NEW ZEALAND JOURNAL OF MATHEMATICS
Volume 42 (2012), 71-82
AN APPLICATION OF THE CONSTANT TERM METHOD TO
RAMANUJAN’S MOCK THETA FUNCTIONS
Bhaskar Srivastava
(Received 7 May, 2004)
Abstract. In this paper we give simple proof of some well known identities using the constant term method. We also show that Ramanujan’s sixth order and
Gordon and McIntosh’s eighth order mock theta functions can be expressed as
constant terms in the Laurent series expansion of rational functions of theta
functions.
1. Introduction
In recent years q-series has found application in Physics, Lie Algebras, Statistics,
apart from the earlier classical applications in classical analysis, combinatorics,
number theory. The most recent break through is the application of constant term
method. Dyson [5] conjectured that the constant term in the expanded form of
a
Y xi i
1−
xj
1≤i6=j≤k
is
(a1 + a2 + .... + ak )!
.
a1 !a2 !....ak !
Wilson [14], Gunson [8], Good [6] provided the proof. Andrews proposed the
q-analogue of Dyson’s Conjecture [2] namely, the coefficient of x01 x02 ....x0k in the
expanded form of
Y xi qxj 1≤i6=j≤k
xj
ai
xi
aj
is
(q)a1 +a2 +....+ak
.
(q)a1 (q)a2 ....(q)ak
In 1985, Zeilberger and Bressoud [15] gave a combinatorial proof of Andrews’
conjecture. Macdonald [12] said such results are a few of the vast family of the constant term identities arising from Lie algebras. The simplest result is the following;
let
q
θ(z, q) = (z; q)∞ ( ; q)∞
z
2010 Mathematics Subject Classification 33D99.
Key words and phrases: Mock theta functions, q- series.
72
BHASKAR SRIVASTAVA
then the coefficient of z 0 in θ(z, q) is
1
1
q 24
=
, q = e2πiτ ,
(q; q)∞
η(τ )
where η(τ ) is Dedekind’s famous modular form [3]. In this paper we specialize
Andrews’ Lemma [3] and give simple proof of some identities. We also show that
Ramanujan’s sixth order [4] and eighth order mock theta functions, recently defined
by Gordon and McIntosh [7], can be expressed as constant term in the Laurent series
expansion of rational expressions of different θ(z, q).
Ramanujan never gave a rigorous definition of the order of a mock theta function
nor explained what he meant by third order, fifth order and seventh order. Andrews and Hickerson wrote that the known identities for the mock theta functions
make it clear that they are related to the number three, five and seven, but we
still have no formal definition of ”order”. However, Andrews and Hickerson have
called the mock theta functions considered in [4] of sixth order by considering the
combinatorial interpretation of the coefficients of the mock theta functions ϕ(q)
and ψ(q). Gordon and McIntosh [7] have given rigorous definition for the order of
a mock theta functions based on their behaviour under the action of the modular
group.
2. Notation
The following q-notations are used: For | q k |< 1,
(a; q k )n = (1 − a)(1 − aq k )......(1 − aq k(n−1) ), n ≥ 1
(a; q k )0 = 1,
(a)n = (a; q)n = (1 − a)(1 − aq)......(1 − aq n−1 ),
∞
X
n
q
(−1)n q (2 ) xn , x 6= 0,
j(x, q) = (x, , q; q)∞ =
x
n=−∞
where (n2 ) =
n2 −n
2 .
