Let
dr
"=o /2«
i>!
Then
(10.1.7) <7B(z)[ex-rBpB(z)] = J"o (f-z)rg*-'A-Jo tn(t + zye-'
= jZo(t-z)"t"e*-'dt
= z2n+lJo (u - l)"uVl_")x rfu .
Now from (10.1.6), for \z\ < l,
"(2/i)! (2w-l)! (2n-2)!
df
(10.1.8) kB(z)|an!
n! (n-l)! 2!(n-2)!
=^4-5-2½
>(2«)!(2-Ve).
Since
r1 i
(10.1.9) Jo (1 - u)"u" du =■£(/! + 1, n + 1) = tt^J
(2n + l)!

10.1 Classical Approximations 319
we have from (10.1.7) and (10.1.8),
(10.1.10) Ik"-V.MM-^(2,)^ + 1).-
Part (b) requires showing that for |z| = 1,
n\n\
(10.1.11) \e* - rB,B(z)|>
8(2n)!(2/i + l)!
which follows from the estimates (Exercise 5)
4 (2n + l)!
(10.1.12)
and
(10.1.13) \q„(z)\*(2n)lVe.
The rest of the argument is analogous to part (b) of Theorem 10.1. If
there were a rational function/?Iq satisfying (b) with the inequality strictly
reversed, then by (10.1.11) and Rouche's theorem,
plq-rnn and e2 - rnn
would both have the same number of zeros, and from (10.1.7) we would
deduce that plq — rnn has at least 2« + 1 zeros and hence is identically
zero. □
The fact that rn n is the Pade approximant is a consequence of (10.1.7),
which shows that
e2-r„,„(z) = 0(z2"+1) as z^0.
In general the (m, ri) Pade approximant to an / (analytic at zero) is the
unique rational function r = plq, where p E Pm and qE Pn, which satisfies
f(z)-p(z)/q(z)=0(zh)
where h (in nondegenerate cases h = m + n + 1) is as large as possible. For
n = 0 this defines the «th Taylor polynomial.
The following is a partial list of the standard series and continued fraction
expansions for exp and log.

320 Other Approaches to the Elementary Functions
(10.1.14) e2=X ""
(10.1.15) e2 = l +
2z
2 - z + 2zz E"_! [1 /(z2 + (2tt«)2)]
(10.1.16) e2 = l+~-^--^-^-#-#-^-
v ' 1-2+ 3- 2+ 5- 2+ 7-
(10.1.17) Mll+f) =^
\n + l_n
log(l + z)=2^^
(10.1.18) " |z|=sl, Z5*-1
log(z) = 2
z-l\ 1 /z-l\3 1 /z-1n5
+ ^ —T +^ ^-T +
,z + l/ 3 Vz + 1/ 5 \z + l,
(10.1.19) re(z)>0,
, (z + l\ Jl 1 1
log =2-+ —^ + —f +
e\z-l/ \z 3z3 5z5
(10.1.20) 1*1^1, Z5^±l
\ _ z z _£_ ij?_ ^£ 2£ —
8 ( -1 _ IT 2+ 3+ 4+ 5+ 6+ 7+ " "
(10.1.21) z^(-oo,-l]
z'-l
(10.1.22) lim —— =logz.
a io o
Comments and Exercises
Pade approximants derive their name from H. Pade, a student of Hermite,
who was one of the first to systematically study such approximations at the
end of the last century. The theory of Pade approximation may be pursued
in Baker and Graves-Morris [81]. The convergence theory for Pade
approximants is far more complicated than the analogous well-known theory for
Taylor series. (See Exercise 9.) Except in special cases, such as exp or
functions given by Stieltjes transforms [J- da{t)l{x + t)], analysis of region
or rates of convergence is only partially understood. Theorem 10.2 can be
sharpened to show that

10.1 Classical Approximations
321
n\n\
R"(e*,Dl)~ (2n)l(2n + l)\
using Pade approximants centered at z : = 1/(2« + 1). This is due to Tre-
fethen [84]. (See Exercise 10.) The discussion of the approximations of exp
follows Newman [79], as do Exercises 7 and 8. These exercises illustrate the
different rates of convergence on disks and intervals. Exercise 8 is the n = m
case of a conjecture of Meinardus, namely, that
Rn,m(ex,[-l,l])
n\m\
2n+m(n + m + l)!(n + m)\
[1 + 0(1)]
as n + m —»°°. This conjecture has been resolved recently by Braess [84].
(See also Nemeth [77].)
For further discussion of the material of this section the reader is referred
to Cheney [66] or Newman [79]. The various expansions may be found in
Abramowitz and Stegun [64].
In Section 11.3 we will use the Pade approximant to exp to derive a
precise irrationality measure for e.
1. (Lagrange interpolation formula) Given n + 1 points in the plane,
(z,., wt), i = 0,. . . , n, so that zt ^ zj for i^j, show that there exists a
unique p E Pn so that
Show that
where
P(?i) = wt » = 0,. .
n
p(z) = X wtlt(z)
/,(*)--=nz~z*
/t=0 Zi Zk
k¥-i
2. (Existence of best approximants) Prove that En and Rn are well
denned, that is, show that the min is achieved in (10.1.2) and (10.1.3).
Hint: A uniformly bounded sequence of polynomials, all of degree n
has a uniformly convergent subsequence whose limit is a polynomial of
degree at most n.
3. a) Suppose that / is entire and that *n is the «th partial sum of / at
zero. Show that

322 Other Approaches to the Elementary Functions
Hint: Show that if f(z) ='■ E™=0 aLz', then for infinitely many m,
kJ(l-e)^
a,z
<|flm|(l+e) |z| = l
Now use the arguments of Theorem 10.1.
b) Suppose / is analytic in a neighbourhood of zero. For fixed n show
that
lim -Tj-7 J,— = 1 .
«10 \\j-Sn\\Ds
Part a) illustrates that the Taylor approximants behave globally
like best polynomial approximants to entire functions on disks.
The story is different on different shaped regions. Part b) shows
that locally the Taylor approximants are always optimal.
4. Prove that (10.1.4) has the representation (10.1.5).
5. Establish the estimates (10.1.12) and (10.1.13).
6. Establish the expansions (10.1.14) to (10.1.22).
7. (Polynomial approximation to exp on [—1,1]) Let sn be the nth
partial sum of exp at zero.
a) Show that if p(z) is a polynomial of degree n, then p(z)p(l Iz) is a
polynomial of degree n in the variable z + 1 Iz,
b) If x is the real part of z, where |z| = l, then ex = ez,2ez'2 =
ez,2e1,2z. On \z\ = 1 approximate ez'2 by s„(z/2) and approximate
e1,2z by sn(l/2z) and estimate the errors.
c) Use part a) to construct polynomial approximations to exp that
satisfy
F^r-1 n^gW2 + Q(1)
En(e,[ 1,1])^ 2> + 1)! •
Note the approximation is
e'-S^iS^i z: = x + iy.
,=o v2 i=0 v2
d) Modify part c) to show that
(This is in fact asymptotically optimal.) Hint; Consider the method
with approximation centered at IIn.

10.1 Classical Approximations
323
8. (Rational approximation to exp on [-1,1]) Show that
«„(«.L M])^^ (2n)!(2/i + l)! '
Hint: Proceed as in 7). First observe that Exercise 7a) holds for
rational functions of degree n. The approximation is given by
^-^.-(5)^-(27)
where \z\ = 1 and z := x + iy. Use estimates like those in the proof of
Theorem 10.2 to prove the result.
9. The (n, 1) Pade approximant pn to /:=E"=0 a,-z' has denominator
an+1z - an and, provided «„ 5^0, satisfies
/>„-/=0(zB+2).
a) Show that if/is entire, then there is a subsequence of {/?„} that
converges to / uniformly on any given compact subset of C.
b) Show that there exists an entire function so that the full sequence
iPn) does not converge uniformly on any open set in C.
Hint: Show that the poles of the pn can be dense in C.
Exercise 9, due to Beardon and Perron, illuminates some of the problems
inherent in uniform convergence questions for Pade approximants. It was
conjectured that subsequential convergence holds for the sequence of (m, k)
Pade approximants (k fixed) and for the sequence of (m, m) Pade
approximants. Much of the conjecture concerning convergence along rows (k fixed)
was recently settled by Buslaev, Gonchar, and Suetin [84]. They show, for
example, that if /is entire, some subsequence of the (m, k) Pade
approximants (k fixed) converges uniformly to / on compact subsets of C. These
conjectures, due variously to Baker, Gammel, Graves-Morris, Wills, and
others are discussed in Baker and Graves-Morris [81]. A more complete
convergence theory is available if one is prepared to settle for weaker types
of convergence, for example, convergence in measure.
10. (More on the Pade approximants to exp) Let
pm.n(*y-=j~at+z)me-'dt
qm,n^Y-=l (t-z)ntme-'dt.

324 Other Approaches to the Elementary Functions
a) Show, as in the proof of Theorem 10.2, that
qmA*V-pmM =zm+n+1 f0 (« ~ i)W1^2 du.
b) Show that, as m + n—»°°,
„ /".A,,2 „ t-\ _ \*-) m-n- nzl(m + n) m+n + 1 ri . /1 s -i
<lm,n(Z)e -Pm,n(Z)~ (m +„ + !)! e Z [1+°(!)]-
Hint: Observe that (u - l)"um is essentially a "spike" at u:=
ml(m + n) and that
fo(u-iyu"
(-l)"m\n\
du~ — —- —
(m + n + 1)! '
c) Show that
_ •£ m\(n + m — k)\ k
Pm>"{Z)'" to (m~k)\k\ Z
a (Z):=f "Kn + m-k)\ kk
qmA) to (n-k)\k\ ( ]
Recall that J0" t"e ' dt=n\
d) Show that
„ . Pm,n , _ "m,n
Pm>n -=-^- and Qm,n-=^-
are polynomials with integer coefficients of degree m and n,
respectively.
e) Show that, as n, m—><*>,
Pm,n(z) = (n + m)\e[m,("+m)]>[l + o(l)]
and
^,,,(^) = ^ + ^)^-^^^1 + 0(1)].
The convergence is uniform on compact subsets.
Hint: Examine the coefficients of pm n(z)/(n + m)\
f) Show that, as m, «—»°°,
g2 _ Pjnjt) = ("J?"7"1"' „ ^0-^^ +1[! + o(1)] .
qm,n(z) (m + n)\(m + n + 1)\
The convergence is uniform on compact subsets that avoid any

10.1 Classical Approximations 325
zeros of the denominator sequence. Observe that, by e), only
finitely many of these zeros lie in any compact set. Further details
may be found in Trefethen [84] or Braess [84].
g) Show that
Hint: Consider the Pade approximant centered at 2p2/(2« + l).
That is, replace z by z - 1p l(2n + 1) in parts e) and f). This gives
the upper bound. Use Rouche's theorem, as in the proof of
Theorem 10.2, to derive the lower estimate.
11. {On the main diagonal Pade approximants to log)
a) Suppose Sn and Tn are polynomials of degree n and suppose that
Tn(x)logx-Sn(x) = 0(l-x)2n+1.
Show that, if Tn(x) := t0 + txx + • • • + tnx", then
[T„(x) log x]( = B+1 Z ———
x ;=o n
"J
and hence
x"+1[r,(x)io8x]<"^_j,(-i)V_
(-ir«! ;e0 /„• ( L)tAx l)
b) Show that
r.w-.t (;)(">
if we normalize so that t„'-=l. [This is the denominator of the
(n, n) Pade approximant to log at the point 1.] Observe that Tn is
of degree n and has integral coefficients.
c) Show (with the above normalization, t„:=\) that dn ■ Sn has
integer coefficients, where dn := LCM(1, ...,«).
d) Let
^) := log *-|r§-
Show that

326 Other Approaches to the Elementary Functions
Hint: Observe that £„ := 0(1 ~xfn+\ Now differentiate to get
1 C T — T ?
Thus
xT\{x)tn{x) = T\- x(SnTn - t„Sn) = 0(1 - xf .
Since the middle term is a polynomial of degree 2« with lead
coefficient 1, we must have
xT2n(x)tn(x) = (l-xf".
Show that
-Efc_„ [fc!fc!/[(fc + n + l)!()k - n)!]](l -*)B+fc+1
s„W =
^0(1)^
e) Show that
■ ^=(2;)
and that for x a 0,
f) Show that, for x £ (1 - S, 1 + S),
where ca > 0 and ds > 0 depend only on 8, 0 < S < 1.
10.2 REDUCED COMPLEXITY METHODS
We are primarily concerned with methods that accelerate the evaluation of
the elementary function by reducing the complexity of evaluating one of the
standard approximants. Most of the approximants listed in the previous
section evaluated by usual methods (such as, Horner's rule) provide between
0(n) and 0(n log n) digits for n arithmetic operations. The slight difference
comes from the more rapid convergence of the Taylor polynomials for exp

10.2 Reduced Complexity Methods
327
than those for log. (See Exercise 1.) We proceed to examine three methods
of complexity reduction. While none of these methods is as fast as those of
Chapter 7, they all have their own particular advantages. The second
method, based on the fast Fourier transform, for example, applies to most
of the special functions.
10.2.1 Acceleration Based on Functional Equations
The exponential satisfies the functional equation
(10.2.1)
f(2z) = [f(z)]2
This allows us to reduce the calculation of the exponential to a small region
about the origin and then to approximate in that region using a Taylor or a
Pade approximant. From estimates like those of Section 10.1 and (10.2.1)
we have
(10.2.2)
•""['■Wl \*nt2* 1'
and
(10.2.3)
e2-
/zYTi i mm
'"■AW J 1 16"2 (2n)!(2n)!
z si
\z =sl
where sn and r„ „ are, respectively, the nth Taylor polynomial and the (n, n)
Fade approximant to exp at zero. Both of these above estimates provide
Oop(Vn) and 0B(VnM(n))
methods for the evaluation of exp.
The functional relation for log is
(10.2.4) , f(z2)=2f(z)
Combined with (10.1.19), this leads to
- 1 (z2-"-l
(10.2.5) 2"log(z2"") = 2"+1 X
k.02k + l \z' " + i
2/fc+l
Truncation after n terms leads to an approximation on \z -1| < \ that has
error 0(4 ~" ) and yields an algorithm for log with complexity
Oap(Vn) and 0B(VnM(n)).
These are good intermediate range estimates of exp and log. For

328
Other Approaches to the Elementary Functions
n '■= 100, (10.2.3) provides in excess of 12,000-digit accuracy at the expense
of roughly 300 full precision multiplications. (300 terms of the Taylor series
provides roughly 600 digits.)
Variations of the above methods can be used to construct 0op(VH)
algorithms for the circular functions and their inverses. (See Exercise 2.)
Since any elliptic function (Section 1.7) satisfies an algebraic "half-angle"
formula, we can, as above, construct Oop(nll2+s) algorithms for elliptic
functions provided we have expansions available at the origin.
(Unfortunately, convenient closed forms of the Taylor series for sn, for example, are
not available.)
10.2.2 Acceleration Based on the FFT
It is possible, based on FFT methods, to evaluate a polynomial or rational
function of degree n at n + 1 distinct points in Oop(«(log n)2). (See Exercise
3 of Section 6.2.) Our aim is to use this observation to accelerate the
evaluation of partial sums or related approximations. We illustrate with log.
Start with
-log (1 - z) = z + \z2 + ^z3 + • • •
(10.2.6)
and consider
(10.2.7)
We can write
(10.2.8)
zk
J„2(z):=X -r
s„2(z) = 2 zknpn(kn)
/t=0
- z
(10.2.9) Pn(y) = l1
i=ii + y
2\
Now, for fixed z, evaluate p„(kn) at k = 0,1,. . . , n - 1 in Oop(«(log n) )
(Exercise 4) and evaluate sn2 in a further 0{n) operation. Since for \z\ ^ \,
sn2(z) provides D,(n2) digits of log z, we have constructed an algorithm for
log which is Oop(nu (logrc)2).
With care, this idea can be extended to produce
(10.2.10) 0op(rc,/2(logrc)2) and 0B(nv2(logn)2M(n))
algorithms for a variety of nonelementary transcendental functions. Almost
any function with regular Taylor coefficients is susceptible to such analysis.

