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 us- ing 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
Y
1≤i6=j≤k
1 − x i
x j
ai
is
(a 1 + a 2 + .... + a k )!
a 1 !a 2 !....a k ! .
Wilson [14], Gunson [8], Good [6] provided the proof. Andrews proposed the q-analogue of Dyson’s Conjecture [2] namely, the coefficient of x 0 1 x 0 2 ....x 0 k in the expanded form of
Y
1≤i6=j≤k
x i
x j
a
iqx j
x i
a
jis
(q) a
1
+a
2+....+a
k(q) a1(q) a2....(q) ak.
....(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 con- stant term identities arising from Lie algebras. The simplest result is the following;
let
θ(z, q) = (z; q) ∞ ( q z ; q) ∞
2010 Mathematics Subject Classification 33D99.
Key words and phrases: Mock theta functions, q- series.
then the coefficient of z 0 in θ(z, q) is 1
(q; q) ∞ = q
241η(τ ) , q = e 2πiτ ,
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. An- drews 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 ), j(x, q) = (x, q
x , q; q) ∞ =
∞
X
n=−∞
(−1) n q (n2) x n , x 6= 0,
where ( n 2 ) = n22 −n . For m ≥ 1,
J a,m = j(q a , q m ), J a,m = j(−q a , q m ), J m = j(q m , q 3m ) = (q m ; q m ) ∞ , θ(z, q) = (z; q) ∞ ( q
z ; q) ∞ , S n =
n
X
j=−n
(−1) j q −j2.
Definition of Ramanujan’s mock theta functions of order six Ramanujan defined the following mock theta functions of order six :
ϕ(q) :=
∞
X
n=0
(−1) n q n2(q; q 2 ) n
(−q) 2n
, ψ(q) :=
∞
X
n=0
(−1) n q (n+1)2(q; q 2 ) n
(−q) 2n+1
,
ρ(q) :=
∞
X
n=0
q
n(n+1)2(−q) n
(q; q 2 ) n+1 , σ(q) :=
∞
X
n=0
q
(n+1)(n+2)2(−q) n
(q; q 2 ) n+1 ,
λ(q) :=
∞
X
n=0
(−1) n q n (q; q 2 ) n
(−q) n
, µ(q) :=
∞
X
n=0
(−1) n (q; q 2 ) n
(−q) n
,
ν(q) :=
∞
X
n=0
q n2(q) n (q 3 ; q 3 ) n
.
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:
S 0 (q) :=
∞
X
n=0
q n2(−q; q 2 ) n
(−q 2 ; q 2 ) n , T 0 (q) :=
∞
X
n=0
q (n+1)(n+2) (−q 2 ; q 2 ) n
(−q; q 2 ) n+1 , U 0 (q) :=
∞
X
n=0
q n2(−q; q 2 ) n
(−q 4 ; q 4 ) n , V 0 (q) := −1 + 2
∞
X
n=0
q n2(−q; q 2 ) n
(q; q 2 ) n
= −1 + 2
∞
X
n=0
q 2n2(−q 2 ; q 4 ) n
(q; q 2 ) 2n+1
,
and
S 1 (q) :=
∞
X
n=0
q n(n+2) (−q; q 2 ) n (−q 2 ; q 2 ) n
, T 1 (q) :=
∞
X
n=0
q n(n+1) (−q 2 ; q 2 ) n (−q; q 2 ) n+1
,
U 1 (q) :=
∞
X
n=0
q (n+1)2(−q; q 2 ) n
(−q 2 ; q 4 ) n+1 , V 1 (q) :=
∞
X
n=0
q (n+1)2(−q; q 2 ) n
(q; q 2 ) n+1 ,
=
∞
X
n=0
q 2n2+2n+1 (−q 4 ; q 4 ) n
(q; q 2 ) 2n+2 .