For m ≥ 1,
Ja,m = j(q a , q m ),
J a,m = j(−q a , q m ),
q
m 3m
m m
Jm = j(q , q ) = (q ; q )∞ , θ(z, q) = (z; q)∞ ( ; q)∞ ,
z
n
X
2
Sn =
(−1)j q −j .
j=−n
Definition of Ramanujan’s mock theta functions of order six
Ramanujan defined the following mock theta functions of order six :
2
2
∞
∞
X
X
(−1)n q (n+1) (q; q 2 )n
(−1)n q n (q; q 2 )n
, ψ(q) :=
,
ϕ(q) :=
(−q)2n
(−q)2n+1
n=0
n=0
n(n+1)
(n+1)(n+2)
∞
∞
X
X
2
q 2 (−q)n
q
(−q)n
ρ(q) :=
,
σ(q)
:=
,
2
2
(q; q )n+1
(q; q )n+1
n=0
n=0
AN APPLICATION OF THE CONSTANT TERM METHOD
∞
X
(−1)n q n (q; q 2 )n
λ(q) :=
,
(−q)n
n=0
2
∞
X
q n (q)n
.
ν(q) :=
(q 3 ; q 3 )n
n=0
73
∞
X
(−1)n (q; q 2 )n
µ(q) :=
,
(−q)n
n=0
The series for µ(q) does not converge, but the sequence of even partial sum
converges, as does the sequence of odd partial sums. Hence we define µ(q) as the
average of these two values, Andrews [4, p. 62 ].
Definition of the mock theta functions of order eight
Gordon and McIntosh defined the following mock theta functions of order eight:
2
∞
X
q n (−q; q 2 )n
,
(−q 2 ; q 2 )n
n=0
2
∞
X
q n (−q; q 2 )n
,
U0 (q) :=
(−q 4 ; q 4 )n
n=0
S0 (q) :=
∞
X
q (n+1)(n+2) (−q 2 ; q 2 )n
,
(−q; q 2 )n+1
n=0
2
∞
X
q n (−q; q 2 )n
V0 (q) := −1 + 2
(q; q 2 )n
n=0
∞
2n
X q 2 (−q 2 ; q 4 )n
= −1 + 2
,
(q; q 2 )2n+1
n=0
T0 (q) :=
and
∞
X
q n(n+2) (−q; q 2 )n
,
(−q 2 ; q 2 )n
n=0
2
∞
X
q (n+1) (−q; q 2 )n
U1 (q) :=
,
(−q 2 ; q 4 )n+1
n=0
S1 (q) :=
∞
X
q n(n+1) (−q 2 ; q 2 )n
,
(−q; q 2 )n+1
n=0
2
∞
X
q (n+1) (−q; q 2 )n
V1 (q) :=
,
(q; q 2 )n+1
n=0
2
∞
X
q 2n +2n+1 (−q 4 ; q 4 )n
=
.
(q; q 2 )2n+2
n=0
T1 (q) :=
3. Specialized Lemma of Andrews
Lemma 1. In the annalus 1 <| z |<| q λ |−1 the coefficient of z 0 in the Laurent
series expansion of
(q B ; q B )∞ (q λ ; q λ )2∞ θ(εz A q C , q B )
θ( z1 , q λ )
(3.1)
is
∞ X
X
j
Aj+1
r+1
(−)j (−1)r+Aj q B [2 ]+Cj+[2 ]λ−[2 ]λ
r=0 |Aj|≤r
Proof. For 1 <| z |<| q λ |−1 Andrews [3, p.49,(2.4)]
(3.2)
74
BHASKAR SRIVASTAVA
r+1
∞
0
X
X
X
r+1
n
(−1)r q [2 ]λ
n
=
z
(−1)r+n q [2 ]λ−[2 ]λ
1 rλ
1 − zq
r=−∞
n=−∞
r≥|n|
+
∞
X
zn
n=1
∞
X
=
X
n=−∞
r+1
n
(−1)r+n q [2 ]λ−[2 ]λ
r≥|n|
X
zn
r+1
n
(−1)r+n q [2 ]λ−[2 ]λ
(3.3)
r≥|n|
The left side of (3.3)
r+1
∞
X
(−1)r q [2 ]λ
1 − z1 q rλ
r=−∞
r+1
r+1
∞
∞
1 X (−1)r q [2 ]λ
1 X (−1)r q [2 ]λ z
=
−
2 r=−∞ 1 − z1 q rλ
2 r=−∞ 1 − zq rλ
r+1
∞
1 X (−1)r q [2 ]λ (1 − z)
=
2 r=−∞ (1 − z1 q rλ )(1 − zq rλ )
λ
q
1
1
λ
τ , z ,z;
lim 3 ψ3
=
;q ,τ
λ
τ, qz ,q λ z
2(1 − z1 ) τ →0
(3.4)
1 (q λ , q λ )2∞
.
