Ramanujan’s Notebooks: Part V [5, eBook ed.] 9781461216247, 9781461272212

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Ramanujan's Notebooks Part V

Springer Science+Business Media, LLC

Bruce C. Berndt

Ramanujan's Notebooks Part

v

Springer

Bruce C. Berndt Department of Mathematics University of Illinois at Urbana-Champaign Urbana, IL 61801-2975 USA

Mathematics Subject Classification (1991): 11-00, 11-03, 01A60, 01A75, 33E05, 33-00,33-03,41-00,41-03, 41A58, 41A60

Library of Congress Cataloging-in-Publication Data (Revised for voI. 4) Ramanujan, Alyangar, Srinivasa, 1887-1920. Ramanujan's notebooks. Includes bibliographies and indexes. 1. Mathematics. 1. Berndt, Bruce c., 1939II. Title. QA3.R33 1985 510 84-20201 ISBN 978-1-4612-7221-2 ISBN 978-1-4612-1624-7 (eBook) DOI 10.1007/978-1-4612-1624-7 Printed on acid-free papcr. © 1998 Springer Science+Business Media New York Originally published by Springer-Verlag New York, Inc. in 1998 Softcover reprint of the hardcover 1st edition 1998

AII rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher Springer Science+Business Media, LLC, except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely byanyone. Production managed by Billlmbornoni; manufacturing supervised by Joc Quatela. Camera-ready copy prepared from the author's AMS-TeX files.

987654321 ISBN 978-1-4612-7221-2

Dedicated to the mathematicians who generously assisted by their many contributions to these five volumes

To the author's knowledge, only three photographs of Ramanlljan are extant. Variations of his passport photo appear in our books, Parts I and IV. A group photo with Ramanujan appearing in cap and gown can be found as the frontispiece of the publication of Ramanujan's lost notebook [11], and has been excised in several cropped versions, often with Ramanujan standing alone. The photograph above is also one of several renditions, the most frequent being one with Ramanujan sitting alone.

Preface

During the years 1903-1914, Ramanujan recorded most of his mathematical discoveries without proofs in notebooks. Although many of his results had already been published by others, most had not. Almost a decade after Ramanujan's death in 1920, G. N. Watson and B. M. Wilson began to edit Ramanujan's notebooks, but, despite devoting over ten years to this project, they never completed their task. An unedited photostat edition of the notebooks was published by the Tata Institute of Fundamental Research in Bombay in 1957. This book is the fifth and final volume devoted to the editing of Ramanujan's notebooks. Parts I-III, published, respectively, in 1985, 1989, and 1991, contain accounts of Chapters 1-21 in the second notebook, a revised enlarged edition of the first. Part IV, published in 1994, contains results from the 100 unorganized pages in the second notebook and the 33 unorganized pages comprising the third notebook. Also examined in Part IV are the 16 organized chapters in the first notebook, which contain very little that is not found in the second notebook. In this fifth volume, we examine the remaining contents from the 133 unorganized pages in the second and third notebooks, and the claims in the 198 unorganized pages of the first notebook that cannot be found in the succeeding notebooks. In contrast to the organized portion of the first notebook, the unorganized material in the first notebook contains several results, particularly about class invariants, singular moduli, and values of theta-functions, which are not recorded in the second and third notebooks. As in the first four volumes, either proofs are provided for claims not previously established in the literature, or citations are given for results already proved in the literature. Urbana. Illinois September. 1997

Bruce C. Berndt

Contents

Preface

ix

Introduction

1

32 Continued Fractions

9

1 The Rogers-Ramanujan Continued Fraction 2 Other q-Continued Fractions 3 Continued Fractions Arising from Products of Gamma Functions 4 Other Continued Fractions 5 General Theorems

33 Ramanujan's Theories of Elliptic Functions to Alternative Bases 1 Introduction 2 Ramanujan's Cubic Transformation, the Borweins' Cubic Theta-Function Identity, and the Inversion Formula 3 The Principles of Triplication and Trimidiation 4 The Eisenstein Series L, M, and N 5 A Hypergeometric Transformation and Associated Transfer Principle 6 More Higher Order Transformations for Hypergeometric Series 7 Modular Equations in the Theory of Signature 3 8 The Inversion of an Analogue of K (k) in Signature 3 9 The Theory for Signature 4 10 Modular Equations in the Theory of Signature 4 11 The Theory for Signature 6 12 An Identity from the First Notebook and Further Hypergeometric Transformations

12 45 50 66 80

89 89 93 101 105 108

116 120 133 145 153 161 165

xii

Contents

13

34

Some Enigmatic Formulas Near the End of the Third Notebook 14 Concluding Remarks

175 180

Class Invariants and Singular Moduli

183

1 2 3 4

Introduction Table of Class Invariants Computation of G n and gn when 91n Kronecker's Limit Formula and General Formulas for Class Invariants 5 Class Invariants Via Kronecker's Limit Formula 6 Class Invariants Via Modular Equations 7 Class Invariants Via Class Field Theory 8 Miscellaneous Results 9 Singular Moduli 10 A Certain Rational Function of Singular Moduli 11 The Modular j-invariant

35 Values of Theta-Functions 0 1 2 3

183 187 204 216 225 243 257 269 277 306 309

323

Introduction Elementary Values Nonelementary Values of ql(e- mr ) A Remarkable Product of Theta-Functions

323 325 327 337

36 Modular Equations and Theta-Function Identities in Notebook 1

353

1 Modular Equations of Degree 3 and Related Theta-Function Identities 2 Modular Equations of Degree 5 and Related Theta-Function Identities 3 Other Modular Equations and Related Theta-Function Identities 4 Identities Involving Lambert Series 5 Identities Involving Eisenstein Series 6 Modular Equations in the Form of Schliifli 7 Modular Equations in the Form of Russell 8 Modular Equations in the Form of Weber 9 Series Transformations Associated with Theta-Functions 10 Miscellaneous Results

354 363 367 373 376 378 385 391 397 403

37 Infinite Series

409

38 Approximations and Asymptotic Expansions

503

Contents

xiii

39 MisceUaneous Results in the First Notebook

565

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

579

References

605

Index

619

Introduction

Knowledge comes, but wisdom lingers. Alfred, Lord Tennyson, Locksley Hall This book constitutes the fifth and final volume of our attempts to establish all the results claimed by the great Indian mathematician Srinivasa Ramanujan in his Notebooks, first published in a photostat edition by the Tata Institute of Fundamental Research in 1957 [9]. Although each of the five volumes contains many deep results, perhaps the average depth in this volume is greater than in the first four. As will be seen in the following paragraphs, several mathematicians made important contributions to the completion of this volume. However, I particularly extend my deepest gratitude to Heng Huat Chan and Liang-Cheng Zhang without whose contributions this volume would have been woefully deficient. This volume, however, should not be regarded as the closing chapter on Ramanujan's notebooks. Instead, it is just the first milestone on our journey to understanding Ramanujan's ideas. Many of the proofs given here and in other volumes certainly do not reflect Ramanujan's motivation, insights, proofs, and wisdom. It is our fervent wish that these volumes will serve as springboards for further investigations by mathematicians intrigued by Ramanujan's remarkable ideas. As in the other four volumes, for each correct claim, we either provide a proof or cite references in the literature where proofs can be found. We emphasize that Ramanujan made extremely few errors, and that most "mistakes" are either minor misprints, or, in fact, they are errors made by the author arising from misinterpretations of Ramanujan's claims, which are occasionally fuzzy. The second notebook is a revised, enlarged edition of the first, and, as with G. N. Watson and B. M. Wilson, who made the first attempts at editing Ramanujan's notebooks, the second was our initial focus. It was therefore quite surprising for us to discover that the unorganized pages of the first notebook contain many beautiful results, especially in the areas of class invariants, singular moduli, and explicit values of theta-functions, that Ramanujan failed to record in his second notebook. The material examined in this volume arises from the unorganized pages in all three notebooks, and we provide now brief descriptions of the contents of each of the eight chapters.

2

Introduction

Ramanujan loved continued fractions, and many of his most beautiful results involve continued fractions. Chapter 32 contains about 70 results on continued fractions scattered among the unorganized pages in his second and third notebooks, and four evaluations of the Rogers-Ramanujan continued fraction from his first notebook. Several modular equations for the Rogers-Ramanujan continued fraction R(q) can be found; in less technical language these are functional equations relating R (q) at two different arguments. Other q-continued fractions were also examined by Ramanujan in these unorganized pages. Several results arise from Ramanujan's beautiful continued fractions for quotients of gamma functions found in Chapter 12 of his second notebook. The present chapter primarily constitutes a reorganized and partially rewritten version of the memoir published by G. E. Andrews, the author, L. Jacobsen, and R. L. Lamphere [2]; a preview and discussion of some of the results was published by the four of us in [1]. The four evaluations of the Rogers-Ramanujan continued fraction from the first notebook first appeared in a paper with H. H. Chan [1], and in Chan's doctoral thesis [2]. In the classical theory of elliptic functions, the ordinary hypergeometric function 2F, 1; x) plays a very important role. In his famous paper [3], in the course of stating without proofs some remarkable series representations for 1/7f, Ramanujan remarked that several of his series arose from alternative theories of elliptic functions wherein the aforementioned hypergeometric function is replaced by either 2F, (~, ~; 1; x), 2F, ~; 1; x), or 2F, (~, ~; 1; x). These theories were never developed, but the first six pages in the unorganized section at the end of his second notebook are devoted to these theories. This is the content of Chapter 33, most of which was first published in a paper with S. Bhargava and F. G. Garvan [1]. A few results from the first notebook have been added to this presentation. The first of the three alternative theories is the most interesting and the most important, and we feel that a large body of work remains to be discovered here. Like Chapter 33, it took several years for us to satisfactorily examine all of the material in Chapter 34, which is devoted to class invariants and singular moduli. Most of this work has appeared in papers with Chan and L.-C. Zhang [1]-[3], [5] and with Chan [3], [4]. A summary, written with Chan and Zhang appears in [6], and several results were established in Chan's Ph.D. thesis [2]. Ramanujan did a prodigious amount of work in calculating over 100 class invariants. For reasons that are unclear to us, he failed to record many of these values in his second notebook. To establish most of Ramanujan's hitherto unproved class invariants, we had to develop methods that were completely unknown to Ramanujan. Thus, Ramanujan's methods and insights into class invariants remain largely a mystery to us. It is also puzzling to us that, except for four values, Ramanujan did not record in his second notebook the more than 30 representations for singular moduli found in his first notebook. For even n, Ramanujan left us a beautiful formula to aid in the calculation of singular moduli, although we are uncertain how he found it, but for odd n, Ramanujan's methods are unknown to us. The values of the classical theta-function E:-oo e- 21rk2 .jii are beautiful algebraic numbers when n is a positive rational number. Chapter 35 is devoted

1, and so we deduce that

Q=

0

T < 1. But clearly

J5 ( 3 + 0 + ~1O + 60) .

(7.14)

By (7.10) and (7.14), it remains to show that

3+0

+ J 10 + 60

5 1/ 4 + 1 5 1/ 4 - 1 .

2

(7.15)

However,

3 + J5 + ) 10 + 6J5

= 3 + J5 + 5 1/ 4 ) 2J5 + 6 = 3 + J5

+ 51/ 4 ( J5 + 1) (3 + 5 1/ 4 + 0 + 5 3/ 4 )(5 1/ 4 -

2(5 1/ 4 + 1) - 5 1/ 4 - 1 '

5 1/ 4 -

1)

1

and thus (7.15) has been shown to complete the proof. Ramanathan [3] gave a more difficult proof of Entry 7 in which class invariants were employed. We have also discovered a proof of Entry 7 that utilizes class invariants. Since our proof is simpler than that of Ramanathan and much different from our proof above, we give it below. Like Ramanathan's proof, our proof requires the value of G25, and so we give a simple derivation of this evaluation next.

Lemma 7.1. G 25

=

1 +0 2

.

Proof. We employ a modular equation of degree 5 found in Entry 13(xiv) of Chapter 19 of Ramanujan's second notebook (Part III [3, p. 282]). Let P

= 21/3{a,B(l

- a)(1 - ,8)}I/IZ

and Q

=

(,8(1 - ,8))1/8, a(l - a)

32. Continued Fractions

23

where f3 has degree 5 over ex. Then

Q+

~ + 2 (p - ~) = O.

(7.16)

Recalling the definition of G n in (7.5), recalling that (Part III [3, p. 37]) X(q) := (-q; q2)ao, and using Entry 12(v) of Chapter 17 of Ramanujan's second notebook (Part III [3, p. 124]), we find that

=

Since it is well known and easy to prove that G 1 reasoning as above,

1, we have, by the same

Hence, it follows that 1 P=-2

G25

Therefore, by (7.16), if x

=

G 25 ,

and

r-

:3 + x3+ 2 (x12 - x2) = (~ + x =

Q

3

=

1

-3 . 25

G

(~ + x) + 2 (~ + x) (~ - x)

(~ + x) {(~ -x)

= (~+x)

2

+ 2 (~-

{(~ -x) + 1

r

x) + 11

=0. Since x

+ 1/x =1= 0 and x

> 0, we conclude that G25 =X

1 + y'5 = --2

and so the proof is complete. We remark that the value of G 25 is given without proof in Ramanujan's paper [3], [10, p. 26].

Second Proof of Entry 7. Recall from our first proof that it suffices to prove (7.10). Set ex = 2rr/5 in (7.3), so that f3 = 5rr/2. After some simplification, we find that

24

Ramanujan's Notebooks, Part V

Ii

Thus,

Q = e4rr / 5 f( -e f( _e- 20rr )

-4rr/S)

=

2 e5rr / 8 2

f( -Srr) -e f( _e- 20rr )·

(7.17)

Since (q;q)oo

2

24

( q 4.,q 4) 00 = (q; q )oo(q ; q

)00'

we deduce from (7.17) that (7.18) where gn is defined in (7.5). We thus must determine g25 and g\Oo. Since G25 was computed in Lemma 7.1, we see from (7.6) and (7.7) that Q can be calculated. For brevity, set x = g~s and a = G~5. Thus, from (7.7), ax 2

a 2x

-

+~

(7.19)

= 0,

Since x > 0, from (7.19), we deduce that a2

+ .../a 4 -

a

x=-----

2a

=

~ (G~5 + G~S!G~; -

G2S12)

=

~G~5 ( ! G~5 + G 25

JG~5

6

By Lemma 7.1, G~5 = (2 + ,,'5)2 by (7.20) and Lemma 7.1,

x

+

- G 2s6 )

2

(7.20)

= 9 + 4.../5, and so G 256 = 9 - 4.../5. Hence,

~ gl, ~ ~ C+2..15)' (,ff8 + /8..15)' =

~ C+2..15)' (3 + 2 . 5'1')' .

Thus, from (7.18), (7.21), and Lemma 7.1,

Q = .J521/4g~5G2S 3/2 =.J5 ( 1 +2.../5) S (3 +

2. SI/4)1/2

=.J5 .Ji (1 + .J5)1/2(3 + 2 . S1/ 4 )1/2 .../5 - 1 {(l + SI/4)4}1/2 = .J5 -'-'----=---'-""'--.../5 - 1 SI/4

+1

= .J5 SI/4 _

1.

(7.21)

32. Continued Fractions

25

Thus, (7.10) is established, and the proof is complete.

Proof of Entry 8. We will again employ Entry 58 of Chapter 25 in Part IV [4, p. 212-213], but now with

_ 81r /5 I (_e- 81r /5) Q -e .:........:...-~...:....

and

I( -e-407r )

By the same reasoning as that used in the first proof of (7.2), in order to prove (8.1), it suffices to prove that Q = a +b.../5, a-b

(8.2)

where a and b are given in the statement of Entry 8. Write (7.11) in the fonn PQ+5 = From Entry 7, P = (5 1/ 4 Q = v'ST, we find that

p2

Q

-2P -2Q+

+ 1)v'S/(5 1/ 4 -

Q2

p.

(8.3)

I). Putting this in (8.3) and setting

5 1/ 4 + 1 _ (5 1/ 4 + 1)2 1 5 1/ 4 + 1 5 1/ 4 - 1 2 51/4 _ 1 ../5T +../5 51/4 _ 1 T - 2 5 1/ 4 _ 1 - 2T + 5 1/4 + 1 T . (8.4) By an elementary verification, it is easily checked that T = 1 is a root of (8.4). Since clearly Q > v'S, this root is not the one that we seek. Writing (8.4) in the fonn a3T3 + a2T2 + a1 T + ao = 0, and dividing by T - 1, we find that (5 1/ 4 _1)2T2 -2(1 +5 1/ 4+../5+5 3/ 4 )T -(9+6.5 1/ 4+3../5+2.5 3/ 4) = O. (8.5) Now set T=a+b a-b in (8.5) to deduce, with the help of Mathematica, that

-(5 + 3../5)a 2 + (6 . 5 1/ 4 + 2 . 53 / 4 )ab + (3 - 3../5)b2 = O. Solving for a. we find that a

=

5 1/ 4 b ± J(7v'S - 15)b2 -------'-~---

v'S

We now set b = 5 1/ 4 ,J2 and choose the plus sign above, because if we had chosen the minus sign, we would find that T < 0, which is impossible. Hence, a=

v'2 + J70 -

= v'2 +

30v'S v'S

J14- 6../5

= v'2 +

J(3 - ../5)2 = v'2 + 3 -../5.

26

Ramanujan's Notebooks, Part V

Hence,

Q = T.../5 = a + b.../5 a- b

= "Ji + 3 "Ji + 3 -

./5 + SI/4"Ji.../5, ./5 - S1/ 4"Ji

and so (8.2) has been shown to complete the proof. Proof of Entry 9. We again employ Entry 58 of Chapter 25 in Part IV, but now we set f( -16rr/5) and Q = e16rr/5 -e . f(-e- 80rr ) By the same argument that we used in the proofs of Entries 7 and 8, to prove (9.1), it suffices to prove that Q

= a +b.../5,

(9.2)

a-b

where a and b are prescribed in the statement of Entry 9. Set A = 3 + "Ji - ./5 and B = (20)1/4. As in the last proof, let Q Thus, by Entry 8 and (8.3), we know that

A+B --.../5T +.../5 A-B

= (A+B)2 - - -1 A-B

T

A+B 2 - - - 2T A-B

= ./5T.

A-B 2 +-T. A+B

(9.3)

Let

a+b a-b

T=--

in (9.3). Clearing fractions and simplifying with the help of Mathematica, we find that

(-10 - 7"Ji + 4.../5 + 20o)a 3

+ (-20 -

+ 5 1/\8 +

IS../2 + 6.../5 + 40o)ab 2

+ 5 1/ 4(10 + 15../2 - 4.../5 - 6.JiO)b 3 Let a

9../2 - 2.../5 - 20o)a 2 b

= o.

= 51/ 4 d, cancel 53/ 4"Ji, and simplify to deduce that (-7 - s../2 + 2.../5 + 20o)d 3 + (9 + if2 - 2.../5 + (4 + 3v'2 - 3.../5 - 20o)b 2d + (-6 - 2../2 + 3.../5 + 0o)b 3 = o.

0o)bd 2 (9.4)

Observe that d = b is a root of (9.4). If this were the root that we are seeking, then Q would equal (SI/4 + 1)./5/(SI/4 - 1). Thus, with P and Q interchanged, we have the same solutions to (8.3) that we had in the proof of Entry 8. Clearly, this is not the solution that we want. Hence, dividing (9.4) by (d - b), we find that

(4 + 6../2 - 2.../5 - 30o)b2 + 2(-1 + (-10 -7../2 + 4.../5 +

+ ../2 +

20o)d 2 =

O.

.../5)bd

32. Continued Fractions

27

Solving for b, we find that

2(1 -./2 - ..;5)d ± 2./2J(116 + 83./2 - S2..;5 - 37.JiQ)d 2

b=

2(4+ 6./2 - 2..;5 - 3v'W)

(9.S)

Since

2(1 - ./2 - ..;5)

(1 - ./2 - ..;5)

2(4 + 6./2 - 2..;5 - 3v'W)

(2 + 3./2)(2 - ",Is) (6 + s./2 + 4..;5 + v'W) 14 (1 +./2 + ..;5)(4 + ./2) 14

we are motivated to set d = 4 - ./2. Therefore, a with what Ramanujan claimed. Thus, by (9.S), b=

= SI/4(4 -- ./2) in agreement

6 - s./2 - 4..;5 +.JiQ ± 25 / 4 J283 + 190./2 - 12S..;5 - 86v'W 4 + 6./2 - 2..;5 - 3v'W

.

(9.6)

Observe that

6 - s./2 - 4-vfs + v'W r;:; c ./2 = 1 + v2 + v5. 4 + 6 2 - 2..;5 - 3.JiQ

(9.7)

We next wish to write

4(283 + 190./2 - 12S.J5 - 86M) = (w + x./2 +

y.J5 + ZM)2,

for certain rational integers w, x, y, and z. Thus, w2

+ 2x 2 + S/ + lO z2 = 1132,

(9.8)

wx + SyZ = 380, wy + 2xz = -2S0,

wz +xy = -172. David Bradley kindly wrote a program to determine the 24 positive solutions to (9.8). We then found the unique solution of the system of four diophantine equations to be w = 20,

x = 9,

y

= -8,

and

z = ·-S.

Thus,

2J283

+ 190./2 -

12S..;5 - 86.JiQ

4 + 6./2 - 2..;5 - 3v'W

20 + 9./2 - 8..;5 - sv'W 4 + 6./2 - 2..;5 - 3.JiQ = 3 - ./2 +

";i -

M.

(9.9)

28

Ramanujan's Notebooks, Part V

Putting (9.7) and (9.9) in (9.6), we find that b = 1+ h

+.J5 ± 21/4(3 - h +.J5 - ",'10).

If we choose the plus sign above, we would find that a - b < 0 and T < 0, which is impossible. Thus, we conclude that b

= 1 + h +.J5 -

21/4(3 -

h +.J5 -

M),

which is what Ramanujan asserted. Thus, (9.2) is proved, and the proof of Entry 9 is complete. Our proof of Entry 9 heavily relies on computation in the later stages. Although Ramanujan possibly used Entry 58 of Chapter 25, he undoubtedly found a less computational proof. Proof of Entry 10. By the same reasoning in the proofs of Entries 7-9, to prove (10.1), it suffices to prove that e6rr/5

f( _e- 6rr/5 ) = f( _e- 30rr )

+ b .J5,

a a -

(10.2)

b

where a and b are specified in the statement of Entry 10. Apply the transformation formula (7.3) with a = 3Jf /5 and f3 some simplification,

=

5Jf /3. After

f( -e -lOrr / 3) =: ~ f( -e -6rr/5) = ~ ~elOrr/9 ~A. 30rr 3 f( _r ) f( _e- 30rr ) 3

(10.3)

Thus, e6rr/5

Because 30 = 9 . ~, we are led to Ramanujan's cubic continued fraction ql /3 G(q):= -1-

+

q+q2

+

q2+q4 1

+

q3+ q 6

+ ... '

Iql

< 1.

From Part III [3, p. 345, Entry l(i)], G(- ) - _

q -

q

1/3

x(q)

X 3(q3)'

where

In particular, (10.4)

32. Continued Fractions

29

by (7.S). Recalling the value G 25 = (1 + .[5)/2 from Lemma 7.1 and the value Gm =

~(1 + .J5)(2 + ,J3)1/3 {J 4 +

J'i5 + (1S)I/4}

from Ramanujan's paper [3], [10, p. 28], or from the table of Section 2 of Chapter 34, but apparently first proved in print by Watson [7], we find that, from (10.4), G(_e- 5rr ) = _

16.../2 (1 + .J5)2(2 + ,J3) 4 + .JT5 + (IS) 1/4}

{J

2.../2(2 - ,J3)(3 (4 + J'i5)3/2 + 3(4 + .JT5)(lS) 1/4 + (2 - ,J3)(3 - - .../2 (../4 +.JT5(1 Now

+.JT5) +

.J5)

3../4+.JI5.JT5 +

(lS)3/4

.J5) (1S)I/4(3 +.JT5») .

j 4 + J'i5(l + J'i5) = J4 + J'i5(1 + J'i5)2 J 94 + 24J'i5 = .../2J ~ (6--/3' + 4.J5)2

=

=

1

.../2(6,J3 + 4.J5).

Thus, G(_e- 5rr ) = _

(2 - ,J3)(3 - .[5) 6,J3 + 40 + (60)1/4(3 +.[5) (2 - ,J3)(3 - 0)(6,J3 +

4./5 -

3(60)1/4 -

8

.JT5(60)1/4) (lO.S)

Now Chan [1, Theorem 1] has shown that G(q) satisfies the modular equation G 2(q)

Replacing

+ 2G 2(q2)G(q) -

G(q2)

= O.

qby -q and solving for G(q2), we find that G( 2) _ 1 -../1 - 8G3(_q) q

-

4G(-q)

.

(10.6)

Set q = e- 5rr , as above, and v = G(e- lOrr ). Recall the definition of A from (10.3). Then Entry l(iv) of Chapter 20 in Ramanujan's second notebook (Part III [3, p. 34S]) can be written in the form (10.7)

Ramanujan's Notebooks, Part V

30

Thus, by (10.7) and (10.6), with w := G( _e- 57f ) given by (10.5),

A 3 = 4 (1- Jl- SW 4w

3)2 +

4w _ 3 1 - .jl - Sw 3

= _ (w + 1)2(2w -1) w2

'

(10.8) by a somewhat lengthy, but straightforward, calculation. Hence, by (10.2), (10.3), and (10.S), it remains to show that - 2W»)1/3 = a + b r,; ( W + 1)2(1 w2 a- bY 3,

(10.9)

where a and b are specified in the statement of Entry 10. We used Mathematica to verify (10.9) and complete the proof. Another proof of Entry 10 has been given by the author, Chan, and Zhang in [3]. The next entry is actually recorded twice by Ramanujan in his notebooks. We quote Ramanujan. Entry 11 (pp. 374, 382). If q > 1, -

q

q2

q3

--

(11.1)

1+1+1+1+'"

oscillates between (11.2)

and q-4

q-l

q-8

q-12

-1-+-1-+-1-+-1-+ .. ··

(11.3)

From the general theory of continued fractions, if all the elements of a divergent continued fraction are positive, then the even and odd approximants approach distinct limits. Thus, since (11.1) diverges for q > 1, Ramanujan is indicating that its odd approximants tend to (11.2) while its even approximants apprQach (11.3). In fact, we shall prove Entry 11 for Iq I > 1. First Proof. Recall the definition of the Gaussian binomial coefficient

[nJk [nJk :=

q '-

(q; q)n (q; qh(q; q)n-k'

where Iql < 1 and nand k are integers with 0 ::S k ::S n. Define

cn(a; q)

aq

aq2

aqn

= - + 1 + -1- + ... + -1-'

32. Continued Fractions

31

Then by a result of Ramanujan (Part III [3, p. 31, Entry 16]), Dn(aq; q) Dn+l(a;q)'

(11.4)

c (a' q) - - - - n

,

-

where

"~ [n -. j] a1qJ. ..,

Dn(a; q) =

J

O~2j~n

We will need two further results of Ramanujan (Part III [3, p. 77, Entries 38(i), (ii))), viz., 00

1

qn'

L - - = (q; q )oo(q ; q )00 n=O (q; q)n

G(q) :=

5

4

5

3

5

(11.5)

and H(q) :=

00 qn'+n

L -- = n=O (q; q)n

1 2

5

(q; q )oo(q ; q )"0

•

(11.6)

These are the famous Rogers-Ramanujan identities first established by Rogers [1]. Lastly, we require the following identities due to Rogers [1] and Watson [11], [14] independently: 00

L

n' q n=O (q; q)m

00

qn'+n

~ (q;qhn+1 00

qn'+n

~ (q;qhn

1 (q; q2)00

G(q4),

(11.7)

=

1 H(-q), (q; q2)00

(11.8)

=

1 G(-q), (q; q2)00

(11.9)

I H(q4). (q; q2)00

(11.10)

and 00

qn'+2n

~ (q;qhn+1

It will be convenient to replace q by l/q in Entry 11. Letting c(q) denote the infinite continued fraction limn~oo C n (1; q), we may rephrase Entry 11 in the following way. For 0 < q < 1,

the odd approximants of C(q-I) tend to l/c(-q),

(11.11)

the even approximants of C(q-I) tend to qc(q4).

(11.12)

while

32

Ramanujan's Notebooks, Part V

We first examine the odd approximants. Using (11.4), replacing j by n - j, and utilizing the fact = [AJ B q-t

[AJ '

q-B(A-B)

(11.13)

B q

we find that C2n (1 ;

q

-I)

D 2n (-I q ; q -I)

=

D2n+I(1;q-l) n

"

[n+/]

!--

2J

J=O n

"

q-I q

-(n- j)2_(n_ j)

[n+j+l]

!-- 2j+1 q-tq

-(n-j)2

J=O

Hence, by (11.8) and (11.9), 00

1

lim C2n(l; q- )

n-+oo

qj2+j

= j~ (q;qJ,2qhj ' +J 00

G(-q) 1 =---=--, H(-q)

j~ (q;qhj+1

c(-q)

where in the last step we employed (11.5), (11.6), and the Rogers-Ramanujan continued fraction (Part III [3, p. 79, Entry 38(iii)]). This establishes (11.11). We next examine the even approximants. Employing (11.4), reversing the order of summation in both the numerator and denominator, and applying (11.13), we find that

L

n-I

[n+j] _(n_I_j)2_(n_l_j) , 2j+1 q-tq )=0

L

n-I

J'--O

[2n++jl]qj2+2j J

=------n---,- - - - - - = q=-----n--,--

L

, J=O

[n+.1] 2J

q-t

L

q-(n-j)2

J'--O

[n;.1]qP J

Thus, by (11.7) and (11.10), 00

.

hm C2n-l(q

n-+oo

_I

)

=q

qj2+2j

j~ (q; qhj+1 00

,2 qJ

j~ (q; qhj

H(q4)

= q G(q4) = qC(q

4 ),

32. Continued Fractions

33

where we used (11.5), 01.6), and the Rogers-Ramanujan continued fraction. Thus, (11.12) is proved. Entry 11 is a truly amazing result. It is very remarkable that the RogersRamanujan continued fraction reappears in determining the limits of both the even and odd approximants of the divergent Rogers-Ramanujan continued fraction. It also should be noted that the continued fractions (11.2) and (11.3) are "near" each other in the sense of Entry 13 below. More precisely, if we set x = 1 in (13.1) and (13.2), we obtain (11.2) and (11.3), respectively, but with q replaced by Ijq. We will now give a second proof of Entry 11. This proof shows that Entry 11 arises from infinitely many Bauer-Muir transformations. Second Proof. The even part of (11.1) is given by (see (64.1» q3

1 + q - 1 + q2

q7

+ q3

- 1 + q4

+ q5

1 + q6 +q7 - ...

The latter continued fraction is a limit I-periodic continued fraction for Iq I > 1. Since the linear fractional transformation t(W) = _

q-l 1 +q-l +W

has the two fixed points _q-l and -1, it follows that(11.14) converges for Iql > 1. Moreover, the modified approximants

also converge to the same value if {w n } does not have a limit point at -1. (For instance, see Jacobsen's paper [1].) Applying the Bauer-Muir transformation (13.7) to the second continued fraction in (11.14), with Wo = 0 and Wn = -1 j q, n ~ 1, we obtain the continued fraction q-l

1

q-4

+ 1 + q-3

q-3

- 1 + q-3

+ q-5

q-3

- 1 + q-3

+ q-7

- ... '

(11.15)

which converges to the same value for Iq I > 1. Again, this is a limit periodic continued fraction. The attractive fixed point of

34

Ramanujan's Notebooks, Part V

is _q-3. Hence, with Wo = WI = 0 and Wn = _q-3 for n :::: 2, the Bauer-Muir transformation of (11.15) is given by q-I

q-4

q-8

+ -1- + q + q-5

-1-

q-5

- 1 + q-5

+ q-7

q-5

- 1 + q-5 + q-9 - ... '

which converges to the same value. Repeating this process k times with Wo =

WI

= ... = Wk-I = 0

n:::: k,

Wn = _q-2k+I,

and

(11.16)

we obtain the limit periodic continued fraction q-I 1

+

q-4

-

1

+

q-8

-

1 + ... q-2k-1

1 + q-2k-1

+

+ q-2k-5

q-4k

q-2k-1

1 +q-2k-1 - 1 +q-2k-1 +q-2k-3

(11.17)

- ... '

which converges to the same value as (11.14). Repeating this process infinitely many times, we obtain (11.3). Since (11.17) converges uniformly with respect to k by the uniform parabola theorem (Jones and Thron [I, p. 99]), and since (11.3) converges, we conclude that (11.3) converges to the same value as (11.14). We have thus proved that the even part of (11.1) converges to the value of (11.3) for Iql > 1. To prove that the odd part of (11.1) converges to the value of (11.2) for Iq I > I, we can use the same idea, and even the same choices (11.16) for W n • We then find that the odd part (Jones and Thron [1, p. 43, eq. (2.4.29)], where the first minus sign is misplaced) of (11.1) equals q q5 q9 1 + q + q2 - 1 + q3 + q4 - 1 + q5 + q6 - ...

1_

= 1_

=

1-

q

-I

q

-I

1 + q-2

= 1-

-

q-I q-I q-I 1 +q-I +q-2 - 1 +q-I +q-4 - 1 +q-I +q-6 - ...

1 + q-2 q-5

q

-5

3

3

+

qq1 + q-4 - 1 + q-3 + q-6 - 1 + q-3 + q-8 - ...

+

q 1 - q-3

-5

+ q-4 +

q

-9

5

q1 + q-6 - 1 + q-5 + q-8

1 +q-5 +q-1O - ...

= 1_

q-I 1 + q-2

+

q-5 1 _ q-3 + q-4

+

q-9 1 _ q-5 + q-6

+

q-13 1 _ q-7 + q-8

+ ...

The last continued fraction is the even part of (11.2). Since (11.2) converges for Iq I > I, the second proof of Entry II is complete by the same argument as above.

32. Continued Fractions

35

K. Alladi [1] has given another proof of Entry 11 that is similar to our first proof. However, he related his proof to the continued fraction III

r(q) := q + q3 + q5 + q7 + .... To be more precise, let Pn(q) and Qn(q) denote the nth numerator and denominator, respectively, of the nth convergent of r(q). Let T(q) := q-l/5 R(q). Then Alladi [1, pp. 225-229] proved that, for iqi < 1, lim P2n - 1(q) n-+oo Q2n-l(q)

= G(qI6) = T(qI6) q3H(qI6)

q3

and lim P2n(q) Q2n(q)

= qH(_q4) = G( _q4)

n->oo

q . T( _q4)

Moreover, in the sense of modified convergence, Alladi proved that r(q) tends to T(q), i.e., lim 1 +

n-+oo

1

1

1

1

-q + -q3 + -q5 + ... + - + q2n+l + 1 = q2n-l

T(q).

For the definition, importance, and historical background of modified convergence, see Jacobsen's paper [2]. For the last entry on continued fractions found in the third notebook, we quote Ramanujan. Entry 12 (p. 383). If ql/5

u:=-

q

q2

-

q3

-

-

1 +1+1+1+'"

,

then u 2 + u - 1 = 0 when qn = 1, where n is any positive integer except multiples of 5 in which case u is not definite. This statement is not quite correct. However, I. Schur [1], [2, pp. 117-136] has established the following theorem. Theorem 12.1. Let K(q)

=

q2

q

1+

'1

+

T

q3

+

T

+ ... '

where q is a primitive nth root of unity. If n is a mUltiple of 5, K (q) diverges. When n is not a mUltiple of5, let A = (~), the Legendre symbol. Furthermore, let p denote the least positive residue ofn modulo 5. Then,for n =1= 0 (mod 5), K(q) = Aq(l-Apn)/5 K(A).

36

Ramanujan's Notebooks, Part V

Note that it is elementary that K (1) = (J5 + 1)/2 and K (-1) We provide a short table of further values of K (q ). n

K(q)

3

_q2V;-1

4

qV;+!

6

q5

7

_q3

= (J5 -

1)/2.

V;+l

°2-1

Ramanujan's lost notebook [11, p. 57] contains some claims on finite, generalized Rogers-Ramanujan continued fractions, and these results have recently been proved by S.-S. Huang [1]. Perhaps the main result gives a formula for evaluating certain finite generalized Rogers-Ramanujan continued fractions at primitive roots of unity x. At the bottom of page 57 is a table of general formulas arranged according to residue classes of n modulo 5, when x is a primitive nth root of unity. However, the table contains some errors. When this table is used in Ramanujan's primary formula, specialized to the ordinary Rogers-Ramanujan continued fraction, we obtain Entry 12 as Ramanujan recorded it. This is evidence that these results in the lost notebook were derived before Entry 12. For more details, see Huang's paper [1]. The next result is stated exactly as Ramanujan recorded it.

Entry 13 (Formula (4), p. 289). qx

q2

q3 x

q4

q5 x

1-1+1--1-+1--1-+ ... q q4 q8 ql2 = x+x+x+x+"·

I

conventional only

nearly

Both continued fractions converge to meromorphic functions of x in C - {OJ for Iq I < 1. We shall indeed prove that

Iql

< 1,

(13.1)

and q x+x+x+x+···

Iql

< 1.

(13.2)

32. Continued Fractions

37

Let f(x; q) and g(x; q), respectively, denote the values of the continued fractions in (13.1) and (13.2). We shall then prove that both f(x; q) and g(x; q) satisfy the same functional equation F(xq2; q)

=

1

(13.3)

+ F( x;q )' xq

for 0 < Iql < 1. Furthermore, we shall prove that f(q; q) follows from (13.3) that

=

g(q; q). It then

(13.4) for every nonnegative integer n. Lastly, we shall describe some work of Zagier [1] which indicates in what sense f(x; q) and g(x; q) are "nearly" equal. The continued fraction on the left side of (13.1) may be identified by employing a continued fraction found in Ramanujan's "lost notebook" [11] and proved by Andrews [5], M. D. Hirschhorn [2], S. Bhargava and C. Adiga [1], [2], Bhargava, Adiga, and D. D. Somashekara [1], and others. Let F(a b J... ) = 1 aq + J...q , "q + 1

+

bq

+ J...q2 1

+

aq2

+ J...q 3

+

bq2

+ J...q4 1

+ ...

and

Then ' ) _ F( a, b ,1I.,q -

G(a, b, J..., q) G(aq,b,J...q,q)

(13.5)

Setting b = 1 and J... = 0, replacing q by q2, and setting a = -x / q, we establish (13.1). The continued fraction on the right side of Entry 13 is equivalent to q/x

+ .. ,,' 1 + + + Thus, (13.2) follows from the corollary to Entry 15 of Chapter 16 (Part III [3, p. 30]). This corollary also appears in Ramanujan's [10, p. xxviii] second letter to Hardy. To prove (13.3) for F(x; q) = f(x; q), we shall use the Bauer-Muir transformation found in Perron's book [1, p. 47]. Briefly, a Bauer-Muir transformation of a continued fraction bo + K(an/bn) is a (new) continued fraction whose approximants have the values Sk(Wk)

al

:= bo + -b 1

a2

ak

+ b2 + .•. + bk + Wk '

k = 0,1,2, ....

Such a transformation exists if (13.6)

38

Ramanujan's Notebooks, Part V

and it is given by

bo + Wo

AI

alAliA1

+ b 1 + WI +

b 2 + W2

-

a2A3/A2

WOA2/AI

+ b3 + W3

WI A 3/ A 2

-

+ ....

(13.7) Suppose that bo + K(an/b n ) converges to a value I E C. Let to i- I be arbitrarily chosen from and define tk = S;:I (to), k ::: 1. Then Sk(Wk) tends to I as well, unless the chordal distance d(wk, tk) has a limit point at O. Hence, the BauerMuir transformation (13.7) converges to I as well, if limk->oo inf d(Wb td > 0 (Jacobsen [5]). We shall apply this to the odd part

C,

= I(x;q)

(13.8)

of (13.1). Let

so that (13.8) can be written as -1 +bo+K(an/bn ) = I(x; q). Let Wk = _q2k for each kEN. Then the modified approximants -1 + SkC wd of (13.8) also converge to I(x; q) by the following argument. The continued fraction -1 +bo+K(an/bn ) is limit periodic since limn-> 00 an = 0 and limn->oo b" = 1. It is then well known (see, for instance, lacobsen's paper [3]) that -1 + bo + K(an/bn ) converges to a value q; and that tk tends to -1 for every tail sequence of -1 + bo + K (an / bn ) with to i- q;. In particular, this implies that limk->oo inf d(wk, tk) = d(O, -1) i- 0, so that limk-+oo Sk(Wk) = I(x; q). Since (13.1) obviously holds for q = 0, we may assume that q i- o. A simple calculation shows that, for k ::: 0,

so that the Bauer-Muir transformation exists, converges to I(x; q), and is given by I(x; q) = -qx

+ 1-

q5 x

1

q3 x

+ 1 + q2(1 -

3

q x)

+ 1+q

Comparing (13.8) and (13.9), we find that I(x; q) which proves (13.3) for F(x; q)

= =

-qx

+ l(q 2x; q)'

I(x; q).

q9 x 4

3

(1 - q x)

+ .,. .

(13.9)

32. Continued Fractions

39

To prove (13.3) for F(x; q) = g(x; q), we just observe that replacing x by q 2x in (13.2) yields g(q2x;q)

= _q

_q4

_q8

+ q 2x + q 2x + .. .

q 2x

x

+x +

q4 x

+

q8 x

+ ...

1

l/q

= x+~g(x;q) =

l/q

xq+g(x;q)'

which is what we wanted to show. For x = q, the continued fractions in Entry 13 reduce to, respectively, q2 f(q;q) = 1- -1

q2

q4

q4

+ -1 - -1 + -1

- .. ,

(13.10)

and I

g(q; q) =

q2

q6

qlO

1 + T + T + -1 + ....

(13.11)

Thus, (13.11) is the even part of (13.10), and so f(q; q) = g(q; q). In order to examine how "close" the continued fractions f(x; q) and g(x; q) are to each other, by (13.1) and (13.2), we are led to examine

Quite remarkably, Ramanujan stated an identity for F(x; q) in his lost notebook [11], namely 00 I:

F(x; q)

(_x)nqn

= n=-004

4

(q ; q )00

2

(q2; q2)00(qx; q2)00(q/x; q2)00 (q4; q4)00

(13.12)

where the last equality is obtained from an application of the Jacobi triple product identity. The identity (13.12) was proved by Andrews in [6, pp. 25-32] and is mentioned by him in his Introduction to the lost notebook [11, p. xxi, eq. (10.6)]. Zagier [1] independently also established (13.12). From (13.12), it is obvious that F(q2n+l; q) = 0 for each nonnegative integer n. We thus have obtained a second proof of (13.4). Of course, the second proof is shorter than the first, but the first proof is more elementary than the second, because (13.12) is somewhat difficult to prove. If x is not an odd power of q, in what sense are f(x; q) and g(x; q) near each other? H. Cohen performed extensive calculations to answer this question, and Zagier [1] established Cohen's conjectures as well as much more. We give a brief summary of some of Zagier's results. We always assume here that 0 < q < 1 and x> O.

Ramanujan's Notebooks, Part V

40

and g(1; q) = ql/sv's2- 1

(1-

v'sQ + 5 -2v's Q2 + 5 - ;v's Q3 _ .. -).

In particular, as q tends to 1-, 1(1; q) - g(1; q)

=

(5 - .J5)ql/SQ

+ 0(Q2).

(13.13)

However, as x tends to 0, I(x; q) - g(x; q) = 0 (exp

Note that the asymptotic behaviors for x In general, Zagier [1] has shown that

=

I(x; q) - g(x; q) = exp (

(~~:)).

(13.14)

1 and x near 0 are different. C(X) + 0(1)) cose logq

as q tends to 1-, where e = (Jr logx)/(210gq) and C(x) = :2

+

+~Li2((JI+X2/4-X/2r) ~ log2 (JI +x2/4+X/2) -logxlog (Jl +X2/4+X/2),

where Li2 (t) denotes the dilogarithm Li2 (t)

=

I:>n /n2, 00

It I

:s

=

Jr2/I5 - log2

1.

n=1

Since Liz(I)

=

Jr2/6 and Li2 (3 - v's)/2)

(0 + v's)/2)

(Lewin [1, p. 7]), we find that c(O) = Jr2/4 and c(1) = Jr2/5, in agreement with (13.13) and (13.14). Zagier's analysis shows that, for instance, I(x; 0.99) and g(x; 0.99) agree to about 85 decimal places if x is near 1, about 96 places if x is near ~, and about 107 places if x is close to O. The function c(x) becomes negative for x larger than about 6.177. Thus, for x larger than this, I (x; q) - g (x; q) becomes exponentially large as q tends to 1-. We do not know the meaning of the words "conventional only" in Entry 13. For Entry 14, we again quote Ramanujan.

32. Continued Fractions

41

Entry 14 (Formula (2), p. 290). 1_

...!!.!..1+q

q 2x

q3x

+ 1 + q2

-

q6x

1 + q3

+ 1 + q4

q

1

q3x

q9x

1 + q5 -I- 1 + q6 - ...

-

nearly.

x+x+x+x+'"

The analysis for Entry 14 is very similar to that for Entry 13. Both continued fractions converge to meromorphic functions of x in C - to} for Iq I < 1. We shall prove that 1_

...!!.!..1+q

q 2x

q3x

+ 1 + q2

-

q6x

1 + q3

+ 1 + q4

q3x

-

1 + q5

q9x

+ 1 + q6

- ...

(14.1)

Iql

< 1,

and 00

x+x+x+x+'"

-

x- 2n q 2n 2 +n

L (2. 1 n=O q, q 2) n

-

X

00

n~o

2

x-2nq2n -n

Iql

,

< 1.

(14.2)

(q2; q2)n

Let f(x; q) and g(x; q), respectively, denote the continued fractions on the left sides of (14.1) and (14.2). We shall prove that f(qn; q) = g(qn; q), for every nonnegative integer n, by invoking the same theorem from Ramanujan's lost notebook that we used in Section 13. Thus, we shall easily see that the "closeness" of f(x; q) and g(x; q) can be determined merely by changing the variables in the analysis of Section 13. Proof. For Iq I < 1 and each nonnegative integer m define 00

P2m(X) = ' "

(_qm x )n q (n 2 +n)/2

00

(_qm+l x )n q (n 2 +n)/2

and P2m+1(x) = ' " . ~ (_q2m; q)n(q; q)n ~ (_q2m+l; q)n(q; q)n

By straightforward calculations, we then find that qm+lx P2m (x) - P2m+1(X) = - (1 +q2m)(l + q2m+l) P2m+2(X)

and

42

Ramanujan's Notebooks, Part V

Since Pm (0) = 1 for each mEN, we thus deduce that qx

Po(x) = 1 _ (1 PI (x)

+

(1

= 1-

+ 1)(1 + q)

+

1

+ q3)(1 + q4) 1

1 {qX 1+q

2"

+ q)(1 + q2)

(1

1

+ ... q 2x

q3 x

+ 1 + q2

-

1 + q3

q6 x

+ 1 + q4

Hence, the first continued fraction in Entry 14 converges for

Iql

= -1 + 2 Po (x) = -1 + 2n~ (q; q)n(-~; q)n PI (x)

00

= -1

+ n~

00

00

n~

(_qx)nq(n +n)/2

n~ (q; q)n(-q; q)n

(_x)n q (n 2 +n)/2(1

+ qn)

(q2; q:)n (_x)nq(n +3n)/2 (q2; q2)n

which immediately proves (14.1). By the corollary of Entry 15 in Chapter 16 (Part III [3, p. 30]),

Iql < 1, and so the proof of (14.2) is completed. From (14.1) and (14.2), f(x; q) - g(x; q)

.

(14.3) < 1 to the value

(_x)n q (n 2 +n)/2

00

f(x; q)

} - ...

32. Continued Fractions

43

(q; q)oo(xq; q)oo(1lx; q)oo

=------~~--~~~~--~~~~00 (_X)n q (n 2 +3n)/2 00 X- 2n q 2n 2 -n'

(q2; q2)00

L

n=O

(2

L

2)

q ; q"

n=O

(2

q; q

2)

n

where we have employed (13.12) with q2 replaced by q and x replaced by x.jij. It follows that f(qn; q) = g(qn; q), for each nonnegative integer n. We also see that Zagier's analysis can be applied mutatis mutandi, after the aforementioned changes of variables are made. It is interesting to note that the continued fractions in (13.1) and (14.1) are connected to the basic hypergeometric functions 21J1l

(a; q)n(b; q)n n (a , b·, c·,q,. Z ) -_ ~ ~ (. ) ( • ) Z , n=O

C,

I;~I

q n q, q n

< 1.

For example, we shall show that the continued fraction in (14.1) can be derived from E. Heine's [1] continued fraction expansion 2CPl (a, b; c; q; z) = 1 _ alZ a2Z a3Z 21J1l (a, bq; cq; q; z) 1 - -1- - -1- -- ... '

(14.4)

where and Let a = 0, c = -1, and Z = xq Ib, and let b approach 00. We then find that and Thus, the continued fraction in (14.4) reduces to the one in (14.3). Likewise, since lim (b; q)nb-n = (_I)n q (n 2 -n)/2

and

b-+oo

lim (bq; q)nb-n = (_I)n q (n 2 +n)/2,

b-+oo

the left side of (14.4) reduces to the left side of (14.3). The identity (14.3) follows then, since the continued fraction (14.4) converges uniformly with respect to bin a neighborhood of b = 00. A rigorous proof of this statement can be given along the same lines as that given for (24.5) below.

Entry 15 (p. 373). Let a, b, and q be complex numbers with Iq I < 1. Define lJI(a) =

q a L . n=O (q; q)n(bq; q)n n2

00

n

Then lJI(a)

aq

--=1+-lJI(aq) 1

Proof. In (13.5), let a

+

aq2 - bq

1

+

aq3

-

1

+

aq4 - bq2

1

+ ....

= 0, replace b by -b, and set A = a. Upon observing that lim (-Ala; q)n an = Anq n(n-I)/2,

a-+O

44

Ramanujan's Notebooks, Part V

we see immediately that Entry 15 follows, since the continued fraction expansion of F (a, b, A, q) in the proof of Entry 13 converges locally uniformly in our domain. Observe that another continued fraction for qJ(a)/qJ(aq) is given in Entry 15 of Chapter 16 (Part III [3, p. 30]). Furthermore, another representation for qJ(a) can be found in Entry 9 of Chapter 16 (Part III [3, p. 18]). With the help of these two observations, Ramanathan [4] has found another proof of Entry 15.

Entry 16 (p. 373). For Iq I < 1, x(-q2)f(-q5) .:...::...-,---=----=--_;:_ f(-q, _q4) -

f(q, q9) 00 qn2 - "'"' f(_q4, _q16) - ~ (q4; q4)n

(16.1)

and qx(-q2)f(-q5)

~----=--:::-'----::::--- f(_q2,_q3)

-

qf(q3,q7)

00

- "'"'

q(n+l)'

f(_q8,_qI2) - ~ (q4;q4)n'

(16.2)

where, as before, x(q) = (-q; q2)00. Moreover, qf(q3,q7) / f(_q8, _qI2)

f(q,q9) q f(_q4, _qI6) =

q2

q3

1 + T + T +....

(16.3)

Proof. First, (16.3) follows immediately from (16.1), (16.2), and a standard representation for the Rogers-Ramanujan continued fraction given in Entry 38(iii) of Chapter 16 (Part III [3, p. 79]). We next demonstrate the first equalities of (16.1) and (16.2). By Entry 19 (the Jacobi triple product identity), (22.2) (Euler'S identity), and Entry 22, all in Chapter 16 (Part III [3, pp. 35, 37, 36]), we find that f(q, q9)

(-q; qlO)00(_q9; q lO)oo(q10; qlO)oo

f(-q4, _qI6)

(q4; q20)00(qI6; q20)00(q20; q20)00 (_q; q 10)00 ( _q9; q 10)00 (_q2; qlO)00(q2; qlO)00(_q8; qlO)00(q8; qlO)oo(_qIO; qlO)oo (_q4; qlO)00(_q6; qlO)oo(_q; qIO)00(_q9; qlO)oo (_q2; q2)00(q; q5)00( -q; q5)00(q4; q5)00( _q4; q5)00

(_q2; q2)00(q; q5)00(q4; q5)00 (q2; q4)00(q5; q5)00 (q; q5)00(q\ q5)00(q5; q5)00 x(-q2)f(-q5) f(-q, _q4)

,

(16.4)

32. Continued Fractions

45

which establishes the first equality of (16.1). The proof of the first equality in (16.2) is completely analogous. By (16.4) and its analogue, in order to prove the second equalities of (16.1) and (16.2), it suffices to show that, respectively, 00

n2

~ (q4~ q4)n

(_q2; q2)00(q; q5)00(q4; q5)00

and q(n+1)2

00

q

~ (q4; q4)n

=

(_q2; q2)00(q2; q5)00(q3; q5)00 .

These lasttwo identities have been proved by L. J. Rogers [1, pp. 330, 331]. Hence, the proof of Entry 16 is complete.

Entry 17 (p. 374). Let a, b, and q be complex numbers with Iq I < 1. Define qJ(a)

=

q(n 2+n)/2 a n

L. .' n=O (q, q)n(-bq, q)n 00

Then qJ(a) = 1 + aq bq aq2 bq2 aq3 bq 3 qJ(aq) 1 + + -1- + -1- + -1- + --1 + ....

T

Proof. The result follows immediately from setting A = 0 in (13.5).

2. Other q-Continued Fractions Entry 18 (p. 373). For Iq I < 1, f(-q, _q5)

q

f(_q3, _q3)

1+

+ q2 1

q2

+

+ q4 1

+ ....

Beneath this continued fraction, Ramanujan writes Num? = qJ( _q3) f(-q)

In fact, he has incorrectly inverted the identifications of the "numerator" and "denominator" on the left side of Entry 18. By Entry 22 and Example (v), Section 31 of Chapter 16 (Part III [3, pp. 36, 51]), f(-q, _q5)

X(_q)1/F(q3)

(q; q2)00(q6; q6)00

(q; q2)00

f(_q3, _q3)

qJ( _q3)

(q3; q6)~(q3; q3)00

(q3; q6)~ ,

46

Ramanujan's Notebooks, Part V

where X (q) [3, p. 36]),

=

(-q; q2)oo. On the other hand, by Entry 22 of Chapter 16 (Part III

ljJ(q3)/f(_q2)

(q; q2)oo

({J( _q3)/f( -q)

(q3; q6)~'

Hence, we have shown that Ramanujan has mistakenly confused the roles of the "numerator" and "denominator." Moreover, we now see that Entry 18 can be written in the more transparent form q

-

1+

+ q2

q2

+

+ q4 1

+ ....

(18.1)

The first proofs of (18.1) in print are due to Watson [2] in 1929 and Selberg [1, p. 19] in 1936. B. Gordon [1] and Andrews [1] found proofs in 1965 and 1968, respectively, while Hirschhorn [3] has shown that (18.1) can be deduced from Ramanujan's continued fraction (13.5). Ramanathan [2], [3] has briefly discussed (18.1). L.-c. Zhang [1] has examined (18.1) when q is a root of unity. A detailed study of G(q) has been made by H. H. Chan [1]. He has derived modular equations relating G(q) with each of G( -q), G(q2), and G(q3). Using these and other modular equations, he has determined values for G(±e-rr.fii) for several positive rational numbers n. The author, Chan, and L.-c. Zhang [1] have found general formulas that enable one to evaluate G(±e-rr.fii) in terms of class invariants. Entry 19 (Formula (3), p. 290). For Iq I < 1, (q2; q3)oo (q; q3)oo -

q

1 - 1+ q

q3 -

1 + q2

q5 -

1 + q3

q7 -

1 + q4

- ....

Proof. We apply Theorem 6 in Andrews' paper [1] with al = wxq, a2 = w-1xq, = _1/(x 2q2), andb = I/(xq), where w = exp(21l'i/3). Thus, I/(ala2) =-a and 1/al + I/a2 = -b, as required in Theorem 6. Accordingly, by the same argument as in the justification of the limiting procedures in Entry 24, using the uniform parabola theorem (Jones and Thron [1, p. 99]), we find that

a

lim H2,I(wxq,w- 1xq;x;q) - 2 - -q~ ~ X-+OHZ,1(WXq,w- 1xq;xq;q) I+q -1+q2 -1+q3 - ... '

(19.1) where the identification of H 2,I will be made shortly. Comparing Entry 19 and (19.1), we see that it remains to show that .

hm

1 -I

x-+o H 2,I(wxq,w xq;x;q) -1 Hz, I (wxq, w-1xq; xq; q)

(q2; q3)oo (q; q3)oo .

(19.2)

32. Continued Fractions

47

By Andrews' paper [1, eq. (1.1)], H 21 (al,a2;x;q) = ,

(xq/al; q)00(xq/a2; q)oo () C2,I(al,a2;x;q). xq;q 00

(19.3)

Furthermore, by [1, eq. (1.1)], lim C 2,I (wxq, w- I xq; x; q)

x~o

(_l)n q 3n(n-I)/2

00

=

~ (q; q)n(w- I ; q)n(w; q)n 1

= (1

1

(00

= - '"" ~

3

00

+~ =

00

-w- I )(1

1

3(0

- w)

~

(1 - w- I q n)(1 _ wqn)(_I)nq3n(n-I)/2

(_ly q 3n(n-I)/2 (q3; q3)n

(q3; q3)n

-

-I

(w

00

(_q)n q 3n(n-I)/2

~

(q3; q3)n

(19.4)

+ w) '"" ...:.........:.:......,..:-~-

(_q2)n q 3n 1. In fact, set q = l/a, so that lal < 1. Then q

1 - 1 + q2 I

1

-

q5

1 + q4

I/a

1 - 1+ -

q3

a

I + q6

- ...

l/a 5

l/a 3

1/a 2 -

--

-

1 + l/a

4 -

a3

1 + 1/a 6

-

...

a5

I - a 2 + 1 - a 4 + 1 - a6 + I

(a 3; a 4)oo

{l/q3; l/q4)oo

(a; a 4 )oo

(l/q; l/q4)oo .

This is, indeed, a beautiful example of symmetry.

(20.1)

32. Continued Fractions

49

It follows more generally from Entry 12 of Chapter 16 that (b q 3; q4)oo (bq; q4)oo

= 1" -

bq

bq 3

bq 5

Iql

1 + q2 - 1 + q4 - 1 + q6 - ... '

< 1.

Now let Iq I > 1 and set q = l/a, so that lal < 1. Then, as in (20.1), 1

bq

1" - 1 + q2

bq 3

-

(ba 3; a 4)oo

bq 5

(b/q3; l/q4)oo

1 + q4 - 1 + q6 - ... = (ba; a 4)oo = (b/q; l/q4)oo .

Although the continued fraction above is symmetric in q and 1/q, the product (bq 3; q4)oo/(bq; q4)oo does not share this invariance. However, if b = -1, then (_q3; q4)oo

(q; q8)oo(q5; q8)oo(q6; q8)oo

(-q; q4)oo = (q2; q8)oo(q3; q8)oo(q7; q8)oo'

and the latter quotient is invariant when q is replaced by 1/q. These observations are due to K. Alladi and B. Gordon [1, p. 298]. The convergence of (20.1) when q is a primitive root of unity has been examined by Zhang [1].

Entry 21 (Formula (5), p. 290). For Iq I < 1, (_q2; q2)oo (-q; q2)oo

1

q

1" + 1 +

q2

+q 1

q3

+T +

q4

+ q2 1

q5

+ 1- + ....

(21.1)

Entry 21 was first proved in print by Selberg [1, eq. (54)]. Another proof has been given by Ramanathan [4]. We provide yet another proof based on Entry 15. Proof. Applying Entry 15 with a

= 1 and b = -1, we find that

(Alternatively, this can also be proved by using (13.5) in Chapter 16 of Part III [3, p. 28] with a = -1 and b = 1.) Using Euler's identity (Andrews [4, p. 19]) once again, we find that the numerator and denominator on the left side above are, respectively, (-q; q2)oo and (_q2; q2)oo. The desired result now follows. If Q(q) denotes the left side of (21.1), then, by using Entries 22(i), (ii) and 25(vii) of Chapter 16 (Part III [3, pp. 36, 40]), we can easily show that 8 (q)

Q

=

l{F4(q2) ifi4(q)

=

X 16q'

in the notation of Entries 5 and 6 of Chapter 17 (Part III [3, pp. 100-102]). Thus, modular equations for Q(q) can be trivially derived from any of Ramanujan's modular equations.

50

Ramanujan's Notebooks, Part V

Entry 22 (Formula (6), p. 290). For Iql < 1, 1

(q; q8)oo(q7; q8)oo

1" +

(q3; q8)oo(q5; q8)oo =

q

+ q2

q4

+T +

1

q3

+ q6 1

q8

+ T + ....

The first published proof known to us is by Selberg [1, eq. (53)]. Other proofs have been given by Andrews [5] and Ramanathan [4]. Entry 22 also appears in Ramanujan's "lost notebook" [11]. Another continued fraction for the left side of Entry 22 has been found by Andrews [1] and Gordon [1]. Chan and Huang [1] have developed an extensive elegant theory for these continued fractions, including modular equations and explicit evaluations.

Entry 23 (p. 373). For Iq I < 1, f(_q,_q7)

1

f(_q3, _q5)

1

+

q

+ q2

q4

+ -1 +

q3

+ q6 1

q8

+ -1 + ....

By the Jacobi triple product identity, we may rewrite Entry 23 in the form (q; q8)oo(q7; q8)oo

1

(q3; q8)oo(q5; q8)oo

+

q

+ q2 1

+

q4

1

+

q3

+ q6 1

+

q8

1

+ ....

Hence, Entry 23 is equivalent to Entry 22.

3. Continued Fractions Arising from Products of Gamma Functions For the first result, we quote Ramanujan [9, vol. 2, p. 281]. We have

1

1

1

L:;- - L x/3 + :;- - log 3 2/3 x2

+

23 - 2 6

+

43 - 4 3x 2

+

53 - 5 6

+

73 - 7 ~

(24.1)

+ ....

The symbol L llx denotes Lk 0 and 1m I < E, If(m) - !k(m) I < Ako

(24.7)

52

Ramanujan's Notebooks, Part V

where Ak tends to 0 as k tends to 00. Also suppose that !k(m) tends to gk as m tends to 0, uniformly with respect to k in a neighborhood of k = 00, i.e., there exists an integer ko such that for k ~ ko,

Igk - fk(m)1 < rem),

(24.8)

where rem) approaches 0 as m tends to O. Now,

If(O) - gkl

:s

If(O) - f(m)1

+ If(m) -

fk(m)1

+ Ifk(m) -

gkl·

Thus, (24.6) follows provided that (24.7) and (24.8) hold. Indeed, these two statements of uniform convergence follow from the uniform parabola theorem (W. J. Thron [2]). Next, the even part (see (64.1) below) of the continued fraction in (24.3) is given by 6(2 3 - 2)(43 6{43 - 4 + 53 - 5

4

CFE(x) . - ----::-------:.- 2 3 - 2 + 6x 2 -

6 2(5 3 - 5)(7 3 - 7) - 6{73 - 7 + 8 3 - 8 + 6· 5x 2 }

-

•.•

62{(3k - 1)3 - (3k - 1)}{(3k

- 6{(3k

+ 1)3 -

4/6

(3k

+ I) + (3k + 2)3 -

+1

-

9k 2(9k 2 - 1)(9k 2 - 4) 3(2k + 1){9k2 + 9k + 2 + 2X2} - ...

-

4 + 53 - 5 +

4(x 2 + 1)/9 - 3{2·

4.2 2(2 2

-

t2

18x 2 - 7 3

+ 2· 1 + 4(x 2 + 1)/9}

!)(22

+ 2·2 +

4k2 (k2 -

-

4(x 2

~)

+ 1)/9}

- ...

(D2) (k2 - n)2)

+ 1){2k2 + 2k + 4(x 2 + 1)/9} - .... Hence, by (24.5), with e = ~,n = %. and x replaced by 2x/3, CFE(x) = H (t O(x + 1)) + t O(x + 2») - t Ox + 1) - t -

- ...

4.1 2 (1 2 - !)(12 - ~)

8/27

- 5{2· 22

+ 1)3 - (3k + I)} (3k + 2) + 6(2k + l)x 2}

5.7(5 2 - 1)(72 - 1) - 7 + 83 - 8 + 30x 2 - ...

x2

43 -

4) + 6· 3x 2 } -

(2k

Since (Abramowitz and Stegun [1, p. 259]) t(z)

(1

= -y + {; k - k 00

1) + 1

z

•

(~x)}. (24.9)

32. Continued Fractions

53

where y denotes Euler's constant, we find that 1/1 Goo

3 log 3.

Using the foregoing calculation in (24.9), we complete the proof.

Entry 25 (Formula (5), p. 292). liRe x > 0, then 1

00

2x 2 + {; (x

1 = :;

1

+ k)2

1 {I 3x

+ 2x 2

+

3 5x

18

+ 7x + ... +

k 2 (k 2

(2k

-

1)/4

}

1)

+ 3k

+ l)x +.. . .

54

Ramanujan's Notebooks, Part V

Proof. Replacing x by 2x in Entry 30 of Chapter 12 (part II [2, p. 149]) we find that T(n, x) : =

II } LCO{ 2x-n+2k+l - -,-----,------2x+n+2k+l k=O

12(12 - n 2)j4 + 3x (nj2) - x + (l - n)F(n,x)' (nj2) x

where 12(1

F(n, x) =

+ n)j4

3x

+

+

22(22 - n 2)j4 5x

22(22 - n 2)j4 5x

+ ...

+ ...

After some elementary algebra, we find that n - 2xT(n, x) 2T(n, x)(1 - n)

..,..---------'-- =

F(n, x).

Letting n tend to 1, applying L'Hospital's rule, and using the facts T(l, x) aT Ij(2x) and -(1, x) =1= 0, we find that an aT 1- 2x-(n,x) lim F(n, x) = lim an aT n~1 n~1 -2T(n, x) + 2(1 - n)-(n, x) an aT 1- 2xa;;-(1,x)

=

-ljx

= -x + 2x2 I

I+ 2k} t;CO{ (2x +I 2k)2 + + 2)2 (2x

= -x + 2 + x 2

1

L (x + k)2· 00

k=1

Since 1 1

F(l, x)

= 2 3x +

22(22 - l)j4 5x

+

32 (3 2 - 1)/4 7x

+ ... '

we see that the proof is completed after a little algebraic manipulation and an appeal to uniform convergence as in the proof of Entry 24. Entry 26 (Formula (6), p. 293). IfRe x > 0, then I ~ 1 I 2x 3 + (x + k)3 = 2x2

b

where,for k ~ I,

Pk

= k 2 (k

I {I

PI

ql

P2

q2

}

+ 4x 3 ~ + ~ + ~ + ~ + ~ + . .. '

+ l)j(4k + 2) and qk =

k(k

+ 1)2 j(4k + 2).

32. Continued Fractions

55

Entry 26 is, in fact, due to T. J. Stieltjes [1], [3, pp. 378-391]. It is also given in Wall's book [1, p. 37], where 4z 2 should be replaced by 4z 3 •

Entry 27 (p. 325). Let n be a complex number such that Re n > 1

00

+ k)2

{ ; (n

(1) =

+ "2

n

{I+ n2

(1 . 1)2

n

(2 . 3f + 9(n 2

4. Then

(1 . 1)2

(2·3)2

+ - 3 - + 5(n 2 + n) + - 7 -

(3 . 5)2

(3 . 5)2

+ n) + -1-1- +

13(n 2

}

+ n) + . .. .

(27.1)

Proot From the corollary to Entry 30 of Chapter 12 of the second notebook (part 11[2, p. 150)), we find that, for x = 2n + 1 and Rex = Re(2n + 1) > 0, 1

00

{ ; (n

+ k)2

= 4(n

1

00

= 4 {; (x

+~) L(4n

2

+ 2k _

+14n

1)2

+ 1) +

=2

{1 ;;

14

24

34

+ 3x + 5x + 7x + ...

}

~4 + 5(4n2:~ + 1) + ~4 + ... }.

(27.2) We shall use the Bauer-Muir transformation (13.7) to prove that this continued fraction converges to the same value as the one presented in this entry. Choose lU2k = 0 and lU2k+ 1 = - (2k + 1)2 for each nonnegative integer k. From (13.6),

for each positive integer k. Moreover, a2k A2k+t! A2k = 4k 2(2k - 1)2, a2k+1 A2k+2/A2k+l

b2k

= 4(k

+ 1)2(2k + 1)2,

+ lU2k -lU2k-2A2k/A2k-l

= 4k - 1,

and b2kH

+ lU2k+l

- (JJ2k-1A2k+l/A2k = (4k

+ 1}(4n2 + 4n).

Hence, the continued fraction in (27.2) is transformed by (13.7) into the continued fraction 4n 2

+ 4n + 1

n2

+n +

4(1 . 1)2

4(1 . 1)2

3 (1.1)2 3

+ 5(4n 2 + 4n) + (1·1)2

+ 5(n 2 + n) +

4(2·3)2

4(2. 3)2

7 + 9(4n 2 + 4n) + ... (2.3)2 (2·3)2 7 + 9(n 2 + n) + ....

Since the even approximants of the two continued fractions coincide and since both the proof of Entry 27 is complete. continued fractions converge for Re n > -

4,

56

Ramanujan's Notebooks, Part V

In the foregoing proof we have seen that the even approximants of the continued fractions in (27.1) and (27.2) are identical. Thus, an alternative proof can be derived by showing that the even parts of (27.1) and (27.2) agree. Such a proof would be shorter and simpler but not as instructive as the constructive approach via the Bauer-Muir transformation. If we let n = 1 in Entry 27, we find that 2 _ 1

]f2 _

6 -

1,"( ) -

~ {~

+ 2 2+

~

3

+

~

22 .

10 +

7

32

+

32 18

}

22 .

+. . . '

where I," denotes the Riemann zeta-function. Letting n = 1 in Entry 28 below, we deduce that 3{I 1- - -

=

log 2

1·1 -

1·1 -

2·3 -

2·3 -

24+ 1 + 4 + 1 + 4 + ...

}

.

Entry 28 (p. 325). Let n be a complex number such that Re n > - ~. Then

t

k=I

= (

(_I)k+I n+k

1) { I I , 1 2n 2 + 2n + -1-

+2

n

2·3 + 2n 2 + 2n

3·5

1. 1

2·3

+ 2n 2 + 2n + -1-

3·5

}

+ 1 + 2n 2 + 2n + . . . .

Proof. From the corollary to Entry 29 of Chapter 12 of Ramanujan's second notebook (Part II [2, p. 149]), it follows that 00

(

-l)k+ I

1

12

22

2{;X+2k-l=~+~+X for Re x > O. Setting x = 2n 00

=

{; n+k = (2n

+

1)

{

_(n + ~)2

-

12

2n+l

I

(2n

2n 2

1)2

32

22

+ 2n+l + 2n+l + 2n+l + ... 12

1

+

+ ... '

+ I, we find, via equivalence transformations, that

1

(_I)k+I

32 +x

22

+ T + (2n +

1

+ 2n + ~ +

-~-'1_2

32

1)2

+ T + ...

}

----:-!~'_._2_2--;-

~ . 32

I

1 + 2n 2 + 2n + ~ + -1- + . . . ' (28.1)

forRen > -~. We apply the Bauer-Muir transformation (13.7). Let -(2k + 1)/2 for each integer k ~ O. Then from (13.6), Al

=

I,

A2k

= k(2k -

1),

and

WZk

A2k+I

= 0 and Wzk+I

= 2k2,

32. Continued Fractions

57

for each positive integer k. Furthennore, a2kA2k+I/A2k = k(2k - 1), a2k+lA2k+21A2k+1

+ CO:zk -

b2k

CO:zk-2A2k/A2k-1

= (k + 1)(2k + 1),

= 1,

and

b2k+1

+ CO:zk+1 -

CO:zk-IA2k+I/A2k

= 2n 2 + 2n.

Thus, the Bauer-Muir transfonnation transfonns the continued fraction in (28,1) into the continued fraction 1 1·1 1·1 2·3 2·3 2n 2 +2n + -1- + 2n 2 +2n + -1- + 2n 2 +2n + .... Since the even approximants of the two continued fractions coincide and since both continued fractions converge for Re n > - &' the proof is complete. Alternatively, the even part of the last continued fraction in (28.1) is precisely the continued fraction in Entry 28. This gives an even shorter proof.

Entry 29 (p. 343). Let x and n be complex numbers such that either Re x > 0 or n = (2k + l)i for some integer k. Then n2 r (x + 1)) nco 1 + (x + 3 + 4k)2 r 2 (!(x + 3») k",O 1 + ____ n2 -:4 (x + 1 + 4k)2 4 n 2 + 12 n 2 + 32 = --x+ 2x + 2x 2

G

Proof. Replacing n by in in Entry 25 of Chapter 12 (Part II [2, p. 140)), we find that, under the conditions specified above,

r

+ in + 1)) r G(x r G(x + in + 3») r G(x 4 n 2 + 12 n2 + 32 G(x

+ 1)) in + 3») in

(29.1)

-

x+ 2x + 2x However, from Euler's product fonnula for the gamma function, rG(x+in+l»)rU(x-in+l))=

r 2 (!(x + 1»)

k~O

4

{

1 + (x

2

•

+ ; + 4k)2 }

Using this fonnula and an analogous fonnula in (29.1), we complete the proof. (A product representation for If(x + iy)1 2 is given in Gradshteyn and Ryzhik's tables [1, p. 945, fonnula 8.326. no. 1].)

58

Ramanujan's Notebooks, Part V

Entry 30 (p. 343). For all complex n, tanh(nnI4) n

4

1+

k-O 1 -

+

1

+

n 2 +12 2

+ ...

= 1 in Entry 29, we deduce that

Proof. Setting x

~ fI

1

(4(k: IS

(

+

)2

n 2(2k+l)

n 2 + 32 ---

2

+

n 2 + 52

2

+ ....

However, by a familiar product representation for tanh z, the left side above equals (lIn) tanh(nnI4). Entry 30 may also be found in o. Perron's book [1, p. 36, eq. (23)]. Entry 31 below is also in Perron's text [1, p. 33].

Entry 31 (p. 343). Let x and n be complex numbers such that either Re x > 0 or n = (2k + l)i for some integer k. Then ~

2L...,.(-1)

k+1

k=1

X +2k-l (x+2k-1)1+n 1

1 n 2 + 12 22 n 2 + 32 42 n 1 + 52 x+ x +x+ x +x+ x

+ ....

Proof. Replacing n by in in Entry 29 of Chapter 12 (Part II [2, pp. 147, 148]), we find that, under the conditions given above, ~

2£:-1(-1)

k+1

X +2k -1 (x+2k-l)1+n2

{ I+I - - -} ---

00 1 = L(-l)k+

k=l

X

+ ni + 2k -

1

x - ni

+ 2k -

1

22 n 2 +3 2 42 1 n 2 +12 x+ x +x+ x +x+····

Entry 32 (p. 344). For all complex numbers n,

L 00

(-l)k+lk

k=l

Proof. Set x

k2 + n 2

1

="1 +

4n 2

+ 12 1

22 4n 2 + 32 + ""1 + - - - +

42 1

+ ....

= 1 and replace n by 2n in Entry 31.

In the continued fraction of Entry 32, Ramanujan inadvertently wrote n 2 for 4n 1 .

32. Continued Fractions

59

Entry 33 (p. 344). Let x and n be complex numbers such that either Re x > 0 or n = (2k + 1)i for some integer k. Then 1

00

2

~ (x + 2k + 1)2 + n 12(n 2 + 12)

1

=

3x

X +

2

32(n 2 + 32 )

22(n2 + 22) 5x

+

7x

+

+ ....

Proof. In Entry 30 of Chapter 12 (Part 11[2, p. 149]) merely replace n by in, and the desired result immediately follows. Entry 34 (p. 343). For every complex number n, 1fn

n2

(1fn) _

2 coth

2

-1+ 1 +

12(n 2 + 12)

3

+

22(n 2 + 22)

5

Proof. It is clear that the identity holds for n Setting x = 1 in Entry 33, we find that

=

+

32(n 2 + 32) 7 + ....

O. Thus, assume that n

i= O.

Upon multiplying both sides by n 2 and rearranging, we complete the proof. Entry 35 (p. 344). Let x and n be complex numbers such that either Re x > with x ;. (- 4,0], or n = ki for some integer k. Then (-ll

00

2

4

~ (x + 1 + k)2 + n2 1 x2 +

X

n 2 + 12 12 + 1 + x2 +

X

+

n 2 + 22

22 + x2

-+ X

+ ....

Proof. In Entry 31 of Chapter 12 (Part II [2, p. 150]), replace n by 2i n and x by 2x + 1. Then, under the proposed hypotheses we find that 1

(_l)k

00

"2 ~ (x + 1 + k)2 + n 2 =

1

4x 2

+ 4x +

4+ 4n 2 1

+ 4x 2 + 4x +

which is equivalent to the continued fraction displayed in Entry 35.

60

Ramanujan's Notebooks, Part V

Entry 36 (p. 344). Let m, n, and x denote complex numbers such that either Rex> 0, or m = ij for an integer j and x i= -(2k + 1) for any nonnegative integer k, or n = i j for an integer j and x i= - (2k + 1) for any nonnegative integer k. Furthermore, let

Then mn

u- v u+V

X

+

(m 2 + I2)(n 2 + 12) 3x

+

(m 2 + 22)(n 2 + 22) 5x

+ ....

Proof. Assume that x, m, and n are positive. In Entry 33 of Chapter 12 (Part II [2, p. 155]), replace m and n by im and in, respectively. Employing a product formula for W(x + iy)1 2 (Gradshteyn and Ryzhik [1, p. 945]), we find that, under the given hypotheses, -mn x

+

(m 2 + I2)(n 2 + 12) 3x

+

(m 2 + 22)(n 2 + 22) 5x

+ ...

Ir H(x + 1) + t(m + n)i)1 2 -Ir (t(x + 1) + t(m Ir H(x + 1) + t(m + n)i)1 2 + Ir H(x + 1) + t(m 00

k~

{

OO{

Do

1 + (x

(m+n)2 }-' + 2k + 1)2 -

1 + (x

(m+n)2 + 2k + 1)2

Do 00

{

}-I + Eloo{

n)i) 12 n)i) 12

1 + (x

(m-n)2 }-' + 2k + 1)2

1 + (x

(m-n)2 }-' + 2k + 1)2

llu - Ilv v- u = Ilu + Ilv v +u This completes the proof for x > 0, m > 0, and n > o. Since the continued fraction converges to a meromorphic function of x for Re x > 0, the entry holds in this half plane by analytic continuation. Furthermore, since the continued fraction converges to a meromorphic function of m and of n, it follows that Entry 36 is true for all complex m and n. That the equality holds if the continued fraction terminates follows by straightforward computation. Entry 37 (p. 344). Ifm and n are complex numbers with m i= n, then m tanh Hrrn)

- n tanh Hrrm) m tanh Hrrm) - n tanh Hrrn) mn 1

+

(m 2 + 12)(n 2 + 12) 3

+

(m 2 + 22)(n 2 + 22) 5

+ ....

32. Continued Fractions

=

Proof. Putting x (u - v)/(u

61

1 in Entry 36, we see that it only remains to show that

+ v) reduces to the left side of Entry 37.

U sing a familiar product representation for sinh z, we see that, when x = 1,

+ n)-l sinh {~1r(m + n)} - (m - n)-l sinh {~1r(m - n)} + n)-l sinh {~1r(m + n)} + (m - n)-l sinh {~1r(m - n)} -2n sinh (41rm) cosh (41rn) + 2m sinh (41rn) cosh (41rm)

u- v

(m

+v

= (m

u

=

2m sinh (41rm) cosh (41rn) - 2n sinh (41rn) cosh (41rm) m tanh (41rn) - n tanh (41rm)

= m tanh (41rm) - n tanh (41rn) ,

and so the proof is complete.

Entry 38 (p. 345). Let m and x be complex numbers such that either Re x > 0, or m = k(1 + i)/2, or m = k(1 - i)/2,for some integer k. Furthermore, set

and

= r (4 (x + 2m + 1)) r (4 (x -

v

Then 2m 2

u- v

u

+V

=

7 +

4m 4

+ 14

3x

+

4m 4

+ 24

5x

2m

+

+ 1»)·

4m 4

+ 34

7x

+ ....

Note that if m = k(1 ±i)/2 for some integer k, the continued fraction terminates.

Proof. We apply Entry 33 of Chapter 12 (part II [2, p. 155]) with m and n replaced by (1 + i)m and (1 - i)m, respectively. We then find that, for m > 0 and x > 0, or for (1 ± i)m = k for some integer k, 2m 2 x +

4m4

+}4

3x

4m4

+

+ 24

5x

4m 4 + 34 + 7x + ...

2

r(!(x+2m+l») r(!(x-2m+l))-r (!(x+l))

2

r(!(x+2m+l)) r(!(x-2m+l))+r (!(x+l))

fi {1+( x+~~+1 )2}-1

k=O

fi {1+(,x+2k+l 2m )2}-1

k=O

u-v u +v' where in the penultimate step, we used the same formula for If (x + iy)I 2 that we used in the proofs of Entries 29 and 36. The result is then valid for all complex m and all complex x with Re x > 0 by analytic continuation.

62

Ramanujan's Notebooks, Part V

Entry 39 (p. 345). For arbitrary complex n, sinh(JTn) - sin(JTn) sinh(JTn) + sin(JTn)

=

Proof. Setting x

1

4n 4 + 14

+

3

4n 4 + 24

+

5

+

4n 4 + 34

7

+ ....

1 and replacing m by n in Entry 38, we deduce that

2n 2

1

2n 2

+

4n 4

+ 14 3

+

4n4

+ 24

\r\ 5

k~O 11 + (~r k~O II + (k : 1

+ ...

r(1

+ n)lr (1 -

+ r (1 + n )Ir (1

n)

- n)

(JTn)-1 sinh(JTn) - (JTn)-1 sin(JTn) - (JTn)-1 sinh(JTn) + (JTn)-1 sin(JTn)' where we have employed a familiar product representation for sinh z and the reflection formula for the gamma function. This completes the proof.

-!;

Entry 40 (p. 345). Let x and n be complex numbers such that either Re x > or pn is an integer, where p is a sixth root of unity, and x is not a negative integer. Furthermore, let

Then

u- v u

+V

n3

-

2X2

+ 2x + 1 +

n6

16 3(2x 2 + 2x + 3) -

+

n6

26 5(2x 2 + 2x + 7) -

+ ....

The constant term within parentheses in the kth denominator is given by k 2 - k + I. In the notebooks, Ramanujan mistakenly indicated that this constant is equal to 2k - 1. Thus, Rarnanujan wrote 5 instead of 7 in the third denominator displayed above. Ramanujan's error can be traced back to a scribal error in recording Entry 40 of Chapter 12. For a discussion of this error, see Part II [2, pp. 163, 164]. Note that if pn is an integer, where p is a sixth root of unity, the continued fraction terminates. Proof. We shall apply Entry 35 of Chapter 12 (Part II [2, pp. 156, 157], Jacobsen [4]) with i = e 2Tri / 3 n, m = e 4rri / 3 n, and x replaced by 2x + 1. Observe that (l2 _

k 2 )(m 2

_

k 2 )(n 2

_

k2)

= n6 _

k6

32. Continued Fractions

63

and thatx 2 _£2 _m 2 -n 2+2k2 - 2k+ 1 is transformed into 4x 2 +4x+2k 2 - 2k+2. Thus, the continued fraction that arises from Entry 35 of Chapter 12 equals

2n 3 4X2+4X+2 n3

4(n 6 - 16 ) 4(n 6 - 26 ) + 3(4x2+4x+6) + 5(4x 2 +4x+14) + ... n 6 _ 16 n 6 _ 26 - 2X2 + 2x + 1 + 3(2x 2 + 2x + 3) + 5(2x 2 + 2x + 7) + ....

(40.1)

We now examine the gamma functions appearing in Entry 35 with the parametric designations given above. Letting w = e 27Ci / 3 and using Euler's product representation of the gamma function, we find that, in the notation of Entry 35 of Chapter 12,

P

=0 2

j=O

.

r(x r(x

.

+ 1 + w1n) + 1 - w1n)

+ 1 + k - n)(x + 1 + k - wn)(x + 1 + k - w2 n) + 1 + k + n)(x + 1 + k + wn)(x + 1 + k + w 2 n) (x + 1 + k)3 - n 3 (x + 1 + k)3 + n 3

=

Om (x m ..... oo k=O (x

=

1]

hm

00

=Iill-(X+~+kY}=!:'.. +( )3 n x+l+k

1

k=O

u

Hence, by Entry 35 of Chapter 12, the continued fraction (40.1) equals

1- P 1 +P

I-vlu u-v =--, 1 + vlu u +v

and the proof is complete.

