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http://www.elsevier.com/locate/aim
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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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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5. Proof of Entry 3.2
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.
6. Proof of Entry 3.3
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.
7. Proof of Entry 3.4
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.),
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[10] G.N. Watson, The final problem: an account of the mock theta functions, J. London Math. Soc.
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