2 θ( z1 , q λ )
=
(3.5)
Hence (3.3) and (3.5) give
∞
X
X
zn
n=−∞
r+1
n
1 (q λ , q λ )2∞
(−1)r+n q [2 ]λ−[2 ]λ =
.
2 θ( z1 , q λ )
(3.6)
r≥|n|
Jacobi’s triple product identity is
∞
X
2
q j z j = (q 2 ; q 2 )∞ θ(−qz, q 2 ),
n=−∞
so
∞
X
q
Bj 2
2
B
(−z A q − 2 +C )j = (q B ; q B )∞ θ(z A q C , q B ).
j=−∞
Multiplying (3.6) and (3.7), we have

∞
X

n=−∞

zn
X
r+1
2
(−1)r+n q [
]
λ−[n
2 ]λ

∞
X
j
(−ε)j q B [2 ]+Cj z Aj
j=−∞
r≥|n|
B
=
B
1 (q ; q )∞ (q λ ; q λ )2∞ θ(εz A q C , q B )
.
2
θ( z1 , q λ )
(3.7)
75
AN APPLICATION OF THE CONSTANT TERM METHOD
Hence the coefficient of z 0 in the Laurent series expansion of the right side will be
equal to the coefficient of z 0 in the left side which we will get when n = −Aj i.e.
∞ X
X
Aj+1
j
r+1
(−)j (−1)r+Aj q B [2 ]+Cj+[2 ]λ−[2 ]λ .
(3.8)
r=0 |Aj|≤r
Hence we have proven Lemma 1.
We can have the lemma by taking the limit as a → 1 in the Lemma 1 of Andrews
[3, p 49]. In our specialization we have used only one 3 ψ3 summmation instead of
two 4 ψ4 and 6 ψ6 summations used by Andrews in the proof of his lemma. Basically
our proof is on the same line.
4. Another Lemma
We prove one more lemma. Rewriting the lemmas 1 and 2 of Andrews [1,p. 452]
enables us to express the mock theta functions of order six and eight as coefficient
of z 0 in a Laurent series expansion of rational function of theta functions.
Writing q λ for q and −z for z in Lemma 1 of Andrews [1, p.452, (1.4)], we have
For 1 <| z |<| q λ |−1 , | q |< 1,
(q λ ; q λ )2∞
=
λ
(−zq ; q λ )∞ (−z −1 ; q λ )∞
∞
X
λ
(−1)r z N q 2 (r
2
−N 2 )+ λ
2 (r+N )
.
(4.1)
N,r=−∞
r≥|N |
Lemma 2. In the annalus 1 <| z |<| q λ |−1 the coefficient of z 0 in the Laurent
series expansion of
(q B ; q B )∞ (q λ ; q λ )2∞ (z A q C ; q B )∞ (z −A q B−C ; q B )∞
(−zq λ ; q λ )∞ (−z −1 ; q λ )∞
(4.2)
is
∞
X
2 j2
) 2
(−1)r+j q (B−λA
j
2
+λ
2 (r +r)+(B+λA−2C) 2
.
(4.3)
j,r=−∞
r≥A|j|
Proof. By Jacobi’s triple product identity
∞
X
j
(−1)j z Aj q B [2 ]+Cj = (q B ; q B )∞ (z A q C ; q B )∞ (z −A q B−C ; q B )∞ .