Entry 41 (p.347). Suppose that m, n, and x are complex numbers such that Re x > 0, or assume that n is an integer or that im is an integer. Then

~ {tan- (x _n: 2k + 1) - tan- (x + n: 2k + 1) } 1

1

= tan- 1 {mn x

+

(12

+ m 2)(12 3x

n 2)

+

(2 2

+ m 2)(22 5x

n 2)

+ ...

}.

Proof. In our proof below, we will temporarily ignore the fact that tan- 1 z is muItivalued. At the end of the proof, we shall show that the correct branches have been chosen.

64

Ramanujan's Notebooks, Part V

Replacing m by im in Entry 33 of Chapter 12 (Part II [2, p. 155]), we find that, for Re x > 0, or n E iE, or i m E iE, r(~ (x+n+ 1)+ ~ im)r( ~ (x-n+ 1)- ~im)- f( ~ (x -n+ 1)+ ~im)r( ~ (x+n+ 1) - ~im) r(!(x+n+l)+ !im)f(~(x-n+l)-1im)+r(~(x-n+l)+~im)r(1(x+n + 1)- ~im)

imn x

+

(12

+ m 2 )(12 -

n2 )

3x

+

(22

+ m 2 )(22 _

n2 )

5x

+ ...

(41.1)

Suppose first that X, m, and n are positive, and let

r (~(x + n + 1) + ~im) r (t(x - n + 1) -

a =

~im) =: a

+ ib,

where a and b are real. Multiplying both sides of (41.1) by -i and then taking the inverse tangent of each side, we see that tan-I {mn

+

x = tan-I

+ m 2)(12 -

(12

n 2)

3x

+

(2 2

+ m 2)(22 5x

n 2)

+ ...

}

{i (: ~:)} = tan-I (b/a) = 1m(log a)

= 1m (log {r (~(x

+ n + 1) + ~im) r (t(x -

(41.2)

n + I) - ~im)}) =: T.

In order to calculate T, we employ Euler's product representation for the gamma function. Hence,

L {1m log 0 (x + n + 1) + k + ~ i m) 00

T = -

k=O

+ 1m log 0 (x = -

-

00

n

+ 1) + k -

~ i m) }

~ !tan- I (~(X +:~ I)+k) +tan-I (~(X _~m~~) +k)}

- {;

{

tan-I

(

m ) - tan-I ( m )} x-n+2k+l x+n+2k+l' (41.3)

Thus, formally, the proof has been completed. To complete the proof, we first apply Stirling's formula to show easily that, for x> 0,

Thus, for the principal branch of tan -I lim tan-I

X"'*oo

z,

{i (~a +a - a)} = 0.

32. Continued Fractions

65

On the other hand,

r

x~'! ~ 00

{

tan-I

-HOO~

-lim

= lim

00

m

x - n + 2k

+1

{m x-n+2k+1

E{

HOO k=O

(

(x

2mn

+ 2k + 1)2 -

n2

)

-

tan-I (

x

m

x+n+2k+1

+0

(

m

+ n + 2k + 1 +0

)}

( (x-lnl+2k+I)3 1 )}

1 )} (x - Inl + 2k + 1)3

= 0,

for the limit as x ~ 00 can be taken under the summation sign, since the series converges uniformly for Inl ~ x < 00. Thus, our calculations in (41.2) and (41.3) demonstrate that Entry 41 is correct for x sufficiently large and positive. However, since both sides of Entry 41 are meromorphic for Re x > 0, the equality of Entry 41 must then be valid for all x with Rex> O. If n or i m is an integer, the continued fraction terminates. A straightforward computation shows that the identity still holds for x > O. Hence, the proposed result follows by analytic continuation. Furthermore, since both sides are meromorphic functions of m and n, the entry holds for all m, nEe by analytic continuation.

Entry 42 (p. 347). Let m. n. and x denote complex numbers. Suppose that either Rex> O. or m = 2ir for some integer r. or n = (2s + l)i for some integer s. Then

~(_I)k {tan-I (x :2:: 1) + tan-I (x: 2~: I)} {m n + m+ n + 3 m+ 4 _

-tan

-I

2

x+

x

12

2

+

x

22

2

2

+

x

2

+

2

x

+ ...

} •

Proof. As in the previous proof, we temporarily ignore the fact that tan -I Z is

multivalued, and assume that x, m, and n are positive. In Entry 34 of Chapter 12 (Part II [2, p. 156], Jacobsen [4]) replace l by im and n by in. Thus, under the given assumptions on x. m, and n,

im n 2 + 12 x + x

+

m2 + x

22

1 - P n 2 + 32 m2 + 42 + x + x + ... = 1 + p'

(42.1)

where r(!(x+1+i(m+n»)rd(x+1+i(m-n)))r( (x+3-i(m-n)))r( (x+3-i(m+n)))

P=~~--------~~~--~--~~~----~~~-~~------~

r(!(x+l-i(m-n)))r( !(x+l-i(m+n»)r( (x+3+i(m+n)))r( (x+3+i(m-n)))

a a+ib = ;j = a - ib'

66

Ramanujan's Notebooks, Part V

where a = a + ibis the numerator of P. Multiplying both sides of (42.1) by -i and then applying the operator tan-I, we deduce that -I{m

-;+

tan

= tan-I i

n 2 +12 x

+

m 2 +22 x

(a/~ - 1) = a/a + 1

+

n 2 +32 x

+

m 2 +42 x

+ ...

} (42.2)

_ tan-I (b/a) = - Im(log a) =: -T.

Employing Euler's product formula for the gamma function, we find that T

= 1m log r U(x + 1 + i (m + n») + 1m log r U(x + 1 + i(m - n») + 1m log r U(x + 3 - i(m - n») + Imlogr (~(x + 3 - i(m + n») 00

= - L(lmlogU(x+l+i(m+n»+k) k=O

+ 1m log U(x + 1 +i(m -n» +k) + 1m log ( ~ (x + 3 - i (m - n» + k) + 1m log (~(x + 3 -

-t; I( OO{

- tan_

tan-

I(

x

x

m+n)

+ 1 + 4k + tan-

+m-n) 3 + 4k -

~ -1 k {tan-I ~()

(

x

tan-

I(

x

I(

x

i (m

+ n» + k) }

m-n)

+ 1 + 4k

)} +m+n 1 + 4k

m + n ) + tan-I + 1 + 2k

(

x

m- n

+ 1 + 2k

)}

.

(42.3)

Using (42.3) in (42.2), we formally complete the proof. To show that we have, indeed, chosen the correct branches in all our calculations, we use the same type of argument as in the proof of Entry 41 for the case Re x > o. Since the details are very similar, we omit them. If mi/2 or (ni - 1)/2 is an integer, the continued fraction terminates. The proposed result again follows, as in the proof of Entry 41.

Entry 43 (p. 347). Let x and n denote complex numbers. Assume either that Re x > 0 or n = j i for some integer j. Then 00 "(_l)k tan-I ( 2 n ) = tan- I { -n

f::o

x+2k+1

x+

n 2 + 12 x

+

n 2 + 22 x

}

+ ....

Proof. Set m = n in Entry 42.

4. Other Continued Fractions Entry 44 (Formula (2), p. 276). Let a and b be complex numbers such that a =1= 0 and I arg(b/a 2 ) I < 1f. Let Bn denote the nth Bernoulli number. For each

32. Continued Fractions

67

nonnegative integer n, define [n/2] ( ) _ ~ -n 2k n-2kbk A n-L." a . k=O

k!

Then, for each positive integer N, as x tends to 0 through values such that Re(bx2) > 0, X

1 ~ -nax-n2bX2! { -+L."e

2

n=l

2b

4b

6b

N-l

a + a + a + a +...

B

2n

2n + f=r ~ - - A 2n - 1 + O(x (2n)! X

2N

).

Our version of Entry 44 is slightly more precise than that of Ramanujan. A proof of Entry 44 has been given by Watson [3] for the case when a > 0, b > 0, and x > O. The extension to bja 2 E C - (-00,0] and Re(bx2) > 0 follows by analytic continuation, since the continued fraction converges to a holomorphic function of (a, b) forbja 2 E C- (-00,0], and the series on the left side converges to a holomorphic function of (a, b) for Re(bx2) > O. The reader should note that the notations of Ramanujan, Watson, and the authors for Bernoulli numbers are different. The next result was communicated by Ramanujan [10, p. 352] in his second letter to Hardy and is simply the case a = 1, b = of Entry 44. We precisely quote Ramanujan below, but, of course, a more accurate version can be formulated as above. Ramanujan tacitly assumed that x > O. However, the result holds for all x E C with Re(x2) > 0, i.e., for I arg x I < T( j 4.

!

Corollary (Formula (3), p. 276). When x is small,

~

-

~

~

~

1+1+1+1+1+···

x

+

2" -

x2

x6

X4

=

x.;e~ e-(l+nx)2/2

f=r

x8

X 10

12 - 360 - 5040 - 60480 - 1710720

nearly.

Entry 45 (Formula (4), p. 276). The formal power series L(x) .-

L oo

. - k=O

(-ll2(2 4k +2 (2k

+

1)B4k +2 I)X 2k + 1 -

has the corresponding continued fraction al a2 a3 CF(x):=x+x+x+···'

where ak > 0 for 1 :s k < 00. In particular, al = a2 = 1, a3 = 30, a4 = 150, and a5 = 493. As before, B j , 0 :s j < 00, denotes the jth Bernoulli number.

68

Ramanujan's Notebooks, Part V

For the definition of correspondence, we refer to the text of Jones and Thron [1, p. 148]. Continued fractions of the form above are called S-fractions or Stieltjes fractions. They have the property that their even and odd parts converge to analytic functions both of which have the asymptotic expansion L(x) (Jones and Thron [I, pp. 136, 342]). Moreover, they converge for all x in the cut plane I arg x I < 17: if and only if they converge for one x in this region. It seems to be very difficult to determine if C F(x) converges.

Proof. To see that C F(x) is an S-fraction, we observe that L(x) can be written in the form 00

L(x) = L(-I)kcdx2k+I, k=O

where

c = 2(2 4k+ 2 k

= -417:4k+2

_

1) B4k+2 2k + 1

1

00

0

u 4k + 1 ---du = 4 sinh u

1

00

0

u 4k+ 1du , sinh(17:u)

(45.1)

where we have used a familiar integral evaluation (E. T. Whittaker and G. N. Watson [1, p. 126]). Since sinh (17: u) > 0 for u > 0, the primary assertion of Entry 45 follows from Stieltjes' theory. (See, for instance, Wall's book [1, p. 363].) To calculate the numerators ak, we may use Entry 17 of Chapter 12 in Ramanujan's second notebook (Part II [2, pp. 124, 125]) Alternatively, Viskovatoff's algorithm (A. N. Khovanskii [1, pp. 27, 28]) can be employed. The calculations in the second method are somewhat easier, and, in either case, they are routine. Hence, we omit them. It is tempting to conjecture that the numerators of C F (x) are integers. However, = 588456/493 and a7 = 10101660478/4029289. Lastly, we find two functions that have the asymptotic expansion L(x) as x tends to 00. From (45.1), for each positive integer n,

a6

as x tends to

00.

Thus, FI (x) has the asymptotic expansion L (x).

32. Continued Fractions

69

From the Laurent expansions of cot t and coth t about the origin (Gradshteyn and Ryzhik [1, p. 42]), we easily find that

cotht - cott =

L

24k +3 B4k

00

(4k

k=O

2

+

+ 2)!

It I < 7f.

t 4k + I ,

Applying Watson's Lemma (E. T. Copson [1, p. 49]), we deduce that

1

00

o

e- xt (coth t - cot t) dt '"

24k +3 B 4k ( + 2)x

L

4k!:2'

00

k=O

as x tends to 00 with Rex> O. Replacing x by 2../iX and by ../iX, where the principal branch of the square root is chosen, we find that, for - 37f /2 < arg x < 7f/2, F 2 (x) := 2i

1

00

(e- 0xt

-

e- 20xt )(cotht - cott) dt '" L(x),

as x tends to 00. Unfortunately, neither FI nor F2 has been of any use to us in determining the convergence or divergence of C F(x).

I

Entry 46 (Formula (6), p. 277). For each complex number x, x coth x

4.5 2 2.3 2 6.7 2 4.5 2 222-X-X-X-X

= 1 + ~ -::... 3

9

~ 5

+

1...:..L 7

+

..1...:.L 9

+

..1...:.L 11

+

~ 13

+ ...

I .

Entry 47 (Formula (7), p. 277). For each complex number x and each complex number n "10,-1,-2,-3, ... , x x x x -

n+n+l+n+2+n+3+··· x

= ;; -

x

{x

n2

n

a2X

a3X

a2kX

a2k+Jx

}

+ 1 + n + 2 + n + 3 + . .. + n + 2k + n + 2k + 1 +... '

where,for k ::: 1,

a

(k + l)(n + k) ken + k - 1)

- ----------

2k -

ken + k - 1) and a2k+1 = (k + 1)(n + k) .

We first remark that the continued fraction on the left side. of Entry 47 is equal to

.ji I n (2i .ji) i

In-I (2i.ji)

for all complex x, where J" denotes the ordinary Bessel function of order v. See Wall's book [1, p. 349] or Part II [2, p. 133, Entry 19]. Next, we show that Entry 46 readily follows from Entry 47.

70

Ramanujan's Notebooks, Part V

Proof of Entry 46. Recall that (Wall [1, p. 349]), for each complex number x,

x2 x2 x2 x coth x = 1 + -3 + -5 + -7 + ... '

(46.1)

which is due to J. H. Lambert [1]. This suggests that we let n = ~ and replace x by x 2 / 4 in Entry 47. Accordingly, we find that x 2 /4 x 2/4 x 2 /4 x 2 /4 3/2 + 5/2 + 7/2 + 9/2 + ... x 2 /4

x 2 /4

----3/2 9/4 x

1.3/2x 2

} 3·7/2x 2 2·5/2x 2 /4 ~4 ~4 -2.-5/-24 -3.-7/-24 5/2 + 7/2 + 9/2 + 11/2 + 13/2 +... ' 2.5/2x2

1

x2

which is equivalent to 1 {X2 x2 x2 23+5+7+'"

=

Ix

2

Ix

2{2 x

23 - 29 5

I

}

4 . 5x 2 2 . 3x 2 6 . 7x2 4 . 5x2 2.3 4·5 4·5 6.7 + - 7 - + - 9 - + -1-1- + -1-3- + . . . .

Multiplying both sides by 2, adding 1 to each side, and employing (46.1), we complete the proof. Proof of Entry 47. We shall employ a lemma of Rogers [2, p. 74]. If /1 = eOel, h = el +e2, hh = e2e3, h + /4 = e3 +e4, /4/5 = e4e S, /s + /6 = es +e6,"" then

eo elx e2x /IX hx hx 1 - 1 - 1 - ... = eo + -1- - -1- - -1- _ ... '

(47.1)

in the sense that both continued fractions correspond to the same (formal) power series. This means that if both continued fractions converge in a neighborhood of x = 0, then they converge to the same value (Jones and Thron [1, p. 181]); that is, (47.1) expresses an identity between their values. Writing the left side of Entry 47 as the equivalent continued fraction x x x n(n + 1) 1

n 1 -

(n

+ 1)(n + 2) I

we see that, in the notation (47.1),

x

eo =-, n

ek

= -

(n

+k -

1 1)(n

+ k)

,

k

~

1.

32. Continued Fractions

We now calculate II

=-

Ib k

~ 1.

x n2(n + I)' h

71

Straightforward calculations show that

=-

n(n

2

+ 2)'

h =-

and

2(n

n

+ 1-)-(n-+-2)-(-n-+-3-)'

By induction, we shall show that hk=-

k(n

+k -

(k+l)(n+k) 1)(n + 2k - I)(n

+ 2k)

,k~l,

(47.2)

~

(47.3)

and k(n +k - 1)

hk+1 = - (k

+ 1)(n + k)(n + 2k)(n + 2k + I)'

We assume that (47.2) and (47.3) hold for k calculations show that hm+2 = e2m+1 +e2m+2- hm+1 = - (m and

e2m+2e2m+3 hm+2

=

12m+3

=

k

1.

1,2, ... , m. Simple algebraic

(m + 2)(n + m + 1) + 1)(n + m)(n + 2m + 1)(n + 2m + 2)

(m + 1)(n + m) + 2)(n + m + l)(n + 2m + 2)(n + 2m + 3)

= - -:----------:--:-----::---:-------::------::-:(m

This completes the proof. For the next result, we again quote Ramanujan.

Entry 48 (Formula (1), p. 290). x 4n

+ 2 + 4n + 6 + 4n + 10 + ... n-l n+l n-2 n+2 + -I- - -x- + -1- - -x- + ...

+ -2n x

,= 1

nearly.

Proof. From our remark after Entry 47, it is not difficult to show that x

--

4n

+2 +

x2 4n + 6

---

x2

+ 4n + 10 + . . .

=

.In-3/2(ix/2) In-lj2(ix/2)

I

4n - 2 x

-- - - ,

(48.1)

in the sense that the continued fraction converges to the function on the right side for all (n, x) E (;2, x =f. O. It will be convenient to write the right side in terms of Bessel functions of imaginary argument (Watson [15, p. 77]). Thus, comparing Entry 48 with (48.1), we must show that _In_-....c 3/_ 2(_x__ /2:-) _ _4n_-_2 _ 1 I n -I/2(X/2)

x

n- 1 2n n+l n-2 n+2 nearly, x + ... x + x + where I v denotes the Bessel function of imaginary argument of order v.

(48.2)

72

Ramanujan's Notebooks, Part V

As x tends to

00

(Watson [15, p. 203]),

I v (x/2) '" (

ex/2

1l' X

(-l)kr(v+k+!)

L k' f( k=O. 00

)1/2

V

_ k

2k + !) 2 x

,

Rex> 0,

where we have ignored the exponentially decreasing terms in the complete asymptotic expansion. Using this expansion in (48.2), we find that the left side of (48.2) is asymptotically equal to the quotient

f: (-l)kr(n + k + 1) k! f(n - k)x x f: (-l)kr(n + k) 2 k=O

2n 2F0(1 - n, n + 1; l/x) x 2Fo(1 - n, n; l/x) ,

k

k=O k! r(n - k)x k where 2FO(a, b; z) denotes the (divergent) hypergeometric series 2 Fo(a,

~ (ah(bh k b; z) = ~ k' z.

k=O

.

From the book of Jones and Thron [1, p. 212], it follows that 2n

n-l x + 1 (1 2n/x 1 2n 2F0(1 =x 2Fo(1

n+l

x

+

n-2

n+2

(n + 1)/x 1 1 n, n + 1; l/x) , - n, n; l/x) - n)/x

x

+ ...

(2 - n)/x

1

(n

+ 2)/x 1

+ ...

(48.3) in the sense of correspondence. From Jacobsen's paper [4, Theorem 2.3 (iii)] , it follows that the continued fraction converges for x E C - [0, (0). For positive x, it is likely to diverge. We are now able to properly interpret the word "nearly" in Entry 48, or, equivalently, (48.2). Replacing the left side of (48.2) by a quotient of asymptotic series as x tends to 00, with Re x > 0, we see, from (48.3), that the continued fraction on the right side of (48.2) equals this quotient of asymptotic series in the sense of correspondence of C-fractions. Observe that, if n is an integer, the power series and continued fraction in (48.3) each terminate. Thus, in such an instance, we have equality in (48.3) in the usual sense.

Entry 49 (Formula (2), p. 292). Let x and n denote complex numbers such that Re x > 0, or such that Re x = 0 and 0 < I 1m x I < 1. Furthermore, let y = {(1 +x 2)1/2 -1}/x andm = n(1 +x2)-1/2. Then x l . 2x 2 ---2 +n + 4 + n

+

2.

3x 2 --6 +n

+

3.

4x 2 --8 +n

+...

(1) L

00 (_I)k y2k =y+m y+Yk=1 m + 2k

.

32. Continued Fractions

73

Proof. Let x i= 0. In Entry 22 of Chapter 12 (Part II [2, p. 136]), Ramanujan offers a continued fraction for a certain quotient of ordinary hypergeometric series. Setting a = v - I, f3 = 0, and y = u, and replacing x by a/ fJ with la/ f31 < I, in that theorem, we readily deduce that

1 -2Fl(1 - v, 1; 1 + u; -a/f3) u f3 l(u+v)af3 - uf3 - av + uf3 - av + 1(f3 - a)

+

(49.1)

2(u+v+l)af3 uf3 - av + 2(f3 .- a)

+ .... (This last result was also established by Preece [3].) We now set r = (1 + X 2 )1/2, a = (r - 1)/x 2 , f3 = (r + 1)/x 2, u = (pr + n)/(2r), and v = (pr - n)/(2r), where p will be specified shortly. Then la / f31 = I(r - 1) / (r + 1) I < 1, since Re r = Re(1 + x 2 )1/2 > 0, and (49.1) takes the simplified form I

-2Fl(1- v, 1; 1 +u; -a/f3) u r+I px 2 2(p + 1)x 2 ---p+n + p+n+2 + p+n+4

+

3(p + 2)x 2 p+n+6

(49.2)

+ ....

By a fundamental result on hypergeometric series (Bailey [1, p. 2, eq. (2)]),

Thus, (49.2) may be recast in the form I -(1 U

+ y2)v+u-1 2 Fl (u + v, u; 1 + u;

r+1 - P+n

We now put p that

y(I

+ y2) u

px 2 p +n +2

_y2)

2(p + l)x 2 p +n + 4

(49.4)

3(p + 2)x 2 p + n + 6-

+ + + + .... = 2, and so u + v = 2. Since y(r + 1) = x, we find from (49.4)

. . 2 x l · 2x 2 2Fl(2,u, 1 +u, -y) = - - + - 2+n 4+n

2 . 3x 2

. + --6+n + ...

(49.5)

An elementary calculation shows that (1 - u)(l

+ y2) 2Fl (1, u; I + u; - i ) =

(1

+ y2) 2Fl (2, u; 1 + u;

_y2) - u.

It follows that

y(1 + y2) . . 2 - - - 2Fl(2, u, 1 + u, -y ) u

=y +

1:

u (y

1

+ ~) i2 Fl(l, u; + u; _y2)

(49.6)

74

Ramanujan's Notebooks, Part V

Combining (49.5) and (49.6) and then simplifying somewhat, we complete the proof. Entry 50 (Formula (4), p. 292). Let x, p, and n be complex numbers such that either Re x > 0, or Re x = 0 and 0 < I 1m x I < 1, or p is a nonpositive integer. Furthermore,let y = {(l + X 2 )1/2 - l}/x and let m = n(l + X 2 )-1/2. Then

x p+n = (1

+

1 . px 2 p+n+2

+

2(p + I)x 2 p+n+4

+ I/x 2 )(p-l)/2(2y)P L 00

(

- I)k(p) y2k

k!(m

k=O

+

3(p + 2)x 2 p+n+6 k

+ p + 2k)

+ ...

•

Proof. Let x 2 E C - (-00,0]. Multiply both sides of (49.4) by y. Note that y(r + 1) = x, u = (m + p)/2, and 1 + y2 = 2ry/x. After some elementary algebraic simplification, we deduce Entry 50 in this first case. If, in addition, p is a nonpositive integer, then both the continued fraction and series terminate. We therefore have an identity between two rational functions of x for Re x > o. Hence, the identity holds for all complex x by analytic continuation, when p is a nonpositive integer.

In his second letter to Hardy, Ramanujan [10, pp. xxix, 353] asserted that a

1+n+3+n+5+n+7+n+··· _ 2a

-

t

Jo

dz

zn(l+a 2 )-1/2

{(I

+ a 2)1/2 + I} + z2{(1 + a 2)1/2 -

I}'

which is a particular case of the continued fraction

a p+n

+

pa 2 p+n+2

+

2(p + 1)a 2 p+n+4

+ ... '

which is a particular case of a corollary to a theorem on transformation of integrals and continued fractions." In Ramanujan's Collected Papers [10, p. 353], the third denominator above appears incorrectly as p + n + 3. Now C. T. Preece [3, p. 99] showed that, for n, p > 0,

a p+n

+

pa 2 p+n+2

+

2(p + I)a 2 p+n+4

+ ...

32. Continued Fractions

75

To see that this result is equivalent to Entry 50, replace a by x and write the right side above as 2P x(1 +x 2)(p-I)/2 tp-I+mdt 2 {(l + X )1/2 + I}P {I + t 2y2}p

f'

10

=

2 Px(I

+X 2)(p-I)/2y p xP

f:

(_I)k(phy2k {I t P+m-l+ 2k dt

k=O

k!

10

L -

I)k(p) y2k = (1 + l/x 2)(p-I)/2(2y)P k . k=O k! (p + m + 2k) 00

(

Entry 51 (Formula (1), p. 292). Let x and n be complex numbers such that Re x =1= 0, or such that Re x = 0 and 0 < I 1m x I < 1. Furthermore, let y = {(I + X2)1/2 - l}/x and let m = n(1 + X2)-1/2, where the principal branch of (I + X 2 )1/2 is chosen. Then x

x2

(2x)2

(3x)2

00

(._I)k y 2k+1

I+n + 3+n + 5+n + 7+n + ... =2t;m+2k+I' An elementary calculation shows that Iyl = 1 if and only if Rex = 0 and 11m xl ~ 1. The choice of the principal branch of (1 +X 2 )1/2 ensures that Iyl < 1, and so the series on the right side above converges. Entry 51 is simply the case p = 1 of Entry 50.

Entry 52 (Formula (3), p. 292). Let n and p denote complex numbers such that either Re n > 0 or p is a nonpositive integer. Then I .P 2(p + 1) 3(p + 2) 4(p + 3) n+ n + n + n + n + ...

= 2P

f:

k=O

(_I)k(ph . k! (n + p + 2k)

Proof. Replace n by nx in Entry 50; thus, now m = nx(1 +x2)-1/2. We then find that x 1 . px 2 2(p + l)x 2 3(p + 2)x 2 p + nx + p + 2 + nx + p + 4 + nx + p + 6 + nx + ... 1 l·p 2(p+1) 3(p+2) = n + pIx + n + (p + 2)/x + n + (p + 4)/x + n + (p + 6)/x + ...

_(1

-

2

~)(P-I)/2 P 00 (_I)k(phy2k + x2 ( y) t ; k!(m + p + 2k)'

(52.1) Now let x tend to 00. Then y tends to I and m approaches n. Thus, we see that the left and right sides of (52.1) approach, respectively, the left and right sides of

76

Ramanujan's Notebooks, Part V

Entry 52. To see that equality still holds, we apply the uniform parabola theorem, just as we did in the proofs of Entries 24, 14, and 19. Entry 53 (p. 342). Let x and y be complex numbers with Re x > 0 and Re y > O. Then (y

x+

+ 1)2 + n

+ (x + 1)2 + n 2x

=y+

2y

(y + 3)2 2x

+n

+

2y

+ (x + 3)2 + n

(y

+ 5)2 + n

+ ... (x + 5)2 + n + 2y + ... 2x

(y + 1)2 + n (x + 1)2 + n (y + 3)2 + n (x + 3)2 + n = x + -=----'----x+y+2 + x+y+4 + x+y+6 + x+y+8 + ....

We remark that, by symmetry, each of the continued fractions above is also equal to

y+

(x + 1)2 + n x+y+2

+

(y + 1)2 + n x+y+4

+

(x + 3)2 + n x+y+6

+

(y + 3)2 + n x+y+8

+ ...

.

The first equality in Entry 53 is actually the same as Entry 27 of Chapter 12 (Part II [2, p. 146]), for x > 0 and y > O. As we remarked there, this elegant identity is found in Ramanujan's [10, p. xxix] second letter to Hardy. The first proof in print is by Preece [2], and the result can also be found in Perron's book [1, p. 37, eq. (31)]. This result has also been proved by Ramanathan [6]. Since both continued fractions converge locally uniformly for Re x > 0 and Re y > 0, the identity follows by analytic continuation for Re x > 0 and Re y > O. The first continued fraction diverges for Re x = 0, while the second diverges for Re y = O. The identity does not hold if Re x < 0 and/or Re y < 0, since the first continued fraction is an odd function of x but not of y, whereas the second is an odd function of y but not of x. The third (and fourth) continued fraction converges to a meromorphic function of x and y, since it is equivalent to a continued fraction K(cdl), where

as k tends to

00.

Proof. As just indicated, it suffices to establish the second equality. In Gauss's continued fraction, Entry 20 of Chapter 12 (part II [2, p. 134]), we set x = 1 and then replace a, p, and y by (x - n - 1)/2, (x + n - 1)/2, and

32. Continued Fractions (x

+ y) /2, respectively. After some simplification, we find that 2FI (4(y + 1 + n), !(x - 1 + n); !(x + )I); -1) (x + y) (I 1 1 ) 2FI l(y + 1 + n), l(x + 1 + n); l(x + y -+- 2); -1 (x + 1)2 - n 2 (y -+- 3)2 - n 2 (y + If - n 2 = x + y + --'----'-----x+y-+-2 + x+y+4 + x+y+6 (x

+

+ 3)2 -

n2

x+y+8

77

(53.1)

+ ...

Second, we use Euler's continued fraction, Entry 22 of Chapter 12 (Part II [2, p. 136]), when x = 1 and a, tJ, and y are replaced by -(y + 1 +n)/2, (x -1 +n)/2, and (x + y)/2, respectively. Upon simplification, we deduce that 2FI

(x

(4(y

+ 1 + n), !(x -

1 -+- n); !(x

+ y); -1)

+ y) 2FI (Il(y + 1 + n), l(X 1 1 + 1 + n); l(X + y -+- 2); -1 ) (x

+ 1)2 -

n2

=2y+----2y

+

(x

+ 3)2 -

n2

2y

(x

+ 5)2 -

n2

+ - - 2 y - - + ... ·(53.2)

Comparing (53.1) and (53.2) and replacing n by i..,fo, we deduce the second equality of Entry 53. It is interesting to note that the third continued fraction in Entry 53 can be obtained from the second one by repeated applications of the Bauer-Muir transformation with modifying factors Wk = x - y + 2k. Also, the first continued fraction can be obtained from the second one by repeated use of equally simple Bauer-Muir transformations.

Entry 54 (p. 342). For all complex numbers x and n, 1 ~ (-xlIx ;; ~ (n

1

x

2

x

3

+ Ih = ;; + T + ;; + T + ;; + T + ;; + ... 1 x 2x 3x n+x - n+x+I - n+x+2 - n+x+3 - ...

Proof. The latter continued fraction in Entry 54 is merely the even part of the former continued fraction, a fact immediately seen from (64.1). To establish the first part of Entry 54, replace x by x / f3 and set y = n in Part II [2, p. 134, Entry 21, eq. (2l.2)]. Since the continued fraction converges uniformly with respect to tJ in a neighborhood of tJ = 00, we may let tJ tend to 00 to complete the proof. Alternatively, we can replace x by - x in the second continued fraction of Corollary 1 of Entry 21 of Chapter 12 (Part II [2, p. 136]) to immediately achieve the desired result, since both continued fractions converge for all x and n.

78

Ramanujan's Notebooks, Part V

Entry 55 (p. 343). For every complex number x, x

1 - e- X Proof. Setting n

-1 -

~

~

~

~

3

x

x

+ 1 + 1 + 1 + 1 + 1 + 1 + 1 + ....

= 1 in Entry 54, we deduce that

1 - e- X = _~ X

f:

(_x)k =

k!

X k=1

f:

k=O

1

(_X)k

(k+l)!

1

x

x

2

3

x

=1+1+1+1+1+1+1+···· Taking the reciprocal of both sides, we complete the proof.

Entry 56 (p. 342). For all complex x, I

z(e

a

x

- 1) =

x

x

x

x

x

1 - 1 + 3 - 1 + 5 - 1 + ....

Proof. If we set n = 1 and replace x by 2x in the first continued fraction of Corollary 1 of Entry 21 of Chapter 12 (Part II [2, p. 136]), we find that, for all complex x, ~(e2X - I) = x

IFI (1;

2; 2x)

2x 2x = x1-2+3-4+5-6+7-'" -4x -4x -6x -6x x

x

x

x

x

x

x

-1-1+3-1+5-1+7-'" . Entry 56 also readily follows from a continued fraction for eZ found in Wall's book [1, p. 348].

Entry 57 (p. 343). For all complex numbers x and n, x - n

n + -----;-xk 00

k~ (n + Ih

x

n

x

1+1+1+

n+l

x

+1+

n+2

1 + ....

Proof. By a straightforward calculation, it is easily shown that the left side of Entry 57 is equal to

32. Continued Fractions

79

To show that F (x) has the given continued fraction, we require the continued fraction

2F,(a, fJ; y; x) _ 1 _ .!.. 2F,(a + 1, fJ; y + 1; x) y

{

fJ(y - a)x

(fJ - a)x + y + 1

(fJ + l)(y -a + l)x - (fJ - a + l)x + y + 2 (fJ + 2)(y - a + 2)x - (fJ - a + 2)x + y + 3

(57.1)

} - ...

'

due toE. Frank [1] and valid for Ixl < 1. Replacing x by x/fJ, letting fJ --+ 00, and using the fact that the resulting continued fraction in (57.1) converges uniformly with respect to fJ in a neighborhood of fJ = 00, we find that, for all x,

,F1(a+l;y+l;x) ,F,(a; y; x) y (y - a)x (y - a + l)x (y - a + 2)x y - x+y+l x+y+2 - x+y+3 Putting a

=

= n + 1, we deduce that, for all x, 1 nx (n + l)x (n + 2)x n + 1 - x + n + 2 - x + n + 3 - x + n + 4 - ...

1 and y

1

~ F(x)

=

By (64.1), this last continued fraction is the even part of

CF(x):=

x

n

1+1+1+

n+l

x

n+2

+ 1 + -1- + ....

It remains to show that C F (x) converges to F (x) / x . The odd part of C F (x) is

1

n

(n

+ l)x

(n

+ 2)x

(n

+ 3)x

l-x+n+l-x+n+2-x+n+3-x+n+4-"" which converges for all x and n. Thus, the even and odd parts of CF(x) both converge to meromorphic functions of x and n. The even part converges to F (x) / x, and so we want to show that the odd part also converges to F (x) / x. Now C F (x) and thus the even and odd parts converge to the same values for x > 0 and n > O. Therefore, by analytic continuation, they are equal for all x and n. This completes the proof. Frank's continued fraction (57.1) can, in fact, be derived from Euler's continued fraction, Entry 22 of Chapter 12 (Part 11[2, p. 136]). Entry 58 (p. 343). Let x be a complex number such that Rex 2 > sinh-' x x 2x 2 = (1 + x 2 )1/2 1+ 1

+

2(1

+ x 2) 1

4x 2

4(1

+ x 2)

-to Then

+ 1 +--1-+ .. ··

80

Ramanujan's Notebooks, Part V

Proof. Using a familiar transfonnation for 2FI (Bailey [1, p. 2, eq. 2]), we find that (Gradshteyn and Ryzhik [1, p. 60]) sinh-I x

=x

2FI (~, ~; ~; _x 2)

= x(1 + X2)1/2 2F I (1, 1;

~; _x 2).

We now apply Entry 21, eq. (21.2), of Chapter 12 (part II [2, p. 134]) with {3 = 0, y = 1/2, and x replaced by x 2 • Hence, for Rex 2 > -~, sinh-Ix _ .3. 2 + X2)1/2 - X 2FI (1,1, 2' -x )

(1

1+x2

x2

x/2

-

1/2 + 1 +

3x 2 3(1 +x 2) 2x 2 2(1 + x 2) + 1 + 1/2 + ... ' + 1 + 1/2

--1/2

which is easily seen to be equivalent to the proposed continued fraction. Entry 59 (p. 343). Let x be any complex number such that Re x 2 > - ~. Then tan-I x

2(1 + x 2) 3x 2 4(1 + x 2) x2 1+1+ 1 +1+ 1 + ....

= -x

Proof. We know that (A. Erdelyi [1, p. 102])

tan-I x = x 2FI (~, 1; ~; _x 2). We again apply Entry 21, eq. (21.2), of Chapter 12 (Part 11[2, p. 134]) but now with {3 = -~, y = ~,and x replaced by x 2 • Thus, for Rex 2 > -~, x/2 x 2/2 l+x2 3x 2/2 2(I+x 2) x = 1/2 + + 1/2 + 1 + 1/2 + ... '

_I

tan

which is equivalent to the proposed continued fraction.

5. General Theorems Entry 60 (p. 339). For 1 nIl I: -=k=1 ak

~

k 2

~

_a_l_

n, assume that ak 2

~

i= O. Then 2

an-I. al - al +a2 - a2 +a3 - ... - an-I +an

(60.1)

In fact, we have stated a finite version of Ramanujan's claim, i.e., Ramanujan's statement is for "n = 00." Proof. Entry 60 is easily established by induction on n. In fact, Entry 60 is a version of an identity

I: b b k=1 n

l

2

bl b2 b3 bn ···bk = ----, 1 - 1 + b2 - 1 + b3 - ... - 1 + bn

(60.2)

32. Continued Fractions

81

due to Euler (Jones and Thron [1, p. 37]), where bk =I- 0, 1 .:-::: k ::; n. To derive (60.1) from (60.2), set b I = l/al and bk = ak-I/ab 2 ::; k:-::: n. After a simple equivalence transformation, we deduce (60.1). In the following three entries, Ramanujan examines the convergence and divergence of limit k-periodic continued fractions of the form al

a2

a3

(61.1)

p+p+p+ ... '

where limn~oo akn+j = aj E i, for 1 ::; j ::; k. The convergence behavior for the special periodic case, akn+ j = aj, 0 ::; n < 00, has been known since the 1880s (0. Stolz [1], Jones and Thron [1, p. 46]). If we think of the continued fraction (61.1) as being generated by the linear fractional transformations sn(w)

= -an- , p+w

n

= 1,2,3, ... ,

such that at

a2

an

p+p+···+p+w

+ An-tw Bn + Bn-Iw

An

-----,

(61.2)

we may deduce the following information: (1) For k = 1, the approximants Sn(O) ofthe continued fraction

a

a

a

p+p+p+ ...

(61.3)

are just iterations of the linear fractional transformation S] (w) evaluated at w = O. Hence, (61.3) converges if SI (w) has one attractive fixed point and it has no repulsive fixed point at O. The fixed points of S] (w) are p(±/1 + 4a/ p2 - 1)/2. Hence, (61.3) converges if and only if

4a/p2

E

C-

(-00,

-1).

(61.4)

(2) For k > 1, we regard the periodic continued fraction as iterations of SkeW), given by (61.2), evaluated at the points (61.5)

Hence, the k-periodic continued fraction converges if and only if SkeW) has an attractive fixed point and it has no repulsive fixed points at any of the points (61.5). It was first pointed out by T. N. Thiele [1] in 1879 that if Sk (w) has a repulsive fixed point at one of the points (61.5), then the periodic continued fraction diverges. This phenomenon is therefore called Thiele oscillation (Perron [1, p. 87]). For more details, we refer to Jones and Thron's book [1, p. 47]. The results quoted above were probably known to Ramanujan who most likely derived them himself, because they are not found in the books of G. Chrystal [1] or G. S. Carr [1], the two primary sources of information about continued fractions for

82

Ramanujan's Notebooks, Part V

Ramanujan. He then must have realized that he could generalize these results to limit k-periodic continued fractions, and Entries 61 and 63 below are the results of his investigations. His first result is on limit I-periodic continued fractions.

Entry 61 (p. 339). -

1-1-1-1-1-···

(61.6)

is intelligible or not according as limn ..... oo an < or > 1/4.

Here we have precisely quoted Ramanujan. By "intelligible," Ramanujan evidently means "convergent." In the periodic case an = a, Entry 61 is true, since the condition (61.4) then reduces to a E and where g(n) is any positive function monotonically decreasing to 0 as n -+ 00, for example, g(n) = cn- a , for some constant c and positive number a. For the next entry, we again quote Ramanujan. Entry 62 (p. 340). The continued fraction 1 p+p+p+p+ ... tends to two limits or one limit according as L

(62.1)

1/Fn is convergent or divergent.

Ramanujan evidently considered an, I ~ n < 00, to be positive and p to be real. From Stieltjes' classical work [2], [3, pp. 402-566], it follows that (62.1) converges if and only if 00

00

ala3··· a2n-1 n=l a2 a 4··· a2n

a2a4··· a2n n=l ala3 ... a2n+1

~

~

L-----+ L

=00;

(62.2)

otherwise, its even and odd parts converge to two distinct values. This coincides with the natural interpretation of Entry 62, except for one matter; the condition (62.2) is not equivalent to Ramanujan's condition (62.3) n=l unless one makes further restrictions. Indeed, (62.3) is a sufficient condition for the convergence of (62.1) (Perron [1, p. 47]), but there exist convergent continued fractions (62.1) with an > 0, p > 0, and L 1/ Fn < 0". For instance, the continued fraction 14 14 34 34 54 54

-----1

+

1

+

1

+

1

+

1

+

I

+ ...

converges since 00

~ ala3 ... a2n-1 L ----n=l a2 a 4··· a2n

00

~

=L

n=l

I

= 00.

Ramanujan was not the only person to have made this mistake; for example, see Khovanskii's book [1, p. 45]. Entry 63 considers limit k-periodic continued fractions for I ~ k ~ 5. We state a rather general version, although Ramanujan probably examined only real continued fractions. Entry 63 (p. 340). Consider C F :=

al

a2

a3

a4

1-1-1-1-···

.

(1) llCF is limit 1-periodic, then CF converges iflimn~ooan = a E C-

[i,oo).

84

Ramanujan's Notebooks, Part V

(2) If C F is limit 2-periodic with limits a and b, then C F converges if ab

~{1---(~a-+~b~)}~2

E C - [1 00) 4"

where a + b ::/= I. (3) IfC F is limit 3-periodic with limn - HXl a3n+1 lim n ---+ oo a3n = c, then C F converges if abc (l-(a+b+c)}2

E

= a, limn -

HXl

a3n+2

= b, and

C - [1 00) 4"

where a + b + c ::/= 0, and iflal > Icl when b = 1, Ibl > lal when c = 1, and Icl > Ibl when a = 1. (4) Suppose that CF is limit 4-periodic with a4n+I, a4n+2, a4n+3, and a4n tending to a, b, c, and d, respectively, as n tends to 00. Then C F converges if {I - (a

C

abcd

+ b + c + d) + (ac + bd)}2

E

I

-

["4,00),

wherea+b+c+d-ac-bd::/= I,andiflabl > Icdlwhenb+c= 1, Ibcl> ladl whenc+d = 1, Icdl > labl whena+d = 1, and ladl > Ibcl when a +b = 1. (5) Suppose that aSn+l, aS n+2, as n+), aSn+4, and aS n approach a, b, c, d, and e, respectively, as n tends to 00. Then C F converges if abcde {l-(a+b+c+d+e)+a(c+d)+b(d+e)+ce}2

---------------------:- E

I

C - [- 00) 4"

where the denominator above is not equal to 0, and if labO - d)1 > Ide(l - b)1, when

b + c + d - bd = 1,

Ibc(l-e)1 > lea(1-c)l,when

c+d+e-ce= 1,

Icd(1 - a)1 > lab(1 - d)l, when

d

+e+a -

da

=

1,

Ide(1-b)1 > Ibc(1-e)l,when

e+a+b-eb= 1,

lea(1 - c)1 > Icd(l - a)l, when

a

and

+b +c -

ac

= 1.

The "extra" conditions on the parameters a, b, ... are not given by Ramanujan. Thus, for example, in (3), Ramanujan says that C F "is intelligible when {1 - (a + b + c)}2 - 4abc is positive." The primary conditions insure that Sdw) has an attractive fixed point and a repulsive fixed point. The "extra" conditions eliminate the cases where the corresponding k-periodic continued fraction diverges by Thiele oscillation. Such limit k-periodic continued fractions were first studied by M. von Pidoll [1] in 1912 and by O. Szasz [1] in 1917. It is interesting to note that Ramanujan probably made his discoveries in the period 1912-1914.

32. Continued Fractions

85

Note that part (1) follows from Entry 61 and the remarks we made following it. Proof. The corresponding periodic continued fraction is given by

-a b c 1 -1-1-'" The corresponding linear fractional transfonnation

-a b * Ak +Ak_IW -- = " 1 - 1 - ... - 1 + W Bk + Bk_IW

Sk(W) = -

has an attractive fixed point and a repulsive fixed point if either (a)

Bk- I =1= 0, Ak-I

+ Bk =1= 0,

and

or (b)

(See, for instance, the book by Jones and Thron [I, pp. 51-52].) (If Sk is singular, i.e., one or more of the elements a, b, ... are equal to 0, then Sk is a constant function whose value we regard as the attractive fixed point of Sk. Its "repulsive fixed point" is then the point W for which Sk is not well defined. For more details, we refer to Jacobsen's paper [3].) From the work of von Pidoll [1], Szasz [1], and, in more generality, Jacobsen [3], it further follows that if condition (a) or (b) holds and none of the points (61.5) is the repulsive fixed point of Sk (Thiele oscillation), then the limit k-periodic continued fraction converges. The results in Entry 63 arise from the application of these criteria when k = 1,2,3,4,5. As noted earlier, the case k = 1 was examined in Entry 61," Let k = 2. Then + W) S2 (W ) -_ • l-b+w

-a(l

For case (a) we require that BI

=

1 =1= 0, Al

+ B2 =

I- a- b

=1=

0, and

larg ( 1 + 4(~~~1+-~~~d) I= larg ( 1 - (l _~~ b)2) I< 7r, i.e., ab/{l - (a + b)f E C - [~, 00). Note that case (b) is impossible. If one of the fixed points in (61.5) is the repulsive fixed point of S2 (w ), then this fixedpointiseitherOor-a.Nowwl = OisafixedpointofS2(w)ifA2 = -a = O. Then S2(W) =

o

1-

b

+W

is singular, and WI = 0 is the repulsive fixed point if and only if 1 - b = O. But this contradicts the requirement 1 - a - b =1= o. Thus, 0 is not a repulsive fixed point. If WI = -a is a fixed point of S2(W), then b = O. It is easily seen that the

86

Ramanujan's Notebooks, Part V

other fixed point of S2(W) is W2 = -1. Now WI = -a is the attractive fixed point of S2(W) (and not the repulsive one) if and only if (63.1) (Jones and Thron [1, p. 52]), i.e., if and only if 11 - a I > O. This is true since 1 - a i= O. Thus, WI = -a also is not a repulsive fixed point, and there is no Thiele oscillation. This completes the proof of (2). Next, let k = 3. Then S3(W)

abc

= --

-

1-1-I+w

where A3 = -a(l - c), A2 = -a, B3 = 1 - b - e, and B2 = 1 - b, by the recursion formulas (0.5) and (0.6), or by direct calculation. For condition (a), we require B2 = 1 - b i= 0, i.e., b i= 1; A2 + B3 = 1 - (a + b + c) i= 0, i.e., a

+ b + e i=

1; and larg (1 + 4(A3(AB 2+- BB3:2») I < :n:, i.e., 2

3)

abc I -{I---(a-+-b-+-e-)}--=-2 E C -Li> 00).

For case (b), we require B2 = 1 - b = 0, i.e., b lal i= 11 - b - el = lei. Now with b = 1

= 1, and

(63.2) IA21

i=

IB31, i.e.,

abc (l-(a+b+e)}2

if and only if la I i= lei. Hence, S3 has an attractive and a repulsive fixed point (in the extended sense if S3 is singular, i.e., if abc = 0) if (63.2) holds. We next need to determine the conditions that yield repulsive fixed points. If one of the fixed points in (61.5) is the repulsive fixed point of S3(W), then this fixed point is either 0, -a, or -a/(l - b). First, WI = 0 is a fixed point of S3(W) if S3(0) = A3 = 0, i.e., if -a(l-e) = O. Case 1. a

= O. Then S3(W)

o

= ------1- b - e

+ (l

- b)w

is singular, and WI = 0 is "the repulsive fixed point" if and only if 1 - b - e = O. But this is impossible by (63.2). Case 2. e

= 1, a i= O. Then S3(W)

=

-aw -b + (l - b)w'

and the other fixed point of S3(W) is given by W2 = (b - a)/(l - b). Hence, in analogy with (63.1), WI = 0 is the attractive fixed point of S3(W) (and not the repulsive one) if and only if IB3

+ B2wII

> IB3

+ B2 w 21

32. Continued Fractions

87

(Jones and Thron [1, p. 52]), Le., Ibl > lal. (The case IB3 + B2W1I = IB3 + B2w21 is excluded by (63.2).) In conclusion, under the condition (63.2), if e = 1, we need to require that Ibl > lal for the convergence of C F. Next, W3 = -a is a repulsive fixed point of S3(W) if and only if WI = 0 is a repulsive fixed point of

~ a 1 -1-1+w

S(I)(W) := -b 3

So, by symmetry, we need to require that lei > Ibl if a = 1 in order for C F to converge. Likewise, W3 = -a/(1 - b) is a repulsive fixed point of S3(W) if and only if WI = 0 is a repulsive fixed point of (2)

._

-e

S3 (w).- -

a

b

--, 1 -l-l+w

which yields the requirement lal > lei if b = 1. This concludes the proof of (3). Cases k = 4 and k = 5 are proved in exactly the same matmer as case k = 3. However, with increasing k, the details become more laborious. For these reasons, we shall provide only brief sketches of the proofs when k = 4 and k = 5. Let k = 4. Then

+ A3W , + B3W where A3 = a(e-l), A4 = a(e+d -1), B3 = I-b-e, and B4 = 1-b-e-d+bd. For condition (a), we require that b + e 1= 1, 1 - (a + b + e + d) + ae + bd 1= 0, S4(W)

=

A4

B4

and

C

abed

1

+ b + e + d) + (ae + bd)}2 E - Li' 00). In case (b), we need b + e = 1 and labll= ledl. Now when b + e = 1, {I - (a

abed

{1 - (a

+ b + e + d) + (ae + bd) F

(63.3)

1 abed {ab+edF EC-Li'OO)

if and only if labll= ledl. Hence, S4 has an attractive fixed point and a repulsive fixed point if (63.3) is valid. We next need to determine when the points (61.5) are repulsive fixed points. Now WI = 0 is a fixed point if A4 = a(e + d - 1) = O. It is easy to see that the case a = 0 is impossible. If e + d = 1, then S4(W)

A3 W

= ---B4

+ B3 W

has the fixed point W2 = (A3 - B 4)/ B 3. Thus, in analogy with (63.1), WI is not a repulsive fixed point if and only if I B 41

i.e., if and only if Ibel > ladl.

> IB4

+ B3 w 21 =

I A 31,

88

Ramanujan's Notebooks, Part V

The remaining three conditions listed in Entry 63 for the case k = 4 arise from the remaining three possible repulsive fixed points in (6l.5) and considerations of symmetry. Let k = 5. Then

Ss(W)

= As + A4w , Bs

+ B4w

where A4 = a(c+d -1), As = a(c+d +e -ce -1), B4 = I-b -c -d +bd, and Bs = 1 - b - c - d - e + ee + be + bd. For condition (a), we require that 1- (b +c +d) +bd =1= 0, 1- (a +b + c +d +e) +ad +ae+ ee+be +bd =1= 0, and

abcde I (l-(a+b+e+d+e)+a(e+d)+b(d+e)+ce}l EIC-[4'OO). (63.4) Forgoing the calculations for condition (b), we conclude that Ss has an attractive fixed point and a repulsive fixed point if (63.4) holds. We now examine the five possible repulsive fixed points given by (61.5). Now, WI = 0 is a fixed point of Ss(w) if Ss(O) = As = a(e + d + e + ee - 1) = O. The case a = 0 is impossible, and so we assume that e + d + e + ee - 1 = O. The remaining fixed point of Ss(w) is (A4 - Bs)/ B4. It follows that WI is not a repulsive fixedpointifandonlyiflBsl > IA 4 1,Le.,ifandonlyiflbe(1-e)1 > lae(l-e)l. The remaining four possible repulsive fixed points yield the additional restrictions listed for the case k = 5 of Entry 63. It may be remarked that the "extra" conditions in Entry 63 can be eliminated if we use the notion of general convergence (Jacobsen [2]).

Entry 64 (p. 342). If n is even, then al az an b l + b1 + ... + bn al

alb 1 + blb1

-

ala3b4 a4a Sbl b6 a 3 b4 + b1 (a4 + b3 b4) - aSb6 + b4(a6 + bsb6) an-lan-Ibn-4bn

This is just the finite form of the even part of an infinite continued fraction, namely (Jones and Thron [1, p. 42]), the even part of

al bl

bo +-

is

al

+ -b z + ...

33 Ramanujan's Theories of Elliptic Functions to Alternative Bases

1. Introduction In his famous paper [3], [10, pp. 23-39], Ramanujan offers several beautiful series representations for 1I rr. He first states three formulas, one of which is 4 -; =

where (a)o

L n=O 00

(6n + 1)(4)~ (n!)34J1 '

= 1 and, for each positive integer n, (a)n

= a(a + l)(a + 2)··· (a + n -

1).

He then remarks that "There are corresponding theories in which q is replaced by one or other of the functions

ql = exp( -rr.J2K;IK 1) , q2

= exp( -27rK~/(K2v'3»),

q3 = exp (-2rr K~I K3)' where KI

=

2FI

n, ~; 1; k 2),

K2

=

2F\

(l, ~; 1; k 2 ),

K3

=

2F\

U, ~; 1; k 2)."

Here Kj = Kj(k'), where 1 ::::: j ::::: 3, k' = ../1 - k 2 , and 0 < k < 1; k is called the modulus. In the classical theory, the hypergeometric functions above are replaced by 2 FI 1; k 2 ). Ramanujan then offers 16 further formulas for lin that arise from these alternative theories, but he provides no details for his proofs. In an appendix at the end of Ramanujan's Collected Papers [10, p. 336], the editors, quoting L. J. Mordell, lament "It is unfortunate that Ramanujan has not developed in detail the corresponding theories referred to in '1[14." Ramanujan's formulas for 1/rr were not established until 1987, when they were first proved by J. M. and P. B. Borwein [1, pp. 177-188], [2], [3]. To prove these formulas, they needed to develop only a very small portion of the "corresponding theories" to which Ramanujan alluded. In particular, the main ingredients in their

(4, 4;

90

Ramanujan's Notebooks, Part V

U, !;

1; x) , to each of work are Clausen's formula and identities relating 2 F J the functions 2FJ G,~; 1; x), 2 F J O,~; 1; x), and 2FJ G,~; 1; x). The Borweins [4], [6] further developed their ideas by deriving several additional formulas for 1/7t. Ramanujan's ideas were also greatly extended by D. V. and G. V. Chudnovsky [1], [2] who showed that other transcendental constants could be represented by similar series and that an infinite class of such formulas existed. Ramanujan's "corresponding theories" have not been heretofore developed. Initial steps were taken by K. Venkatachaliengar [1, pp. 89-95] who examined some of the entries in Ramanujan's notebooks [9] devoted to his alternative theories. The greatest advances toward establishing Ramanujan's theories have been made by J. M. and P. B. Borwein [5]. In searching for analogues of the classical arithmetic-geometric mean of Gauss, they discovered an elegant cubic analogue. Playing a central role in their work is a cubic transformation formula for 2FJ (t, ~; 1; x), which,in fact, is found on page 258 of Ramanujan's second notebook [9], and which was rediscovered by the Borweins. A third major discovery by the Borweins is a beautiful and surprising cubic analogue of a famous thetafunction identity of Jacobi for fourth powers. We shall describe these findings in more detail in the sequel. As alluded in the foregoing paragraphs, Ramanujan had recorded some results in his three alternative theories in his second notebook [9]. In fact, six pages, pp. 257-262, are devoted to these theories. These are the first six pages in the 100 unorganized pages of material that immediately follow the 21 organized chapters in the second notebook. Our objective in this chapter is to establish all of these claims. In proving these results, it is very clear to us that Ramanujan had established further results that he unfortunately did not record either in his notebooks, unpublished papers, or published papers. Moreover, Ramanujan's work points the way to many additional theorems in these theories, and we hope that others will continue to develop Ramanujan's beautiful ideas. The most important of the three alternative theories is the one arising from the hypergeometric function 2 F, (t, ~; 1; x) . The theories in the remaining two cases are more easily extracted from the classical theory and so are of less interest. We first review the classical terminology and theory, which can be found in Part III [3]. In particular, see Chapter 16, pages 34-37, Chapter 17, pages 101-102, and Chapter 18, pages 213-214. The complete elliptic integral of the first kind K = K(k) associated with the modulus k, 0 < k < 1, is defined by

K:=

l

o

dq;

lf!2

J1 -

k 2 sin 2 q;

=!7t

2FJ

(!,!; 1; k

2 ),

(1.1)

where the latter representation is achieved by expanding the integrand in a binomial series and integrating termwise. For brevity, Ramanujan sets (1.2)

33. Elliptic Functions to Alternative Bases

91

The base (or nome) q is defined by (1.3) where K' = K(k'). Ramanujan sets x (ora) = k 2 • Let n denote a fixed positive integer, and suppose that

n

(4, 4; 1; 1 - k 2) 2FI (4, 4; 1; k 2)

2FI

=

(4,4; 1; 1 _l2) , 2FI (4, 4; 1; 12)

2FI

(1.4)

where 0 < k, l < 1. Then a modular equation of degree n is a relation between the moduli k and l which is implied by (1.4). Following Ramanujan, we put a = k 2 and f3 = 12. We often say that f3 has degree n, or degree n over a. The multiplier m is defined by 2FI

m=

(4, 4; 1; a) (I

1.

.

2FI 2' 2' 1,

f3

(1.5)

).

We employ analogous notation for the three alternative systems. The classical tenninology described above is represented by the case r = 2 below. For r = 2, 3, 4, 6 and 0 < x < 1, set I r-I ) z(r) := z(r; x):= 2FI ( ;:' - r - ; 1; x

(1.6)

and

qr .- qr(x) .- exp ( -1f csc(1f /r) ._._

2

F(! 1 r' r-1.1.1-X)) r ' , (I r-I. .) . 2FI r' -r-' I, x

In particular, _

(21f2FIH'~;I;I-X))

q3 - exp -

r;;

",3

( 1 2• .)

2FI 3' 3' I,x

,

(1.7)

(1.8)

and (1.9) (We consider the notation (1.7)-(1.9) to be more natural than that of Ramanujan quoted at the beginning of this chapter.) Let n denote a fixed natural number, and assume that 2FI (~, ~; I; 1 -

n

a)

F(! r-l.l·a) 2 1 r' r ' ,

U. ~; I; 1 - f3) F(! r-I·I·f3) • 2 1 r' r ' ,

2FI

=

(1.10)

92

Ramanujan's Notebooks, Part V

where r = 2, 3, 4, or 6. Then a modular equation of degree n is a relation between a and f3 induced by (1.10). The multiplier mer) is defined by ()

m r

=

2

2

FI

F

I

(I;, (1r'

r-I·I·a) -r-' , r-\. r'

1.

(1.11)

II)'

,fJ

for r = 2, 3, 4, or 6. When the context is clear, we omit the argument r in qr, Z(r), and mer). In the sequel, we say that these theories are of signature 2, 3, 4, and 6, respectively. Theta-functions are at the focal point in Ramanujan's theories. His general theta-function f(a, b) is defined by 00

a n(n+l)/2b n(n-l)/2,

f(a, b):= L

labl

< 1.

n=-oo

If we set a = qe 2iZ , b = qe- 2iZ , and q = e rrir , where z is an arbitrary complex number and Im(r) > 0, then f(a, b) = U3 (z, r), in the classical notation of Whittaker and Watson [1, p. 464]. In particular, we utilize three special cases of f(a, b), namely, 00

qJ(q) := f(q, q) =

(1.12)

L n=-oo 00

1{!(q):= f(q, q3) = Lqn(n+l)/2,

(1.13)

n=O

and f(-q):= f(-q, _q2)

=

00

= 0(1- qn), 00

(_It q n(3n+1)/2

L n=-oo

(1.14)

n=l

where Iq I < 1. The last equality above is Euler's pentagonal number theorem, which is most easily derived from Jacobi's triple product identity (Part III [3, p. 35, Entry 19]). One of the fundamental results in the theory of elliptic functions is the inversion formula (Whittaker and Watson [1, p. 500]; Part III [3, p. 101, eq. (6.4)]) (1.15) We set (1.16)

for each positive integer n, so that ZI = z. Thus, by (1.5), (1.15), and (1.16), m

Zl

qJ2(q)

Zn

qJ2(qn)

= - = --.

(1.17)

In the sequel, unattended page numbers, particularly after the statements of theorems, refer to the pagination of the Tata Institute's publication of Ramanujan's

33. Elliptic Functions to Alternative Bases

93

second notebook [9]. We employ many results from Ramanujan's second notebook in our proofs, in particular, from Chapters 17, 19,20, and 21.

2. Ramanujan's Cubic Transfonnation, the Borweins' Cubic Theta-Function Identity, and the Inversion Fonnula In classical notation, the identity t'Jt(q) = t'J1(q)

+ t'Ji(q)

is Jacobi's famous identity for fourth powers of theta-functions. In Ramanujan's notation (1.12) and (1.13), this identity has the form (Part III [3, p. 40, Entry 25(vii)]) (2.1) The Borweins [5] discovered an elegant cubic analogue which we now relate. For (J) = exp(2Jl'i /3), let 00

a(q) :=

L L

(2.2)

m,n=-oo 00

b(q):=

(J)m-nqm2+mn+n2,

(2.3)

m,n=-oo

and (2.4) m,n=-oo

Then the Borweins [5] proved that a\q) = b 3(q)

+ c\q).

(2.5)

They also established the alternative representations 00

a(q)

= 1 +6~

(q3n+l 1_ q 3n+l -

q3n+2)

l_ q 3n+i

(2.6)

and (2.7) Formula (2.6) can also be found in one of Ramanujan's letters to Hardy, written from the nursing home, Fitzroy House [11, p. 93], and is proved by us in [9]. The identity (2.7) is found on page 328 in the unorganized portions of Ramanujan's second notebook and was proved in Part III [3, p. 462, eq. (3.6)] in the course of proving some related identities in Section 3 of Chapter 21 in Ramanujan's second notebook. Furthermore, the Borweins [5] proved that b(q)

=

~ {3a(q3) - a(q)}

(2.8)

94

Ramanujan's Notebooks, Part V

and c(q)

=! {a(ql/3) -a(q)}.

(2.9)

The Borweins' proof of (2.5) employs the theory of modular forms on the group generated by the transformations t --+ 1/ t and t --+ t + i../3. Shortly thereafter, they and F. G. Garvan [1] gave a simpler, more elementary proof that does not depend upon the theory of modular forms. Although Ramanujan does not state (2.5) in his notebooks, we shall show that (2.5) may be simply derived from results given by him in his notebooks. Our proof also does not utilize the theory of modular forms. We first establish parametric representations for a(q), b(q), and c(q).

Lemma 2.1. Let m

= Zt/Z3,

as in (l.17). Then

a(q) =

,JZiZ3

b(q) =

,JZiZ3

c(q) =

,JZiZ3

m 2 + 6m - 3 4m

(2.10)

'

(3 - m)(9 - m 2 )1/3 4m 2/ 3 '

(2.11)

and 3(m

+ l)(m 2 4m

1)1/3

(2.12)

Proof. From Entry l1(iii) of Chapter 17 in Ramanujan's second notebook (Part III [3, p. 123]), (2.13)

and

where {3 has degree 3 over a. In proving Ramanujan's modular equations of degree 3 in Section 5 of Chapter 19 of Ramanujan's second notebook, we [3, p. 233, eq. (5.2)] derived the parametric representations

a=

(m - 1)(3 + m)3

(2.14)

and {3

=

(m - 1)3(3

+ m)

(2.15)

16m

Thus, by (1.16), (2.7), (2.13), (2.14), and (2.15), a(q) = ,JZIZ3

=

{I + (a{3)1/4}

~{

yZ1Z3

1+

(m - 1)(m

4m

+ 3) }

=

~m2

yZ1Z3

+ 6m - 3 4m

'

and so (2.10) is established. (In fact, (2.10) is proved in Part III [3, p. 462, eq. (3.5)].)

33. Elliptic Functions to Alternative Bases

95

Next, from (2.7) and (2.8),

and, from (2.7) and (2.9), q;(ql/3) ) (l/I(q2/3) ) 2c(q) = cp(q)cp(q3) ( q;(q3) - 1 - 4ql/l(q2)l/I(q6) 1 - q2/3l/1(q6) .

(2.17) By Entry 1(iii) of Chapter 20 (Part III [3, p. 345]), (1.16), and (1.17), 3CP(q9) _ 1 = (9CP4(q3) _ 1)1/3 = cp(q) q;4(q)

(~_ 1)1/3 m2

(2.18)

and cp(ql/3) _ 1 = (q;4(q) _ 1)1/3 = (m 2 _ 1)1/3. cp(q3) cp4(q3)

(2.19)

By Entry 1(ii) of Chapter 20 (Part III [3, p. 345]) and (2.13)-(2.15),

1- 3

2

l/I (q 18 ) = ( 1 _ 9 2 l/I 4 (q 6 ) )1/3 q l/I 4(q2)

q l/I(q2)

=

(I _~!!.-)1/3

= 2ml/3(3 -- m)I/3

m2 a

(m

+ 3)1/3

(2.20)

and 1{!(q2/3)

(

1 - q2/31{!(q6) =

1{!4(q2») 1/3

1 - q21{!4(q6)

= (l_m2~)1/3

=_2(m+l)1/3. (m - 1)2/3

{J

(2.21)

Using (2.13) and (2.18)-(2.21) and then (2.14) and (2.15) in (2.16) and (2.17), we deduce that, respectively, 9

2b(q) = ~ \( m 2 - I

_ -

~

=~

)1/3

- (a{J)1/4

2m l / 3 (3-m)I/3) (m + 3)2/3

{(9 - m 2)1/3 _ (m - I)(m m 2/ 3

(3 - m)(9 - m 2)1/3 2m2/3

4m

+ 3) 2m l / 3(3 _

m)I/3} (m + 3)2/3

Ramanujan's Notebooks, Part V

96

and 2c(q) =

=

=

v'ZiZ3 {(m 2 _

1)1/3

+ (afJ)I/4 2(m + 1;%3} (m -1)

~ {(m2 _ 1)1/3 + (m - l)(m

+ 3)(m + 1)1/3} 2m(m _ 1)2/3

yZIZ3

v'ZiZ3

3(m 2

1)1/\m + 1) 2m .

-

Hence, (2.11) and (2.12) have been established.

Theorem 2.2. The cubic theta-function identity (2.5) holds. Proof. From (2.11) and (2.12), 3

3

b(q)+c(q)= =

(ZIZ3)3/2

64m 3

(ZIZ3)3/2 (

64m 3

3

3

2

{m(3-m) (9-m)+27(m+l)(m 2+6

m

m

2

-l)}

_ 3)3 __ 3( ) a q,

by (2.10). This completes the proof. Our next task is to state a generalization of Ramanujan 's beautiful cubic transformationfor2FI(t,~; l;x).

Theorem 2.3. For Ix I sufficiently small, 2 FI

(c c+!. 3c+ 1'1_ , 3' 2' = (l +2x)

3c

(~)3) 1 +2x

1 3c6+-5; x 2FI ( c,c+ 3;

3) .

(2.22)

Proof. Using MAPLE, we have shown that both sides of (2.22) are solutions of the differential equation

+ x + x 2 )(l + 2x)2y" (l + 2x)[(4x 3 - 1)(3c + 2x + 1) + 18cx]y' -

2x(l - x)(1 -

6c(3c

+ 1)(1 -

x)2y = O.

This equation has a regular singular point at x = 0, and the roots of the associated indicial equation are 0 and (3c - 1)/2. Thus, in general, to verify that (2.22) holds, we must show that the values at x = 0 of the functions and their first derivatives on each side are equal. These values are easily seen to be equal, and so the proof is complete.

33. Elliptic Functions to Alternative Bases

Corollary 2.4 (p. 258). For

97

Ixl sufficiently small,

1 2 ( 1- x ) 2FJ ( 3' 3; 1; 1- 1 +2x

3) =

(1 +2xhFJ

(1 2

3' 3; l;x

3) .

(2.23)

Proof. Set c = ~ in Theorem 2.3. The Borweins [S] deduced Corollary 2.4 in connection with their cubic analogue of the arithmetic-geometric mean. Neither their proof nor our proof is completely satisfactory, because they depend upon prior knowledge of the identity and differential equations. Recently, H. H. Chan [4] has given a considerably more natural proof that depends upon rederiving some of the results in Sections 4-6 of this chapter without appealing to the theorems here in Section 2. The Eisenstein series M(q) and N(q), defined at the beginning of Section 4, play key roles. Chan [4] has also shown that Ramanujan's cubic transformation can be derived from two cubic transformations due to E. Goursat [1]. Our next goal is to prove a cubic analogue of (1.1S). We accomplish this through a series of lemmas.

Lemma 2.5. If n = 3m , where m is a positive integer, then b 3(q») a(q) (1 2 b\qn») 1 2 2FJ ( 3' 3; 1; 1 - a 3(q) = a(qn) 2FJ 3' 3; 1; 1 - ~3(qn) . Proof. Replacing x by (1 - x)/(l 2

2 'I'I-x F ( -1 J 3'3' ,

Setting x

F 21

(2.24)

+ 2x) in (2.23), we find that

3) -1+2x 3 2

X)3) .

2 (1 -- - - F (1- -'1' J 3'3' , 1+2x

(2.2S)

= b(q)/a(q) and employing (2.8) and (2.9), we deduce that

(~ ~.

3 '3"

l' 1 _ b 3(q») _ 3a(q) F a 3(q) -a(q)+2b(q) 2 1

a(q) = a(q3) 2FI a(q)

= a(q3)

2FI

(~ ~. l'

(13' 3;2 (13' 3;2

3'3"

(a(q) - b(q) )3) a(q)+2b(q)

C 3 (q3»)

1; a3(q3)

b\q3») 1; 1 - a 3(q3) ,

by Theorem 2.2. Iterating this identity m times, we complete the proof of (2.24). The next result is the Borweins' [S] form of the cubic inversion formula.

Lemma 2.6. We have (2.26)

98

Ramanujan's Notebooks, Part V

Proof. Letting m tend to 00 in (2.24), noting that, by (2.2) (or (2.6» and (2.3) (or (2.8», respectively, lim a(q") = I = lim b(qn), n~oo

n~oo

and invoking Theorem 2.2, we deduce (2.26) at once. Lemma 2.7. If n = 3m , where m is a positive integer, then 2

F

t

(~ ~. 1· b3(q») _ a(q) 3' 3' 'a3(q)

- na(qn)

2

F

t

(~ ~. 1. b 3 (q,,») 3' 3' 'a3(qn)

(2.27)

.

Proof. By Theorem 2.2, (2.25) with x = c(q)/a(q), (2.8), and (2.9), 2 Ft

b\q») 1 2 ( 3' 3; 1; a3(q) =

(1 2

3' 3; 1; 1 -

2 Ft

3a(q) + 2c(q) 2 Ft

= a(q)

c\q») a3(q)

(13' 3;2 ( 1;

a(q) - c(q) )3) a(q) + 2c(q) (2.28)

Replacing q by q3 in (2.28), and then iterating the resulting equality a total of m times, we deduce (2.27) to complete the proof. Lemma 2.S. Let q3 be defined by (1.7), and put F (x) = q3. Let n = 3m , where m is a positive integer. Then

(2.29) and

Proof. Dividing (2.24) by (2.27), we deduce that 2

F (1 2. 1. 1 t

3' 3' , -

b 3(q») a3(q)

_ - - ' - - : - -_ _--;:;----'-0..:....:......

F (1 2. 1. b\q») 2 t 3' 3' 'a3(q)

= n

2

F (1 2. 1. 1 t

3' 3' , -

b\qn») a3(qn)

.

(2.31)

(1 2. 1. b\qn») 2 F t 3' 3' 'a 3(qn)

Multiplying both sides of (2.31) by -2rr/..(3 and then taking the exponential of each side, we obtain (2.29). Multiply both sides of (2.31) by -..(3/(2rrn), take the reciprocal of each side, use Theorem 2.2, and then take the exponential of each side. We then arrive at (2.30).

33. Elliptic Functions to Alternative Bases

99

We now establish another fundamental inversion formula. Lemma 2.9. Let F be defined as in Lemma 2.B. Then F

(Ca3(q») = q. (q) 3

Proof. Letting n tend to 00 in (2.30) and employing Example 2 in Section 27 of Chapter 11 in Ramanujan's second notebook (part II [2, p. 81]), we find that F

(Ca 33(q») _ r FJ/n (c 3(qn») (q) - n!.~ a 3(qn) 5 c3(qn) . (c3(qn) ( = lim 1+---+ .. · n ..... oo 27a\qn) 9 a 3(qn) =

))J/n

n~~ ((qn + ... ) (1 + ~ (qn + ... ) + ... ) ) \/n

=q,

where in the penultimate line we used (2.6) and (2.9). Theorem 2.10 (p. 258). Let F be defined as in Lemma 2.B. Then

z:= Proof. Let u = u(x) 2.2,

2FJ

0, ~; 1; x) = a(F(x» = a(q3).

(2.32)

= b 3(F(x»/a 3 (F(x». Then by Lemma 2.6 and Theorem a(F(x» = 2F\

H. ~; 1; 1 -

(2.33)

u).

On the other hand, by Lemma 2.9, F(1 - u) = F(x),

or 2FJ

H, ~; 1; u)

2FJ H,~; 1; l-u)

By the monotonicity of 2 F\ 2F\

2F\ H.~; l;x)

~; 1;

(2.34)

.

x) on (0,1), it follows that. for 0 1; 1 - u) = 2F\ H. ~; 1; x).

(1,

H. ~;

2F\ H,~; 1; I-X)
\q) 1/1 2(q)

+ 16q1/1 4(q2)

(9.14)

and 1/1 6(q) _ 6q./,2( )./,2( 3) _ 3 2 1/16(q3) = 4( 3) 1/12(q3) 'I' q 'I' q q 1/12(q) q> q

+ 16q3./,4( 'I'

q

6)

.

(9.15)

Proof. For brevity, set a := a(q) and A := a(q2). Then, squaring (9.12), we find that

4 1/1 6(q) = (a

1/12 (q3)

+ A)2.

(9.16)

From (5.7), (6.3), and (9.12), 12q1/1 2(q)1/1 2(q3) = a 2 _ A2

(9.17)

and (9.18) Hence, from (9.16)-(9.18), 1/1 6(q) 1/12(q3)

+ 18

./,2( )./,2( 3) _ 27 2 1/16(q3) = a 2 + 2aA _ 2A2. q'l' q 'I' q q 1/12(q)

(9.19)

Next, from (9.16) and (9.18), 48q1/1 8(q)

=

(8 ::(~;») (6

q

1/1:~::») = (a + A)3(a -

A)

(9.20)

+ A).

(9.21)

and 3 432q31/18( 3) = (216 3 1/1 9 (q3») (2 1/1 (q») = (a _ A)3(a q q 1/1 3(q) (q3)

1/1

From (5.6) and (6.5), (9.22) Thus, from (9.13) and (9.22), cp8( _q)

=

cp9( -q) cp\ _q3) cp3(_q3) cp(-q)

= !(2A _ a)3(a + 2A) 3

(9.23)

33. Elliptic Functions to Alternative Bases

151

and 918( _q3)

= 919 ( _q3) 913( -q) = ~ (a + 2A)\2A _ 913(_q) 91(_q3)

27

a).

(9.24)

Lastly, we need the elementary identity (Part III [3, p. 40, Entry 25(iv)]) 1/1(q4)91(q2) = 1/12(q2).

(9.25)

Hence, from (2.1), (9.25), (9.23), and (9.20), {914(q)

+ 16q1/1 4(q2)} 2 =

{914(q) _ 16q1/1 4(q2)} 2

+ 64q914(q)1/1 4(q2)

+ 64q1/1 8(q) ~(2A - a)3(a + 2A) + 1(a + A)3(a - A) a 4 + 4A4 + 4aA(a 2 - 2A2) (a 2 _ 2A2)2 + 4aA(a 2 _ 2A2) + 4a 2A2

= 918(_q) = = =

(9.26) Equality (9.14) now follows from (9.19) and (9.26), upon taking the square root of both sides of (9.26) and checking agreement at q = 0 to ensure that the correct square root is taken. The proof of (9.15) is similar. Thus, from (9.16)-(9.18), 6 1/1 (q) _ 6q1/1 2(q)1/1 2(q3) _ 3q2 1/16 (q3) 1/12(q3) 1/12(q)

=

!(2A2 3

+ 2aA _

a 2).

(9.27)

Proceeding as in the proof of (9.14), from (2.1), (9.25), (9.24), and (9.21), we find that {91\q3)

+ 16q31/1\q6)} 2 =

+ 64q31/18(q3) = i7(a + 2A)3(2A - a) + r,(a - A)3(a + A) = ~ (a 4 + 4A4 + 4aA(2A 2 _ a 2») = ~ (a 2 _ 2A2)2 _ 4aA(a 2 - 2A2) + 4a 2A2) 918( _q3)

= ~(a2 _ 2A2 - 2aA)2.

(9.28)

Equality (9.15) now follows from (9.27) and (9.28). Lastly, we need the following lemma connecting 2 F, (~, ~; 1; x) with thetafunctions.

Lemma 9.14. We have 2

( 2F ,

1 3

2 2 )2) = 91

8 '"1n.1,2 q 'I' (q )91 (q)

4' 4; 1; ( 914(q) + 16q1/14(q2)

4

(q)

+ 16q1/1

4

2

(q ).

(9.29)

152

Ramanujan's Notebooks, Part V

Proof. Let

Then

2x

(9.30)

l+x where we have employed (9.25) and the elementary identity

+ 4q1fF2(q4) = q/(q),

q}(q2)

(9.31)

which is achieved by adding Entries 25(v), (vi) of Chapter 16 of Ramanujan's second notebook (Part III [3, p. 40]). Hence, from Theorem 9.1, (9.30), (9.31), (2.1), and (5.19), with _q3 replaced by q, 2F(

(~,~; l;x 2)

=

_

-

1

~x 2F( (~,~; I; 1 ~x)

+ 16q 21fF4(q4) 2 (f{)2(q2) + 4q1fF2(q4)} f{)4(q2)

f{)4(q2)

= f{)\q2)

2

2F t

(1 1..

1fF4(q2») -, -,1, 16q-42 2 f{) (q)

+

16q 21fF4(q4) 4 f{)4(q) f{) (q)

+ 16q 21fF4(q4).

Replacing q2 by q, we deduce (9.29). Lemma 9.14 was first proved by the Borwein brothers [1, p. 179, Prop. 5.7(a)], [5, Theorem 2.6(b)].

Theorem 9.15 (p. 260). If

a .-

64p

+ 6p -

-----"-----::-~

.- (3

p2)2

(9.32)

and

then,for 0:::: p < 1, ../27 - ISp - p2 2Ft

(i, ~;

1; a) = 3./3

+ 6p -

p2 2FJ

(i, ~;

1;

fJ).

(9.33)

Proof. Let 1fF\q3)

(9.34)

P = 9q 1fF4(q) .

Then, by (9.14) and (9.15), respectively,

3 + 6p - p2 =

"fP

.jfi1fF4(q)

{rp4(q)

+ 16q1fF4(q2)}

(9.35)

33. Elliptic Functions to Alternative Bases

153

and 3/2

=

27 - 18p - p2

q3/;1{!4(q3) {(j14(q3)

+ 16q 31{!4(q6)}.

(9.36)

From (9.32) and (9.35), 64q1{!8(q)

+ 16q1{!4(q2)}2

a[q]:= a = {(j14(q)

(

=

8Jq1{!2(q2)(j12(q) )2 (j14(q)

by (9.25). Similarly. by (9.32) and (9.36), ,B[q] :=

f3

(

=

8q3/21{!2(q6)(j12(q3) (j14(q3) 16q 31{!4(q6)

+

+ 16q1{!4(q2)

)2

,

3

(9.37)

(9.38)

= a[q ],

by (9.37). Hence, from (9.36), (9.37), Lemma 9.14, (9.38), (9.35), and (9.34),

J27 - 18p p

p2 2FI

3/4

G, ~; 1; a)

{(j14(q3)

q 3/41{! 2 (q 3)

p3/4 q3/41{!2(q3) 2FI -

.Jji1{!2(q) 13 Jq1{!2(q3) Y

=

3J3 + 6p -

+ 16q31{!4(q6)} 1/2 {(j14(q) + 16q1{!\q2)} 1/2

G, ~; 1;,B)

+ 6p

p2 2FI

- p

2

2

q 1/41{!2 (q) pl/4

J3 + 6p -

p2

3. l' f3) 4' 4' ,

F (I I

G, ~; I;,B).

Thus, (9.33) has been proved. Lastly, it is easily checked that a =: a(p) and,B =: ,B(p) are monotonically increasing functions of p on (0,1). Since a(O) = 0 = ,B(O) and a(1) = I = ,B(I), (9.33) is valid for 0 ~ p < 1.

10. Modular Equations in the Theory of Signature 4 Page 261 in Ramanujan's second notebook is devoted to modular equations in the theory of signature 4. In each case, our proofs rely on (9.7). Thus, we will employ modular equations from Chapters 19 and 20 of the second notebook and convert them via the transformations a

~

2.j(i t;:;' I +va

I-a ~

I -.j(i

t;:;'

I +va

2..{fJ

,B

~ 1+..{fJ'

I-,B~

1- ..{fJ

17i'

1+ vf3

(10.1)

Theorem 10.1 (p. 261). If,B has degree 3, then (a,B)1/2

+ {(l -

a)(1 - ,B)}1/2

+ 4{a,B(1

- a)(1 - ,B)}1/4

=

1.

(10.2)

154

Ramanujan's Notebooks, Part V

Proof. From Entry 5(ii) of Chapter 19 (Part III [3, p. 230]), (afJ)I/4

+ {(l

- a)(1 - fJ)}I/4

= 1.

(10.3)

By (10.1), (10.3) is transformed from the classical theory to .j2(afJ) 1/8 ---..:.......:.....:....----;-:-:+ {(1 + ,Ja)(1 + .JP)} 1/4

{(1 -

-.JP) }1/4 (1 + ,Ja)(1 + .JP) ,Ja)(1

1

in the theory of signature 4, or

=

../2(afJ)1/8

{ (1

+ v1a)(1 + v'fi) } 1/4 -

{

(1 - JiX)(1 -

v'fi) } 1/4 .

Squaring both sides yields 2(afJ) 1/4

+ 2 {(l

- a)(1 - fJ)}I/4 = { (1

+ v1a)(1 + v'fi) }

+ { (1 -

v1a)(l -

1/2 1/2

v'fi) } .

Squaring each side once again, we find that

+ 2 {(l - a)(l 1 + (afJ)1/2 + {(l

2(afJ)1/2

=

fJ)}1/2

+ 4 {afJ(1

- a)(1 - fJ)}1/4

- a)(1 - fJ)}1/2 .

Collecting terms, we deduce (10.2).

Theorem 10.2 (p. 261). If fJ has degree 5, then (afJ)1/2

+ 8 {afJ(1

+ {(I

- a)(1 - fJ)}1/2

- a)(1 - fJ)}I/6 (afJ)I/6

+ {(l

- a)(1 - fJ)}1/6} = 1.

(lOA) Proof. By Entry 13(i) of Chapter 19 of Ramanujan's second notebook (Part III [3, p. 280)), (afJ)I/2

+ {(l

- a)(1 - fJ)}I/2

+ 2 {loofJ(l -

a)(l - fJ)}1/6 = 1.

(10.5)

Transforming (10.5) by (10.1) and simplifying, we find that, in the theory of signature 4, 2(afJ) 1/4

+ 2 { 64JaP(1 -

a)(1 - fJ) }

1/6

=

{

(1

+ v1a)(1 + v'fi) }1/2

- { (1 - v1a)(1 -

1/2

v'fi) } .

Squaring each side and collecting terms, we deduce that 2(a{J)1/2

+ 4 { 64JaP(1

- a)(1 - fJ) }

1/3

+ 8 (a{J) 1/3 {64(1 -

a)(1 - fJ)}I/6

= 2 - 2{(1- a)(1 - fJ)}I/2.

33. Elliptic Functions to Alternative Bases

155

Further simplification easily yields (10.4).