10.2 Reduced Complexity Methods
329
Algorithms for the gamma function and the hypergeometric functions, of
complexity given by (10.2.10), are presented in Exercises 6 and 7.
We can combine the methods of Section 10.2.1 with the above to
construct an
(10.2.11) Oop(rc,/3(logrc)2) and 0B(nu\logn)2M(n))
algorithm for log. This merely requires truncating (10.2.5), say, after n
2
2\
terms and evaluating the truncation using the FFT method in 0 («(log n) )
3\
steps. Note that this approximation provides 0(n ) digits of log.
Likewise algorithms can be constructed for exp and the trigonometric
functions with complexity given by (10.2.11).
The relative difficulty of implementing these algorithms renders them
largely of theoretical interest.
10.2.3 Acceleration Based on Binary Splitting
If we wish to evaluate the constant e by summing the Taylor series at 1, then
we can and should take advantage of the reduced length of each individual
operation. With this in mind consider
(10.2.12) p(a,b):=b\/a\
and
(10.2.13) c(a, b) : = ,(*+£, b)c(a, £**),<*-)« + c(« + * , ,)
where
By construction,
c(a, a + 1) := (a + l)x.
(10 2 14) C(°'2) = X —
With x-=l, (10.2.14) approximates e with an error of less than e/(2")!, and
hence provides fi(«2") digits of e. We can use (10.2.13) to recursively
evaluate e with bit complexity (9B((log n)M(n)).
Brent [76c] shows how to modify this to provide an (9B((log n)2M(n))
algorithm for exp. This is outlined in Exercise 8.
Modifications and variations lead to OB((log n)2M(n)) algorithms for all
the elementary functions. Particular values of various of the nonelementary
special functions are also amenable to this analysis. So, for example, is

330
Other Approaches to the Elementary Functions
Euler's constant. [See Exercises 10 and 11.] Exercise 9 outlines an
0B((logn)2M(n)) algorithm for it based on recursive evaluation of arctan.
It is perhaps worth underlining the observation that these acceleration
methods apply only to the bit complexity. The operational complexity is not
reduced. Nonetheless, the algorithm for the number e implicit in (10.2.12),
(10.2.13), and (10.2.14) is asymptotically as fast as any known and has the
virtue of being implementable using only integer addition and
multiplication, except for a single final division.
Walz [Pr] has studied classes of asymptotic methods using extrapolation
and elimination techniques. These often outperform AGM-based methods
in the "microcomputer range" (less than 20 digits). Interestingly, for the
complete elliptic integral of the first kind he finds the AGM to be always
superior. For incomplete integrals this is not always so.
Comments and Exercises
Much of the material of this section is due to Brent [76c]. In particular,
Exercises 3 and 8 follow Brent closely.
1. a) Show that truncating the series in (10.1.14) and using Horner's
rule leads to an Oop(n/\ogn) algorithm for exp.
b) Show that (10.1.18), (10.1.19), and (10.1.20) all lead to Oop(n)
algorithms for log using usual methods for evaluating the
polynomials or rational functions in question.
c) Show, for n : = 2" and 8 := 1/2", that (10.1.17) and (10.1.22) lead
to Oop(n) algorithms for exp and log.
d) Analyze the complexity of (10.1.15), (10.1.16), and (10.1.21).
Assume in all parts that the method in question is used on a
compact region bounded away from the boundary of the domain of
convergence.
2. a) Use the functional relation
(10.2.15) 2 arctan ( , , Z J = arctan z
Wz2 + 1 + 17
and the expansion
v (—1) z , ,
arctanz = 2j —-,, , .— -2 <1
^o 2/: + 1
to construct on Oop(n ) algorithm for arctan on \z\ ^17 < 1.
b) Use the functional relation
' z
(10.2.16)
cos,2
2 cos z + 1
to construct an 00p(n1/2) algorithm for cos on Dt.

10.2 Reduced Complexity Methods 331
c) Show, in general, that if
CO
f(z)=lLanzn \z\<l + S
n = 0
and f(z/2) is an algebraic function of f(z), then one can construct
an algorithm of complexity
Oop(nin) and 0B(nU2M(n))
on Dr
Instead of approximating ez/2" by sn(zl2") in (10.2.2), approximate
ezn" - 1 by sn(z/2") - 1. Then repeatedly use the relation
(1 + e)2 - 1 = 2e + e2
to evaluate ez - 1. Show that this avoids requiring 0(nin) guard digits.
This modification and its obvious analogue for the Pade approxim-
ant allow the calculation of exp without loss of significant digits beyond
the 0(log ri) loss inherent in performing n112 operations.
Show that p„(y) °f (10.2.9) can be written as
, x wn(y)
p»{y) = ^Jy)
where w„ and vn are polynomials of degree n. Use FFT methods to
show that the coefficients of w„ and vn can all be calculated in
Oop(«(log rif) and hence, that pn{y) can be evaluated at n points
in Oop(«(log rif). (See Exercise lb) of Section 6.2.)
a) Construct an Oop(«1/2(log rif) algorithm for exp by writing
X ^ = ^(0) + --- + ^-1)
where
x x^_
(Nm)\ Lx Nm + 1 (Nm + l)(Nm + 2) + " '
„A/-1
xNm
tN(m):--
+
(Nm + !)••• (N(m + 1)-1)
and evaluating tN{m) for m = 0,. . . , N— 1. Care must be taken
to compute (Nk)\ in the requisite time,
b) Show how part a) can be used to construct an Oop(«1/3(log«)2)
algorithm for exp.

332 Other Approaches to the Elementary Functions
c) Construct an Oop(«1/2(log rif) algorithm for the Bess el function of
order zero,
°° (--\\k72k
<ia2-17) J^:'■s.Wsr-
6. (An 0op(nll2(log n)2) algorithm for the gamma function)
a) From the definition
r(s):= j~ e-T1 dt
show, by breaking the integral at N and expanding, that
(10.2.18) rw^ifc^-^ + JV,---,,.
b) Show for s G [1,2] that
6N
™-«OT&
s2Ne~N
c) Use b) and FFT methods to construct an algorithm for the gamma
function of complexity
Oop(nu\lognf) and 0B(nu\logn)2M(n)).
7. (0op(n1/2(/ogn)2) algorithms for hypergeometric functions) We now
consider a (general) hypergeometric function to be a function
CO
(10.2.19) f(z):=l+% anz"
n-l
where an/an_1'= R(n) for some fixed rational function R. In this
problem R is assumed to have rational (or precomputed) coefficients
and / is assumed to have a nonzero radius of convergence.
a) Show that the Gaussian hypergeometric series F(a,b;c;z) of
(1.3.5) is hypergeometric by the above definition. Show that,
provided a, b, and c, c ¥=■ 0, -1, -2, ..., are rational, F(a, b; c; z)
satisfies the additional assumptions.
b) Show that sin(Vz), e\ log(l-z), EItt, and Klir are all
hypergeometric functions, up to a rational normalization.
c) For fixed n let

10.2 Reduced Complexity Methods 333
n2-l
Sn*-i(z)-= 2 akzk
k=0
/fc-1
T(k):=UR(i) /2(0) := 1
/ = 0
and
Q(fc) :=/?(£) + /?(£)/?(£ +l)z + ---
+ R(k)R(k + l)---R(k+n- l)zn~l .
Observe that
k
ak = U R(k)
/=0
and show that
5„2_! = 6(0) + z"Q(n)T(n) + z2nQ(2n)T(2n) +■■■
+ zn(n~l)Q(n(n - l))T(n(n - 1)).
d) Show that
T(n), 7X2/1),..., T(n(n - 1))
can all be evaluated in Oop(«(log«)2).
Hint: Consider the rational function of y
V(y):=YlR(i + y)
i-0
and observe that
T(kn) = V(0) ■ V(n) ■ • • V((k ~l)n).
Now first compute the coefficients of V by recursively breaking the
problem in half and using a fast multiplication. Then calculate V at
the points K(0), V(n), V(2n),..., V((n - l)n). (See Exercises 1
and 3 of Section 6.2.)
e) Show that
2(0), Q(n),..., Q((n - l)n)
can all be evaluated in Oop(«(log rif).

334 Other Approaches to the Elementary Functions
Hint: Note that Q is a rational function of k of degree bounded by
n ■ degree(/?). Now show that the coefficients of Q as functions of
k can be calculated in Oop(«(log nf) by proceeding recursively.
To do this consider
R(k) + [R(k)R(k + l)]z + ■ ■ ■
+ [R(k)R(k + l)---R(k + 2n- l)]z2n~l
= {R(k) + ■■■ + [R(k)R(k + 1) • • • R(k + n - l)]zn~1}
+
z" I! R(i)
{R(k + «) + ••• + [R(k + n)
xR(k + n + l)---R{k + 2n-l)\zn~1}
and use a fast polynomial multiplication to recombine the pieces.
Finally evaluate 2(0), Q(n),..., Q((n - l)n) as before,
f) Use the preceding parts to construct
Oop(rc1/2(logrc)2) and 0B(nu\logn)2M(n))
algorithms for any hypergeometric function.
(An 0B((log n)2M(n)) algorithm for exp)
a) Show that (10.2.13) can be used to recursively evaluate e with bit
complexity 0B((\ogn)M(n)).
b) Use (10.2.13) for rational p/q, where p2<q<2N, to compute
eplq. Show that the bit complexity is as in Exercise 8a).
c) Suppose x G [0,1) is a binary 2m-digit number. Show that x can be
written as
V Pk
X = Zj
k-0 <ik
where qk- = 22 and 0<pk<22
d) Write
fl ePk'q"
k = 0
with pk and qk as in c). Show that this gives an Os((log n)2M(n))
algorithm for exp on [0,1).
9. (An 0B((log n)2M(n)) algorithm for it)
a) Consider the expansion

10.2 Reduced Complexity Methods 335
and the recursion
c(a, a + 1) := — x2
c(a,b):=P(^,b)c(a/-^y-xy^
I a + b\ (a + b ,\
+ p{a,-2-)c{—,b)
where, for a < b,
p(a, b) := (2a + 1) • (2a + 3) • • • (2b - 1).
Show that
c(0,2")
*a"+%(l,2")
calculates the (2"+1 -l)th partial sum of arctan(l/x).
b) Show that for a fixed integer x > 1 the above recursion computes
arctan(l/x) in
0B((\o%nfM(n)).
This gives an Os((log n)2M(n)) algorithm for it from
tt = 16 arctan ( 5) - 4 arctan ( 535)
or any similar arctan formula.
10. a) Let / be a hypergeometric function defined as in Exercise 7 (with
the same additional assumptions). Show, for fixed plq rational
inside the region of convergence of the expansion (10.2.19), that
f(plq) can be calculated with bit complexity
0B((lognfM(n)).
b) Show, for fixed rational plq, that T(plq) can be calculated with
bit complexity
0B((\ognfM(n)).

Other Approaches to the Elementary Functions
(On the complexity of Euler's constant) Euler's constant or the
Euler-Mascheroni constant y is denned by
, 1 1
1
H log m
m 6
y '■= lim
m—>«
(10.2.20)
It is related to the gamma function by the formula
= 0.5772156649
(10.2.21) ^
a) Use the recursion
= zeyz
n(i+^>-
L„=1\ nl
r<
> + 1) = zY(z)
to prove that if T has an expansion of form (10.2.21), then y is
given by (10.2.20).
b) The exponential integral E1 is denned by
(10.2.22)
Show that
/•CO —t
dt
-E1(x) = y + \ogx+*Z )-½^. X>Q-
(10.2.23)
c) Use (10.2.23) to construct an
0B((\ognfM(n))
algorithm for y by choosing x roughly of size 6«.
This method, suggested by Sweeney [63], is a reasonably efficient
method for computing y. Brent and McMillan [80] present a
number of algorithms for this computation. They calculate over
29,000 partial quotients of the continued fraction for y. As a
consequence they show that if y is rational the denominator of y
exceeds 1015'000

Chapter Eleven
Pi
i
Abstract. The first section of the chapter deals with the history of the
calculation of ir and related matters, while the second section deals with its
transcendence. The third section looks at irrationality measures and includes a
proof of the irrationality of 1(3). This chapter is largely self-contained and
indeed contains considerable related number theory, especially in the
exercises.
11.1 ON THE HISTORY OF THE CALCULATION OF it
The history of ir presumably begins with man's first attempts at estimating
the perimeter or area of a circle of given radius and as such starts at the
dawn of recorded history. The Egyptian Rhind (or Ahmes) Papyrus which
dates from approximately 2000 B.C., gives a value of (16/9)2 = 3.1604.. . for
ir. Various other early Babylonian and Egyptian estimates include 3, 3g,
and 3¾. Implicit in the Bible (1 Kings 7: 23) is a value 3: "And he made a
molten sea, ten cubits from the one brim to the'other; it was round all
about. .. and a line of thirty cubits did compass it round about."
Mathematical interest in ir comes into sharp focus in the classical Greek
period. The Greeks investigated the problem of "squaring the circle." This
question and its final resolution over two millenia later will be pursued in
the next section. Currently we wish to review the primary Western
developments in the calculation of ir.
Archimedes of Syracuse (287-212 b.c.) provided the first major landmark
in the quest for digits of ir. By considering inscribed and circumscribed
polygons of 96 sides, Archimedes gave the estimate
n 10 n 1
3 — <ir<3 - .
71 7
337

338
Pi
A salient feature of Archimedes' method is that it can, in principle, be used
to provide any number of digits of it.
If an denotes the length of a circumscribed regular 6-2"-gon and bn
denotes the length of an inscribed regular 6 ■ 2"-gon about a circle of radius
1/2, then
(11.1.1) aB+1 =
2a„b
n n
(11.1.2) bn+1 = V^Fn.
This two-term iteration, starting with a0 := 2V3 and b0 •'= 3, can be used to
calculate it. (See also Section 8.4.) The fourth iteration yields a4 = 3.1427. . .
and bA = 3.1410. . . and corresponds to estimating it using polygons with 96
sides.
If we observe that
(1M-3> ••"-*■"- (<.., /C)fe^.)(a--tJ
we again see that the error is decreased by a factor of approximately 4 with
each iteration. Variations of this modern formulation of Archimedes'
method provided the basis for virtually all extended precision calculations of
ir for the next 1800 years, culminating with Ludolph van Ceulen (1540-
1610) who correctly computed 34 digits. The limitations of this method stem
from the relatively slow convergence and from the need to extract square
roots. (See Exercise 1.)
Frangois Vieta (1540-1603) gave the first infinite expansion
which he derived by considering a limit of areas of inscribed 2"-gons. (See
Exercise 2.) John Wallis (1616-1703) through a complicated calculation
demanding prodigious numerical insight derived the infinite product
expansion
(1115) £ = 2-2.4.4-6-6.8-8-
^ ; 2 1-3-3-5.5.7-7.9..- '
This appears in his Arithmetica Infinitorum-oi 1655. A few years later Lord
Brouncker (1620-1684), the first president of the Royal Society, recast this
as the continued fraction.

11.1 On the History of the Calculation of rr
339
4
IT =
i + —L
9
(11.1.6) 2+
25
2+
49
2 +
2 +
The Scottish mathematician James Gregory (1638-1675) in 1671 provided
the underlying method for the next era in the history of the calculation of it.
He showed that
/■.I 1 r,x f* dx x3 xs x1
(11.1.7) arctan* = Jo —? = X-J + J-J + ...
and hence, on setting x'-=l, that
(11.1.8) ^ = l-| + |-i + ...
a formula independently discovered in 1674 by Leibniz (1646-1716). By the
beginning of the eighteenth century Abraham Sharp under the direction of
the English astronomer and mathematician E. Halley had obtained 71
correct digits of it using Gregory's series (11.1.7) with x' = Vl/3, namely,
7T = J_ / _J_ _1 1_
6~V3V 3-3 + 32-5 33-7
It is the techniques of calculus that so expanded the scope for calculating,
and it is perhaps not surprising that Isaac Newton (1642-1727) himself
calculated it to 15 digits sometime in 1665-66. He used the series
3V3 /11 1 1 \
(11.1J0) T=„+24(________...)
which is essentially an arcsin expansion. (See Exercise 4.) Newton was later
to write: "I am ashamed to tell you to how many figures I carried these
computations, having no other business at the time." John Machin (1680-
1752) derived the formula which bears his name:
(11.1.11) ^ = 4 arctan (|) - arctan (^) .
Coupled with Gregory's series for arctan this provides a very attractive
method for calculating it since the first term is well suited to decimal
arithmetic and the second term converges very rapidly. Machin calculated
100 digits this way in 1706. In the same year William Jones published his A
(11-1-9) Z = 77?(l-^ + ^-7-^ +

340
Pi
New Introduction to the Mathematics, where he denoted the ratio of the
circumference to the diameter by the Greek letter ir, presumably for the
first letter of periphery. It was, however, Leonard Euler (1707-1783) who
popularized the use of the symbol. Euler derived numerous series and
products for ir and ir2. Among the best-known are
2
(11.1.12) ^-=2^
and
6 t^\ n
(n-1.13) ib"=^^
=1 n
The explicit summation of (11.1.12) had eluded Liebniz and also the
Bernoulli brothers, Jacques and Jean. The method by which Euler derived
his evaluations of E™=1 \ln2k is outlined in Exercise 7. This is to be found in
Euler's Introducio in Analysin Infinitorum of 1748. The Machin-like formula
(11.1.14) ir = 20 arctan (-J + 8 arctan ( —J
coupled with the expansion
V / 2 2-4 , \
(11.1.15) arctanx = - (1 + x y + j^ y +•••)
where y •'= x2/(l + x2), allowed Euler to compute 20 digits of ir in under an
hour.
The next 200 years saw little change in the methods employed to calculate
ir. In 1844 Johann. Dase (1824-1861), a calculating prodogy, used the
formula
(11.1.16) — = arctan (-) + arctan (-J + arctan (-)
to produce 205 digits of ir. (Dase's arithmetical abilities were awesome—he
could multiply 100-digit numbers together in his head, a feat which took him
roughly 8 hours.)
The zenith (or nadir depending on your perspective) in premachine
calculations was achieved by William Shanks (1812-1882), who published
607 purported digits of ir, of which 527 were correct. Later Shanks
published an extension to 707 digits. This was also incorrect after the 527th
digit. These calculations took Shanks years and were performed in an
entirely straightforward fashion using no tricks or shortcuts. (See W. Shanks
[1853].) The mistakes went unnoticed until 1945, when D. F. Ferguson, in
one of the final hand calculations, produced 530 digits. Ferguson produced
808 digits in 1947, using a desk calculator and the formula