(4.4)
j=−∞
Multiplying (4.1) and (4.4) we have
(q B ; q B )∞ (q λ ; q λ )2∞ (z A q C ; q B )∞ (z −A q B−C ; q B )∞
(−zq λ ; q λ )∞ (−z −1 , q λ )∞
∞
∞
X
X
j
2
2 λ
λ
=
(−1)j z Aj q B [2 ]+Cj
(−1)r z N q (r −N ) 2 + 2 (r+N ) .
j=−∞
N,r=−∞
Taking N = −Aj, the coefficient of z 0 in the right side of (4.5) will be
(4.5)
76
BHASKAR SRIVASTAVA
2 j2
) 2
X
(−1)r+j q (B−λA
j
2
+λ
2 (r +r)+(B+λA−2C) 2
,
j,r=−∞
r≥A|j|
which is the coefficient of z 0 in the Laurent series expansion of the left side of
(4.5). Hence the lemma.
5. Mock Theta Function of Order Eight as Coefficients Of z 0
We have given Hecke type series expansion for Mock theta functions of order
eight, Srivastava [13]. They are
P∞
Pn
2
2
(i) J1,4 S0 (q) = n=−∞ q 4n +n j=−n (−1)j q −2j .
P∞ 4n2 +3n
P
2
n
(ii) J1,4 S1 (q) = n=0 q
(1 − q 2n+1 ) j=−n (−1)j q −2j .
j+1
P∞
P2n
2
(iii) J1,4 [1 + V0 (q)] = 2 n=0 (−1)n q 4n +2n (1 + q 4n+2 ) j=0 q −(2 ) .
P∞
P2n −(j+1 )
2
(iv) J V (q) =
(−1)n q 4n +4n+1
q 2 .
1,4 1
n=0
j=0
Theorem 1. J1,4 S0 (q) is the coefficient of z 0 in the Laurent series expansion of
−q 3 (q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(−zq 6 , q 4 )θ(z, q 8 )θ(q 6 , q 4 )
.
3
θ(− qz , q 8 )θ(−q 3 z, q 8 )θ(−q 3 , q 8 )
Theorem 2. J1,4 S1 (q) is the coefficient of z 0 in the Laurent series expansion of
−q(q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(−zq 6 , q 4 )θ(z, q 8 )θ(q 2 , q 8 )
.
θ(− zq , q 8 )θ(−qz, q 8 )θ(−q, q 8 )
Theorem 3. J1,4 [1 + V0 (q)] is the coefficient of z 0 in the Laurent series expansion
q 2 (q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(zq 5 , q 4 )θ(z, q 8 )θ(q 4 , q 8 )
of
.
2
θ(− qz , q 8 )θ(q 2 z, q 8 )θ(q 2 , q 8 )
Theorem 4. J1,4 V1 (q) is the coefficient of z 0 in the Laurent series expansion of
(q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(zq 5 , q 4 )θ(z, q 8 )
.
θ( z1 , q 8 )θ(z, q 8 )
Proof of Theorem 1. Rewriting (i) to enable apply Lemma 1
J1,4 S0 (q) =
∞
X
q 4n
2
+n
n=−∞
=
=
∞
X
n=0
∞
X
n=0
q 4n
n
X
(−1)j q −2j
j=−n
2
+n
n
X
2
(−1)j q −2j +
2
+n
Sn +
∞
X
n=1
j=−n
q 4n
2
∞
X
n=1
q 4n
2
−n
Sn .
q 4n
2
−n
n
X
j=−n
(−1)j q −2j
2
77
AN APPLICATION OF THE CONSTANT TERM METHOD
Using the relation Sn = −Sn−1
J1,4 S0 (q) =
=
=
∞
X
n=0
∞
X
n=0
∞
X
q 4n
q 4n
2
2
+n
+n
Sn −
Sn −
∞
X
n=1
∞
X
q 4n
2
−n
q 4n
2
+7n+3
Sn−1
Sn
n=0
q 4n
2
+n
(1 − q 6n+3 )
n=0
n
X
2
(−1)j q −2j .
