Yesilyurt H - Four Identities Related to Third Order Mock Theta Functions in Ramanujan_s Lost...

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Advances in Mathematics 190 (2005) 278299

Four identities related to third order mock theta

functions in Ramanujans lost notebook

Hamza Yesilyurt

Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA

Received 10 February 2003; accepted 17 December 2003

Communicated by Michael Hopkins

Abstract

We prove, for the first time, a series of four related identities from Ramanujans lost

notebook. The identities are connected with third order mock theta functions.

r 2004 Elsevier Inc. All rights reserved.

1. Introduction

In his last letter to Hardy, Ramanujan introduced mock theta functions

[9, pp. 127131]. Included in this letter were the four third order mock theta

functions:

effq XNn0

qn2

1 q21 q22?1 qn2; 1:1

effq XNn0

qn2

1 q21 q4?1 q2n; 1:2

eccq XN

n1

qn2

1 q1 q3

?

1 q2n

1

;

1:3

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0001-8708/$ - see front matterr 2004 Elsevier Inc. All rights reserved.

doi:10.1016/j.aim.2003.12.007


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ewwq

XNn0

qn2

1 q q21 q2 q4?1 qn q2n: 1:4

They satisfy the equations

2effq effq effq 4eccq q; q1N

XNnN

1nqn2 ; 1:5

4ewwq effq 3q; q1N

XNnN

1nq3n2( )2

; 1:6

where we used the standard notation

a; q

n

a

n :

1

a

1

aq

y

1

aqn1

;

a; qN

YNn0

1 aqn; jqjo1:

Watson [10] proved (1.5) and (1.6). Andrews [1] also gave certain generalizations of

(1.5) and (1.6). Third order mock theta functions are related to the rank of a

partition defined by Dyson [5] as the largest part minus the number of parts. Let us

define Nm; n as the number of partitions ofn with rank m: The generating functionfor Nm; n is given by

XNmN X

N

n0 Nm; nqn

tm

XNn0

qn2

tqnt1qn; jqjo1; jqjojtjoj1=qj: 1:7

The third order mock theta functions defined by (1.1) through (1.4) can be expressed

in terms of this generating function. Third order mock theta functions and their

applications to the rank are detailed by Fine [7]. A comprehensive literature survey

on mock theta functions is given by Andrews [2].

We prove, for the first time, a series of four related identities from Ramanujans

lost notebook. These identities are defined and their connections to (1.5) and (1.6)

are given in Section 3. Proofs of these identities are provided in Sections 47. In

addition, we will show in Section 8 that one of the identities can be used to prove the

following identity:

q2N

tN

t1qN

XN

nN1nq

nn1=2

1 tqn : 1:8

Identity (1.8) was proved by Evans [6, Eq. (3.1)] following a different approach.

Equality (1.8) is also given in a different form by Ramanujan on p. 59 of the lost

notebook [9]. Partition theory implications of the product

q

N

tqN

t1qN

are discussed in [8].

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2. Definitions and preliminary results

We first recall Ramanujans definitions for a general theta function and some of its

important special cases. Set

fa; b :Xp

nNann1=2bnn1=2; jabjo1: 2:1

Basic properties satisfied by fa; b include [4, p. 34, Entry 18]

fa; b fb; a; 2:2

f1; a 2fa; a3; 2:3

f1; a 0 2:4

and if n is an integer,

fa; b ann1=2bnn1=2faabn; babn: 2:5

If n 1; (2.5) becomes

fa; b afa1; a2b: 2:6

Two other formulas satisfied by fa; b are [4, p. 46, Entry 30]

fa; b fa;b 2fa3b; ab3; 2:7

fa; b fa;b 2afba1; ab1a4b4: 2:8

The function fa; b satisfies the well-known Jacobi triple product identity [4, p. 35,Entry 19]

fa; b a; abN

b; abN

ab; abN

: 2:9

Some important special cases of (2.1) and (2.9) are

jq : fq; q XN

nNqn

2 q; q22N

q2; q2N

; 2:10

cq : fq; q3 XNn0

qnn1=2 q; q2N

q; qN

; 2:11

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fq : fq;q2 XN

nN1nqn3n1=2 q; q

N: 2:12

By using (2.10) and (2.11), and elementary product manipulations, we find that

cq q; qNq2; q4N

q2; q2

N

q; q2N

; 2:13

jq q; qNq; qN

: 2:14

Also, after Ramanujan define

w

q

:

q; q2

N

:

2:15

Other basic properties of the functions j; c; f and w are [4, p. 39, Entry 24]

fqfq

cqcq

wqwq

ffiffiffiffiffiffiffiffiffiffiffiffiffiffijqjq

s; 2:16

wq fqfq2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffijqcq

3

s jq

fq fq2cq ; 2:17

f3

q2

jqc2

q: 2:18Combining (2.17) and (2.18), we obtain

f3q cqj2q: 2:19

We will frequently use Eulers identity

q; qN

q; q21N

: 2:20

For any real number a; let

faq :XNn0

qn2

1 aq q2?1 aqn q2n; 2:21

where jqjo1: For jqjo1; jqjojtjojqj1; let

Gt; q :XNn0

qn2

tqnt1qn: 2:22

We need Eulers famous generating function for partitions,

G

1; q

q; q

1N

:

2:23

For a proof of (2.23) see [7, p. 13, eq. (12.311)]. We need variations of two

representations for Gt; q due to Fine [7].

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Lemma 2.1. For jtjo1;

Gt; q 1 tXNn0

tn

t1qn 2:24

1 t1 t1XNn0

tqnt1qn

: 2:25

Proof. Following Fine [7, pp. 1, Eq. (1.1)], we define

Fa; b; t : XNn0

aqnbqntn:

In this notation, Lemma 2.1 can be written as

Gt; q 1 tF0; t1; t 2:26

1 t1 t1F0; t1; tq: 2:27

Eq. (2.26) is Eq. (12.3) on p. 13 of [7] with b replaced by t1; and (2.27) readilyfollows from Eq. (2.4) on p. 2 of [7].

Observe that (2.25) is valid in the region jqjojtjojqj1: Also as noted by Fine [7,p. 51, Eq. (25.6)], Gt; q satisfies a third order q-difference equation. We sketch aproof here since it is stated without a proof in [7].

Lemma 2.2. For jqjo1 and jqjojtjoj1=qj; Gt; q satisfies the q-difference equation

1

1 tq Gtq; q qt3

1 tGt; q 1 qt2: 2:28

Proof. Let

Mt; q :XNn0

tqnt1qn

; 2:29

so that by (2.25),

Gt; q 1 t1 t1Mt; q: 2:30

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Using definition (2.29) and algebraic manipulation, we obtain

Mt; q XN

n0 tq

n

t1qn XN

n0 tq

n

1

t1qn1

t1qn1

XNn0

tqnt1qn1

t1qXNn0

tq2nt1qn1

XNn0

tqnt1qn1

t1q1 t1XNn0

tq2nt1n2

1tq

XNn0

tqn1t1qn1

1tq22t1q1 t1

XNn0

tq2n2t1n2

1tq

Mt; q 1 1 t1

t3q3Mtq; q 1 tq

2

1 t1 : 2:31Now, Lemma 2.2 follows from (2.31) together with (2.30) after rearrangement.

For convenience, define

Vt; q : 11 tGt; q: 2:32

Lemma 2.2 then takes the following form:

Vtq; q qt3Vt; q 1 qt2: 2:33

Observe that

Vt1; q tVt; q: 2:34

The basic property (2.34) will be used many times in the sequel without comment.

The partial fraction decomposition of Vt; q is given by [8, Eq. (7.10)]

Vt; q 1 tqN

XNnN

1nq3nn1=21 tqn : 2:35

We will need the following lemma due to Atkin and Swinnerton-Dyer [3].

Lemma 2.3. Let q; 0oqo1; be fixed. Suppose that Wz is an analytic function of z;except for possibly a finite number of poles, in every region, 0oz1pjzjpz2:

If

W

zq

AzkW

z

for some integer k (positive, zero, or negative) and some constant A; then either Wzhas k more poles than zeros in the region jqjojzjp1; or Wz vanishes identically.