11.1 On the History of the Calculation of rr
341
(11.1.17) I = 3 arctan (~) + arctan (~) + arctan (-^) .
Thus dawns the computer age. In June 1949 ENIAC (Electronic
Numerical Integrator and Computer) was used to evaluate 2037 digits of it using
Machin's formula and 70 hours elapsed time. An analysis of the distribution
of the digits was carried out by Metropolis, Reitwiesner, and von Neumann.
By 1958, Genuys had computed 10,000 digits on an IBM 704 in 100 minutes,
once again using Machin's formula. Felton had performed a 10,000-digit
calculation in 1957; however, due to machine error it was only correct to
7480 digits. In 1961 D. Shanks and Wrench [62] used the identity
(11.1.18) it = 24 arctan ( -J + 8 arctan ( — ) + 4 arctan [^r
and under 9 hours on an IBM 7090 to produce 100,000 digits of it. This was
checked using the formula
(11.1.19) it = 48 arctan (—j + 32 arctan (—j - 20 arctan (^) •
The million-digit mark was set by Guilloud and Bouyer in 1973 on a CDC
7600. The calculation, which took just under a day, used (11.1.19) with
(11.1.18) as a check.
. Kanada, Tamura, Yoshino, and Ushiro [Pr] calculated in excess of 16
million digits using an AGM based algorithm, Algorithm 2.2, and checked
10 million digits using (11.1.19). The 16 million-digit calculation took under
30 hours on a HITAC M-280H and used an FFT-based fast multiplication.
At the end of 1985 the record belonged to W. Gosper. He calculated 17
million terms of the continued fraction expansion for it and so in excess of
this number of decimal digits—after a radix conversion from a binary
computation. His method is based on a very careful evaluation of Ramanu-
jan's series (5.5.23) on a Symbolics 3670. (A remarkable feat considering
the size of the machine.) As is surprisingly often the case with these large
scale calculations, Gosper uncovered subtle design flaws which had not
surfaced in smaller calculations.
In January 1986, D. H. Bailey [88] computed 29,360,000 decimal digits of
tt on the CRAY-2 at the NASA Ames Research Center. This calculation
used only 12 steps of the quartic algorithm (5.4.7) with /-:=4. This results
in computing a(250), which agrees with it-1 to more than 45 million places.
The calculation took less than 28 hours and was verified with a 40-hour
computation of 25 steps of Algorithm 2.1. It is amusing to observe that the
quartic calculation requires well under 100 full precision multiplications,
divisions, and root extractions.
In July 1986, Kanada reclaimed the record with a computation of 225
decimal digits. He again used Algorithm 2.2,. verified in September using
(5.4.7) with r : = 4, but reduced the elapsed time to 5 hours and 56 minutes
on a S-810/20 super computer. This represented a speed-up by a factor of

342
Pi
15. His previous computation now used only 96 minutes of CPU time. Plans
were to compute 227 decimal digits (over 100 million) at the end of 1987.
Nor is the end in sight. It will probably be the case than hundreds or
thousands of millions of digits will be calculated by the end of the century.
(This is now more a matter of will than anything else. Indeed, by the time of
second printing Kanada had similarly computed 201 million digits.) Apart
from observations like "the sequence 314159 appears in the digits of ir
commencing at digit 9,973,760," there is little we care to say about the
digits. They have, however, been subjected to considerable scrutiny. It is an
open question as to whether ir is normal. That is, do all sequences of
integers appear with the same frequency in the digits (are one-tenth of the
digits 7, one-hundredth of the consecutive digit pairs 23, etc.)? On the basis
of the first 30 million digits the answer appears positive. This, of course, is
no great help in deciding the normality issue. (See Wagon [85].)
In terms of utility, even far-fetched applications such as measuring the
circumference of the universe require no more digits than Ludolph van
Ceulen had available—but then utility has had little to do with this
particular story.
Comments and Exercises
This section presents only the highlights of the quest for digits. The matter
may be pursued in detail in Beckmann [77], a most useful though rather
individualistic history, and in Le Petit Archimede [80]. Schepler's chronog-
raphy [50] and Wrench's history [60] are also of interest. Details of the more
recent calculations may be found in Tamura and Kanada [Pr], where a
compendium of Machin-like identities is provided.
There is also a considerable collection of ir-related trivia. For example,
the Indiana House of Representatives attempted to legislate the value of ir
in Bill 246 of 1897. The bill, which appears to proclaim ir to be several
different incredibly inaccurate values, including 4 and 64/25 (see Beckmann
[77], and Singmaster [85]), passed the House and only floundered in the
Senate on the apparently chance intercession of C. A. Waldo, a professor at
Purdue. Keith [86] gives a 402 digit mnemonic for ir.
1. a) Show that the algorithm of (11.1.1) and (11.1.2) calculates ir by
showing that an and bn are as advertised,
b) Prove (11.1.3) and estimate how many iterations of (11.1.1) and
(11.1.2) are required to calculate 35 digits of ir. This should be
compared to Bailey's [88] calculation which uses the same
operations.
2. a) Prove, from the product expansions for sin and cos, that
(1LL20) 6= cos (0/2) cos (0/22) cos (0/23)~ ^<7r'
b) Alternatively deduce (11.1.20) in an elementary fashion by setting

11.1 On the History of the Calculation of rr 343
j)cos(|7 _V2„
ln :=COS I o ) COS \ ^2 J • • • COS ( ^
and showing that
sin 8
2" sin (0/2")-
c) Set 0:=tt/2 and use the formula cos(0/2) = V2 + \ cos 6 to
deduce Vieta's formula (11.1.4)
2 111 /1 1 11 1 /1
^2^2 + 2^2^2 + 2^2 + 2^2
1. a) Prove Wallis's formula (11.1.5) in the form
tt_2-2-4-4-6-6-8-8
2 ~ 1-3-3-5-5-7-7-9 '" '
Hint: Show that
I
■jr/2
o "" ""*" 2 • 4 • 6 • • • 2m
■ 2m , 1-3-5--- (2m - 1) it
sin xdx= —
and
I
vn 2m+i , 2-4-6---2/n
• zm t-1 j
sin xdx =
h l-3-5---(2/n + l)'
b) Establish the corresponding formula for e:
e /2\1/2/2-4\1/4/4-6-6-8N""
2 \1/ V3-3/ V5-5-7-7
c) Show that the volume of the 2«-dimensional unit sphere is ir7n!
while the (2n +1)-dimensional unit sphere has the volume
22n+l[n\l(2n + l)!]ir". Find a unified formula for these two cases.
Deduce (11.1.10) roughly as Newton did. Show that
V3
rl/4
S"32=l V^7*
and that
rl/4 /-1/4
J Vx-x2 dx = Jo V]c(Vl-Jt)
dx
_2 1_ 1
3-23 5-25 28-27

344 Pi
5. a) Deduce Machin's formula
it . . /1\ /_1
-=4 arctan ^ - j - arctan ^ 239
as follows. Let 0 '•= arctan 5. Then
2 tan 0 5
tan 20 =
1 - tan2 0 12
and
Hence
a* 120 -, 1
tan 40 = — = 1 + ^-
ir\ -1 +tan 40 1
tan 40
4 / 1 + tan 40 239
b) Show that arctan satisfies the addition formula
I x + y\
arctan x + arctan y = arctan I -r—— I xy < 1.
c) Show that
(11.1.21) arctan (-) = arctan (—■—) + arctan(-5— )
\p/ \p + q! \p2+pq + l/
and that if 1 + p2 = qr,
(11.1.22) arctan (^) + arctan (^) = arctan (I).
Formula (11.1.21) was known to Euler. Bromwich [26] attributes (11.1.22)
to Charles Dodgson (Lewis Carroll).
6. (Machin-like formulae)
a) Show, for integral a; and bj, that
(bi\ lbi\ (b,
kir = arctan — + arctan —.+••• + arctan —
\a1/ \a2/ \an
where k is an integer if and only if
(a, + ifcJCflj + ib2) •••(«„ + ib„)
has zero imaginary part.

11.1 On the History of the Calculation of rr
345
Hint: Consider (a1 + ib^- ■ -(an + ibn) = re'8, |0| < it, and use the
fact that
1 , (i + iz
arctan z = — log I —
(This gives an algorithmic check for Machin-like formulae.)
b) Show, for positive integral u, v, and k and integral m and n, that
/1\ /1\ kir
m arctan I - I + n arctan I - I = —
if and only if (1 - i)\u + i)m(v + if is real.
c) Verify
^ , /1
— = 4 arctan I -
4 \5
arctan
(±
TT /1 ,
— = arctan I - I + arctan .
V239
1
— = 2 arctan I -
arctan
j = 2 arctan ( - j + arctan ( -
(Machin, 1706)
(Euler, 1738)
(Hermann, 1706)
(Hutton, 1776).
These are, in fact, all the nontrivial solutions of b). This was a
problem of Grave's solved by St0rmer in 1897. The problem can
be reduced to finding integral solutions of 1 + x2 = 2y" or 1 + x2 =
y", n 2 3, n odd. (See Ribenboim [84].) Much related material on
Machin-like formulae occurs in Lehmer [38] and Todd [49].
7. Prove Brouncker's continued fraction by showing that
Hint: If
then
= 1 +
1
2 +
2 +
25
2 +
s '■= a0 + ax + axa2 + a1a2a3 +
s = a0 +
1 + a0
1 + flo

346
Pi
This is a nonsimple continued fraction. The convergents satisfy
a similar recursion to that given in Exercise 2 of Section 11.3.
Apply this to
so that
and
3 5 7
XXX
arctan ^ = ^-- + -^--- +
2 -,2
— X —jX
a0 = 0, a1 = x, a2 = —r~, «3 = —7"
arctan x:
1 +
x2
3-x2 +
9x2
5 - 3x2 +
7 - 5x2 •
Now set x := 1. (According to Beckmann, Brouncker merely announced
his result—the above derivation is essentially due to Euler.)
8. Consider the series
x3 xs
(11.1.23) sin x = x - — + Tj- — ■ • • .
Observe that sin x = 0 exactly when x = ± fcir. Now observe that on
setting y -=x2,
y y2 y3
(11.1.24) 1-^ + ^-^ + ---=0
exactly when y = (fcir)2, k = 1,2,3,. . .. If (11.1.24) were a polynomial,
we would know that the sum of the reciprocals of the roots of (11.1.23)
equals the negative of the coefficient of y and in general the sum of the
reciprocals of the powers would be expressible in terms of the
coefficients and Bernoulli numbers. Thus we would deduce that
tx (kirf 6 3nd £, (kirf 90 '
Use the product expansion for sin [Section 2.2, Exercise Id)] to make
the above argument of Euler's rigorous. (See also Exercise 14 of
Section 11.3.)
9. In computing ir from (11.1.19) one must evaluate arctan(^) and
arctan(^). Use (11.1.15) to observe that

11.2 On the Transcendence of it
347
arctan (h) = 18 (sfe + J^tf + 3^25? + '"')
and
arctan(^) = 57(^ +^^+ ^-^3+••••).
Thus terms of the second series are just decimal shifts of terms of the
first series. (See Ballantine [39].) How does this affect the'complexity
of calculating the two arctans?
10. Prove that the number
0.12345678910111213 • • • /1(/1 + 1) • • •
is normal. A proof may be found in Niven [56].
11.2 ON THE TRANSCENDENCE OF it
The problem of "squaring the circle" is the problem of constructing a square
of the same area as a given circle of radius 1, or alternatively given a line
segment of unit length of constructing a segment of length Vtt. The rules of
construction allow for the use of an unmarked straightedge and an
unmarked compass. A more precise definition of constructible is provided in
Exercise 1. In fact, the constructible numbers are exactly those numbers
which can be obtained from the integers by a finite sequence of rational
operations and extraction of square roots. (See Exercise 1.) Thus
constructible numbers are algebraic and the transcendentality of it shows the
impossibility of the problem.
The Greek notion of number, based on geometric construction, made
consideration of such problems more natural than they perhaps seem today.
Indeed the problem had arisen by the fifth century b.c. Anaxagoras, who
died in 428 B.C., had, according to Plutarch, considered it while in jail. His
contemporary, Hippocrates of Chios, the author of one of the first geometry
texts, also considered the question. The other classical Greek problems of
"duplicating the cube" and "trisecting the angle" also arose in this period.
The "Delian problem" of duplicating the cube (in volume), so named
because the oracle of Apollo at Delos had prescribed duplicating the cubical
altar as a means of halting the plague of 428 b.c, is equivalent to
constructing V2. (The impossibility of solving these problems is also
discussed in Exercise 1.)
By 414 b.c. attempts at constructing it had become so numerous that
Aristrophanes refers to "circle squarers" in his play "The Birds." The term
came to refer to people who attempt the impossible. However, attempting
the futile is not always a waste of time. As Boyer [68, p. 71] points out:

348
Pi
The better part of Greek mathematics, and of much later mathematical
thought, was suggested by efforts to achieve the impossible—or, failing
this, to- modify the rules. The Heroic Age failed in its immediate
objective, under the rules, but the efforts were crowned with brilliant
success in other respects.
It is hard to know whether more energy has been consumed by attempts
at circle squaring or by calculations of it. While doomed attempts to square
the circle flourish to this day, by the eighteenth century it had become
accepted in the mathematical community that the problem was probably
impossible. In 1755 the French Academy of Sciences refused to examine
any more quadratures while as early as 1668 Gregory had attempted to
prove their impossibility. The first substantial step in this direction was due
to Lambert (1728-1777), who proved it irrational in 1761. A subsequent
more rigorous proof was provided by Legendre (1752-1833). Legendre
proved the irrationality of it2 in his Elements de Geometrie of 1794 and
commented:
It is probable that the number it is not even contained among the
algebraic irrationalities . . . But it seems to be very difficult to prove
this strictly.
(See Exercise 3.) This belief was shared by Euler. Liouville established the
existence of transcendental numbers in 1840, and in 1873 Hermite proved e
transcendental. It had been proved irrational by Euler in 1737. Finally in
1882 F. Lindemann [1882] extended Hermite's proof to cover the
transcendence of it, thus laying to rest a 2300-year-old problem. This was simplified by
Weierstrass [67] in 1885, Hilbert [1893] in 1893, and many others.
Lindemann in fact established more generally that
pie"i + --- + pnea»*0
for distinct algebraic numbers alt. . . , an and nonzero algebraic numbers
Plt. .. , Pn. (See Exercise 7.) The transcendentality of it follows since
e,7r -1 = 0. This is the signal achievement of the nineteenth century with
regard to transcendental number theory. Note that Lindemann's theorem
implies the transcendence of cos a, sin a, and tan a for algebraic a^O and
also the transcendence of log p for p algebraic, p 5^0 or 1.
In 1900 Hilbert, as the seventh of his 23 problems posed at the
International Congress in Paris, asked whether a13 is transcendental for a algebraic
(a 5^0,1) and p an algebraic irrational. This was solved independently by
Gelfond and Schneider in 1934. (See Niven [56]). It is interesting to note
that Hilbert had speculated that this problem would probably resist solution
longer than the Riemann hypothesis or Fermat's last theorem. In 1966
Baker substantially generalized the Gelfond-Schneider theorem by showing

11.2 On the Transcendence of rr
349
that any nonvanishing linear combination of logarithms of algebraic
numbers with algebraic coefficients is transcendental. That is, if a;. and /3;. are
nonzero algebraic numbers, then /30 + S"_x a;log j8;.^0. (See Baker [75].)
Note that i' = e-'"72, and so transcendence of e" follows from the Gelfond-
Schneider theorem.
Comments and Exercises
For a discussion of constructibility one might consult Famous Problems in
Elementary Geometry by Klein [1897]. The treatment we give in Exercise 1
follows Clark [71]. The exercises on transcendental numbers follow Baker
[75], Hardy and Wright [60], Hua [82], and LeVeque [77], and for the most
part are simplifications of the original arguments. The particularly simple
proof of the irrationality of it2 (outlined in Exercise 3) is due to Niven [56].
The reader interested in pursuing these matters in depth is directed to Baker
[75], Mahler [67], or Lang [66a],
1. (On constructible numbers, doubling the cube, and trisecting the
angle) Constructible numbers can be defined by:
(i) The points (0, 0) and (0,1) are constructible.
(ii) Lines joining constructible numbers are constructible.
(iii) Circles with constructible centers and constructible radii (that
is, the radius is the distance between two constructible points)
are constructible.
(iv) The points of intersection of constructible lines and circles are
constructible.
It is the points of (iv) that form the constructible numbers.
a) Prove that the constructible numbers are a subset of the algebraic
numbers.
b) Show that perpendicular bisectors are constructible. See the hint
provided in the illustration. Hint:
B