j=−n
3
Taking A = 1, ε = −1, a = −q , λ = 8, B = 4, C = 6 in Lemma 1 of Andrews [3],
we have
∞ X
X
2
2
(−1)j q 4n +n (1 − q 6n+3 )q −2j = J1,4 S0 (q)
n=0 |j|≤n
which by (2.3) of [3] will be the coefficient of z 0 in the Laurent series expansion of
−q 3 (q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(−zq 6 , q 4 )θ(z, q 8 )θ(q 6 , q 8 )
3
θ(− qz , q 8 )θ(−q 3 z, q 8 )θ(−q 3 , q 8 )
,
which proves Theorem 1.
Proof of Theorem 2. Taking ε = −1, A = 1, a = −q, λ = 8, B = 4, C = 6 in
Lemma 1 of Andrews [3], we have
∞ X
X
(−1)j q 4n
2
+3n
2
(1 − q 2n+1 )q −2j = J1,4 S1 (q)
n=0 |j|≤n
which by (2.3) of [3] will be the coefficient of z 0 in the Laurent series expansion of
−q(q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(−zq 6 , q 4 )θ(z, q 8 )θ(q 2 , q 8 )
,
θ(− zq , q 8 )θ(−qz, q 8 )θ(−q, q 8 )
which proves Theorem 2.
Proof of Theorem 3. Using the result
2n
X
j+1
x( 2 ) =
j=0
n
X
2j+1
x(2
),
j=−n
and rewriting (iii), we have
J1,4 [1 + V0 (q)] = 2
∞ X
X
q 4n
2
+2n
2j+1
(1 + q 4n+2 )q −(2
).
(5.1)
n=0 |j|≤n
Taking ε = 1, A = 1, a = q 2 , B = 4, C = 5, λ = 8 in Lemma 1 of Andrews [3], we
have
∞ X
X
2j+1
2
1
q 4n +2n (1 + q 4n+2 )q −(2 ) = J1,4 [1 + V0 (q)],
2
n=0
|j|≤n
78
BHASKAR SRIVASTAVA
which by (2.3) of [3] will be the coefficient of z 0 in the Laurent series expansion of
q 2 (q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(zq 5 , q 4 )θ(z, q 8 )θ(q 4 , q 8 )
2
θ( qz , q 8 )θ(q 2 z, q 8 )θ(q 2 , q 8 )
,
which proves Theorem 3.
Proof of Theorem 4. Again rewriting (iv), using (5.1), we have
J1,4 [V1 (q)] =
=
∞ X
2n
X
j+1
2
(−1)n q 4n +4n+1 q −(2 )
n=0 j=0
∞ X
X
(−1)n q 4n
2
2j+1
+4n+1 −(2
)
q
.
n=0 |j|≤n
Now taking ε = 1, A = 1, B = 4, C = 5, λ = 8 in Lemma 1, we have
∞ X
X
2j+1
2
1
(−1)n q 4n +4n q −(2 ) = J1,4 [V1 (q)],
q
n=0
|j|≤n
which will be the coefficient of z 0 in the Laurent series expansion of
(q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ θ(zq 5 , q 4 )θ(z, q 8 )
,
θ( z1 , q 8 )θ(z, q 8 )
which proves Theorem 4.
6. Mock Theta Function of Order Six as Coefficient of z 0
Andrews and Hickerson [4] have given Hecke type expansion for the sixth order
mock theta functions. They are
∞
X
X
2
2
J1,4 ϕ(q) =
(−1)n+j q 3n +n−j
(6.1)
n=−∞ |j|≤n
J1,4 ψ(q) =
∞ X
X
2
(−1)n+j q 3n
+3n+1−j 2
(6.2)
n=0 |j|≤n
J1,4 λ(q) =
∞ X
X
(−1)n+j q
3n2 +3n−2j 2
2
(6.3)
n=0 |j|≤n
J1,4 µ(q) =
∞
X
X
(−1)n+j q
3n2 +n−2j 2
2
.