Theorem 10.3 (p. 261). If fJ has degree 7, then (afJ)I/2

+ {(I - a)(1 + 8 h {afJ(1 -

fJ)}1/2

+ 20 {afJ(1 -

a)(1 - fJ)}I/4

+ {(I

a)(l - fJ)}I/8 (afJ)I/4

- a)(1 - fJ)}I/4)

=

1. (10.6)

Proof. From Entry 19(i) of Chapter 19 of Ramanujan's second notebook (Part III [3, p. 314]),

+ {(l -

(afJ)I/8

a)(1 - fJ)}I/8

=

1.

(l0.7)

Using (10.1) to transform (10.7) into the theory of signature 4, we find that (1oofJ)ljl6 = { (1

+ v1a)(1 + /fi) } 1/8 -

{

(1 - v1a)(1 -

/fi) }1/8 .

Squaring both sides, we deduce that

+ 2 {(I -

(1oofJ)I/8

+ v1a)(1 + /fi) }

a)(1 - fJ)}I/8 = { (1

+ { (1

- v1a)(I-

1/4 1/4

/fi) } .

Squaring again and simplifying slightly, we find that (1oofJ) 1/4 + 2 {(I - a)(1 - fJ)}I/4

=

{ (1

+ 4 (16afJ(I

+ v1a)(1 + /fi) } 1/2 + { (l -

- a)(l - fJ)} 1/8

v1a)(l -

JP) }1/2 .

Squaring one more time, we finally deduce that 4(afJ)I/2

+ 4{(1 -

+ 8 {afJ(1 + 16 {(l

a)(1 - fJ)}I/2

- a)(1 - fJ)}1/4

+ 32 {afJ(1

- a)(1 - fJ)}1/4

+ 16(afJ)I/4 {loofJ(1

- a)(1 - fJ)}1/8

- a)(l - fJ)}I/4 {16afJ(1 - a)(1 - fJ)}I/8

= 2 + 2(afJ)I/2 + 2 {(l

- a)(l - fJ)}1/2 .

Collecting terms and dividing both sides by 2, we complete the proof of (10.6).

Theorem 10.4 (p. 261). If fJ has degree 11, then (afJ)I/2

+ {(1 -

+ 16 (afJ(1 + 48 {afJ(1

a)(l - fJ)}I/2

+ 68 {afJ(1

- a)(1 - fJ)}1/4

+ {(l - a)(l - fJ) }1/3) fJ)}1/6 (afJ)I/6 + {(I - a)(l - fJ)}I/6) =

- a)(l - fJ)} 1/12 (afJ) 1/3 - a)(l -

1. (10.8)

Proof. By Entry 7(i) of Chapter 20 (part III [3, p. 363]), (afJ) 1/4 + {(l - a)(1 - fJ)}1/4

+ 2 {l oofJ (1

- a)(l - fJ)}I/12

= 1.

(10.9)

156

Ramanujan's Notebooks, Part V

Transforming (10.9) into an equality in the theory of signature 4, we find that

(4M) 1/4 + 2 {64M(l - a)(1 - /3) }

1/12

=

{

(1

+ .,ja)(1 +.[fi) }1/4

- { (1 - .,ja)(l -

.[fi) }

1/4

.

Squaring both sides and simplifying slightly, we deduce that 2(af3) 1/4

+ 2 {(I -

+ 8(af3) 1/6 {(l

=

{(1

a)(l - f3)} 1/4 + 8 (af3) Ijl2 {(I - a)(1 - f3)}I/6

r

- a)(1 - f3)}1/12

+ .,ja)(1 + .[fi)

2

+ {(l - .,ja)(1

-

.[fi)

rz .

Squaring both sides again, we see that

+ 64(af3) 1/6 {(l - a)(1 - f3)}I/3 + 64(af3) 1/3 {(l - a)(l - f3)}1/6 + 8 {af3(1 - a)(1 - f3)}1/4 + 32(af3)I/3 {(l - a)(l - f3)}1/6 + 32(af3)5/12 {(l - a)(l - f3)}1/12 + 32(af3)I/12 {(l - a)(1 - f3)}5/12 + 32(af3) 1/6 {(I - a)(1 - f3)}1/3 + 128 {af3(1 - a)(l - f3)}1/4 = 2 + 2(af3)l/z + 2 {(l - a)(1 - f3)}I/2.

4(af3)I/2

+ 4 {(I

- a)(1 - f3)}1/12

Collecting terms and dividing both sides by 2, we complete the proof. The last six entries on page 261 in Ramanujan's second notebook give formulas for multipliers. By (9.1), m(4)

=

13 zFI (4' 4; 1; a) zFI (~,

~;

1;

=

13)

+ .JP) 1 + via

(1

1/2 F Z I 2Fl

(I2' 2'I·l·~) 'I+v'a .

0, !;

1;

I~~)

(10.10)

Thus, to obtain formulas for multipliers in the theo of si nature 4, take formulas from the classical theory, replace m by (1 + vIa)/(l + .JP) m(4), and utilize the transformations in (10.1).

Theorem 10.5 (p. 261). The multiplier for degree 3 is given by 2

_

m (4) -

(13)1/2 + (1---13)1/2 - -9 - (13(1 - 13»)1/2 m 2(4)

1- a

a

a(t - a)

(10.11)

Proof. From Entry 5(vii) of Chapter 19 (Part III [3, p. 230]), m2=

(t!.) a

1/2

+ (~) 1/2 1- a

_

(13(1 - 13») 1/2 a(l - a)

(10.12)

and

~ m2

=

(~)1/2 + (~)1/2 13 1 - 13

_ (a(l- a»)1/2

13(1 - 13)

(10.13)

33. Elliptic Functions to Alternative Bases

157

Using (10.1) and (10.10), we convert (10.12) into an equality in the theory of signature 4, namely, 1+,Jii I + v'"P

m24

()

(v'"P(1 + ,Jii»)1/2 + ((1- v'"P)(l + ,Jii»)1/2

=

,Jii(1 + v'"P)

(I + v'"P)(1 - ,Jii)

_ (v'"P(1- v'"P)0 + ,Jii)2) 1/2 ,Jii(1 - ,Jii)(1 + v'"P)2 ' or, upon rearrangement, 2 m (4) -

(I - 13) 1/2 _ I _ a

v'"P») 1/2 - (v'"P(1 - v'"P») 1/2 ,Jii(1 _ ,Jii)

(v'"P0 + -,Jii0 + ,Jii)

(10.14)

By (10.13), 00.14), and symmetry, _9__ m 2 (4)

(l-a)1/2 = (,Jii(1 +,Jii»)1/2 _ (,Jii0-,Jii»)1/2 I - 13 -JP(1 + v'"P) v'"P(1 - .f"P)

00.15)

Multiplying both sides of (10.15) by Jf3(l - f3)/(a(1 - a)), we find that

_9_ m 2 (4)

(130- 13»)1/2 _(~)1/2 = a(l - a)

a

v'"P»)1/2 _(v'"P(1 + v'"P»)1/2

(v'"P(I,Jii(l - ,Jii)

,Jii(l +,Jii) (10.16)

Comparing (10.14) and (10.16), we arrive at (10.11). Theorem 10.6 (p. 261). Ifm(4) is the multiplier of degree 5, then

/4 (1_-13 )1 /4- _(13)1 5 (13(1- 13»)1 /4 a + I - a m(4) a(1 - a)

m(4)-

(10.17)

Proof. By Entry 13(xii) of Chapter 19 (Part III [3, pp. 281-282]),

m= (~)1/4 + a

(~)1/4 _ (13(1 I -

a

_13»)1/4 a(1 - a)

(10.18)

and

~ = (~)1/4 + (~)1/4 _ (a(l- a»)1/4.

m

13

I -

13

13(1 - 13)

(10.19)

Transforming (10.18) into the theory of signature 4 via (l0.1) and (10.10), we find, after a slight amount of rearrangement, that

/ _(v'P(1+v'"P»)1/4 + (1_ - -13 )1 4- (v'P(l-v'P»)1/4

m(4) -

I- a

,Jii(I + ,Jii)

,Jii (l - ,Jii)

(10.20)

From (10.19) and symmetry, _5_ m(4)

=

(,Jii(l + ,Jii») v'P(1 + v'"P)

1/4

+

a) 1/4 _ (,Jii(1 _ .ja») 1/4

(I 1-13

v'"P(l

-./"P)

(10.21)

158

Ramanujan's Notebooks, Part V

Multiplying both sides of (10.21) by .:fP(I - p)/(a(I - a» and comparing the result with (10.20), we readily deduce (10.17). Theorem 10.7 (p. 261). Let m(4) denote the multiplier of degree 9. Then

P)1/8 Jm(4)= ( a

(1 -

3

P)1/8 (P(l- P»)1/8 + I-a Jm(4) a(l-a)

(10.22)

Proof. The proof is almost identical to the two previous proofs. By Entries 3(x), (xi), respectively, of Chapter 20 (Part III [3, p. 352]),

.;m = (~)1/8 + (1 a

I -

P)1/8 _ (fJ(l- fJ»)1/8 a a(1 - a)

(10.23)

and _3

Jm

= (~)1/8 + (~)1/8 _ fJ

fJ

I -

(a(1 _ a»)1/8 fJ(l - fJ)

(10.24)

The transforming of (10.23) and (10.24) via (10.1) and (10.10) yields the equalities Jm(4) = (..(P(1 + ..(P»)1/8 + J(i"(1 + J(i")

(1 - fJ)1/8 _ (..(P(l- ..(P»)1/8 a 1-

J(i"(1 - J(i")

(10.25)

and 3 (J(i"(I+J(i"»)1/8 (l_a)1/8 (J(i"(1_J(i"»)1/8 Jm(4) = ..(P(I + ..(P) + 1 - fJ -..(P(l - ..(p) ,(10.26) respectively. A multiplication of (10.26) by .:ifJ(l - fJ)/(a(1 - a» and a comparison of the resulting equality with (10.25) gives (10.22). Theorem 10.8 (p. 261). If m( 4) denotes the multiplier of degree 7, then

(1 -

fJ»)1/2

m2(4) = (~)1/2 + fJ)1/2 _ ~ (fJ(1 a I - a m 2 (4) a(1 - a)

fJ»)1/6 {(~)1/6 + (~)1/6}. a 1- a

_ 8 (fJ(l-

a(1- a)

(10.27)

Proof. By Entry 19(v) of Chapter 19 (part III [3, p. 314]),

m2 = (~)1/2 + (~)1/2 a 1- a and 49 - =

m2

(1 - a)I/2

(a)1/2 + -fJ 1 - fJ

_(fJ(1 - fJ»)1/2 _ 8 (fJ(1- fJ»)1/3 a(1 - a) a(1 - a)

-

(10.28)

(a(1 - a»)1/2 -8 (a(1 _ a»)1/3 . (10.29) fJ(1 - fJ) fJ(1 - fJ)

33. Elliptic Functions to Alternative Bases

159

Transforming (10.28) and (10.29) into the theory of signature 4 via (10.1) and (10.10), we find that, after simplification,

(1 _

/ 1 ( -JP(1 + -JP»)1 2+ - -13)1 2 .J(i(1 + .J(i) 1- a

m 2 (4) =

_ (-JP(l - -JP»)1/2 .J(i(1 - .JCi)

_ 8 (-JP(l - 13»)1/3 .J(i(1 - a)

(10.30)

and

49

m2(4) =

(.J(i(1 -JP(I

+ .J(i») 1/2 (I - a) 1/2 + -JP) + 1 - 13

_ (.J(i(l- .J(i»)1/2 _ 8 (.J(i(I-a»)1/3

"/p(1 - 13)

../P(l - ../P)

respectively. Multiplying both sides of (10.31) by deduce that

,

(10.31)

Jf3(l - f3)/(a(1 - a»,

we

~ (13(1- 13»)1/2 = ("/p(1- '/p»)1/2 + (~)1/2

m2(4) a(1 - a)

a _ ('/p(1 + '/p»)1/2 _ 8 (f3.;r::-p) 1/3 . .J(i(1 + .JCi) aJI=(l (10.32) .J(i(1 - .J(i)

Combining (10.30) and (10.32), we complete the proof of (10.27).

Theorem 10.9 (p. 261). Ifm(4) denotes the multiplier of degree 13, then

_ m(4)-

(I -

/4 (13)1 13)1 14- 13 + - (13(1- 13»)1/4 a 1- a m(4) a(1 - a) - 13») 1/12 {(~) 1/12 + (~) 1/12} . a(1 - a) a 1- a

_ 4 (13(1

(10.33)

Proof. By Entries 8(iii), (iv) of Chapter 20 of Ramanujan's second notebook (Part III [3, p. 376]), m =

(~)1/4 +

a

(1 - 13)1 /4_ (13(1 - 13»)1/4 _ 4(13(1 -- 13»)1/6 1-

a

a(1 -

a)

a(1 -

a)

(10.34)

and 13

m =

(a)1/4 (l-a)1/4 (a(l-a»)1/4 (a(l-a»)1/6 p + 1 - 13 - 13(1 - 13) - 4 13(1 - 13) . (10.35)

160

Ramanujan's Notebooks, Part V

Transfonning (10.34) and (10.35) into the system of signature 4 by means of (10.1) and (10.10), we find, after some simplification, that

..(PO m(4) = ( y'a(l

+ ..(P»)1/4 + (1-- {3)1/4 + y'a) 1- a

_ (..(PO - ..(P»)1/4 _ 4 (..(PO- {3»)1/6 y'a0 - y'a) y'a(l - a)

(10.36)

and

( y'a(1

13

m(4)

"(p(1

+ y'a»)1/4 (~)1/4 + ..(P) + 1 - {3

_ (y'a(1_y'a»)1/4 _4(y'a(l-a»)1/6 ..(P(l - ..(P) ..(P(1 - {3) ,

(10.37)

respectively. Multiplying both sides of (10.37) by ~{3(1 - {3)/(a(l - a» and combining the resulting equality with (10.36), we finish the proof of (10.33). Theorem 10.10 (p. 261). If m (4) denotes the multiplier of degree 25, then {3)1/8 Jm(4)= ( -

a

{3)1/8 + (1-- - -5- ({3(1 1-

a

Jm(4)

_ 2 ({30 - {3») 1/241 (~) 1/24 +

a(1 - a)

_ {3»)1/8

aO - a)

a

(~) 1/24). 1- a

(10.38)

Proof. By Entries l5(i), (ii) of Chapter 19 of Ramanujan's second notebook (Part III [3, p. 291]), {3)1/8 ../iii = ( {3)1/8 + (1--

a

1- a

({3(1- {3»)1/8

a(1 - a)

-2

({3(1- {3»)1/12

aO - a)

(10.39)

and

5

~

_ (a)I/8

- Ii

+

(l_a)I/8 1 - {3

-

(a(1_a»)1/8 {3(1 -

P)

-2

(a(1_a»)1/12

P(l -

{3)

. (10.40)

Transfonning (lO.39) and (lO.40) by means of (10.1) and 00.10) into equalities in the theory of signature 4, we find that Jm(4)

..(PO +..(P»)1/8 (1_P)I/8 = ( + -y'a(1

+ y'a)

1- a

_ (..(P(1 - ..(P»)1/8 _ 2 (..(PO - {3»)1/1 2 y'a(1 - y'a) y'a(l - a)

(10.41)

33. Elliptic Functions to Alternative Bases

and

5

= (-Ja(l

v'm(4)

+ -Ja») 1/8

+

,,[p(l +..fP)

_(-Ja(1 - -Ja»)

(1 - ex) 1/8 1 - fJ

1/8 _

,J""P (l - ,J""P)

161

2

(-Ja(1 - ex»)

1/12

,,[p(1 - fJ)

,

(10.42)

ex»

respectively. Multiplying both sides of (10.42) by ~fJ(l - fJ)/(ex(l and combining the resulting equality with (10.41), we finish the proof of (10.38).

11. The Theory for Signature 6 The most important theorem in this section is Theorem 11.3 below. This result allows us to employ formulas in the classical theory to prove corresponding theorems in the theory of signature 6. To prove Theorem 11.3, we need the following two results. Theorem 11.1 (p. 262). If

ex·-

p(2+ p)

1 +2p

and

fJ

27p2(1 := 4(1 + p

+ p)2

+ p2)3 '

:s p < 1, ~h + 2p 2FI U' ~; 1; fJ) = Jl + p + p2 2FI G, ~; 1; ex).

(11.1)

then,for 0

(11.2)

Proof. From Erdelyi's treatise [1, p. 114, eq. (42)], 2FI U,~; 1; z)

= (1 -

Z/4)-1/4 2FI

(fl,

fi; 1; -27z 2(z -

4)-3),

(11.3)

for z sufficiently near the origin. By Example (ii) in Section 33 of Chapter 11 (Part II [2, p. 95]), 2FI

for

(~, ~; 1; z) =

(l - Z)-1/2 2FI

(~, ~; I; -

(1

~\:i2)'

Izl sufficiently small. Thus, combining (11.3) and (11.4), we find that

(11.4)

162

Ramanujan's Notebooks, Part V

Next, recall the well-known transformation (Erdelyi [I, p. Ill, eq. (2)]),

2F\ G,~; 1; z) = 2FI

(12, -rz;

1; 4z(1- z)}

(11.6)

for z in some neighborhood of the origin. Examining (11.5) and (11.6) in relation to (11.2), we see that we want to solve the equation

27z2(1 - Z)2 . 4(1 - z + Z2)3

4x(1 - x) =

Solving this quadratic equation in x and choosing that root which is near the origin when z is close to 0, we find that

x

=~ 2

=~

2

(1 _{4(1 - z + Z2)3 - 27z2(1 - Z)2 }1/2) 4(1 - z

(1 _ (1

+ z)(2 2(1 - z

+ Z2)3

5z + 2Z2») . + Z2)3/2

(11.7)

Thus, by (11.5)-(11.7), we have shown that

2 F\(~6' ~'I'~ 6' '2

»))

(1_(1+Z)(2-5Z+2Z 2 2(1 - z + Z2)3/2 _

- (1 - z

Now set Z=

(1 I, , ) '

2 \/4 + z) 2 F\ 2' 2' 1, z

(11.8)

p(2 + p) 1 + 2p

Then elementary calculations give 1_

and 1-

(1

2 _

Z+ z -

+ z)(2 -

(1

+ P + p2)2

(1

5z + 2z 2) 2(1 - z + Z2)3/2

+ 2p)2

27p2(1 2(1 + p

+ p)2 + p2)3

= -----:-::-

Using these calculations in (11.8), we deduce (11.2). Lastly, a and /3 are monotonically increasing functions of p on [0, 1] with a(O) = 0 = /3(0) and a(I) = 1 = /3(1). It follows that (11.2) is valid for O::sp 0, (12.43) and N(e- 2ni / T ) = .6N(e2niT ).

(12.44)

As in the previous proof, we set

q = exp( -27ft)

(12.45)

for t > O. Then (12.43) and (12.44) become M(e- 2n / t ) = t 4 M(e- 2nt )

(12.46)

and (12.47) With x defined by (12.40), we consider x as a function of 1. Set z(t) := M(q), so that, from (12.46), (12.48) and, from (12.41),

i; 1; x(t» =

2Fl(i,

(Z(t»l/4.

(12.49)

By (12.40), (12.46), and (12.47), x(1/t) = 1 - x(t).

(12.50)

Hence, from (12.2), (12.50), (12.49), and (12.48),

y=

2Fl

(i, i; 1; 1 -

2 F lCi,

x(t»

i; 1; x(t»

i; 1; x(1/t)) = (ZC1/t»)l/4 = t. zFl(i, i; 1; x(t» z(t)

2 F l(i,

(12.51)

Therefore, by (12.13), (12.51), and (12.45), q6(X(t»

= exp( -27fY) = exp( -27ft) = q,

which is (12.42). It follows from (12.42), (12.41), and (12.1) that IL

(

M3/2(q) - N(q) ~) _ qJ(q) 2M3/2(q) , 6 - Ml/8(q)·

From Entry 25(vi) of Chapter 16 (Part III [3, p. 40]), qJ2(q) =

1(qJ 2 Cyq) + qJ 2 C-yq»),

(12.52)

33. Elliptic Functions to Alternative Bases

173

and so, by (12.33), rp2(q) =.!. (1 rp2 (-.ftj) 2

+ .jY).

Hence, from (12.35), we find that M(q) = rp8(q)

(1

+2.jY) 4 (l _ Y

+ y2).

(12.53)

From (12.40), (12.52), (12.53), (12.37), and (12.38), we find that

/.L

( 6"1) = x,

rp(q) Ml/8(q)

= (l -

f¥. + y

Jv'1+2p + ~

= ,.,fi (l + P + p2)1/4

y2)1/8

'

as claimed. We now rewrite (12.40) as 27y2 x := x(y) := 4(1 y)3'

+

(12.54)

+ pl.

(12.55)

where y = p(l

Since 0 < p < 1, we have 0 < y < 2. Solving (12.55) for p, we deduce (12.31). We have taken the positive square root since 0 < p < 1. Since (12.54) is a cubic equation in y, we may solve it in terms ofradicals. Note that xC-i) = x(2) = 1, x (0) = 0, and dx

27y(y - 2)

4(1

dy

+ y)4

.

i,

Thus, x (y) decreases from 1 to 0 on (- 0), increases from 0 to 1 on (0, 2), and decreases from 1 to 0 on (2,00). Therefore, if 0 < x < I, equation (12.54) has three real roots, and each of the intervals (- 0), (0, 2), and (2, 00) contains exactly one root. We take y to be the unique root that satisfies 0 < y < 2. The root y is given explicitly in (12.32) above.

i,

Some consequences of our derivations of the values of /.L(x, h) are some new hypergeometric transformations. From (2.13) and (2.14), we see that a

=

16qljF4(q2) rp4(q)

(m - l)(m 16m 3

+ 3)3 (12.56)

Hence, from (12.56), (12.30), (5.18), (5.19), (12.9), Lemma 2.6, and (12.8), F 2

1

(~ ~'1' 2'2"

= rp2(q)

(m -l)(m +3)3) _ F 16m3 - 2 1

=

m2

4m 3 / 2

+ 6m _

3 a (q)

(~ ~'1' 1 _ 2'2"

rp4(_q») rp4(q)

174

Ramanujan's Notebooks, Part V

=

4m 3/ 2

m2

+ 6m -

3

(~ ~.

F 2

t

3' 3"

l'

27(m + 1)3(m 2 (m 2 + 6m - 3)3

1») .

(12.57)

This transformation is remindful of Theorem 5.6 but is a different transformation. We have found a generalization of (12.57) via MAPLE, namely,

(12.58) We omit the details. We have also found a generalization of Theorem 5.6 via MAPLE, viz., (12.59)

We have investigated the connection between (12.57) and Theorem 5.6 and found that (12.57) follows from Theorem 5.6 in the following way. Replace m by 1 + 2p, so that (12.57) can be rewritten as P(2+ P )3) (1+2p)3/2 (12 27 P(1+P)4) (1 + 2p)3 = 1 + 4p + p2 2 F t 3' 3; 1; 2(1 + 4p + p2)3 . (12.60) Also, replace p by 2(1 + p + p2)j(2 + 2p - p2) - 1 in Theorem 6.1 to obtain the identity 2F t

( 1 1

2' 2; 1;

(12.61) We now indicate the steps that lead from Theorem 5.6 to (12.60). First, apply Entry 6(i) of Chapter 17 of the second notebook (Part III [3, p. 238]), i.e.,

1 1 P(2+ p )3) (1 1 P3(2+ P 2 F t ( 2'2;1; O+2p)3 =(1+2p)2Ft 2'2;1; 1+2p

») ,

to the left side of (12.60). Second, apply Theorem 5.6. Third, invoke (12.61). Last, after employing Theorem 6.4, we deduce (12.60).

33. Elliptic Functions to Alternative Bases

175

It is interesting to note that (12.61) and Theorem 5.6 give the transformation 2

(2+2p - P hFt

(

1 1. . 2'2,1,

p3 (2 + P») 1 +2p

=2Jl+2p2 Ft

(

1 2 27p4(1 + p) ) 3'3;1;2(2+2p- p 2)3 . (12.62)

We have found a generalization of (12.62) via MAPLE that is different from (12.58) and (12.59), namely, 2 Ft

1 1 p3(2 + P») ( 3d, 2d + 6; 4d + 3; 1 + 2p =

(

»)3d

4(1 +2P (2 + 2p _ p2)2

2 Ft

(

1 1 27p4(1 + p) ) 2d, 2d + 3; 3d + 2; 2(2 + 2p _ p2)3 . (12.63)

Using Theorems 4.2 and 4.3 in (12.41) and then employing Theorem 2.10, we find that (1

4

(1 2

+ 8xhFt 3' 3; 1; x

On replacing x by (x 2

)

=

4 2 Ft

(1 5

1

6' 6; 1; 2 -

(1 - 20x - 8X 2 2(1 + 8x)i

») .

1) /8 we find that

-

(12 . . (X--l)(X+l») ( 1 5. 1. (X-l)(X+3)3)_ '6' 6" 16x 3 3' 3' 1, 8 . - V X 2 Ft (12.64) We have found a generalization via MAPLE, namely, 2Ft

2

F(dd t

,

_

- x

3d

+

~.d 3'

2 Ft

+

~.(X-l)(X+3)3) 6'

(2d 2d ,

16x3

1. d

5. (x - l)(x 8

+ 3' + 6'

+ 1»)

.

(12.65)

Equation (12.64) has an elegant q-version; if we replace x by m in (12.64) and use (12.56) and Lemma 5.5, we find that rp(q 3 hFt

1 5 16q1jr4 (q2») (1 2 c3(q2) ) ( 6' 6; 1; rp4(q) = rp(qbFI 3' 3; 1; - c 3 (-q) . (12.66)

13. Some Enigmatic Fonnulas Near the End of the Third Notebook Near the bottom of page 392 (or possibly at the top of page 392, since the page is reproduced upside down in the published notebooks [9]), Ramanujan recorded

176

Ramanujan's Notebooks, Part V

the following claims: I

+ 5 . -I

2

1·3 .4x (1 - x) 42

+ ...

I ( [4]y

--- 1 - 2x

1- -

-8

noon

~ -eny-- -11 6 ~ e 2ny 00

1

) (13.1)

and

1 1·2 1 + 4· - . -4x(1 - x) 2 32

1 ( [3]

- -- 1 - 2x

1-

-

y

+ ... - 6

no ) 18o n e 3ny - 1 .

-~eny - 1 00

~

(13.2)

These two claims were very difficult for us to interpret. First, Ramanujan did not provide enough terms on the left sides to determine a general term in either case, and it would appear that the series on the left sides do not converge for any x, except trivially for x = 0, 1. Second, although y is not defined, it is reasonable to guess (from Chapter 17) that y=7r

2FI(!,!; 1; I - x) I

I.

.

2FI (2' 2' I, x)

(13.3)

Third, Ramanujan had never before used the notation [], and so we did not know if [4/y] = 4/y and [3/y] = 3/y, or if some other meaning should be attached to the notation [ ]. (Ramanujan never used [ ] to denote the greatest integer function.) Fourth, by (13.3), the right sides of (13.1) and (13.2) have singularities at x = 0, I, but if the left sides are convergent series, they must be analytic at x = 0, 1. Eventually, we determined that the nth terms of the series on the left sides of (13.1) and (13.2) have (n!)3 in the denominators, which is not evident from Ramanujan's formulations. The conjectured definition of yin (13.3) is incorrect, and the expressions [4/y] and [3/y] indicate that the results belong to Ramanujan's alternative theories of signatures 4 and 3, respectively. (See (1.8) and (1.7), respectively, for the definitions of y in these theories.) Moreover, the expressions [4/y] and [3/y] should be deleted from (13.1) and (13.2), respectively. For some inexplicable reason, Ramanujan indicated that his identities arose from the two alternative theories by placing the "symbols" [4/ y] and [3/ y] in the midst of the formulas. We now offer precise renditions of (13.1) and (13.2).

Entry13.1(p.392). Let 0 < x < 1. Ifyisgivenby(l.8)andz = 2FI(~'~; 1; x), then

33. Elliptic Functions to Alternative Bases

- -1- ( 1-8 - 1 - 2x

Ifyisgivenby(1.7)andz

~

f:o

(3n

=

nOOn

-~ -eny---11 6 ~ e2ny - I 00

)

.

177

(13.4)

2Fl(~'~; I;x), then

+ l)(!)n(t)n(~)n

{4x(1 _ x)}n

(n!)3

- -1- ( 1-6 - 1 - 2x

noon

~ -eny---11 8 ~ e3ny 00

)

1

.

(13.5)

Proof. We first prove (13.4). Recall that L(q) = 1 - 24

nqn L --, 1 - qn 00

(13.6)

n=l

where q = e- Y • From the definition (13.6), it is easy to see that

~ n ~ n 1 1 - 8 L..t eny _ 1 - 16 L..t e2ny _ 1 = '3 L (q) n=l

n=l

22

+ '3 L (q

).

(13.7)

From the proof of Theorem 9.11. L(q)

=

(1 - 3x)Z2

dz x)zdx

(13.8)

dz - x)z dx'

(13.9)

+ 12x(1 -

and L(q2) = 1(2 - 3x)Z2

+ 6x(1

Thus, by (13.7)-(13.9), it suffices to prove that

f:

(4n

+ 1)(1)n(i)n(~)n

{4x(1 - x)}"

(n!)3

n=O

= -11 - 2x

( (1 - 2X)Z2

+ 8x(1

dZ) . - x)zdx

(13.10)

In Clausen's formula. Entry 13 of Chapter 11 (Part II [2. p. 58]), put f3 = -~, and y = to deduce that

-i.

1

zF(i.~; l;x)

=

3F2(i,~,

In Entry 12 of Chapter 11 (Part II [2, p. 56]), set p = 4x(1 - x). Accordingly.

2Fl(i,~; 1; 4x(1- x»

=

1; 1, l;x). x=

CI.

=

(13.11)

-i, y = -~. z = 1, and

2Fl(i,~; 1; x).

(13.12)

Thus, from (13.11) and (13.12). 2F(i. ~; 1; x)

=

3F2(i, ~,1; 1, 1; 4x(1 - x».

(13.13)

178

Ramanujan's Notebooks, Part V

Using (13.13), we see that (13.10) is equivalent to the identity

f: n(±)n(~)n(~)n n=1

{4x(l _ xW-1

(n!)3

=

Z dz. 2{1 - 2x) dx

(13.14)

Differentiating (13.13), we find that

2z dz = ~ dx ~

n{±)n(~)n(~)n (n!)3

{4x(l _ xW- 1 (4 - 8x),

which is readily seen to be equivalent to (13.14), and so the proof of (13.4) is complete. The proof of (13.5) is similar. First, from (13.6), we easily find that 00 noon 1 3 3 1-6"---18" =-L(q)+-L(q). ~ eny - 1 ~ e 3ny - 1 4 4 n=1 n=1

(13.15)

From Lemma 4.1,

L(q) = (1 - 4x)Z2

+ 12x(1 -

(1,

dz x)z-, dx

(13.16)

where z = 2FI ~; 1; x). We need a similar representation for L(q3). To that end, from Corollary 3.5 and (4.4), we find that

1 d L(q3) = -q-Iog (q3 f2\-q») 3 dq 1

d

dx

= -q-Iog (zI2r 9x 3(1 3 dx

= ~q 3

(12dZ z dx

= 4x(l

- x») -

1_)

+ 2 __ x

dz - x)zdx

1- x

dq

x(l-x)z2 q

+ ( 1 - 4) j"x z 2.

(13.17)

Hence, from (13.15)-(13.17),

~L(q)

+ ~L(q3) =

(1 - 2x)Z2

+ 6x(1

dz - x)z dx.

(13.18)

Thus, by (13.5), (13.15), and (13.18), it is sufficient to prove that

f:

(3n

n=O

+ l)(±)n(1)n(~)n {4x(1

- xW

(n!)3

=

_1_ 1 - 2x

(0 _

2x)Z2

+ 6x(1

_ x)z dZ) . dx

-i,

-1,

(13.19)

Proceeding as in the proof of (13.4), we put ex = fJ = and y = ± in Clausen's formula, Entry 13 of Chapter 11 (Part II [2, p. 58]). Then we set

33. Elliptic Functions to Alternative Bases

179

-%,

x = -~, y = z = 1, and p = 4x(1 - x) in Entry 12 of Chapter 11 (PartIl [2, p. 56]). Combining these two formulas together, we deduce that Z2

3 F2(%,

=

~,

!; 1, 1; 4x(1 -

(13.20)

x)).

Using (13.20) in (13.19), we find that it suffices to prove that

f: n=1

n(!)n(%)n(t)n {4x(1 _ x)}n-I = Z dz. (n!)3 2(1 - 2x) dx

(13.21)

Differentiating (13.20), we achieve (13.21) to complete the proof. At the very bottom of page 392, Ramanujan [9] wrote 1·2 l+-t+··· 32

2 2 + (gg')1/4

22- (GG')1/4 {t(l - t)}1/3 - ----,..,.-- 3 (GG')1/6

"3

(gg')1/6

.

(13.22)

The notations G, G', g, and g' were not defined by Ramanujan. In view of the appearance of 2 Fl (%, ~; 1; t), it would appear that the latter two equalities pertain to a new (unknown) type of class invariant associated with the theory of signature 3. Chapter 34 is devoted to class invariants, and in the table of class invariants in his second notebook, Ramanujan uses the notations G := G n and g := gn. However, no dependence on n is indicated in (13.22). Also, the appearance of two invariants in each equality should be reflected in the appearance of two distinct moduli on the left side. Thus, we are left with the conclusion that Ramanujan is claiming two q-series identities, one for the pair G, G', and the other for the pair g, g', whatever these "invariants" might be. Now, by Corollary 3.2 and Lemma 5.1, {t(1 _ t)}1/3 = b(q) c(q) = 3ql/3 f2(_q)p(-q:l) a(q) a(q)

= ql/3

(3 -

a 2 (q)

42q

+ 393q2 -

3240q 3 + ... ),

(13.23) where we employed Mathematica to obtain the q-expansion. Ramanujan, in his second notebook, defined G by G 1/ 24 =

2 1/ 4ql/24

,

(-q; q2)oo

and later, in his paper [3], he gave the different definition G = 2 1/ 4q-I/2\ -q; q2)oo.

Using either definition, and any similar representation for G', we do not obtain an expansion for the middle expression in (13.22) that is close to that in (13.23).

180

Ramanujan's Notebooks, Part V

More generally, suppose that GG' = 2a(1

Then

2 2 - 2a/4(1 + (bj4)qC + ... ) 2a / 6 (1 + (bj6)qC + ... ) .

2 2 - (GG')1/4 3

+ bqc + ... ).

3

(GG')1/6

Thus, clearly, if (13.22) holds, we must have a = 4. It then further follows that c = and b = -9 . 22/ 3 • We are unable to use the resulting expansion

1

GG'

=

16(1 - 9. 2 2/3 q l/3

+ ... )

to identify GG' in any meaningful way.

14. Concluding Remarks It seems inconceivable that Ramanujan could have developed the theory of signature 3 without being aware of the cubic theta-function identity (2.5), and in Lemma 2.1 and Theorem 2.2 we showed how (2.5) follows from results of Ramanujan. H. H. Chan [1] has found a much shorter proof of (2.5) based upon results found in Ramanujan's notebooks. H. F. Farkas and I. Kra [1, p. 124] have discovered two cubic theta-function identities different from that found by the Borweins. Let w = exp(2Jrij6). Then, in Ramanujan's notation,

w 2f3 (wq 1f3, wq2/3)

+ f3

(wq 2f3, wql/3) = wf3 (_ql/3, _q2/3)

and f3 (wq 2f3, wql/3) _ f3 ( wq l/3, wq2/3) = ql/3 f3 (wq, w).

H. F. Farkas and Y. Kopeliovich [1] have generalized this to a pth order identity. Garvan [2] has recently found elementary proofs of the cubic identity and the pth order identities, and has found more general relations. Almost all of the results on pages 257-262 in Ramanujan's second notebook devoted to his alternative theories are found in the first notebook, but they are scattered. In particular, they can be found on pages 96, 162,204,210,212,214,216, 218,220,242,300,301,310, and 328 of the first notebook. Moreover, Theorem 9.11 is only found in the first notebook. In Section 8, we crucially used properties of b(z, q), a two variable analogue of b(q). The theory of two variable analogues of a(q), b(q), and c(q) has been extensively developed by Hirschhorn, Garvan, and 1. M. Borwein [1] and by S. Bhargava [I]. Some of Ramanujan's formulas for Eisenstein series in this chapter were also established by Venkatachaliengar [I]. In [3], Garvan describes how the computer algebra package MAPLE was used to understand, prove, and generalize some of the results in this chapter.

33. Elliptic Functions to Alternative Bases

181

Small portions of the material in this chapter have been described in expository lectures by Berndt [10] and Garvan [1].

34 Class Invariants and Singular Moduli

1. Introduction So that we may define Ramanujan's class invariants, set (a; q)oo

=

n 00

Iql

(l - aqn),

< 1,

n=O

and (1.1)

If

q

= exp(-rr.Jn),

(1.2)

where n is a positive rational number, the two class invariants G n and gn are defined by G n :=

r

1/ 4q-l/24 X (q)

gn :=

and

r

1/ 4q-l/24 X (_q).

(1.3)

In the notation of Weber [2], G n =: 2- 1/ 4 f(..;=n) and gn =: 2- 1/ 4 fI (..;=n). The definition of G n employed by Ramanujan in his paper [3], [10, pp. 23-39] is not the same as that used by him in his notebooks [9], while his definition of gn in [3] is that used in his first notebook but not in his second notebook. More precisely, If we replace G and g in the second notebook by H and h, respectively, the relations between the definitions are given by G 24 = _I_ n H.fii

and

As usual, in the theory of elliptic functions, let k := k(q), 0 < k < 1, denote the modulus. The singular modulus kn is defined by k n := k(e- 7r .fii), where n is a natural number. Following Ramanujan, set a = k 2 and an = k~. It is well known that G n and gn are algebraic; for example, see Cox's book [1, p. 214, Theorem 10.23; p. 257, Theorem 12.17]. However, much more is known. Weber [2, p. 540] and, more recently, H. H. Chan and S.-S. Huang [1], using a result of Deuring [1], have proved the following theorem.

184

Ramanujan's Notebooks, Part V

Theorem 1.1. (a) IJn == I (b) Ifn == 3 (e)lfn == 7 (d) Ifn == 2

(mod (mod (mod (mod

4), 8), 8), 4),

then G n and 2an are units. then 2- 1/ 12 G n and 22an are units. then 2- 1/ 4 G n and 24a n are units. then 8n and an are units.

As G. N. Watson [6] remarked, "For reasons which had commended themselves to Weber and Ramanujan independently, it is customary to determine G n for odd values of n, and gn for even values of n." At scattered places in his first notebook [9], Ramanujan recorded the values for 107 class invariants, or polynomials satisfied by them. On pages 294-299 in his second notebook [9], Ramanujan gave a table of values for 77 class invariants, three of which are not found in the first notebook. Since the second notebook is an enlarged revision of the first, it is unclear why Ramanujan failed to record 33 class invariants that he offered in the first notebook. By the time Ramanujan wrote his paper [3], [10, pp. 23-39], he was aware of Weber's work [2], and so his table of 46 class invariants in [3] does not contain any that are found in Weber's book [2]. Except for G 3 25 and G 363 , all of the remaining values are found in Ramanujan's notebooks; twenty-one of these class invariants are found in his second notebook. At scattered places in the second and third notebooks, Ramanujan recorded irreducible polynomials satisfied by four further invariants. In conclusion, to the best of our tallying, Ramanujan calculated a total of 116 class invariants, or monic, irreducible polynomials satisfied by them. In two papers [6], [7], Watson proved 24 of Ramanujan's class invariants from Ramanujan's paper [3]. In the first [6], Watson devised an "empirical process" to calculate 14 of the 24 invariants, while in the second [7], he employed modular equations to prove 10 invariants. In another paper [5], Watson established Ramanujan's value for G l353 , communicated by Ramanujan [10, p. xxix] in his first letter to Hardy, and also stated in his paper [3]. In the introduction to [6], Watson remarked, "It is intended to publish the calculations involved in the construction of the set N + Q (the invariants appearing in both Ramanujan's paper [3] and the second notebook) as part of the commentary on the notebooks by Dr. B. M. Wilson and myself." Although Watson and Wilson's efforts to edit Ramanujan's notebooks have been preserved in the library at Trinity College, Cambridge, Watson's calculations of these twenty-one invariants are not found there. If Watson actually calculated these invariants, it appears that his work has been lost. The twenty-one values of n are: 65,69,77,81, 117, 141, 145, 147, 153,205,213,217, 265,289, 301,441, 445, 505,553, 90, and 198. Watson wrote four further papers [9], [10], [12], [13] on the calculation of class invariants. The values of n considered by Watson depend upon the class numbers for positive definite quadratic forms of discriminant -no In the course of his evaluations, he determined the class invariants for n = 81 [12], 147 [12], and 289 [13]. Thus, after Watson's work, 18 invariants of Ramanujan from his paper [3] and notebooks [9] remained to be verified.

34. Class Invariants

185

Five of these invariants are established in Section 3. For each of these five values, 117, 153,441,90, and 198, n is a multiple of 9, and proofs are effected by formulas relating G9n with G n and g9n with gn, which we establish by using one of Ramanujan's modular equations of degree 3. The latter formula is found on page 318 of Ramanujan's first notebook, but not in his second notebook, while the former formula is not found in any of the notebooks. Since modular equations are crucial in our work on class invariants, we now give a precise definition of a modular equation. Let K, K', L, and L' denote complete elliptic integrals of the first kind associated with the moduli k, k' := Jl - k2, l, and l' :=~, respectively, where 0 < k, l < 1. Suppose that

K' L' n-=-

K

L

(1.4)

for some positive integer n. A relation between k and l induced by (1.4) is called a modular equation of degree n. Following Ramanujan, set

a=e

and

We often say that f3 has degree n over a. As usual, in the theory of elliptic functions, set

q := exp( -7r K' / K).

(1.5)

Since X (q) = 2 1/6{a(l_a)/q}-1/24 and X( -q) = 21/6{a(l-a)-2 /q}-1/24 (Part III [3, p. 124]), it follows from (1.1), (1.3), and (1.5) that and

(1.6)

This formula for G n will be used in certain modular equations. In Sections 4--7, we establish the remaining 13 values, each for G n , n = 65, 69, 77, 141, 145,205,213,217,265,301,445,505, and 553, claimed by Ramanujan. Quite remarkably, the class number for each of the 13 imaginary quadratic fields Q(J=1i) equals 8. Moreover, there are precisely two classes per genus in each case. Our first proofs, given in Sections 4 and 5, employ Kronecker's limit formula, which is used to find representations for certain products of Dedekind eta-functions in terms offundamental units; see Theorems 4.1, 4.2, and 4.7. Each of the 13 values of n is a product of a small prime (3, 5, or 7) and a larger prime. Thus, our proofs also crucially employ certain modular equations of Ramanujan of degrees 3, 5, and 7. It is highly unlikely that Ramanujan was familiar with Kronecker's limit formula and the arithmetic of quadratic fields, and so our proofs certainly are not those found by Ramanujan. However, Ramanujan obviously discerned some unique arithmetical properties in these instances, and it would be fascinating to discover Ramanujan's approach. The methods in Sections 4 and 5 have been further extended by L.-C. Zhang [2], [3] who has rigorously established all the invariants in Watson's paper [6] that were calculated there with his "empirical process."

186

Ramanujan's Notebooks, Part V

Ramanujan used modular equations to calculate only a couple of simple invariants in [3]. This fact and the sentence, "The values of G n and g2n are got from the same modular equation." [3], [10, p. 25] are the only clues to his methods that Ramanujan provided for us. It would seem that if Ramanujan had employed another type of reasoning, he would have dropped some hint about it. As mentioned earlier, Watson [7] used modular equations to establish some of Ramanujan's invariants. However, for his calculations of G n , it is important that n be a square or a simple multiple of a square. We have been able to prove six of the remaining thirteen values for Gn , namely, for n = 65, 69, 77, 141, 145, and 213, by using modular equations. As will be seen in our proofs in Section 6, we need some new ideas to effect proofs of these six invariants via modular equations. To prove the remaining seven invariants by employing modular equations, we would need modular equations of degrees 31, 41, 43, 53, 79, 89, and 101. Apparently, only for degree 31 did Ramanujan derive a modular equation, for he recorded no modular equations for the other six degrees in his notebooks. Thus, Ramanujan's methods appear to be even more elusive. Watson [6, p. 82] opined that "I believe that fourteen were obtained by Ramanujan by means of the empirical process which I described in the discussion of GI353." We are not so confident that Ramanujan used this empirical process, for which Watson offered little explanation. In fact, Watson's "empirical process" is not rigorous. However, in Section 7 we shall use class field theory to make Watson's procedure rigorous for a large class of invariants including those 13 invariants mentioned above, and we use the process to calculate two new invariants as well. Chan [3] has further extended the methods of Section 7 and calculated 27 new class invariants. Section 8 is devoted to some miscellaneous results on class invariants, including two entries to which we have not been able to attach any meaning. Here we also establish Ramanujan's more detailed assertions about the irrreducible polynomials satisfied by G29 and G 79' In Section 9, we tum to Ramanujan's singular moduli. Once G n or gn is known, then it is easy to calculate exn from (1.6) by simply solving a quadratic equation. However, this expression for ex n that one trivially obtains from the quadratic formula is usually not very interesting or attractive. Thus, it is desirable to develop other algorithms for the calculation of exn that will reflect the unit structure of exn described in Theorem 1.1. For the calculation of exn , when n is even, Ramanujan devised a very clever algorithm given in Theorem 9.1. For odd n, we do not have such an inclusive algorithm, and so we had to develop some lemmas to facilitate calculations. In Section 10, a simple function of singular moduli is studied. In the last two pages of his notebooks, Ramanujan studied the j-invariant. He seems to have quoted some results from the literature. However, Ramanujan made some remarkable discoveries, including very simple polynomials satisfied by certain algebraic functions of the j-invariant. Ramanujan's work on the jinvariant is the topic of Section 11.

34. Class Invariants

187

2. Table of Class Invariants Both prior to his table of class invariants in his second notebook and at the beginning of his paper [3, eqs. (5), (7)], [10, p. 23], Ramanujan recorded two simple formulas relating these invariants, and so we first state and prove these here. See also Exercise 5c on page 73 of the Borweins' treatise [1]. Entry 2.1 (p. 294, NB 2). For n > 0,

g4n

= 21/ 4gn G n •

Proof. This identity is an immediate consequence of the definitions of G n and gn in (1.1)-(1.3) and the elementary identity (q2; q4)00 = (q; q2)00(_q; q2)00.

(2.1)

Entry 2.2 (p. 294, NB2). For n > 0, (gn G n)8(G! - g!) = ~.

Proof. Jacobi's identity for fourth powers of theta-functions (part ill [3, p. 40, Entry 25(vii)]) can be written in the form (Whittaker and Watson [1, p. 470]) ( -q,. q 2)800

(.

q, q

-

2)8

00

16q = 16q(-q 2.,q 2)8 = (2. 4)8 q ,q 00

00

'

(22) .

where we used Euler's identity 1

(-q;q)oo = ( . 2)

q. q

00

.

By (2.1), we can write (2.2) in the form r2q-I/3(q; q2):!or2q-I/3(_q; q2):!o

x (r2q-I/\_q; q2):!o - r 2q-I/\q; q2):!o) = ~, that is, by (1.1)-(1.3), when q = exp( - ] l ' v1n), (gn G n)8(G! - g~) = ~.

Recall from Part ill [3, pp. 91, 102] the definition F(x) = exp (

-]l'

2FI(4,

4; 1; I-X»)

I

2 F \(2'

\

2; 1; x)

'

0< x < 1,

(2.3)

(4, 4;

where 2 FI 1; x) denotes the ordinary hypergeometric function. Recall also from Part ill [3, p. 36] or Chapter 33 the theta-function

L 00

qJ(q) =

n=-oo

qn2,

Iql < 1,

(2.4)

188

Ramanujan's Notebooks, Part V

and the fundamental formula (Part III [3, p. 102]) K(k) = ~Jr 2FI(~'

4; 1; k 2) =

4:rrcp2(q),

(2.5)

if q is given by (1.5). (The evaluations of X(q) and X(-q) used in (1.6) depend upon (2.5).) If x = an in (2.3), then F(an ) = exp( -:rr "fii).

(2.6)

Now from (1.6),

(2.7) · and so, smce an

~

I 2'

an =

t (1 - J1 -

1/

G~4) .

(2.8)

In the first notebook, Ramanujan frequently records G n , or equivalently, an, in the form F

(t (1 -JI - I/G~4))

= exp(-:rr"fii),

(2.9)

by (2.6) and (2.8). We have indicated such a representation in the tables by placing "F" after the page number in the first notebook where the corresponding invariant (or, equivalently, singular modulus) is located. We next give a table of all the values of G n and gn found by Ramanujan. We emphasize that formulas for certain values of G n and gn may be different in both the first and second notebooks and Ramanujan's paper [3]. In most cases, it is not difficult to verify that Ramanujan's formulations are equivalent. In particular, when Ramanujan employs (2.9), a modest amount of calculation is needed. For example, such calculations are necessary on pages 284, 287, 292, 293, 296, and 311. In all these instances, the calculations are routine, and there is no need to give them here. In other instances, for example, on page 314 in the first notebook, Ramanujan records the value for G;;24 in the notation "4p(l - P)," that is, he is using (1.6) with an replaced by p. When the invariant is a root of a cubic polynomial, Ramanujan, as well as Weber [2], normally, but not always, only gives the polynomial. There are some instances when Ramanujan calculated the appropriate root but Weber did not. In particular, we establish Ramanujan's values for G 75 and GI75 in Section 8. However, Ramanujan's polynomials are those satisfied by 1/(2 1/ 4 G n ), instead of G n • Thus, on page 345 in his first notebook, Ramanujan recorded the irreducible polynomials satisfied by 1/(2 1/ 4 G n ), for n = 23,31,11,19,27,43, and 67. These polynomials are repeated on page 351 together with monic irreducible polynomials satisfied by 1/(2 1/ 4 G3) and by (erroneously) 1/(2 1/ 4 G7). In the tables in his second notebook, Ramanujan explicitly states G n for n = 3,7, and 27. Lastly, we remark that the tables of Weber [2J contain some errors. Corrections have been made by J. Brillhart and P. Morton [1].

34. Class Invariants

189

For convenience, we use the following abbreviations in citing sources for the listed invariants: Ramanujan's first notebook: Nl, Ramanujan's second notebook: N2, Ramanujan's third notebook: N3, Ramanujan's paper [3]: RP, Watson's paper [5]: WXIV, Watson's paper [6]: WI, Watson's paper [7]: WII, Watson's paper [9]: W3, Watson's paper [10]: W4, Watson's paper [12]: W5, Watson's paper [13]: W6, Weber's treatise [2]: We, Brillhart and Morton's corrections [1]: BM.

Table ofG n

n=l 1

Refs.: N2,RP,W5,We

n=3 Refs.: Nl (282F),N2, W3, We

n=5

Refs.: Nl(287F), N2,RP,We

n=7 Refs.: Nl(287F),N2,W3,W4,We

n=9

Refs.: Nl(284F,287F),N2,RP,We n = 11 2- 1/ 4X, where x3

-

2x 2 + 2x - 2

=0

Refs.: Nl(295F,345,351)N2,W3,We

190

Ramanujan's Notebooks, Part V

Table of G n (Continued)

n = 13

Refs.: N1(292F),N2,RP,We

n = 15

Refs.: Nl(289F),N2,W4,We

n = 17

.j + .ff7 + .j.ff7 5

3

8

8

Refs.: N1(296F),N2,RP,We

n = 19 2- 1/4x where x 3

2x - 2 = 0

-

Refs.: N1(295F,345,351),N2,W3,We

n = 21

Refs.: N1(293F),N2,We

n = 23

2 1/4x,

where x 3

-

x-I = 0

Refs.: N1(295F,345,351),N2,W3,W4,We

n = 25

1 +-V5 2

Refs.: N1(287F),N2,RP,We

n = 27 21112 ( ,ifi -

1)

-1/3

Refs.: N1(305F,345,351),N2,W3,We

n = 29

Gi9 = x,

where x 6

-

9x 5 + 5x 4 + 2x 3

-

5x 2

-

9x - 1 = 0 Refs.: N2(263),We

34. Class Invariants

191

Table of G n (Continued)

n = 31 2 1/ 4 X,

where x 3

-

x2

1= 0

-

Refs.: Nl{296F,345,351),N2,W3,W4,We

n = 33

Refs.: Nl{311F),N2,We

n = 37 (

6 +../37)

1/4

Refs.: Nl(305F),N2,RP,We

n = 39

21/4(~+3r

(J5+;n +J~-3) Refs.: Nl(305F),N2,W4,We

n

= 41 Gil =

x,

where

(x

+

~

y-

5+

j4t (x

+

~)

+ 7+

j4t

= 0

Refs.: N3(382),WE,BM

n = 43

2- 1/ 4X

where x 3

-

2x 2

-

2= 0

Refs.: Nl{313F,345,351),N2,W3,We

n = 45

Refs.: N2,We

n= 47 2 1/ 4X,

wherex s = (1 +x)(l

+x +x

2)

Refs.: Nl(234),N2(263),We

n = 49

Refs.: Nl(293F)N2,RP,We

192

Ramanujan's Notebooks, Part V

Table of On (Continued)

n = 55

2'1'

(,;s +2( (t +! ,;s8- 1) +8,;s

Refs.: Nl(315),N2,W4,We

n

= 57

1/6 ( 3 0 : ; 13 )

(2 +

v'3)

1/4

Refs.: Nl(315),N2,We n

= 63 2'14

(5+;m)'" (/5+

8.;2I

+!~

-3)

Refs.: Nl(305F),N2,W4,We

n = 65

(~+3)'" (,;s2+ 1)'" (!9+~ +/'+~r Refs.: Nl(315),N2,RP

n =67 2- 1/ 4 X

where

X3 -

2x2 - 2x - 2

=0

Refs.: Nl(345,351),N2,W3,We

n

= 69

(5+~r' e"': -'23)'" (! 6+:'" +! 2+: " ' ) 'I' Refs.: Nl(314F,315),N2,RP

n

= 73

!9+v'73 + !1+v'73 8

8

Refs.: N1(313F),N2,RP,We

n =75

3 . 25/ 12

.j5 + 1 (10)1/3 2

+ .j5 2

141/3 .51/6 _

.j5 _ 1 Refs.: Nl(311),We

34. Class Invariants

193

Table of G n (Continued)

n =79 t =

2 1/4 jG 79.

where t 5

-

t4

+t3 -

2t 2 + 3t - 1 = 0 Refs.: N2(263,300)

n = 81

(

;/2(-/3 + 1) +

1)

1/3

;/2(-/3 - 1) - 1 Refs.: N2,WS,RP

n = 8S

Refs.: Nl(31S),N2,RP,We

n =93

(39~-'31r

(-'31;3-'3)'" Refs.: Nl(31S), N2,We

n = 97

J13+ffi + JS+ffi 8

8

Refs.: Nl(30SF),N2,RP,We

n = lOS

;U) ,,4 (2+ -'3) "4 (,;s + 2) ,,6 (6 + ,tjS)''''

( 5+

Refs.: Nl(317),N2,We

n = 117

(3+~) ,,4 (2v'3 + vTI) ,,. ( 3,,4 + ~) Refs.: Nl(31S),N2,RP

194

Ramanujan's Notebooks, Part V

Table of Gn (Continued) n = 121

n = 133

(8 + 3./7) 1/4 ( 5./7 ~ 3.JI9 )

1/4

Refs.: Nl(315),N2,We

n = 141

(4v'3+A7t'(7+~)

1/12(~

~1/2

V¥+V~)

Refs.: Nl(320),N2,RP n = 145

'5

2 (5 + 2m) 1/4(~ 17 + .J145 ~1/2 9+ .J145 8 + 8

(v.J+ )

1/4

Refs.: Nl(315),N2,RP

n = 147 21112

(~+ ~ {I! -(28)II'lf Refs.: N2,RP,W5

n = 153

(/5+;n +J~-3)' (J37+:.m +J33+:.mf Refs.: Nl(315),N2,RP

n = 163 2- 1/ 4 X

where X3

-

6X2

+ 4x -

2

=0 Refs.: N2(300),We

n = 165

1/4 r;; h7)1/6 ( v'15+v'IT )1/6 ) (r;; (4 + v ~)1/4( 15 3v 5 + 2v 11 2 v5+2 Refs.: Nl(317),N2,We

34. Class Invariants

195

Table of Gn (Continued) n = 169

H

(,(l3 + 2)+

x {

CH~,(l3r

(11 +2,(l3 + 3.(3)'" + (11 +2,(l3 - 3.(3)"'}1 Refs.: Nl(294F,317),RP,wn,W5

n = 175

n

= 177 Refs.: Nl(315),N2,We

n =205

(1+ v'S) ev'S;~)'" (t+~ +J~ -I) 2

Refs.: Nl(314),N2,RP n = 213

e./3 ;v'TI)

1/8

(

59

+~v'TI)

1/12

"¥

(~

~1/2

+V~ )

Refs.: Nl(315),N2,RP n =217

(J9+~ +JII +2Wi )'" (J12+.5./7 +P6+:./7)'" Refs.: Nl(314),N2,RP

n = 225

(1+2~)

(2 + .J3)1/3

(J4+ ~ + 4) (15)1 /

Refs.: Nl(293F),RP,WII

196

Ramanujan's Notebooks, Part V

e

n = 253

Table of G n (Continued)

~r (1301; 9v'23)'" Refs.: Nl(315),N2,We

n = 265

(2 +,/5)'" ( 7+2,/53) '" (

89 + 5v'265 8

+

,..-------,. 1/2 81 + 5v'265 8 Refs.: NI(314),N2,RP

n = 273

c5,/7~I03r (~+3)'" (,/7;-t3)'" (2+-t3)'I' Refs.: Nl(317),N2,We

n =289 (

17+m+(17)1/4(5+m) 16

+

r----------.2 1+ + (17)1/4(5 + 16

m

m)

Refs.: Nl(317),N2,RP,W6

n = 301

(8 +3,/7) 'I' (23v'43; 5M) 'I'

(J 46 +;v'43 +;42 +;v'43) 'I' Refs.: Nl(325),N2,RP

n = 325

( 3+.JI3)1/4 t, 2

h 3 2 (1_.JI3)2 (1+.JI3)2 1 were t + t 2 + t 2 + =

~ {3

v5 t -I 2 (I + .JI3) +t (I - .JI3) -I } 2 2 Refs.: RP,WII n = 333

(6 + ../37)1/4(7../3 + 2../37)1/6 (

J7 + 2../3; ./3 + 2..(3) Refs.: NI(314,315),RP,WI

34. Class Invariants

197

Table of G n (Continued)

n = 345

1+ .J5 (1 + -J3) (3-J3 + .J(3) 1/2

2.j2

1/2 (

15.J5 + 7.J(3) 1/6 .j2

2

Refs.: Nl(317),N2,We

n

= 357

(-J3 +2 ../7) (8 + 3",7) Pi

1/4

(.Jfi + 2

../21)1/4

(11 +.j2

.Jffi)1/6

Refs.: Nl(317),N2,We,BM

n = 363

where 2t 3 -

25/12 t,

- t{

12 { (4

+ J33) + v'il+2J33}

1+ J 11 + 2J33} - 1= 0

Refs.: RP,WII n = 385

Ji

(3

+ .JIT)(.J5 + ../7)(../7 + .JIT){3 + .J5) Refs.: Nl(317),N2,RP,We

n = 441

J./3 + ,J7

'I' 12+,J7 + ,/7+ 4,J1 J'/3+,J7 + (f>,/1) 'I' V 2 J3 +../7 - (6../7)1/

(2+ ./3)

2

4

Refs.: Nl(46),N2,RP

n =445

'/2+45e1+~r Un+s./89 +fo+;m)

n =465

(2 + ./3) >/'

x

C+245

r' r' (3./32+ 31

Refs.: Nl(320),N2,RP

(545 + 2./31) 'I"

U2+;m +J6+;mf (Jll+~./31 +J13+~./31f Refs.: Nl(319),RP,WI

198

Ramanujan's Notebooks, Part V

Table of G n (Continued) n = 505

J(2+,/5) ( 1+2,/5) 'I' (10 + ,,'lOf)'/4 113

x (

+ 5.JT05 8

+

105 + 5.J105

1/2

8

Refs.: Nl(344),N2,RP

n = 553

(

96+

x(

11m 4 +

141 + 16m

2

r - - - - - - . 1/2

+ Refs.: Nl(320),N2,RP

n = 765

J3+2,/5 (16 + + ,/fS) '/4 (9+;as) '14 x (J6+;sr +JlO+4v'sff' (J18+:v'sf +J22+:v'sff' .,/225) '02(4

Refs.: Nl(343),RP,WI n =777

,(3)'1'(6 + "/'Yi)'I' ( ,/3 ~ .J1f4 (246.J1 + 107.J37t" x (J6+:.J1 +JlO+:.J1r (J15+26.J1 +l7+26.J1r (2 +

Refs.: N1(319),RP,WI n = 897

./2 +

x

(J

,/3J ~ + 3 (4vTI + 3.,/23) '02 (3,/3 ~ .,/23)

60 + 9v'39 4

+

'/4

.J39 ~ 1/2 (I¥ + J¥l)1/2 56 + 9v'39 4

8 + v'39 4

4+

4

Refs.: Nl(320),WI

34. Class Invariants

199

Table of G n (Continued) n

= 1225

¥ +v'35)'I' (71 +J +.f1f' 24

(6

x(

43 + 15,Ji + (8 + 3,Ji)/i.0J7 8

+v/35 + 15,Ji + (88+ 3,Ji)/i.0J7 Refs.: Nl(46),RP,wn

Refs.: Nl(319),RP,WXIV

Refs.: Nl(320),RP,WI n = 1677

(4414..J13 + 2427.J43) 1/12 (

+2.13)'I' ( x

~+ 3 )

3/4 (

.J43 ;

J39

)1/4

355+;4..143 +/35 [+;4.f4J.l

x

(.JI3

'/2

(/[7+:..143 +/[3+:..I43f Refs.: Nl(320),WI

200

Ramanujan's Notebooks, Part V

Table of gn

n=2 Refs.: Nl(316),N2,We

n=6 Refs.: Nl(316),N2,We n

= 10

Refs.: Nl(316),N2,RP,We

n = 14

Refs.: Nl(316),N2,We

n

=

18 Refs.: Nl(316),N2,RP,We

n

= 22 Refs.: Nl(316),N2,RP,We,BM

n

= 26

~ ( J2 + m

+

/(2 + mhh + m + 3J3(3 + m)

+j(2+mhi2+m -3J3(3 + m)) Refs.: Nl(318,344),W5,We

n

= 30 Refs.: Nl(316),N2,RP,We

n

= 34 3 + ..(f7 4

+

J 10 + 6..(f7 4

Refs.: Nl(316),N2,RP,We

34. Class Invariants

201

Table of gn (Continued) n = 38 g = g38,

where

g3

+ g-./2 = JI + -./2 (1 + g2-./2) Refs.: NI(344),We

n = 42 (2-./2 + .;7)1/6

./3:.;7 )

(

1/2

Refs.: NI(3I6),N2,We,BM

n = 46

Refs.: NI(3I6),N2,We

n = 50

j ( + (5+.,/5)''' (,;t I + 7,/5 +6,/6 + ,;tI + 7,/5 - 6,/6)) Refs.: NI(3I8,344),W5,We

n

= 58

Refs.: NI(3I6),N2,RP,We

n

U

= 62

4+ ,; I +

~+ /9 + 5,/2 + J,; I + ,/2 +

'7

+ 5,/2 - • )'

Refs.: NI(3I9),RP,WII,W6

n = 66

(,/2+ ,/5)'" (7,/2+ 3vTI) 'I"

(t

+;n +

J~ - I) '" Refs.: NI(3I9),RP,WI

n

= 70 (3

+ v's)(l + -./2) 2 Refs.: NI(3I6),N2,RP,We

202

Ramanujan's Notebooks, Part V

Table of gn (Continued)

n

= 78 (

3

+:U13)

1/2

+ .J26)1/6

(5

Refs.: Nl(316),N2,We

n = 82

/13+V4f /5+V4f +

8

8

Refs.: Nl(316),N2,We,BM

n = 90

1(2+ ,/5)(,/5 + ,(6»)'"

(J

3+,,/6

+ / ,/6,- I ) Refs.: Nl(318),N2,RP

(J

n = 94

4+ J7+

.rz.+ J7+ 5,/2 + / J7+ ,/2 + ~7+ 5,/2 - , ) , Refs.: Nl(319),RP,WII,W6

n = 98

(/4+,/2 +

~14+ ,.Jl4 +

,/2 +

JI'; 4v'l4 -,

'

Refs.: N1(318),RP,WII,W6

n = 102

r

Refs.: Nl(316),N2,We n

=

114 ( ,/2 + v'3)'" (3,/2 +

v'f9) '1"

(J

23+

:,/57 + /,5+ :,/57

Refs.: Nl(319),RP,WI

n

=

126

/ v'3 ;./7 (,/6 + ./7) '1'

(J

3+,,/2 + /,/2,- I )' Refs.: Nl(318),RP,WI

34. Class Invariants

203

Table of gn (Continued) n

=

130

n = 138

C+/')"'(3+:ur

:

Refs.: Nl(316),N2,RP,We

e~ ,/23)

1/4

(7S,;2 +

23,/23)""

({5;2.Ji V¥ + V~¥~)

~ 1/2

Refs.: Nl(319),RP,WI n = 142

9 + 5..;2 + ./127 + 90..;2 2 Refs.: Nl(316),N2,We

(J

n = 158

4+ v'o+,;2; ,117+ 13,;2 + Jv'9+,;2+

v'~7 + 13,;2 - 4)'

Refs.: Nl(319),RP,WII,W6 n = 190 3/2 (

1 +2./5 )

(3 +

,JIO)1/2

Refs.: Nl(316),N2,RP,We n = 198

,II + ,;2(4,;2 + v'33)'/6

(J9 \v'33 + PSv'33) Refs.: Nl(318),N2,RP

n = 210

.j,J3

+ ..;2(3y'j4 + 5./5)1/6

J.j7;,J3 J

./52+ 1

Refs.: Nl(320),We,BM

204

Ramanujan's Notebooks, Part V

Table of gn (Continued) n = 238

(/' +:~ +j5+:~) (j1+~~ +j5+~~) Refs.: Nl(319),RP,WI

n=310

(' +2v'S) J1+~ (j7+!,/IO +J3+!,/IO) Refs.: NI(319),RP,WI

n

= 330

J./6 + v'Sj./15 ; v'IT ( v'St

)

(v'IT + ,/10) 0f' Refs.: NI(320),We

n

= 522

j5+ 2./29 (5./29 + 11./6) (j9+ :./6 +j5+:./6) Of'

Refs.: NI(318),RP,WI n = 630

(V14 +VISy/6 ((1 +~) (3+2~) ( x x

v'3: ~) )

(j./15+.'" +2 +j./15+.'" -2) (j./15 +8'" +' + j./15 +8'" -.) Refs.: Nl(318),RP,WI

3. Computation of G n and gn when 91n In this section, we establish Ramanujan's class invariants for n = 117,153,441, 90, and 198. Note that for each such n, 91n. Our starting point is a relation connecting gn and g9n, found on page 318 of Ramanujan's first notebook, but not in his second notebook. K. G. Ramanathan [4] noticed this relation in the first notebook, but apparently he never gave a proof. Also unaware of its appearance in the

34. Class Invariants

205

first notebook, J. M. and P. B. Borwein [1, pp. 145, 149], although not stating the results explicitly, derived a fonnula connecting gn and g9n, as well as a fonnula relating G n and G9n. We use one of Ramanujan's modular equations of degree 3 to establish the aforementioned fonnulas connecting G n and G 9n , and gn and g9n. The fonner fonnula is not found in the notebooks, but it can be proved along the same lines as the latter. The Borweins [1, pp. 145, 146] also derived fonnulas connecting GS ln with G n and G 9n , and gSln with gn and g9n. Of course, the theorems described above can be utilized to establish other class invariants found by Ramanujan when 91n, in particular, for n = 27,45,63,81, 225,333,765,18, 126,522, and 630. Undoubtedly, some of these proofs would be simpler than previous proofs, for example, for 81, 225, 333, 765,126,522, and 630. In particular, Watson's proofs for n = 333,765,522, and 630 [6] were based on his "empirical method." In fact, the Borweins [1, pp. 147, 149, 150] employed the aforementioned fonnulas to calculate the invariants G27, GSl, G225, and g522. Moreover, previously undetennined class invariants, for example, for n = 171, 189, and 279 can be calculated. However, we shall confine ourselves here to the cases, n = 117, 153,441,90, and 98.

Theorem 3.1. Let (3.1)

Then G9" = G n ( p

p2 + ~)1/611 p2 - 1

+

2 + J(p2 - l)(p2 - 4) 2

J

p2 -

4+

J(p2 - 1)(p2 -

4)}

2

1/3

(3.2)

Proof. For brevity, we set G = G n in most of the proof. We shall employ Entry 15(xii) of Chapter 19 of Ramanujan's second notebook (part III [3, p. 231 D. Let P = {16atJ(1 - a)(1 - tJ)}I/S

where tJ has degree 3. Then Q

_ (tJO - tJ»)1/4 Q,

and

+ ~ + 2,,/2

aO - a)

(p - ~)

= O.

(3.3)

(3.4)

Recall from (1.6) that, when q = exp( -Tl.jii) , G" = {4a,,(1 - a n)}-1/24

and

G 9" = {4tJ"O - tJn)}-1/24 .

Hence, by (3.3), P = (G"G9,,)-3 and Q = (G,,/G9n)6. Thus, from (3.4), (G,,/G9,,)6

+ (G n/G 9,,)-6 + 2,,/2 (G nG9,,)-3 -

(G nG 9,,)3) = O.

(3.5)

206

Ramanujan's Notebooks, Part V

Setx

=

(G 9n /G n )3. Then (3.5) can be rewritten in the form

2.fiG 6x 3 + 2.fiG- 6x

x4 -

+ 1 = O.

(3.6)

Rearranging and using the notation (3.1), we find that we can recast (3.6) in the form

Since x > 1,

or

x2

-

.fiG2

(

G4+ JP2=1) x + P + JP2=1 = O.

Remembering that x > 1 when solving for x, we find that x =

~G2 ( G4+ JP2=1) + _1_

IG4 (G8 + 2G4JP2=1 + p2 - 1) - 2p - 2JP2=1.

~V

0~

Now, G 4 =!(p+Jp2_4)

and

Thus, squaring and expanding, we find that

G4(G4 + JP2=1) 2 = ! (p + J p2 - 4) ! (p2 - 2+ pJ p2 - 4) + (p2 _ 2+ pJ p2 - 4) JP2=1 + ! (p + J p2 - 4) (p2 = (p+JP2=1) (p2-2+J(p2- 1)(p2- 4») and

G (G 8 + 2G JP2=1 + p2 - 1) - 2p - 2JP2=1 = ! (p + J p2 - 4) ! (p2 - 2+ pJ p2 - 4) + (p2 _ 2+ pJ p2 - 4) JP2=1 + ! (p + J p2 - 4) (p2 - 1) - 2p - 2JP2=1 = (p + JP2=1) (p2 - 4+ J(p2 - 1)(p2 - 4»). 4

1)

(3.8)

4

(3.9)

34. Class Invariants

207

Using (3.8) and (3.9) in (3.7), we deduce that x

=

~J (p + ~) (p2 - 2+ J (p2 +

~

J+ (p

1)(p2 -

4+ J (p2 -

J p2 - 1) (p2 -

4»)

1) (p2 -

4»).

(3.10)

Recalling that x = (G9n/ G)3, we see that (3.10) is equivalent to (3.2), and so the proof is complete. We next prove the aforementioned result found on page 318 in Ramanujan's first notebook. Theorem 3.2 (p. 318). Let P

= gn4 -

-4

(3.11)

gn .

Then g9n

~)1/6 {J p2 + 4 + J(p22 + 1)(p2 + 4)

= gn ( P + V p~ + 1

+

J +2 p2

+ J(p2 + 1)(p2 +

1/3

4)} .

2

(3.12)

Proof. Set g = gn throughout the proof. Using (1.3) and (Ll), we rewrite (3.5) in the form .Jii( -q; q2)~ (_q3; q6)~

+

(_q3; q6)~ .Jii( _q; q2)~

+

8.Jii (_q; q2)&,( _q3; q6)&,

= o.

_ (-q; q2)~(_q3; q6)~ .Jii

Multiplying both sides by .Jii and then replacing q by -q, we find that q(q; q2)~ -

( 3.

6)6

q ,q ""

(q3; q6)~

+ (q,. q 2)6""

8q -

( .

q, q

2)3 ( 3.

2 3

6)3

"" q ,q ""

-

3

6 3

(q; q )00 (q ; q )""

Using (1.3) and (Ll) again, we find that -(g/g9n)6

Setting x

+ (g9n/g)6 -

2h(gg9n)-3 - 2h(gg9n)3 =

= (g9n/g)3, we deduce that X4 -

2hg 6x 3

-

2hg - 6x - 1 = O.

o.

= o.

208

Ramanujan's Notebooks, Part V

Recalling the notation (3.11), we see that the last equation can be rewritten in the fonn (x 2 _ ../2g6x _ p)2

= 2(p2 + 1) (g2X + ~) 2

It should now be clear that the remainder of the proof is completely analogous to

that for Theorem 3.1, and so we omit the rest of the proof. As a bonus, the fonnulas connecting G 9n with G n and g9n with gn led to closed fonn evaluations of Ramanujan 's cubic continued fraction at the arguments ± exp( -T{,Jii) by the author, H. H. Chan, and L.-C. Zhang [1]. The cube roots in (3.2) and (3.12) are not very attractive, and usually in applications Ramanujan found more appealing expressions for these cube roots. Ramanujan had an amazing ability for denesting and simplifying radicals, and we do not have the insights into radicals that Ramanujan had. However, it seems quite likely that in several instances Ramanujan used the following elementary result from Carr's book [1, p. 52]. Since Carr does not give a proof and since he adds the extraneous hypothesis that (a 2 - b) 1/3 be a perfect cube, we provide a proof here.

Lemma 3.3. Suppose that c := (a 2 (a

-

b)I/3. Then we can write

+ ./b)1/3 = X + ../Y,

(3.13)

where 4x 3

-

3cx

=a

(3.14)

c.

(3.15)

and

y = x2

-

Proof. From (3.13), we easily see that (a _ .jb)1/3

c Suppose we set c

= x2 -

x -../Y x2 - y .

y, so that (a - ./b) 1/3

= X - ../Y.

(3.16)

Cubing both sides of (3.13) and (3.16) and solving for a, we find that a

But since y = x 2

-

= x 3 + 3xy.

(3.17)

c, we deduce (3.14) from (3.17).

Usually, it is best to solve (3.14) by trial or inspection, for if, for example, Cardan's method is used, the value of x so obtained most frequently is the cube root that we originally sought to simplify.

34. Class Invariants

209

Since Carr's book [1] was Ramanujan's primary source for learning mathematics, it seems likely that Ramanujan employed Lemma 3.3 in simplifications. However, because most of us do not possess Ramanujan's ability to discern algebraic relationships, we describe another procedure that rests upon elementary considerations in algebraic number theory and involves less guessing. We see from Theorems 3.1 and 3.2 that it would be advantageous to find a number a such that (3.18) Then

and (3.19) Since

set

u= a

1

+-, a

(3.20)

so that, after squaring, (3.19) takes the shape

u (u 2

2 -

3)2 =

4(b + 1 + c.Jd).

(3.21)

Assuming the relevant expressions above are algebraic integers, we find that, upon taking norms in (3.21), N(u 2 )N 2 (u 2

-

3) = N (4

(b + 1 + c.Jd)).

(3.22)

Using (3.22), we determine u. We then solve (3.20) for a.

Entry 3.4.

Gm =

:or'

~ (3+

(2,,13 + v'i3}'/6

(3'1' + ;.;-+ ,,13) .

(3.23)

Proof. Let n = 13 in Theorem 3.1. From Weber's treatise [2, p. 721], or from the tables of the last section, G13 = (

3+

v'f3)

2

1/4

210

Ramanujan's Notebooks, Part V

By (3.1), p =

.J13. Thus, by a direct application of (3.2),

G117= (3+;0)

1/4

{~

~}1/3

(2~+.J13)1/6 V~+V~

(On page 314 in his first notebook [9], Ramanujan recorded G 117 in the form given above, which is strong evidence that Ramanujan utilized Theorem 3.1 as we have done.) It therefore remains to show that

{/II +26v'3 +/9+:v'3}'" ~ H3114+J4+v'3). (3.24) We apply Lemma 3.3 with

a = / II +26../3

and

Thus, c = 1. We therefore need to solve

4x 3

_

3x =

/11

+26../3.

To solve this by inspection, it perhaps is best to square both sides and set t = x 2 • Thus,

t(4t _ 3)2

= 11 + 6../3. 2

tJ

It is not difficult to see that t = 1 + ../3/4. Hence, x = 4 + ../3, and, from (3.15), y = ../3/4. Thus, (3.24) follows, and the proof is complete. Alternatively, in the notation (3.18) and (3.20), by (3.21), we want to solve u 2 (u 2 -

3)2 = 22 + 12~.

(3.25)

Factoring in Z[.J3] and using (3.22), we find that N(u 2)N 2(u 2 - 3)

= N(22 +

12~)

= 22 . 13.

Then

±2 = N(u 2 - 3) =: N(A Choose A

+ B.J3) =

A2 - 3B2.

= 1 = B. We now observe that N(u z ) = N(4+~) = 13,

as required. It is easily checked that, when u = then find that

J 4 + ../3, (3.25) holds. We readily

34. Class Invariants

211

Thus, (3.24) has been shown once again.

Entry 3.5.

G

=

153

(J 5+ 01 + J01- (J +901 + V- 901)'" 8

8

3 )'

37

{33+

4

4

(3.26) Proof. Set n = 17 in Theorem 3.1. From Weber's treatise [2, p. 721], or from the tables in Section 2, G l7 =

J5+v'f7 8 + Jv'f7-3 8·

(3.27)

Since 4

5 + v'f7

G l7 =

from

4

+

(1

+ v'f7)3/2 4.J2

'

(3.1) we find that p = (5 + v'f7)/2. Also,

~

p+vp--l=

5+ 2v'f7 + JI9+5v'f7 2 .

Thus, from Theorem 3.1,

G", ~

(J5+;m +J.m.- 3)(5+ 01 +F+~01r 2

17 + 5v'f7 +

J(19 + 5v'f7)(13 + 5v'f7> 4

+

13 + 5v'f7 +

J(19: 5v'f7)(13 + 5v'f7)

1/3 }

However, note that

Hence, G

153

=

(J5+01 + J01-3) (5+01 + /,9+501),'0 8

8

2

2

212

Ramanujan's Notebooks, Part Y

x

{J37+:~ +J33+:~r

Comparing the equality above with (3.26), by (3.27), we see that it remains to show that

(3.28) which is rather curious indeed. Note that

(J7+~ + J3+~r ~ 5+~ + 19+5~ 4

4

2

2'

Thus, by (3.28), it suffices to show that

J7+vTI 4

3 + J3+vTI 4 = G 17 ·

In the notation (3.18) and (3.20), we solve U 2 (U 2 -

3)2

= 7+

We now factor in the principal ideal domain Z N(u 2 )N 2 (U 2

We attempt to solve

±4 = N(u 2 - 3) =: N

-

3)

m.

(3.30)

[(1 + vTI) /2] . Thus,

= N(7 + m) = 25.

(1 + A

(3.29)

B

+ 2vTI) = A 2

+ AB -

4B 2 .

Take A = -1 and B = l. Then u 2 = (5 + vTI)/2 and N(u 2 ) = 2. A simple calculation shows that (3.30) indeed holds. Lastly, we find that

a=J5+f1 +J~-3.

By (3.27), we conclude that (3.29) holds to complete the proof.

Entry 3.6.

=

G 441

/2 +

V

v'7 +

J7 + 4v'7 J../3 + v'7 (2

2

2

,J3)1/6

+

./3 + v'7 + 6 1/471/8 ../3 +

v'7 _ 61/471/8

(3.31)

34. Class Invariants

213

Proof. We apply Theorem 3.1 with n = 49. From Weber's treatise [2, p. 723], or from the tables of Section 2,

(3.32) It is easily checked that

J4+./7 + 71/4

=

2

12 +./7 + 2J7 + 4./7.

V

(3.33)

After a somewhat lengthy calculation, we find that G±4 _ 9 + 49 -

4./7 ±

2

(9 + 3./7)7 1/ 4 2.../2'

Thus, P = 9 + 4./7. After a mild calculation, p+

J p2 - 1 = 9 + 4.../7 + 2../3J 16 + 6./7

(3.34)

= 9 + 4.../7 + 2../3(3 +.../7)

= (2 + ../3)(3../3 + 2.../7)

~ (2+ v'3) ( v'3 ~ v'7)'

Using (3.32)-(3.34) in Theorem 3.1, we deduce that G

441

=

J2 +./7 + 2J7 + 4./7 J,.f3 +2 ./7(2 + v:

'3)1/6 1/3

x /

191 + 72./7 + A

2

+

189 + 72./7 + A

2

}

'

(3.35)

where

A: = J(l92 + n.J7)(189 + 72.J7) =

6J2016 + 762.J7 =

.fi(3 + ./7)2.

6· 7 1/ 4

Using this calculation in (3.35), we see from (3.35) and (3.31) that it remains to prove that

.f{(3 + ./7)2

191 + 72./7 + 6· 7 1/ 4 2

+

189 + 72./7 + 6· 7 1/ 4 2

!f(3 + ./7)2

214

Ramanujan's Notebooks, Part V

1 and d2 < o. Let K j = Q(../di), i = 1,2. By (4.7), lim(s - 1)~K, (s)

5 ..... 1

= Ld, (1),

i

=

1,2.

Then, by (4.5) and (4.6), L (1) = 2h)log€) dl

.,fd;

(4.14)

220

Ramanujan's Notebooks, Part V

and

(4.15) where h j is the class number of K j , i = 1,2,1'1 is the fundamental unit of K 1 , and W2 is the number of roots of unity in K 2 • Thus, setting s = 1 in (4.13) and using (4.8), we deduce that Ld, (l)L d2 (1)

= -

4n /QT " L

Wv

Idl

X(A)

(-! loga + log 11/(z)1 2 ).

(4.16)

AECK

Thus, setting

(4.17) where z = (b + Q)/a, with [a, b + Q] E A~I, we conclude from (4.14)-(4.17) that, for X nonprincipal (Siegel [1, p. 72]),

or E~hlh2lW2

=

n

F(A)~x(A).

(4.18)

AECK

We remark that (4.18) was utilized by K. G. Ramanathan [1], [3], [4], [5], [7] to calculate class invariants, values of the Rogers-Ramanujan continued fraction, and certain other invariants of Ramanujan. We next prove the three primary theorems that we need to calculate Ramanujan 's class invariants G m when K = Q(.J-m) has class number 8. Let r = .J-m. Then, by (1.1), (1.3), and (4.3), it is easily seen that 1/ «r + 1)/2) = il4Gm. 1/(r)

(4.19)

Equalities (4.18) and (4.19) are the key ingredients for deriving formulas that will enable us to calculate Gm. We consider three different genus structures, and the first two theorems that we prove can be utilized to determine G m for m = 65, 69,77, 141, 145,205,213,265,301,445, and 505. For m = 217, 553, the genus structure is of a third type. In each case, K = Q(.J-m) has class number 8, and the number of genera equals 4. Thus, each genus contains exactly two ideal classes. Also note that A and A ~I are clearly in the same genus. Throughout the next two sections, for simplicity, we use the notation for a primitive ideal to denote the ideal class containing it; this abuse of notation should not cause difficulty.

Theorem 4.1. Let m == 1 (mod 4), where m is a positive squarefree integer with prime divisor p. Let K = Q(.J-m) be an imaginary quadratic field such that each genus contains exactly two ideal classes and such that the principal genus

34. Class Invariants

221

Go contains the classes [l, 0] and [2p, p + 0]. Let G I be a nonprincipal genus containing the two classes [2, 1 + 0] and [p, 0]. Then

where h, h I> and h2 are the class numbers of K , Q(.J£r.), and Q(.Jd2) , respectively, w and W2 are the numbers of roots of unity in K and Q(.Jd2), respectively, lO I is the fundamental unit in Q( J£r.), and the product is over all characters X (with X(G I ) = -1), associated with the decomposition d = d l d 2, and therefore dl> d 2, hi, h2, W2, and lOl are dependent on x.