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3. Four identities of Ramanujan

We now offer the four identities from Ramanujans lost notebook that we plan to

prove.

Entry 3.1 (Ramanujan [9, p. 2, no. 3]). Suppose that a and b are real with a2 b2 4: Then, if faq is defined by (2.21),

b a 24

faq b a 24

faq b2

fbq

q4; q4

N

q; q2NYN

n

1

1 bqn q2n1 a2b2 2q4n q8n: 3:1

If we take a 0 and b 2; then, by using (2.14) and elementary productmanipulations, we see that (3.1) reduces to (1.5) in the notation of (2.10) as follows:

2effq effq q1Njq:

Entry 3.2 (Ramanujan [9, p. 2, no. 4]). If a and b are real with a2 ab b2 3; then

a 1faq b 1fbq a b 1fabq

3q3; q32

N

q; qN

YNn1

1

1 aba bq3n q6n: 3:2

In (3.2), take a b 1 and use (2.14); then one obtains (1.6) in the notation of(2.10) as

4

ewwq effq 3q1

Nj2q3:

We changed the notation that Ramanujan used in the left-hand side of the next entry

to avoid confusion. Also note that the series on the right side below is f ffiffi3p q in thenotation of (2.21).

Entry 3.3 (Ramanujan [9, p. 17, no. 5]). With faq defined by (2.21),

1 ffiffiffi3p2

f1q 3 ffiffiffi

3p

6f1q

XN

n0

qn2

1

ffiffiffi3p

q

q2

?

1

ffiffiffi3p

qn

q2n

2ffiffiffi

3pcqq

4; q4

N

q6; q6N

YNn1

1

1 ffiffiffi3p qn q2n: 3:3

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Entry 3.4 (Ramanujan [9, p. 17, no. 6]). With effq defined by (1.2),1

21 epi=4

effiq 121 epi=4effiqXNn0

qn2

1 ffiffiffi

2p

q q2?1 ffiffiffi

2p

qn q2n

1ffiffiffi2

pcqq2; q4N

YNn1

1

1 ffiffiffi2p qn q2n: 3:4Note that the series on the right side above is f

ffiffi2

p q in the notation of (2.21).

4. Proof of Entry 3.1

Let a 2 cosy; b 2 siny; and t eiy: Then, it is easy to verify that

faq Gt; q; faq Gt; q; fbq Git; q 4:1

and

b a 24

1 i4t

1 it1 t; b a 24

1 i4t

1 it1 t;b

2 i

2t1 t2; a2b2 2 2cos4y t4 t4: 4:2

Using (4.1) and (4.2), we can rewrite (3.1) as

i 14t

1 it1 tGt;q 1 i4t

1 t1 itGt;q

i2t

1 t2Git; q

q4; q4

Nitq

Nit1q

N

q; q2N

t4q4; q4N

t4q4; q4N

: 4:3

Multiplying both sides of (4.3) by 1 it; we obtain

i 11 t44t

1

1

tGt;q i

1

tGt;q i 1

1

itGit; q

1 itq4; q4

Nitq

Nit1q

N

q; q2N

t4q4; q4N

t4q4; q4N

: 4:4

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Using definition (2.32) and dividing both sides of (4.4) by i 11 t4=4t; we seethat (3.1) is equivalent to the identity

Vt;q iVt;q i 1Vit; q

21 it q4; q4

N1 ititq

Nit1q

N

q; q2N

1 t4t4q4; q4N

t4q4; q4N

21 it q4; q4

Nit

Nit1q

N

q; q2N

t4; q4N

t4q4; q4N

21 it q4; q42

Nfit;it1q

q; q2N

q; qN

ft4; t4q4; 4:5

where in the last step we used the Jacobi triple product identity (2.9). We will verify

that (4.5) is valid for jqjojtjojq1j for any fixed jqjo1: Let

Lt; q : Vt;q iVt;q i 1Vit; q;

Rt; q : 21 it q4; q42

Nfit;it1q

q; q2N

q; qN

ft4;t4q4:

The proof of Entry 3.1 will be complete once we show that Rt; q Lt; q 0: Thiswill be achieved by showing that R

t; q

L

t; q

satisfies a q-difference equation of

the sort stated in Lemma 2.3 and has no poles, thereby, forcing it to vanish

identically.