(6.4)
n=−∞ |j|≤n
Theorem 5. J1,4 ϕ(q) is the coefficient of z 0 in the Laurent series expansion of
q 2 (q 4 ; q 4 )∞ (q 6 ; q 6 )2∞ θ(−zq 5 , q 4 )θ(z, q 6 )θ(q 4 , q 6 )
.
2
θ( qz , q 6 )θ(q 2 z, q 6 )θ(q 2 , q 6 )
Theorem 6. 1q J1,4 ψ(q) is the coefficient of z 0 in the Laurent series expansion of
(q 4 ; q 4 )∞ (q 6 ; q 6 )2∞ θ(zq 5 ; q 4 )
.
θ(−zq 6 , q 6 )
AN APPLICATION OF THE CONSTANT TERM METHOD
79
Theorem 7. J1,4 λ(q) is the coefficient of z 0 in the Laurent series expansion of
(q; q)∞ (q 3 ; q 3 )2∞ θ(zq 2 ; q)
.
θ(−zq 3 , q 3 )
Theorem 8. J1,4 µ(q) is the coefficient of z 0 in the Laurent series expansion of
q(q; q)∞ (q 3 ; q 3 )2∞ θ(−zq 2 , q)θ(z, q 3 )θ(q 2 , q 3 )
.
θ( zq , q 3 )θ(qz, q 3 )θ(q, q 3 )
Proof of Theorem 5. We shall rewrite (6.1) to enable us to use Lemma 1 of
Andrews [3]. Now
J1,4 ϕ(q) =
∞
X
(−1)n q 3n
2
+n
Sn +
n=0
∞
X
(−1)n q 3n
2
−n
S−n .
n=1
Using the relation Sn = −Sn−1 , we have
J1,4 ϕ(q) =
=
=
∞
X
(−1)n q 3n
n=0
∞
X
(−1)n q 3n
n=0
∞
X
2
2
+n
+n
Sn −
Sn +
∞
X
n=1
∞
X
2
−n
2
+5n+2
(−1)n q 3n
(−1)n q 3n
Sn−1
Sn
n=0
(−1)n q 3n
2
+n
(1 + q 4n+2 )
n=0
X
(−1)j q −j
2
|j|≤n
Taking ε = −1, A = 1, a = q 2 , λ = 6, B = 4, C = 5 in Lemma 1 of Andrews [3], we
have
∞ X
X
2
2
(−1)n+j q 3n +n−j (1 + q 4n+2 ) = J1,4 ϕ(q),
n=0 |j|≤n
which is the coefficient of z 0 in the Laurent series expansion of
q 2 (q 4 ; q 4 )∞ (q 6 ; q 6 )2∞ θ(−zq 5 , q 4 )θ(z, q 6 )θ(q 4 , q 6 )
2
θ( qz , q 6 )θ(q 2 z, q 6 )θ(q 2 , q 6 )
,
and this proves Theorem 5.
Proof of Theorem 6.
Taking A = 1, B = 4, C = 5, λ = 6 in Lemma 2, we have
∞ X
X
n=0 |j|≤n
(−1)n+j q 3n
2
+n−j 2
=
1
J1,4 ψ(q),
q
which by (4.2) is the coefficient of z 0 in the Laurent series expansion of
(q 4 ; q 4 )∞ (q 6 ; q 6 )2∞ (zq 5 ; q 4 )∞ (z −1 q −1 ; q 4 )∞
(q 4 ; q 4 )∞ (q 6 ; q 6 )2∞ θ(zq 5 ; q 4 )
=
,
6
6
−1
6
(−zq , q )∞ (−z ; q )∞
θ(−zq 6 , q 6 )
and this proves Theorem 6.