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( 1 z , q λ ) (3.1)
is
∞
X
r=0
X
|Aj|≤r
(−) j (−1) r+Aj q B [
j2] +Cj+ [
r+12] λ− [
Aj+12] λ (3.2)
Proof. For 1 <| z |<| q λ | −1 Andrews [3, p.49,(2.4)]
∞
X
r=−∞
(−1) r q[
r+12] λ 1 − 1 z q rλ =
0
X
n=−∞
z n X
r≥|n|
(−1) r+n q[
r+12] λ−[n2]λ
+
∞
X
n=1
z n X
r≥|n|
(−1) r+n q[
r+12] λ−[n2]λ
=
∞
X
n=−∞
z n X
r≥|n|
(−1) r+n q[
r+12] λ−[n2]λ (3.3)
The left side of (3.3)
∞
X
r=−∞
(−1) r q[
r+12] λ 1 − z 1 q rλ
= 1 2
∞
X
r=−∞
(−1) r q[
r+12] λ 1 − 1 z q rλ − 1
2
∞
X
r=−∞
(−1) r q[
r+12] λ z 1 − zq rλ
= 1 2
∞
X
r=−∞
(−1) r q[
r+12] λ (1 − z) (1 − 1 z q rλ )(1 − zq rλ )
= 1
2(1 − 1 z ) lim
τ →0 3 ψ 3
qλτ
,
1z,z
;τ,
qλz,q
λz ; q λ , τ
(3.4)
= 1 2
(q λ , q λ ) 2 ∞
θ( 1 z , q λ ) . (3.5)
Hence (3.3) and (3.5) give
∞
X
n=−∞
z n X
r≥|n|
(−1) r+n q[
r+12] λ−[n2]λ = 1 2
(q λ , q λ ) 2 ∞
θ( 1 z , q λ ) . (3.6) Jacobi’s triple product identity is
∞
X
n=−∞
q j2z j = (q 2 ; q 2 ) ∞ θ(−qz, q 2 ), so
∞
X
j=−∞
q
Bj22(−z A q −B2+C ) j = (q B ; q B ) ∞ θ(z A q C , q B ). (3.7)
Multiplying (3.6) and (3.7), we have
∞
X
n=−∞
z n X
r≥|n|
(−1) r+n q[
r+12] λ−[n2]λ
∞
X
j=−∞
(−ε) j q B [
j2] +Cj z Aj
= 1 2
(q B ; q B ) ∞ (q λ ; q λ ) 2 ∞ θ(εz A q C , q B )
θ( 1 z , q λ ) .
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
r=0
X
|Aj|≤r
(−) j (−1) r+Aj q B [
j2] +Cj+ [
r+12] λ− [
Aj+12] λ . (3.8)
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 ψ 3 summmation instead of two 4 ψ 4 and 6 ψ 6 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 ∞
(−zq λ ; q λ ) ∞ (−z −1 ; q λ ) ∞ =
∞
X
N,r=−∞
r≥|N |
(−1) r z N q
λ2(r
2−N
2)+
λ2(r+N ) . (4.1)
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
j,r=−∞
r≥A|j|
(−1) r+j q (B−λA2)
j22+
λ2(r
2+r)+(B+λA−2C)
2j. (4.3)
Proof. By Jacobi’s triple product identity
∞
X
j=−∞
(−1) j z Aj q B [
j2] +Cj = (q B ; q B ) ∞ (z A q C ; q B ) ∞ (z −A q B−C ; q B ) ∞ . (4.4) 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
j=−∞
(−1) j z Aj q B [
j2] +Cj
∞
X
N,r=−∞
(−1) r z N q (r2−N
2)
λ
2
+
λ2(r+N ) . (4.5)
Taking N = −Aj, the coefficient of z 0 in the right side of (4.5) will be
X
j,r=−∞
r≥A|j|
(−1) r+j q (B−λA2)
j22+
λ2(r
2+r)+(B+λA−2C)
2j,
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
(i) J 1,4 S 0 (q) = P ∞
n=−∞ q 4n
2+n P n
j=−n (−1) j q −2j
2. (ii) J 1,4 S 1 (q) = P ∞
n=0 q 4n
2+3n (1 − q 2n+1 ) P n
j=−n (−1) j q −2j
2. (iii) J 1,4 [1 + V 0 (q)] = 2 P ∞
n=0 (−1) n q 4n
2+2n (1 + q 4n+2 ) P 2n
j=0 q − (j+12 ).