350
Pi
c) Show that a circle through three given points is constructible. Se*
the hint provided in the illustration. Hint:
d) Show that the constructible numbers form a field. See the hint
provided in the illustration. Hint:
e) A number field C is constructible if C = Q(clt. . . , cn), where
c1;. . . , cn are all constructible. Show that if C is constructible,
then C has degree 2m over Q.
f) Show that real extensions of degree 2 are constructible.
g) (The impossibility of doubling the altar) Use e) to show that V2 is
not construcible and so the Delian problem is not solvable.
h) (The impossibility of trisecting the angle) Show that constructing
an angle 8 is equivalent to constructing cos 8 and show that a 60°
angle is constructible. Show that
cos 38 = 4 cos3 8 - 3 cos 8

11.2 On the Transcendence of rr
351
and hence, that cos 20° is a root of
8x3 -6*-l.
Show that the above polynomial is irreducible over Q and hence, by
part e), that a 20° angle is not constructible. Show that the
constructibility of 8, given a rational cos 38, depends only on
whether 4x3 -3x- cos 38 factors over Q. Show that a 30° angle is
constructible.
The following series of exercises is on transcendental and algebraic
numbers. Recall that a is algebraic of degree n if a is the root of an
irreducible polynomial of degree n with integer coefficients. If a satisfies no
such algebraic equation, it is transcendental.
2. (On Liouville numbers)
a) Prove that if a is algebraic of degree n, then for all e > 0 and for all
c>0.
0<
P
a
9
<
<7
has only finitely many solutions with p and q integral.
Hint: Suppose a satisfies
p(a) '■= ana" + ■ ■ ■ + a0 = 0.
Then, provided plq is not a root of p,
So by the mean value theorem, for plq E [a — 1, a + 1],
P
a
q
\p(a)-p(p/q)\ > D
sup \p(x)\ qn
xe[a-l, a + 1]
for some D, and the result follows.
A much deeper result of Roth (see Baker [75]) shows that a)
holds with qn+e replaced by q2+e.
b) Use a) to show that
a:=£w
and
/3:= 2
1
n-l 10
22"

Pi
are transcendental by showing that the partial sum approximations
would violate a).
c) Modify the construction of b) to exhibit uncountably many
transcendental numbers. (In fact, all infinite subseries of a are
transcendental, which exhibits an uncountable set of such numbers.)
d) Show that the algebraic numbers are countable and hence, that
almost all numbers are transcendental. Numbers that can be proved
transcendental because they can be approximated too rapidly by
rationals are called Liouville numbers. More precisely, a is a
Liouville number if for every m there exists a rational/?Iq, q >*1, so
that
1
0<
P
a
<7
<
9"
e) Prove that Liouville numbers are transcendental.
f) Show that the set of Liouville numbers has measure zero.
g) Show that the Liouville numbers have Baire category II. Show, in
fact, that the complement of the Liouville numbers is of Baire
category I. Actually it is the union of Q and a nowhere dense Fa.
(Recall that a set is category I if it is the countable union of
nowhere dense sets and that a set is nowhere dense if its closure has
empty interior. A set is category II if it is not category I. See
Oxtoby [80].)
{On the irrationality of e and it)
a) Prove that e is irrational.
Hint: Suppose e = qlp and consider
,>(ifca:-i
Show that this is an integer strictly between 0 and 1.
b) Show that for any positive integer n,
has the property that / and all its derivatives are integer valued at
x = 0 and x = l and that for x G (0,1),
0</(*)<^.
c) Prove that ep is irrational for integer p ¥=Q.
Outline: Suppose ep = alb with a,b EN. With / as in b), set
F(x) := p2nf{x)-p2n-Yl\x) + ■■■ +f2"\x).

11.2 On the Transcendence of rr 353
Show, by differentiating epxF(x), that
bjo P2n+1epxf(x)dx = aF(l)-bF(0)
is an integer but that the left-hand side above lies strictly between 0
and 1 for large n.
d) Deduce from c) that eq is irrational for all rational q ^ 0.
e) Show that it2 (and hence it) is irrational.
*• Outline: Suppose it2 = alb with a, b G N. Let / be as in b) and
consider
2n~2k
G(x):=b"^(-l)kf2k\x)7r
Then
— {G(x) sin (irx) - irG(x) cos (ttx)} = Tr2a"f(x) sin (ttx)
and
it J anf(x) sin (ttx) dx = G(0) + G(l).
However, G(0) and G(l) are integers while as in c) the integral on
the left is strictly between 0 and 1 for large n.
4. (On the transcendence of e) This exercise outlines Hilbert's proof of
the transcendence of e, a proof which, as LeVeque [77] puts it, is "as
elegant as it is mysterious."
a) Recall that, for k an integer,
/• CO
/fc:=J0 xke~xdx = k\
and thus if/? is polynomial with integer coefficients,
/• CO
I xkp(x)e"xdx^k\p(0)mod(k + l)\
b) Suppose e algebraic. Then there are integers at so that
«„ + a,e + «-,e2 + • ■ • + a„en =0 «„5^0.

354
Pi
ja:=j xm[(x-l)---(x-n)]m+1e-xdx
f CO /*CO /*CO
S--=a0jo +a1eji +--- + anenjn
T^aiefo+--- + ane"fo.
Observe that S+ T = 0.
c) Show that, for infinitely many m, Slm\ is a nonzero integer. To do
this, observe that if y '■= x - k, then
let
and let
r co
^1 =
fc = 0
«oj0 ymp0(y)e y dy
f CO
«*J0 ^^,()-)^^ fcal
where the /?t are particular polynomials with integer coefficients.
Now use a) to see that the first term of S is divisible by m\; that the
rest are divisible by (m + 1)!; and that for infinitely many m the first
term is not divisible by (m + 1)!.
d) Show that limm^„ \T\/m\=0. To do this let
M := max \(x - l)(x - 2) • • • (x - n)\(x + 1)
Osxsn
and show that
flJo
<yc|ajM"
whence
H- = ° —r
/n! V /n!
•0.
e) Observe that, since S + T = 0, c) and d) lead to a contradiction, and
hence e cannot be algebraic.
5. (On the transcendence of ir) The exercise outlines Baker's [75]
synthesis of Hilbert's proof of the transcendence of ir.

11.2 On the Transcendence of rr 355
a) Suppose 8t '■ = 'nr is algebraic. Let / be the lead coefficient of the
minimal polynomial for iir and let 02,, . . , Bd be the other roots of
the minimal polynomial. Since e'n = — 1,
(1 + ^)(1 + ^)---(1 + ^ = 0.
Show that expanding the above yields a nonzero integer n such that
q + e"1 + • • • + ea" = 0
where q ■ = 2d — n and each ai is a nonzero sum of some subset of
the 0,. of the form E ¢5,0,., S; = 0 or 1. Show that any elementary
symmetric function of lal,. . . , lan with integer coefficients is
integer valued. (See Exercise 6.)
b) Let
I(a):=foea~'f(t)dt
where
fity-=inptp~\t-"i)p---(t-a„y
for some prime p. Let
/:=/(^)+ ••• + /(«„).
Show, by using a) and repeated integration by parts, that for
m :=(« + l)p - 1,
m m n
/ = -^/(/)(0)-XS/(j)K).
/=0 ;-0 /t = l
c) Show that / is an integer. To do this, observe that the right-hand
term is a symmetric function of /a1;. . . , lan.
d) Show that for )¥=■ p - 1,
p!|/(/)(0)
while for j-p- 1,
/("_1)(0) = (p -1)1(-/)¾¾ • ••«„)'.
Thus for p sufficiently large,
(/?-l)!|/(p_1)(0) and p\^f(p~1\0).

356 Pi
e) Show that, for ;' ^p — 1, and p sufficiently large
/(/)K) = o.
Show that
Pi
2 2 /">(«,).
/=0 /fc-1
Hence with d)
\J\>(p-l)\
f) Show, however, that for some M independent of/?,
\J\*\XI(at)\*M".
This contradicts e) for large p and finishes the proof.
6. (On symmetric polynomials) A polynomial in n variables is symmetric
if it remains unchanged by any permutation of the variables. An
elementary symmetric polynomial ft in the variables xlt. . . , xn is defined
by
(y-x1)(y-x2)---(y-xn)=y"-fy-1+f2yn
(-1)7,,-
a) Show by induction that any symmetric polynomial in n variables
with integer coefficients can be written as a polynomial with integer
coefficients in the elementary symmetric functions.
b) Let sk : = x\ + • ■ ■ + xk„. Prove Newton's identities:
sk = (-l)k+lkfk + (-l)k+l 2(-l)'/*-A k< n
/=i
and
sk = (~l)k+1 1 (-l)'/*-A
k>n.
7. (Lindemann's theorem) This proof of Lindemann's theorem follows
Baker [75] and assumes some general familiarity with algebraic integers.
(An algebraic number whose minimal polynomial has lead coefficient 1:
observe that if a is an algebraic number and the lead coefficient of the
minimal polynomial is /, then la is an algebraic integer.)

11.2 On the Transcendence of rr
Lindemann's Theorem
If «j,.. . , an are distinct algebraic numbers and pi,. . ., fin a
algebraic numbers, then
)8^+--- + )8,^^0.
a) Suppose that
p1e°1 + --- + p„ea» = 0.
Show that one can assume that
(i) The jSj are ordinary integers,
(ii) There exist integers 0 = n0 < nx < ■ ■ ■ < ns = n so that,
««T + 1» • • • > anr+1
is a complete set of conjugates and
0v>i = -- = 0^.
b) Let / be an integer so that la1}. . . , lan and /^,.
algebraic integers. For p a large prime, let
/■(*)• = — •
For 1 < j < n, let
where
/,(«):= J V-'/KO*.
Show that
np-l „
/, = - S S &/,(/)(a*)
/=0 4 = 1
is an algebraic integer and that
(p-l)\\f\n(ak)

358 Pi
and for p sufficiently large
pH7iy)(«*). } = P-1, k = i
while
p\\f¥\ak) otherwise.
(An algebraic number a is divisible by h if a/h is an algebraic
integer.)
c) Show that |/i • • • /J is a nonzero integer divisible by (p - 1)! and
hence \JX- ■■ Jn\>(p -1)\.
d) Show that there exists C independent of p so that
\h>->Jn\ = 0{C>)
and that this contradicts c).
8. {Lengths and measures)
a) Suppose P is a polynomial with integer coefficients of length L and
degree D. Suppose a is an algebraic number with minimal
polynomial of degree d and length /. (The length is the sum of the absolute
value of the coefficients.) Show that either
P(a) = 0
or
IP, ,, [max(l,|a|)f
1
- Ld~T ■
Hint: Suppose at ■ = a has minimal polynomial
Q(x):=qdxd + --- + q0
and let a2,. . . , ad be the remaining roots of Q. Show that
qDd-P(ai)P(a2)---P(ad)
is a nonzero integer [if P(a1)^0]. Since
|P(aJ|<L-max(l,k|)D
we have d
l^\qDd\\P(ct1)\U[LmSx(l,Wk\)D]
^Z/-1 maxakD^lPK)!

11.2 On the Transcendence of rr 359
where the last inequality requires showing
d
/£kj II max(l,|aj)
k=\
and follows from part b) of this exercise,
b) Let P and Q be polynomials and let
r <-i
/x(P):=exp
fo\og\P(e2"")\dt .
Show that
i) ^e) = M*V(e)
ii) /x(P) =s length (P)
iii) /ui(x - c) = fi<(x- \c\)
iv) f,((x-c")v") = ^x-c)
v) f^(x- c) = max(l, \c\)
vi) f^(cYl (x- a,)) = |c|Il max(l, |a,|).
(fi is called the measure of P.)
The next exercise constructs a form
pn(x)en* + ■ ■ ■+p1(x)ex+p0(x) = oix^1^1^1)
where the pt are polynomials of degree ^m with integer coefficients. Such
forms can be viewed as higher dimensional Pade approximants. Setting
x'-=l in the above leads to a polynomial in e that can, in conjunction with
Exercise 8, be used to prove the transcendence of e. Setting x '■= iir, so that
e,1T = -1, leads to a proof of the transcendence of it. Such forms were
examined by Hermite. It was, however, Mahler [31] who showed how to
base the transcendence proofs on them.
9. {Another proof of the transcendence of e and it)
a) Suppose that we can find Vn so that
Vn,m := Vn := Pne»* + ■ ■ ■ + Ple* +p0 = o(xin+1^+1^)
where each pt is of degree m and pn '■ = xm + ■ • •. Then
Vf+1) = sne"x + ■■■ + s,ex = 0{xn(m+1)-x)

360 Pi
where each s, is of degree m. In particular
y0n + l) = n^Vn_^
and
r(m + \) — xmgx
1
b) Prove that, for sufficiently smooth G, if
G{x):=~\l{x-tTf{t)dt
then G(m+1)(x) =f(x). Use this to show that
\m + l rx
s \m + l rx
^ = ^1^-^-^
dt
(ni)m+1 cx f1 f2 r'"-1
c) Thus
j- x(n+1)(m+l)-1 Jo M{x)
(ml)'
where
K.m = ?lr^+1)(m+1Hl M(x)dt
fo*:=fo-fofo*>>-*i
and where
M(*> :=[a - tj ■ ■ • (l - oro2.:1^2 • • • <r+B-1
X g'l'g'i^g'i'a's* . . . g'1'2—'«*
d) Use c) to show that Vnm, defined as in a), exists and is uniquely
defined for all m and n.
e) Show that if Cs is a circle of radius 8>n, then
= m!(rc!r+1 f e" A
/.
2iri Jcs [t(t -1)---0 -«)]m+1'
7/mf: Use the residue theorem to see that Vnm is of the right form.

11.2 On the Transcendence of rr
361
Observe by expanding e'x and evaluating on circles of large radius
that
Vn,m = 0(x
(n+l)(m + l)-l
)■
f) Show that, with 8 <1,
Pk(x):
/»!(«!)"
I
e'xdt
2iri Jcs [n"=0 (t + k - J)]'
m.
Hint: Show by the residue theorem that pk is a polynomial of
degree m and that, with e),
2 Pkekx = Vn<m .
k=0
Note that
pk(x) = (nl)m+1
' d~
-dt-
m, jx
i*k
- i)m + i
The n '• = 1 case of e) and f) provides an alternate derivation of
the Pade approximant, equation (10.1.15).
g) Use f) to show that
o*
has integer coefficients [dn := LCM(1,...,«)].
h) Let n be fixed. Show from c) that for m sufficiently large,
Vn(x)^0 for x 5^0.
Note that this is trivial for real x^O.
i) Let D > e and let
From c) show that
W»,mW-=d:v„tm(x).
W„>m(x)= 2 akjxUk
■ Osksn
Qsjsm
where the akj are integers. Use c) and the fact that dn < e"(1+e) for
large n (Exercise 6 of Section 11.3) to show that for m + n large,
\Wm,„(x)\*
\x\(n+1)(m+l)-lenMDnm{n\)m+lm\
[(n + l)(m + l)-l]!