Proof. Each of the ideals [1,0], [2p, p+ 0], [2, 1+0], and [p, 0] is ambiguous. If 21 E A is anyone of these ideals, then 21 "-' 21- 1, A = A -I, and 21 E A -I. For any ideal class B ¢ Go U G I , it is not difficult to see that (Ramanathan [5, p.77])

L

X(B)=O,

x(G\)=-1

which implies that

n

F(B)-x(B) = 1,

x(G\)=-1

where F(B) is defined by (4.17). Therefore, by (4.18),

n

F(A)-x(A) =

n

(4.20) since the number of genus characters equals h/2, and so the number of genus characters with X(G I ) = -1 is h/4. LetAo = [1, 0], A~ = [2p, p+O], Al = [2, 1 +0], and A; = [p, 0]. Then, by (4.20),

n

wh\h2lW2 _ (F(At)/ F(Ao»)h/4

E)

-

x(G\)=-)

I

(4.21)

,

F(Ao)/ F(A I )

By (4.17) and (4.19), F(Ad

F(Ao)

Let

=

TJ 2 (¥)/-J2 TJ 2 (0)

= G2 •

(4.22)

m

or = O/p = J-m/p2. Again, by (4.17) and (4.19), F(A~) = F(AD

TJ 2(¥:-)/..!2P TJ 2(Q) p

=

TJ 2(¥)/-J2 TJ2(0')

The theorem now follows from (4.21)-(4.23).

=

G2 m/p2.

(4.23)

222

Ramanujan's Notebooks, Part V

Theorem 4.2. Let m == 1 (mod 4), where m is a positive squarefree integer with prime divisor p. Let K = Q(J-m) be an imaginary quadratic field such that each genus contains exactly two ideal classes and such that the principal genus Go contains the classes [1, Q) and [p, Q). LetG I be a nonprincipal genus containing the two classes [2,1 + Q] and [2p, p + Q). Then

(Gm Gm/p2 )h/2

-

-

IT

wh 1h2 /W2 EI '

x(G1)=-1

where h, hI. and h2 are the class numbers of K, Q(Jdi"), and Q(.Jil2), respectively, w and W2 are the numbers ofroots ofunity in K and Q(.Jil2), respectively, 101 is the fundamental unit in Q(Jdi"), and the product is over all characters X (with X (G I ) = -1), associated with the decomposition d = d l d2, and therefore d" d2, hi, h 2, W2, and 101 are dependent on X. The proof of Theorem 4.2 is analogous to that for Theorem 4.1, and so we omit it. We say that m is of the first kind or second kind according as it satisfies the conditions of Theorem 4.1 or Theorem 4.2, respectively. It is not difficult to show that [1, Q), [2, 1 + Q), [p, Q), and [2p, p + Q) are representatives of different ideal classes (Mollin and Zhang [1]). Theorems 4.1 and 4.2 need to be combined with three modular equations of Ramanujan (Part III [3, pp. 231, 282, 315]) in order to calculate Ramanujan's class invariants. We have already employed Lemma 4.3 in our proof of Theorem 3.1. Lemma 4.3 (Modular Equation of Degree 3). Let P = {16a{3(1 - a)(1 - {3)}1/8

Then

Q+

and

Q = ( {3(1 - {3»)1/4 a(1 - a)

~ + 2../2 (p - ~) = o.

Lemma 4.4 (Modular Equation of Degree 5). Let P = {16a{3(1 - a)(1 - {3)}1/12

Then Q

+

Q = ( {3(1 - {3»)1/8 a(1 - a)

and

~ + 2 (p - ~) =

O.

Lemma 4.5 (Modular Equation of Degree 7). Let P = {16a{3(1 - a)(1 - {3)}1/8

and

Q = ( {3(1 - {3») 1/6 a(1- a)

34. Class Invariants

223

Then Q

+ ~ + 7 = 2h

(p + ~).

Let q = exp( -rc / .j7i). Since G n = G 1/ n (Ramanujan [3], [10, p. 23]), by (1.6), G n = {4a(1 - a)}-1/24. If f3 has degree p over a, then G n / p 2 = G p 2/n = {4f3(1 - f3)}-1/24. In summary, we can express the equalities of Lemmas 4.3-4.5 in terms of G n and Gn / p 2, p = 3,5, 7, respectively, by employing the formulas G n = {4a(l - a)}-1/24

(4.24)

and

The genus structures for Q(J-217) and Q(J-553) are different from those of the eleven imaginary quadratic fields to which either Theorem 4.1 or Theorem 4.2 applies, and so G 217 and G5 53 must be calculated by another means.

Lemma 4.6. Let m denote a positive integer with 71m. Let r = J-m/7 and Q = (G m /G m / 49)4. Then

(4.25)

Proof. With q = exp( -rc.,(iii/7) , it follows from (4.3) that (

)2

17(r)l1(lil) 17(7r)17 Clil)

f2(_q2)f2(q)

- q3/2 P( -q 14)P(q 7) .

(4.26)

Next, by an entry from Ramanujan's second notebook (Part IV [4, p. 209, Entry 55]), f2( _q)f2( _q2) ......,-'---c---=--=.,-_--=--__ q3/2 P(_q7)P(_q14)

- 8

+ 49

f2(_q2)f2(_q7) ql/2P(_q)P(_q14)

q3/2 f2( _q 7)f2( _q14)

- 8

P(_q)P(_q2) ql/2 f2(_q)f2(_qI4) P(_q2)P(_q7)

=

f6( _q2)f6( _q7)

~~;_=_...:....:...._::_:_--=-_':_:_

q3/2 f6(_q)f6(_qI4)

q3/2 f6(_q)f6(_qI4)

+ -=----7----'-:-'::':"'::':--'----=-'-':" f6(_q2)f6(_q7)·

Multiplying both sides by q3/2 and then replacing q by -q, we find that

224

Ramanujan's Notebooks, Part V

Recall that q = exp( -1f ./iii/7) and recall that G m / 49 is then given by (1.3). Thus, G m = 2- I / 4q-1/24(_q7; qI4)oo. Hence, (

Gm)2 (_q7; qI4)~ G m/ 49 = ql/2(_q; q2)&:, (q2;

f2(-q2)f2(q7)

q2)~(_q1; _q1)~

ql/2(_q; _q)&:,(qI4; qI4)&:, - ql/2 P(q)P(_qI4)· (4.28)

Dividing (4.27) by q3/2 and substituting (4.26) and (4.28) into the resulting equality, we deduce (4.25) to complete the proof. Theorem 4.7. Let m be a squarefree positive integer with 71m and m == 1(mod 4). Let K = Q(.J -m) be an imaginary quadratic field such that each genus contains exactly two classes and such that the principal genus Go comprises [1, Q] and [2, 1 + Q], while [7, Q] and [14,7 + Q]form a nonprincipal genus G I . Then

1v'·, '7

1 ( 7](r)7] (I.:}!) ) }h/2 n(7r)n ., (h2+1)

n

Elwh ,h2/W 2,

(4.29)

x(G,)=-1

where h, hI, and h2 are the class numbers of K, Q(~), and Q(.../di), respectively, wand W2 are the numbers of roots of unity in K and Q(.../di), respectively, EI is the fundamental unit in Q(~), and the product is over all characters X (with X (G 1) = -1), associated with the decomposition d = d 1d2, and therefore dJ, d 2, hJ, h2, W2, and EI are dependent on X.

Proof. Let Ao = [1, Q], A~ = [2, 1 + Q], Al = [7, Q], and A; = [14,7 + Q]. Then by the same reasoning that we used in the proof of Theorem 4.1, F ( F(Ad (AD)h/4

F(Ao)F(A~)

=

n E~h,h2/W2

(4.30)

x(GIl=-1

By (4.17), F(Ao) = 7]2(Q) = 7]2(7r),

/../2 = 7]2 CTiI) /../2, 7]2 (~) /../7 = 7]2 (r) /../7,

F(A~) = 7]2 (gil) F(A I ) =

and F(AD

=

7]2 (OI~7)

/.J14 = 7]2 (Ttl) /.J14.

Substituting these values into (4.30) and recalling that the number of genus characters X with X(G 1) = -1 is equal to h/4, we deduce (4.29) to complete the proof. The class numbers cited below for Idl < 500 can be found in tables in the texts by Z. I. Borevich and I. R. Shafarevich [1, pp. 422-426], H. Cohen [I, pp.

34. Class Invariants

225

503-509], and for 0 < d < 10,000 in the book by D. A. Buell [1, pp. 224-234]. Lists of fundamental units can be found in the book by Borevich and Shafarevich [1] (ford::: 101), the book by M. Pohst and H. Zassenhaus [1, pp. 432-435] (up to d ::: 299), and the tables of R. Kortum and G. McNiel [1] (up to d = 10,000). In Cohen's book [1, pp. 262-274], there is a table providing the ideal class structure for Q(H), d ::: 97 and for Q(.Jti) , d ::: 97.

5. Class Invariants Via Kronecker's Limit Formula Theorem 5.1.

G~= (~+3r (~2+1r U9+~ +l+~r Proof. The following table summarizes the needed information about ideal classes and their characters. dl

d2

X G

c

X(Go) x(Gd X(G 2) X(G 3)

1 -260 XO Go

[1,0] [10,5 + 0]

1 1

1 1

-52 XI G I

[2,1 + 0] [5,0]

1 -1

-1

1

5

1

[3,1 + 0] 13 -20 X2 G 2 [3, -1 + 0]

1

-1 -1

+ 0] + 0]

1 -1

1 -1

65

-4

[6,1

X3 G 3 [6, -1

W2

hi h2

1

2

2

1

2

2

EI

0+1 2

v'i3 +

3

2

Note that 65 is of the first kind. Applying Theorem 4.1 with h = 8 and w = 2, we find that

= (0 + 1)2 (v'i3 + 3)2 ( ~)4 G\3/s 2 2

Let Q

Q=

(5.1)

= (G6S/G\3/s)3 and P = (G 6S G\3/S)-2. Then, by (5.1),

f'

(~2+ 1 (~+3 )'/2 = (~+2)'I'(Sm +18)'1'.

(5.2)

By Lemma 4.4, p_ 1 = (Q

+ Q-l) + J(Q + Q-l)2 + 16 4

.

(5.3)

226

Ramanujan's Notebooks, Part V

Now, by (5.2), (Q + Q-I)2 + 16 = Q2 + Q-2 + 18

= (J5 + = (5

2)(5.Ji3 + 18) +

+ ,J6s)2,

(J5 -

2)(5.Ji3 - 18) + 18 (5.4)

and, by (5.4),

(5.5) Thus, by (5.3)-(5.5), p-I

= ~J74 +

1OJ6s +

~(5 + J6s).

(5.6)

Thus, by (5.2) and (5.6), G 65 = QI/6 p-I/4

~

(./52+ I) 'I' ( ~+ 3) 'I' (lJ74+ 10,/65 + j(5 + ,/65») 'I'

Thus, it remains to show that

which is easily shown by a routine calculation. Theorem 5.2.

G

= (5 + v'23)1/12 (3v'3 + v'23)1/8

2

69.j2

()6+

3./3 4

+

)2+

3./3f' 4

Proof. We summarize the needed information in the following table. dl

d2

X G

1 -276 XO Go

c

X(Go) X(G 2 )

x(Gd

X(G 3 )

hi

h2

W2

t]

[1,0] [6,3 + 0]

1 1

1 1

[2,1 + 0] [3,0]

1 1

-1 -1

1

1

6

24+ 5v'23

[5,1 + 0]

1

1

4

25 + 3vW 2

92

-3

XI G I

69

-4

X2 G2 [5, -1 + 0]

1 -1

-1 1

[7,1 + 0] 12 -23 X3 G 3 [7,-1+0]

1 -1

1 -1

34. Class Invariants We apply Theorem 4.1 with h 69 ( G )

4

G 23/3

Q

= 8 and w = 2, as 69 is of the first kind. Thus,

= (24 + 5.J23)I/3

Let Q = (G 69 /G23/3)6 and P

227

(f£n) + 25

3",,69 2

1/2

(5.7)

= (G69G23/3)-3. By (5.7),

~ (24 + 5-'23)'" (25 +~,,169) '" = (5

v'23) (36v'3 + nJ23)1/2 = (5 + v'23) (3v'3 + v'23)3 /2 .

+ ../2

../2

\

2

(5.8)

By Lemma 4.3, p-I = _l_(Q

4../2

+ Q-I) + _l_J(Q + Q-I)2 + 32. 4../2

From (5.8), Q

+ Q-I = J

Q2

+ Q-2 + 2 =

J16(187 + 108v'3),

(5.9)

(5.10)

and, from (5.10),

Putting these calculations in (5.9), we find that p-I

=

!nJ187 +

108v'3 +

!n(9 +

6v'3).

(5.11)

By (5.8), G69

= QI/12 p-I/6 = ( 5 +~ )

1/12 (

3v'3;

v'23

) 1/8

p-I/6,

and thus, by (5.11), it remains to show that

(J6+ :./3 + J2 +:./3)' ~ :nJ187 + 108./3 + :n(9+6./3). This can be achieved by a straightforward computation. Theorem 5.3.

G" ~ (8+ 3v7)'!' ( ,(IT ; v7)

1/8

(rr:;;rr Y¥ ¥ )

~ 1/2

+V

Ramanujan's Notebooks, Part V

228

Proof. We compose the following table giving needed information about ideal classes and characters. dl

d2

1 -308 XO Go 28 -11

XI

X (Go) X (G2)

c

X G

GI

X(G I )

X (G 3 )

W2

1"1

[1,0] [14,7 + 0]

1 1

[2,1 + 0] [7,0]

1 -1 1 -1

1

1

2

8 + 3../7

1

1

4

9+..;n

1 1

77

-4

[3,1 + 0] X2 G 2 [3, -1 + 0]

1 -1 -1 1

44

-7

X3

[6,1 + 0] [6, -1 + 0]

1 1 -1 -1

G3

hi h2

2

We see from the table that 77 is of the first kind. Thus, by Theorem 4.1, since h = 8 and w = 2,

Q := (

Gn )4 =

G 11 J7

(8 + 3.J1)

(9

+

..;n) 1/2 =

2

(8 + 3.J1)

(.vTI + -J7) . 2

(S.12) If P = (GnG11j7)-3, then, from Lemma 4.S,

P

-I

=

Q + Q-I + 7 +

J(Q + Q-I + 7)2 M

32

.

(S.13)

Now

Q + Q-I + 7

= 8JI1 + 28 = 2../2(252 + 7../2).

Using this in (S.13), we find that p- I =

~ (252+ 7v'2+ J182 + S6JI1) .

By (S.12),

Gn

= Q'I'r'I' = (8 + 3J'i)'1' ( ./IT;

J'i)

(S.14)

'I' r'I',

and thus by (S.14) it remains to show that

(J6+;rr +J2+;rr)'

=

1(2V'n+7h+J182+56./IT),

which is readily shown by a straightforward calculation.

34. Class Invariants Theorem 5.4. r;:;

r;;::; 1/8

G I41 =(4v3+v47)

(7 ./47) + '" v2

229

(f¥ f¥i)

Ijl2

18 + 9..;3 4

+

14 + 9..;3 1/2 4

Proof. We record the necessary information in the following table:

dl

1

d2

X

G

-564 XO Go

X (Go) X (G 2 )

c

X(G I ) hI X (G 3 )

h2 W2

EI

[1,0] [6,3 + 0]

1 1

1 1

1 -1

-1 1

1

1

6

1

1

4 95 + 8.JT4f

188

-3

XI G I

[2,1 + 0] [3,0]

141

-4

X2 G 2

[5,2+ 0] [5, -2 + 0]

1 1

-1 -1

12

[10,3 + 0] -47 X3 G 3 [10, -3 + 0]

1 -1

-1

48 + 7v"47

I

We see that 141 is again of the first kind. Applying Theorem 4.1, we find that, since h = 8 and w = 2,

(5.16)

p_1 = (Q

+ Q-I) + J(Q + Q-I)2 + 32. 4,.f'i

From the last representation of Q in (5.16),

(5.17)

230

Rarnanujan's Notebooks, Part V

and so

Using these calculations in (5.17), we deduce that p-I =

!n!7855 + 4536../3 + ~(7 + 4../3).

Hence, by (5.16) and (5.18), G,,,

C

~ Q'I" P 'I' ~ +~r" (4v'3 X

(v'21V/

+

(5.18)

A'l)'"

7855 + 4536../3 +

9)

v'2 (7 +

4../3)

1/6

It thus remains to show that

which is a straightforward, albeit laborious, task. Theorem 5.5.

G I45

~ + 2) 1/4 = (,,5

(m

2

+ 5)

1/4(~ 17 + v'T45 .~1/2 8 + 9 + v'T45 8

Proof. We compose the following table: dl

d2

X G

c

1

-580 XO Go

[1, Q] [5, Q]

29

-20 XI G I

[2,1 + Q] [10,5 + Q]

5

[7,3 + Q] -116 X2 G2 [7,-3+Q] [11,3 + Q]

145

-4

X3 G 3 [11, -3 + Q]

X(Go) X(G 2 ) 1 1

x(Gd

X(G 3 )

hi h2

W2

tl

1 1

1 -1 1 -1

1

2

2

m+5

1 -1

-1 1

1

6

2

--

1 -1

1 -1

2

v's + 2

1

34. Class Invariants

231

Thus, 145 is of the second kind. Thus, by Theorem 4.2, since h = 8 and w = 2, (G I45 G29/5)

4

=

.J29+5 ( .J29+5 2 )2(../5+1 2 )6 = (--2-- )2 (../5 + 2) . 2

Hence,

(.J29 + 5)

P _I := (GI45G29/5) 2 =

2

(vIr 5 + 2).

(5.19)

By Lemma 4.4, with Q = (G I45/ G29/5)3,

Q = p-I - P

+ J(P-I

- P)2 - 1.

(5.20)

By (5.19), we readily find that p-l _ P = 259 + 5../5,

and so, by (5.20),

Q = 259 + 5../5 + )240 + 20vi145. Thus, by (5.19) and (5.21),

G,,, ~ r"'Q'" ~ (~ + 5

r\

(5.21)

v'5 + 2) 'I'

x (259 + 5../5 + )240 + 20vi145 Hence, it remains to show that

2../29

+ + + 5v'5

J240

20./145

y/6

t

~ ()17 +;145 +)9 +

45

r,

which is readily shown. Theorem 5.6.

G

205

= (../5 + 1) (3v'5 +

2

2

v'4T)

Proof. We record the following table:

1/4

(t ./4l )./4l +8

+

8

1)

.

232

dl

Ramanujan's Notebooks, Part V

X(Go) X(G I ) hi hz Wz X(Gz) X(G 3 )

c

X G

dz

1

-820 Xo Go

[1, Q] [5, Q]

5

-164 XI GI

[2,1 + Q] [10,5+ Q]

1 -1 1 -1

1

8

2

[11,2+Q]

1 -1 -1 1

2

1

4

1 1

205

-4

41

[13,4+ Q] -20 X3 G3 [13, -4+ Q]

W

XZ G z [11, -2 + QJ

1 -1

1:1

1 1

J5+ 1 2 43 + 3v'205 2

1 -1

Note that 205 is of the second kind. Applying Theorem 4.2 with h = 2, we deduce that

=

8 and

(S.22)

Letting Q = (G Z05 /G41/5)3, we deduce from Lemma 4.4 that Q = (p-I - P)

+ J(P-I

From (5.22), P

Thus, from (S.23),

Q=

_I

-P=

- P)Z - 1.

(S.23)

4S + 27.J4f .

~ ( 45 + 7v'4I + J4030 + 630v'4I) .

(S.24)

If follows from (S.22) and (5.24) that G205

= p-I/4QI/6 = ( J52+ 1) (3J5 +2 .J4f) 1/4 X

(! (45 + 7v'4l + J4030 + 630v'4l))

1/6

It thus remains to show that

1( 45+ 1,/41 + j 4030 +630,/41) ~

(/' \,/41 +

J~ -

I )'

34. Class Invariants

233

(t +./4f J./4f -1)' ~ ./4f +3 Jl7+ 3./4f.

This is more readily accomplished if we first note that

8

Theorem 5.7.

+

8

4

+

8

G", ~ (5./3; J7I) '" (59 +~J7I) (J21+;2./3 +JIH;2./3)'" 'I"

x

Proof. We have the following table:

dl

d2

X G

X(Go) X(G 2 )

c

X(G I) hI X(G 3 )

E'I

h2 w2

[1, Q] [6,3, Q]

1 1

1 1

[2,1 + Q] [3, Q]

1 1

-1 -1

1 1 6 3480 + 413.J7f

[7,2+ Q] 213 -4 X2 G2 [7, -2 + Q]

1 -1

-1 1

1 1 4

[14,5+ Q] 12 -71 X3 G3 [14, -5 + Q]

1 -1

1 -1

1 -852 XO Go

284 -3 XI G I

73 + 5,J2!3 2

Observe that 213 is of the first kind. Applying Theorem 4.1 with h = 8 and w = 2, we find that Q2/3 :=

(G2I3) 4 = (3480 + 413-J71)1/3 G 71 / 3

~ so that

Q

~ (59 +~J7I)

(~)1/2 73 + 5-v213 2

(59 :;J7I) (5./3; J7I) , '/3

e./3 ; J7I)'" ~ e9+~J7I) (180./3+

37v'2j)'''. (5.25)

234

Rarnanujan's Notebooks, Part V

Let P = (G213G71/3)-3. Then, by Lemma 4.3, p-I = _1_ (Q

4.J2

+ Q-I) + ./(Q + Q-1)2 + 32).

(5.26)

By (5.25) and moderate calculations,

and IM(Q

4",2

+ Q-I) =

IM./ Q2 + Q-2 + 2 =

4",2

Jt(135619 + 78300.J3).

Thus, by (5.26), p-l =

J

t(135619 + 78300.J3) +

~(87 + 50.J3).

(5.27)

Thus, by (5.25) and (5.27),

G213

= QI/12 p-I/6 =

( 5.;3;.J7f )

X

(

1/8 (

59 +~.J7f

)1/12

3 )1/6 . Jt(135619 + 78300.J3) + .J2(87 + 50.J3)

Hence, it remains to show that

Jt(135619 + 78300.J3) + ~(87 + 50.J3) =

(/2[ + [2./3 + jt9 + [2./3)' 2

which is accomplished by a direct calculation. Theorem 5.S.

Proof. The following table is easily verified:

2

'

34. Class Invariants

dl

d2

X(Go) X(G 2 )

c

X G

X(GI) X(G 3 )

hi

-1060 XO Go

[1,0] [10,5,0]

5

-212 XI GI

[2,1 + 0] [5,0]

1 -1 -1 1

1

6

2

53

-20

X2 G2 [7,-1+0]

[7,1 + 0]

1 -1 1 -1

1

2

2

265

-4

X3 G3 [14, -1 + 0]

1 1

[14,1 + 0]

1 1

:= ( G 265 G 53 / 5

v'5 + 1 2

v's3 + 7 2

1 1 -1 -1

Note that 265 is of the first kind. Applying Theorem 4.1 with h we find that Q4/3

EI

h2 W2

1

235

= 8 and w = 2,

)4 (v'52+ 1)6 (v's32+ ~)2 , =

(5.28) Let P =

(G265G53/5)-2.

Then, by Lemma 4.4,

p_1 _ (Q + Q-I) +

-

J(Q + Q-I)2 + 16 4

By using (5.28) and the identity Q + Q-I = that p-I =

By (5.28) and (5.30),

Gun ~

Q'I'r'I'

J Q2 +

.

(5.29)

Q-2 + 2 in (5.29), we find

1M ) 6917 + 425../265 + ~ (85 + 5../265). 2v2 4

(5.30)

~ (,/5 +2)'1' (~+7f' X

1 / 1 ) ( 2../2 Y 6917 + 425.J265 + "4 (85 + 5.J265)

Hence, it remains to show that 1",)6917 + 425../265 + ~(85 + 5../265) 2v2 4

~

(

.------". 2

89 + 5../265 ----+ 8

81

+ 5../265 8

1/4

.

236

Ramanujan's Notebooks, Part V

which is easy to establish. Theorem 5.9.

Proof. We compose the following table: dl

dz

X (Go) X (G 2 )

c

X G

1 -1204 Xo Go

x(Gd hi h z Wz X (G 3 )

[1, Q] [14,7, Q]

1 1

1 1

[2, 1 + Q] [7, Q]

1 -1

-1 1

1 1 2

8 + 3../7

1 1 4

22745 + 1311,J301 2

28

-43

301

-4

[5,2+ Q] XZ G 2 [5, -2 + Q]

1 1

-1 -1

172

-7

[10,3+Q] X3 G 3 [10, -3 + Q]

1 -1

1 -1

XI

GI

10 1

Thus, 301 is of the first kind. Applying Theorem 4.1 with h we find that

Q := ( G

301)4 =

G~n

(8 + 3../7) (22745 + 1311.J361) liZ 2

r.:; (23,J43 2+ = (8 + 3v7)

Let P

=

= 8 and W = 2,

57../7) .

(5.31)

(G301G 43/7 )-3. Then, by Lemma 4.5 and (5.31), p-I

1 = _(Q +

4.J2

= ~(301

1 -J(Q +

Q-I +7) +

4.J2

Q-I +7)2 - 32

~------

+ 46v43) +

~j7(25941 + 3956v43).

Therefore, by (5.31) and (5.32), G301

= QI /8p-I/6 = (8 +

1/8

3../7)1/8

(

23,J43: 57../7 )

(5.32)

34. Class Invariants

x

237

(~(301 + 46.J43) + ~J7(25941 + 3956.J43)y/6.

It remains to show that

~(301 + 46J43) + ~J7(25941 + 3956J43) =

(J46+ 7.[43 + J42+7.[43)' 4

4

'

which is a routine task. Theorem 5.10.

~5 + 2) 1/2 (J44S+21)1/4 (J13+~ 2 8 + J5+~) 8 .

G 445 = (v.J

Proof. We form the following table: d2

dl

X G

X(Go) X(GI) X(G2) X(G 3) hi h2 W2

c

1

-1780 XO Go

[1, OJ [5, OJ

5

-356 XI G I

1 1

1

[2,1 + OJ [10, S + OJ

1 -1

-1 1

1 12 2

[13,6+ OJ

1 1

-1 -1

4

[19,7+ OJ

1 -1

1 -1

445

-4

X2 G2 [13, -6 + OJ

89

-20

X3 G 3 [19, -7 + OJ

10 1

I

1

4

J5+1 2 J445+21 2

Thus, 445 is of the second kind. Applying Theorem 4.2 with h = 8 and w = 2, we deduce that (G G )4 _ (J5 + 1)12 P -2._ .445 89/5 2

(J445

+21)2 2'

so that p-I = (9

+ 4V5) ( ~ + 21) .

(5.33)

Let Q = (G445/G89/S)3. Then, by Lemma 4.4 and (5.33),

Q = (p-I - P) + V(P-I - P)2 - 1 = 189 + 20J89 + J71320 + 7560J89. (5.34)

238

Ramanujan's Notebooks, Part V

Therefore, by (5.33) and (5.34), G445

= p-I/4QI/6 = (9 + 4J5)1/4 ( ~ + 21 ) x

1/4

(189 + 20v's9 + J71320 + 7560v's9Y/6

It thus remains to show that

189+ 20../89+ /71320+ 7560../89

~ (/t3 +8../89 +)5 +~)'

By first squaring the binomial on the right side and then cubing the resulting expression, we can easily verify the desired equality. Theorem 5.11. G'05

~ (,/5 + 2)'1' ( v's2+ 1) 'I' (v'iOl + 10)'1' x

(

113 + 5./505 8

+

105 + 5./505 8

IP

Proof. We compose the following table: dl

X G

d2

X(Go) x(Gd X(G 2 ) X(G 3 )

c

1

-2020 Xo Go

[I, Q] [5, Q]

1 1

5

-404 XI GI

[2,1 + Q] [lO,5+Q]

1

-20

X2 G2 [11, -1 + Q]

505

-4

X3 G3 [22, -1 + Q]

[22,1 + Q]

-1

1

1 -1

1

1 14 2 1

2

Hence, 505 is of the second kind. Applying Theorem 4.2 with h we find that P

-2

4

:= (G505GIOI/5) =

(

W2

£1

1

1 -1 -1

h2

1

1 -1 -1

[l1,I+Q]

101

hi

v's + 1 ) -2-

2

J5+ 1 2

.v'1Of +

10

= 8 and w = 2,

14

(Fol + 10) 2 ,

34. Class Invariants

239

so that

p-'

~ ( ./52 + 1)' (v'W\ + to) ~ (./5 + 2) (./52 + 1)' (v'W\ + to) (.J5 + 2)

=

(7

+

3./5) (V'101 +

2

10). (5.35)

Let Q = (G505/GIOI/5)3. Then, by Lemma 4.4 and (5.35),

Q = (p-I - P) =

+ J(P-I

- P)2 - 1

(13o.J5 + 29v'101) +

J

169440 + 7540.J505.

(5.36)

Therefore, by (5.35) and (5.36),

x

(130.J5 + 290(1) +

J

169440 + 7540.J505) 1/6

Thus, it remains to show that

(130.J5 + 29.JiOi) +

J169440 + 7540.J505 ,------" 3

105+5.j505 8 which is straightforward. Theorem 5.12. G 217

= (

/¥ 11 + 4.J7 2

9 + 4.J7 +~ 2

1/2

/¥ f¥)1/2

(16 + 5.J7 4

+

12 + 5.J7 4

Proof. We set up a table to summarize some information that we need.

240

Ramanujan's Notebooks, Part V

X G

dz

dl

X(G o) X(G 2 )

C

X(G I ) X(G 3)

hI

hz

Wz

EI

1 -868 XO Go

[1, 0] [2,1 + 0]

1 1

1 1

Xl G I

[7,0] [14,7+ 0]

1 -1

-1 1

1 1 2

1 1 4 3844063 + 260952.J2i7

124 -7 217

[11,5 + 0] -4 X2 G z [11, -5 + 0]

1 -1

1 -1

28

[13,2+ 0] -31 X3 G 3 [13, -2 + 0]

1

-1

1

-1

1520 + 273&

It is clear that 1QJ(v' -217) satisfies the conditions of Theorem 4.7. Thus, since h = 8 and w = 2, we deduce that

~7 ( 1/(7r)1/--T1/(r)1/ ~~~ ))4 = (1520 + 273.J3i)(3844063 + 260952v'217) 1/2 = (1520 + 273.J3i)(524.J7 + 249.J3i), so that

where E = (1520 + 273.J3i)1/2(524.J7 + 249.J3i) 1/2.

(5.37)

It follows from (4.25) that Q3/2

where Q

=

8Q-I/2 _ Q3/2 = 7(E - E- 1),

(5.38)

(G217/G31f7)4. By an elementary calculation,

(E - E-I)2

Let x

+ 8QI/2 _

= E2 + E- 2 - 2 = 4(1053643 + 398240.J7) = 4(27 + 1O.J7)2(367 + 140.J7).

= QI/2 - Q-I/2. Then (5.38) can be recast in the form x3

+ llx = =

14(27 + 1O.J7)/367 + 140.J7 (367 + 140.J7)/367 + 140.J7 + 11 /367

+ 140.J7.

It is now obvious that

x = /367 + 140.J7.

(5.39)

34. Class Invariants

Solving (5.39) for Q1/2, we find that QI/2 =

241

4

! ()367 + 140.J7 + )371 + 140

r

=

! ()367 + 140.J7 + (14 + 5v7»)

=

(J 16 +4S./7 + J'2 +4S./7

(5.40)

Now let P = (G2\7G31/1)-3. Using Lemma 4.5 and (5.39), we deduce that I

P + p-l = _ (Q + Q-I + 7) 2../2 1",(X 2 + 9) =

=

2v2

1,r,;-(376 + 140v7). 2v2

Solving for p-I, we find that p-l = 94+35v7 +

../2 =

17409 + 6580.J"7 2

(J'1+24./7 +JH;./7)'

(S.41)

Thus, from (5.40) and (5.41),

G217 = p-1"QI/8 =

(JIl +24./7 +J9 +:-n-) x

J

(J' 6 +.S./7 + 12 +.S./7 )''' .

which completes the proof. Theorem 5.13. G~3= (

100 + 11m

4

II'

,----~

+

96 + 11m

4

:1/2

242

Ramanujan's Notebooks, Part V

Proof. We set up the following table to summarize the information that we need. dl

X G

d2

1 -2212 XO Go

X(G o) X(G 2 )

C

[1, Q] [2,1 + Q]

1 -1 -1 1

[17,5+Q]

1 -1 1 -1

[19,6 + Q]

1 1 -1 -1

-79

553

-4

X2 G 2 [17, -5 + Q]

316

-7

X3 G 3 [19, -6+ Q]

GI

hI h2

W2

EI

1 5

2

8 + 3v7

1 1

[7, Q] [14,7+Q]

28

XI

1 1

X(G I ) X(G 3)

1 1 4

624635 837407 + 26 562 217 704,J553

It is clear that Ql(./ -553) satisfies the hypotheses of Theorem 4.7. Thus, since

h

= 8 and w = 2, 1 72

(

17(')17 17(7r)17

= (8

('-¥)

C't)

4 )

+ 3./7)5(624, 635, 837, 407 + 26, 562, 217, 704./553)1/2,

so that (5.42) where

(' = (514,088 + 194,307./7)1/2(211,227./7 + 62,876./79)1/2.

(5.43)

Then an elementary calculation gives

«(' _ ",-1)2 = ('2 + ",-2 _

2

= 4(143,650,096,411 + 16, 161,898,544./79) = (391 + 44./79)(19170 + 2156./79)2.

(5.44)

By Lemma 4.6 and (5.42)-(5.44), with Q = (G 553 /G 79j7 )4, Q3/2

+ 8QI/2 _

8Q-I/2 _ Q-3/2

=

7«(' _

",-I)

= 7/391 + 44./79(19170 + 2156./79). If x

=

QI/2 - Q-I/2,

x3

then the foregoing equality may be written in the form

+ llx = 7/391 + 44./79 (72(391 + 44./79) + 11) ,

34. Class Invariants

243

from which it is obvious that x

=

QI/2 - Q-1 / 2

= 7/391 + 4459.

(5.45)

Solving for Q 1/2, we readily find that QI/2

=

t (7/391 + 4459 + (98 + 1159») 100 + 11m 4

_ ( Now let P

=

2..fi(P

(G553G79/7)-3.

+ p-l) =

~----..

Q

+

96 + 11m 4

2

(5.46)

Then, by Lemma 4.5 and (5.45),

+ Q-l + 7 = x 2 + 9 = 19168 + 215659.

Solving for p-I, we find that p-I =

~ (4792 + 53959 + /45,914,421 + 5,165,77659)

= (

143+;6./79 +

+ 16m

141

(5.47)

2

Thus, by (5.46) and (5.47),

G 553

= QI / 8 p-l/6 =

(

100 +411m

+ J96 + 411v~ _ ) 1/2

143+ 16m 2

~------..

+

141

+ 16m

1/2

2

and the proof is complete.

6. Class Invariants Via Modular Equations In this section we establish six of Ramanujan's class invariants by using tools well

known to Ramanujan, in particular, modular equations. Second Proof of Theorem 5.1. From (1.1) and (4.3) it is easily seen that f(-q) f(_q2)

=

X(-q)·

(6.1)

244

Ramanujan's Notebooks, Part V

U sing this equality, we rewrite two of Ramanujan's eta-function identities in terms of X. Thus (Part IV [4, pp. 206, 211]) f(-q)f(-q2) q3/2 f(_q13)f(-q26)

+ 13 q3 / 2 f(_qI3)f(-q26)

= (q_I/2 X (-q13»)3

f(-q)f(-q2)

_4(q-I/2X(-qI3») _4(ql/2 X(-q) ) X(-q) X(_qI3)

X(-q)

+ (ql/2

X(-q) X(_q13)

)3

(6.2)

and

(6.3) Replace q by -q in (6.2) and then set q

= exp( -T( .J5/13). If

f(e- x ,J57T3) f( _e- 2x ,J57T3)

A .- e(3rr/2l,J57T3 '::"""'--..........,=-''"---=-.f(e- rr .;65)f(-e- 2rr .;65)

(6.4)

and ( -rr.;65)

B.- (rr/2l,J57T3....::.X.::.......-e_=~ .- e X (e-rr,J57T3) ,

(6.5)

then (6.2) can be recast in the form

A - 13A- 1

=

B3

+ 4B -

4B- 1 - B- 3.

(6.6)

Next, replace q by -q in (6.3) and then set q = exp( -T( .J13/5). If

f

(e- rr v'IT75) f (_e- 2rr v'IT75)

A' .- e(rr/2lv'IT75 _ _ _=-_ _ _-=:-.f (e- x .;65) f ( _e-2x .;65)

(6.7)

and

(6.8) then (6.3) takes the shape (6.9)

We shall prove that B=B'

(6.10)

and

Now, G n = 2-1/4errv'n/24x(e-rrv'ii) and G 1/ n (1.3). Since G n = G 1/ n , we find that

=

2-1/4err/(24v'nlx(e-rr/v'n),

by

(6.11)

34. Class Invariants

245

(This could also be proved by using (6.1) along with the transformation formula for f.) In particular, if n = 5/13, (6.11) yields the equality (6.12) The aforementioned transformation formula for f (-q) is given by (part ill [3, p. 43, Entry 27(iii))) e- a /12 a l/4f(_e- 2a ) = e-b/12bl/4f(-e-2b),

where a, b > 0 with ab deduce from (6.13) that

(6.13)

= 1f2. If a = 1fJ5/13, so that b = 7rJI3/5, then we

f(-e-7:rr../5Tf'J) = (13/5)1/4 e -21Z'/(3.J65) f(-e-7:rr,,/ITj5).

(6.14)

First, from (6.5) and (6.12), B

=

e(1Z'/2)../5Tf'J X (e-1Z'.J65) e1Z'/(3.J65) X (e-1Z'../I375)

= e(1Z'/6)./f3T5

X (e-1Z'.J65) X (e-1Z'-tTI75)

= B',

by (6.8). Thus, the first equality of (6.10) has been demonstrated. Second, by (6.4), (6.1) with q = -exp(-1fJ5/13), (6.12), (6.14), and lastly (6.1) with q = exp(-1fJ13/5),

by (6.7). Thus, the second equality of (6.10) has been established. Employing (6.10) in (6.9), we find that

fj;A - J65A- 1 = B3 - B- 3 = (B - B- I )3

+ 3(B -

B- 1).

Dividing both sides by u := B - B- 1 (I- 0), we find that

v'65

8(u 2 +7) = u2 +3. Solving for u 2 , we find that u 2 =

(v'65 -

B - B- 1 =

1) /2. Thus, since, clearly, B > 1,

J~-

1.

246

Ramanujan's Notebooks, Part V

Now solving for B, we find that B=

J~-1 8 + J~+7 8'

(6.15)

where in solving the quadratic equation we took the plus sign since B > o. If q = exp(-n,Ji375), then q5 = exp(-Jr~). Hence, from (1.3) and (6.5), we readily see that B = G 65 /GI3/5. Furthermore, from (1.6), GI3/5 = {4a(1- a)}-1/24. Hence, if fJ has degree 5 over a, then G65 = {4fJ(l- fJ)}-1/24. We now employ Lemma 4.4, where it is to be noted that P = (G65GI3/5)-2 and Q = B- 3 = (G 65 /GI3/5)-3. We already know Q from (6.15). To determine P from Lemma 4.4, we first calculate

Q + Q-I = B3 + B- 3 = (B + B-1){(B + B-l)2 - 3}

=J~+7 (~+7 -3) =J74+1Ov65.(6.16) 2 2

Thus, using (6.16) in Lemma 4.4 and solving for p-l, we find that p-l =

~ (J74 + 1Ov65 + J90 + 1Ov65),

(6.17)

since P > O. Hence, by (6.15) and (6.17),

G,,=B'I'r'I'= (J~+7 +J~-lr x

(~ (J74 + 1Ov65 + J90 + lOKs)) 1/4

(6.18)

We must show that (6.18) can be transformed into the form of Theorem 5.1. First,

!

(J74 + 1Ov65 + J90 + 1Ov65)

1/2

I + v65)(1 + v65) + 5 + v65)1/2 = =! ( ,,(9

J9+~ 8 +

Jl +-/65

8' (6.19)

Second,

(J~+7 +J~-lr =i(HM+J58+6M) =

~ (3 + v65 + 3~ +

v'D)

34. Class Invariants

= ( ~ + 3) (v's2+'-!').

247

(6.20)

Putting (6.19) and (6.20) in (6.18), we complete the proof. Before commencing our second proof of Theorem 5.2, we establish a general principle. Let p and r denote coprime, positive integers. Set q = exp( -7r -J P / r) and q' = exp( -7r...(jii), and let fJ have degree rover a. Then, by (1.6), G p/ r

= {4a(1 -

and

a)}-1/24

G pr

= {4fJ(1 -

fJ)}-1/24.

(6.21)

Furthermore, from (1.2) and (1.5), K(yT=a) = K(.jii)

fE,

(6.22)

V;:

and from the definition (1.4) of a modular equation,

r

K(yT=a) K(~) K(.jii) = K(.jp) .

(6.23)

If we solve (6.22) for r and substitute this in (6.23), we find that K(.jii) P K(yT=a)

=

K(~)

K(.jp)

.

From the last equality we conclude: If fJ has degree r over a, then fJ has degree p over 1 -- a.

(6.24)

Furthermore, from (2.5) and (6.22), ~

K(.jii)

f{)2(e->r../PTr)

rp2(e->r.rr7P) = K(yT=a) =

Vp.

(6.25)

Second Proof of Theorem S.2. We need two of Ramanujan 's modular equations of degree 23 (Part III [3, p. 411, Entry 15(i), (ii))). If fJ has degree 23 over a, then (afJ)1/8

+ {(1 -

a)(1 - fJ)}1/8

+ 22/3{afJ(1 -

a)(1 - ,B)}1/24

=I

(6.26)

and

1+ (afJ)1/4 + {(I = {2 (1

a)(1 - ,B)}1/4

+ (afJ)I/2 + {(l -

+ 24/3{afJ(1 -

a)(1 _ fJ)}1/12

a)(1 - fJ)}1/2)} 1/2.

(6.27)

We also need two of Ramanujan 's modular equations of degree 3. The first is given by Lemma 4.3, while the second is given by (Part III [3, p. 231. Entry 5(ix))) {a(1 - fJ)}I/2

+ {(l -

a)fJ} 1/2 = 2{afJ(1 - a)(l - fJ)}I/8.

=

=

(6.28)

We shall apply (6.24) with r 3 and p 23. Thus. fJ has degree 23 over (1 - a). Thus. replacing a by (l - a). from (6.26) and (6.27), we find that.

248

Ramanujan's Notebooks, Part V

respectively, {(l - a)p}I/8

+ {a(l -

+ 22/3{aP(1 -

p)}1/8

a)(l - p)}I/24

=1

(6.29)

and 1 + {(l - a)p}I/4 + {a(1 - p)}I/4 + 24/3{aP(1 - a)(1 _ p)}I/12

= {2 (1

+ {(l

+ (a(1

- a)p}I/2

_ p)}I/2)} 1/2.

For brevity, in the remainder of the proof, set G we can rewrite (6.29) in the form {(l - a)tJ}1/8

+ {a(l

(6.30)

= G69 and G' =

G23/3. By (6.21),

- p)}1/8 = 1 - -li(GG')-I.

Setting u = (GG,)-I and squaring both sides, we deduce that {(I - a)p}I/4 + (a(l - p)}I/4

= 1 + 2u 2 -

2-1iu - -liu 3.

(6.31)

Substituting (6.31) into (6.30), we find that

2+4u 2 - 2-1iu - -liu 3 =

-Ii (1 + {(l -

a)p}I/2

+ {a(l

- p)}I/2f/2. (6.32)

Then, using (6.28) in (6.32), we deduce that

2 + 4u 2

-

2-1iu - -liu 3 = -li(1

+ -liu3) 1/2.

Squaring both sides and simplifying, we arrive at 2 - 8-1iu

+ 24u 2 -

22-1iu 3 + 24u 4 - 8-1iu 5 + 2u 6 = 0,

which, with x = u + I/u, is equivalentto

22-1i = 2(u 3 + u- 3) - 8-1i(u 2 + u- 2) + 24(u = 2(x 3

-

3x) - 8-1i(x 2

-

2)

+ u- I )

+ 24x.

Simplifying, we find that

x 3 - 4..tix 2 + 9x - 3-1i =

o.

By inspection, we verify that .../i is a root. Now G n is a monotonically increasing function of n, and it is not difficult to check numerically that the root that we seek is greater than .../i. Thus,

x2 and so x

-

= (3 + ..(3)/.../i. Since x

~u = since u < l.

+ 3 = 0, = u + l/u, we find that

3..tix

J+ 6

3../3 4

+

J+ 2

3../3 4'

(6.33)

34. Class Invariants

We now apply Lemma 4.3. Noting that P

u- 3 _ u3 = ../(u-3 + U 3 )2 = J(x 3

=

-

249

= u 3 , we see that we want to calculate 4

3x)2 - 4

J374 + 216.J3.

Thus, by Lemma 4.3,

(~r + (~,r = 2../2(u-

3 -

u3 ) = 2../2J374 + 216.J3.

Solving for G / G', we deduce that

~, = (J748 + 432.J3 + J747 + 432.J3) 1/6

(6.34)

a

Thus, by (6.33) and (6.34), G

=

u- 1/ 2 = (J748 + 432.J3 + J747 + 432~ 1/12 x

(J6+:./3 +/2+:./3) '"

To complete the proof, it suffices to show that

(/748 +432./3 +/747+ 432J3)' = (5+~)' C./3 ~ ,/23)'. which is a straightforward task. Second Proof of Theorem 5.3. We need two of Ramanujan's modular equations of both degrees 7 and 11. If {J has degree 7 over a, then (Part III [3, pp. 314, 315, Entry 19(i), (viii))) (a{J)1/8

+ {(l -

a)(1 - {J)}1/8 = 1

(6.35)

and m -

2. = 2 (a{J) 1/8 m

{(I - a)(1 - {J)}1/8)

x (2 + (a{J)1/4 + {(l - a)(1 - {J)}1/4) ,

(6.36)

250

Ramanujan's Notebooks, Part V

where m = rp2 (q)jq;2 (q7). If {J has degree 11 over a, then (Part III [3, p. 363, Entry 7(i), (ii)]) (a{J) 1/4

+ {(l

- a)(1 - {J)}1/4

+ 2{16a{J(l

- a)(l - {J)}1/12

=

I

(6.37)

and 11

m ' - m' = 2 (a{J) 1/4 - {(I - a)(l - {J)}1/4)

x (4 + (a{J) 1/4 + {(l - a)(1 - {J)}1/4) ,

(6.38)

where m ' = rp2(q)/rp2(qll). If q = exp( -7f .JIT77), by (6.21), G ll j7 = {4a(1 - a)}-1/24

G 77

and

= {4{J(1 -

{J)}-1/24.

Thus, setting u = (G 77 G 11 /7)-I, we deduce from (6.35) that

+ {(I -

(a{J)1/8 - {(I - a) (1 - {J)}1/8)2 = (a{J)1/8

a)(1 _ {J)}1/8)2

- 4{a(1 - a){J(1 - {J)}1/8

= 1-2hu 3 and (a{J)1/4

+ {(I

- a)(l - {J)}1/4 = (a{J)1/8

+ {(l

- a)(1 _ {J)}1/8)2

- 2{a(l - a){J(1 - {J)}1/8

= 1 - hu 3 • Thus, from (6.36),

7 = 2 ( 1 - 2v2u r;;; 3)1/2 r;;; 3 m- m (3 - v2u ), where m = q;2(e- rr ../IT77)/q;2(e- rrffi ). Let q = exp( -7f ..j77IT), and note that u Thus, by (6.37), (a{J) 1/4

+ {(l

= (GnGll/7)-1 =

- a)(l - {J)}1/4

=I

(6.39)

(G77G7/1 1)-1.

- 2u 2

and (Ca{J)1/4 - {(l- a)(l - {J)}1/4)2 = (Ca{J) 1/4 + {(l - a)(l - {J)}1/4)2 - 4{a(1 - a){J(l - {J)}1/4

= (1 - 2u 2 )2

-

2u 6 .

Hence, from (6.38),

m' -

.!..!. = 2 ((l m'

2u 2 )2

-

2u 6 ) 1/2 (5 - 2u 2 ),

(6.40)

34. Class Invariants

251

From (6.25), we see that

m' = !lim. Since

m-

~ = (7m' _ {IT ~ = (7 (m' _ .!..!.) , m VJj V1" m' VJj m l

we deduce from (6.39) and (6.40) that

2 (1 - 2-./2u 3 )

1/2

(3 _ -./2u 3 )

= ~2 (1 _ 2U 2 )2 _

2u 6 ) 1/2 (5 - 2u 2 ).

Squaring both sides and simplifying, we find that 4u lO

-

11-./2u9

-

98u 8 + 327u 6

-

322u 4

-

66-./2u 3 + 210u 2

-

19 = O.

Isolating the tenns involving ..ti on one side of the equation, squaring both sides, simplifying, and factoring, we deduce that (u 8 X

-

8u 6

+ 7u 4 -

(196u 12

8u 2 + 1)

1418u lO + 6044u 8

-

-

13262u 6 + 13073u 4

-

5092u 2 + 361) = O.

(6.41)

Now x := u 2 is an algebraic integer (see Lemma 7.2) and so must be a root of a monic irreducible polynomial. The latter polynomial in (6.41) is irreducible, and so x must be a root of the fonner polynomial in (6.41). Alternatively, we used Mathematica to check numerically that x is not a root of the latter polynomial on the left side of (6.41). Thus, X4 -

Since x

8x 3 + 7x 2 - 8x + 1 = x 2 (x + llx)2 - 8(x + l/x) +

+ llx

5) = o.

> 1,

1 x

x + - = 4+.Jil. Thus,

Since u < 1, we find that

~=/6+v'Tf /2+v'Tf. u 4 + 4

(6.42)

Lastly, we apply Lemma 4.5. Since P = u- 3 , we deduce, by (6.42), that Q

+

Q-I

= 2-./2(u- 3 + u 3 ) -7 = 2-./2 (u + u- I )3

= 2-./2(3 +

"fli)J6 + "fli -

= 21 + 8v'Tf.

-

3(u + u- 1»)-7

7 = 2(3 + ,Jii)(1 +

"fli) -

7

252

Ramanujan's Notebooks, Part V

Hence,

In conclusion, by (6.42) and (6.43),

and the proof is complete. Second Proof of Theorem 5.4. We need two of Ramanujan 's modular equations, one of degree 3 and one of degree 47. If P has degree 3 over a (Part III [3, p. 231, Entry 5(ix)]), {a(1 - p)}I/2

+ {P(1

- a)}1/2

= 2{aP(1 -

a)(1 - p)}I/8.

(6.44)

If P is of degree 47 over a (Part III [3, p. 444, Entry 23(i)]),

+ (aP)I/2 + {(l - a)(1 _ p)}I/2»1/2 1 + (aP)I/4 + {(l - a)(1 _ p)}I/4 + 41/3{aPO - a)(1 - p)}1/24 (1 + (aP)I/8 + {(l -

2 (HI =

a)(1- p)}I/8).

(6.45) Let q = exp( -7r ../47/3). Then, by (6.21),

G' := G 47/ 3 = {4a(1 - a)}-1/24

and

Applying (6.24) withr = 3 andp = 47, we find thatp has degree 47 over (I-a) when P has degree 3 over a. Thus, by (6.45),

+ {(l - a)p}I/2 + (aO _ p)}I/2» 1/2 1 + {(l - a)p}I/4 + {a(1 _ p)}1/4 + 41/3{a(1 - a)p(I - p)}I/24 (1 + {(l -

2 (HI =

a)p}I/8

+ (a(1

_ p)}I/8). (6.46)

If u := (GG')-I, by (6.44), {a(I - p)}I/2

+ {P(1

- a)} 1/2 = .../2u 3 •

(6.47)

34. Class Invariants

253

Hence, (la(1 - ,8)}l/4

+ {,8(1 -

a)}I/4)2

+ {,8(1 -

= {a(1 - ,8)}1/2 =

a)}I/2

+ 2{a(1

- a),8(1 - ,8)}1/4

../2u3 + u6

(6.48)

and ({a(1 - ,8)}1/8

+ {,8(1 -

a)}I/8)2

= {a(1 - ,8)}1/4 + {,8(1 - a)}I/4 + 2{a(1 - a),8(1 - ,8)}1/8 = ( ../2u 3 + u 6)

1/2

+ ../2u 3.

(6.49)

Substituting (6.47)-(6.49) into (6.46), we find that

20(1

+ ../2u 3»)'/2 = 1 + (../2u 3 + U6)1/2 + ../2u

(I + (

../2u 3 + U6)1/2

+ ../2u3) 1/2) . (6.50)

Using Grabner bases, A. Strzebonski denested (6.50) and obtained a polynomial of degree 48 for u. The value of u that we seek is a root of the factor u 8 - 32u 6 + 15u 4 - 32u 2 + 1 of this 48th degree polynomial. If x = u 2, then x4 - 32x 3 + 15x 2 - 32x

Since x

+ I/x >

+1=

x 2 (x

+ l/x)2 -

32(x

+ I/x) + 13) =

O.

I, we find that

x

+ .!. = x

16 + 9./3.

Hence,

so that

(6.51) Lastly, we apply Lemma 4.3 with P = u 3 and Q = (G' /G)6 to deduce, from (6.51), that 1 Q+ Q

= 2../2(u- 3 -

u 3) = 2../2 (u- I - u)3

+ 3(u- 1 -

u»)

= 2../2(14 + 9./3)1/2(17 + 9./3).

Solving for 1/ Q, we find that

~

= ../2(14 + 9./3)1/2(17 + 9./3)

+ )31419 + 18144./3.

(6.52)

254

Ramanujan's Notebooks, Part V

Thus, by (6.51) and (6.52),

G I41 = Q-I/12 U -I/2 = (v2(14 + 9·J3)1/2(17 + 9.J3)

x

+ /31419 + 18144.J3Y/12

(l8+:.f3 +l4+ 9./3r 4

It remains to show that

v2(14 + 9.J3)1/2(17 + 9.J3) + /31419 + 18144.J3

= (M

+

../47)3/2 ( 7

+~)

,

which is easily accomplished via Mathematica. Second Proof of Theorem 5.5. We need two modular equations, one of degree 5 and the other of degree 29. The first is found in Ramanujan's second notebook. If 13 has degree 5 over a, then (part 1lI [3, p. 281, Entry 13(x)]) {a(1 - f3)}I/4

+ {f3(1

- a)}I/4 = 22/3{af3(1 - a)(1- f3)}1/24.

(6.53)

The second is found in Ramanujan's first notebook on page 304, but curiously not in his second. R. Russell [2] established this modular equation in 1890, but his formulation is imprecise; in particular, it has a sign ambiguity. We give Ramanujan's formulation as stated in Entry 65 of Chapter 36. Let P = 1-

Q

M - J(1- a)(1 -

= 64 {M + J(1

- a)(l -

f3),

13) -

Jaf3(1 - a)(1 - f3)},

and R = 32Jaf3(1 - a)(l -

13)·

Then, if 13 has degree 29 over a,

+ 17 P R 1/ 3 - 9R2/ 3) = RI/6(9P2 + Q - 13P R 1/ 3 + 15R2/ 3 ). (6.54) = exp( -7r J29/5) , so that we may apply (6.21) and (6.24) with r = 5

"fP(p2

Let q and p = 29. If u

=

{a(l - f3)}I/2

(G I4SG29/S)-I, then, by (6.53),

+ {f3(l

- a)}I/2 = ({a(1 - f3)}I/4

+ {f3(l

_ a)}I/4)2

- 2{a(l - a)p(l - f3)}I/4 (6.55)

34. Class Invariants

255

Thus, by (6.55), with a replaced by (1 - a), P=1-2u 2 +u 6 ,

Q = 128u 2

64u 6

-

-

16u 12 ,

and R = 8u 12 .

Substitute these values into (6.54), square both sides, simplify, and factor, with the help of Mathematica. We then find that (u 2 + 1)(u 4 - u 2 - 1)(u 4 - u 2 + 1)(u 8 - 20u 6 - 43u 4 - 20u 2 + 1) X (U 12 -

9U lO

+ 181u 8 -

126u 6

-

181u 4

9u 2

-

-

1)

= O.

In numerically checking the roots of each of these polynomials, we find that x := u 2 is a root of 20x 3

X4 -

Thus, x

-

43x 2

+ l/x =

20x

-

+ 1 = x 2 (x + 1/x)2 -

+ l/x) -

45)

= O.

10 + .Jl45, and so u - l/u = ./8 + J145. Hence,

~= u

; 8

+ J145 +; 12 + J145. 4

~ = 2 (~ -

(6.56)

4

Lastly, we apply Lemma 4.4 with P

Q+

20(x

= u 2 and Q = (G 29/ 5 /G I45)3. Then

p) = 2 (~ + u) (~ -u) = 2J 12 + = 2J241

.J14s~ J145

+ 20J145.

Hence,

~

= J241

+ 20J145 + J240 + 20v'14s.

From (6.56) and (6.57), G145

= Q-l/6U -l/2 =

(J241

X

(J

r r

+ 20J145 + J240 + 20J145) 1/6

12+;145 +;8+ ~145

To complete the proof, we must show that

(v's+2)

(6.57)

(~+5) ~ (J12+;145 +;8+~145

256

r

Ramanujan's Notebooks, Part V

and

(/17 \J145 +/9+ ;'45 ~ J241+ 20./145 +J240 +20./145. Both equalities are easily verified. Second Proof of Theorem 5.7. In addition to the modular equation of degree 3 given by (6.44), we need Ramanujan's modular equation of degree 71. If {3 has degree 71 over a, then (Part III [3, p. 444, Entry 23(ii)])

+ {(l - a)(1 - {3)}1/4 - H(I + (a{3)1/2 + {(l- a)(1 _ {3)}1/2))1/2 = (a{3) 1/8 + {(I - a)(1 - {3)}1/8 - {a{3(1 - a)(1 - {3)}1/8 + 22/3{a{3(1 - a)(1 - {3)}1/24 (I - (a{3)1/8 - {(I - a)(1 - {3)}1/8).

1 + (a{3) 1/4

(6.58) Let r = 3 and p = 71 in equalities (6.21) and principle (6.24). Thus, {3 has degree 71 over (I-a). Replacing a by I-a in (6.58) and employing (6.47}-(6.49), but now with u = (G 213 G71/3)-I, we deduce that

1 + (../i u3

+ u6 ) 1/2 _

= ( ( .J2u 3

(!(1

+ .J2u 3») 1/2

+ u 6 ) 1/2 + .J2u 3) 1/2

1 3 (( 1/2 - ../i u +.J2u 1 - (.J2u 3 + u6 ) + .J2u 3 )1/2) = (

(.J2u 3 + u 6 )

1/2

+ .J2u 3

)1/2

(1 - .J2u) -

1

",u 3 ...;2

+ .J2u .

(6.59)

Using resultants, A. Strzebonski and M. Trott denested (6.59) and found a polynomial that factors into several polynomials of degrees 8,12, and 28. Numerically eliminating all factors except one, we find that u satisfies

u8

-

80u 6

-

126u 4

-

80u 2 + 1 = O.

Letting u 2 =: x, and solving for x + l/x, we find that x then follows that u + l/u = J 42 + 24.J3. Hence,

.!.. u

=

/21 +

12.J3

2

+

/19 +

+ l/x =

12J3.

(6.60)

2

Lastly, apply Lemma 4.3 with P = (G213G71/3)-3 (G71/3/G2I3)6. So, by Lemma 4.3 and (6.60),

Q+~ Q

40 + 24.J3. It

u 3 and Q

= 2.J2(u- 3 - u 3) = 4(19 + 12·J3)1/2(41 + 24J3).

=

34. Class Invariants

257

Solving for 1/ Q, we find that

..!.. = Q

2(19 + 12·J3)1/2(41

+ 24../3) + J542, 475 + 313, 200../3.

(6.61)

Hence, by (6.60) and (6.61), G213

=

Q-I/12 U -I/2

=

( 2(19 + 12../3)1/2(41

x

+ 24../3) + J542, 475 + 313, 200../3)

1/12

(J21+;M +l9+;2./3)'"

It thus remains to show that

2(19 + 12../3)1/2(41

~ (5./3;

+ 24../3) + J542, 475 + 313,200../3

v'7f)'" (59 +~v'7f) ,

which can be verified via Mathematica.

7. Class Invariants Via Class Field Theory In [6], Watson employed an "empirical process" to evaluate 14 of Ramanujan's class invariants. Motivated by Watson's idea, we succeeded in formulating theorems which give rigorous evaluations of G pq and G p/q when p and q are distinct primes satisfying pq == 1 (mod 4) and h (.J - pq) = 8. Let K = Q(.J -m) (m squarefree) be an imaginary quadratic field, and let 0 K be its ring of integers. By class field theory (1. Janusz [1, p. 228, Theorem 12.1]), there exists an everywhere unramified extension K(l) IK such that the Galois group Gal(K(l) I K) :::::: C K , where C K is the ideal class group of K. The field K(l) is known as the Hilbert class field of K . A Hilbert class field of K is usually defined as the maximal unramified abelian extension of K. Let a = [t'\, t'2] be an 0 K -ideal. Define

.()

}a where

.([

= }

t'1, t'2

])

=

1728

g23([ t'\,

g~([t'I' t'2]) t'2

])

-

27g32([ t'\,

t'2

])'

258

Ramanujan's Notebooks, Part V

and

L 00

g3([t'] , r2]) = 140

1

m,n=-oo

(m,n)#(O,O)

It is clear from the definitions of g2 ([ r], t'2]) and g3 ([ r], t'2]) that j([t'], t'2]) = j([I, t']) =: j(r),

where t' = t'z/ t']. We also let

with the cube root being real-valued when j ( a) is real. It is well known that K(I) = K(j(OK» (D.A. Cox [1,p. 220, Theorem 11.1]). If DK is the discriminant of K and 3 f D K , then K(I) = K(n(rK» (Cox [1, p. 249, Theorem 12.2]), where

l

.j-m

t'K =

DK

3+J-m 2

,DK

== 0 ==

(mod 4),

1 (mod 4).

Lemma 7.1. Let a and b be two OK-ideals. Define O'a(j(b» by (7.1)

O'a(j(b» = j(ab), where an is aprincipalideal. Then O'a is a well-defined elementofGal(K(I) and a ~ 0'a induces an isomorphism

I K),

Proof. See Cox's book [1, p. 240, Corollary 11.37]. Lemma7.2. Let K = Q(.j-pq), where pandq are two distinctprimes satisfying pq == 1 (mod 4), and let y

=

{

4,

if 3 f pq,

12,

if3

I pq.

Then G~q is a real unit generating the field K(l).

Proof. From a paper by B. J. Birch [1, p. 290], we find that G~~ is a real unit of K(I). Since (Cox [1, p. 257, Theorem 12.17]) j(OK) = j(.j-pq) =

(I6G 24 _ 4)3 ~;4 pq

'

(7.2)

we conclude that (7.3)

34. Class Invariants

259

Next. suppose that 3 t pq. Then 3 t DK and Y2(rd generates K(l). From the equality (Cox [1. p. 257. Theorem 12.17])

Y2CJ-pq)

=

16G 24

~:

-

4

pq

(7.4)

and (7.3). we find that G~q E K(l). Hence. G!q E K(l). by (7.3). In [1. p. 290]. Birch quoted Deuring's results [1. p. 43] and indicated that G pq is a unit when pq == 1 (mod 4). A more elaborate proof of this statement was given in a paper by Chan and Huang [1. Cor. 5.2] and is contained in Theorem 1.1 in this chapter. In fact. from the treatment given in their paper. one can show that G p/q is also a unit. This fact will be needed in our main theorem. From class field theory. we know that if H is a subgroup of C K • then there exists an abelian and everywhere unramified extension L IK such that Gal(K(I)IL) :::::: H.

cl

:= the subgroup of squares in C K. the corresponding In particular. when H = field M IK is known as the genus field of K. One can show that M is the maximal unramified extension of K which is abelian over Q (Cox [1. p. 122]).

Theorem 7.3. Let K and Y be defined as in Lemma 7.2. If the order ofC K is 8, then

and {3

._ ( Gpq)Y p,q'- G p/q

+(

G pq )-Y

G p/q

are algebraic integers which belong to the real quadratic .field R, where R E {Q(,JP). Q(,Jq). Q(,Jpq)}' and where R is afield such that none of the prime ideals (2). (p). or (q) are inert. Proof. From the hypothesis. we deduce that al = [1, J-pq]. a2 = [q, J-pql. a3 = [2. 1 + J-pql. and a4 = [2q. q + J-pq] are DK-ideals lying in distinct equivalence classes (see Section 4). This implies that C K contains the Klein fourgroup generated by the ideal classes [ad and [ajl for i > j > 1. Using the isomorphism described in Lemma 7.1. we conclude that Gal(K(l) I K) contains a Klein four-group V generated by ani and a nj for i > j > 1. To show that CXp,q and {3p,q belong to a field with degree 2 over K. it suffices to show that ani and a nj fix CXp,q and (3p,q' More precisely. if F := Fix(V) is the field fixed by V. then by Galois theory (1. Rotman [1, p. 49. Theorem 63]). IF : KI = IGal(K(1) I K) : VI = 2 (since IC K I = 8). which implies that F is of degree 2 over K. Since CXp,q and {3p,q are real numbers in F. they belong to R := F n JR. and R is clearly a real quadratic field over Q. The fact that they are algebraic integers follows from the fact that G~q and G~/q are units (see Lemma 7.2).

260

Ramanujan's Notebooks, Part V

At this stage, we will assume that 3 I pq. From Cox's text [1, p. 257, Theorem 12.17], (7.5) By Lemma 7.1, we find that aa, (J (ad)

= l(a2 a \) = l(a2).

From (7.2), (7.5), and (7.6), we find that (16G 24

p/q

-

(7.6)

4)3

(7.7)

G 24 p/q Simplifying (7.7), we deduce that (a - b)(a

where a

+ b) {64(a 2 + b2)a 2b 2 -

= aa,(G1~) andb = 64(a 2

48a 2 b 2 + I}

= 0,

G1)q. But

+ b 2 )a 2 b2 -

48a 2 b 2

+ I =I 0,

for otherwise it would contradict the fact that a and b are algebraic integers. Thus, we deduce that

aa2(G1~) = ±G~]q.

(7.8)

Similarly, corresponding to (7.8), or i.e., aa,(G1)q) may have the same or opposite sign as aa2(G1~). Since aJ, the latter is inadmissible. Hence,

aa2(G~~q) = ±G~~.

=

1,

(7.9)

From (7.8) and (7.9), it is now clear that

and aa,(fJp.q)

= {3p.q.

Next, from Cox's text [1, p. 263], we find that .( ) = .

] a3

and

.( ) = ].

] a4

]

(3 + ~) 2

= G 24

(3 + FJilq) = 2

pq

G 24

p/q

(~ 4)3 _

(7.10)

(~_ 4)3

(7.11)

G24 pq

G24 p/q

34. Class Invariants

261

Now, applying Lemma 7.1 again, we have CTIl3 (j (ad) = j (a3). By (7.10) and (7.2), we find that (16o';3(G1;) - 4)3 = G 24 2 (GI2) CT113 pq pq

(~ _ G24 pq

4)3

which implies that CTIl3(G~!) = ±G;~2.

Similarly, since a3a2 is equivalent to 114, 12 I2 ) = ±GCTIl3 (G p/q p/q

or

12 =F G-p/q'

by (7.11) and (7.5), i.e.,CTIl] (G1~q) may have the same or opposite sign as CTIl3 (G1~). We will show that the latter case is inadmissible. If

CTIl2(G~~q) = ±G~!

and CTIl3(G~~q) = =FG;;:,

then CTIl2 CTIl3 (G~!)

= ±G;};

and CTIl]CTIl2 (G~!)

= =FG;};.

This is clearly a contradiction since CTIl2 CTIl ] = CT1l3 CT1l2 . Hence, and CTIl3 ({Jp,q)

= {Jp,q'

Collecting our results, we see that both CTIl2 and CTIl3 fix ap,q and {Jp,q, and this implies that ap,q and {Jp,q are real quadratic algebraic integers. The proof for the case when 3 f pq is similar. In this case, G!q generates K(l) , and so CT11, (G!q) is well defined for i > 1. Hence, we may deduce from (7.7) that 8)2

16 ( CTIl2(G pq )

-

4

8

CTIl2 (G pq)

16

= 16G p/ q -

4

--8-'

G p/q

(7.12)

Simplifying (7.12), we have (a - b)(4a 2 b

+ 4ab2 + 1) =

0,

where a = CTIl2(G~q) and b = G~/q' But 4a 2b + 4ab 2 + 1 ::j:. 0,

for otherwise it would contradict the fact that a and b are algebraic integers. Hence, we deduce that

262

Ramanujan's Notebooks, Part V

Now, since O'B2 E Gal(K(l)IK) and G~q generates K(I) (see Lemma 7.2), O'B2(G!q) = ±G!/q'

The rest of the arguments are similar to those of the previous case, and we shall omit them. We have already seen that ap,q and {Jp,q lie in a real quadratic field R. Our next task is to give a necessary condition for R. First, we observe that R = F n JR, where F = Fix(V) is an abelian, everywhere unramified extension of K (see the paragraph before the statement of Theorem 7.3). Hence, R E (Q(JP), Q(Jq), Q(.Jpq)}. Next, we will show that none of the prime ideals (2), (p) or (q) are inert in R. Suppose the contrary holds. Then without loss of generality, we may assume that (p) is inert in R. This implies that pin K is inert in F, where pl(p) and the Frobenius automorphism [

F~K] has order 2 (see the books by Cox [1,

pp. 106-107] or Janusz [1, pp. 126-127]). K(I)IK]

On the other hand, we know that the Frobenius automorphism O'p = [ --phas order 2 and that

[K(~ IK] IE =

1, where

. ([K(l)IK]) E=F,X --. p

Since (Janusz [1, p. 127, Property 2.3])

[ ElK]

we find that -p- has order 1. Consequently,

This clearly contradicts the last statement of the previous paragraph. Thus, (p) is not inert in R. Our next step is to determine a p,q and {J p,q using the numerical values of G pq and G p/q' To achieve this, we need the following result. Theorem 7.4. Let R = Q( ..jm) be the field which contains ap,q and {Jp,q' If (7.13) and

(7.14) then ai, a2, b l , and b 2 are positive integers.

34. Class Invariants

263

Proof. Let [a] E A := {[a2], [a3], [a4]}, and let H be the group generated by [a]. From the paragraph before the statement of Theorem 7.3, we know that there exists an abelian and everywhere unramified extension L IK such that Gal(K(\)IL) ~ H.

In fact, from the isomorphism of Lemma 7.1, we find that L Gal(K(l)IK) ~ Z2 E9 Z4, the group Gal(LIK) ~ Z2 E9 Z2

or

=

Fix(O'a). Since

Z4.

The first case can only happen for exactly one element in A, and the field L in this case is the genus field M of K. As for the second case, Gal(LIQ) ~ D g , the dihedral group of eight elements, since L is generalized dihedral over Q (Cox [1, p. 191]). Hence, Gal(LIQ) is nonabelian. Now, rewrite (7.13) as (7.15) where 11 = (GpqGplq}Y· NotethatO'a2 fixes 11 andO'a3 (11) = 11- 1 • Therefore, the field L := K (11) = Fix( < O'a2 » is of degree 4 over K. Suppose L is the genus field of K. Since a a 2 1L = 1, we conclude that the ideal [a2] lies in an ideal class belonging to the principal genus. Hence, by Theorem 4.2, we may write (7.16) where

el = Since w

{

Whlh2/W2, 3wh 1h2l w 2,

if3 t pq, if3lpq.

= 2 and W2

={

t

2 or 4,

if 3 pq,

2,40r6,

if3lpq,

we conclude that el must be of the form e;/2, where e; rewrite (7.16) as 11

=

n

€~'1/2

E

N. Hence, we may (7.17)

X(Gil=-1

Now, it is known that a fundamental unit of a real quadratic field takes the form + v./d with u, v > 0 (Borevich and Shafarevich [1, p. 133]). Furthermore, if J u + v./d = u'...,!d; + v' ..,Iili, then u' , v' ~ O. Collecting these observations, we deduce that 11 is of the form UI +U2'[p +U3"fii +U4"fiiZi, where Uj ~ 0 for each i. Using (7.15) and (7.17), we conclude that a 1 and a2 are positive integers. Next, suppose L is not the genus field. Then from the beginning of our discussion, Gal(LIQ) ~ Dg is nonabelian. We claim that there exists an element a in Gal(LIK) such that 0'(11) is complex. Suppose the contrary holds. Then

U

264

Ramanujan's Notebooks, Part V

L n IR = Q(1]) would be Galois over Q, and hence Gal(LIQ(1])) is a normal subgroup of Gal(LIQ). On the other hand, Gal(Q(1])IQ) ~ Gal(LIK), a normal subgroup of Gal(LIQ) (Cox [1, p. 191]). Hence, Gal(LIQ) is isomorphic to the direct sum of Gal(LIQ(1])) and Gal(Q(1])IQ) and is therefore an abelian group, and this contradicts our initial assumption. Next, we will show that a(.jm) = -.jm. Suppose that the contrary holds. Then

and therefore a(1]) =

1]

or

1]-1.

This shows that a(1]) is real, which contradicts our choice of a. Now, applying a to (7.15), we deduce that 2(a(1])

+ a(1])-I) = al

-

a2.Jiii.

(7.18)

From (7.15), (7.18), and the fact that a (1]) is complex, we find that (al

+ a2.Jiii)2

::: 16

and (al - a2.Jiii)2 < 16.

This implies that 4ala2.jm > O. Since 1] > 0, we deduce that al and a2 are positive. The integrality of al and a2 follows easily from Theorem 7.3. In a similar way, we can show that hI and b2 are positive integers in (7.14). The argument given here for the case when Gal(LIQ) is nonabelian is due to H. Weber; see Cox's book [1, p. 269]. Let!}tK be the subset of (Q(,JP) , Q(y'q), Q(ffi)} satisfying the last statement in Theorem 7.3. Note that, since I!}tK I is finite and 2a p ,q and 2f3 p ,q lie in a discrete subset of the ring Z(.jm) for some Q(.jm) E !}tK, we can therefore determine their exact values, based on the numerical values of G pq and G p/q, in a finite number of steps. This will in tum lead to exact values of G pq' Except for K = Q(J -217) and Q(.J -553), in all of our calculations, I9t KI = 1. We illustrate our computations with two examples. Before we proceed, we let u := GpqG p/ q , v := Gpq/G p/q, U j := (u i + u- i )2, and Vj := (v j + v- j )2. Example 1. Let p = 5 and q = 13. In this case, y = 4. By Theorem 7.3, a5,13 and f35,13 are real quadratic algebraic integers. Since the primes 2, 5, and 13 are not inert in Q(.J65), we deduce that they are in Q(.J65). Now, evaluating u and v using the product representation of G n (see (1.3», we find that a5,13 = 81.311288 ... and f35,13 = 57.186772 .... We know that these numbers are of the form a + b.J65, and, by Theorem 7.4, we conclude that a5,13

=

41

+ 5.J65 2

and

f35,13

=

33

+ 3.J65 2

.

34. Class Invariants

265

Therefore, 45 + 5.J65

37 + 3.J65 2

and

V2=---,

U)=---

and

V)=---.

u=J~+9 +J~+l

and

U2 = - - 2 - -

which implies that .J65 + 9 2 This further implies that

.J65 + 7 2

Hence,

G65

)/2~~1/2 .J65+1 (.J65+7 .J65-1 ~~ 8 + 8 8 + 8

=

(.J65+9

=

(.J65+9

and G

13/5

1/2~~1/2 .J65+1 (.J65+ 7 _ .J65-1 ~~ + 8

8

8

8

Example 2. Let p = 3 and q = 23. In this case, y = 12. Using the numerical values of u and v and Theorems 7.3 and 7.4, we find that U6 = 281344 + 162432.J3

and

V6

= 2992 + 1728.J3.

The first equality implies that u6

+ u- 6 =

8(47 + 27.J3).

Since

we conclude that

u2

+ u- 2 = 4 + 3.J3

and

Collecting our results and simplifying, we deduce that

G 69

=

(

J748

x

+ 432.J3 + J747 + 432.J3)

1/12

(J6+:./3 +J2+:./3)'"

The following table summarizes our further calculations. The values for and G 793 are new.

G697

266

Ramanujan's Notebooks, Part V

U I = 6 +.JIT

n = 77 Gn

(j6+flf j2+flf)'"

-

(

~f~r' 23+:flf 19 + :flf

{¥if¥f UI

Gn

= 125680 + 72576.J3

+ 18144v'3J31419 + 18144v'3) 1/12

= 12 + ,Ji45

VI

=

17 + .J145 2

~!~\r

~ ( 17 + ./145 9 + ./145 8

x (

U2 =

Gn =

V6

18 +49../3 14 +49../3

x (J31420

n = 145

+ 8.JIT

4

UI=18+9.J3

Gn =

n = 205

+

4

x (

n = 141

V2 = 23

2025

8

12 + ;'45 8 + :'45 ~~'/2

+ 315.J4f 2

VI =

25 + 3.J4f 2

( 2025 +:15.;41 2017 +: 15.J4f ) I

x (

~~'/2 25+:.;41 17+:.;41

1/4

34. Class Invariants n

= 213 Gn =

=

Uj

42 + 24.J3

1252800.J3

~/~r

= 217

UI=

313200.J3) 1/12

=

(J +48~l8 +48~)'12 8,J7

VI

16 + 5,J7

22

x (

= 265

313200.J3J542475 +

= 22 +

UI

Gn =

n

= 2169904 +

( 21 + ;2~ 19+ ;2~ x ( J 542476 +

n

V6

16+45~ 12+45~ f¥if¥f

89 + 5v'265 2

VI

=

16 + v'265

r 1/2

( 89+ ;'/265 81 + !v'265 )

Gn =

I

x (J16+;'265l2+;'265

n

= 301

UI

Gn =

= 46 +

7v'43

V2

=

~I¥)'/'

( 46 + :./43 42+:./43 x

(~1199 +484v'43

1199 + 184v'43

r

1195 +484v'43

267

268

n

Ramanujan's Notebooks, Part V

= 445

U2

= 71325 + 7560.J89

Gn = (

85 + 9.J89 2

VI=

71325 +:560./89 71321

1/4

r

+:560.J89 )

x(J 85+:./89 J77 +:./89 n

= 505

= 292 + 13.J505

UI

292+ ~3./505

Gn = (

x( n

= 553

113 + 5.J505 8

VI

288

113

=

+ 5.J505 2

1/2

+ ~3.J505

)

, 1/2

105 +:.J505 ) /

UI =286+32m

V2

= 19163 + 2156m 1/2

Gn = (

286 + :2./79 282+:2m)

x( n

= 697

UI = Gn = (

19163 + 2156m 4

19159 +:156m

769 + 29.J697 2

769+ ~9';697

x(

661

+ 25J697 8

VI

761

=

661

+ !9.J697

653

I

+ 25.J697 2

1/2

)

+ !5.J697

1/4

1/2

) J

34. Class Invariants

269

VI = 452 + 16J793

n = 793

1/2

1/2

452+ 16J793 4

8. Miscellaneous Results In this section we collect together some miscellaneous results of Ramanujan on

class invariants. We have been unable to provide meaningful interpretations for two of these entries.

Entry 8.1 (p. 311, NB 1). We have

3.25/12 G 75 = ~=-----------=-----------------

.../5 + 1(10)1/3 + .../5 2

141/3.51/6

2

-.../5 _ 1

(8.1)

and G3/25

3.25/ 12

= -----------------------.../5 + 141/35 1/6 _ .../5 - 1(10)1/3 + .../5 _ 1·

2

(8.2)

2

Proof. We apply Lemma 4.4 with q = exp( -Jr Jii), and so by (1.6), P

= Gn-2G-2 25n'

Q = G~/G~5n'

and

G~ + (j3 Gb" + 2 (G" - 2 -2 G 25n -

-3G 25n

n

Let n = 3. From our table in Section x = G75,

2,

2

2

)

Gn G 25n =

G3 =

21/12.

o.

(8.3)

Thus, by (8.3), with

(8.4) This sextic polynomial has two real roots, and, via Mathematica, we easily checked that the expression on the right side of (8.1) satisfies (8.4) and is the correct real root.

270

Ramanujan's Notebooks, Part V

Alternatively, from Weber's book [2, p. 724], G75 is a root of a certain cubic polynomial. In fact, (8.4) factors over Q(.J5) into a product of two cubic polynomials, one of which is Weber's cubic polynomial, and so we have given another derivation of Weber's cubic polynomial for G 75 . To prove (8.2), we again use (8.3), but now we set n = Thus, G 25n = G 3 = 21/12. Hence, with y = G3/25,

fs.

l 21/4 -21/4 + -y3 + 2(2-1/6 y -2 -

21/6 y 2)

=0

(8.5)

'

which is exactly the same as (8.4), but with x replaced by y. We used Mathematica to verify that the right side of (8.2) is a root of (8.5) and indeed is the correct one. Of course, G 3/ 25 is the "other" real root of (8.4) that we mentioned above. Entry 8.2 (p. 316, NB 1). We have 3.2 1/ 4 G175=--------------~~------------------------------

./52- 1 + ( 5 -4./5) '/3 (,0/8 - 3./5 + 3,/2\ + ,0/8 - 3./5 - 3,/2\) (8.6)

and

1/3

.J52+ 1 + ( 5 +4.J5)

(V'8 + 3.J5 + 3ffi + V'8 + 3.J5 -

3ffi)

.

(8.7)

Proof. We employ (8.3) with n = 7. From our tables in Section 2, G 7 = 21/4. Thus, with G = G 175, we find that 23/ 4 G3 + + 2(T 1/ 2 G- 2 G3 2 3/ 4

-

21/2G 2)

= O.

(8.8)

Now, by straightforward algebra, it is easily checked that (8.8) yields 2G 3 - 4· 21/4G 2 +

.JiG -

3 . 23/ 4 =

±.J5(2 . 21/4G 2 - .JiG +

23/ 4). (8.9)

We shall show that the right side of (8.6) is a solution of the cubic equation in (8.9) where the plus sign is chosen on the right side. Thus, with the plus sign chosen in (8.9), we find that, after simplification, 2G 3

-

2 1/ 4(4+ 2.J5)G 2 +

(.J2 +

v'iO)G - 2 3/\3 +

.J5) = o.

(8.10)

The form of Ramanujan's formula (8.6) suggests that he set G = 1/x and solved for x. Thus, by (8.10), 23/4(3 + .J5)x 3

-

(.J2 +

00)x 2 + 21/4(4 + 2.J5)x - 2

= O.

(8.11)

34. Class Invariants

271

We solve (8.11) by employing Cardan's method (Hall and Knight [1, p. 480]). Thus, set

x

=y+

v2 +,JTO + ../5)

3 . 23/4(3

in (S.II), and, after dividing out the common factor 47 + 21../5, we find that

l + 18· 21/4../5y + !(-55 + 23.J5) = O.

27· 23 / 4

The solution of the general cubic equation y3 +qy +r oL/r 4 /4 + q3/27. With 23../5 - 55 27.27 / 4

r=--:-;:-;-;-

and

q

(8.12)

= 0 requires the calculation

=

,JTO

-3-'

we find, after much simplification, that

J+ r2

4

q3 = .J35(../5 - 1). 27 3.J3211/4

Thus, from Hall and Knight's text [1, p. 480], the real root of (8.12) that we seek is y

= 155 - 23../5 27.2 11 / 4

=

1») 1/3

+ 155 -

23../5 _ .J35(../5 - 1») 1/3 27.2 11 / 4 3.J3211/4

3.J3211/4

3.2111/12 ({55 - 23.J5 + 15v'2T - 3.Ji05} 1/3

+ {55 =

+ .J35(../5 -

23.J5 - 15v'2T + 3.Ji05} 1/3)

(53~ ~~~/3 ({ 8 _ 3.J5 + 3v'2T} 1/3

+ {S _3.J5 _

3v'2T} 1/3) .