Note that if we define kz : fcz; c1z1q; then by (2.6) we have

kzqkz

fczq; c1z1fcz; c1z1q

c1z1fcz; c1z1qfcz; c1z1q cz

1: 4:6

Following the same reasoning of (4.6), we obtain

Rtq; qRt; q

tq

t

fitq;it1

fit;it1qft4q4;t4ft4;t4q4

q it1

t41 iqt3:

Let us verify now that Lt; q also satisfies the same q-difference equation. To thatend,

Ltq; q iqt3Lt; q

Vtq;q iVtq;q i 1Vitq; q iqt3fVt;q iVt;q i 1Vit; qg

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fVtq;q qt3Vt;qg ifVtq;q qt3Vt; qg

i

1

fV

itq; q

iqt3V

it; q

g 1 qt2 i1 qt2 i 11 qit2

1 qt2 i1 qt2 i 11 qt2 0;

where we employed (2.33). Now Lemma 2.3 implies that Rt; q Vt; q either hasat least 3 poles in the region jqjojzjp1; or vanishes identically. But Rt; q Vt; qhas at most 3 poles, namely at t 1;1; and i in that region, and they are allremovable as we shall demonstrate. It suffices to show that t 1 is a removablesingularity. Thus,

limt-1

1 tLt limt-1

1 t fVt;q iVt;q i 1Vit; qg

limt-1

1 t 11 tGt;q

i

1 tGt;q i 11 itGit; q

ilim

t-1Gt;q iq; q1

N; 4:7

by (2.23).

Next, by two applications of (2.9) and (2.20),

limt-1

1 tRt

limt-1

1 t 21 it q4; q42

Nfit;it1q

q; q2N

q; qN

ft4;t4q4

( )

21 i limt-1

1 tt q4; q42

Nfit;it1q

q; q2N

q; qN

t4; q4N

t4q4; q4N

q4; q4N

21 i limt-1

1 tt q4; q4

Nfit;it1q

q; q2

N

q; q

N

1

t4

t4q4; q4

N

t4q4; q4

N

1 iq4; q4

Nfi;iq

2q; q2N

q; qN

q4; q4N

q4; q4N

1 ifi;iq2q; q2

Nq; q

Nq4; q4

N

1 ii; qNiq; qNq; qN2q; q2

Nq; q

Nq4; q4

N

1 i1 iiq; qNiq; qN2q; q2

Nq4; q4

N

i q2; q2

N

q; q2N

q4; q4N

iq; q2N

q4; q4N

q2; q4N

i

q; q4

N

q3; q4

N

q4; q4

N

q2; q4

N

i

q;

q

N

: 4:8

Hence, by (4.7) and (4.8), Lt; q Rt; q has a removable singularity at t 1: Byour earlier remarks, this completes the proof of Entry 3.1.

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Our proof of Entry 3.2 is similar to our proof of Entry 3.1. Since 3

a2 ab b2 a b2 3ab a b2 ab; we must have jabjo4: Assume with-out lost of generality that jajojbj; and let a 2 cosy: Solving a2 ab b2 3 forb gives b cosy8ffiffiffi3p siny: We will take b cosy ffiffiffi3p siny 2 sinyp=6; since replacing y by y gives the other value for b:

Let t eiy and r e2pi=3: Using this parametrization we obtain

a t t1; b r1t rt1; and a b rt r1t1;

which, in turn, implies that

faq Gt; fbq Gr1t; and fabq Grt:

One can easily verify that

a 1 1 t3

t1 t; b 1 r1 t3

t1 r1t; and a b 1 r11 t3

t1 rt :

Now, the left side of (3.2) which we recall below, becomes

a 1faq b 1fbq a b 1fabq

1 t3

t1 tGt r1 t3

t1 r1tGr1t r

11 t3t1 rt Grt

1 t3

tVt rVr1t r1Vrt:

While the right-hand side of (3.2), after observing that

aba b 2 cos3y t3 t3;

reduces to

3q3; q32N

q; qN

t3q3; q3N

t3q3; q3N

:

Thus, Entry 3.2 is equivalent, by (2.9), to the identity

Vt rVr1t r1Vrt 3tq3; q33

N

fqft3;t3q3: 5:1

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Let Nt and Dt denote the right and left sides of (5.1), respectively. We will verify thatNt Dt satisfies the q-difference equation Ntq Dtq qt3Nt Dtwithout any poles in

jq

jo

jt

jp1: Then using Lemma 2.3, we conclude that N

t

Dt 0:We employ (4.6) with c 1; and t and q replaced by t3 and q3; respectively, to

deduce that

NtqNt

tq

t

1

t31 qt3:

Next,

Dtq qt3Dt

Vtq rVr1tq r1Vrtq qt3fVt rVr1t r1Vrtg

Vtq qt3Vt rfVr1tq qt3Vr1tg r1fVrtq qt3Vrtg

1 qt2 r1 qr1t2 r11 qrt2

1 r r1 qt21 r1 r 0;

where we used (2.33). Lemma 2.3 now implies that either Nt Dt vanishes or hasthree more poles than zeros in jqjojtjp1: But Nt Dt has at most three poles,namely at t 1; r;r1; and they are all removable as we demonstrate. It suffices toshow that t 1 is removable.

By (2.23),

limt-1

1 tDt limt-1

1 tfVt rVr1t r1Vrtg

limt-11 t1

1 tGt r1

1 r1tGr1t r11

1 rtGrt lim

t-1Gt 1

fq:

By the Jacobi triple product identity (2.9),

limt-1

1 tNt limt-1

1 t 3tq3; q33

N

f

q

f

t3;

t3q3

lim

t-11 t 3tq

3; q32

N

fq1 t3t3q3; q3N

t3q3; q3N

1fq:

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We have shown that Nt Dt has a removable singularity at t 1: By ourearlier remarks this completes the proof of Entry 3.2.


If a 1; b ffiffiffi3p in Entry 3.1, thenffiffiffi

3p 1 2

4f1q

ffiffiffi3

p 1 24

f1q ffiffiffi

3p

2f ffiffi3p q

q4; q4

N

q; q2NYNn1

1

ffiffiffi3p

qn

q2n

1 q4n q8n :

Multiplying both sides by 2=ffiffiffi

3p

; we find that

3 ffiffiffi3p6

f1q 1 ffiffiffi

3p

2f1q

f ffiffi3p

q

2

ffiffiffi3pq4; q4

N

q; q2

NYN

n1

1

ffiffiffi3

pqn q2n

1 q4n

q8n

:

We need to show then that

2ffiffiffi3

pcqq4; q4

N

q6; q6N

YNn1

1

1 ffiffiffi3p qn q2n 2ffiffiffi3p q4; q4

N

q; q2N

YNn1

1 ffiffiffi3p qn q2n1 q4n q8n :

6:1

Recall that c is defined by (2.11). Now,

q6; q6N

q; q2N

YNn1

1 ffiffiffi3p qn q2n1 ffiffiffi3p qn q2n1 q4n q8n

q6; q6

N

q; q2N

YNn1

1 q2n q4n1 q4n q8n

q6; q6

N

q; q2

NYN

n

1

1

1

q2n

q4n

q6; q6

Nq2; q2

N

q; q2N

q6; q6N

q2; q2

N

q; q2N

cq;

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where in the last step (2.13) is used. Equality (6.1) now follows, and so the proof of

Entry 3.3 is complete.


Let a eip=4: Clearly, using the notation of (2.22), we have effq Gi; q; andf ffiffi

2p q Ga; q: We can then restate Entry 3.4 as

1 a2

Gi; iq 1 a1

2Gi;iq Ga; q 1

ffiffiffi2

p cqq2; q4

N

aqN

a1qN

:

Dividing both sides by 1 a=2 and employing (2.9), we arrive at

Gi; iq a1Gi; iq 21 aGa; q

ffiffiffi2

pcqfqq2; q4N

fa; a1q :

7:1

If we replace q by iq; (7.1) becomes

G

i;

q

a1G

i; q

2V

a; iq ffiffiffi2

pciqfiqq2; q4N

fa; aq:

7:2

The following identities will be needed for the remainder of the proof:

fa; a1qfa;a1q 1 iq4; q4N

f2q; 7:3

fa; aqfa;aq 1 iq4; q4N

f2iq; 7:4

fa; a1q ciq aciq; 7:5

fa;a1q ciq aciq; 7:6

fa; aq cq acq; 7:7

fa;aq cq acq: 7:8We now offer proofs for all six identities.