362
Pi
Show from f) with 8 -=1/n, that for m + n large,
n2(m+1)Dnm(ml)2n(m + 1)
|a*jl -
(n-l)m+1
j) Let x 5^0 and « be fixed. Then for m sufficiently large,
nm
o<|w„,m(*)|<-3
m
and
l I _- nm m
where cx and c2 are constants, independent of, n and m.
k) Show that e is transcendental.
7/wtf: Let Pm(e) = Wn m(l) for fixed large «. Show that as m^-o°,
Pm(e) is sufficiently small that, by Exercise 8, e must be
transcendental.
1) Show that ir is transcendental.
Hint: Let x'-= iir in i) so that e,1T = -1. Now proceed as in k).
Observe that W (wr) is a polynomial of degree m in ir.
11.3 IRRATIONALITY MEASURES
We examine the rate of approximation of e, ir, and log 2 by rationals. For
example, we show, for p and q integers and q sufficiently large, that
(11.3.1)
<7
>
<7
Estimates such as (11.3.1) are termed irrationality measures. The expected
rate of rational approximation is as follows: If f(x) is a positive nonincreas-
ing function, then
(11.3.2)
<
<7
has infinitely many integer solutions in p and q for almost all a exactly when
CO
(11.3.3) S /(<?) = «>.
¢=1

11.3 Irrationality Measures
Thus, with probability 1, we expect
363
(11.3.4)
a
q
<
1
q log<?
to have infinitely many solutions, while
(11.3.5)
P
a
<7
<
1
<r(iog <?r
usually has only finitely many solutions. This result is due to Khinchin [64].
(See also Exercise 2 of Section 11.2 and Exercise 1 of this section.)
It is standard to the theory of continued fractions that if
(11.3.6)
P
a
<7
<
2q2
p, gEZ
then p/q is a convergent of the simple continued fraction for a, while of any
two consecutive convergents of the continued fraction, at least one of them
satisfies (11.3.6). (See Exercise 2.) Roth's theorem states that for algebraic
a,
(11.3.7)
a
<7
<
1
2+e
e>0
has at most finitely many solutions. (See Baker [75].)
For specific transcendentals exact estimates are usually unknown, but e
can be analyzed very precisely.
Theorem 11.1
If
(11.3.8)
then
(11.3.9)
e~T
y (In - k)\
y (In - k)\
1 log log t„
~2 t>gtn
(-l)k
[1 + 0(1)].
Proof. From Exercise 10 of Section 10.1, with sn'= Pnn(l) and
^ = 6,,.,,(1) we have

364
(11.3.10)
and
(11.3.11)
The result now follows, as an application of Stirling's formula gives
21ogf„
t»
Pi
nlnl r\\ I
(2n)!(2n + 1)! L '
= ^!e-[l + 0(l)].
(11.3.12)
Theorem 11.2
If p, gEZ and
2n + l =
log log f „
[l + o(l)]. D
9
<
4^
then, for some n, plq = sjtn, where sn and f„ are as in Lemma 11.1.
Proof. The continued fraction for (e - 1)/2 is given by
(11.3.13)
e-1
= [0,1, 6,10,14,18,...] := K, «!,...].
If /?„ := (•?„ ~ t„)/2 and q„'-= tn, then one can verify that
(11.3.14) P„ = a„p„-1+pn-2 Po = Q />i = 1
(11.3.15) q„ = an(ln-i + (ln-2 4o = 1 ¢1 = 1-
This, with Theorem 11.1, shows that /?„/<?„ are the convergents for the
simple continued fraction for (e - 1)/2 and that expansion (11.3.13) holds.
(See Exercise 2.) In particular, if
<7
< —2 then
4q
e-1 p-q
2q
<
1_
8q2
and by Exercise 2i), (p - q)llq is a convergent of (11.3.13). □
Corollary 11.1
Let 0 < 8 < 1 and let v be a positive integer. Then
(11.3.16)
<7
1 + S log log g
2u q2log g

11.3 Irrationality Measures
365
has infinitely many integer solutions, while
1 - S log log q
(11.3.17)
2v q log q
has at most finitely many integer solutions.
The above corollary due to Davis [79] shows that ellv is atypical with
respect to the rate of rational approximation [compare (11.3.4) and
(11.3.5)]. For v = l, Corollary 11.1 can be deduced from Theorem 11.1
without reference to Theorem 11.2 on continued fractions. (See Exercise 4.)
For general v the proof is left as Exercise 5.
We now turn to irrationality measures for it, £(2), and £(3). The
approach follows Beuker's [79] elegant treatment of Apery's startling proof
of the irrationality of £(3).
Let
(a)
(b)
Lemma 11.1
r and s be nonnegative integers.
f1 f1 xrysdxdy _
Jo Jo 1 - xy
( n
—2 for some n in Z
U(2)
f1 f1 - xrys log xy dx dy
Jo Jo 1 - xy
■
f n
—5 for some n in Z
■2
k2(
/(3) l3 23 ••• r3
r(3)
r>5
r = s = 0
r> s
r = s>0
r = s = 0
where d,:=LCM(l,2,. . . , r)
Proof. Consider
(11-3.18) I'-\l\lA
1 rl „r+S s+5
^
dx dy-
If we expand (1 - xy) 1 and integrate term by term, we get
(11.3.19)
/=2
;r0 (/c + r + S + l)(/c + s + s + 1)

366 Pi
which on setting 8 = 0 establishes the r = s case of (a). For r > s,
1 1
/-fr1
(11.3.20)
k=o r s
1
k + s + 8+1 k+ r + 8 + 1
r- s
1
•s + l + S
+ ...+
1
r+8
which establishes the rest of (a). If we differentiate I with respect to 8 and
set 8 = 0, v/e get
(11.3.21)
Jo Jo 1-
log*y
•*y
dxdy
and part (6) follows from differentiating (11.3.19) and (11.3.20). D
Theorem 11.3
£(2) = it2/6 is irrational.
Proof. Let pn be the "shifted" Legendre polynomial
(11.3.22) p^y-hiih^1-^
and note that
is a polynomial of degree « with integer coefficients. Consider
rp<w»,w
" Jo Jo
(11.3.24)
xy
dxdy
^ ' Jo Jo (l-xy)n+1
dxdy
where the equality follows on integrating n times by parts with respect to x.
Since for 0<x, y ^1,
(11.3.25)
xy(l-y)(l-x) ^(V5-l\5
1-xy \ 2 /
[with equality for x = y = (V5 - 1)/2] we have, by (11.3.24) and Lemma
11.1(a),
(11.3.26) 0<|/J<(^-^)5
£(2).

11.3 Irrationality Measures 367
On the other hand by the same lemma applied to the first form of /„,
(11.3.27)
where
(11.3.28)
Pn£&)-f
r =Y (n\(n\(n + k)l
Pn £0\k)\kJ n\k\ '
an is an integer and dn = LCM(1,2,. . . , n). It is a simple consequence of
the prime number theorem (see Exercise 6) that, for any 17 > 1,
(11.3.29) dn = 0(ev").
Furthermore, Stirling's formula and a little calculus lead to the estimate
(11.3.30)
c,n
1 + V5
< /3„ < c3nc*
1 + V5
2 I ~<-n -a- v 2
where cx > 0, c2, c3, and c4 are constants. In fact, van der Poorten [79] gives
A,=
[(l + V5)/2]4 [(l + V5)/2]
2irV5 + 2V5 »
5n
l+O
From (11.3.26), (11.3.27), (11.3.29), and (11.3.30) we deduce that for
sufficiently large n,
(11.3.31)
0<
£(2) - ^
in
l
Jn
where yn '■= d\ /3n and an are integers, and where
Thus (11.3.31) proves the irrationality of £(2) = ir2/6. D
Theorem 11.4
For p and g integers and q sufficiently large,
(11.3.33)
and
(11.3.34)
,r2 P
p
IT- -
1
> ^11.86
1
23.72
<7

368
Pi
Proof. From (11.3.30), (11.3.31), (11.3.32), and Exercise 3 we deduce
that, given e>0, if
2
TT p
~6~q
<
1
<7
for sufficiently large q, then plq = ajyn for some n. [Here an, yn, and
8 :=0.092. . . are as in (11.3.31).] Thus we need only verify (11.3.33) for
plq = an/yn. This follows from the observation that, for small 17 > 0 and for
large n,
r /V5-1 \5"
[See (11.3.24) and (11.3.25).] In conjunction with (11.3.27) this leads to
[(V5-1)/2-tj]5"
«2> - 7
in
In
One now finishes the result by using estimates (11.3.29) and (11.3.30) to
show that
[(V5-1)/2-t,]5" 1 1
■v ~~ 3 1:
•n J n J n
3 > ..11.86
for large n and small 17.
The irrationality measure for it follows from the irrationality measure for
it since, for large q,
ir-P-
1
■ + p/q\
9
>
TT + p/q\ q*"-
86.
n
The first irrationality measure for it was due to Mahler [53] (see also
Mahler [67]), who showed that
TV ■
1
>
<7
Q
>2
This was later refined by Mignotte [74] to
TV ■
<7
>
1
<7
<?>2
and by Chudnovsky and Chudnovsky [84] to
9
> ^^65 q large
1
(which can be marginally sharpened).

11.3 Irrationality Measures 369
Theorem 11.5
£(3) is irrational.
Proof. Consider
(11-3.35) /„ := fQ fQ =^- Pn(x)Pn(y) dx dy
where, as in (11.3.22),
We observe that
-log xy _ f1 1_
1-xy Jo 1-(1-.
and rewrite (11.3.35) as
■ xy Jo 1 - (1 - xy)z
" Jo Jo Jo l-(l-*y)z y
An rc-fold integration by parts with respect to x yields
Jo Jo Jo [\-{\-xy)z\n '
which, on substituting w '■= (1 - z)l[\ - (1 - xy)z]> becomes
>W1 - vWn
dx dy dw
■ f f f [x(l~x)w(l-w)y(l -y)]n
Jo Jo Jo [1-(1- xy)w]"+1
where the last equality follows from an rc-fold integration by parts with
respect to y. For 0 < x, y, w ^ 1,
x(l -x)w(l-w)y(l-y) ,^ p*
and from (11.3.36),
(11.3.37) 0</„<2£(3)(V2-l)4".

From Lemma 11.1(6),
(11.3.38)
ft £(3)
dl
where a^, ft, and d\ are integers. Since e3(V2~ 1)4 <1 and ft grows at
most geometrically [see (11.3.42)], we deduce that there exists S' > 0 so that
(11.3.39)
0<
£(3)
K<
(P'ndl)1+S
for large n. This proves the irrationality of £(3). □
Corollary 11.2
For p and q integers and q sufficiently large,
(11.3.40)
£(3)
<7
>
1
<7
13.42 •
Proof. The proof is similar to the proof of Theorem 11.3 and 11.4. One
first calculates ft explicitly from (11.3.35) and (11.3.23),
(11.3.41)
k=22(:
k**0
2 / i i \ 2
n + k
From Stirling's formula one can show that
(11.3.42) Clnc\l + V2)4" < ft < c3rcc*(l + V2)4
for constants c^O, c2, c3, and c4. In fact,
2(1 + V2)2
(11.3.43)
ft
«3/2(2ttV2)3
(1 + V2)4
l + OI-
The remainder of the proof follows much as in Theorem 11.4, and the
details are left as Exercise 8. □
Similar considerations lead to the irrationality measures for log.
Theorem 11.6
Let p, q, and n be integers. Then
log 2
<7
>
(11.3.44)
and for any e > 0, and fixed n> Ne
n + 1 p
1
<7
q large
(11.3.45)
log
<7
>
2+e
q large .

11.3 Irrationality Measures
371
The proof of the theorem is sketched in Exercise 9 or in Alladi and
Robinson [79]. Similar results may be found in Baker [75]. The best known
irrationality estimate for log 2 is also due to Chudnovsky and Chudnovsky
[84], who show that the constant can be reduced to 4.13... in (11.3.44).
Comments and Exercises
Results on the rational approximation of e may be found in Adams [66],
Bundschuh [71], and Davis [79]. Adams shows that the number of relatively
prime integer solutions of
e-P-
< —2 and
<7
\<q<n
behaves like 3 log n/(log log n) (rather than the expected clogn that holds
for almost all numbers).
Transcendence estimates of type
(11.3.46)
|ir-j8|>
1
^cd log d
h large
where /3 is an algebraic number of degree d and height h (the height is the
modulus of the maximum coefficient of the minimal polynomial) due to
Feldman are discussed in Baker [75].
Irrationality measures are often difficult. The best known estimate for ew
is
e —
q
>
1
<7
c log log q
c a constant.
Apery's proof of the irrationality of £(3) does not obviously extend to
other values of £. It is not known whether £(2« + 1) is irrational for n > 1.
Equally, whether (log ir)/ir, e + it, Catalan's constant, and Euler's constant
are irrational is unknown.
1. Let / be a positive nonincreasing function so that
2 f{q)<™.
¢=1
Show that the set of a for which
<
/(<?)
has infinitely many solutions has (Lebesque) measure zero. (The other
half of Khinchin's theorem is more delicate.)

Pi
Hint: Fix N. For each q > N consider intervals of radius f(q)Iq around
the q points 0/q, llq,. . . ,(q-l)lq. The measure of the union of all
these intervals is bounded by
(On simple continued fractions) For integral an a0 > 0, at > 0, i ¥=■ 0,
let
1
[a0, a1,...,an\:=a0+ -
a, H
•i
a2 +
1
«3+ •
and let
1
+ —
a„
[a0, «!,...] := lim[a0, «!,...,«„].
Unless otherwise specified, we assume the continued fraction is
infinite. The nth convergent is defined by
-* := [a0,fl„...,flB].
Hn
a) Show that the convergents satisfy
(11.3.47) Pn = a»Pn-l+P»-2 pt = 0^+1 p0 = «0
(11.3.48) <?" = an<ln-x + <7«-2 qx = «i <7o = 1
(11.3.49) P„q„-i-p„-iq„ = (-l)n-1
(11.3.50) P„9„-2-P»-29» = (-l)Bfl„
/"11 a CI \ ^2" > Pln-2 Pln + 1 < Pln-1
lln 1ln-2 1ln + l 1ln-l
Deduce that pn and qn are relatively prime. Note that {/?„} and
{qn} are increasing sequences.
b) Show that continued fractions are well defined, that is, show that
the limit in the definition always exists.
c) Rational numbers have two representations since
[fl0. •••.«»] = [«<>.•--.0,.-1 + 1] fl« = 1
and
[a0,. . . , a„] = [a0,..., a„_„ a„ ~ 1,1] a„*l.
Show that a number is rational if and only if it has a finite simple
continued fraction. [See e).]

11.3 Irrationality Measures 373
d) Let a'n '■= [an, an+1,. . .] where a := [a0, a1}. . .]. Show that
«-£ = til
<ln 1n(a'n + l<ln+<ln-l)
Thus {p2n/q2n} f a and {p2„+1/?2n + 1} I a.
e) One constructs the simple continued fraction to a as follows
(L J := integer part). Let
«o:= LaJ and ao:= a
and proceed inductively to let
a"+l := a - [a J 3nd a"+1 :== ^+1-1
unless a„ - \_an\ = 0, in which case the algorithm terminates.
Show that
a = [a0,at,. . .]
and if the continued fraction is infinite, then the representation is
unique.
f) Show from d) that
Pn
In
<
1
1
< —
Inln+l q„
g) (Best approximation property) Show that if 0 < q < qn and if
plq^P„lq„, then
Pn
In
- fV
<-.
P
q
- fV
Hint: Prove \pn ~ qna\ < \p - qa\. Show first that one may
assume q„-1<q<q„. Now write q-= uqn +vq„_t, p:=upn +
vPn-u and Pn~ «»« = u(Pn~ 9»«) + "(P„-i ~ ?„-!«)• Show
that u and u are nonzero integers and that u(pn- qna) and
v(Pn-i ~ qn-ia) nave the same sign,
h) Show that if for z>l,
pz + r
qz + s
p, q,r,s<
and
q>s>0 ps — qr—±l

374
Pi
then rls and p/q are consecutive convergents of the continued
fraction for a.
Hint: Write p/q as a continued fraction and show that if
f-"»-
then
a = [a0;. . . ,an,z]
and
z = [an+1,an+2,...].
[There are two representations for p/q [part c)]. Choose the one
for which n satisfies ps — qr = (-1)"- •]
i) Show that if
<7
<
2q2
then p/q is a convergent of a.
Hint; Suppose
P
a
2
<7
0<8<
and
Write
<7
= [a0, au...,an\.
PnZ+Pn-
q»z + ?»-
where pilqi is the /th convergent to /?/^. Show that z > 1 and
apply h).
j) Of two consecutive convergents at least one satisfies
In
<
Hint; By a), if the above fails, then
qn<in
+i
Pn-i
Qn-i
Pn
In
-
Pn
(In
+
Pn + l
(ln +
+
2<ln ' 2^ + I
Hurwitz has shown that of any three consecutive convergents at
least one satisfies

11.3 Irrationality Measures
375
<7„
<
1
V5<?2'
This cannot be sharpened since, for example,
V5-1
1
(Vl-S)q2
8>0
for q sufficiently large. (See, for example, Hua [82].)
k) (Fibonacci numbers) Let
Show that {Fn+1/Fn} are the convergents to (l + V5)/2. Show
that
1 + V5
= [1,1,1,1,...].
1) Show that
Vl + fl2 = [a, 2a, 2a, 2a,. . .].
m) Show that a periodic continued fraction (one where ak+l = ak for
some / and all large k) represents a quadratic irrational (the root
of a quadratic equation with integer coefficients).
n) (Lagrange) Show that a quadratic irrational has a periodic
continued fraction. Thus with m) this characterizes quadratic
irrationals.
Hint: Suppose ra2 + sa + t = 0 and a = [a0, alf a2,. . .]. Then by
d),
P„-l<+Pn-2
a = TT ■
Substitute this into the quadratic equation for a to obtain integers
An, Bn, and Cn with
(11.3.52)
An(a'n)2 + Bna'n+Cn = 0.
Express An, Bn, and Cn in terms of r, s, and t and show that
A „5^0. Show that
^-4,4,,^=/-4^
and that
C„ .A,

376
Pi
Show, using f), that
and
\An\ ^2\ra\ + |r| + |*|
|Cj<2|ra| + |H + UI
\Bn\2<4\AnCn\ + \s2-4rt\.
Hence \An\, \Bn\, and \Cn\ are bounded independently of n. Thus
(Ani, Bni, Cni) = (A„2, B„2, Cn) = (A„3, B„3, Cn) for three distinct
indices n1, n2, and n3, and by (11.3.52) one of
a„. = ot„
or
or
Oil = ot'
«9 n
which implies the periodicity of a.
o) Show, using i), that any integral solution of Pell's equation
n - dm2 = 1
d a positive integer
has nlm a convergent of the continued fraction of Vd.
This outline of the basic theory follows Hardy and Wright [60].
3. Suppose there exists a sequence of rationals {pnlqn} and 8 >0 so that
(11.3.53)
and
(11.3.54)
Then either
Pn
In
<
<ln
+ o(.l)
<ln<<ln + l<<ln
— = — for some n
1 In
or for e > 0 and for q > ce
a —
>
l+i/S+e
Hint: Let \a -p/q\ = l/q1+lls+e , e'>0, and choose n so that
USn + l^q< Wn- Then

11.3 Irrationality Measures
377
and if p/q*pjq„,
l<-3* +
<7„
S ' llS+e'
In <l
The right-hand side, however, is less than 1 for q large.
4. Refine Exercise 3 to deduce Corollary 11.1, with v'=l, directly from
Theorem 11.1.
5. (Irrationality measure for e1/v) Let u be a positive integer. As in
Exercise 10 of section 10.1, let
£0(n-k)lkl\v
and
a) Show that
£0(n-k)\k\ \ v,
ellv--±
t„
n\n\
and that
Hence
t=v
llv an
6 ~Z
1
(2/i +l)!(2n)! " \v
„ (2n)! _,/2„
[1 + 0(1)]
/l!
e-"» [1 + 0(1)].
1 log log f„
[1 + 0(1)].
2u '>§'*
b) As in Exercises 3 and 4, show that if plq^s Jtn for some n then
<7
>
<7
for large g, where cv is a positive constant depending only on v.
c) Deduce Corollary 11.1.
6. Let dn:=LCM(l,...,n).
a) Show that there is a constant C so that
d. < C".