Thus,

This is easily seen to be equivalent to (S.6). To prove (8.7), we set n = in (8.3). Since G n = G I / n , it follows that G I / 7 = G 7 = 21/4. Thus, with G = G25j7, we deduce (S.S) once again. Hence, G25/7 is the other real root of (8.8). Therefore, taking the minus sign on the right side of (8.9), we find that G satisfies the equation

t

2G 3

-

21/4(4 - 2.J5)G 2

+ (v2 -

0o)G - 23/ 4(3·-.J5)

= o.

272

Ramanujan's Notebooks, Part V

Now repeat the calculations from the proof of (8.6), but with 0 replaced by -0. We then deduce (8.7) to complete the proof. We also calculated G 175 from (8.10) by using Cardan's method and found that 21/4

G175

x

=3

(2+ v'5 + (5+;v'5)'" ({ 3511'" + { 3511"')) , 17+

17 -

(8.13) which is a slightly more elegant representation than (8.6). By combining (8.6) and (8.13), we deduce that

1 (5 - 0)1 /3({

0 2- - + (-

-4-

~

~}1/3

8 - 3v5 + 3v21

+ {8 - 3../5 _ 3h1} \/3)) x

(2+ v'5 + (5+;v'5)'" x ({ 17 + 3h1r3

+ { 17 - 3h1r3) ) = 9.

(8.14)

We are unable to establish (8.14) directly. At scattered places in the second notebook, Ramanujan discusses a few additional class invariants. Entry 8.3 (p. 263, NB 2). Let t of

x6 Then x

= t 4 , where t

=

+ 9x 5 + 5x 4 -

IjGi9 and let x denote the positive real root 2x 3

-

5x 2

+ 9x -

I

= O.

(8.15)

> O. Furthermore,

t6 + t 2 = 1 - t4

J..ti9 - 5 2

(8.16)

and (8.17)

Lastly, if (8.18)

34. Class Invariants

273

where IL > 0, then

(8.19) Proof. It is not difficult to show that the first claim is equivalent to a result in Weber's book [2, p. 722]. We now prove (8.16). By straightforward algebra, it is readily verified that (8.1S) is equivalent to the equation

~ (~)2 -x (~)2 =S. x

l+x

(8.20)

I-x

Let

y_- -I

x

(I-X)2 -1 +x

Then, from (8.20),

y2 _ Sy - 1 = 0, which has the roots be taken. Hence,

! (5 ± v'29) . Since x = ~=X(I+X)2 = y

1 -x

0.1192S2 ..... the plus sign must

v'29-S.

(8.21)

2

Taking the square root of each side and recalling that x = (4, we complete the proof of (8.16). We next prove (8.17). Employing (8.21) and (8.16) and remembering that x = t 4 , we see that we are required to prove that

l+x I-x

(8.22)

Again, from (8.21),

l-X±JXl+X)2={jv'29+S± ( _1 ,Jil+x I-x 2

G'229-sj2 y-

=.J29± 2, i.e.,

J

I I-x I +x ~ ---±JX--= v29±2. ,Jil+x I-x Hence, from (8.23),

JX+

Jv29-2= ~ -2x

3-

x2

-

2x

,Ji(1- x 2 )

+ I.

(8.23)

274

Ramanujan's Notebooks, Part V

and

J

1+.Ji 59+2= Thus,

( .;x + ./.tI9 - 2 )2 1 + .;xJ-J29 + 2

x3

+ 2x2 I-x

X+2 2'

(-2x 3 - x 2 - 2x + 1)2 x(x 3 + 2x2 - X + 2)2 .

By (8.22), we want to show that the right side above is equal to (1 + x)/(l - x). Thus, it suffices to prove that (1 - x)( -2x 3 - x 2

2x + 1)2 = x(1 + x)(x 3 + 2X2 - X + 2)2.

-

Expanding both sides and collecting tenns, we find that the foregoing equation is equivalent to the equation

o=

x 8 + 9x 7 + 6x 6 + 7x 5 + 7x 3 - 6x 2 + 9x - 1 = (x 2 + 1)(x 6 + 9x 5 + SX4 - 2x 3 - Sx 2 + 9x - 1).

Since x 2 + 1 i= 0, it suffices to prove that the second factor above equals O. But this is true by (8.1S), and so the proof of (8.17) is complete. Next, we prove (8.19). From (8.18),

4~

2

(8.24)

y-;-x=l+lL. This equality and the symmetry in (8.1S) suggest that we set p calculation shows that

x 6 + 9x 5 + SX4 - 2x 3 Since x

= x -\ -

x. A brief

Sx 2 + 9x - 1 = _x 3 (p3 - 9p2 + 8p - 16).

-

i= 0, by (8.1S), p3 _ 9p2

+ 8p - 16 = O.

From (8.24) and the definition of p,

lL=jJ~-X-1=J~-1. Thus, 1L3

Since clearly IL 3

+ IL

+ IL = ~J~

- 1.

> 0, it therefore suffices to prove that

v'/i(~ - 1)

= 2,

and since p > 0, it is sufficient to prove that

4

4

v'/i--=-+l.

.JP

p

(8.2S)

34. Class Invariants

275

Squaring both sides and simplifying, we find that it suffices to prove that

p3 _ 9p2

+ 8p -

16

= O.

But this is precisely (8.2S), and so the proof is complete. Parts of the proof of Entry 8.3 were taken from the notes of V. R. Thiruvenkatachar and K. Venkatachaliengar [1]. Entry 8.4 (pp. 263, 300, NB 2). Let t = 21/4/ G 79 . Then t is the real root of

(8.26) Furthermore,

if

J~

- =~,

(8.27)

t

then ~5 _ 2~4

+ ~3 + 2~ - 3 = O.

(8.28)

Proof. The class equation for n = 79 was not computed by Weber [2] and is not otherwise given by Ramanujan in his paper [3] or notebooks [9]. However, Russell [2] and Watson [10] determined the class equation, which is easily shown to be equivalent to that of Ramanujan. Now (8.28) is valid if and only if

(8.29) since the square root of each side is positive. Also, (8.29) holds if and only if, by (8.27),

o = ~1O _ =

(~

= _t 5

2~8 +S~6

tY -

-

2

- 8~4+4~2 - 9

(~- tY + S (~- tY - 8 (~- tY + 4 (~- t) - 9

2t 4 + t - S - t- I

-

= -

(t 10 +

= -

(t 5

-

2t 9

-

t4 + t3

-

2r4 + t- 5

t 6 + St 5 + t 4 + 2t -

2t 2 + 3t -

1) t- 5

1) (t 5 +

3t 4 + 2t 3 + t 2 + t +

1) t- 5 •

Since t > 0, t 5 + 3t 4 + 2t 3 + t 2 + t + 1 > o. Hence, (8.28) holds if and only if t satisfies (8.26), and this is what we wanted to prove. Entry 8.5 (p. 382, NB 3). Let z = x 2

z - z

+ 1/x, where x

S+.J4f 2

+

= G~I. Then

7+.J4f 2 =

o.

(8.30)

276

Ramanujan's Notebooks. Part V

Proof. From Weber's book [2, p. 722], as corrected by Brillhart and Morton [1], iff2(J-41) = -/2u, then

(U+~)2 _5+~ (u+~) + 7+~ =0.

(8.31)

Since u = G~I' (8.30) and (8.31) are equivalent. Ramanujan's formulation of Entry 8.5 is, in fact, slightly enigmatic. In particular, in contrast to the notation used throughout the second notebook [9], Ramanujan employed the conventional notation in the theory of elliptic functions and wrote l/x = ':/2kk'. We now make a few remarks about two entries possibly related to invariants. On page 294 of the first notebook, Ramanujan claims that "F

(1 -J~ - X24)

lI

= e- .jir

where

x + .!.. = (9J3 + 1)1/3 + (9J3 - 1)1/3 (.!..!.)1/6 X J3 2

,3/' {; Jx + '~Jx + , _ ;r-.jX-X-+-l-~-.jX-X---I }

{v'f+A ± Y'A}

x

3 A = - --;=:::::==;::::::;=:::;:::::;=2 _x_+_1-,-/x_+_l _ 1 x + l/x - 2

where

As intimated in Section 2 prior to the tables, it is not difficult to show that the indicated formula for G 121 is equivalent to the one given in the tables. Parts of the last two lines of the entry above are difficult to decipher, especially the definition of A. Moreover, we are uncertain that these two lines pertain to the first two lines, that is to say, that the value of x in the first two lines is the same as in the second two lines. Even more enigmatic is that an equality sign seems to be missing in the last two lines. With the two values of x arising from the second line, we calculated the expressions in the last two lines and could not account for a missing equality sign. In conclusion, we are unable to supply any meaningful interpretation to the incomplete entry offered in the last two lines. On page 343 in the first notebook, Ramanujan wrote "y

(

03 2

3

)

1/4

03 - 03 + 2(03 + 1)2 (03 - 1) _

(

where,.f5 y3 + y2

±

(Y3+y

2

2

1

+ y

+y

2

2

1

)

- 1

1)2 +

- 0. ..

34. Class Invariants

277

We have no explanation for this mysterious fragment. The definition of y is not given, but perhaps Ramanujan intended to write y = J(.Jf3 - 3)/2. However, this value of y is not a solution of either of the indicated polynomial equations. The top half of the page comprises results discussed in Chapter 37, and underneath the fragment is Ramanujan's (equivalent) representation for G 765 ; neither topic appears to be connected with this fragment.

9. Singular Moduli Recall from the Introduction that the singular modulus kn is defined by kn

.-

k(e-rr.fii), where n is a positive integer. It is clear from (1.6) that if the value of G n (or gn) can be determined, then an := k~ can be computed by solving a

quadratic equation. For example, see (2.S) or (9.1) below. However, the expression that one obtains generally is unattractive and does not evince the fact that an can be expressed in terms of units in certain algebraic number fields. (See Theorem 1.1.) Thus, formulas for an that facilitate their representations via units are desirable. In his second letter to Hardy, Ramanujan [10, p. xxix] asserted that k2JO

=

(.../2 - 1)4(2 - .../3)2(..;7 - ,J6)4(S - 3..;7)2 X (V'iO - 3)4(4 - ~)4(~ - JI4)2(6- J35)2.

This was first proved by Watson [4], who used the following remarkable formula which he found in Ramanujan's first notebook [9, vol. 1, p. 320] and which enables one to calculate an for even n.

Theorem 9.1. Set 6_

gn -

u2

+ l/u 2 = 2U,

UV,

v2

W = JU2

+ V2 -

+ l/v2 =

2V,

I,

and 2S = U

+ V + w + 1.

Then an

= {vis - JS'=-ll{v'S=17 -

Js - U - 1}2

X {Js - V - Js - V _1}2{JS -

w-

~- W _1}2.

Watson's proof of Theorem 9.1 is a verification; it does not shed any light on how Ramanujan might have discovered the formula. K. G. Ramanathan [1], [5] stated Ramanujan's Theorem 9.1 but did not find another proof. Heng Huat Chan has found a much more motivated proof of Theorem 9.1, and we present his proof below. Later we show that the algorithm implicit in Theorem 9.1 can be adopted to determine an for odd n as well.

278

Ramanujan's Notebooks, Part V

Proof. From (1.6) and the notation above,

2 2 .;a .;a = 2u v = 2(U + JU2 -

_1__

I)(V

+ .JV2=1)

= 2(U.JV2=1 + VJU2 - 1 + UV = 2 (J(a

+ I)(b -

+ J(U2 -

1)(V2 - I»

I) +../ib),

where we set

../ib = UV

+ J(U2 -

I)(V2 - I)

and

J(a

+ l)(b -

I) = u.JV2=1 + VJU2 - 1.

Squaring each of the last two equalities, we find that, respectively,

and

ab+b -a -1 = 2U 2V 2 - U 2 - V 2 +2UVJ(U2 -1)(V2 - 1). These two equalities imply that a = b. Thus,

.Ja -.;a

= 2 (J(a

+ l)(a -

1) + a).

Solving for,Ja, we find that

.;a =

(.Ja+l - .jQ.) (.jQ. - v'll=l").

It thus suffices to compute .;0.. From the definition of a and the fact that a = b,

a = UV

+ J(U2 -

1)(V2 - I)

=

4(2UV + 2JU2 V2 -

U2 - V2

=

4(2UV + 2JU2V2 -

W2)

=

4(JUV -

W

+ Juv + W

+ 1)

r.

(9.1)

34. Class Invariants

279

We write the last equality in two ways as follows:

~

V(U+ +J +

V)'

(U

a=

~

-~' -

~' -

V)' -

(J-(u -

V)'

V' -2W

V'

+ 2W) 2

+~' +V' -2W

+~' + V' +2W

+J -(U - V)'

(9.2)

r

Now,

+W (U + V + W + 1) (U + V -

(S - W)(S - 1) = S(S - W - 1) =

2

=

=

(U

+ V)2 -

(U

+ V)2

4

(W

+ 1)2

+W

+W

- U 2 - V2 - 2W

4

(U + V)2 - U 2 - V2

W - 1)

2

+W

+ 2W

4 Thus, from the first equality in (9.2),

a= (JS(S-W-l)+J(S-W)(S-I»)2 = S(S - W - 1) + (S - W)(S - 1) = S(S - W)

+ (S -

W - l)(S -

+ 2J,.--S(----:S,.---W---l.,,-)-----:(S,----,-W:c-)(-:-:cS,----I:-) 1) - S + (S - 1)

+ 2JS(S - W - 1)(S - W)(S - 1),

i.e., a

+ 1 = S(S -

W)

+ (S -

W - I)(S - 1)

+ 2J S(S -

W)(S - W - l)(S - 1)

= (JS(S-W)+J(S-W-l)(S-1)r·

Hence,

Ja+:l-..ra =

JS(S - W)

+ J(S -

W - I)(S -1)

- JS(S - W - 1) - J(S - W)(S - 1)

=

(~- v's-=1)

(JS - W- ~- W-

1).

280

Ramanujan's Notebooks, Part V

Next, (S - U - I)(S - V - I) = (S - U)(S - V) - W

W+I+V-UW+l+U-V 2 2 - W (W

+ 1)2 -

(U - V)2

----------------W U2

+ V2

4

- 2W - (U - V)2

4 Thus, from the second equality of (9.2), a

=

(J(S - U -1)(S - V-I)

=

(S - U - 1)(S - V-I)

+ J(S - U)(S - V)r

+ (S -

U)(S - V)

+ 2J(S - U - 1)(S - V - I)(S - U)(S - V)

=

(S - U - I)(S - V)

+ (S -

U)(S - V-I)

+1

+ 2J(S - U - I)(S - V)(S - U)(S - V-I), i.e., a-I = (J(S - U - I)(S - V)

+ J(S -

r.

U)(S - V - 1)

Hence,

Ja -

~ = J(S - U -1)(S - V-I)

+ J(S -

U)(S - V)

- J(S - U - I)(S - V) - J(S - U)(S - V-I)

=

(Js-V-v'S-V-l)(v'S-U-JS-U-I).

Using our just calculated formulas for.Ja+T we complete the proof.

Va and Va -,Jli'-=-I in (9.1),

Furthermore,Watson [4] inexplicably claimed, " ... this is the sole instance in which Ramanujan has calculated the value of k for an even integer n." In fact, 20 additional values of kn for even n are found in the first notebook. Theorem 9.2 gives 13 of these values. On page 82 of his first notebook, Ramanujan offers three additional theorems for calculating otn when n is even. The first (Theorem 9.3) expresses ot4p as a product of units involving G p . The second (Theorem 9.5) expresses otl6p as a product of units involving G p • The third (Theorem 9.6) enables one to determine ot8p as a product of two fourth powers of units, provided that ot2p can be expressed as a product of units of a certain form. We calculate eight examples of Ramanujan as illustrations. The calculation of otn when n is odd is slightly more difficult. On page 80 in his first notebook, Ramanujan recorded the values of ot21, ot33, and ot45 in terms

34. Class Invariants

281

of units. This list is repeated, with the addition of al5, at the bottom of page 262 in his second notebook. On pages 345 and 346 in his first notebook, Ramanujan recorded units that appear in representations of an when n = 3,5,7,9, 13, 15, 17, 25, and 55. (Inexplicably, the units for a7 and al5 are recorded twice.) Ramanujan also indicated that he had intended to calculate a39, but no factors are given. Of course, the result for n = 15 is superseded by the complete formula given on page 262 in the second notebook. It is unclear to us why Ramanujan only listed portions of an and not complete formulas. Initially in our investigations we employed computational "trial and error" to "guess" the complete formulas for an, n = 5, 9,13,17,25, and 55. We remark that the values fora3 anda7 are easily determined from (2.8), i.e.,

an =

!G2n

12

(G n12 - yn fG24 -

1) .

(9.3)

For further values of n, however, (9.3) becomes unwieldy, and so better algorithms were sought. We adopt the algorithm of Theorem 9.1 and reformulate it in Theorem 9.8 in terms of G n to calculate some values of an when n is odd. Theorem 9.9 provides a list of all of Ramanujan's values for odd n. Although Theorem 9.8 yields a systematic procedure for calculating an when n is odd, the calculations are often cumbersome and the representations that we obtain, although expressed in terms of units, are frequently more complicated than we would like. Thus, we establish three simple lemmas, Lemmas 9.10--9.12, that provide an alternative procedure for calculating all of Ramanujan's singular moduli for odd n. We conclude Section 9 with two further algorithms of Ramanujan for computing an. These were cryptically stated by Ramanujan in his first notebook and are rather different from his other algorithms. The first provides a method for determining a2n from a certain type of modular equation of degree n. The second also arises from modular equations and gives a formula for a3n. Ramanujan likely learned about singular moduli from a brief discussion in A. G. Greenhill's book [3, p. 331]. It would be extremely difficult to assess the priority of each singular modulus that has been determined. Ramanathan [1] and J. M. and P. B. Borwein [1] previously calculated some of Ramanujan's values for an, and we shall cite their specific determinations in the sequel.

Singular Moduli/or Even n We begin with a list of 13 values for an found on scattered pages in the first notebook. Theorem 9.2 (pp. 214, 288, 289, 310, 312, 313, NB 1). We have a2

=

(../2 -

1)2,

282

Ramanujan's Notebooks, Part V

r:;;;,

Z

r,:;

Z

alO=(vlO-3) (3-2v2) =

3 v'2 -

v'5 - 2 v'2 v'5 '

3 2+

S+2

a22 = (10 - 3.JU)z(3.JU - 7h)z, a30

= (S -

a42

=

2,J6)2(4 - v'tS)z(-V6 - ,v'Si(2 - ,J3)Z,

(8 - 3.J7)2(7 - 4,J3)2(3 - 2h)z(.J7 - -V6)2,

aS8 = (13v'58 - 99)2 (99 - 70h)z, a70 =

(1S -

4v14i(8 -

3.J7)2(3v'14 - Sv's)z(6 - ,J35)z,

a78 = (2 - ,J3)6(3,J3 - 56)2(03 - 2,J3)\S - 2-V6)2,

au"

~ ( v'5~ 7) ,(5 -

al30

= (sv'13O -

2,/6)'(Jsl - 5J2)'(2 - ,(3)',

S7)z(v'iQ - 3)4(56 - S)4(3 - 2h)4,

and

Proof. The value of az was, in fact, established in Example 1, Section 2 of Chapter 17 in the second notebook (Part III [3, p. 97]). Ramanathan [1] and the Borweins [1, p. 139] also determined az. All of the remaining values for an are easily determined from Theorem 9.1. The required values for gn can be obtained from the tables of Weber [2] or the table in Section 2. In each instance, we list the values for u, v, U, V, W, and S in the table below. The reader can easily verify the calculations.

34. Class Invariants

283

n

u

v

U

V

W

S

6

1

1 + v'2

1

3

3

4

10

1

2+.J5

1

9

9

10

18

1

5 + 2../6

1

49

49

50

22

1

7 + 5v'2

1

99

99

100

30

2+.J5

3 + v'IO

9

19

21

25

42

2v'2 +-J7

3v'3 + 2-J7

15

55

57

64

58

1

70 + 13.J29

1

9801

9801

9802

70

9+ 4.J5

7 + 5v'2

161

99

189

225

78

18 + 5.Jf3

5 +.j26

649

51

651

676

102

7 + 5v'2

35 + 6.J34

99

2449

2451

2500

130

38 + 17.J5

18 + 5.Jf3

2889

649

2961

3250

190

38 + 17.J5

117 + 37v'IO

2889

27,379

27,531

28,900

The second and third fonnulas for a6 and aiS. and the second fonnula for alO can be easily verified by direct calculations. The Borweins [1, p. 139] calculated an for 1 ::: n ::: 9. Ramanathan [1] also established al30 by using Theorem 9.1. Recall the definition of F(x) given in (2.3). Theorem 9.3 (p. 82, NB 1). If P > 0, n e

-rr"/p_

- F

~

1, and

(I-Jl-l/n2) 2

'

then

From (2.9), n = G~2. Hence, in Theorem 9.3 Ramanujan provides an algorithm for detennining a4p from the value of a p, or from G p. namely, a4p

=

(JG~2 + 1 -

fG1j)

4

f.

(JG~2 - JG~2 - 1

(9.4)

Before proving Theorem 9.3, we verify four examples recorded by Ramanujan.

284

Ramanujan's Notebooks, Part V

Examples 9.4 (p. 82, NB 1). We have = (./2 - 1)4,

a4

al2 = (.J3 - ./2)4(./2 - 1)4, a28

= (./2 - 1)8(2./2 _ ../7)4,

and

The value of a4 was also recorded in the second notebook (Part III [3, p. 97]). Both a4 and a28 were also determined by Ramanathan [1 J, and the Borwein brothers have determined a4 and al2 [1, pp. 139, 151J. Proof. Let p = 1, so that trivially G I = 1. Then, from (9.4), a4

= (./2 -

1)4(1 - 0)4

= (./2 -

Let P = 3, so that, from the table in Section 2, G3 from (9.4),

1)4.

= 21/12 and n = 2. Thus,

al2 = (.J3 - ./2)4(./2 - 1)4. Let P = 7, so that, from the table in Section 2, G7 from (9.4), a28

= (3 -

2./2)4(2./2 - ../7)4

= (./2 -

= 21/4 and n = 8. Thus,

1)8(2./2 _ ../7)4.

Let p = 15. From the table in Section 2, GIS = 2- 1/ 12 (1 + .J5)1/3. Thus, n = (1 + ../5)4/2 = 4(7 + 3../5). Hence, from (9.4),

a60 = ( /29 + 12./5 - 2/7 + 3./5y (2/7 + 3./5 - /27 + 12./5y To denest these radicals, we employ the following denesting theorem (Landau [1]). If a 2 - qb2 = d 2 , a perfect square, then

/ a + b.jij = Y/a+d {Q-d ~-2- + (sgn b)y ~-2-'

(9.5)

where we have corrected a misprint. To that end, from (9.5),

a60= ( J29+1l 2

x =

+ ~9-1l -2 2

f¥+2 R-2)4 ---2 -2 2

(21';2 +21';2 ~r:3 ~r;3r

(..no + 3 -

3./2 -

M)\3./2 + M -.Ji5 -

= {(M - 3)(./2 - 1)}4{(,J6 - ./5)(.J3 - ./2)}4.

vTI)4

34. Class Invariants

285

Proof of Theorem 9.3. From Part III [3, p. 215, eq. (24.21)], we find that

f3 where

=

(1 _~)

~2

(9.6)

4,

v'n+l -

f3 has degree 2 over a. To prove Theorem 9.3, we must show that f3 = (.In=l -

~/1i)4(

.j1i) 4 •

(9.7)

From our hypothesis, with n := G~2 and a := a p ,

-.J1i2"=l

n

a=----

2n

which implies that a-I

= (In(n

+ 1) + In(n -

1»2.

(9.8)

Since 4a(1 - a) = n- 2 , we deduce that

_1_ = (In(n I-a

+ 1) -

In(n _ 1»2.

(9.9)

Hence, from (9.6), (9.8), and (9.9),

f3 = (In(n + 1) + In(n _

1»4 (1 _v'n(n + 1) -1./n(n - 1) )4

~ (4n+T + v'n=i)' ( .jii _ 4n+T ~ v'n=i) , + n + Jn 2 - n - n - Fn2=1)4 = (.rn-=-I - .j1i)4(v'n+1- .j1i)4,

= (Jn 2

and so (9.7) has been shown. Theorem 9.5 (p. 82, NB 1). Under the same hypotheses as Theorem 9.3,

e- 4Jf ';p = F

x

(v'n+l + {..n;; + .j1i)8

{..n;; -

1 - J2.j1i(Vn

1 - J2.j1i(v'n+1 -

h)} 4) .

+ 1 + h)

r (9.10)

Thus, together Theorems 9.3 and 9.5 yield the formula

a"p

~ (,jGl' + 1 + ,jGl')' {,J2G1' + 1 - J2fG'je,jGi' + 1 +./2) x

{,j2Gi' - 1 - J2,jGl'e,jGl' + 1 -./2{

r

286

Ramanujan's Notebooks, Part V

For example, if p

= 1, then n = G I = 1, and a simple calculation shows that 0!16 = (../2 + 1)4(21/4 - 1)8.

Proof. From Theorem 9.3,

e- 4rr .jP = F2 (( In+l -

v'il) 4 (v'il - In=1) 4) =: F2 ((1 :xx)2) . (9.11)

Also, by Entry 2(v) of Chapter 17 in the second notebook (Part III [3, p. 93]), for 0< x < 1,

F(x2)

= F2 (

(1 +4x

X)2

) .

(9.12)

Thus, from (9.12), (9.10), and (9.11), it suffices to show that

x

= (In+l + .fi/)4 X

{5n -

{5n +

1 _ j2.fi/(Jn+l +../2)} 2

1 - j 2.fi/(In+l -

../2)

r

(9.13)

From (9.11), it follows that 2 1 ---===--------==,--- = (..In+l - y'n)2(y'n - ~)2 Jx

Let u = seek is

u

Jx.

Since u tends to 0 as n tends to

00,

+ Jx.

the solution of (9.14) that we

(J ~)2 + ~ -J2(vn+! - .Jiij:(.jii ~ = ! (In+l + (j (.fi/ + In=1)2 + (In+l-J(.jii + (vn+! - .jii)')' =

(9.14)

2(..Jn+l - y'n)!(y'n _

,;n::I)' -

),

.fi/)2

,;n::I)' -

(vn+l +.fi/r (j2n + ./n(n -

-J./n(n -

1)

+ In(n + 1) _

1) -

1)2

( vn+l + .fi/f (2n - 1 + 2Jn(n -

1)

In(n + 1)

.fi/)2

34. Class Invariants

-2j(2n + l)Jn(n + 1) + (2n - 1)Jn(n - 1) - 4n) .

287

(9.15)

Comparing the proposed value of u from (9.13) with that of (9.15) above, we see that it suffices to show that

2j(2n + l)Jn(n + 1) + (2n - Ih/n(n - 1) - 4n =

(../2n + l)j2~(..!n+l- h) + (../2n -l)Jr-2~-n-(-,;-n-+-l-+-h-2-). (9.16)

If we square both sides of (9.16), it is a routine matter to show that (9.16) indeed is a correct equality. This therefore completes the proof.

The next theorem enables one to determine asp from the value of a2p. Theorem 9.6 (p. 82, NB 1). lin

~

1, P > 0, and

e-7r ./2p = F (..!n+l- ~)2(~ - "1n~1)2),

(9.17)

then

e-"'''''' = F ({ -'" + 1~,;;i+T X

{ ,In -

1 + In"TI

...ti 2

J(-'" +

1)(-," + ,;;i+T)

r

- V/(~- 1)(~ +..!n+l) }4) .

(~1~

Observe that

{ ,In±1+Jn"TI ...ti - V/(~ ± 1)(~ + ..!n+l) } x

/ } { ,In±I+Jn"TI ...ti + V (~ ± 1)(~ +..!n+l) =

1.

(9.19)

Thus, ifJ2 P can be expressed as a product of units of the form (In"TI - ,In)2 (,In - n - 1)2, then asp can be expressed as a product of two fourth powers of units. Before proving Theorem 9.6, we present three examples recorded by Ramanujan. Examples 9.7 (p. 82, NB 1). We have

as = (

a24 =

(

J3 + 2h - J2 + 2h) j 6 + 3J3 - J5 + 3J3)

4,

4 (

J

2+

J3 -

J

1+

J3) 4,

288

Ramanujan's Notebooks, Part V

and a40

= (2./2 +

.J5 -

2J3 + JIO) 4 (./2+

.J5 -

J6 + 2JIO)

4

Proof. Let p = 1. Then from Theorem 9.2, a2 = (./2 - 1)2. Thus, n = 1, and from Theorem 9.6,

(9.20) But from (9.5),

/

V 3 + 2./2 =

~ Vf3+T ~-2- + V~-2- =

./2 + 1.

Using this in (9.20), we achieve the desired representation of ag. Let p = 3. From Theorem 9.2, a6 = (2 - ,J3)2CJ3 - ",fif. Thus, n = 3, and from Theorem 9.6,

au

~

r

e:: -

J5+ 3v3)' (' : : - JI + v3

(9.21)

But from (9.5),

and

J2 +

J3 =)2 +

2

1 +)2 - 1 2

= J3 +

",fi

1.

Using these calculations in (9.21), we complete the verification of Ramanujan's representation for a24. Let p = 5. From Theorem 9.2, alO = (v'iO - 3)2(3 - 2",fi)2. Thus, n = 9, and from Theorem 9.6, a40

= (2./2 +

.J5 -

J4(3 + JIO) )

4 (

./2 +

.J5 -

J2(3 + JIO) )

4 ,

which is what is claimed. Proof of Theorem 9.6. From (9.17),

e- 2rr .J2jj

=

F2

(~ _

In)2(Jn _

v'n=-lY) =: F2 ( (1 +4xx)2 ).

Hence, as in (9.14),

2

----;:==---==----=--==- = (,In+T - .jn)(.jn - .;n-=-T)

1

Jx + -IX

-

(9.22)

34. Class Invariants

289

and, by (9.18) and (9.12), it suffices to prove that

x

~ {.;n + I;'v'ftTI - J(.;n + Il(.;n + .;n+I{ X

+v'n+T - y/(.,fn { -J1I- 1.Ji

1)(.,fn + In+i)

12 (9.23)

Let u = ,.fX. In view of the form (9.22), it is natural to assume that u

=

b l )(a2

(al -

b2 ),

(9.24)

- b~.

(9.25)

-

where

a~ - bf

= 1 = a~

Then, by (9.22) and (9.24),

(v'n+! + .,fn)(.,fn + v'n=l) = ~ If s := are

(u +~) =

ala2

+b]b2.

(9.26)

-J1I + v'n+T, the values of ai, b l , a2, and b 2 that satisfy (9.25) and (9.26) a2

=

s- 1

.Ji'

Then, as already observed in (9.19), (9.25) is satisfied. Furthermore, ala2

+ b l b2 = n + In(n + 1) + (.,fn + v'n+!)v'n=l = (In"+l + .,fn)(-J1I + In"=i),

and so (9.26) is satisfied. Hence, (9.23) has been shown, and the proof of Theorem 9.6 is complete.

Singular Moduli for Odd n From (1.6), we find that, in the notation of Theorem 9.1, 1 2g12 = 2U 2 U 2 = - - - .fii;..

(9.27)

Fn

n

By elementary manipulation, we find from the other equality of (1.6) that .

12

i

2lGn = 20t G12 n

2otnG~2

(9.28)

n

If we set

and

r,;;; '=

YU n

·

2otnG~= ., I

290

Ramanujan's Notebooks, Part V

then (9.28) takes the fonn 2(g~)12 =

1

..;a;. -

.;a:..

(9.29)

Comparing (9.27) and (9.29), we deduce the following theorem from Theorem 9.1. Theorem 9.8. Set (g~)6 = UV,

u 2 + l/u 2 = 2U, W

v 2 + l/v2

= J U2 + V2

= 2V,

- 1,

and 2S = U

+ V + W + 1.

Then a~ =

{vis X

~/S=-1}2{VS=U - ~S - U - l}2

{~ - ~ S

- V - l} 2 {~ S - W - -.;'-=S----=W:-:-----:-l} 2 .

The next theorem gives the twelve values of an, when n is odd, that are found in Ramanujan's notebooks. In those instances when two representations are given, the former one is that which is in the notebooks, or that which contains the units provided by Ramanujan in his notebooks. The Borweins [1, pp. 139, 151] calculated an, for n = 3, 5, 7, 9, and 15, and Ramanathan [1] detennined an, for n = 3, 7, 9, and 15.

.J3

Theorem 9.9 (pp. 80, 345, 346, NB 1; p. 262, NB 2). We have a3

=

a, ~

H-'52-1r(J3+

2--4-'

4-'5

-

j -'54-I) ,

=~(0-1)3(-'5+I_j-'5+lr 2 2 2 2' 8 - 3-/7 16

a 9

_

=~(.J3_1)4( ~ ~8 2.j2

y~

V~)

34. Class Invariants

291

292

Ramanujan's Notebooks, Part V

and a55

=

116

x

(../5 - 2)

2

(10 - 3m)(3../5 - 2m)

(t+80 -)°8-1)" (J4+20-)2+2°)'

Proof of Theorem 9.9 for n = 3, 5, 7, 9, and 13. These five values are easily computed by using Theorem 9.8. The required values for G n may be found in the table in Section 2. In each instance, we list the values for u, v, U, V, W, and Sin the table below. The reader should easily be able to verify the calculations. n

u

v

U

V

3

exp(Jri/4)

.fi

0

5

exp(Jri/4)

J2 +-/5

4

0

7

exp(Jri/4)

23 / 2

0

16

16

2

9

exp(Jri/4)

2+ v'3

0

7

4v'3

4+ 2v'3

13

exp(Jri/4)

J18 + 5v'I3

0

5v'I3

18

!(19 + 5v'I3)

5

W 3

S 3

4

2

-/5

2

!(3 +-/5)

65

63

9

Except for n = 55, we have also used Theorem 9.8 to calculate the remaining values in Theorem 9.9. However, the following lemmas lead to simpler calculations. Lemma 9.10. lfr is any positive real number and t

= .../(r + 1)/8, then (9.30)

Proof. The equality (9.30) can be readily verified by elementary algebra. Lemma 9.11. lfr and t are as given in Lemma 9.10, then

(9.31)

r

Proof. It is readily verified that

(JJt

\! + 1 -

JJt

34. Class Invariants

~ Jt +! - Jt - !.

+} - 1

293

(9.32)

Using Lemma 9.10 in (9.32), we deduce (9.31). We frequently set G = G n below, when the value of n is understood. Proof of Theorem 9.9 for n table in Section 2,

= 5, 9, 13, 15, 17, and 25.

Let n

= 5. From the

e~2 ~ ( ,/52+ 1)' ~,/5 + 2.

(9.33)

Ifr = G~2inLemma9.IO,then

t=/3+v'S = 8

v'S+1 4

and

e" - ,fe2L 1

~ (1,/54+ 3 _/,/54- ~)

4

(9.34)

Thus, the given value for as follows immediately from (9.3), (9.33), and (9.34). Let n = 9. From the table in Section 2,

el2 ~ (

"':,n ~ 4

7+ 4.'3.

(9.35)

Applying Lemma 9.11 withr = G~2, we find that t

Jt + ~ and

e" - ,fe2L

= /8 + 4../3 = ../3 + 1. 8

=

2

/2 +2../3 = 1+2../3,

1~ (/3 +4.13 -/.134- 1)'

Thus, by (9.3), (9.35), and (9.36), we deduce Ramanujan's value for a9'

(9.36)

294

Ramanujan's Notebooks, Part V

Let n

=

13. From the table in Section 2, 12

Gl3 = Then in Lemma 9.10, set r =

t=

(

.Jf32 + 3)

= 18 + 503.

(9.37)

Gg to deduce that

J

19 + 5.Jf3 =

and

a" -

3

Jc"-l =

8

5 + .Jf3 4

(J7+;U -J3+;Ur

(9.38)

Hence, the given value for al3 follows from (9.3), (9.37), and (9.38). Let n = 15. From the table in Section 2,

GJ; = 4(v'S + 1)4 = 28 + 12v'S. Apply Lemma 9.10 with r

(9.39)

= Gg. Then

t=J29+;2y'5, and so

aU - "Ia2L

1= (

=

(J29+;2./5 _~2

= 28 + 12v'S -

J

(29 + 12v'S)(27 + 12v'S)

= 28 + 12v'S - 16J3 = (2 - J3)2(4 -

-7M

M).

Hence, by (9.3), (9.39), and (9.40), the desired result follows. Let n = 17. From the table in Section 2,

(9.40)

34. Class Invariants

295

after a lengthy calculation. We now apply Lemma 9.10 with r = 20 + 5./Pi. Then

t= and so G- 12 =

u

Next, set r

J21 + 5./Pi = 5 + ./Pi, 8

(t+..m -J3+..m), 4

= 20 + 5./Pi + 21

t=

2

=

JG"

(20 + 5../f7)2 - 1 in Lemma 9.11. Then

J

(20 + 5../f7)2 - 1

42 + 1O./Pi +

4J206 + 50./Pi 16

4

J5 +./Pi + 2J4+./Pi = 1 +J4+ ./Pi. 4

Thus,

G" -

J

+ 5./Pi +

=

It + ~

(9.41)

4

8

=

and

4

-I =

2

(J3+ J:+..m -JJ4+:?

r

-I

r

(9.42J

Using (9.41) and (9.42) in (9.3), we complete the proof. Let n = 25. From the table in Section 2,

Gll = (02+ 1

= 161+ 720.

With r = G~~ in Lemma 9.11,

t= and

162 + 72./5

8

6 + 3./5 2

(9.43)

296

Ramanujan's Notebooks, Part V

Hence,

G ,JG24 -\ (/5+/5 -l +4../5)' =

12 -

(9.44)

Putting (9.43) and (9.44) in (9.3), we complete the proof. The next lemma will enable us to calculate a210 a33, a4S, and ass. Lemma 9.12. Let r = UV, where v > u and u is a unit. Set u = u\ u I, U2 > 0 and u~ = 1. Furthermore, let

ur -

a 2 = 1 + 2vul

+ v2

b 2 = 1 - 2vu\

and

+ U2, where

+ v 2,

where a, b > O. Then

r _~ =

(

x

fa +b+2 + 1 fa +b+2 _! V 16 "2 - V 16 2

(Na-b+2 16

4

~_Na-b+2-~r 16 2

+2

Proof. The right side of (9.45) equals

Set

r' a ~ =

2

b

2

+/

(a ~ 2

b

2

Y

Then, by an elementary calculation,

a J(a--2- b)2

r;;--; +b vrl2 -1 = -2-

-1

a- J(a--2+b)2 -1.

+ -2-b

(9.45)

34. Class Invariants

Hence, we see that the right side of (9.46) equals r' remains to show that r = r'. In fact, from the definitions of a 2 and b2 ,

r' =

VUI

= VUI

+

JV 2UI

+ vJur -

+ 1 - (1 + 1 = VUI

';r ll -

297

I, and it therefore

v 2)

+ VU2 = VU =

r,

which completes the proof. Proof of Theorem 9.9 for n = 21, 33, 45, and 55. Let n = 21. From the table in Section 2,

12 r := G 21

=(3 . ,+f i../7)2 (../7 +2 ../3)3 = (8 + 3./7)(2./7 + 3../3).

= 2../7 and v = 8 + 3../7 in Lemma 9.12. Then a 2 = 1 + 2(8 + 3./7)2./7 + (8 + 3./7)2 = 212 + 80./7 = (10 +

(9.47)

Set UI

4./7)2

and

= 1 - 2(8 + 3./7)2./7 + (8 + 3./7)2 = 44 + Hence, a = 10 + 4../7 and b = 4 + 2../7. Moreover, b2

/a+b+2

V

16

=

J16 + 6../7 = 16

16./7 = (4 + 2./7)2.

3+../7 4

and

Ja-b+2 =J8+2../7 16

16

= 1+../7

4'

Thus, by Lemma 9.12,

G" - JG24 - b

(15+/ -J1+/7)' (J3+/ -J.;74-

1) '

(9.48) On using (9.47) and (9.48) in (9.3), we complete the proof. Let n = 33. From the table in Section 2,

G:l ~ (2 + ,(3)3 ( v'1~ 3)' ~ (26 + 15,(3)(10 + 3,fU).

(9.49)

Apply Lemma 9.12 with Ul = 10 and v = 26 + 15../3. Then

a2

=

1 + 2(26 + 15./3) 10 + (26 + 15./3)2

=

1872 + 1080./3 = (30 + 18./3)2

298

Ramanujan's Notebooks, Part V

and b 2 = 1 - 2(26 + 15,J3) 10 + (26 + 15,J3)2 = 832 + 480,J3 = (20 + 12,J3)2.

Thus, a

= 30 +

18v'3 and b

= 20 +

Ja+b+2 = 16

12v'3, so that 52+30v'3 = 5+3v'3 16 4

and

Thus, by Lemma 9.12,

G"-JG24-1

~ (J7+:./3 -/3+:./3)' (JS+4./3 -/I+4./3)' (9.50)

Upon substituting (9.49) and (9.50) in (9.3), we complete the proof. Let n = 45. From the table in Section 2,

GlJ ~ (,fS + 2)3 (

,fS:n../3)' ~ (38+ 17,fS)(31

+8,(15).

(9.51)

We apply Lemma 9.12 with Uj = 31 and v = 38 + 17J5. Thus, a 2 = 1 + 2(38 + 17.J5)31 + (38 + 17.J5)2 = 5246 + 2346.J5 = (51 + 23J5)2

and b 2 = 1 - 2(38 + 17.J5)31 + (38 + 17.J5)2 = 534+ 238.J5 = (17 + 7.J5)2.

Hence, a = 51 + 23J5 and b = 17 + 7J5, so that Ja+b+2= 16

70 + 30J5 16

Ja-b+2 = 16

36+ l6J5 2+J5 = 16 2

5+3J5 4

and

So, by Lemma 9.12,

G"-./G'4-1 ~

(J7+:,fS _/3+:,fS)' (J3+ ,fS - /I+2,fS) , 2

(9.52) The desired evaluation now follows immediately from (9.3), (9.51), and (9.52).

299

34. Class Invariants Let n

= 55. From the table in Section 2,

GIl ~ S(-is + 2)2

(t +S-is +v ~)

~12

~ S(-is +2)2 ( 99+ ;S-is + J (99 +;S-is)' _}

(9.53)

Apply Lemma 9.12 with UI = (99 + 45v's)/4 and v = 8(v's + 2)2. Thus,

a 2 = 1 + 2· 8(.J5 + 2)2~(99 + 45.J5)

+ 64(.J5 + 2)4

= 17469 + 7812.J5 = (93 + 42.J5)2 and b2 = 1 - 2· 8(.J5 + 2)2~(99 + 45.J5)

+ 64(.J5 + 2)4

= 3141 + 1404.J5 = (39 + 18.J5)2. Thus, a

= 93 + 42v's and b = 39 + 18v's, so that /a

V

+b +2

=

16

/67 + 30v's 8

and

r

Thus, by Lemma 9.12,

G" - ,fG2L]

~( x

/67+ :O-is +

4- J/67+ :O-is - 4

(/4+2-is _/2+2-is)

4

(954)

Now, by Lemma 9.10,

(

/67+:0-is+~-J~-~r

= 66 + 30.J5 =

J(66 +

66 + 30.J5 - 9v'55 -

30.J5)2 - 1

20~

= (10 - 3~)(3.J5 - 2~).

(9.55)

300

Ramanujan's Notebooks, Part V

Using (9.55) in (9.54) and then (9.53) and (9.54) in (9.3), we complete the proof. We close this subsection by showing how different modes of calculating an can lead to interesting identities between radicals. Entry 9.13 (p. 311, NB 1). We have

~1 ± Y 1 /

(a)

J5-1.:t5±1

G

('" 5 - 2)8 = - - - = 2 .fi

and (b)

Proof. From the table in Section 2, G 25 = (J5 (9.3), we easily find that (2a 25 )1/8 =

/1- J1 - (-./5 -

On the other hand, from Theorem 9.9,

(2a,,)

+ 1) /2, and using this value in

'I' = ~ 161 - 72./5

(J

J

5+4./5 -

=

J./5 - 2 ( 5+4./5

=

(./52- 1)'"

-l

2)8.

J1+4./5) +4./5)

(J5+ ./5 - J1+,./5) 4

= J52- 1 (:;; _

~).

Combining these two calculations, we deduce (a) with the minus signs chosen on each side. From the table in Section 2, G9 = + .j3)/.fiy/3 , and with this value in (9.3), we readily find that

((1

(2a9)1/8 = /1 -

J1 - (2 -

,,/3)4.

34. Class Invariants On the other hand, from Theorem 9.9, ('la,)l/'

~ /.r~ 1 =

2:/

4 (

301

(/3+•

./3 _ / ./3.- 1)

J2./3 - (./3 - 1)) .

Combining the last two calculations, we obtain (b) with the minus signs chosen on both sides. We verified via Mathematica that both (a) and (b) also hold with the plus signs chosen on each side.

Calculating a2n and a3n with the Help of Modular Equations Ramanujan obscurely described two further methods for calculating an in his first notebook [9]. In the first, Ramanujan indicated that a2n may be calculated by solving a certain type of modular equation of degree n. For several prime values of n, the desired form of modular equation exists; many of these modular equations can be found in Ramanujan's notebooks and are proved in Part TIl [3]. This very novel method is the only known method that does not require a priori the value of g2n. Thus, the method is a new, valuable tool in the computation of a2n. In the second, Ramanujan disclosed a method for determining a3n arising from the definition of a modular equation of degree n. We give a rigorous formulation of this formula and prove it by using a device introduced in Section 6 to calculate certain class invariants. The results in this subsection are due to the author and Chan [3]. In this paper we also devise a method for calculating aSn that is similar to Ramanujan's method for determining a3n' In the middle of page 292 in his first notebook, Ramanujan claims that, "Changing fJ to 4B /(1 + B)2 and a to 1 - B2 we get an equation in 4B(l - B)/(l + B) and the value of B2 is for e- n .f2ii." We now state and prove a rigorous formulation of this assertion.

Theorem 9.14. Let fJ have degree n over a, and suppose a and fJ are related by a modular equation of the form F (afJt, {(l- a)(l- fJ)Y) = 0,

(9.56)

for some polynomial F andfor some positive rational number r. Ifwe replace a by 1 - x 2 and fJ by 4x/(l + x)2, then (2.1) becomes an equation of the form G(x) := g({4x(l - x)/(l

for some polynomial g. Furthermore, x =

+ x)}') =

~.

0,

is a root of this equation.

302

Ramanujan's Notebooks, Part V

Proof. Under the designated substitutions,

F( (afJ)"

{(l - a)(1 - fJ)}')

= F (((I-X 2)

(1

:xX)2

y,

(X2

(1-

(1 :xX)2)

y)

= F ((4X(1 - X»)r ,_1 (4X(1 - X»)2r) l+x 42r l+x

=0. Hence, the first part of Theorem 9.14 follows. For brevity, let 2F\(~,~; 1; x) = 2F\{X). Setting fJ that zF\

2F\ (l - fJ) =

(1 _

(l

F ( 2

4x

+ X)2

)

4x ) (1 + x)2

\

=

=

4x/{l

+ X)2,

1 2F\ (l - x2) 2F\ (x 2) ,

2:

we find

(9.57)

by a fundamental transformation for F(x) (part III [3, p. 93]), which actually arises from a special case of Pfaff's transformation. With a replaced by 1 - x 2 , we find from (1.4), (2.5), and (9.57) that

Therefore, 2 F\{l-x 2 ) ~ --:::--:---;:--- = v 2n. 2 2F\ (x )

Now recalling the definition of a singular modulus and (1.2), we deduce that = a2n, and the proof is complete.

x2

Example 9.15(a). Let n = 3. Then (afJ) \/4

+ {(l

- a)(1 - fJ)}\/4

=

1,

(9.58)

which is originally due to Legendre and was rediscovered by Ramanujan (Part III [3, pp. 230, 232]). With Ramanujan's substitutions, (9.58) takes the form u

\ 2 + 2u =

1,

(9.59)

where u = (4xl-X)I/4 l+x

(9.60)

Solving (9.59), we find that u = ,.j3 - 1. Then solving (9.60), we find that = 2,.j3 - 3 - 2,.,fi + ../6. Using two different modes of calculation, we find

x

34. Class Invariants

303

that a6 =

X2

= (2 _ .J3)2(.J3 _ ../2)2 = 2.J3 +..(6 - 3 -

2..ti.

2v'3 + ..(6 + 3 + 2..ti

The fonner representation for a6 can be found in Theorem 9.2.

Example 9.15(b). Let n = 5. Then we have Ramanujan's modular equation of degree 5 (Part III [3, p. 280]), (afJ)1/2

+ {(l -

a)(1 - fJ)}I/2

+ 2{16afJ(1 -

a)(1 - fJ)}1/6 = 1.

(9.61)

Using Ramanujan's substitutions, we find that (9.61) can be put in the shape 12 + 3 1 2+2U = 4U 1, (9.62) u + 4U U = where

I-X)I/2

u= ( 4x--

(9.63)

I+x

Solving (9.62), we deduce that U = -6 + 2v'iO. Next, solving (9.63), we find that x = 3v'ill- 9 - 4J5 + 6..ti. Lastly, by two distinct routes for calculation, alO = x 2 =

(50 _ 3f(3 _

6..ti - 9 3v'ill + 6..ti + 9 +

2../2)2 = 3v'ill +

4J5.

4J5

The fonner representation is given in Theorem 9.2. We next derive Ramanujan's formula for a3n. Let q be given by (1.5), and suppose that fJ has degree n over a. Thus, (1.4) holds. Now suppose also that fJ has degree 3 over 1 - a =: a'. Then, by (1.4), 3

__ 2FI (l - fJ) __ n 2FI (I - a)

2FI (a) 2FI (I - a)

2FI (fJ)

(9.64)

2FI (a)

Hence, (9.65) and, from (9.64) and (9.65), 2 FI ( l -

fJ)

2FI(fJ)

= 3 ~ =-J3n.

Next, from Part III [3, p. 237], since parametrizations

a'=

p

(~)3 1 +2p

(9.66)

V"3

fJ

has degree 3 over

and

fJ

a', we have the

P ) = P3(2+ 1 + 2p

,

(9.67)

where 0 < p < 1. It follows from (9.67) that (1 _ a')fJ = (1 - p)3(1 + p) (1+2p)3 P

3( 2 + p ) 1+2p

= (

1- P PI+2p

)3 (1 + p)(2 + p) 1+2p

(9.68)

304

Ramanujan's Notebooks, Part V

and (1 _ fJ)a'

= (1 -

(2 + )3 =

p)(1 + p)3

(1 +2p)

P 1 +2p

P

1- P

PI +2p

((1 +

p)(2 + p»)3 1 +2p (9.69)

Next, set

I-p 2t=p-1 +2p

(9.70)

and observe that

=

2(1 _ t)

(1

+ p)(2 + p) .

(9.71)

1 +2p

It follows from (9.68}-(9.7I) that

afJ

= (1 -

a')fJ = 16t 3(1 - t)

(9.72)

and

fJ)

(1 - a)(1 -

= (1 - fJ)a' = 16t(I - t)3.

(9.73)

Now, set

k = 4t(I - t).

(9.74)

Observe from (9.72), (9.73), (1.6), (9.65), and (9.66) that k =

(2G~/3G~nrl

(9.75)

.

We determine fJ (a3n) as a function of k. From (9.74), we find that, with no loss of generality in choosing the minus sign in the first equality below, t

1 -,Jf"="k

= ---'-2--

and

I-t= I+,Jf"="k. 2

Thus, by (9.72) and (9.73), we find that (1 - a')fJ

=

(1 - .J1=k)3(1

+ .J1=k) =

k(1 - v'T=k)2

(9.76)

+ v'T=k)2.

(9.77)

and

+ .J1=k)3 =

(1 - fJ)a' = (1 - v'T=k)(I

k(I

Subtracting (9.76) from (9.77), we find that

a' =

4kv'T=k + fJ.

(9.78)

Substituting (9.78) into (9.77), we deduce that

fJ2 + fJ(4k.J1=k -

1) + k(1 - .J1=k)2

= o.

34. Class Invariants

305

Thus,

a3n =

f3

=

1 - 4k,Jf=k ± ../(4k,Jf=k - 1)2 - 4k(1 - ,Jf=k)2

2

1 - 4k,Jf=k ± (1 - 2k)~ 2 =

1-JI-4k(1-2k±v'(1-k)(l-4k»)2 2

(9.79)

This last formulation was that given by Ramanujan. We now resolve the sign ambiguity in (9.79). It is clear that both a' and f3 satisfy the equation

y2

+ y(4kv'l'=k -

1) + k(1 - v'l'=k)2 = O.

(9.80)

Since a' = 1 - a, it follows from (9.65) and (9.66) that the solutions of (9.80) are a3n and a3/n' Thus, it suffices to show that (9.81) for n > 1. From the definition of rp in (2.4), it is obvious that

rp2(e- n.fJTn) > rp2(e- n.J3il). From (2.5), it follows that

2F\ (a3/n) > 2F\ (a3n)' Since 2F\ (x) is increasing on (0, I), (9.81) follows. Thus, we have proved the following theorem.

Theorem 9.16. Let q be given by (1.5), suppose that f3 has degree n over a, let f3 have the parametrization (9.67), and define t by (9.70). Then, if k is defined by (9.74), ll3n has the representations given in (9.79), where the minus sign must be chosen. Ramanujan's formulation of Theorem 9.16 at the bottom of page 310 in his first notebook is a bit different. He first gives (9.72) and (9.73), but with the left sides switched. He then states (9.70), followed by the equality

F (p3 2 + p ) = e-n.J3il 1 +2p , which is a consequence of (9.66) and (9.67), where F is defined in (2.3). He concludes by defining kin (9.74) and by claiming that a more complicated, somewhat ambiguous version of the right side of (9.79) equals e- n.J3il, i.e., he forgot to write "F" in front of the right side of (9.79). (In recording specific values of F(a,,), Ramanujan frequently omitted parentheses about the arguments.)

306

Ramanujan's Notebooks, Part V

Example 9.17(a). Let n = l. Then, from the table in Section 2, G 3 = 21/12 = G 1/3 , since G n = G lln (Ramanujan [3], [10, p. 23]). Hence, by (9.75), k = ~, and by (9.79), (\(3

=

2 -,J3 --4-'

which is given in Theorem 9.9. Example 9.17(b). Let n = 5. Then, from the table in Section 2, GIS 2- 1/12 (v's + 1)1/3. It can also be verified that G S/3 = 2- 1/12 (v's - 1)1/3. Hence, it is easily seen from (9.75) that k = rt;. Therefore, from (9.79),

16 - 7,J3 - y'f5 32 which is simpler than the formula given in Theorem 9.9.

10. A Certain Rational Function of Singular Moduli Set b 2 := b~ := fJ, where fJ has degree 2n and b > O. On page 312 in his first notebook, Ramanujan defines

u:=

Un

:=

b(l - b)

and

U := Un :=

(10.1)

1+b

~ (U + ~ -

2) .

(10.2)

(In fact, Ramanujan uses the notation fJ, instead of b above, and he has no notation for the right side of (10.2).) He then provides the following table. n

Un

1 3 5

1

7 9

3 9 (v2 + 1)2(1

+ 2v2)

11

49 99

15

3(5 + 4v2)2

23

9(1

+ v2)4(3 + 4v2)

29

99 2

35

+ 5v'sP 9(1 + v2)1O(2v2 + 1)2(6v2 + 1)

71

63(8v2

34. Class Invariants

307

In this section, these eleven values are established. It is unclear to us why Ramanujan studied this particular function Un. Recall from (1.6) that g := g2n = (

(l - fJ)2) 1/24

(10.3)

4f3

It follows that (lOA)

Hence, from (1O.l}-(1OA), U = ~ (b(1 - b) 4 1+b =

2..(fJ (l/b -

1 - f3

2)

1+b _ = (1 + b2)2 = ~ (1 + b2)2 + b(1 - b) 4b(1 - b2 ) 1 - b2 8b 2 b)2 + 4 =! -12 (4 24 + 4) = ! ( 12 + -12) 8g

8

g

2

g

g

.

(l0.5)

Equality (10.5) will now be utilized to calculate Un for eleven values of n. All values for gn below can be found in the table in Section 2. First, g2 = 1, and so, by (10.5), UI

Second, since g6

=

!(1

+ 1) =

= 2- 1/ 4(4 + 2./2)1/6,

1.

we find that

g±12 = 3 ± 2./2.

Thus, from (10.5), U3

Third, gIO = J(.j5

= ! (3 + 2./2) + (3 -

2./2»)

= 3.

+ 1)/2, and so g±12

= 9±

4.J5.

Thus, from (10.5), Us

Fourth, gl4 =

= ! (9 + 4.J5) + (9 -

J(1 +

./2 +

g~12 =

11

J2./2 -

ifs») = 9.

1) /2, and so

+ 8./2 ± (8 + 6./2)J 2./2 -

1.

Thus, from (10.5), U7

=

11

+ 8./2 = (./2 + 1)2(1 + 2./2).

Fifth, gl8 = (./2 + .J3)1/3. Thus, g±12 = 49 ± 20.J6,

308

Ramanujan's Notebooks, Part V

and so, from (10.5),

Sixth, g22

= J,.[i + 1, and so g±12 = 99 ± 70h.

Hence, from (10.5), UII =99. Seventh,

g30

= (2 + .J5)1/6(3

+ vl0)1/6. It follows that

g±12 = (9 ± 4.J5)(19 ± 6M). Hence,

UI 5 Eighth, g46 =

= 171 + 120h = 3(5 + 4h)2.

)(3 +,.[i + J7 + 6,.[i)/2. Thus,

g±12 = {18

+ 13h ± (5 + 3h)/7 +6h}2

It follows that

U23 = (18 + 13h)2 = Ninth,

g58

+ (5 + 3h)2(7 + 6h) = 9(147 + 104h) 9(17 + 12h)(3 + 4h) = 9(1 + h)4(3 + 4h).

=

)(5 + m)/2. Thus, g±12 = (70 ± nJ29)2,

and so U29

Tenth, g70 = )(3

= 702 + 13 2 ·29 = 9801 = 992.

+ .J5)(1 + ,.[i)/2. It follows that g±12 = (9 ± 4.J5)2(7 ± 5h)2.

Thus, from (10.5), U 35 = 63(253

+ 80M) =

Eleventh, gl42 = )(9 + 5,.[i + J127

63(Sh + 5.J5)2.

+ 90,.[i)/2. So,

g±12 = {1026 + 725h ± (65

+ 45h»)127 + 90h}2

34. Class Invariants

309

Thus, from (10.5),

U7l = (1026 + 725.J2)2

+ (65 + 45.J2)2(127 + 90.J2)

= 9(467539 + 330600.J2) = 9(3363 + 2378.J2)(57 + 58.J2) = 9(1

+ .J2)1O(2.J2 + 1)2(6.J2 + 1).

11. The Modular j-invariant Except for four entries discussed in Section 13 of Chapter 33, the last two pages, 392-393, in Ramanujan's second notebook are devoted to values of the modular jinvariant. Recall (K. Chandrasekharan [I, p. 81], Cox [1, p. 224]) that the invariants 1 Cr:) and j ( r), forr E 1HI, are defined by l(r)

where, for q

= gi(r)

= 17281('r),

and

j(r)

(mr

+ n)-4 = 3

8(1')

= exp(27ri1'), 00

L

47r4

M (q),

m,n=-oo (m,n)'I(O,O) 00

L

(mr

+ n)-6 =

87r 6 27 N(q),

m,n=-oo

(m,n)'I(O,O)

and 8(r)

= gi(r) -

27gi(1')

M3(q) N2(q)'

= 1728 M3(q) _

(11.1)

Here M (q) and N (q) are the Eisenstein series defined at the beginning of Section 4 of Chapter 33. (See also the beginning of Section 7 in this chapter.) Furthermore, the function Y2(r) is defined by (Cox [I, p. 249]) Y2(1') =

l j (r),

(11.2)

where that branch which is real when r is purely imaginary is chosen. At the top of page 392, which inexplicably is printed upside down, Ramanujan defines 1 := 1n and U := Un by 1 6a n....:..(I_-_ct"""n:7) ], _ _1_-_ ----C n - 8(4a n (1 - ctn»I/3

and

(11.3)

where n is a natural number. (To avoid a conflict of notation later, we have replaced Ramanujan's tn by un.) For 15 values of n, n == 3 (mod 4), Ramanujan indicates the corresponding values for 1n , although not all values are explicitly given. In

310

Ramanujan's Notebooks, Part V

each case, the value had been given in the literature or can be readily deduced from results in the literature. Now, from (1.6) and (11.3), we easily find that

s 24 ) Jn = IG S n (1 - 4Gn'

(11.4)

We now identify I n with Y2. From Cox's text [1, p. 257, Theorem 12.17], for q = exp(2:n'i-r), Y2(-r) = 2

Setting -r

= (3 +

Fn)/2, + Fn)

sq2/3 j16(_q2) fI6(_q)

fS(-q)

+ ql/3 fS(_q2),

(11.5)

we deduce from (11.5), (6.1), and (1.3) that

3 Y2 (

2

= 28

26G~4

24GI6

= -4G8(1 _ 4G-24). n

n

n

Hence, from (11.4) and (11.2),

J

=_~ (3+Fn)=_~r(3+Fn). 32 Y2 2 32 VJ 2

n

(11.6)

There are 13 cases when the class number of the order in an imaginary quadratic field equals 1 (Cox [1, p. 260]). In such an instance, the value of the j-invariant is known to be a rational integer (Cox [1, p. 261]), and we first tum to Ramanujan's seven values in these cases. Formula (11.3) can be used to calculate some values, but in most instances the value of G n is unavailable.

Entry 11.1. h = O. Proof. From the table in Section 2, G3 = 2 1f\2. Thus, by (11.4),

h = k(G~ - 4G 316 )

= k(22/ 3 -

4· r

4/ 3 )

= O.

Entry 11.2. h7 = 5 . 31/ 3 • Proof. From the table in Section 2, G27 = 21/12(21/3 _1)-1/3. Thus, from (11.4), fl.7 =

;3

(4(21/3 -

1)-S) (1

- (21/3 -

I)S)3 =

375,

with the help of Mathematica. For the last five cases of degree 1, we simply use the relation (11.6) in conjunction with the table in Cox's book [1, p. 261].

Entry 11.3. Jll = 1. Entry 11.4. J I 9 = 3. Entry 11.5. J43

= 30.

34. Class Invariants

311

= 165.

Entry 11.6. 167

Entry 11.7. 1 163 = 20,010. There are a total of 29 cases when the degree of j (3 + ..;=-:ti) /2) equals 2 CW. E. Berwick [1]). Ramanujan gives the values of 1n in six of these cases. We will verify Ramanujan's values by simply referring to the tables in Berwick's paper [I, pp. 55, 57-59].

Entry 11.S. h5

v'5+1)4 = v'5 ( 2-

Entry 11.9.

= 3 ( 07 + 4)

151

2/3

(5 +07)

2'

Entry 11.10. h5 = 3 .5 1/ 6 ( 69 + 231v'5) .

r

Proof. In Berwick's table on page 58, we need the values of E (given on page 55) and y (given at the top of page 57). Thus,

1"

= 3·5'/6 (,/52+ I

(3 ,/52+ I- I)

= 3 . 51/ 6(9 + 4v'S) (I + ~v'5) = 3.51/6 (69 +~1v'5) . Entry 11.11. 191 = ~(227 + 63-J'I3). Proof. Proceeding with the same tables as in the preceding proof, we find that J" =

3(

Entry 11.12. 199

~+ 3)' ('~ - 7) =

= ~(23 + 4-J'3'3)2J3(77 +

j(227+ 63,,'l3).

15.J33).

Proof. From the work of Berwick [1, pp. 55, 58],

199 = (23 + 4v'33)2/3 ~ - 5 (2.J33 + 11)(.J33 + 4)

=

~(23

+ 4v'3'3)2/3(77 + 15-J'33).

The value of 199 was not given by Ramanujan; he wrote" 199 = .... "

312

Ramanujan's Notebooks, Part V

Entry 11.13. 1115 = ~(785

+ 351.J5).

Proof. From pages 58,55, and 57 in Berwick's paper [1],

1115 = 3.J5 (

.J5 + 1 2

) 10 (

3.J5 2

1

)

3

= 2(785

+ 351.J5).

Ramanujan misrecorded the value of 1115 as

~(785 + 341.J5). Lastly, Ramanujan discusses two values of gree 3.

In when j (3 + Fn)/2) has de-

Entry 11.14. Let Un be defined by (11.3) and define p > 0 by U59

=

(1 _ p8)3

Then p9 _7 p s + 22p7 _ 34 p 6 +40p 5 -28 p 4+22 p3 _lOp2 + IIp -1 = O. (11.7) Furthermore, u~'3 satisfies an irreducible cubic polynomial over Q. Proof. We turn to the work of A. G. Greenhill [1] who sets (with notation altered so as not to conflict with that of Ramanujan) an

=

(1 - 16an (1- a n

»3

(11.8)

Comparing (11.8) and (11.3), we see that an = 1281;. Thus, from (11.3), Un = Then Greenhill sets (p. 311)

an.

and shows that p is a root of (11.7). Greenhill [1] claims that a := a59 satisfies a cubic equation but does not give it. However, in a subsequent paper [2, p. 404], Greenhill gives the equation

a - 392· 21/3a 2/3

+ 1072· 4 1/3a 1/3 -

2816

Entry 11.15. Let Un be defined by (11.3). Define s > 0 by US3

=

(1 - 256s 24 )3 256s 24

and set {J=

1 - 2s - 2s2 - 2s 3 . 2s 3

= O.

(11.9)

34. Class Invariants

313

Then

fJ3 + 4fJ2 + 2fJ - 5 = O. Furthermore, U~~3 satisfies an irreducible cubic polynomial over Q. Proof. Ramanujan's claim, which is incomplete as stated on page 392, is a quotation from Greenhill's paper [I, pp. 312; 313]. In fact, except for US3, the notation is the same. Greenhill [1] indicates that a l/3 := u~;3 satisfies a cubic polynomial, which he gives in his second paper [2, p. 405], namely, (11.10)

which he claims is due to Russell. On page 393, Ramanujan considers various polynomials satisfied by In. The work is divided into three parts. In the first part, Ramanujan gives the following table. Entry 11.16. n

In

11 19

1

43

81n

641; - 241n + 9

+3

3

11 27

49 =7 2 513 = 27·19

30

243 = 27.3 2

56, 889 = 27 . 43 . 7 2

67

165

1323 = 27.7 2

1,738,449 = 27.31 2 .67

163

20010

160,083 = 27.77 2

25,625, 126, 169 =: 27 . 163.2413 2

Table 11.1 Note that (81n

+ 3)2 =

(641; - 241n + 9)

+ 721n.

Observe that, except for n = 11, 641; - 241n + 9 contains n as a factor. Also note that, except for n = 11, both 81n + 3 and 641; - 241n + 9 contain 27 as a factor. The factorizations in the last column are presented as Ramanujan recorded them. Perhaps he failed to observe that 2413 = 19· 127. In the second part of the page, Ramanujan recorded the following remarkable theorem. Entry 11.17. For q

= exp( -T( ,.fii), define t:= tn := -/3q1 / 18f(q1 / 3)f(q3). J2(q)

(11.11)

314

Ramanujan's Notebooks, Part V

Then

t" =

(

2)641; - 241"

+9 -

(161" - 3) )

I~

(11.12)

Entry 11.IS. For the values of n given below, we have the following table of polynomials Pn(t) satisfied by tn: n 11 35 59 83 107

Pn(t) t- 1 {l.+t-l

+ 2t - 1 + 2t1. + 2t 2t1. + 4t tJ

tj t

1 1

J -

Table 11.2 The simplicity of these polynomials is remarkable, since the corresponding well-known polynomials of the same degrees satisfied by In are considerably more complicated, especially in the latter three instances. If n is squarefree, n ::;; 11 (mod 24), and the class number of the Hilbert class field K(l) is odd, thent" and I" satisfy irreducible polynomials of the same degree; for a proof, see a paper by the author and Chan [4]. The form of Entry 11.17 suggests hitherto unknown connections between the jinvariant and Ramanujan's cubic theory of elliptic functions to alternative bases developed in Chapter 33. We defer a proof of Entry 11.17, as a similar result commences the third part of page 393. A proof of the latter claim also depends upon Ramanujan's cubic theory.

Proof of Entry 11.IS. Using Entry 11.3 and Table 11.1, we find that tll

= (2·7 -

13)1/6

=

1,

as desired. Second, from Entry 11.8,

r (,/52+ 1r+9 - (16,/5 (,/52+ 1r-3)r

'" ~ (2 64· 5(,/52+ 1

-24,/5

= (2)7349 + 3276...(5 _ 117 _ 56...(5) 1/6

34. Class Invariants

315

Now, 73492

-

5 . 32762 = 5892 •

Thus, by the denesting theorem (9.5), J7349 + 3276.../5 = J7349; 589 + J7349; 589 = J3969 + J3380 = 63 + 26.../5.

Hence, t35

= (2 ( 63 + 26.../5) - 117 - 56.../5) 1/6 = (9 -

4.../5) 1/6 =

.../52- 1 .

Hence, t35 is a root of t 2 + t - 1, and the second result is established. For n = 59, recall that a59 = -yi/256 is a root of (11.9), where )12 )/2 (3 + J - 59) /2) . Thus, )/2 satisfies (11.13)

yi + 3136yi + 68608)12 + 720896 = O.

Setx:= t 6 := t:andJ:= In.From(I1.12),weseethatxisarootofaquadratic polynomialx 2 +bx+c, withb = 2(16J -3). Sinceb2 -4c = 16(64J 2 -24J +9), we easily calculate that c = -27. Thus, x 2 + 2(16J - 3)x - 27 = O.

(11.14)

This suggests that we make the substitution Y2

= x - 6 - 27x- 1

(11.15)

in (11.13). Upon turning to Mathematica, we find that x 6 + 3118x 5 + 31003x 4 + 25355x 3 =

(x 3

+

13x 2

+ 115x -

l)(x 3

-

837081x 2 + 2273022x - 19683

+ 3105x 2

-

9477x + 19683) = O.

Numerically checking the roots, we see that x is a root of the first factor above, i.e., t l8

+ 13t 12 + 115t6 -

1 = O.

We use Mathematica to factor this polynomial and find that t l8 + 131 12 + 1151 6 x (1 6

-

1 = (1 3 + 21 - 1)(t 3 + 2t + 1) 214 + 21 3 + 41 2 - 2t + 1)(1 6 - 2t 4 - 21 3 + 41 2 + 2t + 1). -

Again, numerically calculating the roots, we find that 13

+ 21 -

1 = 0,

as claimed by Ramanujan. For n = 83, recall that a83 = -yi/256 is a root of (11.10), where )12 = )/2 (3 + J-83)/2). ThUS,)/2 satisfies yi + 13920yi + 128000)/2 + 8192000 =

o.

316

Ramanujan's Notebooks, Part V

As in the last example, set x = t 6 and )'2 = x - 6 - 27x- l . Using Mathematica, we find that

x 6 + 13902x 5 + 39013x 4 = (x 3

+ 7174196x 3 + 1053351x 2 + 10134558x 3x 2 + 515x - 1)(x 3 + 13905x 2 + 2187x + 19683) = o.

-

19683

A numerical examination of the roots shows that t is a root of the first factor, i.e.,

t l8 =

+ 515t 6 - 1 (t 3 + 2t 2 + 2t - 1)(t 3 - 2t 2 + 2t + 1)(t 6 + 2t 5 + 2t 4 + 6t 3 + 6t 2 x (t 6 - 2t 5 + 2t 4 - 6t 3 + 6t 2 + 2t + 1) = o. 3t 12

-

2t + 1)

Numerically inspecting the roots, we conclude that

t 3 + 2t 2

+ 2t -

1 = 0,

which is what Ramanujan claimed. For n = 107, Greenhill [2, p. 405) proved that a := a I07 satisfies

a -79·80· 2 1/ 3a 2/ 3

69·800· 4 1/ 3a l / 3

-

-

17·16000 = O.

As in the previous two examples, setting a = -yi /256, we find that

yi + 50560yi As before, set x = t 6 and Y2

x6

+ 50542x 5 -

=X

4139493x 4

3532800)'2 + 69632000 = O. 6 - 27x- 1 . Using Mathematica, we find that

-

+ 89919476x 3

+ 111766311x 2 + 36845118x -19683 = (x 3 - 83x 2 + 1875x - l)(x 3 + 50625x 2 + 60507x + 19683) = o. A numerical examination of the roots indicates that the first factor equals 0, i.e.,

t l8

-

83t l2

=

(1 3 -

x

+ 1875t6 212 + 41 -

([6 -

2t 5

-

1

1)(1 3 + 2t 2

6t 3 + 14t 2

+ 4t + 1)(t 6 + 2t 5 + 6t 3 + 14t 2 + 4t + 1) 4t + 1) = o.

-

Checking the roots of each polynomial, we conclude that [3 _

2t 2

+ 4t -

1 = 0,

which is what Ramanujan asserted. In the third portion of page 393, Ramanujan first sets

tn =

1~

"3'1 1 + "3Jn

and then gives the following table of values for tn.

(11.16)

34. Class Invariants

317

Entry 11.19.

n 19 43 67 91 115 163

tn 1 3 7 7 + 2.J13 (t 2 - 14t - 3 = 0) 13 + 6.J5 (t 2 - 26t - 11 = 0) 77 Table 11.3

Proof. The values of tn for n = 19,43, and 67 follow trivially from Entries 11.4-11.6, respectively. From Entry 11.11,

Now 9092

-

13.2522 = 272 •

Thus, by the denesting equality (9.5), t91

=t

(../468 + ../441)

=t

(6m + 21) = 7 + 2m,

which proves Ramanujan's assertion. Next, by Entry 11.13 and another application of the denesting theorem (9.5), 1115

= tJ3141 + 1404Js = t(39 + 18Js) = 13 + 6Js,

which establishes the desired result. Lastly, the value for t163 follows very readily from Entry 11.7. Return to the definition of tn given in (11.11) and set H(n) = 27t;;12. For the four values 19, 43, 67, and 163 in Entry 11.19, H. H. Chan found that

+ 20.../57, 16855 + 1484../129, 515095 + 36332../201,

H(19) = 151

H(43) = H(67) =

and H(163) = 7592629975

+ 343350596-../489.

Amazingly, these numbers are fundamental units of their respective real quadratic fields. Essentially, these same observations were also made by H. M. Stark [1].

318

Ramanujan's Notebooks, Part V

Entry 11.20. Let tn be defined by (11.16). Then, as n tends to tn '"

00,

err .../li/6 + 6e- rr .../li/6

6.J3

Proof. From the definition (1.3), as n tends to 00, Gn Hence, from (11.4),

In '" k2-2err.../li/3(1

,....

2-1/4err.../li/2\1

+ e-rr.../li).

+ e- rr .../li)8 (1 - 4· 26e -rr.../li(l + e- rr .../li)-24) ,

which implies that

Thus,

tn =

~Jl + ~ln

=

~Jl + flerr.../li/3 + 0

= _1_err .../li/6

6.J3

"

(r2rr.../li/ 3)

/1 + 12rrr .../li/3 + 0

= 6~err.../li/6 (1 +6e- rr .../li/3 + 0

(rrr.../li)

(e- 21r .../li/3)) ,

from which Ramanujan's asymptotic formula for tn follows. Entry 11.21. Let q = exp( -7r In) and put R

.= R .= 31/4q1/36 •

n •

f(q) f(q I/3)·

(11.17)

Then (11.18)

We now use Ramanujan's cubic theory from Chapter 33 to prove both Entries 11.17 and 11.21. Proof of Entry 11.17. We shall prove a stronger version of Entry 11.17 by removing the condition q = exp( -7r In), i.e., we interpret (11.12) as a q-identity in the following way. The more general definition of t is clear from (11.11). To extend the definition of In, note that, from (1.3), G n = 2- 1/4q-I/24 f(q)lf(-q2), where q = exp( -7r In). Removing this stipulation on q, in view of (11.4), it is natural to define J by J - 1... -1/3 f8(q) _ 8q2 / 3 fI6(_q2) - 32 q f8( _q2) fI6(q) .

(11.19)

34. Class Invariants

319

To prove (11.12), recall that in proving Entry 11.18 we showed that it suffices to establish (11.14). Replacing q by _q3 in (11.14) and using (11.11) and (11.19), we find that it suffices to prove that

(11.20) Setting h(q)

fl2( _q3) qf6(_q)f6(_q9)

=

(11.21)

and using (11.5) with q replaced by q3, we find that (11.20) assumes the more palatable form 27h- 2(q)

+ 2h- 1 (q) (!Y2(3r) + 3) -

1 = 0,

or Y2(3r) = h(q) - 6 - 27h- 1 (q).

(11.22)

Set (11.23) Then upon cubing Entry l(iv) of Chapter 20 of Ramanujan's second notebook with q replaced by q3 (Part III [3, p. 345]), we find that h(q) = s

27 s

+ 9 + -.

(11.24)

Substituting (11.24) into (11.22) and utilizing elementary algebra, we now find that we are required to prove that 3r _ (s Y2()-

+ 9)(s + 3)(s2 + 27)

(11.25)

s(s2+9s+27)

We now recall some basic facts from Ramanujan's cubic theory. From (1.7) of Chapter 33, ._

._

q .- q3 .- exp

(_ 21f

M

v3

2Fl

(t, I~; 21; 1 -

a))

2F l(3' 3; 1; a)

'

O 0, we detennine the proper square root on the right side above to be

In solving this quadratic equation, we again let n tend to 1 and use the fact that p > 0 to detennine the correct square root. To that end, -(1 p =

+~) + J2 + n + 2.Jf=1l + 2Jn 2 + 2(1-~) 2

. (8.8)

We shall use (9.5) of Chapter 34 to denest the second rightmost inner radical above. Here, d 2 = n 2 (n + 2)2, and so /

V n 2 + 2(1

-

r;---::;

v 1-

n3 ) = =

n 2 + 2 + n(n

+ 2)

_ jn2

2

Jl + n + n 2 -~.

+2-

n(n

+ 2)

2 (8.9)

36. Modular Equations

361

Using (8.9) in (8.8) and then simplifying, we deduce (8.2) to complete the proof. The quotient p3 (2 + p) / (1 + 2 p) occurs in the theory of modular equations of degree 3 (Part III [3, pp. 237, 238]). Taking the parametrizations of fJ and 1 - fJ in terms of p on page 237, by an elementary calculation, we can show that n 2 = 4fJ(l- fJ), where fJ has degree 3. Thus, by (1.6) of Chapter 34, n 2 is related to class invariants. Entry 9 (p. 283). Let fJ have degree 3 over a. 11m

= a l/8 and n = fJ1/8,

then (9.1)

In fact, Ramanujan does not indicate that fJ has degree 3, i.e., that (9.1) is a modular equation of degree 3.

Proof. Ramanujan's equation (9.1) is equivalent to the equation

(9.2) Using (4.5) and (4.6), we find that (9.2) takes the form

4m m-l 1 + (m - 1) - - - - - - m = 0, 3+m 3+m which is trivial to verify. The modular equations in Entry 10 do not seem to be easily deducible from Ramanujan's modular equations of degree 3 listed in Entry 5 of Chapter 19 (Part III [3, pp. 230-231]). However, they can be easily verified by using the parametrizations given in (4.5) and (4.6). Since the details are simple and straightforward, we do not provide them. Entry 10 (p. 290). We have

-1

+(

1_fJ)3)1/4 I-a

m - 1 - {(I - a)(l - fJ)J1/ 4 -

1-

(fJ3)1/4

3

1 + (fJ 3(1 - fJ)3)1/8 a(l - a)

---~--~--~~=

1

+

a

1 - (afJ)I/4

and m2

-

---''--~-:-

(a\l - a)3)1/8' fJ(l - fJ)

362

Ramanujan's Notebooks, Part V

Entry 11 (p. 295). Let fJ have degree 3 over a, and define x and y by

a=

1±

v'f=X3 2

Then

./2y = Xl/8 (Jl +x +x 2 - x -

and

fJ =

1-

J1=Y8 2

Jo- x)(2Jl +x +x2 -

(11.1)

.

1- 2X»). (11.2)

The following proof by H. H. Chan supplants the author's more ad hoc proof. Proof. Routinely solving 01.1) and (11.2) for x and y, we find that x = (4aO - a»1/3

and

y

= (4fJ(l -

fJ»1/8.

Replacing G n by (4a(l-a»-1/24 = x- 1/8 and G 9n by (4fJ(I- fJ»-1/24 in Theorem 3.1 of Chapter 34, we find that y-1/3

where

p=..(X+

= X-1/8

Jx

= y-1 / 3

(p + JP2=1) 1/6 (Ju +.JU=l) 1/ 3,

and

u=!(p 2 -2+J(p2 -I)(p2 -4»).

After straightforward elementary algebra, we find that

./2y = X- 3/8 (x + 1 -

JI

+ x + x 2y/2 (

J + x2+ 1

(1 -

x)JI + x + x 2

-Jo-x)(l-x +Jl +x +x2») = Xl/8 (J 1 + x + 2x2 - 2xJl + x + x 2 -J(1-X)(-I-2X+2JI+X+X 2)) = x 1/8 ( J 1 + x + x 2 - X - J (1

-

x) ( 2J 1 + x + x 2 - 1 -

2x ) ) ,

since

Entry 12 (p. 297). If fJ has degree 3, then (afJ5) 1/8

+ {(I

~--------------~

- a)(l - fJ)5}1/8 = JI - {afJ(1 - a)(1 - fJW/4.

(12.1)

36. Modular Equations

363

Proof. Comparing (12.1) with Entry 5(viii) of Chapter 19 (Part III [3, p. 231)), we find that we must show that

1- {afJ(l -

a)(1 - fJ)}I/4 =

t (1 +..;ap + /(1 - a)(1 -

fJ») ,

or I = (afJ) 1/4 + {(l - a)(1 - fJ)}I/4.

(12.2)

But (12.2) is identical to Entry 5(ii) of Chapter 19 (part III [3, p. 230)), and so the proof is complete.

Entry 13 (p. 297).

If fJ has degree 3, then

3 1 3) 1/8 = 1 + ( fJ ( - fJ)

a(1 - a)

m/l - {afJ(1 -

a)(1 - fJ)}I/4.

(13.1)

Proof. We are unable to simply deduce (13.1) from Ramanujan's modular equations of degree 3 found in Entry 5 of Chapter 19 (part III [3, pp. 230-231)). Thus, we use (4.5) and (4.6) to verify (13.1). Thus, (13.1) is equivalent to the equation m2 - I

1 + -4- = m 1 -

(m 2 - 1)(9 - m 2) 16m 2 ,.

which is easily verified.

2. Modular Equations of Degree 5 and Related Theta-Function Identities Entry 14 (p. 222). We have rp5(q) + 41{1\q) = 5 rp2(q) . rp(q5) 1{I(q5) rp(q )rp3(q5) + 4q21{1(q )1{13 (q5) rp2(q5)

(14.1)

Proof. Let fJ have degree 5 over a, and let m denote the multiplier of degree 5. By Entries 1O(i) and l1(i) of Chapter 17 (Part III [3, pp. 122-123]), (14.1) is equivalent to the modular equation Zl5/2 1/2

ZI5/2 (

a 5) 1/8

+ 172 Ii Z5 Z5 fJ Z:/2Z~/2 + z:/2z~/2(afJ3)1/8

_

-

5~

Z5'

or 1 + (a 5JfJ)I/8 1 + (afJ3)1/8

5 m

(14.2)

364

Ramanujan's Notebooks, Part V

However, (14.2) is one of Ramanujan's modular equations of degree 5 (Part III [3, p. 281, Entry 13(vi))), and so the proof is complete.

Entry 15 (p. 284). We have 1/1\_q5) 1/1 (-q) -

1/15 (q5)

1/1\q10) f5(_qW) 1/1(q) =4q3 1/1(q2) +2q f(_q4) .

(15.1)

Proof. By using Entries I I (i)-(iii) and 12(iv) of Chapter 17 (part III [3, pp. 123124]), we can easily show that (15.1) can be translated into the modular equation of degree 5, 5 ( 1- 13)5)1/8 _ 1 = (13 )1/8 I - a a

+ 2 1/ 3 (13\1

- 13)5)1/24, a(1- a)

which is Entry 13(iii) of Chapter 19 (Part III [3, p. 280]). This completes the proof.