(iv) J 1,4 V 1 (q) = P ∞
n=0 (−1) n q 4n
2+4n+1 P 2n
j=0 q − (j+12 ).
Theorem 1. J 1,4 S 0 (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 ) θ(− q z3, q 8 )θ(−q 3 z, q 8 )θ(−q 3 , q 8 ) .
Theorem 2. J 1,4 S 1 (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(− z q , q 8 )θ(−qz, q 8 )θ(−q, q 8 ) .
Theorem 3. J 1,4 [1 + V 0 (q)] is 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 )
θ(− q z2, q 8 )θ(q 2 z, q 8 )θ(q 2 , q 8 ) .
Theorem 4. J 1,4 V 1 (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( 1 z , q 8 )θ(z, q 8 ) .
Proof of Theorem 1. Rewriting (i) to enable apply Lemma 1 J 1,4 S 0 (q) =
∞
X
n=−∞
q 4n2+n
n
X
j=−n
(−1) j q −2j2
=
∞
X
n=0
q 4n2+n
n
X
j=−n
(−1) j q −2j2+
∞
X
n=1
q 4n2−n
n
X
j=−n
(−1) j q −2j2
=
∞
X
n=0
q 4n2+n S n +
∞
X
n=1
q 4n2−n S n .
Using the relation S n = −S n−1
J 1,4 S 0 (q) =
∞
X
n=0
q 4n2+n S n −
∞
X
n=1
q 4n2−n S n−1
=
∞
X
n=0
q 4n2+n S n −
∞
X
n=0
q 4n2+7n+3 S n
=
∞
X
n=0
q 4n2+n (1 − q 6n+3 )
n
X
j=−n
(−1) j q −2j2.
Taking A = 1, ε = −1, a = −q 3 , λ = 8, B = 4, C = 6 in Lemma 1 of Andrews [3], we have
∞
X
n=0
X
|j|≤n
(−1) j q 4n2+n (1 − q 6n+3 )q −2j
2 = J 1,4 S 0 (q)
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 ) θ(− q z3, 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
n=0
X
|j|≤n
(−1) j q 4n2+3n (1 − q 2n+1 )q −2j
2 = J 1,4 S 1 (q)
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(− q z , q 8 )θ(−qz, q 8 )θ(−q, q 8 ) ,
which proves Theorem 2.
Proof of Theorem 3. Using the result
2n
X
j=0
x(
j+12) =
n
X
j=−n
x(
2j+12),
and rewriting (iii), we have J 1,4 [1 + V 0 (q)] = 2
∞
X
n=0
X
|j|≤n
q 4n2+2n (1 + q 4n+2 )q − (
2j+12 ). (5.1)
Taking ε = 1, A = 1, a = q 2 , B = 4, C = 5, λ = 8 in Lemma 1 of Andrews [3], we have
∞
X
n=0
X
|j|≤n
q 4n2+2n (1 + q 4n+2 )q − (
2j+12 ) = 1
2 J 1,4 [1 + V 0 (q)],
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 )
θ( q z2, 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 J 1,4 [V 1 (q)] =
∞
X
n=0 2n
X
j=0
(−1) n q 4n2+4n+1 q − (
j+12 )
=
∞
X
n=0
X
|j|≤n
(−1) n q 4n2+4n+1 q − (
2j+12 ).
Now taking ε = 1, A = 1, B = 4, C = 5, λ = 8 in Lemma 1, we have
∞
X
n=0
X
|j|≤n
(−1) n q 4n2+4n q − (
2j+12 ) = 1
q J 1,4 [V 1 (q)], 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( z 1 , 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
J 1,4 ϕ(q) =
∞
X
n=−∞
X
|j|≤n
(−1) n+j q 3n2+n−j
2 (6.1)
J 1,4 ψ(q) =
∞
X
n=0
X
|j|≤n
(−1) n+j q 3n2+3n+1−j
2 (6.2)
J 1,4 λ(q) =
∞
X
n=0
X
|j|≤n
(−1) n+j q
3n2 +3n−2j22