Pi
Hint:
av=rU)
(2n)! . (2n)! .„
v ' and ^-4-<4
where the product is taken over the primes pt, n<pt<2n. Thus
b) Let ir(«) denote the number of primes less than or equal to n. The
prime number theorem asserts that
(See, for example, Hardy and Wright [60].) Use this to prove that
dn = 0(eSn) for any 8 >1.
Hint:
where the product is taken over the primes <n and where at is the
largest integer so that p? ^n.
a) Show that
v> a,a7- ■ ■ at_. 1
fc=I (x + «j) • • • (x + ak) x x(x + «j) • • • (x + aK) '
b) Show that
"v (-l)*"1^-!)!]2 1 2(-l)"_1
1
^i{n-l2)---{n-k2) n /2n
n I
V n
Hint: Use a) with x '■= n and ak '■= —k2.
c) Set
* -=I W)2(n-k)\
"•*" 2 k\n + k)\ '
Show that
)*''[(*-I)']2
(n* - 12) • • ■ (n2 - k2)
(-1) "(s„,/t _ °n~i,k) ~ 73—Ta 73—3^

11.3 Irrationality Measures
379
and
N n-l N 1 N /-1 \«-
A/
= 2(-1)^,,-8,,,)
A/
= 2
(-i)* , i v (-1)^1
/t=l
N + k\fN\ 2 k"1 ,3(2k
2k ' . A*,
d) Let iV^>°° in c) to deduce that
«3) = |2
(-1)-
2«
(This exercise follows van der Poorten [79] and is essentially due
to Apery.)
8. Finish the proof of Corollary 11.2.
9. {Irrationality measures for log)
a) Let p„(x):= {d/dx}V(l - x)". Then as in (11.3.23),
'.w-scx:*)*-1**-
Estimate pn(—m). In particular show that for mE.N there exist
Cj >0, c2, c3, and c4 so that
cInC2(Vm + l + Vmf" <p„(-m) < c3nq(VmTl + Vm)2" .
ifwtf: By Stirling's formula, up to a power of n
n\ln + k
m behaves like
ma(l + a)1+a 1
a\i~a la
— a) a ->
where k-- an. It is now a calculus exercise to maximize the above
by differentiating with respect to a. The maximum occurs at
a '■= yjm/(m + 1).
b) Prove that for large q,
(11.3.55)
log 2
>
1

Pi
Outline: Let pn be as in (11.3.22) and let
(11.3.36) /.-J^*
= r1 pn(x)-pn(-i) dx+r £,,(-1)
Jo 1 + x Jo 1
+ x
dx
= ^+p„(-l)log2
where a„ is an integer and dn is as in Exercise 6. (Note that
(1 + x)\ [pn(x) -p„(-l)].) Furthermore, integrating /n by parts n
times with respect to x yields
(11.3.57) |/J = J-
^^.
(1 + *)
Estimate |/| from above and below by computing that the
maximum of x(l — x)/(l + x) occurs at x '■- V2 - 1. For large n,
(11.3.58) (V2-1-)2"<|/„|<(V2-1+)2" .
From (11.3.56), (11.3.58), and part a) prove (11.3.55).
c) Prove, for e> 0 and n> Ne, that
n + 1 p
log
° n q
> ~2T7 Q lar§e •
<7
Compare (11.3.4) and (11.3.5).
Hint: Consider
Jo 1 + x/m Jo 1 + x/m
and proceed as in b). Note that
■" 1 f1 X"(1~Xr J,~ 1 1
m" Jo /-1 + -/-^+J ax
n /\n
1 "' mn Jo (l + x/m)n+1 mn 4
a) Show that 83(q) and 84(q) are irrational for rational q :=1 In,
n = 1,2, 3,. . . .
b) (Euler) Show that m4 + nA = p2 with m, n, p integral implies
mn = 0.
c) Use b) to* show that 82, 83, and 8A are never all nonzero and
rational.

11.3 Irrationality Measures 381
a) Show that
2 » 1
(11.3.59)
Outline: Let
(-1)"
2n + l
= 2
(2fc + 1)2'
N N / ixm+n A/
i\ i\ / -* \m+n ^v ^
a .= y y (-1) y 1
°N' ~~ (2m+ 1)(2/1 + 1) t^(2)t + l)2
and
%'■= 2
" (-1)"
m=-N m-n
Show that
and that
^ ^, (_ir + n
dN ~~ (2/i + l)(m-/i)
1
JV-rc + 1
Thus prove (11.3.59) by showing that 8N^>0.
b) Evaluate £(2) = ir2/6 from a) and Gregory's formula.
a) Show that for 0 ^ x ^ 1, one has the following functional equation
for the dilogarithm E"=I x7n2:
(11.3.60) log(l-x)logx+2 ^+2 (1~2X)"=^.
n=I n n = I n o
7/mf: Show by taking derivatives that the left-hand side is zero.
Evaluate the constant by letting x-»0.
b) (Euler) Show that
2 1 ™ 1
T2-2^2)2=„?l2V-
c) Let Li3(x) denote the trilogarithm E"=I x7n3. Show that, for
0<x<l, the following identity due to Landen holds:
U3(x) + Li3(l -x) + Li3(y^) - Li3(l)
= y log (1 - x) - ! log (*) log2 (1 - *) + ^ log3 (1 - x)

Pi
d) Deduce that Li3(l) = £(3) and that
i) Li3(|) = ^Li3(l)-^-log2+ilog3(2)
.., T. /3-V5\ 4T. .„, tt\ /3-V5
e) Show that
1 , 3/3-V5
12 log ("1-
log[(V5+i)/2]
£(3) = Li3(l) = 10 J t2 coth (t) dt
and combine this with Exercise 17a) of Section 5.5 to provide
another verification of Exercise 7d). This is discussed in Lewin
[81, Sec. 6.3].
(Euler) Establish
CO
a) ir[cot (irx) - cot (ira)] = 2
a - x
(x - n)(a - n)
1
b) ir3fcot(irx)cosec2 (ttx)] = 2 rr
-»(x-n)
c) 7/
cosec4 (ttx) - - cosec2 (ttx) = 2 r
3 J -„ (x - n)
n=i An -1 2
e)
f)
2
CO
2
CO
V
1
(4rc2-
1
(4rc2-
1
I)2
I)3
tt2-;
16
32-3tt2
64
„=1 (4« - 1)
Hint: Use x := 3 and
77-4 + 3077-2 - 384
768
In - 1 2« + 1 An -I'

11.3 Irrationality Measures 383
(Evaluation of £(2m)) Define {Bn}, the Bernoulli numbers, by
ez-l ' 2 ~0~2m (2m)\
5, :=-¾ and 52B+1 : = 0,/1 = 1,2,3,...
(11.3.61) -^ + | = X B2m -^- \z\ =£2,
a) Show that
/fc=0 \ K /
b) Show, from (11.3.61), that
. . 2iriz
irz cot (irz) = -2^jj—- + iriz
e — 1
= Z (-1) (2ir)
m=o (2m)!
and from the product expansion for sin that
, x 1 V 2z
IT COt (iTZ) = 2j
Z n = :
c) Show that
2 2 •
« - Z
d) Thus
aim)-=1 ^=(-1) 2(2m)! •
«2) = T «4) = ^ «6) = ^
(Evaluation of /3(2m + 1)) Define {£2n}> ^e £u/er numbers, by
\«r _2n
J_ = V (-1) ^2 , . . ir
OS z „=(
a) Show that £0 = 1 and
b) Show that
(1L3'62) c^I = i^l2#- N<2
Z ( o?J£2«-2/t-°

Pi
IT
= X 4k+1p(2k + \)z
2k
cos (irz) k=0
where, as before,
)8(2/: + 1):=2 J"^" +i ■
„=o (2« + 1)
c) Show that
If I , _\ik+i
«2* + ')-5^f(f) ■
d) Thus
3 c 5
■"■ „,„* IT „,„. JIT
)8(1)=- )8(3)=- )8(5) =
4 ^v ' 32 ^w 1536 '
There are similar evaluations for more general L functions.
(Series involving ())
a) Show that
2x arcsin x _ y (2x)2m
vr^5
* m=i /2m
ml
V m
7/mf: Let /:= (arcsin x)l\l - x2. Show that
(l-x2)/=l + ;t/.
Show that
z* ■"=' '2m
ml
V m
also satisfies the above differential equation. (Compare Exercise
16 of Section 5.5.)
b) Show that
» /^ \2m
2(arcsin x)2 = 2 -±-^ .
m=1 2/2m
m
m

11.3 Irrationality Measures
Hint; Use a) and the fact that
d . . .2 2x arcsin*
x -j- (arcsin x) =
2
dx Vl-x
c) Show, by differentiating in a), that
Y (2x)2m _ x2 x arcsin x
«=i (2m\ 1~x (1_* )
\m/
d) Specialize the above series or the derivatives of the above se
show that
Y 1 2ttV3 + 9
^ 2m^= 27
2-^-
m=1 /2m
ml
V m
co 1
E -^-
m
ttV3
2
IT
i 2/2m^ 18
m I
m
m 2
2—= £(W5+9)
m=i /2m x ^7
m
E -^- = ^ + 3
m=1 /2m
\ m,
E 3
2/2m
m
m i
V 3m 4ir „
Z, = -7?+3
—' /2m ^ V3
V
m
(-l)"1"1 _ 1 , 4V5 , V5 + 1
V m
v (-1) i Wi
m-i /2mN 5 25 2

386
Pi
= Alo /Vs + i
-=i /2m\ V5 °gV 2
ml
V m
e) Show that
and that
■
Thus
2 ("ir =2
m=1 2/2m\
V m 1
M*;1)]
2
1 y (2n\ „
Vl - 4* ~0 V n r
01 /l-Vl-4x\
21oH 2x )
„=i n \ n 1
, A v ! /1•3 -••2« ~1 \
f) From e) deduce that
and that
- H 1
V ^ " ' In 1 / • „ x
^ 1 , 1 * = o (arcsin 2x)
„=i 2« + 1 2x v '
■ (2")
ttx (2n + !)16" 3 '
Further material is available in Lewin [81], Lehmer [85], and
Zucker [85]. There is an interesting evaluation of Comtet's, namely,
E
l
17ir4
m i
/2m \ 3240 '
m
g) We conclude with Ramanujan's
E
l
IT ,
(2m + l)2(2m
\ m
-G- 3-log(2 + V3)
with G denoting Catalan's constant.
Hint: Use part a), and parts b) and h) of Exercise 10 in Section
5.6.

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K(k)
Symbol List
Symbol Description Section
AGM Arithmetic-Geometric Mean Iteration 1.1
Af(-,-) Common limit of the AGM 1.1
AG(-, •) Common limit of the AGM 1.1
a„,bn,cn Variables in the AGM 1.1
kn Legendreformofthe AGM 1.1
k' k':=yl~kz, Complementary modulus 1.1
/ the derivative of / 1.1
J "nil in /*1 Jf
o Vl - k2 sin2 9 ~ J" V(l - <2)(1 ~ kY)
complete elliptic integral of the 1st kind
p'2 . r1 vT-HtV
E(k) E(ky-"]o Vl - fc2 sin2 6 M = Jo —r==f- <* 1.3
complete elliptic integral of the 2nd kind
K'(k) K'(k):=K(k') 1.3
E'(k) E'(k):= E(k') 1.3
F(a,b;c,z) Hypergeometric function 1.3
(2i-l)!! (2i-l)!!:=l-3-5-...-(2i-l) 1.3
1(-,-) Ha.by.-r ., 2de 2 , -IW*) 1.4
Jo Va2 cos2 61 + 62 sin2 61 « ™'
/(•,•) /(a,6):=Jo Va2 cos2 6" + 62 sin2 6" <# = aE'^-j 1.4
G G(k)-=klnk'K(k) 1.5
G* G*(jfc) = klnk'K'(k) 1.5
T T(a;):=Jo e~V_1df, Gamma function 1.6
396

Symbol
Description
Section F
p
sn(u, k)
cn(u, k)
dn(u, k)
P
»2
%
%
fo,yy.=^f-\\-ir>dt.
™<«.*) d(
Jo . V(i-<2)(i-^2)
r<--*> dt
h V(i - ?W2 + kV)
(•*.<«,*> dt
h V(l - t2)(t2 - k'2)
P(z)-~ 2 + E (, ,2 J
2 ,61' V(Z - W) W /
Weierstrass function
rt = — ™
rt = — ™
04(<7):= 2 (-l)V2
Beta function 1.6
1.7
2.7
1.7
2.7
1.7
2.7
1.7
2.1
2.1
2.1
r2 r2(n) -~ number of representations of n
as a sum of two squares 2.2
0,(.5) $(.5) := 0,(9) where q:=e~" 2.2
k(q) k(q):=k = e\{q)iel(q) 2.3
*'(*) *'(«):=*'=»J(«)/»U«) 2.3
*W AX*)-f »?(9) 2.3
9 , = «,-*'<*>'«<*> 2.3
yt(s) fc(i):=yt(?) where q:=e~" 2.3
0,(z, g) General theta functions 2.6
0y(<7) Theta functions in q, 2 := 0 2.6
dj(z) Theta functions in 2, g suppressed 2.6
0(2) Theta functions in 2, q and / suppressed 2.6
Qo e0(?):= n (I-?2") 3.1
C, 6,(9):= n(l + <72") 3.1
62 e2(<7):=n(l + <72"-1) 3.1
e3 &(*)=-n a-*ta~!) 3.1
A* A*(r) :=*(?) where 9 = e_w7 3.2
0f 0*(9): = 2E (-l)"(2n + l)9("+!,2)2 3.2
n-0

Symbol
Description
Section F
05 8,(,):-2 2 (-l)V2-"2)2 3.2
e; »6*(?):=2i;(-ir("-i'>+i)?("+"i'! 3.2
/.=0
1? 91/!2Go, Eta function 3.2
/ <r"24e3 3.2
/, . V291,I2e1 3.2
a <r"24e2 3.2
2' 2' am.n-= 2 <*„,„ 3.2
G„ G„:=(2A;A;')~"I2=2~"4/(V=n) 3.2
,«"2^'2
Sn - g,
'-:-(|t) =2-"4/1(V-H) 3.2
) Gaussian or g-binomial coefficients 3.3
(«). («).=- fi(i-<r)/(i-<rm) 3.3
r4 r4(n) := number of representations of n as a sum
of four squares 3.5
<rx ^(/1):=^(1 3.5
w(n) w(n) ••= a-jin) + tr^odd (n)) 3.5
¢-1
S(P,1) S(p, <?):= S^'" 3.5
r-0
Af(/) Af(/):=Afs(/):=Jo f{x)^~ldx, Mellin transform 3.6
I i(s) := 2 n"! , Riemann zeta function 3.6
n = l
g(t) g(t) : = («,(») -1)/2 3.6
/3"
L(/3) L(j3):° E n„ , |/3|<1. Lambert series
n=l 1 P
3.7
F„ F0: = 0, F, :=1, F„+I:=F„ + F„_1, Fibonacci numbers 3.7
(Ex.3)
I<„ L0 := 2, Lj := 1, Ln+1:= L„+ L„_, , Lucas numbers 3.7
ST Sif:={im(T)>0} 4.3
sr sr ••= ar u {;<*>} u {Q> 4.3
T-group Inhomogeneous modular group 4.3
A-group 4.3
Fr Fundamental set of T-group 4.3
FA Fundamental set of A-group 4.3
398