Entry 16 (p. 285). We have l/F 5(q)

1/15 (-q)

f5(_q4)

-+ + 2 f(_q20) 1/1 (q5) 1/1 (_q5)

1/15 (q2)

= 4--. l/F(qlO)

(16.1)

Proof. With the use of Entries 11(i)-(iii) and 12(iv) of Chapter 17 (Part III [3, pp. 123-124]), it is easily shown that (16.1) is equivalent to the modular equation of degree 5,

1

(0+

a)5)1/8 1 - 13

+

2 1/ 3 (a 5 (1 - a)5)1/24 = (a 5 )1/8 13(1 - 13)

13'

which is Entry 13(ii) of Chapter 19 (Part III [3, p. 280]), and so the proof is complete.

Entry 17 (p. 286). We have rp\-q) +4 f5 (-q) = 5rp3(_q)rp(_q5). rp( _q5) f( _q5)

Proof. Replacing q by -q, we are required to show that rp5(q) rp(q5)

+ 4 f5(q) f(q5)

= 5 3( ) ( 5). rp q rp q

(17.1)

Now by Entry 9(ii) of Chapter 19 (part III [3, p. 258]), 4q f5(q5) f(q)

+ rp5(q5) rp(q)

=

() 3( 5). rp q rp q

(17.2)

36. Modular Equations

365

We shall apply transformation formulas to the functions in (17.2) to deduce (17.1). If a, f3 > 0 and af3 = Jr2, then, by Entry 27(iv) of Chapter 16 (Part III [3, p. 43]), e-a/24al/4f(e-a)

= e-fJ/24f31/4f(e-fJ)

and

Thus, e_af5(e-5a) f (ra)

=

= _1_ (~)

(e-5a/24f(e-5a»)5 e- a/ 24 f (e- a )

55/ 2

a

f 5(e-fJ!5). f (e- fJ )

Next, from Entry 27(i) of Chapter 16 (Part III [3, p. 43]), if a, f3 > 0 andaf3 a l / 4cp(e- a )

=

(17.3)

= Jr2,

f3 1/ 4cp(e-fJ)

and

Hence, (17.4) and cp(e- a )cp\e- 5a )

=

_1_ 5 3/ 2

(~) cp(e-fJ)cp\e-fJl5) a .

(17.5)

Therefore, using (17.3)-(17.5) in (17.2), we deduce that 1 (f3) f5 (e-fJ/ 5) 4 55/ 2 -;; f(e- fJ )

1 (f3) cp\e- fJ / 5) cp(rfJ) -;;

+ 55 / 2

1 (f3)

= 5 3/ 2

-;;

Cancel the common expression 5- 5/ 2f3/a and set e- fJ / 5 simplifies to the equality

4 f5(qt> f(q~)

+ cp\qd = 5 cp(qi)

=

_ 3 -fJ 5 cp(e fJ)cp (e /). (17.6) ql. Hence, (17.6)

( 5) 3( )

cp ql cp ql ,

which is (17.1) with q replaced by qt. Entry 18 (p. 295). We have y,2(_q)

+ qy,2(_q5) =

f(q5)cp(q5). X(q)

Proof. Replacing q by -q, we find that it suffices to prove that y,2(q) _ qy,2(q5) =

f(

5) (

-q cp -q X(-q)

5)

=

(5.

q ,q

5)2 (5.

q ,q (q; q2)oc 00

10)

00,

(18.1)

366

Ramanujan's Notebooks, Part V

by (22.4) and Entries 22(iii), (iv) of Chapter 16 of Part III [3, pp. 36, 37]. Now by Entry lO(v) of Chapter 19 and the Jacobi triple product identity (Part III [3, pp. 262, 35]), 1/J 2(q) - q1/J2(q5)

= f(q, q4)f(q2, q3) = (_q; q5)00( _q4; q5)00( _q2; q5)00( _q3; q5)00(q5; q5)~ (_q; q)00(q5; q5)~

(q5; q5)~ (q5; q 10)00

(_q5; q5)00

(q; q2)00

(18.2)

by Euler's identity, (22.3) of Chapter 16 (Part III [3, p. 37]). The desired result now follows from (18.1) and (18.2). Entry 19 (p. 295). We have 1/J2(-q)+5q1/J2(_q5)=

rp2(q) . X(q)X(q5)

(19.1)

Proof. After replacing q by -q and employing Entries II(i), lO(ii), and 12(vi) in Chapter 17 (Part III [3, pp. 122-124]), we can translate (19.1) into the modular equation of degree 5, 5)1/8 5 _ -(a 3f3)1/8 (~ f3 m

= 41/3 ( a \1 _

)5)1/12 a f3(1 - f3)

(19.2)

On the other hand, by Entry 13(iv) of Chapter 19 (Part III [3, p. 281 D,

1(5) =

-

2

- - 1 m

2 1/ 3

(aSC1_a)5)1/24 f3(1 - f3)

(19.3)

Squaring (19.3) and subtracting the result from (19.2), we see that it suffices to prove that

(~5)

1/8

5

- m (a 3f3)1/8

1(5 )2

="4

(19.4)

m - 1

Recall from Part III [3, p. 284, eq. (13.3)] the notation p

=

(m 3 _ 2m2

+ 5m)I/2.

By Part III [3, p. 284, eq. (13.4); p. 285, eq. (13.10»), (

a 5 ) 1/8 _ ~(a3f3)1/8 f3 m

= 5p + m2 + 5m

_ 5(p

4m 2

+ 3m 4m 2

5)

= ! (~ _ 4

m

1)2

Thus, (19.4) has been proved, and the proof of Entry 19 is complete. Entry 20 (p. 297). If f3 has degree 5 over a, then (af33)1/8

+ {(l -

a)(1 - f3)3}1/8 =

Jl - {l6af3(1 -

a)(1 - f3)}1/6.

(20.1)

36. Modular Equations

367

Proof. Comparing (20.1) with Entry 13(vii) of Chapter 19 (Part III [3, p. 281]), we find that it suffices to show that

1-

{16a~(1

- a)(l -

~)}1/6 = ~ (1 + N + JO= a)(1

-

~»),

or 1 - 2{16a~(1 - a)(1 - ~)}1/6

= N + J(l -

a)(l - fi).

(20.2)

But (20.2) is the same as Entry 13(i) of Chapter 19 (Part III [3, p. 280]), and so the proof is complete.

Entry 21 (p. 325). We have f(-q, _q14)f(_q4, _qll)f(-q6, _q9)f(_q5)

= f(-q,

_q4)f3(_qI5)

and

Proof. Both of these identities are readily established by employing the Jacobi triple product identity.

3. Other Modular Equations and Related Theta-Function Identities Entry 22 (p. 246). We have

y,' (q)

y,' (q2)

y,(q)

y,(q2)

2-- - 2q--

q/ (q) =cp(q)

(22.1)

cp' (q) =-. cp(q)

(22.2)

and cp' (_q)

cp' ( _q2)

cp(_q)

cp(_q2)

---4q

Proof. By using the product representations for y,(q), y,(q2), cp(q) , cp( -q), and cp( _q2) (Part III [3, p. 36]), we can easily show that y,2(q) y,(q2)

= cp(q) =

cp2( _q2) cp( _q)

(22.3)

Taking the logarithmic derivatives of both equalities in (22.3), we deduce (22.1) and (22.2) at once. The following two entries can be found in Section 24 of Chapter 18. Regrettably, in Part III [3, p. 216], we claimed that the two results are false.

368

Ramanujan's Notebooks, Part V

Entry 23 (p. 300). The equation

Jm(l -

a)1/8

+ {31/4 =

1

(23.1)

is a modular equation of degree 8.

Proof. By using Entries lO(iii) and l1(iii) of Chapter 17 (part III [3, pp. 122123]), we find that (23.1) is equivalent to the theta-function identity qJ(_q2)

+ 2q2y,(qI6) =

qJ(q8).

By Entries 25(ii), (i) of Chapter 16 (Part III [3, p. 40)), qJ(_q2)

+ 2q2y,(qI6) =

qJ(_q2)

+ ~{qJ(q2) _

= Hcp( _q2)

cp(_q2)}

+ qJ(q2)} = qJ(q8),

and the proof is complete.

Entry 24 (p. 300). If {3 has degree 16 over a, then

Jm =

21

1 + {31/4 + (l _ a)1/4'

(24.1)

Proof. By Entries lO(i), (ii), and II(iii) of Chapter 17 (Part III [3, pp. 122-123]), (24.1) is equivalent to the theta-function identity qJ(q)

+ qJ(_q) =

2qJ(qI6)

+ 4q4y,(q32).

By Entries 25(i), (ii) of Chapter 16 (Part III [3, p. 40]), 2qJ(qI6)

+ 4q4y,(q32) =

{qJ(q4)

= 2qJ(q4)

+ qJ(_q4)} + {cp(q4) _

qJ(_q4)}

= qJ(q) + qJ(_q),

and so the proof is complete.

Entry 25 (p. 297). If {3 has degree 7 over a, then 1 {37)1 / 8 (7)1/8 ( \ __ ; -~ = m

(1 -

{a{3(1 - a)(1 -

{3)}1/8).

Proof. The result follows by combining the first part of Entry 19(iii) with the second part of Entry 19(i) of Chapter 19 (part III [3, p. 314]).

36. Modular Equations

Entry 26 (p. 296).

-3 (

369

If fJ has degree 7 over a, then

fJ7(1 _ fJ)7)lfl 2 _ - O.

(26.1)

a(1 - a)

Proof. Let (26.2) and R := where afJ

= t8, t

{(2 -

3t + 2t 2 )(2 - t + t 2 )(1 - t + 2t 2 ) J1/2 ,

> O. From Part III [3, pp. 316, 318, eqs. (19.2), (19.3), (19.15)],

A=

! {2 - 7t + 11t 2 -

8t 3

+ 4t 4 -

(l - 2t)R}.

(26.3)

From Entry 19(vii) of Chapter 19 in Ramanujan's second notebook (Part III [3, p. 314]),

( 1-

fJf)I/8 1- a

+

(fJ7 )1/8 + 2 (fJ 7(l a

fJ)7) 1/24 =

~ + m2 .

a(1 - a)

(26.4)

4

Thus, using (26.2) and (26.4), we may recast (26.1) in the fonn

1 + (3 +4m 2

_

2A) 2

2A 3

_

_

2 (A3 + 3 +4m 2

-

2A)

2 7+ - A(4m- - 2A ) - 3A 2 = 0,

or (m 2

-

1)2 -

20Am 2

-

12A

+ 48A 2 -

64A 3 == O.

(26.5)

We also know that (Part III [3, p. 319, eq. (19.20))) m

= -3 + 8t -

6t 2 + 4t 3 + 2R.

(26.6)

Putting (26.3) and (26.6) in the left side of (26.5) and simplifying with the aid of Mathematica, we verify that indeed (26.5) holds to complete the proof. We have altered Ramanujan's notation in the next entry so that it is consistent with that in Entry 3 of Chapter 20 (Part III [3, pp. 352-353]).

370

Ramanujan's Notebooks, Part V

Entry 27 (p. 286). Let y have degree 9, and put m Then

=

Zi/Z3

and m'

= Z3/Z9.

, 3 _(a)1/8 +(1_a)1/4 (-a)1/8(I_a)1/4 ../mm + - - - -y 1- y ..jmm' y I- y

Proof. Using Entries 3(x), (xi) of Chapter 20 (Part III [3, p. 352]), we find that

, 3 (-a)1/8(1_a)1/4 -../mm +-y l-y ..jmm'

(~)1/4 ((~)1/8 + (~)1/8 _(y(l _ y»)1/8) ( ~)1/8 y 1- y a 1- a a(l - a)

+ (~)1/8 + (~)1/8 _ (a(1-a»)1/8

y

l-y

C=~y/4 + (~y/8

y(1-y)

which completes the proof.

Entry 28 (p. 296). If m denotes the multiplier of degree 9 and y has degree 9, then

(1I-a _y)1/2 + (y(ly»)1/2 a a(1-a) _4(y(l- y»)1/411 + (~)1/4 + (1- y)1/4) = m 2. a(1-a) a I-a

( ~)1/2 +

Disproof. Write

S.- (~)1/2

a

+

(1 - y)1/2 + (y(1 _y»)1/2 a(1-a) I-a

_4(Y(I - y»)1/4I1 + (~)1/4 + (1 _y)1/4) a(I-a) a I-a

I( Ia

~)1/8 + (~)1/B)4 _4 (y(1 a

I-a

- y»)l/B

aCl-a)

y»)1/4 a(1-a) + (Y(I - y»)1/2 _4(y(l - y»)1/4I1 + (~)1/4 + (1 _y)1/4) aCI - a) a(1 - a) a 1- a x

(~)1/4 + (~)1/4)_ 6 (y(1I-a

36. Modular Equations

l(

r)1/8 a

_41

tim"

l(

(1 -

+

y)1 / 8j4

1- a

(Y(1- y»)1/8 a(1 - a)

-10

371

(Y(1- y»)1/4 + (Y(1- y»)1/2 a(1 - a) a(1 - a)

+ (y(1

- y»)1/4j a(1 - a)

II (f)'" c=:rr - (:~: =~~r} +

r)1/8 a

2

+ (~)1/8j4 _2(y(1_y»)1 /4 1- a

+ 8 (y(1

_ y»)3 / 8 a(1 - a)

+ (y(1- y»)1/2 a(1 - a)

_41(yO-

a(1 - a)

y»)1/8

a(l-a)

+

(YO- y»)1/4jl(r)1/8 + (~)1/8j2 a I-a a(l-a)

(28.1) From our study of Ramanujan's modular equations of degree 9 in Part III [3, pp. 352-356], each of the expressions above can be expressed in terms of a parameter I. In particular, from page 356, ( r)1/8 +(I_y)1/8 a

=

I-a

1+21, 1-1

and from equations (3.7) and (3.9) on page 354, ( y(1 - y») 1/8 = t 1 + 21 . 1- t a(1 - a) Thus, from (28.1),

s=

(1+2t)4 _2t 2 (1+21)2 +t4(1+21)4+8/3(1+2t)3 1-1 I-t I-t 1-1

_41 ~ ~I + t\

=

(1 + 2t)2 {(1 (1 - t)4

+8t\1

=

+ 2t)2 -

+ 2t)(1

(I + 2t)2 (1 - t)4

12 ( \

{I -

~ ~t Yj(\ ~ ~t Y 2t 2(1 - t)2

- t) - 4 (t(1

+ t 4(1 + 2t)2

+ 2t)(1 -

6t 2 + 4t 3 - 9t 4 - 12t 5

t)

+ t 2(1 + 2/)2)}

+ 4t 6 }.

Now from equations (3.10) and (3.11) on page 354 of Part 1II [3], m 2 = (1

+ 2t)4.

(28.2)

372

Ramanujan's Notebooks, Part V

Thus, if Ramanujan were correct, the expression in curly braces on the far right side of (28.2) should equal (1 + 21)2(1 - t)4. But (1

+ 2t)2(1 -

t)4 = 1 - 6t 2 + 4t 3 + 9t 4 - 12t 5 + 4t 6.

(28.3)

Thus, there is a discrepancy between (28.2) and (28.3) in the terms -9t 4 and +9t 4 , respectively. It seems therefore that Ramanujan made a sign error in his calculations. This analysis is evidence that Ramanujan also used parametric representations.

Entry 29 (p. 230). We have 1{t(q)1{t(q II) -1{t( -q)1{t( _qll) = 2qf(q2, qlO)f(q44, q88) + 2q 15qJ(q6)1{t(q 132).

(29.1) Proof. In (36.8) of Chapter 16 (Part III [3, p. 69]), we set J.L find that

=

6 and v

=

5 to

+ qf(q44, q88)f(q2, qlO) + ql4 f(q22, q 1I0)f(q20, q-s) + q391{t(q l32)f(q30, q-18).

1{t(qll)1{t(q) = qJ(q66)1{t(qI2)

Replacing q by -q and subtracting the resulting equality from that above, we find that 1{t(q)1{t(qll) -1{t(-q)1{t(_qll)

= 2qf(q44, q88)f(q2, qlO)

+ 2q391{t (q 132)f(q30, q-18).

But by Entry 18(iv) of Chapter 16 with n = 2 (Part III [3, p. 34]), f(q30, q-18) = q-24 qJ (q6).

Using this above, we deduce (29.1).

Entry 30 (p. 298). If {J has degree 11 over a, then II 1 {J)1I)1/24 1 + 2 10/3 ({J ( = m {(a{J)1/4

a(1 - a)

x {(a{J)1/4 - {(l- a)(1- {J)}1/4

+ {(l

- a)(1 - {J)}1/4}

+ (2 + 2vfaP + 2J(1- a)(1 -

{J»),/2J. (3 .1)

Proof. By adding Entries 7(vi), (vii) of Chapter 20 (Part III [3, p. 364]), we find that

~ ( 1 + 210/ 3 ( {J II (I m

(J) a(1-a)

+ v'2 {(a{J) 1/4 + {(l -

\I )

1/24)

= vfaP -

J(1 - a)(1 -

a)(1 - {J)}1/4} ( 1 +.;afi +

(J)

J(1 - a)(1 -

(J) )

1/2

.

(30.2)

36. Modular Equations

373

Comparing (30.1) and (30.2), we find that the proof is complete.

4. Identities Involving Lambert Series Entry 31 (p. 266). 1 2

2

2

"4q1 (q)q1 (q ) =

1

4+L k=l 00

(2k - l)q2k-l 1 _ 2k-l q

+L 00

(2 + (_I)k)kq2k 1 + 2k

k=l

q

.

Proof. Using successively Entries 25(vi) and 25(iii) of Chapter 16 and Entry 8(ii) of Chapter 17 (Part III [3, pp. 40, 114]), we find that ~q12(q)q12(q2) = lq12(q) (q12(q)

1

=

1

kqk

00

l (q14(q) + q14(_q2»)

+ q12(_q») =

00

k( _q2)k

8 + {;; 1 + (-q )k + 8 + {;; 1 + q2k 1 (2k - 1)q2k-l (2 + (-1 )k)kq2k 00

=-+'" 4 b

00

l_ q 2k-l

+'" b

1 +q2k

,

and the proof is complete.

Entry 32 (p. 267). 4 _

L 00

q1(q)q1(q ) - 1 + 2

k=O

(_I)k q 2k+l _ 00 (_I)k q 4k+2 1 _ 2k+l 2 1+ 4k+2 . q k=O q

L

(32.1)

Proof. By Entries 25(i), (iii) of Chapter 16 (Part III [3, p. 40]), q1(q)q1(q4) = ~q1(q) (q1(q)

+ q1(_q»

= ~ (q12(q)

+ q12( __ q2»).

(32.2)

Recall from Entry 8(i) of Chapter 17 (Part III [3, p. 114]) that 2

q1 (q) = 1 + 4

L 00

k=O

(_I)k q 2k+1 1 _ 2k+l . q

Using this twice in (32.2), we deduce (32.1) at once. Note that Entry 32 is a companion to Entries 8(iii), (iv) of Chapter 17 (part III [3, p. 114]), and, in fact, is placed after these entries in the first notebook.

Entry 33 (p. 274). If n is real. then

r::-oo(

~ qk cos(nk) 3 2 _1)kq 2k 2 cos(2kn) 1 + 4 £2k = q1 (-q ) 2• k=l 1 +q (r:~_00(-1)kqk2cos(kn»)

(33.1)

374

Ramanujan's Notebooks, Part V

Proof. Comparing (33.1) with Entry 33(iii) of Chapter 16 (Part III [3, p. 53]), we find that it suffices to prove that, with z = e in , rp

2(

-q

2)

f(zq, q/z) f(-zq, -q/z)

rp3(_q2)f(_z2q2, _q2/Z2) j2(-zq, -q/z)

or

f(zq, q/z) =

rp(_q2)f(_Z2 q 2, _q2/Z2) f( / ) . -zq, -q z

(33.2)

By the Jacobi triple product identity and (22.4) of Chapter 16 (Part III [3, pp. 35, 37]), the right side of (33.2) equals

(q2; q2)00(q2; q4)00(Z2 q 2; q4)00(q2/Z2; q4)00(q4; q4)00 (zq; q2)00(q/Z; q2)00(q2; q2)00 = (q2; q2)00(-Zq; q2)00(-q/z; q2)00

=

f(zq, q/z),

again, by the Jacobi triple product identity. This proves (33.2), and so the proof is complete.

Entry 34 (p. 284).

If(~)denotes

the Legendre symbol. then

Proof. By Entry 4(iii) of Chapter 19 (Part III [3, p. 226]), 1/J3(q) 00 (q6n+l q6n+5) 1/J(q3) = 1 + 3 ~ 1 _ q6n+l - I _ q6n+5 •

(34.1)

and by Entry 3(i) of Chapter 21 (Part III [3, p. 460]), 1/J3(q)

1/J3(q3)

1/J(q3)

1/J(q)

00

f::t

(n) -qn- .

--+3q--=1+6'" -

3

1 - qn

(34.2)

Thus, from (34.1) and (34.2),

We now use the elementary equality

2q2m 1 - q2rn

qm qrn ----1 - qm I + qrn

(34.3)

36. Modular Equations

with m = 3n + 1, 3n 1/13(q3) q 1/1(q)

+ 2. Hence, after some cancellation, (q3n+1

L 00

=

1+

n=O

=

375

00

+ 1-

q3n+1

q3n+1

(

~

q6n+2

1+

q6n+2 -

q3n+2

_

1 - q6n+2

q3n+2

q6n+4)

1_

q3n+2 -

q6n+4

)

1 _ q6n+4

'

where we used the elementary equality 1 _q2m with m = 3n + 1, 3n

+ 2. This completes the proof.

Entry 35 (p. 284). 1/(1) denotes the Legendre symbol, then qJ3(q3) _ 1 _ 2 qJ(q) -

?; (~)3 100

qn (_q)n'

(35.1)

Proof. From Entries 3(i), (ii) of Chapter 21 (Part III [3, p. 460]),

+

qJ3(q) qJ(q3)

3

4(1 + 6~ (~) ~ 3 1_

qJ3(q3) = qJ(q)

q2n ) , q2n

and from Entry 4(iv) of Chapter 19 (Part III [3, p. 227]), qJ3(q)

(q6n+1

- - 1 + 6" qJ(q3) ~ 1_ 00

q6n+2

q6n+1

+ 1 + q6n+2 -

q6n+4

1 + q6n+4

-

q6n+S) ---=-----:-__=_

I _ q6n+s

•

Hence, using (34.3), we find that qJ3(q3)

-- - 1+8" qJ(q)

(q6n+2

00

1- q6n+2

~

-

00

- 2 "~

-

(q6n+1

1-

ql2n+8

(ql2n+l

q12n+7

1 - ql2n+7

q6n+2

q 12n+8

=1+2" - 1~

-

1 _ q6n+4 q6n+2

+21q6n+4

00

q6n+4) ---=------:--:-:-

+ 1-

1 - q6n+1

q6n+4

-

q6n+5)

- -------:-----= 1 _ q6n+5

+3 1 -

ql2n+8 ql2n+8

q12n+4

2---=------:-::--:-:I _ q 12n+4

ql2n+2

ql2n+l

+ 1-

-

-

q12n+2

3

ql2n+4

- 1-

ql2n+\O

1 - ql2n+\O

q12n+4

ql2n+S

+-----:-::,..--.,--::1 _ ql2n+S

ql2n+I1)

+ --=-----:-::--:-:1 - ql2n+11

'(35.2)

376

Ramanujan's Notebooks, Part V

On the other hand, the right side of (35.1) equals, by (34.3), 00

1 - 2 ""

-f=o

=

00

-f=o

1 - 2 ""

+

(q6n+l

1 + q6n+l q6n+l

(

1 - q6n+l

q6n+4

q6n+2

1 - q6n+2

- 2

q6n+4

+

q 12n+2

1 - q12n+2

q6n+5

-

1 - q6n+4

-

1 - q6n+4 -

q6n+2 ---=------,--___::_

1 _ q6n+2

q 12n+1O

1 - q6n+5

q6n+5)

- ---:----::1 + q6n+5

)

+2~~~=

1 _ q12n+1O

(35.3)

.

If we now compare (35.2) and (35.3), we find that they are equal, and so the proof is complete.

Entry 36 (p. 284). We have 1jI\_q3) 1jI3(q3) ljJ3(q6) ------=2q--. 1jI (_q) ljJ (q) 1jI(q2)

Proof. From Entry 34, 1jI3 (_q3) 1jI(-q)

since (~)

=

1jI3 (q3)

--1jI(q)

-1.

Entry 37 (p. 285). We have ljJ3(q) ljJ(q3)

1jI3(_q)

+ 1jI(_q3)

ljJ3(q2)

= 2 1jI(q6) .

Proof. This result easily follows from (34.1) and (34.3).

5. Identities Involving Eisenstein Series Recall that L := L(q) := 1 - 24

L 00

k=l

kqk --k'

1- q

Iql

< 1.

(38.1)

From Entry 12(ix) of Chapter 15 (Part II [2, p. 326]), or from the fourth entry on page 264 of the first notebook, L

_ L~o(-ll(2k (q) -

+ 1)3 q k(k+1)/2

L~o(-I)k(2k +

1) q k(k+l)/2 .

(38.2)

36. Modular Equations

377

Ramanujan, in fact, expresses the next entry in terms of the right side of (38.2).

Entry 38 (p. 264). We have 4L(q4) - L(q)

= 3fP4(q).

(38.3)

Proof. From Entries 13(viii), (ix) of Chapter 17 (Part III [3, p. 127]), 4L(q4) - L(q)

=

(4L(q4) - 2L(q2»)

= 2z 2(1 - tx)

+ (2L(q2) -

+ z2(1 + x)

L(q»)

= 3z 2 = 3fP4(q),

by Entry 6 of Chapter 17 (Part III [3, p. 101]). Suppose that, in the summands on the left side of (38.3), we expand 1/(1 _ qk) and 1/(1 - q4k) into geometric series. Collecting coefficients of like powers of qn, n ~ 0, we find that 00

= 3 - 96 I:O'(n)q4n + 24

4L(q4) - L(q)

L a (n)qn , 00

n=!

n=!

where a (n) is the sum of the positive divisors of n. Thus, equating coefficients of qn, n ~ I, on both sides of (38.3), we find that, if r4(n) denotes the number of representations of n as a sum of four squares, r4(n) =

I

ifn¢.0(mod4),

8a(n),

8a(n) - 32a(n/4), if n

== 0 (mod 4),

=8Ld. din

4fn

Thus, Entry 38 yields a very short proof of a famous result of C. G. J. Jacobi [1], [2].

Recall that 00

k 3q k

Iql

M:= M(q):= 1 +240L--k' k=! 1 - q

< 1.

Entry 39 (p. 264). We have E~o(-I)k(2k + 1)5 qk(k+1)/2 = M _ 480 ~ k 2q k • E~O(_1)k(2k + l)qk(k+ 1)/2 (1 - qk)2

6

Proof. By Entry 12(v) of Chapter 15 (Part II [2, p. 326]), 00

M _ L2

k2qk

~ (1 -

qk)2

=

288

'

(39.1)

378

Ramanujan's Notebooks, Part V

where L is defined in (38.1). By Example (ii) in Section 35 of Chapter 16 (Part III [3, p. 65]),

+ 1)5 q k(k+1)/2 L~o(-I)k(2k + l)qk(k+1)/2

L~o(-1)k(2k

Entry 39 thus easily follows from the last two equalities. Entry 40 (p. 271). If M (q) is defined by (39.1), then M(q) = rps(_q)

+ 256ql/ls(q).

Proof. From Entry 13(iii) of Chapter 17 (Part III [3, p. 127]), M(q) = z4(1

+ 14x + x 2 ).

(40.1)

On the other hand, by Entries lO(ii) and 11(i) of Chapter 17 (Part III [3, pp. 122-123]), rps( _q)

+ 256ql/ls(q) = z\1 -

X)2

+ 16z4 x = z\1 + 14x + x 2 ).

(40.2)

Combining (40.1) and (40.2), we complete the proof.

6. Modular Equations in the Form of SchHifti This section contains some of the deepest results in the chapter. As indicated at the beginning of the chapter, Ramanujan's methods for some entries have remained hidden from us. Entry 41 (p. 90). Let

Q'.-

and Then,

~(1- /3))1/24 ( a(1a)

if ~ has degrees 11, 13, 17, and 19, respectively, over a,

(2. _2..!.. +

Q6

+ _1 _ Q6

Q7

+ ~7 + 13 ( Q5 + ~5) + 52 ( Q3 + ~3)

2J2

p5

p3

+ 78 ( Q + ~) - 8 (~6 Q9

+ ~9 _

-

(~ pS

34 ( Q6

22 _ 22P P

- p6) =

+ 11 p 3 _

= 0,

0,

+ ~6) + 17 ( Q3 + ~3)

(;4 +

_ 136 _ 340 _ 136P4 + 16P S) = 0 p4

2P5)

'

7 + 4P4)

36. Modular Equations

379

and

10 +-+114 1 (6Q +Q6-I) -190",2~ (4Q +Q4-I) (-p31- P 3) I) (--5+8P 8 6). ~ (-+--19P 4 19 3-4P9) =0. +19 (Q 2+- O. We thus need to calculate lim

1

-2rrik/fJ+i
2 that can be represented by (15.3), and where, for each fixed l, the sum is also over all distinct solutions (c, d) of (15.3). The theorem now follows after a small amount of simplification.

442

Ramanujan's Notebooks, Part V

Entry 19 (Formula (12), p. 278). ~ ~

(-1)k(2k

+ 1)n-1

n

-----:,--------:- = (2 -

k=O cosh {(2k

+ 1)1f../3/2 }

1) -IBnl

. (1fn)

S10

-

{1

-

6

n

+ 2 cos (ntan- 1 (../3/2»)

2cos(1fn/6)

---:::-3n / 2

_ ... }

7n/2

.

(19.1)

As before, we assume that Ramanujan intended n to be an even positive integer. If n = 6m, where m is a positive integer, then we deduce from (19.1) that (-1)k(2k

+ 1)6m-1

L { } =0. k=O cosh (2k + 1)1f../3/2 00

(19.2)

The evaluation (19.2) was first achieved by Cauchy [1, p. 317]. Ramanujan recorded (19.2) as part of Entry 18(iii) of Chapter 17 of his second notebook. For two proofs of (19.2), see the author's paper [7, Corollary 7.6] and book [3, pp. 140-141]. For references to other proofs, generalizations of (19.2), and further results of this sort, see the last two cited references. Before stating Theorem 19.1, we need to say a few words about solutions to gcd(c, d) = 1.

(19.3)

Each solution (c, d) of (19.3) generates four solutions, namely,

±(c, d),

±(-c, d).

(19.4)

We shall say that (cJ, d l ) and (C2, d 2 ) are distinct solutions to (19.3) if they belong to different sets of four solutions given by (19.4). We remark that the number of positive odd solutions (c, d) to 4n = 3c 2 + d 2 , where n is odd, equals d1,3(n) - d 2 ,3(n) (L. K. Hua [1, p. 309]). Theorem 19.1. Let m be a positive integer such that m ¢ 0 (mod 3). Then

f:

(-1)k(2k

k=O cosh { (2k

+

1)2m-1

+ 1)1f../3/2 } B

- _(2 2m -1)~ 2m i

L

even

(_l)(c+d)/2 sin (2m tan- 1 (c../3/d») -----~-::-------'-

(l/4)m

(19.5)

where the sum on the right side 0/ (19.5) is over all even positive integers l which can be represented by (19.3), and where,for each fixed l, the sum is also over all distinct solutions (c,d) 0/(19.3).

37. Infinite Series

443

Proof. By a calculation like that in (18.5) and by Lemma 18.1 with T = i.J3, we deduce that

+ 1)2m-l L (l)k(2k 00

k=O cosh {(2k

+ 1)1r.J3/2}

= 2

L( _1)r G2m-l (2r + 1)e-(2r+l)rrv'3/2 00

r=O

(_I)j+k

00

L j,k=-oo

«2)

(2j+l,2k+1)=!

+ l)i.J3 + 2k + 1)2m

L

B 00 (_I)(c+d)/2 = - i22m(22m - 1) 2m 8m c,d=-oo (ci.J3 + d)2m (c,d)=! c,d odd

.

(19.6)

Now, (ci../3 + d)-2m

= rm exp (-2mi tan-!(c../3jd») ,

(19.7)

where i is given by (19.3). We group terms according to increasing values of i. The sum of the four terms arising from (19.4) equals, by (19.7), 2( _l)(c+d)/2

2( _1)(-c+d)/2

(ci.J3 + d)2m

(-ci.J3 + d)2m

--=,...--- + ---=--(_1)(C+d)/2) . = 4, 1m (ci.J3 + d)2m

= _4i(_1)(c+d) /2 r m sin (2m tan- 1 (c../3/d) ) .

(19.8) Thus, from (19.6) and (19.8), we complete the proof of Theorem 19.1. Comparing (19.5) with (19.1), we see that the trigonometric sums on the right sides have quite different shapes. The first three terms in (19.1) evidently arise from the values c, d = 1,1; 1,3; 1,5, respectively. However, we note that 28 has two representations, 3 . 12 + 52 and 3 . 32 + 12. Easy calculations show that the first two terms on the right side of (19.1) agree with the first two terms on the right side of (19.5). However, there is a discrepancy between the third terms. This discrepancy exists if we take either term arising from the two representations of 28, or if we take the sum of the two terms, from our sum (19.5). We shall now establish an alternative representation for the sum on the right side of (19.5). This will give a result which is "close" to that of Ramanujan and perhaps indicate where Ramanujan erred. An elementary calculation shows that

Ramanujan's Notebooks, Part V

444

and so tan -I (ev'3) d

+ tan -I

(v'3(d - de») -_ ~ 3e+ 3

+ k 7r,

for some integer k. Thus, . ( 2m tan-I (ev'3)). sm d = sm ( 2m {7r"3

-

tan-I (v'3(d 3e +- de»)})

= sin (2~7r) cos (2m tan-I ( ~~d+-de») ) _ cos

(2~7r) sin (2m tan-I (~~d+-de»))

= ( - l) m-I sm. (m7r) cos (2 m tan -I (v'3(d - e»)) 3 3e+d + 4 sin (2m tan-I

(~~d+-de»)).

(19.9)

Thus, we obtain "half" of what Ramanujan probably found, because for e, d = 1,1; 1,3; 1,5, the first term on the far right side of (19.9) yields precisely the trigonometric functions given by Ramanujan in (19.1), except for an additional factor of 2 in the second and third terms in (19.1). For e = d = 1, the second term on the far right side of (19.9) vanishes. For c, d = 1, 3, a simple calculation shows that the first and second terms on the far right side of (19.9) are equal, while for e, d = 1, 5, the second term does not equal the first term. Entry 20 (Formula (15), p. 278). If n is a positive integer, then 00

~

(_I)k-Ik6n 2nv'3 (B6n cosh(k7rv'3) - (-I)k = -7r- 12n

where B j, j

~

00

(_1)k- I k 6n -

+ ~ ek1r ./3 -

1 )

(-I)k

'

0, denotes the j th Bernoulli number.

Proof. We shall easily show that Entry 20 is equivalent to a result of the author [6, p. 163, Cor. 2.22], namely,

6n

k6n-1 7rv'3 k B L +-L -~ k=1 (-I)ke ./3 - 1 12n k=! sin (k7rp) - 12n' 00

00

2

k1r

(20.1)

where n is a positive integer and p = exp(27ri/3). By a straightforward calculation, sin 2 (k7rp)

= 4(1 -

cos(27rkp»

= 4(1 -

(_1)k cosh(k7r.J3».

Using this and elementary manipulation in (20.1), we complete the proof.

37. lnfinite Series

445

Ramanujan's statement of Entry 20 does not contain an equality sign. Ramanujan next offers two puzzling transfonnations for doubly exponential series. We shall state them exactly as Ramanujan wrote them and then refonnulate them. If afJ = 21r, then 00

~ e-n~a

a

where qJ(fJ) =

{

1

00

(

I)k-lnk }

= a 2 + {; k!~eka _

I)

- y - log n

00

+ 2 {; f{J(kfJ) ,

(21.1)

+ . .. ) .

(21.2)

J

. 7r cos (fJf logJ - - fJ - -7r - -B2fJ sinh(7rfJ) n 4 1 ·2fJ

where 1/F(fJ) =

7r . ( fJ VIfJ sinh(7rfJ) sm fJ log;; -

fJ -

7r

"4 -

B2

1 . 2fJ

B4

+ 3 . 4fJ3

- ...

)

.

(21.4) In (21.1), y denotes Euler's constant, and, in (21.2) and (21.4), B j , j ::: 0, denotes the jth Bernoulli number. We emphasize that in Ramanujan's notation, all even indexed Bernoulli numbers are positive. The definitions of qJ(fJ) and 1/F(fJ) given in (21.2) and (21.4), respectively, are certainly enigmatic. Appearing in the arguments of the trigonometric functions are apparently asymptotic series as fJ tends to 00. Thus, the definitions of f{J(fJ) and 1/F(fJ) are imprecise, and so Ramanujan's claims are unclear. Nonetheless, we shall show that (21.1) and (21.3) are correct, if (21.2) and (21.4) are properly interpreted. The work on Entry 21 which follows was done jointly with J. L. Hafner [2]. We begin by defining functions G(fJ) and B(fJ) by r(ifJ

+ 1) =

(ifJ)iP+l/2 e-iPv'2iiG(fJ) = (ifJ)i P+l/2 e-i Pv'2iie- iB (P).

(21.5)

Then (Whittaker and Watson [1, pp. 252-253], as fJ tends to 00, 00 (_I)k-l B2k B(fJ) '" { ; (2k _ 1)(2k)fJ2k-1

and 1

1

139

571

G(fJ) "" 1 + 12ifJ - 288fJ2 - 51840(ifJ)3 - 2488320fJ4

+... .

(21.6)

Ramanujan less explicitly gives the asymptotic expansion for B(fJ) in the arguments of the trigonometric functions in (21.2) and (21.4).

446

Ramanujan's Notebooks, Part V

We now state rigorous formulations of (21.1) and (21.3), and then immediately show that Ramanujan's aforementioned claims are consequences. Entry 21 (Formulas (4), (5), p. 279). Let n, a, and f3 be positive with af3 = 2rr. Then (21.1) holds, where rp(f3)

=

1 ~

1m {ofJ n- I r(if3

+ 1) }

= Pfe- JTfJ/2 {sin (f310g

+

cos (f310g

~-

~n -

f3

f3

{1 __1_ + ... } +... }}

+ ~)

288f32

4

+~) {_l_ 4 12f3

as f3 tends to 00. Let n, a, and f3 be positive with af3 1/1(f3)

+ ~) Re {G(f3)}

~ - f3 +~) 1m (G(f3)}}

~ V{2; e-JTfJ/2 {sin (f3 10g ~ 7i n -cos (f310g

f3

'

(21.7)

= rr /2. Then (21.3) holds, where

= - ~ Re {n- ifJ r(if3 + 1)} - Pfe- JTfJ/2 {cos (f310g

- sin (f310g

~-

+

~) Re (G(f3)}

~ - f3 + ~) 1m (G(f3)}}

JTfJ/2 {cos (f310g ~ ~ - V{2;e7i n

+ sin (f3 10g ~n as f3 tends to

f3

f3

f3

+ ~) 4

{I __1_ + ... } 288f32

+ ~) {_1_ + ... } } 4 12f3 '

(21.8)

00.

We first show that Ramanujan's definitions (21.2) and (21.4) are compatible with the far right sides of (21.7) and (21.8), respectively. As f3 tends to 00, . rr cos f3 smh(rrf3)

(f3 10g ~n -

f3 -

~4 -

B(f3»)

= Pfe- 1rfJ/2 (1 - e- 21rfJ )-1/2 sin (f310g

~-

f3

+~-

B(f3»)

37. Infinite Series

.....

~e-rrp/2 {sin (p log ~ -

P

+ ~) cos

447

B(P)

(p log ~ - P + ~) sin B(P)}.

- cos

Thus, (21.2) and (21.7) are in agreement. The argument showing that (21.4) and (21.8) agree is similar. We now proceed to prove Entry 21. Proofs of (21.1) and (21.7). First,

00 (_l)k-Ink k! (eka _ 1)

00 (_1)k-I n k 00 -kja k! eka e

L

=L k=1

k=1

L

J=O

L (1- e-

= L L -'----'---c--- = 00 j=1 k=1 k. j=1 ~~ (-I)k-l n , ke-kja

ne - ja ).

Thus, the proposed identity may be written in the equivalent form

L 00

a

(e- neka

+ e-ne- ka

1) - 4a + ae-

-

k=1

n

= -y -log n + 2

L rp(kP). 00

(21.9)

k=1

Second, we apply the Poisson summation formula (Titchmarsh [2, p. 60)) to the function (21.10) Observing that f(O) = 2e- n

-

a (4(2e- n

-

=

1, we find that, for a, P > 0 with ap = 21r, 1)

+

f: (ek=1

1

00 +2L

00 f(x) dx o

neka

1

k=l

00

+ e-ne - k•

-

1))

f(x) cos(kpx) dx.

(21.11)

0

Comparing (21.9) and (21.11), we see that it remains to proVI~ that

-y -logn + 2 ~rp(kP) =

1

00

f(x) dx

+2

t;; 1

00

f{x)cos(kpx) dx.

(21.12) Observe from (21.10) that f(x) is even. Setting u = eX, we find that

1

00

o

1100 f(x)dx = f(x) dx

11

2

= _

2

-00

0

00

(e- nu

+ e-n/ u -

du 1)-

= ~ (_ {I/n 1 - e- nu du 2 10 u

u

+ ["10 e-"~du

11/n

u

448

Ramanujan's Notebooks, Part V

l 11

= -1 ( 2

1 - e- X dx+

0

+

1

00 1 - e- nlu lln e-nlu ) --dudu o u lin U

+

X

1

00 e-X

-dx-

Since (Part 1[1, p. 103]) y

X

we find that

1

I(x) dx = -1 ( -y o 2 00

+

1

00

I

I

e- X -dx X

X

e- X X

0

1

' 1 - e- dx = 10o x,

e- X -dx

00

10,,2 1 -

X

,,2

1

dx

)

.

e- X -dx,

00

X

11 0

1 - e- X dx X

l

n2

dX)

-

OX

= ~ (-y - y -10gn 2 ) (21.13)

=-y-logn. Using (21.13) in (21.12), we find that it suffices to prove that

L tp(kf3) = L 10 00

00

[00

k=1

k=1

I(x) cos(kf3x) dx.

(21.14)

0

Set /:= /(13):=

10

00

f(x)cos(f3x) dx.

By (21.14), it now suffices to prove that / (13) = tp(f3), where tp(f3) is defined by (21.7). Letting u = eX, we find that

11

/ = 2

00

-00

l(x)cos(f3x) dx = -110 2 0

00

(e- nu

du + e-nlu -1)cos(f310gu)-. u

Integrating by parts, we find that

[00 (e- nu 213 10

/ = .!!..--

= .!!..-- (

213

=

n

~e-nlu) sin (13 log u) du u2

[00 e-"" sin(f310gu) du _

10

213 (It

_

- 12 ),

10

[00

e- n;" sin(f310gu) dU) u

37. Infinite Series

say. Setting t = l/u in

I

449

h we deduce that /z = -h Hence, n (oo

n

= ~11 = ~ Jo

e- nu sin(,8logu) du

=

~ {oo (e-nuui.B _ e-nuu-i.B) dll 2,8.

Jo

=

2~i

(n- i.Br(i,8

=

~Im {n- iP r(i,8 + 1)}.

+ 1) - ni.Br(-i,8 + 1»)

Hence, I = 1(,8) = rp(,8), by (21.7). This completes the proof of (21.1). Lastly, from (21.5),

rp(,8) =

~Im (i,8)1/2 (~yP v'iHG(,8»)

= f3j.lm (e i(Jr/4+Plog(iP/(ne»)G(,8»)

=

J2; e-Jr.B/2Im (e i(Jr/4+Plog(P/n)-.B)G(,8»).

Hence, the second equality in (21.7) follows, and the asymptotic formula follows by using (21.6). Proofs of (21.3) and (21.8). First,

Thus, the proposed identity (21.3) can be recast in the form 00

U

L(-I)k (e- ne(2k+l)a

-

e- ne -(2k+l)a

+ 1) =

1u + L(-ll1fr«2k + 1),8).

k=O

0 and afJ = rr/2,

00 100 I(x) sin«2k + l)fJx) dx. L(-ll

=

k=O

(21.16)

0

We recall that rr

a

1

00 (-It

"2= 413 =1i~2k+l·

(21.17)

Using (21.16) and (21.17) in (21.15), we find that it suffices to prove that

00

L(-I)k1/l«2k + 1)13) k=O

(rOO l(x)sin«2k + 1),8x) dx _

= f(_1)k

10

k=0

Set

1 := 1 (13) :=

1

00

(2k

1

+ 1)13

).

1 I(x) sin(fJx) dx - -.

o 13 By (21.18), we now see that it suffices to prove that 1(13) = 1/1(13), where 1/1(13) is defined by (21.8). Setting x = eU and integrating by parts, we find that 1= = =

1 (x 1 (e00 e- ne

o

00

nu -

o

-

e- ne

e-n / u

-x) + I sin(fJx) dx -

-I

13

du I + 1) sin(fJ logu)- - u 13

-~ foo (e- nu + ~e-n/u) cos(fJlogu) du 2 13

= -

u

1

~13 10t (e- nt + ~e-n/I) cos(fJ log I) dt 12 '

where we set u = 1/ t. Hence, 1

n roo (nl = - 213 10 e- + tI2 e-nit) cos(fJ 1og t) dt = -

n 2,8 (11

+ h),

say. Letting t = l/u in /z, we easily find that h = It. Consequently,

1 = -~It = -

13

=

~

roo e-nt cos (,8 log t) dt

13 10 _.!:.. roo (e-nttip + e-n't-iP) dt 213 10

37. Infinite Series

~ (n-iPr(ifJ +

-

2fJ

1) + niPr(-ifJ +

451

1))

~Re {n-itJr(ifJ + 1)}.

= -

Thus, we have shown that I (fJ) = l/J' (fJ), by (21.8). This completes the proof of (21.3). The remaining two claims in (21.8) follow as in the proof of (21.7).

Entry 22 (Formula (2), p. 280). Let, as usual, l/J'(x) = r/(x)j r(x). Then, if 0< x, rr 3

1 1 rr cot(rrx) rr log 12 sin(rrx) 1 - -- + + 2x 4rrx2 e2rrx - 1 2sinh2(rrx)

l/J'(x + 1) = -logx + 00

f=r

+2"

rr 00 log In4 _ x 41 -" (e 2rrn - I)(n 2 - x 2) 2 sinh2(rrn) n

f=r

~

-2rrnx ( 2 ~ sin(2rrkx)

n=1

k=l

-2rrL.,e

n L.,

k

2

+n

2

3

~£., COS(2rrkx»)

-n L.,

k==1 k(k

2

+n

2

)

•

(22.1)

Using Entry 8 of Chapter 30 (part IV [4, p. 374]), we obtain a formula for Euler's constant y, which is equivalent to (22.1), namely,

L

oo rr 1 rr log 12 sin(rrx)1 x2 -y=-logx+--+ --3 4rrx 2 2sinh2(rrx) k=lk(k2+x2)

+2

oo n L 2 n=1 (n2 + x )(e2rrn -

L

rr oo log In4 -- x 41 -1) 2 n=1 sinh2(7irn)

~

-2rrnx ( 2 ~ sin(2rrkx)

n=1

k=1

-2rrL.,e

n L.,

k

2

+n

2

3

~ COS(2rrkx»)

-n L.,

k=1 k(k

2

2

+ n)

•

(22.2)

In collaboration with J. M. Borwein and W. Galway, the author originally proved (22.2) by showing that the derivatives of both sides of (22.2) equal 0 and then letting x tend to 00 to show that both sides of (22.2) are equal to -yo Shortly thereafter, D. Bradley [1J found a more natural proof by working directly with the double series on the right side of (22.1), and therefore we shall give! Bradley's elegant proof. Bradley has shown that Entry 22 has several interesting consequences. In particular, Ramanujan's famous formula for ~(2n + 1) (Part II [2, pp. 275-276, Entry 21(i)]) follows as a corollary.

452

Ramanujan's Notebooks, Part V

Proof of Entry 22. We begin with the partial fraction expansion from Entry 3 of Chapter 30 (part IV [4, p, 359]), 7r 2 csc 2 (7rx) 7r 1 e21rX _ 1 = 3x - 2x2

+

1 '( 27r x3 - 1/1 x

t; (e'llrk -

4kx

00

+

+ 1)

t;

27r x3

00

1)(x2 - k 2)2 -

sinh2(7rk)(x4 - k 4)' (22.3)

By the product rule for differentiation,

7r 2 CSC2(7rx) e 2rrx - 1

d 7rcot(7rx) = dx e21rX - 1

7r 2 cot(7rx) 2 sinh2(7r x)

and d7rlogI2sin(7rx)1 7r 2 cot(7rx) dx 2sinh2(7rx) = 2sinh2(7rx)

7r1

+"2

og

12'( )Id h2( ) sm 7rX dx csc 7rX,

Hence, we can rewrite (22.3) in the form I

1/1 (x

+ I) =

7r l I d 7rcot(7rx) - - +-+3x 2x2 27rX 3 dx e 2rrx - 1

8

4kx

00

+

8

d 7rlogI2sin(7rx)1

+ ----='----=---dx 2sinh2(7rx) 27r x3

00

(e 2rrk - 1)(x 2 - k 2)2 -

sinh 2(7rk)(x4 - k4)

7r . d h2 - "2 log 12 sm(7rx)1 dx csc (7rx).

(22.4)

But,

7r d ---csch2(7rx) 2 dx

27re 21rx dx (e21rX _ 1)2

d = --

= ( -d

dx

= (~)2fe-21rkx = dx

k=!

)2 -:;,.----e 2rrx - 1

f(-27rk)2 e -21rkX. k=!

Using this in (22.4) and integrating both sides with respect to x, we deduce that 1/I(x

+ 1) =

7r I - ogx 3

I

1

+ -2x - -+ 47rX2 2k

+L 00

k=1 (e 21rk - 1)(k2 -

7rcot(7rx) e'llrx - 1

7rlogI2sin(7rx)1

+ -~---=--2sinh2(7rx)

7r log Ik4 _ x L - Sex) +c X2) 2 k=! sinh2(7rk) (22:5) 00

41

where C is a constant of integration to be determined and

Sex) :=

1

00

x

=

[00

h

log 12 sin(7rv) I f(-27rk)2 e-21rkV dv k=!

log 12sin(7r(x + u»1 E(_27rk)2 e -21rk(X+U) duo b!

~~

37. Infinite Series Using the Laplace transforms

1

00

o

e-

kt

sin(nl) dt =

n

2 2 k +n

1

00

and

0

e

-kt

cos(nl) dt = k 2

453

k

+n

2'

and the well-known Fourier series (Gradshteyn and Ryzhik [1, p. 46, formula 1.441, no. 2])

~

cos(2nky)

k=1

k

~

= - 1og 12' sm (ny )1 ,

y real,

in (22.6), we find that S(x) = (2n)2

f

k 2e- 2Jrkx

k=l

{'XJ e-27rku log 12 sin(n(x + u))l du,

10

1 1 1

2 ~k2 -brkx = - (2n) ~ e

k=1

00

e

-brku

0

k2 -2T 0 with af3 = 7r 2 /2, F(a) = F(f3).

(31.1)

This is a particularly beautiful result. Ramanujan evidently did not possess a complete proof, for he recorded the result (in abbreviated notation) as "The difference between the two series (af3 = 7r 2 /2) F(a) and F(f3) is O?" As we shall see, upon examining our proof below, (2n + 1)3 can be replaced by (2n + 1)4m+3, for any nonnegative integer m. Proof. Let

fez) :=

Z3

(cosh(az)

I

I·

+ cos(az» COS(27rZ) cosh(27rz)

Observe that fez) has simple poles atz = 2n+ 1, (2n+ l)i, (2n+ 1)7r(1 ±i)/(2a), for each integer n. Let {eN) ,N ~ 1, denote a sequence of rectangles having vertical and horizontal sides and centers at the origin. We shall choose the rectangles so that, as N tends to 00, the sides tend to 00 but remain at a bounded distance away from the poles of fez). It is then easy to see that lim

N-+oo

n,

f

leN

fez) dz

= O.

(31.2)

We apply the residue theorem. By straightforward calculations, for each integer

RZn+I

2( _1)n(2n = - 7r (cosh {(2n

R(2n+I)i

2( _1)n (2n 7r (cosh {(2n

=-

+ 1)3 sech B7r(2n + 1)} + l)a) + cos{(2n + l)a})

=

R-(2n+1),

+ 1)3 sech B7r(2n + 1)} + l)a} + cos {(2n + l)a)) = R-(2n+l)i,

(31.3)

(31.4)

and R(2n+ l)rr(l+i)/(2a)

«2n + l)rr(1 + i»3 sec {(2n + 1)rr2(l + i)/(4a)} sech {(2n + l)rr2(l + i)/(4a)} 8a 4 (sinh {4(2n + I)rr(l + i)} - sin {4(2n + I)rr(l + i)}) (-INj 2i(l + i)(2n + 1)3 sec {4(2n + l)tl(1 +i)}sechB(2n + l)tl(1 +i)} rra 2 (i - I) cosh {~rr(2n + I)}

=

2( _I)" tl 2 (2n + 1)3 rra 2 cosh {~rr(2n + I)} (cosh {(2n + I)tl} + cos {(2n + I)tl})·

(31.5)

Furthermore, we see that R(2n+I):rr(l+i)/(2a)

= R-(2n+I):rr(I+i)/(2a) = R(2n+I):rr(l-i)/(2a) =

R-(2n+l):rr(l-i)/(2a).

(31.6)

37. Infinite Series

463

Hence, applying the residue theorem to the integral of f (z) over eN, employing the calculations (31.3)-(31.6), letting N tend to 00, and using (31.2), we deduce that 8 - -;

00

~ cosh {(2n +

8/32

1)}

(_1)n (2n + 1)3 sech H1l'(2n +

I)a} + cos {(2n + I)a}

(-l)n(2n+I)3 sechH1l'(2n+I)} 1)/3} + cos {(2n + I)/3} = O.

00

+ 1l'a2 ~ cosh {(2n +

Multiplying both sides by 1l'a2 /8, we arrive at (31.1) to complete the proof. If we divide both sides of (31.1) by /3 2 and let a tend to 00, and therefore let /3 tend to 0, and replace (2n + 1)3 by (2n + I)4m+3, as we may, for any nonnegative

integer m, we deduce that (-I)n(2n + I)4m+3

?; cosh {~1l'(2n + I)} = O. 00

This result is due to Cauchy [1, pp. 313, 362] and is a special case of another theorem of Ramanujan (Part II [2, p. 262]). Entry 32 (Formula (3), p. 288). Ifa, /3 > 0 with a/3 (-I)n

00

~ (2n +

= 1l' 2 /4, then 1l'

I)(cosh {(2n + I)a} + cos {(2n + I)a})

= "8

(- on cosh {(2n +

I)/3} cos {(2n + I)/3} n=O (2n + 1) cosh (t1l'(2n + I)} (cosh {(4n + 2)/3} + cos {(4n + 2)/3})'

~ -2~

(32.1) Proof. Let fez) :=

1

I

z (cosh(az) + cos(az» cos(21l' z)

.

Note that fez) has simple poles at z = 0, 2n + 1, and (2n + 1)1l'i(l ± i)/(2a), for each integer n. Let {eN} denote the same sequence of rectangles as described in the proof of Entry 31. We now calculate the residues of fez). First, - I R0-2'

(32.2)

Second, by an easy calculation,

2(-on 1l'(2n + 1) (cosh {(2n + I)a} + cos {(2n + l)a}) =

R_(2n+I).

(32.3)

464

Ramanujan's Notebooks, Part V

Third, by straightforward calculations, (_I)n

R(2n+I)Jri(l±i)/(2a)

= - ]l'(2n + I) cos {(2n + I)(=FI + i)tJ} cosh HH(2n + 1)} =

(32.4)

R-(2n+I)Jri(l±i)/(2a).

By (32.4) and an elementary calculation, R(2n+I)Jri(l+i)/(2a)

=

+ R(2n+I)Jri(l-i)/(2a)

+ l)tJ} cos {(2n + l)tJ} H(2n + 1) cosh {~:rr(2n + 1)} (cosh {(4n + 2)tJ} + cos {(4n + 2)tJD· 4( _1)n cosh {(2n

(32.5) We now apply the residue theorem to the integral of f (z) over show that lim

N .... oo

1 eN

eN. It is easy to (32.6)

f(z) dz = 0.

Hence, from (32.2)-(32.6),

1

4

°= 1: - -; ?; -

(_I)n

00

+ l)(cosh {(2n + l)a} + cos {(2n + l)a}) 8 ~ (- nn cosh {(2n + l)tJ) cos {(2n + l)tJ} -; ~ (2n + 1) cosh {~:rr(2n + I)} (cosh {(4n + 2)tJ) + cos {(4n + 2)tJ})· (2n

Equality (32.1) now readily follows. If we let a tend to

00,

?;

and hence tJ tend to 0, in (32.1), we find that (_I)n

00

(2n

:rr

+ 1) cosh H:rr(2n + 1)} = 8'

which is a special case of another theorem of Ramanujan, Entry 15 of Chapter 14 (Part II [2, p. 262]). If we let tJ tend to 00, and therefore a tend to 0, we find that (32.1) yields the well-known evaluation (32.7)

Entry 33 (Formula (4), p. 288). [fatJ = ]l'2/2, where a, tJ > 0, then ~

f:o

(-I)ncos{(2n+l)a) :rr]l'3 (2n + 1) (cosh {(2n + l)a) - cos {(2n + l)a}) + "8 - 32a 2 =

t

n=1

sin(ntJ) sinh(ntJ) coth(n:rr) . n (cosh(2ntJ) + cos(2ntJ»

(33.1)

37. Infinite Series

465

Proof. Let fez) :=

cos(az)

z (cosh(az) - cos(az»

. cos(~1rz)

We observe that fez) has a triple pole at z = 0 and simple poles at z = 2n + 1, for each integer n, and at z = n1r(±1 + i)/a, for each nonzero integer n. Let { eN} ,N ~ 1, denote a sequence of rectangles having centers at the origin and horizontal and vertical sides approaching 00 as N tends to 00. The rectangles are also chosen so that the sides remain at a bounded distance from the poles of f (z) as N approaches 00. We now calculate the residues of fez). First, after a modest calculation, 1r2 I Ro=---. 8a 2 2

(33.2)

Second, for each integer n,

+ l)a} = R-{2n+l). + I)a} - cos {(2n + l)an

2( _1)n cos {(2n

R2n+l=

----..-:-'---~--'--'-----

1r(2n + I) (cosh {(2n

(33.3)

Third, straightforward calculations yield, for each nonzero integer n, ±( _l)n cos {n1r(±l

Rmr(±l+i)/a

+ i)}

= 21rin sinh(n1r) cos {nJ3(±1 +i)} = R-mr(±l+i)/a.

(33.4)

Using (33.4), we find that R±mr{l+i)/a

+ R±mr(-l+i)/a =

2 coth(n1r) sin(nJ3) sinh(nJ3) 1rn (cosh(2nJ3) + cos(2nJ3» ,

(33.5)

for every positive integer n. Lastly, we apply the residue theorem. It is easy to show that lim (

N.-+oo

leN

fez) dz = O.

(33.6)

Thus, by (33.2), (33.3), (33.5), and (33.6),

_! _ ~ ~

0- 1r2 - 8a 2

2

1r::O (2n

(-on cos{(2n + I)a} + 1)(cosh{(2n + l)a) - cos{(2n + l)an

4 ~ sin(nJ3) sinh(nJ3) coth(n1r)

+L... 1r n=l n (cosh(2nJ3) + cos(2nJ3»

.

After a slight amount of rearrangement, we deduce (33.1). If we let a tend to 0, and therefore J3 tend to well-known result

00,

in (33.1), we deduce the

466

Ramanujan's Notebooks, Part V

At the top of page 289, Ramanujan writes, "The difference between the series

() L:

-+ 8rr

oo

sin(n2()

n=l n(e Zm,. -

1~ (-IYB 4n + 2 ()2n+l

and

1)

"

4 ~ (2n + I)! (2n + 1) .

The sentence is not completed. Observe that, by Stirling's formula, the series on the right diverges for all () i= O. Most likely, Ramanujan realized that the series diverges and stopped here, or that he proceeded formally and could not evaluate the requisite integrals. These integrals do not appear to have evaluations in closed form. Note below that Ramanujan has a (possible) misprint in the first expression of the quote above. This entry should be compared with formula (3) on page 274 (Part IV [4, p. 298]).

Entry 34 (Formula (1), p. 289). Formally,

() 4rr

00

+~

sin(n2() n(e 2n:rr

1

- 1)

00

= 4~ +2

where B j, j

~

(_I)n B 4n + z()2n+I (2n

+ 1)! (2n + 1)

cos(2rrnx) L: 100 sin(x2() dx, 2rrx x(e - 1) oo

n --l

0

(34.1)

0, denotes the j th Bernoulli number.

"Formal Proor'. Apply the Poisson summation formula (Titchmarsh [2, p. 60]) to sin(x2() I(x) := x(eZrrx _ 1) .

Note that 1(0) = () / (2rr). Hence,

e I: -+ oo

4rr

n=I

sin(n2()

-

n(e 2nrr - 1) -

100 0

+2

sin(x2() x(e 2rrx - 1)

dx

cos(2rrnx) L: 100 sin(x2() dx. 2rrx x(e - 1) oo

n--I

0

(34.2)

Now, by Entry 16(iv) of Chapter 13 (Part II [2, p. 220]),

1

00

o

sin(x2() ---:----dx x(e 21l"X - 1)

00

= '" ~ 00

=

~

(_lye 2n + I (2n

+ I)!

(_l)n()2n+I

(2n

+ 1)!

1

X 4n + I

00

0

e2rrx - 1

B 4n + 2 8n

+4

dx (34.3)

Note that the inversion in order of summation and integration has not been justified. Indeed, the series on the far right side of (34.3) diverges. Putting (34.3) in (34.2), we complete the "proof."

37. Infinite Series

467

Ramanujan begins page 312 by stating two series identities involving the Mobius function f.L(n). He then offers a general claim, which contains the previous two results as special cases, and which is an analogue of the Poisson summation formula, with f.L(n)fn as coefficients. We first formally state Ramanujan's general claim. We then show that the two examples follow from the general claim. Next, numerical calculations show that Ramanujan's two examples are false but that the errors are numerically very small. Lastly, we indicate how G. H. Hardy and J. E. Littlewood corrected Ramanujan's claims. In our calculations below we employ the prime number theorem in the form L~l f.L(n)fn = O. Perhaps Ramanujan used a stronger, but incorrect, version of the prime number theorem, which would account for the excellent numerical agreement in the two applications. As we saw in Chapter 24 (part IV [4]), Ramanujan made several errors in deriving approximations to 1l'(x) that involve f.L(n).

The author [8] has established a general arithmetical Poisson summation formula which can be applied to series with f.L(n) as the coefficients. However, it is considerably more complicated than Ramanujan's claim. Similar arithmetical summation formulas were derived via contour integration by the author [5] and by P. V. Krishnaiah and R. Sita Rama Chandra Rao [1]. Entry 35 (p. 312). Let af3 = 21l', where a, f3 > O. Let 1jf(n) = Then

~ 2

f:

1

00

qJ(x)cos(nx) dx.

f.L(n)qJ(afn) =

n

n=1

f:

f.L(n)1jf(f3In) .

n

n=1

Entry 36 (p. 312). If p > 0,

1;

f.L(n)n

00

p2

+

1l'

n2

=

p 1; 00

f.L(n)e- 2lf !(n P)

n

Proof based on Entry 35. Apply Entry 35 with qJ(x) = If(x2 + 1). Now, by contour integration or tables (Gradshteyn and Ryzhik [1, p. 445, formula 3.723, no. 2]), 1jf(n) =

1

00

o

cos(nx)

x

2

+1

1l' -n dx = -2 e .

Thus, by Entry 35, a

~ f.L(n)n

2" f:r a 2 +n 2

a

=

~

f.L(n)

2" f:r n (afn)2 + 1)

= ~ f.L(n). "!...e- fJ1n , n 2

f:r

468

Ramanujan's Notebooks, Part V

or

Entry 37 (p. 312). If p > 0,

Proof based on Entry 35. Replace a and fJ in Entry 35 by ,.fiX and ..!p, respectively, so that ,.fiX

2 whereafJ

t

JL(n)f{J(,.fiX/n) =

n

n=l

t

JL(n)1/!("!p/n) ,

(37.1)

n

n=l

= 4Jl'2. Letf{J(x) = exp(-x 2). Then 1/!(n) =

1

00

o

2

e-x cos(nx) dx

.jii n2/4 = _e• 2

Thus, by (37.1),

or

We now numerically examine Entries 36 and 37. If 0 < p < 1,

- t;( _

00

_

2

p)

k

00

/L(n) _

~ n 2k+l

-

t; ~(2k + 00

(_I)kp2k

1)'

where we have used the fact that L~l/L(n)/n = 0, which is equivalent to the prime number theorem.

37. Infinite Series

469

Next, using this same fact, we find that

~

E

E f: ~ (_~rr)k = ~ f: f: ~ (_ = ~ f:(t =~

/L(n)e- 2rr /(n p ) p n=1 n

p n=!

/L(n) n k=O k!

np

p n=1

/L(n) n k=1 k!

2rr)k np

2rr)k_l_ P k! {(k 1)

p k=1

Thus, Entry 36 may be rewritten in the fonn (-llp2k {(2k + 1)

00

~ Setting p we find that

rr

(2rr)k

00

P~ -p-

=

1 k! {(k + 1)

(36.1)

= ~ and summing the first 50 tenns of each series via Mathematica, (_1)k

00

"

2k

=

+ 1)

6 2 {(2k

-0.1600806325 ...

and 2rr

L (-4rr i k=1

1

00

k! {(k

+ 1)

=

-0.1600806298 ....

Since the series are alternating, and since 1

2102{(1O3) < 1.98

x 10-

31

2rr(4rr)5I (51)!{(52) < 4.65 x 10- 10 ,

and

=

these calculations show that (36.1) is false for p If p > 0, since L:I /L(n)/n = 0,

E n=1

/L(n) e- p/ n2 n

=

E E n=l

/L(n) n

(-ll k!

k=l

Also, by a similar calculation,

!?i /L(n)e- rr2 Vp ~ n 00

/(n 2p )

_ -

(.!!.-

n2

~.

rE =

(-I)k pk . k=1 k! {(2k + 1)

!?i (_I)k rr Vp ~ k! pk{(2k + 1). 2k

00

Thus, Entry 37 may be rewritten in the fonn

(_I)k pk !?i (_1)k rr ~ k! {(2k + 1) = Vp b k! pk{(2k + 1). 00

00

2k

(37.2)

Setting p = 1 and summing the first 50 tenns of each series above by Mathematica, we find that

b 00

(-ll

k! {(2k

+ 1) =

-0.4805338008 ...

470

Ramanujan's Notebooks, Part V

and

..[ii

L 00

k=1

(_1)k;rr2k

kl (2k •

~

+ 1)

= -0.4805622889 ....

Since the series are alternating, and since

(51)!

1

~(103)

...fii;rrICYl

< 6.45 x 10- 67

~'---~ < 5.85 X 10- 16 ,

and

(51)!

~(103)

we conclude that (37.2) is false for p = 1. Now, in fact, during his stay in Cambridge, Ramanujan told G. H. Hardy and J. E. Littlewood about Entry 37, and they discuss the formula and the more general claim, Entry 35, in their paper [1] (Hardy [3, pp. 20-97, especially pp. 57-63]). Assuming that all of the zeros of ~ (s) are simple and that the series on the far right side below converges, Hardy and Littlewood proved that 00

~

J.L(n)e- p / n2 =

n

t

ff. J.L(n)eVPn=1 n

1f2 /(n 2p )

__ 1_ 2...fii

L

(~)P

p

f'(4 - 4p)

p

~/(p)'

(37.3)

where the latter sum is over all complex zeros p of ~(s), arranged according to increasing moduli. Thus, Ramanujan's claim in Entry 37 must be altered by the latter expression on the right side of (37.3). This is then another instance where Ramanujan's ignorance of the complex zeros of ~(s) led him astray. Hardy and Littlewood also briefly address the more general formula in Entry 35. Although they do not give a complete proof, they clearly demonstrate that Ramanujan's claim must be modified by a similar sum over p. Returning to (37.3), Hardy and Littlewood [1, p. 161] showed that the estimate

~

(_p)k

_

f:r k! ~(2k + 1) -

oc

-1/4+,)

p,

(374) .

as p tends to 00, where € is any positive number, is equivalent to the Riemann hypothesis. Titchmarsh [3, pp. 186-187] also provides a proof of the corrected version (37.3) of Ramanujan's Entry 37 and briefly mentions (37.4) (with the exponent € unfortunately missing) as well [3, p. 328]. We are very grateful to Richard Brent for correcting some inaccuracies in our discussions of the past two entries in an earlier version of this chapter. Although Entries 35-37 are false, it would be exceedingly interesting to discover Ramanujan's heuristic arguments, since Hardy and Littlewood's approach is through contour integration. The next result generalizes Entry 35 and is stated in the notebooks without the latter series on the right side of (38.2). If rp(z) == 1 and we omit the latter series on the right side of (38.2), we obtain (36.1).

37. Infinite Series

471

If we set ~(z) = 1/ r(~{z + I}), replace p by .;p, and again ignore the second series on the right side of (38.2), we deduce (37.2). To see this, we first note that if n is odd, ~(-n) = O. If n = 2m is even, we observe that

22m

22m r(~{2m

(2m)! r(~{-2m

+

+ I})

(2m)! 1f(_l)m

I})

(-2)m(2m - 1)(2m - 3)·· ·1 (2m)!

(_l)m

.fii

m!

.fii'

Hence, (37.2) readily follows. Entry 38 (p. 312). Let ~(z) denote an entire function, and put fez) :=

~(z)pZ

,

cos(~1fz){(z)

where p is any fixed nonzero complex number, and where {(z) denotes the Riemann zeta-function. For simplicity, assume that each nonreal zero of {(z) is simple. Let eN denote the positively oriented circle of radius N + ~ centered at the origin, where N is a positive integer. Suppose that

r

lim

N ...... OO]CN

fez) dz

= O.

= 1f

(-I)n~(-n) (21f)n

(38.1)

Then 00

~

(-1)n~(2n

{(2n

+

I)p2n+l

+ 1)

00

~

+ 1)

n! {en

+:: L 2

p

~(p)pP cos(~1fp){/(p)'

P

(38.2)

where the sum on p is over all nonreal zeros of {(z) arranged according to increasing moduli. Proof. We apply the residue theorem to the integral in (38.1). Observe that fez) has simple poles at z = 2n + 1, for each integer n, at z = - 2n, for each positive integer n, since {( -2n) = 0 (Titchmarsh [3, p. 19]), and at each nonreal zero p of {(z). By a straightforward calculation, for each integer n, R 2n +1 =

2( -1)n~(2n

+ 1) p2n+l

1f{(2n + I)

-

(38.3)

From the functional equation of {(z), (0.4), for each positive integer n, . Z + 2n hm - z--+-2n {(z)

(_l)n(2rr)2n+l

= 1f r(2n + 1){(2n + 1)

(38.4)

Hence, R_2n

(21f)2n+l~( -2n) p-2n

= ------~----~-rr(2n)!{(2n

+

1)

(38.5)

472

Ramanujan's Notebooks, Part V

Lastly, for each nomeal zero p of ~ (z), a simple calculation gives R _ rp(p)pP P - cos(~ll'PW(p)'

(38.6)

Thus, applying the residue theorem, using (38.3), (38.5), and (38.6), letting N tend to 00, and employing (38.1), we find that 2

- -; ?; +2

00

+ 1)p2n+l 2 (-I)nrp(-2n _ ~(2n + 1) + -; ~ ~(-2n -

(-1)nrp(2n

f:

00

rp(-2n) (21l')2n n=l ~(2n + 1)(2n)! p

l)p-2n-l 1)

+L

rp(p)pP - 0 P cos(~1l'p)~/(p) - ,

(38.7)

where the sum on p is over all nonreal zeros p of ~(z) arranged according to increasing moduli. Now, by the functional equation (0.4) of nz), ~(-2n - I) = 2( _1)n+l (21l')-2n- 2r(2n

+ 2)~(2n + 2).

Using this in the second sum in (38.7), we deduce that 00

~

(-1)nrp(2n

+ l)p2n+l

~(2n + 1)

= -1l'

rp(-2n - 1) (21l')2n+l 2)(2n + 1)! p

?; ~(2n + 00

00

rp(-2n)

+ 1l' ~ ~(2n + 1)(2n)!

+:: L

rp(p)pP 2 P cos(~1l'p)~/(p)'

(21l')2n

p

(38.8)

The first two series on the right side of (38.8) can be combined into one series, and since 1/ ~ (1) = 0, the proof of (38.2) is complete. In Ramanujan's formulation of Entry 39, the latter series on the right side of (39.2) does not appear. Entry 39 (p. 312). Let rp(z) be an entire function, and put fez) := f(~(z

rp(z)pZ , cos(~1l'z){(z)

+ 1»

where p is any fixed nonzero complex number, and where ~ (z) denotes the Riemann zeta-function. Assume,for simplicity, that all nonreal zeros p of ~(z) are simple. Let eN be the same circle as in Entry 38, and assume that

lim

N-+oo

1 eN

fez) dz = O.

(39.1)

37. Infinite Series Then

f:

(-l)n(ji(2n

n=l

{(2n

+ l)p2n+l

+ 1)n!

-.J1if: -

(-l)n(ji(-2n)

+

n=l {(2n

rr ' "

+-L...2

P

473

(~)2n

l)n!

p

(ji(p)pP

1

f(2{P

1 ' + I}) cos(2rrp){'(p) (39.2)

where the sum on p is over all nonreal zeros p of {(z) arranged according to increasing moduli.

Proof. The proof follows along the same lines as the proof of Entry 38. The function fez) has simple poles at z = 2n + 1, for each nonnegative integer n, at z = - 2n, for each positive integer n, and at z = P, for each nonreal zero p of {(z). Note that the zero of cos(4rrz) at z = 2n + 1 is cancelled by the zero of II r(4cz + 1) at z = 2n + 1 when n is a negative integer. A simple calculation yields 2(-W(ji(2n + 1)p2n+l R2n 1 - - ---'-----'-----'----'....:..-+ rr{(2n l)n! '

n:::: O.

+

(39.3)

By (38.4) and the duplication formula (or the functional equation) of the gamma function, for each positive integer n, 2( - W(ji( --2n) (rr )2n

(2rr)2n+l(ji( -2n)p-2n R_2n

=

rrr( -n

+ 4)r(2n + 1){(2n + 1)

= .J1in! {(2n

P

+ 1)

(39.4)

For each nonreal zero p of {(z), we easily see that R P-

(ji(p)pP

f(Hp

(39.5)

+ I}) cos(4rrp){,(p)·

Invoking the residue theorem, using (39.3)-(39.5), letting N tend to ploying (39.1), we conclude that

_~

f:

rr n=l

f:

+ l)p2n+l + ~ + l)n! .J1i n=l +L

(-l)n(ji(2n {(2n

(-1)n(ji(-2n) {(2n

+ 1)n!

and em-

(~)2n p

(ji(p)pP

P r(4{p

00,

+ l})cos(4:rrp){'(p)

-0 -

,

where the sum on p is over all nonreal zeros p of {(z) arranged according to increasing moduli. The last equality is equivalent to (39.2).

Entry 40 (p. 324). For each positive integer n, n I l

~ (4k -

2)3 - (4k - 2)

1

n

= "2 ~

2n

+ 2k -

1.

(40.1)

474

Ramanujan's Notebooks, Part V

Proof. The following proof is due to R. Sitaramachandrarao [1]. For any nonzero number a and positive integer n, define n 1 rp(a, n) = 1 + 2 {; (ak)3 _ ak'

Then it is easy to see that

6 n

1 (4k - 2)3 _ (4k _ 2) =

4{rp(2, 2n) -

rp(4, n)}.

(40.2)

By Entry 1 of Chapter 2 and Example 4 in Section 5 of Chapter 2 (part 1[1, pp. 25,31]), rp(2, 2n) - rp(4, n) = 1 + 2

1

1

1

1

L--- - - L -k - L -k 2n + k 4n + 1 2 2n

4n

2n

k=n+1

k=1

1

1

1

1

411+1

k=2n+1

1

=-+2 L --- L -- L 4n + 1 k=2n+1 k 2 k=n+1 k k=2n+1 k 411

=

kf..1

411

1

1

1

2n

k - 2: k~1 k =

2n

t;

411+ I

1

2n-1

2k + l'

(40.3)

Putting (40.3) in (40.2), we obtain an equality that is easily seen to be equivalent to (40.1). In the entry immediately following Entry 40, Ramanujan proposes an equality between two finite sums of inverse tangent functions, ttan- I (2n

+ ~k _

1) =

ttan- I (2k 31+

1)'

However, a line has been drawn through the right side. Indeed, it is easily checked that this claim is false, in general.

Entry 41 (p. 324). For each positive integer n,

n-I tan-I (

~

1

) (2n + 2k + 1).v'3 -

n-\

~

tan-I

1

(

(2k

+ 1).v'3

)3

(41.1)

Proof. We induct on n. For n = 1, (41.1) is trivial. Assume that (41.1) is valid. We thus will prove that (41.1) holds with n replaced by n + 1. Recall that, for 0 ~ x y < 1, tan- I

x+ tan-I y= tan-I (x1 -xy + y).

(41.2)

37. Infinite Series

Thus, by induction, n I Ltan-

k=O

1

(

(2k

-t; -t; -

n-I

-

n-2

+ 1).J3

tan-I

(

tan-I

(

+ tan-I

)3 1 ) + (1)3 + l).J3 1 ) + (1) + + + l).J3 + 1).J3 )3 + l).J3

(2n

+ 2k + l).J3 2

(2n

(

(2n

475

tan-I

(2n

tan-I

2k

(2n

1

Thus, it remains to show that

(41.3)

By (41.2),

tan-I

C4n +1l).J3) + tan-I C4n +\).J3) _ tan-I (

-

3(4n

.J3(8n + 4) + 1)(4n + 3) -

1

)

_ tan-I ( .J3(2n + 1) ) 3(2n + 1)2 - 1

-

= tan-I (.J3(2n

+ 1) (1 + 3(2n + 1)2)) + 1)4 - 1

9(2n

= tan-I

C2n +\).J3) + tan-I C:zn +1l).J3Y ,

by another application of (41.2). Thus, (41.3) has been proved, and the proof is complete. On pages 334, 335, 340, and 341, Ramanujan offers four related claims about products of certain alternating series. In particular, on page 335, he asserts that "af - 2ala2 + (2ala3 + ail - (2ala4 + 2a2a3) + ... oscillates between (al a2 + a3 - ... )2 ± (rr /2) limn--+ oo na;. For example,

1- ~+(~+~)-(~+ ~)+(:S+ 58+~)-'"

476

Ramanujan's Notebooks, Part V

oscillates between

and :n:

2 We state this claim more precisely. Suppose that, for 0 ~ x ~ 1, f(x) has the power series representation L~I (_l)n-I anxn, where an > 0, 1 ~ n < 00. Let g(x) denote the power series obtained by forming the Cauchy product of f(x) with itself. If L := limn~oo na; exists and is finite, then the even and odd indexed partial sums of g(1) tend to f2(1) + (:n:/2)L and f2(1) - (:n:/2)L, respectively. In particular, if an = 1/../ii, then the even and odd indexed partial sums of g (1) tend

{(1 -

r

{(l -

r-

.J2)~(~) + :n:/2 and .J2R(~) :n:/2, respectively, where ~ to denotes the Riemann zeta-function. The case when L = 0 implies that the series g(l) converges. In the case when L = +00, the result states that the even and odd indexed partial sums diverge to +00 and -00, respectively. On page 340, Ramanujan states a generalization of the foregoing result for the kth power of f(x), where k is any positive integer exceeding 1. Finally, on page 341, Ramanujan offers a similar theorem for the product of k (possibly distinct) alternating series L~I (_1)n-I anl , L~I (_1)n-I an2 , ... , L~l (_l)n-I ank under the hypothesis that limn~oo n k- 1a nlan2 ... ank exists or is +00. The statement on page 334 is the special case k = 2 of the last claim. We shall prove Ramanujan's assertions under appropriate assumptions. Our results are possibly not as general as Ramanujan intended. Slightly stronger theorems can undoubtedly be established at the cost of additional technical details in the proofs. (See the remarks following our proof of Entry 42.) Ramanujan's results are quite remarkable for their explicit description of the behavior of the partial sums of certain alternating divergent series. We know of no other comparable results in the literature. The results in this section first appeared in a paper by the author and J. L. Hafner [1]. We begin with a simple lemma concerning the asymptotic behavior of a certain finite sum. Lemma 42.1. Let a and {J denote constants with 0 < a, (J < 1, 1 fn(x) = x«(n - x)P'

n::::

1,

and n-I

c(n) =

L

k=l

fn(k),

n:::: 2.

(42.1)

37. Infinite Series Then, as n tends to

00,

r(1 - aW(1 - fJ)

c(n) = r(2 _ a _ {J)na+/H

{(fJ)

{(a)

+ -;p + --;;;- + 0

477

(1) + (1) na+1

0

nP+I

'

where { denotes the Riemann zeta-Junction. Proof. Define, for x > 0 and n ?: 1, qI(x)

=

1 1 1 - -a x a (1 - x)P x (1 - x)P

+1

(42.2)

and

Note that g(j)(x) = _1_, qI(j) (~) n na+P+J n

j ?: O.

,

Since, for j ?: 0, (')

qI J

(x) =

{

I

0< x _ < 2' I I 1 2 SX < ,

O(X -j+l-a) ,

0«1 _ X)-HI-p),

unifonnly for x in the given ranges, it follows that

0 (x-jp:;a) ,

g(j)(x)

=

o

0 < x:5 n12,

(nn_X)-HI-P) na+1 ,n12 :5 x < n.

(42.4)

We now apply the Euler-Maclaurin summation formula (0.5) to gn (x). Recalling that H2 (x) denotes the second Bernoulli polynomial, we find that

n-l

Lgn(k) = k=1

/n-I

gn(x) dx

+ 4gn(1) + 4gn(n -

(42.5)

1) -I- Rn,

I

where

I

I/n-I

Rn = 12 {g~(n - 1) - g~(l)} - 2 = 0 (na:

l )

+0

(n p1+1)

I

+0

H2 (x - [xDg;(x) dx

(42.6)

(na;p+I) '

as n tends to 00, by (42.4). (Note that the notation Rn has a meaning here different from that in (0.6).) Using (42.4) and (42.6) in (42.5), we deduce that

n-I

{;gn(k) = as n tends to 00.

10{n gn(X) dx + 0

(

1) + (1)

n a+1

0

n P+1

'

(42.7)

478

x

Ramanujan's Notebooks, Part V

Recalling the definitions (42.2) and (42.3) of ({J and gn, respectively, and setting

= nt in the integral above, we find that

1 n

gn(x) dx = na;P-I

11

({J(t) dt

_ _1_ { r(l - a)r(1 - fJ) _ _ 1_ _ _ 1_ r(2 - a - fJ) 1- a 1 - fJ

- n a+p- I

+

I}

'

where we have used the classical integral representation for the beta function. Hence, by (42.7), as n tends to 00,

1 {r(l - a)r(1 -

n-I k "8n( ) = - n a+p- I

8

II} +

fJ)

- -- - -1- a 1 - fJ

r(2 - a - fJ)

+ 0 (n a1+1 )

+0

1

(42.8)

(n:+ I ).