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To prove (7.3) we employ (2.9) to find that

f

a; a1q

f

a;

a1q

a

N

a1q

N

f

q

a

N

a1q

N

f

q

i; q2

Niq2; q2

Nf2q

1 iq4; q4N

f2q:

Clearly, (7.4) is obtained by replacing q by iq in (7.3). Recall that cq fq; q3:From (2.7) and (2.8),

fa; a1q fa;a1q

2fa2q; a2q3 2fiq;iq3 2ciq; 7:9

fa; a1q fa;a1q

2afa2q; a2q3 2afiq; iq3 2aciq 7:10

Equalities (7.9) and (7.10) readily imply (7.5) and (7.6). And finally we obtain (7.7)

and (7.8) by replacing q by iq in (7.5) and (7.6), respectively.

We now return to (7.2) and use (7.4), (7.8), and (2.13) with q replaced by iq todeduce that

Gi;q a1Gi; q 2Va; iq

ffiffiffi

2pciqfiqq2; q4

N

fa; aq

ffiffiffi2

pciqfiqq2; q4N

fa; aqfa; aqfa;aq

ffiffiffi2pciqfiqq2; q4Ncq acq1 iq4; q4N

f2iq

aciqq2; q4

Ncq acq

q4; q4N

fiq

aq2; q4

Ncq acq

q4; q4N

q2; q4N

aq2; q4

Ncq acq

q2; q2N

aq2; q42Ncq acq: 7:11

It suffices now to prove (7.11).

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Let

K

t; q

:

aV

it; iq

aV

it; iq

iV

t; iq

iV

t; iq

1 iVat;q 1 iVat;q: 7:12

The identity,

Kt; q 4a1t f3q4fa1t; at1q

fiqfa; a1qft4;t4q4

21 it cq2f2qfa1t; at1q

f

t; t

1qf

it; it

1q

f

it2; it

2q2

; 7:13

together with Entry 3.1 will be used to verify (7.11). We will not prove (7.13), because

its proof is very similar to that of (5.1). The q-difference equation satisfied by Kt; qis Ktq; q aqt3Kt; q: It then suffices, by Lemma 2.3, to verify that the residuesat four of the six singularities match those of the two representations (7.12) and

(7.13) of Kt; q: It is easily verified that t a is a zero for the two representations(7.12) and (7.13) of Kt; q: Therefore, one only needs to check the residues at anythree of the six singularities. If we knew the two other zeros whose existence is

guaranteed by Lemma 2.3, we then would be able to reduce the right-hand side of

(7.13) to a single product, but we are unable to determine these two zeros.Let us define, by using (4.5),

Et; q : Vt;q iVt;q i 1Vit; q 7:14

21 it q4; q42

Nfit;it1q

q; q2N

q; qN

ft4;t4q4: 7:15

We will verify by using (7.12) and (7.14) that

Gi;q a1Gi; q 2Va; iq

1a1 iEa; iq Ea; iq

1

2iKa; q 1

2aKa;q: 7:16

Equalities (7.13) and (7.15) will then be used to verify that (7.16) reduces to the right-

hand side of (7.11), which will complete the proof of Entry 3.4.