Symbol
Description
Section Formula
Sr
Si
u
(I
c
(J
G
m
/(0
m-
!)
"!)
?)
J)
:=fc2(0 =
._ 4 (1-
' 27 A2(
= 1728/(0
k(«)J '
A(0 + A2(0)3
:o(i-a(o)2 :
{(c bd}-<«l-bc = p;a,b,c,
transformations of order p
V
A,:-
V
Br-
-(!!)
-G i).
■(S3
■(! ?)■
i = 0,l,...
/ = 0,1,..
,:=«"'
Kle
dez]
,,p-l
-,p-i
4,3
(Ex. 1)
4.3
(Ex. 1)
4.3
(Ex. 1)
4.3
(Ex. 1)
4.3
Klein's absolute invariant 4.3
/ /(0:=1728/(0 4.3
Tp {(" ^j:ad~bc = p;a,b,c,d&ZJ 4.4 4.4.1
transformations of order p
v-(5 ?)
A .-4,--(5 p)> / = 0,1,..., p-1 4.4
B, 5,:=(J J'), / = 0,1,..., p-1 4.4
W„ W,(*. A) :=11(^-^), A,:=A = B,. 4.4 4.4.2
i=0
pth order modular equation for A
p
FP Fp(x,j)-Yl(x-Jd, /,:=/-/1,. 4.4 4.4.6
i=0
pth order modular equation for /
Qp Q adjoin the pth roots of unity 4.4
fip np(u, a) := 11 («- «i) 4.5 4.5.1
i=0
pth order modular equation for u
up «„:=(-l)<"2~!)'8(A(9''))"8:=('-l)<''2"I)'8"(9''), <7 = «''"' 4.5 4.5.1
«i:=(A(a8i9'"'))1'8:=«(a8A:91"'), *=0,1,... , p-1 4.5 4.5.1
„ = /»a-'"
Mp MpV<ky-= e*fq"')=lm' MultiPlieroforderP 4-6 4-6-1
k kp := yt(e_w?) pth singular value 4.6
/, /,:=^-^-^) 4.6
a(r) «W:=f-^p, k:=k(e'^) 5.1 5.1.1

Symbol
Description
Section Formula
8(r):=V7-2o(r) 5.1 5.1.9
^¢:=¾¾^ 5.2 ,5.2.7
5.2
5.2
5.4
5.4
5.4
5.4
5.4
5.2.11
5.2.12
5.4.1
5.4.1
5.4.1
5.4.2
P(<7):=1-24S -^-j- • 5.2 5.2.8
/i-i 1- q
nick'2 dM
^•^:=w)"^(/,")+m;2(/,")/2~m2
<r(p):=Rp(k',k), k:=e~"^
For ȣN, compute kn+l by solving Wp{k\, k2n+i) = 0
m„-~M~\kn,kn+l)
rn:-Rp(kn,kn+1)
«. := [«.»■. + «;(1 + *2) -p(l + ^+1)]/3
^,(^:^):=2^=^^, where 5.5 5.5.1
n-0 (C)„ «!
(a)„ := Ye ■) ~o(o + 1) ■ ■ ■ (" + " ~ 1), rising factorial 5.5
*.(*):-f - a^i(§ -*.§ + *;!; **) 5.5
£,(*) =- f ' a^.(- § - *, | + *; 1; **) 5.5
G:=2 7^ rt5 , Catalan's constant 5.6
.-o (2« + D (Ex.10)
b„ = fl(a„) if b„ = 0(a„) and a„ = 0{bn) 6.1
Bit complexity 6.1
Operational complexity 6.1
Fast Fourier Transform 6.2
Bit complexity of multiplication 6.4
6.3
(Ex, 3)
Bit complexity of division 6.4
Bit complexity of root extraction 6.4
a a 6:=min(a, b) 8.1 8.1.1
a v b ■ max(a, b) ' 8.1 8.1.1
tM{x) : = M(x, 1), M a mean, the trace 8.1
R-{0> 8.1
Mf(a,b):=rlM(f(a),f(b)) 8.1
Mp denotes Mf when f(x) = xp 8.1
5.5.3
5.5.4
400

Symbol
Description
Section Form
pM
Lp
Gs,
SP
M>N
M~N
M®N
M®gN
[M,N]g
M®„N
W, N]„
N*
M
R(a;8,8';x\y2)
F2
F%
F*
arcsl(;c)
arcslh(*)
la" + b"\Up
Hp(a,b):=y— J , a,b>0, Holder means
M(ap, b") _ Mp(a,b)
pM(a, b) :=
Lp(a,b) =
M(ap-\bp~l) MpZ\(a,b)
ap + bp
p-\ , ,P~i
+ bp
Lehmer means
[s . ii-jl/i-r
p -\up-i
p(«-bj-
P 7*0,1,
8.1
i,(,t):»«(^«.[-^]
Stolarsky means
j?(a, b) : = S0(a, b) , Logarithmic mean
$(a, b) : = Si (a, b) , Identric mean
„ , L, rs(ar- br)V'ir-s)
M(<t>(a), <t>(b)) = <t>(N(a, b)) , dominance 8.2
if M > 0jiV and N > <j>2M, equivalence 8,2
Compound Mean of M and N 8,3
M comparable to N 8.3
The Gaussian product (Def. 8.2) 8.3
The Gaussian mean iterative process (Def. 8.2) ■ 8.3
The Archimedean product (Def. 8.2) 8.3
The Archimedean mean iterative process (Def. 8.2) 8.3
N*(a,b) = N(M(a,b),b)
M(a, b) := M(a a b, a v b)
«(a;S,S';^/):=-^_/o «■-«(» + x2ys(t + y2)~s' dt 8.
Cv:=Ft®Fj (/,/ = 1,2,3,4)
ul 1.1. a + b
FM b) •= —J—
F2(a, b):=Vab
Ua,b):=^±±7a
F4{a,b):=^±^b
arcslW:=|o'(l-/)-"2
arcslh(*):=Jo (1 + s4)~in ds
ds
8.3
5
8.5
8.5
8.5
8.5
8.5
8.5
8.5
401

Symbol
Description
Section F
a a :=(<Jj,... ,aN) 8.7
LAS) LAS) : = (S af )/(2 af_1) , Lehmer means 8.7
I \ E, \VP
Hp(S) Hp(S):={—Z,a!) , Holder means 8.7
® Af, The common limit (when it exists) oi [M ,.. ., M ] 8.7
[M\ ..., MN] a'„+1 := M'(S„), or vectorially a„+1 = M(S„) 8.7
®gAf' iV-dimensional Gaussian product 8.7
N
®„Af' iV-dimensional Archimedean product 8.7
;-i
• t , % >t / x. a + Vac + c
He(a, c) He(a, c) := , Heronian mean 8.7
S(a,c) S(a,c): = He®sG(a,c) 8.7
MSNf %
(«,,«„....«„)-/ [2oiy^)j 8.7
TG(/: F) The algebraic transformation group of / over F 8.8
bA2s) 6,(2*):- 2' J"1] ' -av 9.2
b2(2s) 6,(2«):- £' -¾¾
64(2»):
a (*)'■ =
«*)■'-
£**(*)
(«; ?).
v' (-
i,y,t,i— 0 +J
y (-1)"
.ti (2»+iy
:=S(±rf|«Ks
t-0
jy+;'+*+/
r2 + yt2 + /2)s
-2^)^),
9.2
64(2*) 6,(2.):- 2 t*l* + tz + #v 9.2
(,y,t,i—« (' +/ + * +' )
«(*) <»(*)■■- 2 ^1¾^ = (1 -2^)^), ' 9.2
m m:-f. /ijPiv. 9.2
L±,(s) L±,(s):=2(±<i|nKs 9.2
(Ex. 6)
(«;?)- (a; 9).:=11(1-V) 9.4
(Ex.7)
«,(») M,(*-*a) 9-5
Np Np := i?2(o)/i?2(o"") , i? given by 3.2.9 9.5
Ds D6:={2eC||z|sS> 10.1
P„ Algebraic polynomials of degree n 10.1
ll/L ll/L:=sup|/W| 10.1
£„(/,/1) e„(/,/l):=rnm||/-/>L 10.1
402

Symbol
Description
Section Form
RnU.A) «.(/,i4)=-«.(/)= miri ||/-p/q\\A 10.1 10.1
P>QePn
= lim 1 + - + - + • • • + log m , Euler's constant 10.2
m-»«L 2 3 m J
y -y : = Urn 1 + - + -+---+ log m , Euler's constant 10.2 10.2.
m-.«.|_ 2 3 m J
d„ d„: = LCM(l,2, ...,n) 11.3
[0,,,0,,...,0,,] [o0, a,,..., o„] := o0+ — ^-^
0, + —— v " ;
'• 1
+ —
simple continued fraction
[a0,a„...] Jim [a0, a,, . .., o„] 11.3
(Ex. 2)
Z 22"1
B. 7^ + 2=2^(2^ ^S2w 1L3 1L3
Bi = ~2 and B2«+i:=0' n = l,2,3,
Bernoulli numbers
F 1 __ y (-l)nE2nz2" „
2" cosz~~£0 (2n)! ' |2|<2>
Euler numbers 11.3 11.3
403

Author Index
Numbers in parenthesis indicate references in the Bibliography. "Pr" indicates references in press
Abel, 5, 28
Abramowitz and Stegun (64), 15, 321
Adams (66), 371
Aho, Hopcroft and Ullman (74), 201, 203, 206,
211
Alladi and Robinson (79), 371
Almqvist (Pr), 99
Almqvist and Berndt (Pr), 3, 196, 371
Andrews (76), 67, 86
Andrews (86), 286
Apery, 365, 371, 379
Apostol (74), 37, 40
Apostol (76a), 86
Apostol (76b), 116
Arazy et al. (85), 269, 272
Archimedes, 243, 249, 337, 338
Aristophanes, 347
Askey (80), 89, 308
Backstrom (81), 98, 99
Bailey, D. (88), 211, 341, 342
Bailey, W., 181
Bailey, W. (35), 184
Baker, 323, 348
Baker (75), 349, 35J, 354, 356, 363, 371
Baker and Graves-Morris (81), 320, 323
Ballantine (39), 347
Baxter, 80
Beckenbach (50), 236, 269
Beckmann, 346
Beckmann (77), 342
Beeler et al. (72), 220, 222
Bell (27), 36
Bellman (61), 39, 55, 56, 86, 89
Benson, 301, 302, 303
Berndt (Pr), 142, 150, 177, 196, 309
Bernoulli, Jacques, 340
Bernoulli, Jean, 340
Beukers, 195
Beukers (79), 365
Bhargava and Chandrashekar Adiga (84), 81
Biagioli (Pr), 146
Birkhoff.(73), 10, 18, 116
Birkhoff and Rota (69), 9
Bohr, 89
Borchardt, 250, 256, 257, 263
Borchardt (1888), 272
Borodin and Munro (75), 206, 208
Borwein, Borwein and Taylor (85), 290, 291,
292
Borwein, D. and Borwein, J. (86), 291
Borwein, J.M. (85), 163, 164
Borwein, P.B. (85), 218
Borwein, P.B. (Pr), 254
Borwein and Borwein (84a), 44, 50, 51, 222,
223
Borwein and Borwein (84b), 108, 174
Borwein and Borwein (84c), 174
Borwein and Borwein (84d), 222, 225
Borwein and Borwein (86), 174
Borwein and Borwein (Pr), 313
Bouyer, 341
Bowman (53), 29
Boyer (68), 347
Braess (84), 325
Brent, 51
Brent (76a), 48, 222, 229
Brent (76b), 222
Brent (76c), 213, 216, 217, 222, 329, 330
Brent and McMillan (80), 336
Bressoud, 78
Bressoud (83), 77
405

406
Author Index
Bring, 136
Bromwich (26), 189, 291, 344
Brouncker, 338, 345
Bundschuh(71), 371
Buslaev, Gonchar and Suetin (84), 323
Carlitz(71), 71
Carlson, 248, 257, 261
Carlson (71), 7, 256
Carlson (75),261
Carlson (78), 262
Catalan, 66
Cauchy, 76, 77, 292
Cayley (1874), 106, 133, 138, 140, 142
Cayley (1895), 22, 102, 138
Chandrasekharan (85), 116
Cheney (66), 321
Chudnovsky and Chudnovsky (84), 368, 371
Clark (71), 349
Clausen, 179, 188, 189, 190
Cohen and Nussbaum (87), 273
Comtet, 386
Cook, 69, 213
Cook and Aanderaa (69), 69
Cooley, Lewis and Welch (67), 65
Cooley and Tuckey (65), 65
Cox (85), 15
Dase, 340
Davenport (81), 226
Davis (79), 365, 371
Denninger (84), 89
Dickson (29), 293
Dickson (71), 55, 66, 85, 141, 284, 287, 292
Dirichlet, 86
DuVal (73), 23, 30, 133
Eagle (58), 29
Edwards (79), 252
Epstein, 303
Erdelyi et al. (53), 10, 30, 55, 178, 179, 181
Euler, 3, 18, 64, 67, 77, 189, 264, 286, 306,
307, 340, 344, 346, 348, 381, 382, 383
Ewell (81), 66
Ewell (82), 306
Ewell (83), 86
Ewell (86), 151
Feldman, 371
Felton, 341
Ferguson, 340
Fermat, 82, 286
Foster and Phillips (84a), 260
Foster and Phillips (84b), 247, 252
Gammel, 323
Gauss, 1, 3, 5, 8, 10, 28, 43, 51, 85, 86, 250,
256, 284, 286
Gauss (1866), 5, 7, 35, 44, 48, 52, 65
Gelfond, 348
Gini (38), 233
Glasser and Zucker (80), 68, 290, 293, 303
Goldfield (85), 295
Gordon (61), 306, 308
Gosper, 341
Gould and Mays (84), 263
Goursat(1881), 185, 186
Gradshteyn and Ryzhik (80), 10
Grave, 345
Graves-Morris, 323
Greenhill (1892), 133, 167
Gregory, 339, 348, 381
Guilloud, 341
Hancock (09), 29
Hanna (28), 133
Hardy, 80, 187
Hardy (40), 141, 314
Hardy, Littlewood and Polya (59), 235, 242,
269
Hardy and Wright, 80
Hardy and Wright (60), 66, 67, 85, 86, 95,
281, 284, 349, 376, 378
Hermite, 135, 264, 348, 359
Hermite and Stieltjes (05), 264
Hilbert, 348, 353, 354
Hilbert (1893), 348
Hirschhorn (85), 285
Holloway, 188
Householder (70), 215, 216
Hua (82), 294, 349, 375
Hughes (84), 175
Ivory (1796), 12, 196
Jacobi, 5, 10, 20, 22, 28, 38, 52, 65, 68, 73,
74, 85, 106, 143, 180, 252, 261, 284, 301,
306
Jacobi (1829), 36, 106
Jones, 339
Joubert, 142
Kaltofen and Yui (84), 133
Kanada, 224, 225, 341
Karatsuba, 211
Keith (86), 342
King (24), 16, 18, 44, 52,59,60
Klein, 116
Klein (1897), 349

Author Index
407
Klein (79), 7
Klein and Fricke (1892), 116, 126
Knuth(81), 203, 211,212, 218
Kronecker, 135, 293, 296, 297
Kummer, 179
Kung and Traub (78), 216
Lagrange, 3, 82, 321, 375
Laguerre, 99
Lambert, 99, 348
Landau, 268, 269
Landau (1899), 94
Landau (58), 86, 294
Landen, 381
Lang (66), 349
Lang (73), 29, 30, 116, 123
Leach and Scholander (78), 237
Legendre, 3, 10, 23, 26, 152, 153, 154, 178,
348
Lehmer, 264, 265
Lehmer (38), 345
Lehmer (71), 247, 263
Lehmer (85), 386
Lehner(66), 116
Leibniz, 339, 340
Le Petit Archimede (80), 342
LeVeque (77), 294, 349
Lewin (81), 382, 386
Lindemann, 274, 348
Lindemann(1882), 348
Liouville, 10, 36, 116, 348
Lipson(81), 216
Lorenz, 287, 289
Machin, 339, 341, 344
MacMahon, 67
Mahler (31), 359
Mahler (53), 368
Mahler (67), 349, 368
Meil (83), 250
Meinardus, 321
Metropolis, 341
Mignotte (74), 368
Mobius, 100
Mollerup, 89
Mordell, 187
Mordell (16), 286
Nemeth (77), 321
Newman (79), 321
Newman (82), 7, 222, 223
Newman (85), 7
Newman, M., and Shanks (84), 193
Newton, 123, 207, 208, 212, 339, 343, 356
Niven (56), 347, 348, 349
Nyvoll (78), 196
Oxtoby (80), 352
Pade, 320
Pfaff, 250
Phillips (81), 252
Picard, 119
Poincarfi, 116
Poisson, 38
Polya, 86
Rademacher (73), 56, 71, 85, 90, 286
Rainville (60), 179
Ramanathan (84), 81
Ramanujan, 69, 80, 81, 84, 100, 138, 140, 144,
146, 150, 154, 157, 158, 161, 164, 179, 184,
186, 187, 188, 194, 195, 197, 287, 295, 306,
307, 308, 309, 310, 311, 312 313, 314, 315,
386
Ramanujan (14), 158, 163, 179, 184, 186, 191,
193, 195, 313
Ramanujan (62), 80, 187, 281, 286
Rankin (77), 116
Reitwiesner, 341
Ribenboim (84), 345
Ribenboim (85), 95
Riemann, 89
Ritt (48), 226
Rogers, 80
Rouche, 317, 319
Salamin, 51, 220, 222
Salamin(76), 48, 49, 51
Sasaki, 225
Sasaki and Kanada (82), 224, 225
Schepler (50), 342
Schlafli, 147, 297
Schlomilch, 268, 269
Schneider, 348
Schoeneberg (76), 116, 123, 126
Schoeneberg (77), 247, 269
Schoeneberg (82), 247, 250
Schonhage, 211
Schonhage and Strassen (71), 211
Schroter, 111, 133
Schur, 80
Schwab, 250
Selberg and Chowla (67), 296
Shanks, D. (82), 193, 194
Shanks, D. and Wrench (62), 341
Shanks, W., 340
Shanks, W. (1853), 340