Now, from the definitions (42.1) and (42.3), we deduce immediately that n-I 1 n-I 1 1 n-I 1 n - 1 C(n)=L8n(k)+pL k a+----;;L( -k)P- a+p k=1 n k=1 n k=1 n n n-I = Lgn(k) bl

1

+p

nIl L

n

bl

ka

n

1

n

+ ----;; L

bl

kP -

Recall (Part I [1, p. 150]) that for any complex number r n r (;k

n r +1

~~(-r)+r+l

n+1 n

a+p·

(42.9)

i= -1, as n tends to 00,

nr 00 B2k r(r + l)n r - 2k+1 +2+(;(2k)!r(r-2k+2)'

(42.10)

where B j, j ?: 2, denotes the jth Bernoulli number. Employing (42.8) and (42.10) in (42.9), we conclude that c(n) =

1 (1 - - + -1) - - -1 (I)} n + 2 + (I)}

r(1 - a)r(1 - fJ) --r(2 - a - fJ)n a +p- I n a+p- I

1{ 1{ + (n

1- a

+_

I-a -a {(a)+_n_+~+O 1- a 2

+ na

{(fJ)

nP

0

u1+ 1 )

1

P

+ 1n _- fJ +0

(n

n- P

0

n a+p

__

a +1

nP+1

p1+ 1 )

_ r(l - a)r(l - fJ) + {(a) - r(2 - a - fJ)n u+P- 1 nP

1 - fJ

+ ~(fJ) + 0 nU

(_1_) + (_1_)

which completes the proof of Lemma 42.1. Our next lemma extends the result to multiple sums.

n u+1

0

n P+I

'

37. Infinite Series

479

Lemma 42.2. Let k ~ 2, let ai, az • ... ,ak be constants such that 0 < ak :::; ak-l :::; ... :::; al < 1 and al + az + ... + ak ~ k - 1, and let Yk az + a3 + ... + ak - k + 3. Then Yk > 1 and there exist constants bkj and fJkj such that for each j, 1 :::; j :::; 3k - 1 - 1, al + az + ... + ak - k + 1 < fJkj < 1 and

=

1

(42.11) as n tends to 00.

Proof. Naturally, we induct on k. The case k = 2 is just a restatement of Lemma 42.1 with a = al and fJ = az (assuming fJ :::; a). In this case, we have b 21 = ~(a2)' b22 = ~(al). fJ21 = ai, fJ22 = a2. and Y2 = a2 + 1. Suppose that Lemma 42.2 is valid with k replaced by k - I. where k - 1 ~ 2. Then fOT some constants bk-I,j. fJk-l,j. with 1 :::; j ::: 3k- 2 - 1 and 0 < al + a2+···+ak-I-k+2oo nanbn is zero, infinite, or finite, when an and bn do not contain any logarithmic functions." It seems that Ramanujan is tacitly assuming that an = (a + Enl)n~al, and bn = (b + En2)n- a2 , where a, b, aI, and a2 are constants with aI, a2 > 0, and where Enl and En 2 tend to 0 as n tends to 00. Ramanujan evidently assumes similar hypotheses for the aforementioned claims on pages 335, 340, and 341. Indeed, our theorem can undoubtedly be generalized by replacing lIn? by (a j + En,})n ja j , where, for each j, 1 :::: j :::: k, aj is a constant and En,j is a suitable function approaching 0 as n j tends to 00. We have been able to show that this claim is correct under the assumption that En,j, 1 :::: j :::: k, is monotonic, though this is probably not the weakest assumption under which Ramanujan's assertion would hold. We forego giving the details of the proof of this more general result. Adolf Hildebrand has kindly pointed out to us that Ramanujan's weak assumption is not sufficient for the conclusion of Entry 42 to hold, even in the case k = 2. He observes that the series in (42.12) cannot oscillate if ck(2n) - ck(2n - 1) > 8n for sufficiently large n, with 8n > 0 and E:I 8n = 00. We reconstruct his example here. In the notation of the previous two paragraphs, set

IIn,

In(l

a/l = bn =

where En tends to zero,

E:I E; =

00,

Ek = En

+ En),

if n is even, if n is odd,

and En satisfies the relation

+ O(E;),

(42.14)

uniformly for ..(ii :::: k :::: n, as n tends to 00. For example, we can take En = (log log n)-1/2. Define, in analogy with (42.11), c2(n) = c(n) = akan-k. If n is even, then

EZ::

c0)=

~ (1 ~ k=1

k even

=

, C

+ Ek)(l + En-d .jk(n - k) n-I

(n)

2 ""

+ f;;t

1

k=1 k odd

.jk(n - k)

/I-I

Ek

.jk(n - k)

k even

= c'(n)

~

+~

+ 2d(n) + e(n),

""

+ f;;t

k even

EkEn-k

.jk(n - k)

37. Infinite Series

483

say, where n-I

1

k=1

Jk(n - k)

L

c'(n) =

= 7r

+ O(n- I / 2),

according to Lemma 42.1 and the fact that r(~) = ..j7i. To estimate e(n) we proceed as follows: e(n)

=

{" " "I +

L

l:sk:S.;n k even

L

.;n 1, is a sufficient condition for convergence. For complex an, the most general result is due to W. J. Thron [1] who proved that (50.2) converges if lanl ::5 e lle for n sufficiently large. We now state Ramanujan's claim on page 390. He asserts that (50.2) is convergent when

! {~ + (n logI n)2 + (n log n log 1 + ... } ' n2 log n)2

1 + log log an < - 2

(50.5)

37. Infinite Series

491

and is divergent when the left-hand side is greater than the right side when any 1 is replaced by 1 + E. This statement needs some clarification. First, the series on the right side of (50.5) is finite for each n and persists as long as the iterated logarithms remain positive. Ramanujan evidently had in mind a test for the convergence of (50.2) when an ::: 1. If for n sufficiently large, (50.5) holds, then Ramanujan claims that (50.2) converges. On the other hand, if one of the numerators 1 in the series on the right side of (50.5) is replaced by 1 + E, for some fixed number E > 0, and if there exists a subsequence ani tending to 00 for which the left side of (50.5) is greater than or equal to the right side with one of the numerators 1 replaced by 1 + E, then Ramanujan claims that (50.2) diverges. An easy calculation shows that Barrow's theorems (50.3) and (50.4), even in the stronger form when the inequality < in (50.3) is replaced by ~, are contained in Ramanujan's assertion (50.5) with the right side of (50.5) truncated after the first term. In the remainder of this section we follow the analysis in Bachman's paper [1]. We shall establish a version of Ramanujan's claim for complex an. So that the exponentiation is unambiguous, we assume that the sequence of complex numbers Ibn}, 1 ~ n < 00, is given and set n:::1.

(50.6)

n:::

(50.7)

With this definition, ,an

1,

is well defined. We first give the following test for the convergence of {En} for complex exponents.

Theorem 50.1. Let {an} and {En} be given by (50.6) and (50.7), respectively. Set

n:::

(50.8)

1,

and define {En}, n ::: 1, by (50.7) with an in place ofan. Then, if{En} converges, {En} must converge as well. The test above is of independent interest. In particular, Thron's result [1] follows from Barrow's theorem for real exponents an, 1 ~ an ~ el/e, and Theorem 50.1. To state Bachman's results concerning Ramanujan's test for convergence, we introduce the following notation for iterated logarithms. Setting XI = e and LI (x) := L(x) := logx,

X :::

e,

we recursively define Xk and Lko for k ::: 2, by Xk := eXk - 1 , and Lk := Lk-I (L(x»,

Entry 50a (p. 390). Let {an} and {En} be defined by (50.6) and (50.7). respectively. Then {En} converges if there exist positive integers ko and no, such that for all

492

Ramanujan's Notebooks, Part V

I

+ log I logan I = 1 + log Ibnl 1 { -1 - 2 n2

< -

+

1 (nL[(n»2

1 + ------:(nL[(n)L 2(n»2

+ ... + _____1_ _ _~} (nL[(n)L2(n)··· Lko(n»2

. (50.9)

Entry 50b (p. 390). Let {En} be defined by (50.7), where the sequence {an} is real, an > 1, for every integer n, and

1{ 1 1 + log logan ~"2 n2

+

1

(nL[(n»2

1 +(nL[(n)L2(n) ... Lko_[(n»2

+

1

(nL[(n)L 2(n»2

+ ...

1+E } + (nL I (n)L2(n)···L ko (n»2

' (50.10)

for n ~ no, for some positive integers ko and no, and for some infinite exponential {En} diverges.

E

> O. Then the

We first set some convenient notation and establish three useful lemmas. The first lemma reduces the principal case of our problem to an equivalent problem that is easier to attack. Set and Also set I fo(x) := -

x

and

fn(x) =

I

xLi (X)L2(X)'" Ln(x)

,

n

~

l. (50.11)

Lemma 50.2. Let {xn}, n ~ 1, be a sequence of real numbers such that Xn > 1. Define another sequence {XII}, n ~ 1, by Xn = exp

C

+e Xn ).

(50.12)

Then [XI, X2, ... ] converges if and only if there exists a sequence {Yn }, n such that Yn ~ -1 and such that the inequality 1 + Yn ~ (1

+ Xn)eYn+l

~

1,

(50.13)

holds.

Proof. Since Xn > I, the sequence [XI, X2 • ... ] is monotonically increasing. Hence, to prove that it converges, it suffices to show that it is bounded. By (50.12)

37. Infinite Series

493

and (50.13),

In the opposite direction, suppose that [Xl, X2, ••• ] converges. Since, Xn > 1, then [x n , Xn+ 1, •.. ] also converges for each n ::: 1. Denoting the limit of [x n , Xn+J. .•• ] by el+ Yn , we observe that Yn ::: -1 and that

Thus, we deduce (50.13) with equality, and this completes the proof of the lemma. The next two lemmas are the primary ingredients in the proofs of Entries 50a and50b.

Lemma 50.3. Let T!: , C~, and X! be defined by k

(50.14)

T!: := Llj(n - 1), j=O

(50.15) and

(50.16) where l j (n) is defined in (50.11), and where k ::: 0 and n ::: 2 are any integers for which the right sides of (50.14) and (50.15) are defined. Then there exists a sequence of integers (nd such that,for n ::: nb

(50.17) Proof. Let k ::: 0 be fixed. For brevity, set Tn = Tnk and Xn = X!. By (50.14) and (50.11), Tn = Ok(1/n) < 1, for n ::: nb say, and where the notation Ok indicates that the implied constant is dependent on k. For such n, we can expand the right side of (50.16) in a Taylor series about 0 and so find that 1 + Xn

+ Tn)e- Tn + , = (1 + Tn) (1 - Tn+l + 4(Tn +d 2 - i(Tn +d 3 + Ok(n- 4 ») = I + Tn - Tn+l + 4(Tn+d 2 - TnTn+1 + 4Tn(Tn+l)2 - i(Tn +d 3 + Ok(n- 4 ). =

(l

(50.18)

Now, by (50.14) and (50.11), expanding Tn about n ' T.n = T.n+l - Tn+l

+ 21 Tiln+l -

+

I, we find that

(;1 T.II! ; ,

(50.19)

494

Ramanujan's Notebooks, Part V

for some number g, such that n < k

g < n + 1. Note that k

j

T~+1 = Llj(n) = - Llj(n) Ll;(n), j=O

j=O

2 (1)

= -+Ok n3

(50.20)

;=0

n 3 10gn

(50.21)

,

and

=L k

Tt'

l'J'(g - 1)

= Ok(n-4 ).

(50.22)

j=O

Substituting (50.19)-(50.22) into (50.18) and simplifying the resulting expression, we find that

1 + Xn

'121 + n

=1-

-33

Tn+1 - 2:(Tn+1)

+ Ok

(1) n ogn 31

•

Hence, by (50.20), (50.14), and (50.15), we deduce that

, 121 + n

Xn = -Tn+l - 2:(Tn+1)

t

-33

+ Ok

(-:3:-/-)

=

~

=

c! + 3n13 + Ok ( n 3 / ogn )

j=O

l 7(n)

+ 313 + Ok n

(1) 31 n ogn

n ogn •

(50.23)

Thus, for n :::: nk, (50.23) implies (50.17), and this completes the proof of Lemma 50.2.

Lemma 50.4. Let Tnk and X! be defined by (50.14) and (50.16), respectively. Moreover, let x! be defined by

x! =

exp (

1 +Xk) n • -e-

(50.24)

Then (50.25)

Proof. We begin with the observation that it suffices to show that there exists a sequence of integers {nD such that (50.25) is valid for each n :::: n~. Indeed,

37. Infinite Series

assuming this, we have, for each l

· I1m

[k

m~oo

k k Xn'Xn+1""'Xm

1=

495

n k,

~

[k k k Xn'Xn+1' ••• 'Xlk' I'1m [k Xe+I'Xlk+2" " ' Xm m~oo

1]

by (50.24) and (50.16). To exhibit the existence of such a sequence Ink}' we first observe that, by (50.24), Lemma 50.2, and Lemma 50.3, any infinite exponential [x!, X!+I' ... 1, with n ~ nb is convergent, where {nk} is a sequence defined in the statement of Lemma 50.3. Denote the limit of such an infinite exponential by e1+S!. Then (50.25) will follow if we can show that (50.26) for all n ~ nk ~ nk. To this end, we define, for integers k ~ 0 and n ~ nb the numbers

t! by (50.27)

We will deduce (50.26) from the three inequalities,

s! > S! > 0,

k > l; n

~

sup(nk' ne),

tnk > - 0, and tk

> tk (

m -

n

L (m k

-

Lk(n)

1»)t!/2 l

k

(m - 1)

(50.28) (50.29)

,

m > n ~ n~,

(50.30)

where nk is sufficiently large, where in the case k = 0, Lo (x) =: x. Indeed, assume that (50.26) fails for k = 0 and some n ~ n~. Then, by (50.29), t~ > 0, and so, by (50.30) and (50.14), we find that

m - 1)1~/2 t~ ~ t~ ( - n lo(m - 1) > io(m - 1) =: T~, for some m > n, where m is sufficiently large in terms of t~. But, by (50.27), this implies that ~ < 0, which contradicts (50.28). Thus, (50.26) is valid with k = 0 for all n ~ n~. Proceeding by induction on k, assume that (50.26) holds up to k - 1. Assume, to the contrary, that (50.26) fails to hold for some k > 0 and n ~ nk' By the same argument as used above, we find that

t! > ik(m -

I),

for some m > n that is sufficiently large in terms of and (50.14) shows that

t!. This. together with (50.27) (50.31)

496

Ramanujan's Notebooks, Part V

by the inductive hypothesis, provided that m ~ n~_I' which we may assume. But since (50.31) contradicts (50.28), we conclude that (50.26) and therefore (50.25) hold. Thus, it remains to prove (50.28}-(50.30). To that end, assuming that Ink} is increasing, as we may, we find that, for k > l andn ~ nk ~ ni, and so (50.32) by (50.17), (50.15), and (50.24). Recall that Euler [1], [2] showed that the infinite exponential with constant exponents e1/e converges to e. This fact, together with (50.32), yields (50.28). To prove (50.29), we first observe that, for m > n ~ nk, [k k k I+T!+I] _ e I+T: , m < xn,xn+1, ... ,xm,e

k k k] [ xn,xn+1"",x

T;,

by (50.24) and (50.16). Hence, S~ :::: and so (50.29) holds by the definition (50.26) of t~ . For the proof of (50.30), we first observe that, by the definition of S~ and (50.24), k

el+S~ = [X~' el+S~+I] = e(l+x~)in+1

.

Hence, S~ satisfies (50.16) with Tnk replaced by S~. We now fix k and write Sn, Tn, and tn for S~, Tnk , and t~, respectively. From our last observation it follows that

+ Sn)e- Sn + = (1 + Tn)e- Tn + tm and m = n, n + 1 into the last identity, we find that

(1

Substituting Sm

= Tm

-

1

t

_n_

1 +Tn

= 1_

l •

e-tn+l.

(50.33)

By (50.29), (50.28), (50.14), and (50.11), (50.34) Hence,

provided that n ~ n~, where n~ is sufficiently large. Using this bound on the right side of (50.33), we deduce that

Hence, for any integers m > n

~

n",

tm > tn

mn-I 1 + ti+l/2 . i=n 1 + T;

(50.35)

37. Infinite Series

497

We use (50.35) twice. First, by (50.35) and (50.34), tm > In

for m > n

~

n +1

m-I

I=n

}"T.:" = In exp

1) > tn exp (-?: Ii LI=n log }"T.:" + I=n

(m-I

I

m-I

)

,

I

n", where n" is sufficiently large. Now, by (50.14) and (50.11),

t; Ii

m-I

m-I

~

=

ko k

j; k

i j (; - 1)
tn exp ( -

j+1 (m

- 1») = In (m _ I)LI (m _11) ... Lk(m _ 1) (50.37)

for any integers m > n ~ n k. We now reiterate the argument above but this time using (50.37) instead of (50.29) on the right side of (50.35). Employing also (50.36), (50.34), and the inequalities tn i k (i)/2 < Ii/2 «k Iii, we find that 1m > tn

mn-I

i=n

1+ +

tn i k (i)/2

1

Ii

= tn exp (~l L...J og

> tn exp

(~Htnik(i) -

> tn exp

(~tn (Lk+1 (m -

= tn (

Lk(m Lk(n)

i=n

Ii)) 1) - Lk+1 (n» -

1»)10/2ik(m -

(1 +

tn i k (i)/2))

1 + Ii

t

}=o

L j+1 (m - I»

I).

where the second inequality above holds for m > n ~ nk' where n" is sufficiently large. Thus, (50.30) is established, and the proof of Lemma 50.4 is complete. Proof of Theorem 50.1. We assume at the outset that an =1= ]. for each n ~ 1, for otherwise both [ai, a2, ... ] and [al. a2, ... ] converge trivially. Fix a positive integer n and, for any complex number z. set d f(z) := dz [a" a2 • ... ,an, z]

(50.38)

and (50.39)

498

Ramanujan's Notebooks. Part V

Foreachm > n. [al. a2 • ...• am] - [al. a2 • ...• a,,] = [al. a2 • ...• a". [a,,+l. a,,+2 • ...• am]]

(50.40) Setting (50.41) and (50.42) we find, upon estimating the right side of (50.40), that I[al, a2, ... ,am] - [aI, a2, ... , a,,] I =

=

Ilu I(z) dzl 111 1(1 +

(u - l)t)d(l

::: lu - 1111 1/(1 + (u

-

+ (u

- l)t)1

1)1)1 dt.

o

(50.43)

Thus, by (50.38), (50.6), (50.39), and (50.8),

and g(z) =

n"

Ibkl[ako ak+J. ... , an, z].

k=1

Hence, by (50.6), (50.8), (50.41), and (50.42), we obtain the inequalities

1/(l - t + ut)1

=

n" n"

Ibk[ako ak+J. .•. , a", (1 - t

+ ut)]l

Ibkl[ako ak+l, ... , a", (1 - t

+ lult)]

k=1

:5

k=!

::: g(1 - t

+ wt),

(50.44)

37. Infinite Series

499

which is valid for 0 ::: t ::: 1. Applying (50.44) to the right side of (50.43), we find that I[al, a2,···, am] - [ai, a2, ... , an]1 :::

lu - 1111 g(l

lu -1111 =- g(l + (w

°

w - 1

- I)t)d(l

+ (w -

+ (w -

I)t) dt

I)t)

= lu - IIll w g(z) dz w-

I

lu - 11 ([A A A] [A A A ]) = - - al,a2, ... ,arn - al,a2, . ··,an , w -1 by (50.39) and (50.42). Observe that w > 1, since an :f= 1, and so Moreover,

lu - 11 = =

If: ~(bn+l[an+2' k.

1

n

> 1.

an+3,···, arnDkl

::: L ,(lb +l l[a +2, a +3,.·., a k=1 k. 00

an

11

lebn+,[an+2,an+3, ... ,aml -

k=1

(50.45)

n

m ]/

n

= w - 1.

(50.46)

Theorem 50.1 now follows from (50.45), (50.46), and the Cauchy criterion for convergence.

Proof of Entry 50a. By Theorem 50.1, it suffices to consider real exponents an :::: 1. In such a case, [a1' a2, ... ,an] is monotonically increasing, and so it suffices to show that it is bounded. Define the sequence {cn} == {C~o+1} by Cn

=exp (

1 + C~+I)

e

,

where C!o+! and nko+1 are defined in the statement of Lemma 50.3. Setting C n C!o+1, we find that, by (50.15), (50.11), and (50.9), 1 + log log Cn = log(l

+ Cn)

1 = Cn - 2C;

=

+ O(C~)

to

>

4L (] (n) :::: 1 + log log an, j=O

for n :::: no, sufficiently large in terms of ko, as we may assume. Therefore, for n :::: no, we have an ::: Cn, and so Thus, it suffices to show that the infinite exponential [c no ' c no + I, ... ] converges. By Lemma 50.2, this, in tum, is equivalent to the existence of a sequence Sn, n =

500

Ramanujan's Notebooks, Part V

no, no

+ 1, ... , such that Sn

~

-1 and

1 + Sn ~ (1

+ Cn)e sn +

(50.47)

1•

But, by Lemma 50.3, 1 + Cn = 1 + C!o+l < 1 + X!o+l = (1

+ T;o+l)e- Tn+

kO+1

1 •

Hence, (50.47) is satisfied with Sn = Tnko+ 1 , n ~ no. This completes the proof of Entry 50a. Proof of Entry SOb. We proceed by contradiction. Suppose then that the infinite exponential [aI, a2, ... ] is convergent. Then, since an > 1, so is [an, a n+), ... ] convergent for any n ~ 1. Denote the limit of the latter infinite exponential by e 1+S•• Define also a sequence {An} by 1 + An) . an = exp ( --e-

(50.48)

Then (50.49) In the remainder of the proof, n always denotes an integer such that n ~ no. For such n, it follows immediately from (50.10) that An > 0, since an > e1/e. Moreover, by (50.48), (50.10), (50.11), (50.15), and (50.17), An ~ 10g(1

for n

~

+ An} = 1 + log log an

~ C!o

+ 4Eiio (n)

> C~+l > X!o, (50.50)

no, where no, which depends on k and E, is sufficiently large. Thus, ko

an > Xn '

x!o

where is defined by (50.24). Therefore, by the definition of Sn and Lemma 50.4, we find that Sn > Tnko .

For brevity, set Tn

(50.51)

= T;o, Xn = X!o, (50.52)

and by (50.51). By (50.50), (50.23), (50.15), and (50.11),

Bn

~ C!o + 4Ei~°(n) -

X!o =

4Ei~(n) -

3

13 + Ok ( n 3/ogn ) > 4Eii o(n),

n

(50.53) where n ~ no, and where no is sufficiently large. The remainder of the argument in Bachman's paper [1] is incorrect. We are very grateful to A. Hildebrand for supplying the following elegant argument to complete the proof. We begin with a lemma.

37. Infinite Series

501

Lemma 50.5. For M, N :::: no,

Proof. From (50.11), we observe that Lj(x) ij(x) = - - . Lj(x)

Thus,

L M

S·-

n=N+l

in

Since ij(x) = Lj(x)/Lj(x) is decreasing for x:::: no, L'.(n - 1)

> Lj(n - 1) 1

n-l

L'.(x)

-l-dx. Lj(x)

Thus,

and so the proof is complete. Recall from (50.18) and (50.49) that (50.54)

and Thus, from (50.52) and (50.54), 1+ ~ 1 + Xn

= 1 + An = 1 + Sn e-Rn+1 = 1 + Xn

1 + Tn

(1

+ ~_) e-Rn+l. 1 + Tn

Hence, R n + 1 = log (1

Since

Rn+l ::::

+ ~) 1 + Tn

log (1

+ ~-) . 1 + Xn

0 by (50.52), equality (50.55) first implies that Rn Bn -->--

1 + Tn - 1 + Xn'

and then, with the use of the inequality, h

logO +x +h) -log(l +x)::: 1 +x'

x,h:::: 0,

(50.55)

502

Ramanujan's Notebooks, Part V

secondly implies that

R n+ l

.::;

(

Rn Bn) ( Bn) -I 1 + Tn - 1 + Xn 1 + 1 + Xn 1

= 1 + Xn + Bn

(1 ++ 1

1 + Xn .::; 1 + Tn Rn - Bn =

Xn ) Tn Rn - Bn

e

-T. n+1

Rn - Bn,

where, in the last inequality, we used the fact that Rn+ 1 where, in the last equality, we used (50.54). Thus,

~

0 from (50.52), and

Iterating this inequality, we find that > ...

>R eTn+1+,,+Tn+m _ n+m

m

+ '" Bn+}-I . eTn+I+··+Tn+J ~

j=1 m

> ' " Bn+}-I . eTn+I+··+Tn+j • _~

(50.56)

j=1

By Lemma 50.5, for n

~

no,

eTn+1+.+Tn+j

> ik(n)>> - ik (n + j) i k (n

1

+j

- 1)

(50.57)

Hence, using (50.53) and (50.57) in (50.56), we conclude that m

Rn

» Lik(n + j

- 1).

j=1

Since m > 0 is arbitrary and since L~I ik(n + j -1) diverges, we have reached the desired contradiction, and so the proof is complete.

38 Approximations and Asymptotic Expansions

One of the primary areas to which Ramanujan made fundamental contributions, but for which he received no recognition until recent times, is asymptotic analysis. Asymptotic formulas, both general and specific, can be found in several places in his second notebook, but perhaps the largest concentration lies in Chapter 13. Several contributions pertain to hypergeometric functions, and an excellent survey of several of these results has been made by R. J. Evans [1]. The unorganized pages in the second and third notebooks also contain many beautiful theorems in asymptotic analysis. This chapter is devoted to proving these theorems and a few approximations as well. On pages 270-273 of the second notebook, Ramanujan examines some related functions that can be considered as hybrids of the Riemann zeta-function and hypergeometric functions. Some of these results were established in a paper with Evans [2]. Entries 2-8 contain accounts of Ramanujan's findings described on these pages. In Entries 12 and 13, Ramanujan determines the asymptotic behavior of some multivariate exponential series. These results are in the spirit of several theorems that can be found in Chapter 15 (part II [2, pp. 303-314]). In Entry 16, Ramanujan derives the asymptotic expansion of 00

2L 1 and a > 0 by l;(s, a)

=

00

L(k +a)-s. k=O

Thus, from (5.2) and (6.2), un(a)

=

(-It f(a ~o

+ k)-n-2 ts(n + 1, r + 1) =0

= (-on ts(n + 1, r + 1) r=O

t

t

(~)(_l)r-J (k + ay-Jaj

j~ J

(~)(_1)r-jajl;(n + 2 + j

j=O J

- r, a).

(6.3)

By the Euler-Maclaurin summation formula, (0.5) of Chapter 37, as a tends to forRes > 1,

00,

l;(s, a)

rv

~(_1)m Bm{m + s - 2)! a'-s-m,

!;;;Q

m! (s - 1)!

where Bm , m ;:: 0, denotes the mth Bernoulli number. Employing this asymptotic expansion in (6.3), we deduce that, as a tends to 00, un(a)

rv

~ t(-I)n+r+ j s (n + 1, r + x

l)C)

~(_l)m Bm(m + n + j - r)! a- n+r- m- I.

!;;;Q

m! (1

+n +j

t

(6.4)

- r)!

t

Inverting the order of summation on j and m in (6.4), we are led to the inner sum -1 J(r)(m+n+j-r)! = (m+n-r)! (-r)}(m+n-r+1)j J=O( ) j (l+n+j-r)! (l+n-r)! j=O j!(n-r+2)j (m

+ n - r)!(1 + I)! (n -

(n - r

- m),. r + 2)r '

514

Ramanujan's Notebooks, Part V

by Vandennonde's theorem (Bailey [1, p. 3]). Observe that if 0 < m :::: r, where r ~ I, then the sum above equals O. Extracting the term for m = 0, we find that, from (6.4), as a tends to 00, n (n -r)! r! un(a)'" L(-It+rs (n+l,r+l) , a- n+r- l r=O (n + 1). n

+ 1, r + 1)

+ L(-l)n+rs(n r=O X

~ ~

m=r+l

m Bm

+ n - r)! (l - m)r a -n+r-m-l . r + I)! (n - r + 2)r

(m

(-1) -

m! (n -

(6.5)

We now calculate the coefficients of a- N , 1 :::: N :::: 8. First, the coefficient of 1/a equals sen

+ 1, n + 1)

n! (n+l)!

1 = --, n+l

in agreement with (6.1). For the remainder of the calculations, it will be convenient to use the following formula (c. Jordan [1, p. 150, formula (3)]): s(n, n - m) =

where

Cl,D

=

-1, C1,k

= 0 if k =1= 0,

Cm+1,k

= -(2m -

~ Cm,k(2mn_ k).

(6.6)

and k

+ I)(Cm ,k + Cm,k-]).

(6.7)

The coefficients Cm,k for 1 :::: m :::: 6, 0 :::: k :::: 5, are found in a table on page 152 of Jordan's book [1]. For n ~ 1, the coefficient of a- 2 equals

(n-I)! n(n+l) 1 -s(n+l,n)(n+l)!= 2 (n+l)n

1 2

If n = 0, the series in (6.5) also yields! for the coefficient of a- 2 .

For n ~ 2, the coefficient of a- 3 equals sen

+ I, n -

2(n - 2)! {(n + 1) (n + I)! = 3 4

n-2

1) +

2

2

n

(n

+ 3

I)}

(n

2

+ l)n(n _

1)

1

= -4- + "3 = "4 + 6' For n = 0, I, we obtain the same value. For the remainder of the calculations, we assume that n ~ N - 1. In each case, the same formula for the coefficient of a- N also holds for 0 :::: n :::: N - 2, from an examination of the double sum in (6.5).

38. Approximations and Asymptotic Expansions

515

The coefficient of a -4, if n ~ 3, is equal to

We employed MACSYMA in (6.5) to calculate the remaining coefficients as polynomials in n. We then collected terms to write Ramanujan's coefficients as polynomials in n to verify that the polynomials agree for each coefficient. The coefficients of a- 5 , a- 6 , and a- 7 are, respectively,

+ 1, n -

sen

4! 4)! (n -

3)---(n + I)!

1) + 2IOe; 1) + 130(n; 1) + 24(n; I) }4~~:-1:;! 15n + 15n IOn -

= {I05(n; 3

-sen

240

+ I, n -

= {945(n

+

8

2 -

5! (n - 5)! 4)---,-----,-(n + I)!

~ I) + 25220(n; I) +

924(n

+ 7

1) + 120(n + I)} 6

2380e; I)

5! (n - 5)! = 3n 4 (n + I)!

+ 2n 3 -

96

7n 2

-

6n

'

and 6! (n - 6)! sen + l,n - 5 ) - - - (n

{10395(n

+ I)!

~ 1) + 34650(n ~ 1) + 44IOO(n ~ 1) + 26432e; 1)

+ 7308(n +

I) + no(n + I)} 6! (n+- I)!6)!

63n 5

-

8

-

315n 3

7

224n 2

4032

(n

+ 140n + 96

The coefficients of sen + I, n - 6) are not found in Jordan's book [1]. Thus, we used (6.6) and (6.7) to calculate the needed coefficients. Therefore, the coefficient

516

Ramanujan's Notebooks, Part V

of a -8 equals - s(n

=

7! (n -7)! 6)--'---'(n + 1)!

+ 1, n -

1) + ~ 1) + 303660(n + 1) 64224(n + 1) + + + {135135(n ~

54054O(n

10

9

23n 3 + 114n2 1152 This completes the proof. 9n 6

-

9n 5

-

75n 4

-

1) + + I)} (n +

866250(n ~ 5040(n

705320e ~

7!

8

(n

1)

7)! I)!

+ 80n

Although our first proof is a natural one, Ramanujan's formulas for the coefficients of a- N indicate that another approach utilizing less calculation was employed by him. In our second proof, calculations lead to coefficients in the form given by Ramanujan, but the calculations are even more difficult to perform by hand than in our first proof. Second Proof. For n

~

0 and a

(a +k)-n-2 =

+k > (n

0,

1

+ I)!

1

00

e-(a+k)ttn+1dt.

0

Using also (5.2) and inverting the order of summation and integration by absolute convergence, we find that un(a)

=

roo f: (k + l)n(e-t )k e-attn+1dt

10

1 1

(n

k=O

e-al t n + 1

00

=

o

00

=

+ I)!

(n

+ 1)(1 -

rl)n+l

dt

(6.8)

e-al({Jn(t) dt,

where we have used the generalized binomial theorem and set ({In(t) = n

~1

C y+l _t e- I

Applying Watson's Lemma (Olver [1, p. 71]), we deduce that

L ({J~k) (O)a -k-l , 00

Un (a) ""'

k=O

as a tends to 00. Thus, it remains to calculate ({J~k)(O), 0 ((In(t)

1

= n+1(

?; Br 00

7!"(-t)

:s k :s 7. Since

r )n+!

,

38. Approximations and Asymptotic Expansions

517

where Br denotes the rth Bernoulli number, we may readily calculate the coefficients. With the help of MACSYMA, we easily verified Ramanujan's coefficients of a- N , I ~ N ~ 8.

Entry 7 (Formula (5), pp. 27'J,..273). Let a, p > O. As p tends to

S(a p ) ',

00,

I -2 E (_l)np2n E (2p(a+n)n-l '" - - e + a + n)n+I 2ap (a + p)2n+1 ' oo

p

. - n=O

oo

n=O

where P2n := P2n(p), n

(7.1)

I, is a polynomial in p of degree n - 1. In particular,

~

1 2p

Po(p) = - ,

P2(p) =

1

6' 1

P4 (p) = 30

1 42

p

+ 6' 5p2

P

+ 6 + 18'

P6 (p)

=

Ps(p)

= 30 + 10 + 9"'" + 54'

I

5

P\O(p) = 66

3p

7p2

5p

I7p2

35 p 3

35p4

+ 6" + -6- + -9- + --yg'

691

Pdp)

35 p 3

= 2730 +

69Ip 210

6I6p 2

45I p 3

385p4

385p 5

+ ~ + ~ + ~ +~,

and P ( ) _ ~ 14 P - 6

+

Moreover, for n

35p 2 ~

+

7709p2 90

+

26026p 3 2002p4 135 + 9

+

7007 p 5 54

+

5005 p 6 162 .

I,

P2n (p)

_ (2n)! (n-1 - 2. ~n! p

+

n(n - 1) n-2 10 P

+

n(n - I)(n - 2) { 200 (n - 3)

+ n(n - I)(n - 2)(n - 3) {en _ 4)(n _ 5) 6~

+

+

60 (n _ 4) 7

+

20} n-3 p

+ '1

90} pn-4 7

n(n - I)(n - 2)(n - 3)(n - 4) { 240000 (n - 5)(n - 6)(n - 7)

120 + 7(n - 5)(n - 6)

3720 + "'49(n -

5)

6OOO} p n - 5 +... ) . + 77

(7.2)

518

Ramanujan's Notebooks, Part V

Lastly, for n ~ 2 and n even,

(_I)n/2-1 Pn(p)

+ (n + l)BnP + {

= Bn

+{

+{ +

+

+ l)(n + 2) 3

(n

+ l)(n + 2)(n + 3)

(n

+ l)(n + 2)(n + 3)(n + 4)

18

120

(n

Bn -

Bn -

n 2 (n - 1)

9

Bn -

180

n(n - l)(n - 2)(n - 3)

+{ -

(n

}

Bn -4 P

n 2 (n 2

1)(n

-

270

+ 2)

Bn -

6

} 2 Bn-2 P

} 3 B n-2 P

n 2 (n 2 - 1)

36

Bn-2

4

+ 1)(n + 2)(n + 3)(n + 4)(n + 5) 2700

n(n - 1)

Bn

2

n(n - l)(n - 2)(n - 3)(23n - 25) } 5 5400 Bn -4 P

+... ,

(7.3)

where B j' j ~ 0, denotes the j th Bernoulli number and where P2n (p) has degree n -1,forn ~ O.

Before proving Entry 7, we shall show that the case (}oo = ~ of Entry 4 follows from Entry 7.

Completion of the Proof of Entry 4. Let

P

A(p) = e (S(1, p/2) _

l-;-P).

In Entry 4, Ramanujan is claiming that, if the rational function within the large parentheses of (4.1) is expanded in powers of 1/ p when () = ~, then the first five terms coincide with the asymptotic expansion of A(p) in powers of 1/ pas p tends to 00. Expanding this rational function in powers of 1/ p, we therefore must prove that I 2 16 3712 A(p) '" - - - + - -56 + -45 + ... , (4.4) P p2 3 p3 3p4 p5 as p tends to 00. Now let a = 1 and replace p by p/2 in (7.1). Then, as p tends to eP

(S(I, p/2) _

.!.) p

= _

-

~

(_l)n P2n(P) ~ (1 + p/2)2n+l

2

+ p2(1 +2/p)

00,

+ 0 (~) p6

4/3 p3(1 +2/p)3

32(3~ + ~) - ----;:-'-:--::---:-:-p5(1 +2/p)5

38. Approximations and Asymptotic Expansions

519

(4.5)

Proof of Entry 7. From Entry 5 and (6.8), for 0 < p < a,

Sea, p)

(_2p)n

L 00

=

n=O

1 2p

- -

- -1

2p

1 1 n!

00

0

00

0

= - 1 - -1 2ap

unCal

e- at

00

(n

(-2pt)n+l dt e-t)n+l

+ 1)! (1 -

t) } (2 Pt)

e- at { exp (-_-2P e t - 1

1

00

2p

?;

0

- 1 dt

e -at exp - - - dt.

(7.4)

e- t - 1

Referring to the definition of Sea, p) in (7.1), we see that Sea, p) represents an analytic function of a and p for Re a > 0 and Re p > O. Likewise, the right side of (7.4) is analytic for Re a > 0 and Re p > O. Thus, by analytic continuation, (7.4) is valid for all a and p with Re a > 0 and Re p > O. Multiplying both sides of (7.4) by -e 2p , we see that (7.1) is equivalent to the asymptotic expansion _1 2p

roo e-(a+p)t exp (2P + pt + ~) dt t

10

rv

e- - I

as p tends to

00.

(7.5)

roo e-(a+p)t dt,

10

the asymptotic expansion (7.5) is equivalent to

1

00

0

(-l)n P2n(P) , ~ (a + p)2n+l

Since the first term on the right side of (7.5) is equal to _1_ 2p

-1 2p

~

e-(a+p)t {exp(p w(t) - l}dt

rv" f=: - + 00

(

l)n p 2n ( P ) , (a p)2n+l

(7.6)

where w(t):= 2

u

+t +-- = 2 e-

t -

1

tcoth(t/2)

= -2 L 00

n=l

B

~t2n, (2n)!

It I < 21l', (7.7)

where B j , j :::: 2, denotes the jth Bernoulli number.

520

Ramanujan's Notebooks, Part V

Let us now define the polynomials P2n(P), n ePw(t) - 1

~ (_1)n P2n (p)

- - - = L..J 2p

n=1

(2n)!

I, by the expansion

~

t2n

'

It I < 21l'.

(7.8)

We therefore shall prove that the polynomials P2n(p) have the properties enunciated in Entry 7. By employing MACSYMA, we verified that P2(p), P4(P), ... , P I4 (P) are indeed given by the formulas displayed in Entry 7. Second, we remark that it is easy to see from (7.8) that the polynomial P2n (p) has degree n - 1, n ~ 1. Next, we prove (7.3). From (7.8),

w" pn-I L (_l)n(2n)!P2n(p) 2n -- -21 L---"-n! ' 00

00

t

n=1

n=1

In particular, by (7.7), I2(-I)"P2n(O)t 2n = W n=1 (2n)! 2

=-I2

B2n t 2n , n=! (2n)!

It I < 21l'.

(7.9)

It I < 21l'.

(7.10)

Equating coefficients of t 2n ,n ~ 1, we find that P2n (0) = (_l)n-1 B 2n , as claimed by Ramanujan in (7.3). Next, differentiating (7.9) with respect to p and setting p = 0, we find that

=4'

(7.11)

(-I)np" (0) w3 L t 2n - n=1 (2n)! 2! - 12'

(7.12)

00

~

(-I)np~n(O) (2n)!

2n

t

w2

00

2n

?; 00

00

~

(-I)np~(o) 2n (2n)! 3!

t

(-I)nPi:)(o) 2n (2n)! 4!

t

= =

w4 48'

(7.13)

w5 240'

(7.14)

and (-I)n p(5) (0) w6 L 2n t2n - - n=1 (2n)! 5! - 1440· 00

(7.15)

Since the coefficient of pm, m ~ 0, in the Taylor series of Pn (p) about p = 0 equals p~m)(O)/m!, we can calculate the coefficients of pm on the right side of (7.3), for 1 ::::: m ::::: 5, by equating coefficients of t 2n on both sides in each of the foregoing five equalities. These calculations are facilitated by observing from (7.7) that

38. Approximations and Asymptotic Expansions

521

Hence, w 3 = 2W2

+ t(W 2)' + wt 2,

+ 2t(w 3 /3)' + w 2t 2, w 5 = 2W4 + 2t(w 4/4)' + w 3 t 2,

W4 = 2w 3

and w6

= 2w 5 + 2t(w 5 /5)' + w 4t 2.

U sing these five equalities in (7.11 )-(7.15), utilizing (7.10), and employing MACSYMA, we can readily verify that each of the coefficients given by Ramanujan in (7.3) is correct. Lastly, we prove (7.2). Replacing p by 1/P and t by t,Jii in (7.8), we find that 00

(

~p(exp(w(tv'P)/P)-I)=~ or exp (w(t v'P)/ p) - 1

=2

_l)n P2n (1/ p ) (2n)! _(tv'P)2n,

f: (- on

n=l

Q2n(P)t 2n , (2n)!

(7.16)

where Q2n (p) := pn-l P2n (1/ p) is a polynomial in p of degree n - 1. Thus, the coefficient of pn-m in P2n(P) equals the coefficient of pm-l in Q2n(P). Now, by (7.7),

2~B2n 2 ,~B2n 12n ~ --(tv'P) n = -2 ~ _ _ pn- t .

u(p, t) := w(tv'P)/p = - P

n=l

(2n)!

n~l (2n)!

(7.17)

In particular, (7.18) Thus, from (7.16), 2

f:

n=l

(-ltQ2n(0) t 2n = (2n)!

f: k=l

uk(O, t) = k!

f: (__ k=l

I)k t 2k. 6kk!

Equating coefficients of t 2n , n ~ 1, on both sides, we find that Q

2n

(0) _ (2n)! - 2-~n!'

as claimed by Ramanujan. Next, by (7.17), (7.19)

522

Ramanujan's Notebooks, Part V

Thus, from (7.16), (7.18), and (7.19), 2

t

=

(-WQ;n(O) t 2n

n=l

(2n)!

t

=

uk-leO, t)up(O, t) (k - I)!

k=l

t

(-t 2/6)k-l(t 4/360).

k=l

(k - 1)!

Equating coefficients of t 2n , n :::: I, on both sides, we find that (2n)! n(n-I) 2·6nn! 10

(2n)! Q ' (0) _ 2n - 2. 6 n- 2360(n - 2)!

which again agrees with Ramanujan. Next, by (7.17), upp(O, t)

4B6t6

= -(5"! = -

t6

(7.20)

63 .35'

Thus, from (7.16) and (7.18)-(7.20), 2t(-ltQ2n(0)t 2n n=l

(2n)!

- t; _

00

= t(uk-2(0,t)U~(0,t) + Uk-l(O, t)upp(O, t)) (k - 2)!

k=l

(k - I)!

(_t 2/6)k-l(_t 6 /(6 3 • 35)))

(_t2/6)k-2(t4/360)2

+

(k - 2)!

(k - I)!

Thus, equating coefficients of t2n, we find that Q2n(0)

(2n)!

-2-!- = -4-

=

(1

6 n- 4(360)2(n - 4)!

+6

1l

1)

35(n - 3)!

(2n)! (n(n - I)(n - 2)(n - 3) 2 . 6n n! 200

.

+

n(n - I)(n 70

2))

'

as claimed by Ramanujan. The remaining two coefficients recorded by Ramanujan are similarly calculated, and we omit the details. (We used MACSYMA to effect the calculations.) We now return to the task of establishing (7.6). We would like to employ Watson'sLemma, but we cannot do so becauseexp(pw(t)) - I depends upon p. Thus, a completely new procedure seems necessary. We prove a very general theorem (Theorem 7.1 below) from a paper by the author and R. J. Evans [2] that includes (7.1) as a special case. Define ~ r(n

T(x):= L n=O

+ s)

n!

(a

(x

+ n)n-r

+ a + n)n+s

'

(7.21)

where r is a fixed positive integer, and a and s are fixed complex numbers with positive real parts. Note that when r = s = I, T(2p) = S(a, p), where S(a, p) is defined in (7.1).

38. Approximations and Asymptotic Expansions

Theorem 7.1. Let N denote a positive integer. Then as x --+ I Argxl ~ 8, where 8 > 0,

!iT r~l

T(x) = '"' A

20

kX

~k~s

_

e~x

(N~l

C ( )

'"'

m X

!;;:o (a + x/2)m+l

00

523

in the sector

+ O(x~1-r~N/2) )

'

(7.22)

where Ak =

t(_l)k~jr(S + k)(~)(a + jl~r,

(7.23)

J

j=O

and where the functions Cm(x) (defined in (7.32)) have the estimate Cm(x) as x tends to

=

(7.24)

O(x[m/21~r),

00.

The arbitrarily small positive number [, is fixed throughout the sequel. Observe that (7.22) is a genuine asymptotic expansion, in view of (7.24). Note also that x can be replaced by x + b in (7.22), for any constant b. Thus, for example, if the sign of a is reversed in the denominator of (7.21), then Cm(x)/(a + x/2)m+l is replaced by 2m+1Cm (x - 2a)/xm+l. If r = s = 1, the leading sum on the right side of (7.22) equals l/x. Furthermore, assuming the validity of (7.22), we conclude that Cm (2p) = {

0,

ifm is odd,

(_l)n P2n(P),

if m = 2n is even.

Before beginning the proof of Theorem 7.1, we need to define several functions and prove an auxiliary lemma. Consider the confluent hypergeometric function

Izl This function is related to U(s, s of the second kind, by

+ r; z),


O. The proof of Lemma 7.2 below makes heavy use of Faa di Bruno's formula (J. Riordan [1, p. 36J, S. Roman [1])

~h(g(t)) = '"' n! hk(g(t)) (gl dtn

L...Jkl!k2!···kn!

1!

)kl (g2)k2 ... (gn )kn , n!

2!

(7.40)

where the sum is over all integers kl' k2' ... , kn for which

k;

~

0, 1 :::: i ::: n,

(7.41)

526

Ramanujan's Notebooks, Part V

and where k = kl hk(z)

+ k2 + ... + kn, dk

= dz k h(z),

and

Lemma 7.2. Fix an integer N ?: 1. As x --+ f(N)(t,x) = 0

di gi := gj(t) := d-:-g(t). t' with I Arg x I ~ ~ 1l'

00

-

(7.42) 8,

(x-r+[N/21~lxtlj),

(7.43)

uniformly for t in [0, 1].

Proof. Let 0 ~ t ~ 1 and n ?: O. We shall obtain uniform estimates for the nth derivatives of each factor (_t)r-I, w S , e-t(l-W+I/2), and U(r, s +r; wx) of f(t, x) in (7.31) and then combine them to deduce (7.43) from Leibniz's rule. First, for each n ?: 0, d n (-t) r dt - I=o (1), n

(7.44)

since r is a positive integer. Next, by (7.30), we find that, for each k ?: 0, dk dt k W = 0(1).

(7.45)

= ZS and g(t) = w,

Consequently, by (7.40) with h(z)

dn -d W S = 0(1). tn

(7.46)

For IArg zl ~ 41l' - 8, U(r, s + r; z) is analytic (Olver [1, p. 257, eq. (10.04)]), and so we can differentiate (7.27) (Olver [1, pp. 9-10, Theorem 4.2]) to obtain, for k ?: 0 and Izl sufficiently large, k -d U( r, s dz k

+ r; Z)

L oo

~

m=O

(r)m+k(-l)m+k(l - s)m = O( z-k-r) . m! zm+r+k

(7.47)

Now apply (7.40) with h(z) = U (r, s +r; z) and g(t) = wx to deduce from (7.45) and (7.47) that, as x tends to 00 with IArg xl ~ 41l' - 8,

dn

-U(r, r dtn

+ s; wx) =

O(x- r ),

(7.48)

uniformly for 0 ~ t ~ 1. A final application of (7.40) with h(z) = eZX and g(t) = 1 + t/2 - w yields dn dtn e X (I+t/2-w) =

exg(l)

L B(k l , k2, ... , kn)g~l g~2 ... g!" Xkl +k2+·+k.,

(7.49)

where the sum is over all integers ki satisfying (7.41), where the coefficients B(k l , k 2 , ••• , kn ) are independent of x, t, and where gj is defined by (7.42). By

38. Approximations and Asymptotic Expansions

527

(7.30), g

( t) - _ -

~

B2m t 2m

~ (2m)!

'

and so g; = O(t),

for all odd i :::: 1

(7.50)

=

for all j :::: 1.

(7.51)

and gj

0(1),

Since get) ::: 0 for 0 ::: t ::: I, exg(t) = 0(1).

(7.52)

By (7.41), k2

+ k4 + k6 + ... ::: i(k l + 2k2 + ... + nkn )

= in.

(7.53)

Combining (7.49)-(7.53), we see that

d ex(l+t/2-w) _ dtn n

«

~

~

Ixlkl+k2+··+kntkl+k3+·· n

« L Ixtlkl+k3+"'lxlk2+k4+" «lxl[n j 2] L

Ixtli.

;=0

(7.54)

The result now follows from (7.44), (7.46), (7.48), (7.54), and Leibniz's rule. Proof of Theorem 7.1. By (7.26) and (7.39), (7.55)

T(x) = A(x) - B(x),

where

res)

10roo e-attr-I(-w)'U(s, s + r; -wx) dt

10

e-at(_ty-IwSe-wxU(r, s

A(x) = r(r)

(7.56)

and B(x) =

00

+ r; wx) dt,

(7.57)

irr

where < arg(-wx) < ~rr. We first examine A(x), which yields the dominant part of the asymptotic expansion of T(x). Using (7.28) in (7.56), we find that A(x)

res) r-I =- L(_l)k(sh(1rh r(r)

1

k=O

r , =res) - - L- (-V(sh(1 - rh r(r) k=O x.,+k

00

e-attr-I(_wy(_wx)-s-kdt

1 0

00

0

e -at t r-k-l( e -t - l)kd t

528

Ramanujan's Notebooks, Part V

= l(s) ~ ~ ~ (-I)i(s}t(1 ~ s+k 1( )

r

(k) 1

rh

00

•

x

k=O i=O

where we have expanded (e- t from (7.58) that

-

]

0

e

-t(a+i)tr-k-1d

t,

(7.58)

l)k by the binomial theorem. It follows easily r-l

A(x) =

L Ak

X - k- s ,

k=O

in agreement with (7.22) and (7.23). Now (7.24) follows by putting t = 0 in (7.43). Thus, by (7.22), (7.31), (7.55), and (7.57), it remains to show that

roo e-t(a+x/2) /(t, x) dt = !;;o ~ Cm(x) + O(X- 1- r- N/ 2). (a + xj2)m+1

Jo

(7.59)

By (7.27) and (7.31), /(t, x) «eX (I-w+t/2)t r- 1w s (wx)-r,

and so e- t (a+x/2) /(t, x)

« e-at ex(l-w) x- r t s- I ,

(7.60)

uniformly for t ~ 1. Since, for t ~ 1, we know that 1 - w < - ~, it follows from (7.60) that

foo e- (a+x/2) f(t, x) dt « e- x/2x-r foo e- t ReatRes-ldt «e-x/2. t

(7.61)

In view of (7.59) and (7.61), it remains to show that

t

e- t (a+x/2) /(t x) dt

Jo

=

I:

' m = o (a

Cm(x)

+ xj2)m+l

+ O(X- 1- r- N/ 2).

(7.62)

Integrating by parts N times, we find that

1 1

o

e- t (a+x/2) f(t x) dt

=L' N-l

'm=o

(a

+ (a +X/2)-N

By Lemma 7.2, e-(a+x/2) f(m)(l, x)

,

f(m)(o x) - f(m)(1 x)e-(a+x/2)

«

+ xj2)m+1

11

e- t (a+x/2) /(N)(t,X) dt.

e-(a+x/2)x 3m / 2

« e- x/ 3.

Thus, to prove (7.62), it remains to prove that

l'

e- t (a+x/2) /(N)(t, x) dt = O(x N/ 2- r-').

38. Approximations and Asymptotic Expansions

Again, by Lemma 7.2,

11

e- l (a+x/2) f(N)(t, x) dt

o

« X N/ 2- r

11

e- I Re(a+x/2)

0

X N / 2- r

-

t

i=O

//

"x

529

L Ixtlidt N

.i=0

(Re(a

Ixl i j!

+ X/2»j+1

I Ii" I li+

N N/2-r ~ ~ / / N/2-r-1 1 "x L...J . i=O x

This completes the proof of Theorem 7.1. We show now that Cm(x) possesses an asymptotic expansion in descending powers ofx. As in the proof of Lemma 7.2, we shall estimate Cm (x) = f(m) (0, x) by combining Leibniz 's rule with formulas for the nth derivatives of ( - 1)' -I , W S , ~(I-W+I /2) , and U(r, s +r; wx). The nth derivatives of (-ty-l and WS att = 0 are constants. Since, for the function g(t) in (7.49), we have g(O) = 0, the nth derivative of ex (l+1/2-w) att = 0 is, by (7.49), a polynomial inx.ltremains to show that the nth derivative of U (r, s + r; wx) at t = 0 has an asymptotic expansion in descending powers of x. By (7.40) with h(z) = U(r, s + r; z) and g(t) = wx, we find that

d" d"iU(r, s t

+ r; wx)

I ="L 1=0

k

d Ekxkd""iU(r, s

k=O

Z

+ r; z)

I

(7.63)

Z=X

for some constants Ek. Using the asymptotic formula (7.47) in (7.63), we obtain the desired result. If s is a positive integer, we can deduce the stronger result that em (x) is a Laurent polynomial. To see this, note that when s is an integer, U(r, s + r; z)

~ (-I)k(rh(1 -

= L...J k=O

Z

k+r

S)k

by (7.28) with r and s interchanged. Thus, U (r, s + r; z) and its derivatives with respect to z are Laurent polynomials in z, and the result follows from (7.63) as before. After stating (7.3), Rarnanujan provides what is evidently a hint for proving (7.3). However, we have been unable to use Ramanujan's advice in establishing (7.3). Correcting three misprints, we state Ramanujan's "hint" as a separate entry.

530

Ramanujan's Notebooks, Part V

Entry 8 (p. 273). Let n be a nonnegative integer and suppose that 0 < p < a. Then e

2p (a - P + n)n-l I ( 2ap = exp - (a + p + n)n+l (a + n)2 a +n p4 + 2(a

+ n)4

1

=

(

(a+n)2 - 2(a

2np5 - 5(a + n)5

p6

a+n

}

- ...

a 2 + 1. 2 (a+n)2

2p4 {a 4 + 5a 2 + (a + n)4

+ n)2 + -3-

~

4 p 3 {a 3 + 2a +_

3

- (a

+ 4 p 5 {a 5 + 1Oa 3 + 23aj2 _ 5a 4 + 4 } 15 (a + n)5 (a + n)4

Proof. Write

+ n)n-l e 2p (a + p + n)n+l (a - p

2np3 3(a + n)3

-~--=

)

+ 3(a + n)6

1+ 2 pa- - + 2p2

I

p2

+ (a + n)2 -

(a+n)3

2a}

+ n)3

+ ... ) .

(8.1)

)n-l ----=- ....;'-------'---:-:(a + n)2 ( p )n+l 1+-p ( 1--a +n

e2p

a+n

= =

=

(a

e

2p

+ n)

2

exp

(en - 1) log (1 - -p-) - (n + 1) log (1 + -p-)) a +n a +n

e2p ( {p p2 p3 p4 exp (n - 1) - - - - --"--~ (a+n)2 a+n 2(a+n)2 3(a+n)3 4(a+n)4 -

... } - (n

(a

+ n)2 exp

1

p4 + 2(a

+ n)4

+ 1) {a: n ( 2ap -a +n

- 2(a: n)2

+ 3(a: n)3

- 4(a: n)4

+ ... })

p2 2np3 + (a + - -----:n)2 3(a + n)3

2np5 - 5(a + n)5

p6

+ 3(a + n)6

) - ...

.

This proves the first equality in (8.1). If A(a, p, n) denotes the expression in large parentheses on the far right side above, then exp (A(a, p, n»

~ =1+ ( --+ a+n

~ (a+n)2

1 ( 2ap a +n

+ . . . ) + 2:

-

~~ 3(a+n)3 p2

+ (a + n)2

~ ~~ + 2(a+n)4 ---'--5(a+n)5

2np3 - 3(a + n)3

p4

+ 2(a + n)4 -

) ...

2

38. Approximations and Asymptotic Expansions

1 ( 2ap +"6 a + n

+~( 24

=

2ap a+n

+

p4 (a + n)4

+ (a

p5

2np 3 - 3(a + n)3

p2

+ n)2

(a

(1+2a 2)+

(1"2 - 3 +"21+ 4an

(2n

2a

2n

+ n)5 -5 + a - 3" -

1+ 2 p a- - + 2p2 a+n

a2 + ! 2 (a+n)2

2 p4 \ a 4 + 5a 2 + (a +n)4

+ -3-

+ 4 p 5 {as 15

+ ...

)3

p2 _ ... )4+_1_( 2ap (a+n)2 120 a+n

2ap p2 1+--+ a+n (a+n)2

+

=

+

~

2

+ ...

)5 + ...

p3 (2n 4a 3 ) --+2a+(a+n)3 3 3 2a 4 )

+3

4a 2n -3-

4a 3

4a 5 )

+ a + 3 +'15 + ...

1}

3 + -4 p 3 { a + 2a - - - -

3

2a - (a +n)3

+ lOa 3 + 23aj2 (a + n)5

531

I

(a+n)3

_ 5a 2 + 4 } (a + n)4

2(a+n)2

+ ....

This concludes the proof. The next result is somewhat enigmatic. Ramanujan offers an asymptotic series for L~=l 11k as n tends to 00, but he expresses the asymptotic expansion in powers of 11m instead of 1jn, where m = !n(n + 1). We cannot find a "natural" method to produce such an asymptotic series. Therefore, we take Ramanujan's expansion, convert it into powers of lin, and show that it agrees with Euler's well-known asymptotic series for a partial sum of the harmonic series.

Entry 9 (Formula (5), p. 276). Let m = !n(n + 1), where n is a positive integer. Then as m approaches

"11

00,

~ k ~210g(2m) +

1

1

y + 12m - 120m 2

191 - 360360m 6

29

+ 30030m 7

1

+ 630m3 2833

-

1166880m 8

+

1

1

1680m4 + 231Om 5

140051 17459442m 9

-

where y denotes Euler's constant.

Proof. We rewrite (9.1) in the form n i l

~k~10gn+y+210g

(1)

1

1

1+;; +6n(n+1)-30n 2 (n+1)2

•.

'(9.1)

532

Ramanujan's Notebooks, Part V

4

+ 315n 3 (n + 1)3 -

-

8·191 45045n 6 (n

+ 1)6

1 105n 4 (n

+

+ 1)4

16

+--~----:: 1155n 5 (n 1)5

64·29 15015n 7 (n

+ 1)7

+

8·2833

-

----~----~ 36465n 8 (n 1)8

+

256·140051

+ 872972ln 9 (n + 1)9 + . .. .

(9.2)

Using Mathematica, we expand log(l + 1In) and (n + 1) -k, I ~ k ~ 9, in powers of lin and collect coefficients of like powers of lin. We then find that (9.2) can be put in the shape nIl

1

{; k ~ logn + y + 2n I

I

+ 24On 8 3617

-

1

+ 120n 4

-

691

132n lO

+ 8160n16

I

- 12n 2

+ 32760n l2

43867

14364n 18

252n6

1 -

12n 14

+... .

(9.3)

On the other hand, from a result of Euler found in Entry 2 of Chapter 8 (Part I [1, p. 182]), B L -kI ~ log n + y - L ----;., kn n

00

k=1

k=l

(9.4)

as n tends to 00, where Bb k :::: I, denotes the kth Bernoulli number. Calculating the first 18 terms in the sum on the right side of (9.4), we find that they are in agreement with those in (9.3). D. W. DeTemple and S.-H. Wang [1] found an analogue of (9.4) with n replaced by n +~. We quote Ramanujan in the next entry.

Entry 10 (Formula (13), p. 284). The property of the function ~

logn 2 +x 2 ~ n n=l and the integral

rOO

10

t

dt

(e 2rrt - I)(t

+ x)·

Ramanujan did not inform us what property he had in mind. Since these two functions are not equal, it would seem that he is claiming that they are asymptotically equal as x tends to 00. However, as we shall demonstrate, this is not the case.

38. Approximations and Asymptotic Expansions

Theorem 10.1. As x tends to

00,

~ F(x) := ~ n=1

logn

2

n +x

Proof. By partial summation,

~ logn

+

n=1 n 2

10gN! N2 + x 2

-_

~ --"-~

x2

533

_

[':>0

11

(10.1)

2x

[N 2t log ret + 1)dt.

+

(t 2

I

Letting N tend to 00, we deduce that F(x) - 2

7r log x

2 ~ --.

+ x 2 )2

ret

log + 1) d (t 2 + X 2 )2 t,

t

(10.2)

since, by Stirling's formula,

+

logr(u as u tends to

00.

1) ~ (u +~) logu - u

(t

2

+x

t 2 2

+ 0 (1) }

)

1 + + 1 + 21 + _ 1 __ + 1 + +

= 2 = 2

I

00

t(t

o

1) log t - t 2

(t

00

o

;

2 2

+x )

logt -d t

x2

t2

o

-

dt

00

= II

(10.3)

Using (10.3) in (10.2), we find that

_ [00 t {(t + ~ )log t -

F(x) - 2

+ 0(1),

t2

x2

0(x- 2 )

dt

00

2x

0

dt

logt

(t2

X 2 )2

dt

+

1

00

0

00 ----:------:--.". dt

x2

0

(t 2

t logt (t 2

+ x 2 )2

dt

X 2 )2

+ /z + ... + Is,

(10.4)

say. First, from elementary considerations, 7r 7r 7r 14+ Is = - - + - = - - . (10.5) 2x 4x 4x Next, from Gradshteyn and Ryzhik's Tables [1, p. 564, formula 4.231, no. 8], 11

2

12

(7rlOg X + [00 logu dU) = 7rIOg x, (10.6) 2 10 u 2 + 1 x log(xu) 2 ([00 log u [00 du ) (u 2 + 1)2 du = -:; 10 (u 2 + 1) 2 du + logx 10 (u 2 + 1)2

= ~ [00 log(xu) du = ~ x 10 u 2 + 1 x [00

= -:; 10 _

7r~X + 0 (~),

(10.7)

and

h

=..!.- [00 ulog(xu) du = 0 x2

10

(u 2

+ 1)2

(IOgx). x2

(10.8)

534

Ramanujan's Notebooks, Part V

Putting (10.5)-(10.8) in (lOA), we complete the proof of (10.1). So that we may find an asymptotic expansion of the integral in Entry 10, we first establish an analogue of Watson's Lemma (Olver [1, p. 71]).

Lemma 10.2. Suppose that

L ant n, 00

f(t) ~

(10.9)

n=l

as t approaches O. Then, as x tends to

00,

provided that the integral converges for x sufficiently large, where (z) denotes the Riemann zeta-function. Proof. Let m-I

fm(t):= f(t) - Lant n. n=1

Then

1

00

o

1 -1

f(t) ---dt = ext - 1

-

00

0

fm(t) ---dt ext - 1

00

O

fm(t)

+ m-l an

----dt+ ext - 1

~

1

00

0

tn ---dt ext - 1

mL-I ann! (n + 1) . n=1

xn+1

(10.10)

The integral on the right side converges for x sufficiently large, because the corresponding integral with fm(t) replaced by f(t) converges for all x sufficiently large. As t tends to 0, fm(t) = o(tm). Thus, for some positive constants km and K m,

Hence,

1

km

1

o

fm(t) I ---V---I dt ~ Km e -

l

0

km

tm m! (m + 1) ---V---l dt < Km m+l • e x

(10.11)

Let X be a value of x for which the integral on the right side of (1 0.1 0) converges. Now

38. Approximations and Asymptotic Expansions

is continuous and bounded on [km, (0). Let Lm = x> X,

1

00

km

1m (I) dl = eXt - 1

1

_11

= Fm(t) eXt ext - 1

1

00

k

IFm(t)I. Then, for

1m (t) eXt - 1 dt eXt - 1 ext - 1

00

km

= -

SUPkm :9 1. Suppose further that 2nrri loga

i=

2mrri 10gb'

Jor every pair oJnonzero integers m, n. Then,for x > 0,

~

~ e m,n=O

_an bm X

=

log2 X +logx 2 log a log b

1 (10 g b + 12 log a

+

+L

(_x)n

oo

n=1

n!(a n

-

loga log b

+

(y

loga log b

rr 2 +6 y2 ) y log a log b - 2"

1

1) 1) + 4

- -- ---

2 log a

(1

log a

210gb

+ log b

1

r (_ 2mri)x2mri/l0ga oo 1 loga + ----'---=:-:::-'--:--;-;--1) loga n=-oo 1 - b2mri/loga

L

l)(b n

-

n;60

+-

1

L

oo

log b 11=-00

r (_ 2nrri) x2mri/logb 10gb

1 - a 2mri / 10gb

n;60

where y denotes Euler's constant.

'

(12.1)

38. Approximations and Asymptotic Expansions

537

Equality (12.1) is a double series analogue of Hardy's theorem (11.2). The three series on the right side of (12.1) do not appear in Ramanujan's formulation. There is one further discrepancy in that Ramanujan omitted the expression

1210ga 10gb

on the right side of (12.1).

Proof. Set

L 00

f(x):=

m,n=O

Then, for u = Re s > 0,

By Mellin's inversion formula, 1

x -f( ) - 2ni

i

a +oo

a-oo

Let

I

.-

M,T·-

i

C

r(s)x- S ds (I - a-S)(I - b- S ) ,

M,T

r(s)x-S ds (1 - a- S )(1 - b- S ) ,

a> O.

(12.2)

(12.3)

where CM,T is a positively oriented rectangle with vertices ata±iT and -M ±iT, where T > 0 and M = N + ~, where N is a positive integer. We choose T = Tn, n ~ 1, tending to 00 so that ITn log a - knl

nj3

(12.4)

ITn 10gb - knl ~ nj3,

(12.5)

~

and

for every positive integer k. We evaluate 1M, T by the residue theorem. The integrand in (12.3) has a triple pole at the origin and simple poles at s = -n, for each positive integer n. Furthermore, there are simple poles at s = - 2nn i j log a and s = - 2mn i j log b, where m and n are nonzero integers. The latter two sets of poles are simple poles by hypothesis.

538

Ramanujan's Notebooks, Part V

To calculate the residue at 0, we use the expansions 1 s

r(s) = - - y

+ (n2 -12 + -y2) s2

1

1

1-a-S

(-sloga)

slogae-sloga-1

2

+ -yn + -y3) s2 + ... (126) 12 6 ,.

(t;"(3) 3

1

1

log a

=- + -2 + -12s + .. . sloga

1 slog b

1

log b 12

- -s = - - + - + -s+·" 1 - b-

2

(12.7) (12.8)

and

(12.9) The expansion (12.6) can be deduced from a well-known formula found in the Tables of Gradshteyn and Ryzhik [1, p. 944, formula 8.321, no. 1]. After a lengthy calculation, we find that

Ro =

log2 x 2 log a log b

+ logx

1 (10 g b loga

+ 12

1

(y

log a log b

loga

n 2 +6

1)

- -- - -2 log a 2 log b

y2 ) y(

+ 10gb + loga 10gb -"2

1 loga

1)

1

+ 10gb + 4' (12.10)

The remaining residues are much easier to calculate. For each positive integer

n, R_ n

(-x)"

= - - - - -n- n! (all - l)(b - 1)

For each nonzero integer n, R

. -21111'! jlog a

2nni)

r (_ x2nIri/ toga = _-'---_lo.....:g:::....a-'--,----c-:-__ loga(l _ b2nIrijloga)

and

r (_ R_ 2nIri /logb

2nni) 10gb

= log b(l _

[M.T

=

Ro

+ ~ n! (an

a2nIrijlogb) .

(_x)n _ l)(bn - 1)

L loga 12nIri/logal 0, 00

~

k,!;:=O e

-anb'"c!'x

I

= -

og 3 X 6 log a log b log e

+ _log_2_x {~ ( 2

-

2

1 log a log b

+

1 log b log e

+ __1_ _ ) log clog a

IOgal~blOge} -Iogx {~ (-lo-~-a + -IO-~-b + -lo-~-e)

_!.( 2

1 log a log b

+

1 log b log e

+

I ) log clog a

1 (lOge a 10g b ) ++ loglog +-~12 log a log b b log e log clog a

540

Ramanujan's Notebooks, Part V

1(2 y2) 1 } + ( -+12 2 log a log b log e

1 (IOgb+10ge + loge+loga +----"' IOga+logb) +_ _ _..:0.24 loga 10gb loge 1(2+ y2) + (24 4

(Ia

log log b

+

1

log b log e

1)

+--log clog a

y (lOge a 10g b ) - -12 + loglog + .,-------=-log a log b b log e log clog a y

- '4

(I

1 1)

loga + 10gb + loge

00

+ ~ n! (an

+ y1(2 + 2y 3 1210galogbloge +

4{(3)

-

1

8

(_I)n-l xn

- l)(bn - 1)(cn - 1)

r (_ 2n1(i) x2mri/loga 1 ~ log a + loga n~oo (1 - b2mri/loga)(1 - e2mri/loga) n;iO

r (_ 2n1(i) X2n1ri/logb 1 00 10gb + 10gb n~oo (1 - e2n1ri/logb)(1 - a2n1ri/logb) n;iO

r (_ 2n1(i) X2n1ri/logc 1 ~ loge + log e n~oo (1 - a2n1ri/logc)(1 - b2n1ri/logc) ,

(13.1)

n;iO

where y denotes Euler's constant, and {(s) denotes the Riemann zeta-function.

The four infinite series on the right side of (13.1) do not appear in Ramanujan's formulation. Furthermore, the terms 1210ga 10gb log e and (1 1) +-y1(2 -1(2 + 1 + log clog - 1210ga -4{(3) - -10gb -loge 24 logalogb 10gb loge a are not found in Ramanujan's version. The proof of Entry 13 follows along the same lines as the proof of Entry 12. In particular, (12.6}-(12.9) are needed to calculate the residue of the quadruple pole at the origin. Therefore, we forego the proof. In Entry 14 we quote Ramanujan.

38. Approximations and Asymptotic Expansions

Entry 14 (p. 314). Let perimeter of ellipse

541

= n(a + b)(1 + h), then

b)2 _ 4h _ _ --;3h=2== ( a___ a+b 2+../l=3h

(14.1)

very nearly. According to the above approximation, the perimeter of a parabola = 3.99944(a + b) for 4(a + b). Proof. Set)... = (a - b)/(a + b). If L denotes the perimeter of the ellipse given by x = a cos t, Y = b sin t, 0 S t S 2n, then (part III [3, p. 146]) L = n(a + b) 2FI (-~, -~; 1; )...2) =: n(a + b)(1 + h).

(14.2)

Hence, h

= ~)...2 + ~)...4 + ~)...6 + 25)...8 + 4

43

44

47

49)...10 + 441 )...12 + .... 48 410

(14.3)

From (14.1), we find that, according to Ramanujan, 3h 4 + (16 - 4)...2)h 2 + ()... 4 - 8)..2)h + ).. 4 ~

o.

(14.4)

Beginning with the approximate solution h = )...4/4, we solved (14.4) by the method of successive approximations to deduce that

h~~)...2+~)...4+~)..6+25)..8+ 49)..10+ 439)..12+ .... 4

43

44

47

48

410

(14.5)

Comparing (14.3) and (14.5), we find that the two power series agree up to the coefficient of ).. 12, where the difference is remarkably only 2)" 12/410. Thus, Ramanujan's claim is certainly justified. The second claim in Entry 14 remained an enigma to us for a few years before R. J. Evans deciphered the proper interpretation. Since the eccentricity for a parabola equals I, set b = 0 in ( 14.1). Solving (14.1) via Mathematica when the left side equals I, we find that h = 0.2730576913. Using this value of h in (14.2), we find that L

~

n(a + b)(l + h) = 3.99943(a + b).

Taking h = 0.27306, we would obtain the approximation 3.99944(a +b) claimed by Ramanujan. Lastly, we know that L If b

["/2

= 4 Jo

Ja

2

sin 2 t

+ b 2 cos2 t dt.

= 0, then ["/2

L=4Jo

asintdt=4a=4(a+b).

This concludes the explanation of Ramanujan's strange claim.

542

Ramanujan's Notebooks, Part V

The work in the following section first appeared in a paper by the author and Evans [1]. Before stating and proving Entry 15, we introduce some notation and offer a preliminary lemma. Ramanujan defines the logarithmic integral Li(x) by Li(x) = PV

l

x

dt -1-'

x> O.

o ogt

Define the unique positive number JL by Li(JL) = O.

(15.1)

Ramanujan [9] and Soldner (N. Nielsen [1, p. 88]) numerically calculated JL. We used MACSYMA to also calculate JL. The table below summarizes the three calculations: Ramanujan

1.45136380

Soldner

1.4513692346

MACSYMA

1.4513692349

Lemma 15.1. Let JL be defined by (15.1 ), and let y denote Euler's constant. Then, for x> 1, .

Ll(X)

~ logk X

= Y + log log x + 6~'

Lemma 15.1 can be found in Ramanujan's notebooks on the same page as Entry 15, and a proof is given in Part IV [4, p. 126]. For another proof, see Nielsen's book [1, pp. 3, 11]. We first give Ramanujan's version of Entry 15: If S :=

4)

4· 4]. 4

Ln k1 ( 1 + -Pl)k = log p,

(15.2)

k=!

then n = (p + log JL We might interpret Ramanujan's statement as giving an estimate for S when n = [(p + log JL With such an interpretation, the error made in the approximation by log pis 0(1/ p), as p tends to 00. However, if p is chosen so that n = (p + log JL is a positive integer, then, as stated in our version of Entry 15 below, the error term is 0(I/p2). Amazingly, Ramanujan found the precise linear function of p that yields an error term of 0(1/p2). Thus, if the constant log JL in the definition of n is replaced by any other constant, the error term is 0(1/ pl.

4)

4)

4

4

Entry 15 (p. 318). Let S be defined by (15.2), and let n a positive integer. Then, as p tends to 00, S = log p

+ 0(p-2).

=

(p

+ 4) log JL -

4be (15.3)

38. Approximations and Asymptotic Expansions

543

Proof. Setting y = log/L, applying Lemma 15.1, and using (15.1), we find that yk

00

O=y+logY+L-=y+logy+ k=1 k! k

lY

el

1

-

--dt. t

0

(15.4)

Setting x = 1 + 1/ p and using the definition of S, a familiar estimate for a partial sum of the harmonic series (Olver [1, p. 292]), and (15.4), we deduce that, as n tends to 00, "xk_l "1

S={;-k-+t;k

1 (1)

II xk-l = {; - k - + log n + y + 2n + 0

=

t

~_ p

xk - 1 + log (~) + y

k

k=1

10

2n

n2

el

-

1 dt

t

+0

(-;) . P

(15.5)

Next, applying the Euler-Maclaurin summation formula from (0.5) of Chapter 37, we deduce that, as p tends to 00, II xk - 1 1 x" - 1 L--=--Iogx+--+ k=1 k 2 2n

1" 0

1

Xl -

( 1)

--dt+0"2' 1 P

(15.6)

Employing (15.6) in (15.5), we find that S= =

r 10

1 dt _ (Y e l

Xl -

-

1 dl _

~ logx + log (~) + xn + 0 (~)

10 t 2 y 2n p2 rlOg:x; f(t)dt - ~logx + log (p + ~ -~) + ~~ + 0 (~), 1y 2 2 2y 2n p (15.7) t

where f(l) = (e l

-

1)/t. Now, I 1 ( -1 ) , --logx=--+O

2

( 1 1)

log p + - - 2

2y

2p

p2

= log p + -

1 1 (1)

2p

- -

2py

+ 0

-

p2

,

and xn 2n

=

(1

+ l/p)Py+(y-I)/2 = ~ + 0 2py + (y - 1) 2py

(~_) p2'

as p tends to 00. Substituting these estimates in (15.7), we find that S=

h

rlOg:x;

f(t) dt + f(y) + logp + 0 2p

(-~), p

as p tends to 00. Thus, it remains to prove that

i

II

y

lop

f(t) dt = /(y) 2p

+0

(-;) . P

(15.8)

544

Ramanujan's Notebooks, Part V

Now,

n log x = y _ _ 1 2p

+ 0 (~) .

(15.9)

p2

By the first mean value theorem for integrals and (15.9),

l

y

nlogx

f(t) dt

=

f(u)(y - nlogx)

= -f(u) + 0 2p

( 2" 1) ' P

(15.10)

for some value u such that n log x < u < y. By the mean value theorem, there exists a value v such that u < v < y and f(u) = f(y)

+ (u -

y)f'(v) = f(y)

+ 0 (~) ,

(15.11)

where the last equality follows from (15.9), since f' is bounded on (u, y). Using (15.11) in (15.10), we complete the proof of (15.8). So as not to interrupt the proof of Entry 16 below with two calculations, we now set them aside in two lemmas. Lemma 16.1. If a and (J are positive and n is a nonnegative integer, then ._

1

In .- - .jii(j Proof. Setting u =

l

°o+ai

-oo+ai

z2n e-z2/9d Z -_

(}n(2 )' 2

n .

2 nn !

•

z/..;e and applying Cauchy's theorem, we find that

Lemma 16.2. If a and (J are positive, and n is a nonnegative integer, then

38. Approximations and Asymptotic Expansions

Proof. Setting z =

545

u,JO and applying Cauchy's theorem, we find that

Entry 16 (p. 324). As t tends to 0+, 00

2 ~)_l)n n=O

--=+

(1

1

'"

t)n(n+l)

t

1+t

+ t 2 + 2t 3 + 5t 4 + 17t5 + ....

(16.1)

The function on the left side of (16.1) is not a theta-function but is a false theta-function in the sense of Rogers [3). In fact, we shall obtain a more explicit asymptotic expansion, which enables the calculation of further terms in the asymptotic series. It is interesting that Ramanujan appended an abbreviation for "asymptotically" after the series on the right side. We are unaware of any other instance in the notebooks where Ramanujan used this word. Usually, he wrote "nearly" or ''very nearly."

Proof. Let e

1- t .---,

-0._

1+t so that () is small and positive. If a > 0 and n is a nonnegative integer, by Lemma 16.2, e-(n+l/2)2 0 = __ 1 l oo+ai e- z2 / oe(2n+l)iz dz .

..;;ro

-oo+ai

Multiply both sides by 2(-1)" and sum on n, 0 ::: n < 00. Upon inverting the order of integration and summation, we see that the resulting series on the right side converges absolutely and uniformly on (-00 + ai, 00 + ail and that the resulting integral also converges absolutely, and so the inversion is justified. Hence, 00 2 loo+ai z2 00 2 L(_1)n e -(n+l/2)20 = - . e- /0 L(_l)n e (2n+l) iZ dz

l

.jiilj -oo+a.

n=O

1

= .jiilj

,,=0

oo+ai e- z2 / 9

-oo+ai

cos z dz.

Now recall that (Abramowitz and Stegun [1, p. 804]) ~ (_I)n E2n 2n

secz = ~

(2n)!

z ,

Izl

< rr/2,

(16.2)

546

Ramanujan's Notebooks, Part V

::s rc /4,

where E j , j ::: 0, denotes the jth Euler number. Thus, for Izl

~ (-1)" E2n z2n I < C ~ (2n)! -

sec z -

I

I

Iz1 2N + 2

,

where the positive constant C 1 depends on N but not on z. If 0 < a < 1 and Izl ::: rc /4 on the contour (-00 + ai, 00 + ai), then a

cosz'

~ (_I)n E2/l 2n

1

and

Z2N+2 ~

(2n)!

z

are bounded functions of both a and z. In particular, observe that cos«2n - l)rc/2 + ia) = i(-1)n sinha, where n is an integer, and so a/ cos z remains bounded as a tends to O. Hence, for all points on the contour, a

IZ2N+2

~ (_1)n E 2n 2n) I (2n)! z ::s C2,

(

(16.3)

secz - ~

for some positive constant C2, which is independent of both z and a. Therefore, by (16.2) and Lemma 16.1, 2

L(_I)lt e- O. We now apply this theory to f(t) = fP(x - n + 1 + 2t). Since S(x, n) is a sum of n terms, we want da(t) to be a discrete measure weighted at the integral points 0, 1, 2, ... , n - 1. This leads us to the Hahn polynomials, which were introduced by P. L. Tchebychef [1] in 1875 and which are constant multiples of Qk(t; a, fl, N):=

3F2

[

-k, k

+ a + fl + 1, -t ] ; 1 a + 1,-N

= t(-k)j(k+a+ fl +l)j(-t)j j=O (a + l)j(-N)jj! '

o ::s k
O. Then (20.2) is valid, when cis given (approximately) by (20.9) below.

Proof. Taking the logarithm of each side of (20.2), we find it suffices to show that I

00

2

k=!

a)

-loga + Llog 1 + - (

qJ(k)

= logc

+

1

qJ-I(ax) ---dx. qJ(O)/a x(l + x) 00

(20.3)

Since qJ(O) > 0 and qJ(x) is monotonically increasing,

rOO log

11

(1 +

_a )dX:::: flOg(I qJ(X) k=1

+~):::: roo log

10

qJ(k)

(1 + ~)dX. qJ(X)

By the intermediate value theorem, there exists a number Xa , 0 :::: Xa :::: 1, such that

Sa :

(1 + -T) = 1 (1 + _a_') (1 + ~) 10r· (1 -~)

= flog

00

qJ( )

k=]

=

10roo log

qJ(X)

x.

log

qJ(X) I

log

dx _

+

dx

qJ(X)

dx

= It - 12 ,

(2004)

say. By examining the inverse function oflog(l II =

Setting x = I/(e Y

1

Iog (l+a N (0))

o

qJ-I

+ a/qJ(x», we see that

(a) --

eY - 1

dy.

1), we find that

-

It = Thus, from (2004),

Sa =

1

00

qJ-I(ax)

qJ(O)/a x(I

1

00

+ x)

qJ-I(ax)

qJ(O)/a x(l

+ X)

dx.

dx - 12 •

(20.5)

Comparing (20.5) with (20.3), we see that (20.3) has been proved with logc

= -/z + ~ loga.

(20.6)

We shall make the determination of c slightly more explicit. For 0 :::: x :::: Xa :::: 1,

= log a-log qJ(x)+log(l+qJ(x)/a) = 10ga-logqJ(x)+O(l/a),

10g(l+a/qJ(x» as a tends to

/z =

00.

Xa log a

Thus,

_l

x

a 10gqJ(x)

dx

+ O(I/a) = Xa log a - h + O(I/a),

(20.7)

562

Ramanujan's Notebooks, Part V

say, Since q; is increasing, Xa 10gq;(0)

:s h

:S Xa logq;(xa),

so that by the second mean value theorem for integrals, xa q;(x) q;(xa) o :S h - Xa log rp(O) = o log --dx = (xa - ~a) log - - , q;(0) q;(0)

l

for some number ~a, 0 :S

:S

~a

Xa'

(20,8)

Combining (20,6)-(20,8), we deduce that

c = via (q;(o»)xa (q;(xa»)xa-~a (1 + OO/a», a

(20,9)

q;(0)

This completes the proof. It seems likely that, in many cases,

Xa

as a tends to

00,

= ~ +0(_1 ) 2

and

log a

Xa -

~a = 0(1),

In such cases, (20,9) yields c

=

Jq;(O) {1

+ 0(1)} ,

The next entry is recorded in Ramanujan's Quarterly Reports and is discussed in detail in Part 1[1, pp, 311-312], Entry 21 (p. 351). Consider the equation 00

(-x3l

F(-x) := { ; (3k)!

=

e-x

+ e-wx + e- w2x 3

(21.1)

= 0,

where w : = exp(2;r i /3), Then there exist an infinite number ofpositive roots, They are close to the zeros ;r(2n + 1)/.J3 of cos(x.J3/2) , where n is a nonnegative integer. More precisely, if h

= e-rr(2n+I)~/2,

then these roots are given by _ ;r(2n + 1) _ ~ (h2 x .J3 2

+

+ _(__I_)n

(h

+

~

3! h

4

+

28,31 6 49,52,57 h 8 5! h + 7!

76,79,84,91 h lo 9!

+ .. ,)

+ 2h 3 + _19_'_2_1 h 5 + 37,39,4\7 + _61_,_6_3_,_67_,7_\9

.J3

3!

+

5!

7!

91 ,93 ' 97 ' 103 ' 111 h II II!

Lastly, all roots of(21,1) are given by x, and w = exp(2;ri/3),

WX,

+ .. ,) ,

9! (21.2)

and w 2x, where x is given by (21.2)

38. Approximations and Asymptotic Expansions

563

Entry 22 (p.351). LetxO,X\,X2, ... be the realrootsof(21.1}. Then

J] 00

F (x) =

1+

(

3

X ) x~

(22.1)

.