80
BHASKAR SRIVASTAVA
Proof of Theorem 7. Taking A = 1, B = 1, C = 2, λ = 3 in Lemma 2, we have
∞ X
X
3n2 +3n
2
(−1)n+j q 2 −j = J1,4 λ(q),
n=0 |j|≤n
which by (4.2) is the coefficient of z 0 in the Laurent series expansion of
(q; q)∞ (q 3 ; q 3 )2∞ (zq 2 ; q)∞ (z −1 q −1 ; q)∞
(q; q)∞ (q 3 ; q 3 )2∞ θ(zq 2 ; q)
=
,
(−zq 3 ; q 3 )∞ (−z −1 ; q 3 )∞
θ(−zq 3 ; q 3 )
and this proves Theorem 7.
Proof of Theorem 8. Rewrite (6.4) and proceed as in (6.1), to get
2J1,4 µ(q) =
∞ X
X
(−1)n+j q
3n2 +n
−j 2
2
(1 + q 2n+1 ).
n=0 |j|≤n
Taking ε = −1, A = 1, a = q, B = 1, C = 2, λ = 3 in Lemma 1 of Andrews [3] , we
have
∞ X
X
3n2 +3n
2
(−1)n+j q 2 −j (1 + q 2n+1 ) = 2J1,4 µ(q),
n=0 |j|≤n
which is the coefficient of z 0 in the Laurent series expansion of
q(q; q)∞ (q 3 ; q 3 )2∞ θ(−zq 2 , q)θ(z, q 3 )θ(q 2 , q 3 )
,
θ( zq , q 3 )θ(qz, q 3 )θ(q, q 3 )
and this proves Theorem 8.
7. Applications
Hecke [1] studied double theta type series. Subsequently Kac and Patterson [11],
Zeilberger and Bressoud [15] and Andrews [2] studied such identities.
(i) We give a very simple proof of the following identity.
(q; q)2∞ =
∞
X
(−1)n+m q
(n2 −3m2 )
(n+m)
+ 2
2
(7.1)
m,n=−∞
Kac and Peterson [11] proved the identity by considering certain modular
functions which arise in affine Lie Algebra. Andrews proved by using his
lemma 1 [3].
Proof of (7.1). Taking A = 2, = 1, λ = 1, B = 1, C = 2 in Lemma 1, the
right side is the coefficient of z 0 in
(q; q)3∞ θ(z 2 q 2 , q)
j(z 2 q 2 , q)
= (q; q)3∞
1
j(zq, q)
θ( z , q)
(q, q)3∞
j(−zq, q)j(z 2 q 3 , q 2 ), by Hickerson [10, (1.14)]
J1 J2
∞
∞
2
(q, q)3∞ X n2 +n n X
=
q 2 z
(−1)m q m +2m z 2m ,
J1 J2 n=−∞
m=−∞
=
AN APPLICATION OF THE CONSTANT TERM METHOD
81
We get the coefficient of z 0 when n = −2m and is
∞
2
(q, q)3∞ X
(−1)m q 3m +m = (q; q)2∞ ,
J1 J2 m=−∞
using Jacobi’s triple product identity. This proves (7.1).
(ii) We now prove the following identity
∞
X
r+1
2
(−1)r+j q [2 ]−j = (q; q)∞ (q 2 ; q 2 )∞ .