408
Author Index
Singmaster (85), 342
Slater (66), 178, 179
Stickel (85), 222, 224
Stieljes, 264
Stolarsky (75), 233
Stolarsky (80), 237
Stormer, 345
Strassen, 202, 211
Sweeney (63), 336
Tamura and Kanada (Pr), 211, 342
Tannery and Molk (1893), 138, 141
Todd (49), 345
Todd (75), 394
Todd (79), 7
Toom, 211
Trefethen (84), 321, 325
Tricomi (65), 262
van Ceulen, 342
van der Poorten, 379
Vieta, 338, 343
von Neumann, 341
Wagon (85), 342
Waldo, 342
Wallis, 338, 343
Walz (Pr), 330
Watson (29), 306
Watson (32), 140
Watson (33), 3, 59, 261
Watson (35), 286
Weber, 168, 295
Weber (08), 68, 123, 138, 140, 300
Weierstrass, 29, 30, 141
Weierstrass (67), 348
Whittaker and Watson (27), 10, 26, 27, 29, 55,
59,72
Wills, 323
Wimp (84), 7, 242, 260, 269
Winograd (80), 206
Wrench (60), 342
Wright, 80
Zucker, 100, 141, 296,297
Zucker(77), 140, 297
Zucker (79), 100, 154
Zucker (84), 75
Zucker (85), 386
Zucker and Robertson (76a), 293, 294
Zucker and Robertson (76b), 293

Subject Index
Algebraic addition theorem, 29, 30, 32
Algebraic function, 273, 276
Algebraic functional relation, 273
Algebraic integrals, 5
Algebraic number, 351, 352
Algebraic series for 1/ir and 1/K, 181-190
Algebraic transformation, 273, 274, 278
Algebraic transformation group, 273
Alpha, singular value of second kind, 152
cubic, 160
monotonicity, 153
quadratic, 158
quartic, 161
quintic, 310
recursions, 156
septic, 311
theta function form, 154
Alternating series test, 291
Archimedean mean iterations, 246, 249, 251, 252
Archimedean means, 259
Archimedean product, 246, 249, 250,
253,262
Archimedes' method, 249, 338
Arclemniscate, 25
Arclemniscate sine, 259
Arithmetic-geometric mean (AGM), 1
AGM relation, 1
complex starting values, 15
calculation of it, 341
computational performance, 330
continued fraction, 188
Gaussian, 181, 241, 257
Jacobi's identity, 35
Legendre form, 3, 45
matrix AGM, 223, 226
multidimensional AGM, 272
quartic AGM, 17, 51, 254
theta forms, 146
variant of, 254
Benson's formula, 301, 303
Bernoulli numbers, 383
Bessel function transform, 39
Best approximants:
existence, 321
uniform polynomial, 316
uniform rational, 317
Best approximation property, 373
Beta function, 24, 88
Bit complexity, 200
Borchardt's algorithm, 250, 256, 263
Brouncker's continued fraction, 339, 345
Calculating powers of x, 208
Carlson's integrals, 256
Carlson's log, 248, 257
Catalan's constant, 198, 371, 386
Cauchy's binomial theorem, 76
Class number, 141, 295
Clausen's hypergeometric product, 179,188, 189,
190
Comparable means, 244, 246, 247
Complement, 3, 178
Complementary integral, 8
Complementary modulus, 8
Complete elliptic integrals, 7
complementary integrals, 8
cubic algorithm for K, r07
differential equations, 9
E in terms of theta functions, 42
of first kind (K), 7
higher order transformations, 102ff
homogeneous forms of K and E, 12
K in terms of:
gamma function, 297, 298
theta functions, 35
moments of K and E, 198
quadratic transformations, 12
quadratically convergent algorithms, 14
409

410
Subject Index
Complete elliptic integrals (Continued)
of second kind (£), 7
series expansions of, 8
Complexity:
of algebraic functions, 215
bit, 200
elementary functions, 226
elliptic integral calculations, 227
Jacobian elliptic functions, 227
Lambert series, 98
log, 219
lower bound for log and exp, 227
operational, 201
17, 219
of theta computation, 95
transcendental functions, 219, 226
Complexity of algebraic functions, 215
Compound of means, 243
Conjugate divisors, 297
Constructibility, 347, 349-351
Continued fraction, 372-377
Contra-harmonic mean, 255
Convergents, 372
Cubic invariants, 173
Cubic modular indentities, 142
Cubic recursions for Ramanujah's invariants, 145
Diagonal mapping, 230
Dilogarithm, 381
Dirichlet class number formulae, 294, 295
Dirichlet L functions, 289
Discrete Fourier transform, 204
Discriminants, 294
Disjoint discriminants, 293
Domination, 239
e:
irrationality of, 352
transcendence of, 353, 359
Elementary symmetric mean function, 270
Elementary symmetric polynomial, 356
Ellipse, arc length of, 8
Elliptic functions, 29, 30, 31
en, 29
degenerate elliptic functions, 29
dn, 29
fundamental parallelogram, 29, 53
Jacobian, 29
lattice, 29
odd elliptic functions, 31
order, 30
period of, 29
sn, 29, 57
Elliptic integrals
(see Complete and Incomplete)
Equivalence of operations, 213
Equivalent iterations, 252
Equivalent mean, 231, 239
Estimates of e, 363
Eta-multiplier, 311, 314
of orderp, 311
Euler-Mascheroni constant, 336
Euler numbers, 383
computation of, 336
Euler's addition theorem, 18
Euler's formula, 189
Euler's identity, 64
Euler's pentagonal number theorem, 64, 66, 80,
306
Euler's totient function, 301
Evaluation problem, 204
Exponential:
algorithms for, 227
approximations to, 317, 318, 320, 321, 322,
323, 327, 330, 331, 335
series for, 320
Exp(ir), quadratic algorithm for, 50
Factorial calculation, 218
Fast base conversion, 218
Fast computation, 50
Fast Fourier transform (FFT), 204, 206, 211,212
Fast matrix multiplication, 202
Fast multiplication, 209
Fast polynomial division, 207
Fast polynomial evaluation, 208
Fast polynomial multiplication, 206
Fibonacci numbers, 94, 151, 287, 375
Fibonacci sequences and series, 91, 97. 98, 100
Finite Fourier transform, 204
Fully monotone, 291
Fundamental limit theorem, 5
first proof, 5
second proof, 6
third proof, 12
fourth proof, 21
Fundamental regions, 113
Fundamental sets, 113
Fundamental unit, 294
Galois group:
for Fp, 123
for W., and W7, 136
Gamma function, 24, 27, 28, 87, 332, 336
duplication formula, 88, 90
evaluation of K and £, 25, 27, 189, 191, 297,
298

Subject Index
411
Gauss's multiplication formula, 90
Gaussian binomial coefficients, 76
Gaussian mean iterations, 246, 249, 251, 252
Gaussian product, 246, 248, 253
Gaussian sums, 83, 86
Gelfond-Schneider theorem, 277, 278, 348
Generalized complete elliptic integrals:
of first kind, 178
of second kind, 178
transformations, 179, 180, 185
Generalized Legendre symbol, 293
Generalized singular value function, 185
Genera, 295
Geometric series, 91
Gini's means, 233, 264
Gregory's series, 339
Halley's method, 216
Height, 371
Heronian mean, 237, 269, 270
Hexagonal lattice, 292
Hexagonal sum, 292
Holder's means, 232, 235, 263, 264
Homogeneous mean, 231
Homogeneous multidimensional mean, 266
Horner's rule, 204
Hurwitz zeta function, 303, 304
Hyberbolic arclemniscate, 259
Hyberbolic function identities, 74, 75, 100
Hypergeometric functions, 332-334
Hypergeometric functions and series:
Gaussian hypergeometric series, 8
generalized, 178
hypergeometric differential equation, 11
Identric mean, 234
Incomplete elliptic integrals, 10
Eulers addition theorem, 18
of first kind, 10, 58, 60
of second kind, 10
of third kind, 10
Inhomogeneous modular group, T, 113, 117
Interpolation problem, 204
Invariance principle, 245, 246, 260
Invariants, evaluating, 293
Irrationality measures, 362
for exp(l/v), 363, 364, 371, 377
for exp(ir), 371
for log, 370, 379
for 17, 367, 368, 371
for i72, 367
for 5(3), 369, 370
Irrationality of e, 352
Irrationality of it, 352
Isotone mean, 231
Isotone multimimensional mean, 266
Jacobian elliptic function, 29
en, 29
dn, 29
half angle formula for en, 262
half angle formula for sn, 30
sn, 29, 57
Jacobi's differential equation, 20, 22
Jacobi's duplication formula for arcsl, 260
Jacobi's formula for r4(n), 81
Jacobi's identity, 35
Jacobi's imaginary quadratic transformation, 73,
180
Jacobi's triple-product identity, 62, 65, 66, 72,
301, 306
corollaries, 64, 65
finite form, 77
first proof, 62
Khinchin's theorem, 362, 371
Klein's absolute invariant, 115, 179
Kronecker symbol, 293
Kummer's indentity, 179
Lagrange interpolation formula, 321
Lambert series, 91, 99, 281, 286, 288, 300
Landen transform, 17, 57, 59
in terms of elliptic functions, 58
in terms of theta functions, 57
Laplace transform, 39
Lattice sums, chemical, 288, 290
Ugendre's relation, 24, 26, 27, 49, 152, 153
generalized, 178
theta function proof, 43
Lehmer's means, 232, 235, 263
Lemniscate sine, 5, 28
Length, 358
Lengths and measures, 358
Lindemann's theorem, 348, 357
Liouville numbers, 351, 352
Liouville's function, 100
Liouville.'s principle, 54
Liouville's summation principle, 36
Logarithm:
approximations to, 325, 327, 328, 330
bit complexity, 219, 222
calculation, 220, 221, 222, 223
complex, 222
operational complexity, 219, 222
series for, 320
theta function algorithms for, 224

412
Subject Index
Logarithmic mean, 234, 248, 261
Lucas numbers, 95, 97, 100, 287
Machin's formula, 339, 340, 341, 344
Madelung's constant, 288, 292, 301
Matrix logarithm, 224, 225, 226
Mean, 230, 231
discontinuous, 230, 238
equivalent mean, 231
homogeneous, 231, 266
isotone, 231, 266
strict, 230, 266
symmetric, 231, 266
trace of, 231
Mean iterations, 243
algebraic, 273-280
Archimedean, 246
arithmetic-harmonic, 4
Carlson's, 257
Gaussian, 246
harmonic-geometric, 4
Lehmer's 265
multidimensional, 267
pathology, 253
rates of convergence, 251, 267, 272
Schlomilch's, 268
Tricomi's, 265
Means:
classes:
Gini, 233, 264
Heronian, 237, 269, 270
Holder, 232, 235, 263, 264
identric, 234
Lehmer, 232, 235, 263
logarithmic, 234, 248, 261
Neo-Pythagorean, 255
series expansions of, 263-266
Stolarsky, 233, 236, 264
Measure, of polynomial, 359
Meinardus conjecture, 321
Mellin transform, 87, 90, 289, 290, 301
Modular equations, 103, 315
cubic, 104, 107, 109, 110
degree of, 125
of degree 15, 314
of degree 23, 133
elliptic, 112
endecadic, 106
forj, 123
for K,/(i and K,/4, 315
for X, of orderp, 121, 140
octicity of u-v form, 126, 134
quadratic, 109
quintic, 105, 107, 109, 135, 136, 297, 313
septic, 106, 112, 313
solvable, 310, 311, 313
u-v form, 126, 127-132, 134
Modular functions:
T-modular functions, 114, 118
X-modular function, 114, 118, 121, 133, 134
X-modular group, 113, 117
Modular transformations, 103
Modulus, 8
Multidimensional Archimedean iteration, 267,
269
Multidimensional Gaussian iteration, 267, 268,
270
Multidimensional invariance principle, 269
Multidimensional mean, 266, 269
Multiplier, 103, 105, 106
Multiplier of order p, 136
cubic, 138, 144, 149
of degree 13, 138
of degree 17, 138
quintic, 138, 309
septic, 138, 311
Neo-Pythagorean means, 255
Newton's identities, 356
Newton's method, 212, 214, 215, 216, 217, 218,
223
N-monotone, 291
Nome, 41
Non-prime invariants, 295
Nth convergent, 372
Operational complexity, 201
Order of convergence, 2
Pade approximant, 319, 320, 323, 325, 327, 331,
359
Partially comparable iterates, 247
Partition congruences, 84, 308
Partition function, 67, 307
Partitions of natural number, 67
Pell's equation, 376
Pendulum, period of, 8
Pentagonal number, 66, 67
Perimeter of ellipse, 8, 168, 195-197
Periodic continued fraction, 375
Pi:
AGM identities, 48, 52, 169, 197
algorithms, 46, 48, 170, 171, 175', 222, 310,
315, 335
approximations to, 168, 191, 192, 195, 197
complexity of calculating, 219
computation of, 337-342
cubic iteration, 171, 174

Subject Index
413
general iteration, 169
irrationality, 348, 352
irrationality measure, 367
legislation on, 342
mnemonic for, 342
normality of, 342
quadratic algorithm for exp(ir), 50
quadratic iteration, 170, 174
quartic iteration, 170, 174
quintic iteration, 175,. 313
septic iteration, 171, 174, 175
series for 1/ir, 181-190
transcendence of, 347, 348, 352, 354
Picard's theorem, 119
Piecewise monotone, 234
Pochhammer symbol, 178
Poisson summation formula, 36, 89
Prime number theorem, 378
Primitive binary form, 141
Properly equivalent forms, 141
Q-binomial coefficients, 76
Q-binomial theorem, 308
Quadratically attractive transformation, 277
Quadratic computation, 50
Quadratic modular equation, 102
Quadratic reciprocity, 86
Quintic multipliers, 309
Quintuple-product identity, 143, 146, 306
Ramanujan's continued fraction identity, 81
Ramanujan's invariants, 70, 179
Ramanujan's modular identity:
of order 3, 287
of order 5, 310, 312
of order 7, 312
Ramanujan's ,^, sum, 308
Ramanujan's multiplier of second kind, 157, 159
Rational mean iteration, 278
Recursion, 201
Recursion formula for r2(n), r,(n), 151, 306, 307
Reduced complexity methods, 327
acceleration based on:
binary splitting, 329
on FFT, 328
on functional equations, 327
Reversion of power series, 208
Riemann hypothesis, 90
Rising factorial, 178
Rogers-Ramanujan identities, 65, 78, 80
Bressoud's "easy proof," 78
Roth's theorem, 363
Schlafli's equation, 147
Schonhage-Strassen multiplication, 211
Schroter's formula, 111
Schwarz derivative, 20, 118
Septic multipliers, 309
Simple continued fractions, 372
Singular invariants, 141
Singular moduli, 139
Singular value function, 152
of first kind, 152
generalized, 185
of second kind, 152ff
Singular values, 26,49, 139, 140, 141, 172, 173,
296, 297
Solvability of quintic in modular terms, 135
Solvable modular equations, 310, 311, 313
Solvable numbers:
of type one, 293
of type two, 293
Square-free invariants, 295
Stirling numbers, 305
Stirling's formula, 90
Stolarsky's means, 233, 236, 264
identric, 234, 271
logarithmic, 234, 271
multidimensional, 271
Strict mean, 230
Strict multidimensional mean, 266
Strong equivalence, 241, 243
of iterations, 252
Sums of squares:
of two squares, 82, 285, 290
of three squares, 66, 151, 286
of four squares, 81
of others, 71, 287, 292, 293
Supremum norm, 316
Symmetric mean, 231
Symmetric multidimensional mean, 266
Symmetric polynomial, 356
Tan, computation of, 227
Theta functions, 10, 33, 91, 93
basic identities, 64, 67, 68,70,71, 73,74, 111
finite transformation, 86, 87
general theta functions, 52
one-dimensional heat equation, 56
theta transformation formulae, 38, 44, 54, 87
Theta series, 91
Trace of mean, 231
Transcendence of e, 348, 353, 359
Transcendence of it, 347, 348, 352, 354
Transcendental functions, 274
Transcendental number, 351, 352
Transformation of order p, 119
Transformations of complete elliptic integrals:

414
Subject Index
Transformations of complete elliptic integrals
(Continued)
algebraic transformation for K, 21
of generalized integrals, 179, 180, 185
higher order, 102f
quadratic transformations for E and K, 12, 13
quadratic transformations for K, 36
Triangular numbers, 286, 288
Trilogarithm, 381
Triple-product identity, 62, 65, 66, 72, 306
Tschirnhaus transformation, 136
Turing machines, 201
Vieta's formula, 338, 343
Wallis' formula, 338, 343
Weierstrass function, 27, 30, 141
5(2), irrationality of, 366
5(3), irrationality of, 369
£(3), series for, 190, 379
Zeta function, 87, 88
functional equation for, 89
relation to prime distribution, 90
Ultimately monotone, 235, 241
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