Proof. The real roots of F(x) = 0 are -Xo, -x\, -X2, ...• Then -Xn, -XnW, -Xnw2, 0 ::s n < 00, constitute all the roots of F(x) = 0, wherew = exp(2Jri/3). Now (x

+ xn)(x + xnw)(x + Xn(2) =

x3

+ x~.

We now apply the Weierstrass product formula. From the definition of F in (21.1) and the fact that the infinite product in (22.1) converges, since Xn '" 1f(2n + 1) /../3 as n tends to 00, we easily deduce that F(x) has the representation (22.1).

Entry 23 (p. 370). As x tends to 1-, 00

~

( X 3n +1 x 3n +2 ) 1 + X4n+1 - 1 + x 4n+3

1

1

~ 4 + 4.J2 log (1 + ../2) -

1f

g.J2.

Proof. First, 00

~

(X3n+1

x3n+2) 1 + x4n+1 - 1 + x 4n +3 =

00

~

(1 _ x)(1 _ x4n+2)x3n+1 (1

+ X4n+1 )(1 + x 4n +3)

.

(23.1)

We apply the Euler-Maclaurin summation formula, (0.5) of Chapter 37, to (1 - x)(1 - x 4t +2 )X 3t+1 f(t):= (1 + x4t+1)(1 + x4t+3)

Thus, 00 Lf(n) =

n=1

Let xt

1

00

0

f(t) dt - -1 (-X

2

- -X2) 1+x 1 + X3

+

1

00

0

(I -- [I] -

1.) f'(/) dl 2

(23.2)

= u. Since 4X2 4x 4X3) f'(t)=f(t)logx ( 3-1_u4x2 -1+u 4x -1+u 4x 3 '

we find that

1

00

(I - [I] -

- x(x - 1) -

Hf'(t) dt

1 1

0

(logu -- log x

x ( 3-

[IOgU] -log x

1)

--

2

U

u 2(1 - 4X2) (1 + u 4x)(1 + U 4X 3)

4x2 4x 4X3) - - - - du 1 - U4X2 1 + u 4x 1 + U4X3

564

Ramanujan's Notebooks, Part V

(23.3)

= 0(1),

as x tends to 1-. Setting Xl = u also in the first integral on the right side of (23.2) and using (23.3), we find that, as x tends to 1-, 00

x(1 - x)

n=1

log x

LI(n) = -

t

11 0

u 2 (1 - u 4x 2 )

(1 + u 4 x)(1 + u 4x 3 )

u 2 (1 - u 4 ) (1 + u4)2 du

= 10

du +0(1)

(23.4)

+ 0(1).

Using Mathematica, we find that

t 10

u 2 (1 - u 4) du

(1

+ u 4 )2

=

~ _ ..Ji tan-I 4

8

..Ji

- - log(2 16

..Ji) _ ..Ji tan-I (2 + ..Ji) ..Ji 8 ..Ji h) + -..Ji log(2 + h)

(2 -

1

..Ji 1(

4

8 8

..Ji 31( 8

1

1(

1

16

..Ji

= - - - - - - - + -log(1 + h) =- - -

4 8..Ji

8

8

r,:;

+ -log(1 +"12).

4..Ji

(23.5)

Using (23.5) in (23.4) and combining the result with (23.1), we complete the proof of Ramanujan 's asymptotic formula.

39 Miscellaneous Results in the First Notebook

In this last chapter we collect together some miscellaneous results from the unorganized portions of the first notebook. Most are from analysis, with some pertaining to hypergeometric functions. We use the familiar notation associated with hypergeometric functions; e.g., see Part 11[2, p. 8]. In particular, for each nonnegative integer n, (a)n =

r(a + n) r(a) .

Page numbers after entries refer to the pagination of the Tata Institute's publication of the first notebook [9].

Entry 1 (p. 72). If 00

f(x) := L

fP(nx) ,

(1.1)

n=1

then 00

00

LfP(n 2x) = L(-l)o(n) f(nx), n=1

(1.2)

n=1

where 0(1) = 0 andJor n > 1, O(n) denotes the total number ofprime factors ofn counting multiplicities.

Proof. By (1.1), 00

00

00

oo(~(_l)O(d) ) fP(rx).

?;(_l)o(n) f(nx) = ?;(_l)o(n) ~ fP(mnx) = ~

(1.3)

566

Ramanujan's Notebooks, Part V

It is easy to see that (_l)O(n) is completely multiplicative. Hence, Ldlr< _1)O(d)

is multiplicative. For each prime p and positive integer n, if n is even,

L(-l)O(d) = { 1,

ifn is odd.

0,

dip'

Hence, by multiplicativity, if r is a perfect square,

= { 1,

L(_l)0(d)

otherwise.

0,

dlr

Using the equality above in (1.3), we deduce (1.2).

Entry 2 (p. 94). If n is a positive integer,

1

00

o

n

•

SlD n

x

X dx =

where the plus sign is taken

±:

( l)n

L

[(n-l)/2]

1l

2 r(n)

(-n)k(n _ 2k)n-l,

k=O

if n is even, and the minus sign is chosen if n is odd.

In fact, Ramanujan claimed different values for the integral and crossed out the entry. The evaluation above appears to have been given first by O. Schlomilch [1] in 1860, as pointed out by M. L. Glasser. In the book by D. S. Mitrinovie and J. D. Keekie [1], the integral of Entry 2 is evaluated by contour integration. In 1980, E. T. H. Wang [1] submitted Entry 2 as a problem; solutions by R. L. Young and T. M. Apostol and several references to the problem's appearances in the literature were given.

Entry 3 (p. 94). For 0 < x < 1,

2Fl

{ I

2.1.

)_

3' 3' 2' -x -

{ ~

r,;:\1/3

v1+x+v X )

(~

r,;:\1/3

+ v 1 + x - V x)

2v'f+x

(3.1)

and 2Fl

{I

2.3.

)_

3' 3' 2' -x - 3

~

(v 1 +

X

+.Ji)

- v 1 + x -yx) 2.Ji

1/3

(~

r,;:\1/3

(3.2)

Proof. From Entry 35(iii) of Chapter 11 (part II [2, p. 99]) with x 2 replaced by

-x,

(3.3)

39. Miscellaneous Results in the First Notebook

567

Now, cos (~sin-l(i.Ji») = cos

(~IOg {.Jf+X - .Ji})

=! ( .Jf+X -

.Ji)

-1/3

(

)1 13

+!.Jf+X - .Ji

_ )1 / 3 =! ( .Jf+X+.Ji) 1/3 +! ( 0+x-.Ji .

(3.4)

Using (3.4) in (3.3), we deduce (3.1). Next, Entry 35(ii) of Chapter 11 (Part 11[2, p. 99]), with x 2 replaced by -x, states that F 21

12.3. )_ 3 . (1 . -1(. r;;\) ( 3'3'2'-x - . r;;sm 3 sm ,,,,x,.

'yX

The remainder of the proof of (3.2) is similar to that for (3.1). Entry 4 (p. 104). Let x and y be complex numbers such that x/y is not purely imaginary. Let qJ (z) be an entire function such that fez) := ~1l" sec (!1l"xz) sech (!1l"Yz) (qJ(xyz) - qJ( -xyz)} tends to 0 as z tends to

00.

Then

~1l" sec (!1l" x) sech (!1l"Y) (qJ(xy) - qJ( -xy)} 00

= x

~

(_1)n sech (1l"(2n + I)Y) 2 (2n + 1)2 _ {qJ «2n + l)y) - qJ (-(2n + l)y)}

:2

(_1)n sech (1l"(2n + 1)x) 2 -iYL 2 1)2 Y2 (qJ«2n+1)ix)-fp(-(2n+l)ix)}. n=O ( n + +Y (4.1) 00

Proof. Observe that f(z) has simple poles atz = (2n+I)/xandz = (2n+l)i/y for each integer n. The poles do not coalesce because x / y is not purely imaginary. Letting Ra denote the residue of fez) at a pole a, we easily find that (_1)n (1l"(2n + I)Y) R(2n+l)/x = -~ sech 2x {qJ «2n + l)y) - qJ (-(2n + 1)y)}

and (_1)n (1l"(2n + 1)X) R(2n+l)iIY = --.- sech {qJ «2n + l)ix) - 'P (-(2n + l)ix)}. 2y, 2y

568

Ramanujan's Notebooks, Part V

We thus find that the sum of the principal parts arising from the poles ±(2n+ l)/x, n 2: 0, is (_1)n xz sech (

(2n

n(2n

+ 1)2 _

+

I)Y)

X;;2

«2n

{qJ

+ l)y) -

qJ

(-(2n

while the sum of the principal parts arising from the poles ±(2n equals

(-ltyzi sech (n(2n

-

22; (2n + 1) + y z

{qJ

Since fez) tends to 0 as z tends to theorem that

00

- iyz Letting

L

(_l)n sech (n(2n

(2n

n=O

z=

2: 0,

00,

qJ

(-(2n

+ l)ix)}.

we conclude from the Mittag-Leffler

l)y) {qJ

«2n

+ l)y) -

qJ (-(2n

+ l)y)}

+ l)x)

2;

2

+ l)ix) -

«2n

2+ (2n + 1)2 _ x:z 2

(_l)n sech (n(2n

fez) = xz ~

+ l)i /y, n

+ l)X)

2

00

+ l)y)} ,

+ 1) + y z

2

{qJ

«2n

+ l)ix) -

qJ

(-(2n

+ l)ix)}.

1, we deduce (4.1) to complete the proof.

Entry 5 (p. 108). If x is not an integer and 0 n

- {cot(nx) cos (ex) X

:s e :s 2n, then

+ sin(ex)} =

~ cos(ne)

1

2" - 2 ~ X

n=l

2

n -x

2·

(5.1)

Proof. By Entry 34(i) in Chapter 13 (Part II [2, p. 237]), for Ie I :s n, ~ cos(ex) x sin(n x)

Replacing

e bye -

=

~

+2~

(_l)n-l

x2

~

n2 -

n=l

cos(ne). x2

(5.2)

n in (5.2), we readily deduce (5.1) to complete the proof.

Entry 6 (p. 120). The maximum value of aX / r (x

+ 1) is equal to

r~:-:~) exp (1152a3 ~ 323.2a)

(6.1)

"very nearly." This entry is the same as Example (i) in Section 25 of Chapter 13 in the second notebook. However, after proving a slightly weaker version of the result in Part II [2, p. 228], we unfortunately claimed that the appearance of "323.2a" in

39. Miscellaneous Results in the First Notebook

569

Ramanujan's expression (6.1) is erroneous. We are very grateful to Richard Brent for pointing out to us that Ramanujan, indeed, is correct. Moreover, he kindly provided the proof below.

Proof. We shall prove more precisely that the maximum value equals aa-l/2 exp r(a +

4)

(_1_ _41 4720a 101 1152a 3

+ 0 (~)) a7

5

(6.2)

'

as a tends to 00. Since I 101 1152a3 - 414720a5

( 1) 1 ( 101 a7 = 1I52a3 1 - 360a 2

+0

+0

( 1 )) a4

1

1152a 3

(1 + 3~~~2 + 0 (:4) )

1 - 1152a 3 + 323.2a

+ O(1/a) ,

this would confirm Ramanujan's claim. Let I(a) := log r(a + By Corollary 1 in Section 6 of Chapter 8 in the second notebook (Part 1[1, p. 184]),

4).

,

1 7 + (1)

+ 24a 2

I (a) = loga

-

0

960a 4

a6

.

(6.3)

Further differentiations of the asymptotic expansion of f' (a) are valid by a general theorem on the differentiation of asymptotic expansions (F. W. J. Olver [I, p. 21 and we have

n,

I

" = -1- - 1 + (1) (a)

as

(6.4)

-~ + 0 (~). a2 a4

(6.5)

a

0

12a 3

and

I"'(a) = Now, sup

aX

X~O r(x

=

+ 1)

where R(a) =

aa-l/2 r(a

1

+ 2)

R(a),

+ 4) ' f'(a + 2 + E) a'r(a

(6.6)

1

and where (see Part II [2, p. 228]), if the maximum is obtained at x = x(a),

E

:= x(a)

+ 4- a =

-

2~ + 6~3 + 0

(:5)'

(6.7)

570

Ramanujan's Notebooks, Part V

as a tends to 00. Using (6.3)-(6.5), we find from (6.6) that, as a tends to 00, log R(a) =

€

=€ = -

log a

+ f(a)

- f(a

log a - €f'(a) -

t€2

+ €) f"(a) - i€3 fill (a) - ...

(2:a2 +~) +(9~4 + 2~3 +~2) +0 (:7) ·(6.8)

From (6.7),

and €3

= -

(2~)3 + 0

(a

I5 ) ,

as a tends to 00. Using these expansions and (6.7) in (6.8), we conclude that log R(a)

=

1 1152a 3

-

101 414720a5

+0

( 1) a7

'

which is the required result in (6.2).

Entry 7 (p. 138). Let n > m > 0 and let p be real. Put fj (x) = cos x, sin x, for j = 1, 2, respectively. Then,for j = 1, 2,

f(m)r(n - m) ff(n) J

(t {tan-

n{I +

k""O

I

p

n- m

+k

- tan- I

_P_}) +

m

k

=-----r==~====~==~====~==--~

(n _

~2+ k)2 } {I +(m ~ k)2 }

(7.1)

Proof. It is not difficult to identify I j in terms of beta functions. More precisely (A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev [1, p. 450, eq. 11]), 1 II = --Re {f(m f(n)

+ ip)f(n -

m - ip)}

(7.2)

h = --1m {f(m + ip)f(n - m - ip)}.

(7.3)

and 1 f(n)

39. Miscellaneous Results in the First Notebook

571

Now, by Euler's product formula for r(z),

rem + ip)r(n -

m - ip) Nm+ip N!

= J~

(m

Nn-m+ip N!

+ ip)(m + ip + 1)·· . (m + ip + N) N mN! Nn-m N! exp

•

(-i t

tan-I

k=O

(n - m - ip)··· (n - m - ip

+

-pm +k

it

tan- I

k=O

p n- m

+k

+ N)

)

= hm -r====~-,====~~~~==~===i~-r========7~~ N .... oo Jm 2 + p2 ... J(m + N)2 + p2J(n - m)2 + p2 ... j{n - m + N)2 + p2 r(m)r(n - m) exp

!J

(i f

k=O

00

1 + (m

{tan- I

p

I

p2

n - m +k

+ k)2 VI + (n -

-

tan-I

~k })

m+

(7.4)

p2 m

+ k)2

Using (7.4) in (7.2) and (7.3), we readily deduce (7.1). We consider the next result as a formal identity.

Entry 8 (p. 158). If

Jor

h

In

~(x) cos(2nx) dx = T1/t(n),

then

Proof. We have

roo e-

Jo

x2

1/t(x)dx = = =

2

{'XJ

In Jo

e- x2 dx

r

h

Jo ~(u)cos(2xu)du

Jrr lh ~(u) 1 2r:;;

yTr

lh 0

du

In

00

~(u)-e-U

2

2

e- x2 cos(2xu) dx du,

upon using a familiar integral evaluation (Gradshteyn and Ryzhik [1, p. 515, formula 3.896, no. 4]). This completes the proof. The next entry is somewhat obliquely stated by Ramanujan, who used a notation peculiar to him (see, e.g., Part 1[1, p. 138]).

Entry 9 (p. 184). IfRe fJ > Re a, then

roo rex + a + 1)

Jo rex + fJ + 1) {1/t(x + fJ + 1) -1/t(x + a + 1)}

d

x

rca + 1)

= r(fJ + 1)'

572

Ramanujan's Notebooks, Part V

where y,(x) = r/(x)/ rex). Proof. We have 00 r(x +a + 1) o r(x+P+l) {y,(x+P+l)-y,(x+a+l)}dx

1

=-

1

d {r(x + a + 1) } d dx r(x + ,1 + 1) x

00

0

r(a + 1) 1) .

= rep +

Entry 10 (p. 194). For each nonnegative integer n, F

3 2

[!,!,:> 1-n] -_

..j1ir(n + 1) ~ (4)~ L

3

2'

2r(n + 2)

k=O (

k

!)

2'

Proof. This result follows in a straightforward manner from Entries 29(b), (c), (d) of Chapter 10 of Ramanujan's second notebook (Part II [2, pp. 39-40)). We are grateful to R. A. Askey for providing the proof of Entry 11 given below.

Entry 11 (p. 206). If 0 < Ren < inf(Rea + 1, Rep + 1), then

1

00

o

n-I -IX 2FI(a, ,1; a +x

+ ,1; -x) dx

rCa - n + 1)r(p - n + 1) r(a + ,1 - n) = r(a - n + 1)r(n) r(a)

f:

k=O

f: k=O

(ah(ph (a + ,1 - n + k)(a + phk!

(ah(nh . (a + k)(a + phk!

(11.1)

Proof. Let I denote the integral on the left side of (11.1). Using Pfaff's transformation (part II [2, p. 36, Entry 19)), we find that

1

x n- l (1 + x)-I-a 2FI (a, a; a + ,1; _x_) dx. o x+1 Making the change of variable t = x / (x + 1), we deduce that 1=

00

I = =

=

11 L

tn-I (1 - t)-n+a 2FI (a, a; a (a)2

00

k

+ phk!

11

t n+k - l (1 -

+ ,1; t) dt

t)-n+adt

k=O

(a

k=O

(a)~ r(n + k)r( -n + a + 1) (a + phk! r(a + k + 1)

f:

0

= r(n)r(-n + a + 1) r(a)

E k=O

(ah(n)k

(a + k)(a + phk!

,

39. Miscellaneous Results in the First Notebook

573

which proves the second equality in (11.1). To prove the first equality in (11.1), we first establish a general transformation for 3F2(a, b, c; d, e; O. Suppose that Re(d+e-a-b-c) > OandO < Re c < Re e. Then, by inverting the order of integration and summation, we find that J:= =

=

10 1xc-I(1-xt-c-12FI(a,b;d;x)dx

f:

(ah(bh r(c + k)r(e - c) k=O (dhk! r(e + k) r(c)r(e - c) 3F2(a, b, c; d, e; 1). r(e)

(11.2)

On the other hand, by a fundamental transformation for hypergeometric functions (W. N. Bailey [1, p. 2]) and by the same argument as given above, J = =

=

10 1xc-I (1 -

f

x)d+e-a-b-c-I 2FI (d - a, d - b; d; x) dx

(d - ah(d - b)k r(c + k)r(d + e - a - b - c) k=O (d)kk! red + e - a - b + k) r(c)r(d + e - a - b - c) [ d - a, d - b, c ] 3 F2 red + e - a - b) d, d + e - a - b

•

(11.3)

Combining (11.2) and (11.3), we deduce that

b, c] = r(e)r(d + e - a - b - c) 3F2 [ d - a, d - b, c ]. (11.4) d, e r(e - c)r(d + e - a - b) d, d + e - a - b

3 F2 [a,

Now set a = a, b = n, c

1=

=

r(n)r(a - n + I) [ a, a, n ] 3F2 rea + 1) a + (3, a + 1 r(n)r(a - n + 1) rea + 1)r({3 - n + 1) [{3, a + (3 - n, a ] 3~ rca + 1) r(1)r(a + (3 - n + 1) a + {3, a + (3 - n + 1

+ I)r({3 - n + 1) rea + (3 - n + 1)

_ r(n)r(a - n

-

= a, d = a + (3, and e = a + 1 in (11.4). Then

3 F2

[ a , {3, a + {3 - n ] a + {3, a + (3 - n + 1

,

which establishes the first equality in (11.1). One might surmise that Entry 11 can be found in integral tables. However, we are unable to find these evaluations in the tables of Gradshteyn and Ryzhik [1] or Prudnikov, Brychkov, and Marichev [2], although on page 315 of the latter tables, some similar evaluations are given.

Entry 12 (p. 208). Define F(n) (x), n = 2m , where m is a nonnegative integer, by

574

Ramanujan's Notebooks, Part V

and F(2n) (x) = F(n) (F(2)(x». Then F(n)(x) =

4xn {(I

+ .Jf-=-x)n + (1 - .Jl="X)n}

Proof. It is easily checked that (12.1) is valid for m tion, we find that F(2n)(x) = F(n) (

2'

(12.1)

= 0,1. Proceeding by induc-

x2 ) (2 - x)2 x2n

4--....,...(2 - x)2n

-

{

(2 - x + 2.Jl=X) n + (2 - x - 2.Jl=X) n}2 4x 2n

=

{(I +.Jl=X)2n + (1 - .Jl=X)2n

r'

which completes the induction.

Entry 13 (p. 265). Ifcp(x) is any polynomial, thenformally

(~~n

E(-on (~~n

E(-I)n cp(n) = n=O n. n=O

n.

cP (_ 2n + 1). 2

(13.1)

Proof. Let cp(x) = CPm(x) := x(x - 1)(x - 2)· .. (x - m

+ I),

where m is any nonnegative integer. Since {CPm(x)} ,0 ~ m ~ j, form a basis for the vector space of all polynomials of degree ~ j, it suffices to prove (13.1) for cP (x) = CPm (x). With this choice of cP (x), the proposed identity becomes

(4~n

f(-l)"+m (4~n

E(-l)n+m (-n)m = n=m n. n=O

First,

n.

(2n

+

2

1). m

(13.2)

39. Miscellaneous Results in the First Notebook

575

Second, :E(-I)n+m (!)n n=O n!

(2n + 1) 2

m

= :E(_l)n+m (!)n (!)m(7 + !)n . n=O n! (2)n

(13.4)

Comparing (13.3) and (13.4), we see that we have established (13.2), as desired. Observe that the series in Entry 13 do not converge.

Entry 14 (p. 265). Let a, fJ > 0 with afJ = 7r 2 • Then

,Ja

{

I

00

_l)n}

(

4 + ~ e(2n+l)a -

1

{1

00

(

_1)n }

# 4 + ~ e(2n+l).8 -

=

(14.1)

1 .

Proof. Let 00

L(s) =

L(-lt(2n + 1)-S,

Res> O.

It is well known that this Dirichlet L-function can be analytically continued to an entire function. We apply Entry 21(iii) of Chapter 14 (part II [2, p. 277]) in the case n = 0 to find that

I

,Ja { '2 L (O)

00

(-W

+ ~ e(2n+l)a _

1

}

=

1

1

1

'2#?; cosh(fJn) + 4#' 00

(14.2)

where a, fJ > 0 and afJ = 7r 2 • From the functional equation (H. Davenport [I, p. 71]) L(s) = cos (!7rs) (!7r)'-1 r(1 - s)L(1 - s)

and the value L(1) = 7r/4, found as (32.7) in Chapter 37, we easily deduce that L(O) = Thus, (14.2) may be rewritten in the form

!.

I

,Ja { 4 +

L 00

n=O

(_I)n}

e(2n+l)a _ 1

=

# 4+ L e.8n + e-.8n n=l {1

1

00

}

(14.3)

.

But 00 1 00 00 ""' = ""' e-.8 n ""'(_I)je- 2.8nj L e.8n +e-.8n L L n=l n=! j=O

= ~(-I)j ~ e-(2j+l).8n =

~

n=l

j=O

L

j=O

L

Using (14.4) in (14.3), we deduce (14.1). We consider the next result in a formal sense only.

L

~-l)j

e(2}+l).8 - 1

. (14.4)

576

Ramanujan's Notebooks, Part V

Entry 15 (p. 265). If

1/1 (n) then,

if r

=

1

00

(15.1)

qJ(x) cos(nx) dx,

is an even nonnegative integer, 1/I(r)(n) = cos

(t1fr)

and

1f

2"qJ(r)(n)

1

00

xrqJ(x)cos(nx) dx

(i)

roo xr1/l(x)cos(nx)dx.

= cos (t1fr) 10

(ii)

Ramanujan put no restrictions on r. He also wrote 1/I r (n) and qJr (n) for 1/I(r)(n) and qJ(r) (n), respectively. Proof. Formally differentiating (15.1) r times, we deduce (i). By Fourier's inversion theorem (Titchmarsh [2, pp. 16-17], Part 1[1, p. 333]),

roo 1/1 (x) cos(nx) dx.

::qJ(n) =

10

2

Formally differentiating r times, we deduce (ii).

Entry 16 (p. 267). Let a, 13 > 0 with af3 = 1f / 4. Then a 3/ 2

L(-It(2n + 1)e-(2n+l)2 2 = 13 3/2 L( -1)n(2n + 00

00

a

I)e-(2n+l)2jl2.

(16.1)

n=O

n~

Proof. We apply Poisson's summation formula for Fourier sine transforms (Titchmarsh [2, p. 66], Part II [2, p. 236]) to the function x exp (_x 2 ) • Thus, if a, 13 > 0 and af3 = 1f /2, then

But (Gradshteyn and Ryzhik [1, p. 529, formula 3.952, no. 1])

1

00

Replace af3 =

xe- x2 sin {(2n

13 by 213

1f/4,

and use the evaluation above in (16.2). Thus, for a, 13 > 0 with

a 2 I:(-lt(2n n=O

Since ..j(i =

+ l)f3x} dx = ~.Ji(2n + l)f3e-(2n+l)2P2/ 4 •

+ 1)e-(2n+l)2 a 2 = ~f3 I:(-I)n(2n + l)e-(2n+l)2p 2. n=O

-J7i/ (2.f/J), we find that the proof of (16.1) is complete.

39. Miscellaneous Results in the First Notebook

577

Entry 17 (p. 339). If Iv denotes the Bessel function of imaginary argument of order v, then, ifn is an integer, (17.1) Ramanujan's recording of Entry 17 is incomplete. He writes the difference of two series on the left side but does not indicate what this difference equals. The series are Bessel functions of imaginary argument, as we have indicated (Watson [15, p. 77]), and (17.1) is a well-known, easily proved equality (Watson [15, p.

79]). Entry 18 (p. 350). '" I ( I { 1+ 100 I ( 1 + 35 1 { 1+ 90 I +I90.25 })}) . e -2Jr '" 540 1 + 120

Most of Ramanujan's approximations for exp( -a1r) arise from modular equations or class invariants. However, it appears that this approximation to exp( - 21r) was empirically derived by some method of successive approximations.

Proof. Using Mathematica, we find that

5~ (I + I~O {I + I~ ( 1 + 315 {I + 9~ + 90 ~ 25}) }) = 0.001867442731726 .... On the other hand,

e- 27r = 0.001867442731707 .... Thus, Ramanujan's approximation agrees to 13 decimal places. We conclude by briefly mentioning two claims made by Ramanujan on page

112 that apparently have no precise meanings. Entry 19.

Since Ramanujan does not specify the function qJ or the constants cn , we are unable to offer a definite interpretation of this formula. Possibly, Ramanujan applied the Euler-Maclaurin summation formula to the function qJ(logx)/x on the interval (1, 00). A simple change of variable then gives the integral on the right side above. The constants Cn are therefore those that appear in the Euler-Maclaurin summation formula, and the series on the right side should probably be interpreted as an asymptotic series.

578

Ramanujan's Notebooks, Part V

Entry 20. 2n

+L 00

k=2

kn

+ k-n

(k - l)k

1

= - -

n

1f

cot(1fn)

+ ....

Since the series on the left side diverges and since the meaning of the "dots" on the right side is not divulged by Ramanujan, we are unable to offer a meaningful interpretation of this formula. We think that it pertains to material in the first part of Chapter 7 of the second notebook. In particular, see Part I [1, p. 161].

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

In Part IV [4] we provided the locations in the second and third notebooks of all entries in the 16 organized chapters of the first notebook. A small minority of these entries cannot be found in the second or third notebooks, and so we provided proofs for these results in [4]. Like the second notebook, the first notebook contains much unorganized material, in fact, considerably more than in the second notebook. The unorganized pages also contain a higher proportion of entries not found in the second notebook than the organized part of the first notebook. In the sequel, we indicate where proofs can be found for each correct result in the unorganized portions of the first notebook. Page numbers of the first notebook are given in boldface at the left margin. We assign numbers, in order, to each formula on each page. If the result appears in the second or third notebooks, we indicate where in these notebooks, and where in Parts I-V, it can be located. If an entry cannot be found in the second or third notebooks, we inform readers where a proof can be found in the present volume. 20

1. This is a version of Entry 1 of Chapter 3 [1, p. 45]. 2.,3. These are Corollaries 1 and 2, respectively, in Section 2 of Chapter 3 [I, p.46].

26 1. This is contained in Entry 10 of Chapter 3 [I, pp. 57-59].

46 1. Entry l1(i) of Chapter 18 [3, p. 162]. 2.,3. The two radical expressions are equivalent to the formulas for G 1225 and G 441, respectively, given in the table of class invariants in Section 2 of Chapter 34.

48 1.,2. Entries 24 and 24(i), respectively, of Chapter 14 [2, pp. 291-292]. 3. Entry 36, Chapter 13 [2, p. 238].

580

Ramanujan's Notebooks, Part V

50 1.,3. These are equivalent to Entries 36(i) and (ii), respectively, of Chapter 11 [2, p. 100]. 2.,4. Entries 36(iii) and (iv), respectively, of Chapter 11 [2, pp. 100-101].

52 1. Entry 49, Chapter 12 [2, p. 184]. 2. Corollary 2, Section 44, Chapter 12 [2, p. 170].

54 1.,2. Entries 2(i), (ii), Chapter 13 [2, p. 188]. 3. Entry 3, Chapter 13 [2, p. 188]. 4.,5. Entry 4, Chapter 13 [2, p. 189]. 6. Entry 5, Chapter 13 [2, p. 190].

56 1. Part (i) is the same as Entry 20, Chapter 5 [1, p. 123]. 2. Part (ii) follows from Entries 19(i), (ii), Chapter 5 [1, pp. 122-123]. The word "multiple" on page 56 should be replaced by "factor."

58 1. Although this formula is not in the second notebook, it is formula 14 of Table 1 in Ramanujan's paper [7], [10, p. 141]. 2. The value of the Bernoulli number B38 can be found in Ramanujan's paper [1], [10, p. 5] and on page 53 of his second notebook [9]. 3. This deleted entry is a partial version of formula 15 in Table 1 mentioned above. 60

1.-3.,5.-9. ThevaluesoftheBernoullinumbersBn , n = 22,24,26,28,30,32, 34, 36 can be found in Ramanujan's paper [I], [10, p. 5] and second notebook [9, p.53]. 4. This is formula 12 in Table 1, mentioned in our commentary on page 58.

62 1. This is a trivial statement about the factorization of polynomials. 2.-4. These three formulas are renditions of the same result. The first two contain unexplained asterisks and are deleted by Ramanujan. The third version is imprecise and contains an error term that is not completely specified. See Entry 27(ii) of Chapter 7 [1, p. 178] for a correct version of these three formulas. 64

1.,2. Entry 11, Chapter 11 [2, p. 54]. 3. This geometrical figure can be found in Section 19 of Chapter 18 [3, p. 190]. 4.,5. Entries 19(iv), (iii), respectively, of Chapter 18 [3, pp. 184, 181].

66 1. See Section 24 of Chapter 18 [3, p. 211]. 2. See Section 19 of Chapter 18 [3, p. 194].

Location of Entries in the Unorganized Portions of Ramanujan 's First Notebook

581

3. Part of the corollary in Section 3 of Chapter 18 [3, p. 151]. 4. Part of Entry 3, Chapter 18 [3, p. 146].

68 1.,2. Corollaries (i), (ii), Section 19 of Chapter 18 [3, pp. 185, 190]. 70

1. Entry 14, Chapter 4 [1, p. 107]. This is also a special case of Entry 27 of Chapter 13 [2, p. 230]. 2.,3. Both drawings appear to be versions of a figure of Chapter 18 [3, p. 194]. 72

1. Entry 29, Chapter 5 [1, p. 130]. 2. Entry 1 of Chapter 39. 3. Entry 6, Chapter 12 [2, p. 111].

74 1.-3. Entries 21(i), (iii), (iv), Chapter 18 [3, pp. 200--201].

78 1. Entry 4(i), Chapter 6 [1, p. 136]. 2.,3. Examples 1,2 in Section 4 of Chapter 6 [1, p. 137]. 4. See Section 4 of Chapter 6 [1, p. 137].

80 1.-3. These three singular moduli are given in Theorem 9.9 of Chapter 34.

82 1.-4. Examples 9.4 of Chapter 34. 5.-7. Examples 9.7 of Chapter 34. 8.-10. Theorems 9.6, 9.3, and 9.5, respectively, in Chapter 34.

84 1. Entry 14, Chapter 15 [2, p. 332].

86 1. Entry II(xv), Chapter 20 [3, p. 385]. 2.-10. Entries 45, 42, 43, 44, 46, 48, 49, 47, 50, respectively, of Chapter 36.

88 1. Special case of Entry 11, Chapter 6 [1, p. 143]. 2.,3. Two elementary abelian theorems for power series with no hypotheses. See Titchmarsh's treatise [1, pp. 229-231] for general abelian theorems. 4.,5. Entries 52 and 51, respectively, of Chapter 36. 90

1. Entry 5(xii), Chapter 19 [3, p. 231]. 2. Entry 13(xiv), Chapter 19 [3, p. 282]. 3. Entry 19(ix), Chapter 19 [3, p. 315]. 4.-7. Entry 41, Chapter 36.

582

Ramanujan's Notebooks, Pan V

92 1. Entry 5(xiii), Chapter 19 [3, p. 231]. 2. Entry 13(xv), Chapter 19 [3, p. 282]. 3. Entry 19(x), Chapter 19 [3, p. 315].

94 1.,2. See Entry 2 of Chapter 39 for comments on these deleted formulas. 3. This is a special case of Entry 35(iii), Chapter 11 [2, p. 99]. 4.,6. See Entry 3 of Chapter 39. 5. This is a special instance of Entry 35(ii), Chapter 11 [2, p. 99]. At the bottom of the page, Ramanujan apparently indicates that he has found the generalizations given in Entries 35(i}-(iii) of Chapter 11 of his second notebook (Part II [2, p. 99]).

96 1. Theorem 2.12 of Chapter 33. 2.-5. These are contained in Theorem 8.7 of Chapter 33. 6. All of Section 12 of Chapter 33 is devoted to an examination of this claim. 7.-9. See Theorem 8.1 of Chapter 33.

98 1.,2. Entries 9(iv), (iii), respectively, of Chapter 19 [3, p. 258].

100 1. Entry 35, Chapter 9 [1, p. 294]. 2. Entry 11(iii), Chapter 13 [2, p. 217]. 3. Entry 22(ii), Chapter 14 [2, p. 278]. 4. Corollary, Section 22, Chapter 14 [2, p. 279].

102 1. Entry 18, Chapter 14 [2, pp. 267-268]. 2. Ramanujan, in essence, states a general formula for the multiplication of two Laurent series. 3. Special case of Corollary 1, Section 18, Chapter 14 [2, p. 268].

104 1. Entry 19(iv), Chapter 14 [2, p. 273]. 2. See Entry 4 of Chapter 39.

106 1. Entry 20(i), Chapter 14 [2, p. 274]. 2. Entry 19(i), Chapter 14 [2, p. 271]. 3. Corollary, Section 20, Chapter 14 [2, p. 274]. 4. Entry 20(iv), Chapter 14 [2, p. 275]. 5. Corollary of Entry 20(iv), Chapter 14 [2, p. 275]. 6. Special case of Entries 29(i), (ii), Chapter 13 [2, p. 231].

108 1. Entry 35, Chapter 12 [2, pp. 156-157].

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

583

2. See Entry 5 of Chapter 39. 3. Entry 34(ii), Chapter 13 [2, p. 237].

110 1. Entry 40, Chapter 12 [2, p. 163].

112 1.,2. Entries 19 and 20, respectively, in Chapter 39. 3.-5. Entry 14, Corollaries 1,2, Chapter 7 [1, pp. 166-168].

114

1. Entry 6, Chapter 13 [2, p. 193]. 2. Entry 10, Chapter 13 [2, p. 207]. 3. A more precise version of this entry can be found in Entry 9 of Chapter 11 [2, p. 51].

116 1.,2. Entries II(i), (ii), Chapter 13 [2, pp. 215-216].

118 1. The entry is deleted and is a forerunner of Entry 7 of Chapter 13 [2, p. 195].

120 1. The first statement

f(::

1) =

~ exp (f ~da )

is incorrect. 2. The second claim is a version of Example (i), Section 25, Chapter 13 [2, p. 228]. See also Entry 6 of Chapter 39 for a correction to our claim made in Part IT [2].

122 1. Entry 26(ii), Chapter 13 [2, p. 229]. 2.,3. Entries 17(iii), (iv), respectively, in Chapter 18 [3, p. 176].

124 1.,2. Entries 11 and 12, respectively, in Chapter 16 [3, pp. 21, 24].

126 1. Entry 31, Chapter 10 [2, p. 41]. 2. Entry 7, Chapter 16 [3, p. 16].

128 1. Entry 8, Chapter 16 [3, p. 17]. 2. Entry 4, Chapter 16 [3, p. 14].

130 1. Version of the corollary in Section 36 of Chapter 13 [2, p. 239]. 2. Entry 5, Chapter 16 [3, p. 14].

584

Ramanujan's Notebooks, Part V

132 1. Entry 6(vi), Chapter 9 [1, p. 247]. 2. Entry 6, Chapter 16 [3, p. 15]. 3. Entry 8, Chapter 13 [2, p. 202]. 4. Corollary of Entry 48, Chapter 12 [2, p. 181].

134 1. Entry 9, Chapter 16 [3, p. 18]. 2. Entry 15, Chapter 16 [3, p. 30]. 3. This entry is very vague. It is possibly a less specific version of Entry 16, Chapter 16 [3, p. 31]. 136 1. Corollary (i) in Section 9, Chapter 16 [3, p. 18]. 2. Entry 17, Chapter 16 [3, p. 32].

138 1. Entry 21, Chapter 13 [2, p. 2241. 2. See Entry 7 of Chapter 39 for a proof. 140 1.,2. Both entries are versions of Entry 21 of Chapter 13 [2, p. 2241.

142 1. Version of Entry 17, Chapter 13 [2, pp. 220-221]. 2. Deleted by Ramanujan. 3. Entry 20, Chapter 13 [2, p. 224]. 4.,5. Entry 16 (Second Part) (iii), (i), respectively, of Chapter 18 [3, p. 1741.

144 1. Deleted by Ramanujan. 2.,3. Entries 17(ii), (i), Chapter 18 [3, p. 176]. 4.,5. Entry 16 (Second Part) (iv), (ii), respectively, of Chapter 18 [3, p. 1741. 146 1. Entry 24, Chapter 12 [2, p. 139]. 2.,3. Entries 45(i), (ii), Chapter 12 [2, p. 171]. 4. Entry 16, Chapter 16 [3, p. 31]. 5. Entry 7(vii), Chapter 17 [3, p. 106].

148 1. Entry 6(viii), Chapter 9 [1, p. 2471. 2.-11. Entry 14, Chapter 18 [3, pp. 168-1691. 12. Entry 18(iv), Chapter 18 [3, p. 1791. 150 1. Corollary, Section 12, Chapter 18 [3, p. 164]. 2. Special case of Entry 14, Chapter 12 [2, p. 121]. 3. Corollary, Section 34, Chapter 12 [2, p. 1561.

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

152 1.,2. Entry 47, Chapter 12 [2, p. 179]. 3. Entry 13(i), Chapter 18 [3, p. 165]. 4. Entry 22(iii), Chapter 18 [3, p. 207]. 5. Deleted by Ramanujan.

154 1. Example (iv), Section 2, Chapter 15 [2, p. 305]. 2. Entry 9, Chapter 13 [2, p. 205]. 3. Entry 48, Chapter 12 [2, p. 181]. 4. Entry 7, Chapter 13 [2, pp. 195-196].

156 1. Entry 12(ii), Chapter 18 [3, p. 163]. 2. Entry 22(ii), Chapter 18 [3, p. 207]. 3.,4. Entries 13(ii), (iii), Chapter 18 [3, p. 165].

158 1. Entry 39(i), Chapter 16 [3, p. 83]. 2. Entry 38(iv), Chapter 16 [3, p. 80]. 3. Entry 11 (iii), Chapter 19 [3, pp. 265-266]. 4. See Entry 8 of Chapter 39.

160 1. See Part 11[2, p. 147] for comments. 2. Corollary, Section 15, Chapter 16 [3, p. 30]. 3. Entry 3, Chapter 16 [3, p. 14]. 4.,5. Entries 38(i), (ii), Chapter 16 [3, p. 77].

162 1. Entry 7(iii), Chapter 17 [3, p. 105]. 2. Corollary 3.4 of Chapter 33. 3. Theorem 9.9 of Chapter 33. 4. Theorem 11.6 of Chapter 33. 5. Theorem 9.10 of Chapter 33. 6. Corollary 3.5 of Chapter 33. 7. Theorem 4.4 of Chapter 33. 8. Theorem 4.5 of Chapter 33.

164 1.-3. See Section 21, Chapter 13 [2, p. 225]. 4. Entry 13, Chapter 16 [3, p. 27]. 5. See Section 13, Chapter 16 [3, p. 28].

166 1.,2. Entries 31(i), (ii), Chapter 13 [2, pp. 233-234].

168 1. Entry 23, Chapter 12 [2, pp. 137-138].

585

586

Ramanujan's Notebooks, Part V

2. Entry 7(viii), parts (a), (c), and (d), Chapter 17 [3, p. 107].

170 1. Entry 7(ii), Chapter 17 [3, p. 105]. 2. Entry 7(ix), Chapter 17 [3, p. 108]. 3. Entry 7(x), Chapter 17 [3, p. 110]. 4. Entry 12(i), Chapter 18 [3, p. 163]. 5. Entry 22(i), Chapter 18 [3, p. 206]. 6. Entry 7(i), Chapter 17 [3, p. 104].

172 1. Entry 7(xii), Chapter 17 [3, p. 112]. 2.,3. See Entries 79 and 80, respectively, of Chapter 36. 4. Entry 7(xiii), Chapter 17 [3, p. 113]. 5. Entry 7(iv), Chapter 17 [3, p. 105].

174 1. Deleted by Ramanujan. 2. Entry 6, Chapter 10 [2, p. 12].

176 1.,2. Deleted by Ramanujan. 3. Entry 7(vi), Chapter 17 [3, p. 106]. 4. See Entry 1 of Chapter 36.

178 1. Entry 14, Chapter 13 [2, p. 219]. 2. Quarterly Reports [1, p. 331]. 3. Special case of Entry 32(i), Chapter 13 [2, p. 235]. 4. Entry 16, Chapter 15 [2, p. 338].

180 1. Entry 15, Chapter 13 [2, p.220]. 2. Entry 22(i), Chapter 13 [2, p. 225]. 3. Entry 13, Chapter 13 [2, p. 219]. 4. Corollary of Entry 13, Chapter 13 [2, p. 219]. 5. Corollary of Entry 21, Chapter 13 [2, p. 224].

182 1. Quarterly Reports [1, p. 298]. 2. Entry 23, Chapter 13 [2, p. 226]. 3. Entry 14, Chapter 16 [3, p. 29]. 4. Entry 22(ii), Chapter 13 [2, p. 225]. 184 1. See Entry 9 of Chapter 39. 2. See Part II [2, eq. (17.3), p. 265]. 3.-5. Examples (i), (ii), (iii), Section 30, Chapter 13 (2, p. 233].

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

587

186 1.,2. Entries 36(i), (ii), Chapter 16 [3, pp. 65-66].

188 1. Entry 34, Chapter 12 [2, p. 156]. 2. Entry 39, Chapter 12 [2, p. 159]. 3.,4. Entry 38, Chapter 12 [2, p. 158].

190 1. Entry 20, Chapter 10 [2, pp. 36-37]. 2. Part of second part of Entry 35, Chapter 12 [2, pp. 156-157]. 3. Entry 27, Chapter 12 [2, p. 146].

192 1. First part of Entry 35, Chapter 12 [2, pp. 156-157]. 2.,3. Entries 36 and 37, Chapter 12 [2, p. 158].

194 1. See Entry 10 of Chapter 39. 2. Entry 29(c), Chapter 10 [2, p. 40]. 3. Entry 32, Chapter 10 [2, p. 41].

196 1. Entry 8, Chapter 11 [2, p. 51]. 2. Entry 22, Chapter 11 [2, pp. 64-65]. 3. Corollary, Section 22, Chapter 11 [2, p. 68]. 4. Example, Section 17, Chapter 12 [2, p. 131].

198 For the meaning of the geometrical figure, see Section 7 of Chapter 19 [3, pp. 244-245]. 1. Entry 22, Chapter 12 [2, p. 136]. 200

For the significance of the geometrical figure, see Section '7 of Chapter 19 [3, p.243]. 1. Entry 17, Chapter 12 [2, pp. 124-125]. 2.,3. Corollaries (i), (ii), Section 17, Chapter 12 [2, pp. 130-131].

202 1. In essence, Entry 28, Chapter 13 [2, p. 231]. 2.,4.-6. See Section 24 of Chapter 13 [2, pp. 226-227]. 3. Entry 20(iii), Chapter 18 [3, p. 197].

204 1. Entry 21, Chapter 11 [2, p. 64]. 2. See Theorem 7.3 of Chapter 33. 206 1.,2. See Entry 11 of Chapter 39.

588

Ramanujan's Notebooks, Part V

3. Example 2, Section 12, Chapter 11 [2, p. 58). 208 Except for one example, which we establish in Entry 12 of Chapter 39, all results on this page are found in Section 15 of Chapter 15 [2, pp. 335-337].

210 1. The first five lines on the page continue material from page 208 and can be found in Section 15 of Chapter 15. 2. Theorem 11.5 of Chapter 33. 3.,4. Theorems 4.2 and 4.3 of Chapter 33. 5.,6. Theorems 9.5 and 9.6 of Chapter 33. 212 1.,2. Entries 23(i), (ii), Chapter 18 [3, p. 208]. 3. See Example (iii) in Section 17 of Chapter 9 [2, p. 266). 4. See Theorem 6.4 of Chapter 33. 214 1. This is a definition. 2.,3.,5. These are, respectively, Theorems 10.1, 10.3, and 10.2 of Chapter 33. 4. Theorem 9.11 of Chapter 33 6.,7. Part of Theorem 9.2 of Chapter 34. 8. Corollary 2.4 of Chapter 33. 9. Theorem 10.4 of Chapter 33.

216 1. Theorem 11.4 of Chapter 33. 2. Theorem 11.1 of Chapter 33. 3. Theorem 5.6 of Chapter 33.

218 1. Theorem 6.1 of Chapter 33. 2.-6. Theorem 7.1 of Chapter 33. 7.,8. Theorems 7.6 and 7.8 of Chapter 33. 9. Deleted by Ramanujan.

220 1. This follows from (2.6) and Theorem 2.10 in Chapter 33. 2. This is an incorrect version of part of Entry 3(i) of Chapter 21 [3, p. 460). 3. This is a misstatement of the first part of Entry 5(i) of Chapter 21 [3, p. 467). 4.,5. Theorem 2.13 of Chapter 33. 222 1.-4. Entries 8(i)-(iv), Chapter 19 [3, p. 249). 5. See Entry 14 of Chapter 36. 224 1.-3. Corollary, Section 37, Chapter 16 [3, p. 74).

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

589

4.,5. Entries 4(i), (ii) Chapter 19 [3, p. 226].

226 1.,3. Deleted by Ramanujan. 2. Part of Entry 34(iii), Chapter 11 [2, p. 97]. 4. Entry 7(xi), Chapter 17 [3, p. 111]. 228 1. This result is essentially equivalent to the example in Section 37 of Chapter 16 and can be found explicitly in (37.7) on page 76 of Part III [3]. 2. Entry 37(iii), Chapter 16 [3, p. 73]. 3. This result follows from Entries 37(i), (ii), Chapter 16 [3, p. 73]. 4. Entry 32(v), Chapter 11 [2, p. 93].

230 1.-3. Entries 3(iii), (iv), (i), Chapter 19 [3, p. 223]. 4.,8. Entries 17(i), (ii), Chapter 19 [3, p. 302]. 5. Entry 9(i), Chapter 20 [3, p. 377]. 6. See Entry 2 of Chapter 36. 7. Entry 29 of Chapter 36. 232

1. An incomplete version of Entry 28, Chapter 11 [2, p. 83]. 2. Deleted by Ramanujan.

234 1. A deleted version of Entry 30, Chapter 11 [2, p. 87]. 2. See the table in Section 2 of Chapter 34.

236 1. Corollary, Section 28, Chapter 11 [2, p. 85]. 2. Special case of Entry 30, Chapter 11 [2, p. 87]. 3. Entry 31(ii), Chapter 11 [2, p. 88]. 238

1. Entry 30, Chapter 11 [2, p. 87]. 2. Entry 14, Chapter 11 [2, p. 59]. 240

1. Part of Corollary (ii), Section 31, Chapter 16 [3, p. 49]. 2. Trivial algebraic identity. 3. Example (i), Section 31, Chapter 16 [3, p. 50]. 4.,5. Corollary, Section 28, Chapter 16 [3, p. 44]. 6. With the use of Entry 22(ii) and (22.4) in Chapter 16, it can easily be shown that this entry is equivalent to Example (v), Section 31, Chapter 16 [3, p. 51]. 242

1. See Theorem 9.3 of Chapter 33. 2. Entry 9(v), Chapter 19 [3, p. 258].

590

Ramanujan's Notebooks, Pan V

3. Entry 31, Chapter 16 [3, p. 48]. 4. Corollary, Section 30, Chapter 16 [3, p. 47]. 244 1. This result is easily seen to be equivalent to the first part of Example (iv) in Section 31 of Chapter 16, although Ramanujan, in the numerator, neglected to write the factors (x 3 ; x 8 )oo(x 5 ; x 8 )oo(x 8 ; x 8 )oo [3, p. 51]. 2. An incomplete version of the second part of Example (iv), Section 31, Chapter 16 [3, p. 51]. 3. See Entry 81 of Chapter 36. 4. This is equivalent to the example in Section 9 of Chapter 17 [3, p. 122]. 5. See Entry 82 of Chapter 36. 6.,7. Example (ii), Section 31, Chapter 16 [3, p. 50]. 246 1.,2. See Entry 22 of Chapter 36. 3.-11. Entry 10, Chapter 17 [3, p. 122]. 12.-17. Entries II(i), (iii), (iv), (v), (vi), (viii), Chapter 17 [3, p. 123]. 248 1. Example (i), Section 6, Chapter 17 [3, p. 103]. 3. Example (iii), Section 6, Chapter 17 [3, p. 104]. 1.-16. See Entry 1 of Chapter 35.

250 1.-9. See Entry 2 of Chapter 35. 10. Entry 27, Chapter 11 [2, p. 80]. 11. Deleted by Ramanujan.

252 1. Entry 7, Chapter 15 [2, p. 313]. 2. Special case of Theorem 6.1 of Chapter 15 [2, p. 310].

254 1.,2. See Entries 3 and 4 of Chapter 36. 3.-5. Entries 13(ii)-(iv), Chapter 15 [2, p. 330].

256 1.-3. Entries 12(ii)-(iv), Chapter 15 [2, p. 326]. 4. Entry 13(i), Chapter 15 [2, p. 330].

258 1.-4. Entries 12(v)-(viii), Chapter 15 [2, p. 326].

260 1.,2. Entries 35(ii), (i), Chapter 16 [3, p. 63, 61]. 3. Entry 33(i), Chapter 16 [3, p. 52].

262 1. Entry 34(i), Chapter 16 [3, p. 54].

Location of Entries in the Unorganized Portions of Rarnanujan's First Notebook

591

2. Entry 33(ii), Chapter 16 [3, p. 53]. 3. Corollary (i), Section 34, Chapter 16 [3, pp. 57-58]. 4. Corollary (ii), Section 34, Chapter 16 (3, pp. 59-60].

264 1.-3. Entries 32(i}-(iii), Chapter 16 [3, pp. 51-52]. 4. Entry 12(ix), Chapter 15 [2, p. 326]. 5.,6. See Entries 38 and 39 of Chapter 36. 265 We use the numbering given by Ramanujan. 1. Entry 31, Chapter 12 [2, p. 150]. 2. See Entry 13 of Chapter 39. 3.,4. Entries 8(i), (ii), Chapter 17 (3, p. 114]. 5. See Entry 14 of Chapter 39. 6. Entry 12(i), Chapter 4; Quarterly Reports [1, pp. 107,321). 7(i). Special case of the preceding entry. 7(ii). Entry 12(ii), Chapter 4 [1, p. 107); special case of the corollary of Entry 21 of Chapter 13 [2, p. 224). 8. See Entry 15 of Chapter 39. 266 1.-4. Entries 8(ix), (xi), (xii), (x), Chapter 17 [3, pp. 114-115]. 5. See Entry 31 of Chapter 36. 6. Special case of Entry 18(i), Chapter 13 [2, p. 221]. 7. We have found no meaningful interpretation of this equality. 8. See Entry 5 of Chapter 36. 9. Entry 9, Chapter 19 [3, p. 257]. 267 The numbering is continued from page 265. 9.,10. Entries 8(iii), (iv), Chapter 17 [3, p. 114). 11. See Entry 32 of Chapter 36. 12. Entry 24(ii), Chapter 16 [3, p. 39). 13. See Entry 16 of Chapter 39. 14. Special case of Entry 31(i), Chapter 13 [2, p. 233]. 15.-17. Entries 16(i}-(iii), Chapter 13 [2, p. 220]. 18. Special case of Theorem II in Ramanujan's Quarterly Reports [1, p. 313).

268 1. Entry 9(i), Chapter 17 [3, p. 120). 2. This is an incorrect version of Entry 9(iv) in Chapter 17 [3, p. 120]. 3.,4. Entries 13(iii), (iv), Chapter 17 [3, p. 127). 5. Principle of duplication, Chapter 17 [3, p. 125).

269 1.,2. Entries 13(i), (ii), Chapter 17 [3, p. 126]. 3.-6. Entries 14(i}-(iv), Chapter 17 [3, pp. 129-130).

592

Ramanujan's Notebooks, Part V

7. Entry 13(viii), Chapter 17 [3, p. 127].

270 1.-5. Entries 17(i}-(v), Chapter 17 [3, p. 138]. 6. Entry 11, Chapter 14 [2, p. 258].

271 1.-4. Entries 17(vi}-(ix), Chapter 17 [3, p. 138]. 5. See Entry 40 of Chapter 36. 6. See Entry 83 of Chapter 36. 272

1.-4. Entries 14(v}-(viii), Chapter 17 [3, p. 130]. 5.-8. Entries 15(i}-(iv), Chapter 17 [3, p. 132].

273 1.-4.,8.,9. Entries 13(viii), (x), (xi), (ix), (v), (vi), Chapter 17 [3, p. 127]. 5.-7. Entries 14(ix}-(xi), Chapter 17 [3, p. 130].

274 1. Entry 21 (iii), Chapter 14 [2, p. 277]. 2. Entry 34(ii), Chapter 16 [3, p. 56]. 3. See Entry 33 of Chapter 36.

275 1. Entry 33(iii), Chapter 16 [3, p. 53]. 2.-6. Entries 16(i}-(v), Chapter 17 [3, p. 134].

276 1. Entry 14, Chapter 14 [2, p. 262]. 2. Corollary, Section 12, Chapter 14 [2, p. 260]. 3. Entry 12, Chapter 14 [2, p. 260].

277 1.,2. Entry 15, Chapter 14 [2, p. 262]. 3. Entry 16(vii), Chapter 17 [3, p. 134]. 4. Entry 12(iii), Chapter 17 [3, p. 124]. 5. Entry 9(iv), Chapter 17 [3, p. 120].

278 1.-4. Entries 15(ix}-(xii), Chapter 17 [3, pp. 132-133]. 5.-7. Entries 8(viii), (vi), (vii), Chapter 17 [3, p. 114].

279 1. Entry 11, Chapter 15 [2, p. 323]. 2. Entry 21(ii), Chapter 14 [2, p. 276].

280 1. Entry 6, Chapter 18 [3, p. 153]. 2.,3. See Entries 77 and 78 of Chapter 36.

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

593

4. Entry 7, Chapter 18 [3, pp. 154-155].

281 1. Example, Section 7, Chapter 18 [3, p. 156]. 2. See Entry 76 of Chapter 36. 3. This is a formal application of the "change of sign" process. 4. The principal of "change of sign," as described in Section 13 of Chapter 17 [3, p. 126]. 282

1. Entry 8(i), Chapter 17 [3, p. 114]. 2.,3. Entries 4(iii), (iv), Chapter 19 [3, pp. 226-227]. 4. This is equivalent to the second part of Entry 5(iii) of Chapter 19 [3, p. 230]. In particular, see the first equality in the proof of (iii) [3, p. 232]. 5. This is equivalent to the first part of Entry 5(i) of Chapter 19 [3, p. 230]. In particular, see the equality at the top of page 232 [3], where l,03( _q6) should be replaced by l,O( _q6). 6. This is equivalent to the formula for G3 in the table in Section 2 of Chapter 34. 7.-9. Entries 5(i), (iv), Chapter 19 [3, p. 230]. 10. See Entry 6 of Chapter 36.

283 1.,2. See Entry 7 of Chapter 36. 3. This is equivalent to Entry 5(vi) of Chapter 19 [3, p. 230], with a replaced by 1 - f3 and f3 replaced by I-a. 4. In essence, this is contained in Entry 5(vi) of Chapter 19 [3, p. 230]; in particular, see (5.4) on page 233 [3]. 5.,6. See Entries 8, 9 of Chapter 36. 284 1. Equivalent to the formula for G9 in the table in Section 2 of Chapter 34. 2. See Entry 4 of Chapter 35. 3. Deleted by Ramanujan. 4.-6. See Entries 34-36 of Chapter 36. 7. See Entry 15 of Chapter 36. 8. Second part of Entry 5(i), Chapter 19 [3, p. 230]. 285

1. Corollary, Section 23, Chapter 18 [3, p. 209]. 2.,4. Examples, Section 23, Chapter 18 [3, pp. 209-210]. 3. Ramanujan gives the value l/l,O\e- lr ) = 0.71777, which is correct and can be verified using the evaluation of l,O(e-:rr) given in Example (i) of Section 6 in Chapter 17 [3, p. 103]. 5. See Entry 37 of Chapter 36. 6. See Entry 16 of Chapter 36. 7. Entry 13(iii), Chapter 19 [3, p. 280].

594

Ramanujan's Notebooks, Part V

286 1. Entry 13(ii), Chapter 19 [3, p. 280]. 2. See Entry 27 of Chapter 36. 3. See Entry 17 of Chapter 36. 287 1. Multiply the two equalities of Entry 5(iii), Chapter 19 [3, p. 230]. 2. Multiply the two equalities of Entry 13(iv), Chapter 19 [3, p. 281]. 3.-5. Equivalent to formulas for G 9 , G25, and G s given in the table in Section 2 of Chapter 34. 6. The value of Ot7 is found in Theorem 9.9 of Chapter 34. 7. Entry 14(v), Chapter 19 [3, p. 289]. 8. Entry 5 of Chapter 35. 9. Entry 14(iv), Chapter 19 [3, p. 289].

288 1.-4. Deleted by Ramanujan. 5. See Theorem 9.2 in Chapter 34.

289 1. We have not been able to discern the meaning of this entry. 2. Deleted by Ramanujan. 3. Equivalent to the formula for GIS in the table in Section 2 of Chapter 34. 4.,5. See Theorem 9.2 of Chapter 34. 6. Definition of a modular equation of degree 7. 7. Entry 19(i), Chapter 19 [3, p. 314]. 8. Entry 5(ii), Chapter 19 [3, p. 230]. 9. This modular equation of degree 1 is trivial since Ot = {J. 10. Ramanujan evidently intended to write a modular equation here, but there is no equality sign. Furthermore, the degree is not given. The proposed modular equation has the unusual feature that, in the first term, Ot and {J appear with no radical signs about them.

290 1. Part of Entry 5(viii), Chapter 19 [3, p. 231]. 2. Part of Entry 5(v), Chapter 19 [3, p. 230]. 3. Entry 19(ii), Chapter 19 [3, p. 314]. 4. Entry 13(v), Chapter 19 [3, p. 281]. 5.,7. See Entry 10 of Chapter 36. 6. This modular equation of degree 7 follows from Entry 19(iii) of Chapter 19 [3, p. 314) by dividing the first part of Entry 19(iii) by the second part of the same theorem. 8. This modular equation of degree 7 follows from Entry 19(ii) of Chapter 19 [3, p. 314] by dividing the first part of Entry 19(ii) by the second part of the same theorem.

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

595

291 1. This modular equation of degree 3 follows from Entry 5(v) of Chapter 19 [3, p. 230] by dividing the first part of Entry 5(v) by the second part of the same theorem. 2. Entry 7(i), Chapter 20 [3, p. 363]. 3. Entry 15(i), Chapter 20 [3, p. 411]. 4. Entry 13(i), Chapter 19 [3, p. 280]. 5. Part of Entry 5(vii), Chapter 19 [3, p. 230]. 6. Part of Entry 13(xii), Chapter 19 [3, p. 281]. 7. Entry 3(x), Chapter 20 [3, p. 352]. 8. Entry 15(i), Chapter 19 [3, p. 291]. 9. Entry 8(iii), Chapter 20 [3, p. 376]. 10. Part of Entry 19(v), Chapter 19 [3, p. 314]. 11. Entry 11(vi), Chapter 20 [3, p. 384].

292 1. Entry 5(viii), Chapter 19 [3, p. 231]. 2. Part of Entry 13(vii), Chapter 19 [3, p. 281]. 3. Entry 22(i), Chapter 20 [3, p. 439]. 4. Equivalent to a formula for G 13 given in the table in Section 2 of Chapter 34. 5. Theorem 9.14 of Chapter 34. 6. Entry 3(i), Chapter 20 [3, p. 352]. 7. Part of Entry 5(ix), Chapter 19 [3, p. 231]. 8. Part of Entry 13(x), Chapter 19 [3, p. 281]. 9. Entry 5(xiv), Chapter 19 [3, p. 231]. 10. Entry 14(i), Chapter 19 [3, p. 288]. 11. This entry, sin(2u) =

4Sin(~V)JcOS(2v) + 3sin2(~v),

is difficult to read. Although not apparent, by using elementary trigonometry, it can be shown that this modular equation of degree 7 is equivalent to Entry 19(xi) of Chapter 19 [3, p. 315].

293 1.-3. Equivalent, respectively, to formulas for G21, G 49 , and G225 given in the table in Section 2 of Chapter 34. 4. Part of Entry 11(ii), Chapter 20 [3, p. 383]. 5. Entry 2(i), Chapter 20 [3, p. 349]. 6.-10. Entries 3(iv), (vi), (v), (ii), (iii), Chapter 20 [3, p. 353].

294 1.,3. Entries 2(ii), (iii), Chapter 20 [3, p. 349]. 2.,4. Entries 9(v), (vi), Chapter 20 [3, p. 377]. 5.,6. Equivalent to Entry 9(vii), Chapter 20 [3, p. 377]. 7. Equivalent to the formula for G I 69 in the table in Section 2 of Chapter 34; for more details, see Watson's paper [7, p. 195].

596

Ramanujan's Notebooks, Part V

8. Equivalent to the formula for G 121 in the table of Chapter 34; for more details, see Watson's paper [7, p. 190]. 9. This entry is difficult to read and is evidently incomplete. We offer some comments on it at the end of Section 8 of Chapter 34.

295 1.-3. See Entries 18, 19, and 11, respectively, of Chapter 36. 4. Deleted by Ramanujan. 5.-8. Equivalent, respectively, to formulas for G 45 , Gil, G23, and table in Section 2 of Chapter 34.

G19

in the

296 1. Equivalent to the formula for G31 in the table of Chapter 34. 2. Entry 12(iii), Chapter 20 [3, p. 397). 3. Equivalent to the formula for GI7 in the table of Chapter 34. 4. Deleted by Ramanujan. 5.,6. Entries 28 and 26, respectively, of Chapter 36. 7. Deleted by Ramanujan.

297 1.-4. See Entries 12, 20, 13, and 25, respectively, of Chapter 36. 5. See Entry 7 of Chapter 35. 6. Entry 9(iii), Chapter 20 [3, p. 377). 7.-11. Entries II(i}-(v), Chapter 20 [3, pp. 383-384). 12. Part of Entry 19(iv), Chapter 19 [3, p. 314).

298 1. Entry 18(i), Chapter 20 [3, p. 423). 2. This is equivalent to Entry 12(iii) of Chapter 20 [3, p. 397); the right-hand side, K, in the formulation given on page 298 is in the customary notation for the multiplier m. 3. See Entry 75 of Chapter 36. 4. Entry 13(viii) of Chapter 19 [3, p. 281). 5. See Entry 30 of Chapter 36. 6. Part of Entry 19(vii), Chapter 19 [3, p. 314).

,.;m,

299

1. Entry 18(ii), Chapter 20 [3, p. 423). 2. Entry 4(iv), Chapter 20 [3, p. 359]. 3.,4. Entries 17(i), (ii), Chapter 20 [3, p. 417). 5.-7. Entries 19(i), (iii), (ii), Chapter 20 [3, p. 426). 8.,9. Entries 17(iii), (iv), Chapter 20 [3, p. 417]. 300 1. Part of Entry 5(ix), Chapter 19 [3, p. 231]. 2. Part of Entry 13(x), Chapter 19 [3, p. 281). 3.,4. Part of Entry 5(x), Chapter 19 [3, p. 231]. 5.,6.,10. Entry 24(ii), Chapter 18 [3, p. 214).

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

597

7.,8.,11. Entry 24(iii), Chapter 18 [3, pp. 214-215]. 9.,12. See Entries 23 and 24 of Chapter 36. 13.-15. See, respectively, Theorems 7.2, 7.7, and 7.5 of Chapter 33.

301 1.-6. See Theorems 10.5-10.10, respectively, of Chapter 33. 7. Theorem 9.4 of Chapter 33. 8. Theorem 9.15 of Chapter 33. 9. This is an incorrect version of Entry 3(iii) , Chapter 21 [3, p. 460]. 10. This is an incorrect version of Entry 5(iii), Chapter 21 [3, p. 468]. 302 1. Entry 19(i), Chapter 19 [3, p. 314]. 2. Entry 15(i), Chapter 20 [3, p. 411]. 3.,4. Deleted by Ramanujan. 5.-9.,12-14. See Entries 56, 53, 54, 55, 57, 58, 60, and 59, respectively, in Chapter 36. 10. Entry 5(ii) of Chapter 19 [3, p. 230]. 11. Another version of Entry 7(i) of Chapter 20 [3, p. 363].. 303 A discussion of the material on this page is given in the last part of Section 7 of Chapter 36.

304 1. The top left comer of the page in the original notebook has been tom away, and so the degree of the first modular equation is unknown. However, the form of the equation is exactly that of the modular equation of degree 23 in Entry 15(i) of Chapter 20 [3, p. 411]. 2.-6. Entries 62, 63, 64, 65, and 61, respectively, of Chapter 36. 305 1. Incomplete version of Entry 4(ii) of Chapter 21 [3, p. 464], and is deleted by Ramanujan. 2.-5.,7. Equivalent to formulas for G 27 , G 37 , G 39 , G 97 , and G63, respectively, in the table in Section 2 of Chapter 34. 6. Deleted by Ramanujan. 306 1.-3. Deleted by Ramanujan. 4. Part of Entry 19(vi), Chapter 19 [3, p. 314]. 5. Entry 7(iv), Chapter 20 [3, p. 363]. 6. Part of Entry 5(xi), Chapter 19 [3, p. 231]. 7. Part of Entry 13(xiii), Chapter 19 [3, p. 282]. 8. Entry 19(viii), Chapter 19 [3, p. 315]. 9. Entry 7(ii), Chapter 20 [3, p. 363]. 10. Entry 19(xi), Chapter 19 [3, p. 315]. II. Part of Entry 19(i), Chapter 19 [3, p. 314].

598

Ramanujan's Notebooks, Part V

307 1.-3. Entries II(ix), (viii), (xiv), Chapter 20 [3, pp. 384-385]. 4. Entry 15(v), Chapter 19 [3, p. 291]. 5.,6. Entry 19(iv), Chapter 20 [3, p. 426]. 7.,8. Entries 3(xii), (xiii), Chapter 20 [3, pp. 352-353].

308 1.,2. Entries 13(i), (ii), Chapter 20 [3, p. 401]. 3. Entry 5(ii), Chapter 20 [3, p. 360]. 4. Deleted by Ramanujan. 5. Entry 15(iv), Chapter 19 [3, p. 291]. 6. Deleted by Ramanujan.

309 1. Entry 15(iii), Chapter 19 [3, p. 291]. 2. Entry 5(i), Chapter 20 [3, p. 360]. 3.,4. Entries 69 and 68, respectively, of Chapter 36. 5. Entry 73 of Chapter 36. 6. Entry 13(iii) of Chapter 20 [3, p. 401]. 7. Entry 14(i) of Chapter 20 [3, p. 408]. 8. Entry 11(x) of Chapter 20 [3, p. 384]. 9. Entry I3(i) of Chapter 20 [3, p. 401]. (There is a sign error; replace -4 by 4.) 10.-13. Entries 72, 74, 71, and 70, respectively, of Chapter 36.

310 1. See Entry 66 of Chapter 36. 2.,3. See Theorems 7.10 and 7.9, respectively, of Chapter 33. 4.-11. These singular moduli are given in Theorem 9.2 of Chapter 34. 12.,13. Theorem 9.16 in Chapter 34. 311 1. Equivalent to the formula for G 33 given in the table in Section 2 of Chapter 34. 2.,3. See Entry 8.1 of Chapter 34. 4.,5. Entry 9.14 of Chapter 34. 6.-9. Entries 7-10, respectively, of Chapter 32.

312 1. See Entry 6 of Chapter 35. 2.-12. See Section 10 of Chapter 34. 13.,14. Deleted by Ramanujan. 15.,16. Part of Theorem 9.2 of Chapter 34. 313

1. Equivalent to the formula for G73 in the table in Section 2 of Chapter 34. 2. Deleted by Ramanujan. 3. Equivalent to the formula for G 43 in the table of Chapter 34.

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

599

4. Entry 7(iii), Chapter 21 [3, p. 475]. 5.-7. Part of Theorem 9.2 of Chapter 34.

314 1.-9. Equivalent to formulas for G69, GI\7, G 333, GSio and G265, respectively, in the table of Chapter 34.

G147, G 363 , G217,

G205,

315

1.-15. AtableofvaluesforGn , n =57,93,177,85,133,55,65,253,145,117, 333, 153, 77, 69, 213. See the table of Chapter 34.

316

1. This is the definition of gn. 2.-13. These twelve values for gn, n = 2, 6,10,16,18,22,30,58,70,46,142, 82, are found in the table of Chapter 34. 14.,15. See Entry 8.2 of Chapter 34. 16. Deleted by Ramanujan. 17.-22. These six values for gn, n = 42, 78, 102, 130, 190, 34, are found in the table of Chapter 34.

317

1.-9. The nine values of G n , n = 289,121, 169, 105, 165,345,385,273,357, are given in the table of Chapter 34.

318

1.-3.,5.,6.,8.-10. These eight values for gn, n = 98, 90,198,522,630,50,126, 26, can be found in the table of Chapter 34. 4. See Entry 3.2 of Chapter 34. 7. This formula for gll70 has been deleted by Ramanujan.

319

1.,2.,5.,8.,9. Five values for gn, n = 66,138,238,154,310, are given. These can be found in the table of Chapter 34. 3. This formula for gl54 has been deleted by Ramanujan, but it is correct, except for one misprint; see the table of Chapter 34. 4. This formula for gl14 is incorrect; see the table of Chapter 34 for a correct version. 6.,7.,10. These three values for gn, n = 62,94,158, are also given in the table of Chapter 34, but the formulations are somewhat different. 11.,12.,13. Three values for G n , n = 465,777, 1353, are given. Some calculations are needed to show that the formula for G 1353 given here is equivalent to that in the table of Chapter 34.

320 1.--6. The values for G n , n = 1645,897, 1677, 141,445,553, are given in the table of Chapter 34. 7.,8. The values for gn, n = 210, 330, are given in the table of Chapter 34. 9. See Theorem 9.1 of Chapter 34.

600

Ramanujan's Notebooks, Part V

10. See the introduction to Section 9 of Chapter 34.

321 1.,2. Entry 11(iii), Chapter 19 [3, pp. 265-266]. 3.-6. Parts of Entries l(iv), (v), (ii), (iii), Chapter 20 [3, pp. 345-346]. 7. Entry I, Chapter 18 [3, p. 144]. 8. Entry 2, Chapter 18 [3, p. 145]. 9. Entry 8(i), Chapter 18 [3, p. 157]. 10. Entry 9, Chapter 18 [3, p. 159].

322 1. Entry 18(vi), Chapter 19 [3, p. 306]. 2.,3. Part of Entry 3(i), Chapter 21 [3, p. 460]. 4.,5. Entry 3(ii), Chapter 21 [3, p. 460]. 6.,7. Entry 4(i), Chapter 21 [3, p. 463]. 8. Entry 4(ii), Chapter 21 [3, p. 464]. 9. Part of Entry 5(i), Chapter 21 [3, p. 467].

323 1. Entry 5(ii), Chapter 21 [3, p. 468]. 2. Part of Entry 7(i), Chapter 21 [3, p. 475]. 3. Entry 7(ii), Chapter 21 [3, p. 475]. 4.-6. Entries 2(vii), (v), (viii), Chapter 20 [3, p. 349]. 7.-9. Entries 8(i), (ii), (iii), Chapter 21 [3, p. 480]. 10. Entry 9(i), Chapter 21 [3, p. 481].

324 1.,2. Entries 9(ii), (iii), Chapter 21 [3, p. 481]. 3.,4. Entry 4(iii), Chapter 21 [3, p. 464]. 5.-7. Entry l(i), Chapter 20 [3, p. 345]. 8.-10. Entry 1(ii), Chapter 19 [3, p. 221].

325 1. The formula for G301 can be found in the table of Chapter 34. 2.,3.,6.,7. Entries lO(i)-(iv) of Chapter 20 [3, p. 379]. 4.,5. Entry 21 of Chapter 36. 8. Entry I, Chapter 19 [3, p. 221]. 9. Entry 18(i), Chapter 19 [3, p. 305]. to. Entry 6(iii), Chapter 20 [3, p. 363].

326

1.-4. Entry 18(i), Chapter 19 [3, p. 305]. Note that the definitions of u, v, and

w are different in the first notebook.

5.-10. Entry 8(ii), Chapter 20 [3, pp. 372-373].

327 1.-4. Entry 12(i), Chapter 20 [3, p. 397]. 5. Entry 5(iv), Chapter 20 [3, p. 360].

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

601

6.,7. Entry 18(v), Chapter 20 [3, p. 423].

328 1. Theorem 7.11 of Chapter 33. 2.-4. Entries 8-10, Chapter 30 [4, pp. 374-377].

329 1.-3. Entries 62, 69, and 63, respectively, of Chapter 25 [4, pp. 221, 236, 223]. 4.,5. Entries 25 and 26 of Chapter 32.

330 1.-4. Entries 51,49,52, and 50, respectively, of Chapter 32.

331 1. Entry 21, Chapter 28 [4, p. 309]. 2.-6. Entries 48,14,19,20, and 21, respectively, of Chapter 32.

332 1. Entry 23 of Chapter 32. 2. Deleted by Ramanujan.

333 1.,2. Entries 6 and 7 of Chapter 30 [4, pp. 364, 370]. 334 1. Entry 5 of Chapter 30 [4, p. 363]. 2. An incomplete version of Entry 34 of Chapter 37.

3. Deleted by Ramanujan but proved in Entry 31 of Chapter 37. 4. Entry 20, Chapter 28 [4, p. 309]. 5. See Entry 32 of Chapter 37.

335 1. See Entry 33 of Chapter 37. 2.,3. Entries 6 and 4, respectively, of Chapter 29 [4, pp. 338, 336]. 336

1.,2. Entries 5 and 3, respectively, of Chapter 29 [4, pp. 337, 336]. 3. Entry 18, Chapter 28 [4, p. 307]. 4. See Entry 27 of Chapter 37. 337

1.,2. Entry 29 of Chapter 37. 3. Entry 30 of Chapter 37. 4. Entry 17, Chapter 28 [4, p. 306]. 5. Deleted by Ramanujan; for a correct version see the tables ofGradshteyn and Ryzhik [1, p. 546, formula 4.113, no. 4].

338 The results on this page comprise the contents of Section 3 of Chapter 35.

602

Ramanujan's Notebooks, Part V

339 1.-3. These results are also contained in Section 3 of Chapter 35. 4. A definition. 5. A triviality. 6.-8. These results are essentially Entries 4, 3, and 7, respectively, of Chapter 35. 9.-11. Entries 23, 24, and 22, respectively, of Chapter 25 [4, pp. 154, 155, 153]. 12. See Entry 17 of Chapter 39.

340 1. Entry 12 of Chapter 35. 2. Entry 3, Chapter 30 [4, p. 359]. 3.-8. Entries 1-6, Chapter 26 [4, pp. 245-255].

341 1.-3. Entries 7-9, Chapter 26 [4, pp. 255-257]. 4. Entry 4, Chapter 30 [4, p. 360]. 5. Entry 24, Chapter 32. 6. Entry 20 of Chapter 37. 342 1.-3. Deleted by Ramanujan. 4. Entry 23 of Chapter 37. 5. Entry 10, Chapter 26 [4, p. 258]. 6. Entry 1, Chapter 29 [4, p. 335]. 343 1.-3. Entries 24-26, respectively, of Chapter 37. 4. Entry 22 of Chapter 37. 5. An incomplete entry. We offer some comments on it at the end of Section 8 of Chapter 34. 6. The value for G765 is given in the table in Section 2 of Chapter 34. 344 1. Entry 6 of Chapter 32. 2. The value of G 505 is given in the table of Chapter 34. 3.-5. Monic irreducible cubic polynomials satisfied by gn, n = 38,26, and 50. See the table in Section 2 of Chapter 34. 6. Entry 13 of Chapter 32. 345 1.-7. Monic irreducible cubic polynomials satisfied by G n , n = 23, 31, 11, 19, 27, 43, and 67. See the table in Section 2 of Chapter 34. n = 3, 5, 7, 9,13,15,17,21, and 25. 8.-16. Factors of singular moduli However, for n = 21, Ramanujan fails to record any factors. See Theorem 9.9 of Chapter 34.

..;a;.,

Location of Entries in the Unorganized Portions of Ramanujan's First Notebook

603

346

1.-4. Factors ofsingular moduli ...;a;, n = 7, 15,39, and 55, but for n = 39, Ramanujan left a blank space. See Theorem 9.9 of Chapter 34.

347-349 All the results on these three pages are found in Ramanujan's paper on Bernoulli numbers [1J, [10, pp. l-14J.

350 For comments on Ramanujan's notes at the top of the page, see Berndt and Rankin's book [1, p. IOJ. 1. See Entry 18 of Chapter 39. 351 1. Ramanujan gives the first 22 digits of ../2. 2.-10. These are irreducible polynomials for the reciprocals of the invariants G n , n = 3, 7,11,19,23,27, 31, 43, and 67. See the table of Chapter 34. Ramanujan had evidently intended to calculate several further polynomials, as indicated by vacant spaces beside certain other values of n.

References

Abramowitz, M. and Stegun, I. A., editors [1] Handbook ofMathematical Functions, Dover, New York, 1965. Alladi, K. [1] On the modified convergence of some continued fractions of Rogers-Ramanujan type, J. Combin. Theory, Ser. A 65 (1994), 214-215. Alladi, K. and Gordon, B. [1] Partition identities and a continued fraction of Ramanujan, J. Combin. Theory, Ser. A 63 (1993), 275-300. Andrews, G. E. [1] On q--difference equations for certain well-poised basic hypergeometric series, Quart. J. Math. (Oxford) 19 (1968), 433-447. [2] On the general Rogers-Ramanujan theorem, Memoir, American Mathematical Society, No. 152, Providence, RI, 1974. [3] On Rogers-Ramanujan type identities related to the modulus 11, Proc. London Math. Soc. 30 (1975), 330-346. [4] The Theory of Partitions, Addison-Wesley, Reading, MA, 1976. [5] An introduction to Ramanujan's "lost" notebook, Amer. Math. Monthly 86 (1979), 89-108. [6] Partitions: Yesterday and Today, New Zealand Mathematical Society, Wellington, 1979. [7] Ramanujan's "lost" notebook III. The Rogers-Ramanujan continued fraction, Adv. in Math. 41 (1981), 186-208. Andrews, G. E., Berndt, B. C., Jacobsen, L., and Lamphere, R. L. [1] Variations on the Rogers-Ramanujan continued fraction in Ramanujan's notebooks, in Number Theory, Madras 1987, K. Alladi, ed., Lecture Notes in Mathematics No. 1395, Springer-Verlag, New York, 1989, pp. 73-83. [2] The continued fractions found in the unorganized portions of Ramanujan's notebooks, Memoir, American Mathematical Society, No. 477, Providence, RI, 1992. Andrews, G. E. and Bowman, D. [1] A full extension of the Rogers-Ramanujan continued fraction, Proc. Amer. Math. Soc. 123 (1995), 3343-3350. Askey,R.A. [1] Gaussian quadrature in Ramanujan's second notebook, Proc. Indian Acad. Sci. (Math. Sci.) 104 (1994),237-243.

606

Ramanujan's Notebooks, Part V

Atkin, A. O. L. [1] Note on a paper of Rankin, Bull. London Math. Soc. 1 (1969), 191-192. Atkin, A. O. L. and Swinnerton-Dyer, P. [1] Some properties of partitions, Proc. London Math. Soc. (3) 4 (1954), 84-106. Bachman,G. [1] On the convergence of infinite exponentials, Pacific J. Math. 169 (1995),219-233. Bailey, W. N. [1] Generalized Hypergeometric Series, Stechert-Hafner, New York, 1964. Barrow,D.F. [1] Infinite exponentials, Amer. Math. Monthly 43 (1936), 150-160. Berndt, B. C. [1] Ramanujan's Notebooks, Part I, Springer-Verlag, New York, 1985. [2] Ramanujan's Notebooks, Part II, Springer-Verlag, New York, 1989. [3] Ramanujan's Notebooks, Part III, Springer-Verlag, New York, 1991. [4] Ramanujan's Notebooks, Part N, Springer-Verlag, New York, 1994. [5] A new method in arithmetical functions and contour integration, Canad. Math. Bull. 16 (1973), 381-388. [6] Modular transformations and generalizations of several formulae of Ramanujan, Rocky MountainJ. Math. 7 (1977),147-189. [7] Analytic Eisenstein series, theta-functions, and series relations in the spirit of Ramanujan, J. Reine Angew. Math. 303/304 (1978),332-365. [8] An arithmetic Poisson formula, Pacific J. Math. 103 (1982),295-299. [9] On a certain theta-function in a letter of Ramanujan from Fitzroy House, Ganita 43 (1992), 33-43. [10] Ramanujan's theory of theta-functions, in Theta Functions, From the Classical to the Modern, M. Ram Murty, ed., Centre de Recherches Mathematiques Proceedings and Lecture Notes, Vol. 1, American Mathematical Society, Providence, RI, 1993, pp.I-63. Berndt, B. C. and Bhargava, S. [1] Ramanujan's inversion formulas for the lemniscate and allied functions, J. Math. Anal. Appl. 160 (1991), 504-524. Berndt, B. C., Bhargava, S., and Garvan, F. G. [1] Ramanujan's theories of elliptic functions to alternative bases, Trans. Arner. Math. Soc. 347(1995),4163-4244. Berndt, B. C. and Bialek, P. [1] Five formulas of Ramanujan arising from Eisenstein series, in Number Theory. Fourth Conference of the Canadian Number Theory Association, K. Dilcher, ed., Canad. Math. Soc. Conf. Proc., Vol. 15, American Mathematical Society, Providence,RI,1995,pp.67-86. Berndt, B. C. and Chan, H. H. [1] Some values for the Rogers-Ramanujan continued fraction, Canad. J. Math. 47 (1995),897-914. [2] Ramanujan's explicit values for the classical theta-function, Mathematika 42 (1995), 278-294. [3] Notes on Ramanujan's singular moduli, in Number Theory, Fifth Conference of the Canadian Number Theory Association, R. Gupta and K. S. Williams, eds., Canad.

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Heine,E. [11 Untersuchungen fiber die Reihe

1+

(1 - qa)(l - qfJ) (1 - q)(1 - qY)

x+

(1 - q")(1 - q,,+l)(l - qfJ)(1 - qfJ+ 1 ) (1 - q)(1 - q2)(1 - qY)(1 - qY+l)

2

x+···

,

J. Reine Angew. Math. 34 (1847), 285-328.

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