Using (7.14), we have

Et; q Et; q 1 ifVt;q Vt;q Vit; q Vit; qg:

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Setting t a; we find that

E

a; q

E

a; q

1 ifVa;q Va;q Via; q Via; qg

1 ifVa;q Va;q Va1; q Va1; qg

1 ifVa;q Va;q aVa; q aVa; qg:

Replacing q by iq and dividing by a1 i; we obtain

1

a1 iEa; iq Ea; iq a1Va;iq a1Va;iq Va; iq Va; iq: 7:17

By (7.12),

Ka; q aVia; iq aVia; iq iVa; iq iVa; iq

1 iVi;q 1 iVi;q

aVa1

; iq aVa1

; iq iVa; iq iVa; iq 1 iVi;q 1 iVi;q

iVa; iq iVa; iq iVa; iq iVa; iq

i1 iVi;q 1 iVi;q

2iVa; iq 2iVa; iq 2iGi;q: 7:18

Combining (7.17) and (7.18), we find that

1

a1 iEa; iq Ea; iq 1

2iKa; q 1

2aKa;q

a1Va;iq a1Va; iq Va; iq Va; iq

Va; iq Va; iq Gi;q

a1Va;iq a1Va;iq a1Gi; q

Gi;q a1Gi; q 2Va; iq:This proves our first claim that (7.16) holds.

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Using (7.13), (2.3), (2.6), and (2.19) with q replaced by q4; we find that

Ka; q 4f3

q4

f

1; q

fiqfa; a1qf1; q4 2a1 i cq

2f2qf1; qfa; a1qfia; ia1qf1; q2

4 f3q4cq

fiqfa; a1qcq4 2a1 icq2f2qcq

fa; a1qfa1; aqcq2

4 f3q4cq

fiqfa; a1qcq4 21 icq2f2qcqf2a; a1qcq2

4j2

q4

c

q

fiqfa; a1q 21 ic

q2

f2

q

c

q

f

a;

a1q

f2a; a1qcq2fa;a1q :Using (7.3) and (7.6) above, we deduce that

Ka; q 4 j2q4cq

fiqfa; a1q

21 i cq2f2qcqciq aciq

fa; a1qcq21 iq4; q4N

f2q

4j2

q4

c

q

fiqfa; a1q 2c

q2

c

q

c

iq

fa; a1qcq2q4; q4N 2a cq

2cqciqfa; a1qcq2q4; q4

N

4 j2q4cq

fiqfa; a1q 2cq2cqf2q2

fiqfa; a1qcq2q4; q4N

2a cq2cqf2q2

fiqfa; a1qcq2q4; q4N

; 7:19

where we used (2.17) in the form fqcq f2

q2

with q replaced by iq and iq;respectively. But by (2.16),

cq2f2q2cq2q4; q4

N

fq2cq2fq2

cq2q4; q4N

q2; q2

Nq2; q2

N

q4; q4N

q2; q4

Nq4; q4

Nq2; q4

Nq4; q4

N

q4; q4N

q4; q8

Nq4; q42

N

q4; q4N

q4; q42

N

q4; q4

2N

j2q4; 7:20

where we used Eulers identity (2.20), and (2.14). Using (7.20) in (7.19), (2.17) in the

form fqcq f2q2 with q replaced by iq and iq; respectively, (7.5), and

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(2.14), we deduce that

Ka; q

4

j2q4cqfiqfa; a1q

2j2q4cq

fiqfa; a1q 2a

j2q4cqfiqfa; a1q

2j2q4cq

fa; a1q1

fiq a1

fiq

2 j2q4cq

fa; a1qf2q2ciq aciq

2j2q4cq

f2q2 2q4; q42

Ncq

q4; q42N

q2; q22N

2 q4; q4

2Nc

q

q4; q42N

q2; q42N

q4; q42N

2c

q

q2; q22N

2q2; q42Ncq; 7:21

where in the last step we used (2.20). Thus, by (7.21),

1

2iKa; q 1

2aKa;q iq2; q42

Ncq aq2; q42

Ncq: 7:22

Let us evaluate now Ea; q: By (2.3), (2.6), and (7.15),

Ea; q 21 ia q4; q42Nfia; ia1qq; qN

q; q2N

f1; q4

21 ia q4; q42

Nfa1;aq

q; qN

q; q2N

f1; q4

1 iq4; q42

Nfa;a1q

q; qN

q; q2Ncq4: 7:23

Employ (2.11) and (2.20) to deduce that

q4; q42Nq; q

Nq; q2

Ncq4

q4; q42Nq2; q2

Nq; q2

Nq; q2

Nq4; q42

Nq4; q4

N

q4; q4

N

q2; q2N

q2; q4N

q4; q42N

q2; q2

Nq2; q22

N

q2; q2N

q4; q42N

q2; q22

N

q4; q42N

q2; q42

Nq4; q42

N

q4; q42N

q2; q42N

:

7:24

Using (7.24) in (7.23), we obtain

Ea; q 1 iq2; q42N

fa;a1q: 7:25

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Similarly, we obtain

E

a; q

1

i

q2; q4

2N

fa; a1q

:

7:26

Combining (7.25) and (7.26), we arrive at

1

a1 ifEa; q Ea; qg

1 ia1 iq

2; q42

N fa; a1q fa;a1q 2aq2; q42

Nciq;

7:27

by (7.9). Finally, replacing q by iq in (7.27), we deduce that

1

a1 ifEa; iq Ea; iqg 2aq2; q42

Ncq: 7:28

Adding (7.22) and (7.28) together, we find that (7.16) reduces to the right-hand side

of (7.11), i.e.,

12

iKa; q 12aKa; q 1

a1 ifEa; iq Ea; iqg

iq2; q42Ncq aq2; q42

Ncq 2aq2; q42

Ncq

aq2; q42N

cq acq:

This completes the verification of (7.11), since we have already verified (7.16). Hence,

the proof of Entry 3.4 is complete.

8. Proof of 1.8

Let us recall Eqs. (2.35) and (5.1), which is the equivalent form of Entry 3.2. Thus,

Vt rVr1t r1Vrt 3tq3; q33

N

qN

ft3; t3q3; 8:1

Vt 1 tqN

XNnN

1nq3nn1=2

1 tqn ; 8:2

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where r e2pi=3: Using (8.2) in (8.1), we obtain

3t

q3; q3

3N

qN

ft3;t3q3

1 tqN

XNnN

1nq3nn1=2

1 tqn

r rtqN

XNnN

1n q3nn1=2

1 r1tqn

r1 r1t

qN

XNnN

1nq3nn1=2

1 rtqn

tqN

XNnN

1nq3nn1=2 11 tqn r1 r1tqn r

1

1 rtqn 3tq

N

XNnN

1nq3nn1=2

1 t3q3n:

Then, we have

q3; q33N

f

t3;

t3q3

XN

n

N

1nq3nn1=2

1

t3q3n

: 8:3

Now, (1.8) follows if one replaces q3 by q and t3 by t; respectively, and employs (2.9)in (8.3).

Acknowledgments

I would like to thank my adviser Professor Bruce C. Berndt for his guidance and

assistance at all stages of this work.

References

[1] G.E. Andrews, On basic hypergeometric series, mock theta functions, and partitions. I, Quart.

J. Math. Oxford Ser. 17 (2) (1966) 6480.

[2] G.E. Andrews, Mock theta functions, Theta functionsBowdoin 1987, Part 2, Proceedings of

Symposia in Pure Mathematics Vol. 49, Brunswick, ME, 1987, pp. 283298.

[3] A.O.L. Atkin, P. Swinnerton-Dyer, Some properties of partitions, Proc. London. Math. Soc. 4 (4)

(1954) 84106.

[4] B.C. Berndt, Ramanujans Notebooks, Part III, Springer, New York, 1991.

[5] F.J. Dyson, Some guesses in the theory of partitions, Vol. 8, Eureka, Cambridge, 1944,

pp. 1015.

[6] R.J. Evans, Generalized Lambert series, in: B.C. Berndt, H.G. Diamond, A.J. Hildebrand (Eds.),

Analytic Number Theory Allerton Park, IL, 1995), Vol. 1, Birkha user, Boston, 1996, pp. 357370.

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[7] N.J. Fine, Basic Hypergeometric Series and Applications, Mathematical Surveys and Monographs,

Vol. 27, American Mathematical Society, Providence, RI, 1988.

[8] F.G. Garvan, New combinatorial interpretations of Ramanujans partition congruences mod 5,7 and

11, Trans. Amer. Math. Soc. 305 (1988) 4777.[9] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 1988.

[10] G.N. Watson, The final problem: an account of the mock theta functions, J. London Math. Soc.

11 (1936) 5580.

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