(7.2)
j,r=−∞
Proof of (7.2). Taking = 1, A = 2, B = 2, C = 2, λ = 1 in Lemma 1 the
right side of (7.2) is the coefficient of z 0 in
(q 2 ; q 2 )∞ (q; q)2∞ θ(z 2 q 2 , q 2 )∞ (z −2 , q 2 )∞
(q 2 ; q 2 )∞ (q; q)2∞ θ(z 2 q 2 , q 2 )
=
1
θ( z , q)
(qz, q)∞ ( z1 , q)∞
1
= (q 2 ; q 2 )∞ (q; q)2∞ (−qz; q)∞ (− ; q)∞
z
= (q 2 ; q 2 )∞ (q; q)∞ j(−qz; q)
2
2
= (q ; q )∞ (q; q)∞
∞
X
q
n2 +n
2
zn,
n=−∞
We get the coefficient of z 0 , when n = 0 and is
(q 2 ; q 2 )∞ (q; q)∞ .
This proves (7.2).
(iii) We give a very simple proof of the famous formula of Jacobi [9, Th. 357, p.283]
∞
X
(−1)r (2r + 1)q
r(r+1)
2
= (q; q)3∞
(7.3)
r=0
Proof of (7.3). Take = 1, λ = 1, A = 1, B = 1, C = 1 in Lemma 1. The
right side is the coefficient of z 0 in
(q; q)3∞ θ(zq, q)
(q; q)3∞ θ(zq, q)
=
1
θ(zq, q)
θ( z , q)
= (q; q)3∞ .
This proves (7.3).
(iv) We now prove the following identity
(q; q)2∞ =
∞ X
X
(−1)j q
−j(3j+1)
2
q 2n
2
+n
(1 − q 2n+1 ).
(7.4)
n=0 |j|≤n
Proof of (7.4). Taking ε = −1, λ = 4, A = 1, B = 1, C = 2, a = −q in
Lemma 1 of Andrews [3]. The left side of (7.4) is the coefficient of z 0 in the
82
BHASKAR SRIVASTAVA
Laurent Series expansion of
−q(q; q)∞ (q 4 ; q 4 )2∞ θ(−zq 2 , q)θ(z, q 4 )θ(q 2 , q 4 )
θ(− zq , q 4 )θ(−qz, q 4 )θ(−q, q 4 )
4
=
−q(q; q)∞ (q 4 ; q 4 )2∞ (−zq 2 ; q)∞ (− zq ; q)∞ (z; q 4 )∞ ( qz ; q 4 )∞ (q 2 ; q 4 )2∞
3
(− zq ; q 4 )∞ (−q 3 z; q 4 )∞ (−qz; q 4 )∞ (− qz ; q 4 )∞ (−q; q 4 )∞ (−q 3 ; q 4 )∞
2
4
−q(q; q)∞ (q 4 ; q 4 )2∞ (−z; q 4 )∞ (−zq 2 ; q 4 )∞ (− z1 ; q 4 )∞ (− qz ; q 4 )∞ (z; q 4 )∞ ( qz ; q 4 )∞ (q 2 ; q 4 )2∞
=
zq(1 + z)(−q; q 4 )∞ (−q 3 ; q 4 )∞
4 4
2 4 2
(q; q)∞ (q ; q )∞ (q ; q )∞ j(z 2 , q 8 )j(−zq 2 , q 4 )z −2
=−
(−q; q 4 )∞ (−q 3 ; q 4 )∞ (q 8 ; q 8 )∞
∞
∞
X
X
2
(q; q)∞ (q 4 ; q 4 )∞ (q 2 ; q 4 )2∞
n 4(n2 −n) 2n−2
=−
(−1)
q 2m z m .
q
z
4
3
4
8
8
(−q; q )∞ (−q ; q )∞ (q ; q )∞ n=−∞
m=−∞
We get the coefficient of z 0 when m = 2n − 2 and is
=
(q; q)∞ (q 4 ; q 4 )∞ (q 2 ; q 4 )2∞
(−q; q 4 )∞ (−q 3 ; q 4 )∞
= (q; q)2∞ .
This proves (7.4).
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AN APPLICATION OF THE CONSTANT TERM METHOD
83
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Bhaskar Srivastava
Department of Mathematics and Astronomy
Lucknow University,
Lucknow, India.
[email protected]