# A Mock Theta Function of Second Order

Authors: Bhaskar Srivastava
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We consider the second-order mock theta function𝒟5 (q), which Hikami came across in his work on mathematical physics and quantum invariant of three manifold. We give their bilateral form, and show that it is the same as bilateral third-order mock theta function of Ramanujan. We also show that the mock theta function 𝒟5 (q) outside the unit circle is a theta function and also write h1(q) as a coefficient of z0 of a theta series. First writing h1(q) as a coefficient of a theta function, we prove an identity for h1(q).


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## Body

## 1. Brief History of Mock Theta Functions

The mock theta functions were introduced and named by Ramanujan and were the subjects of Ramanujan’s last letter to Hardy, dated January 12, 1920, to be specific [1, 2]. Ramanujan gave a list of seventeen functions which he called “mock theta functions.”  He divided them into four groups of functions of order 3, 5, 5, and 7. Ramanujan did not rigorously define a mock theta function nor he define the order of a mock theta function.  A definition of the order of a mock theta function is given in the Gordon-McIntosh paper on modular transformation of Ramanujan’s fifth and seventh-order mock theta functions [3] Watson [4] while constructing transformation laws for the mock theta function found three further mock theta functions  of order 3.In 1976, Andrews while visiting Trinity college, Cambridge, discovered in the mathematical library of the college a notebook written by Ramanujan towards the end of his life and Andrews called it “Lost” Notebook. In the lost notebook were six more mock theta functions and linear relation between them. Andrews and Hickerson [5] called these mock theta functions of sixth-order and proved the identities.In the “Lost” Notebook on page 9 appear four more mock theta functions which were called by Choi of tenth-order. Ramanujan also gave eight linear relations connecting these mock theta functions of tenth-order and these relations were proved by Choi [6].Gordon and McIntosh listed eight functions in their eighth-order paper [7], but later, in their survey paper [8], classified only four of them as eighth-order. The other four are more simple in their modular transformation laws and therefore are considered to be of lower order.We now come to the second-order mock theta functions. McIntosh [9] considered three second-order mock theta functions and gave transformation formulas for them. Hikami [10] in his work on mathematical physics and quantum invariant of three manifold came across the q-series:
(1.1)𝒟5(q)=∑n=0∞qn(-q;q)n(q;q2)n+1(1.2)=1(q;q2)∞2∑n=0∞(q;q2)n2q2n
and proved that 𝒟5(q) is a mock theta function and called it of “2nd” order.He further showed that𝒟5(q) is a sum of two mock theta functions h1(q) and ω(q) where h1(q) is of second-order and ω(q) is Ramanujan’s mock theta function of third-order.  This 𝒟5(q) will be the basis of our study in this paper.Before we  begin with the study of𝒟5(q) and h1(q) it will be appropriate to mention the work done earlier.Gordon and McIntosh in their survey paper [8] have shown that h1(q) is essentially the odd part of the second-order mock theta function B(q),which appears as β(q) in Andrews’ paper on Mordell integrals and Ramanujan’s lost notebook [11] and also in McIntosh paper on second-order mock theta functions [9]. In particular,(1.3)h1(q2)=B(q)-B(-q)4q,
where(1.4)B(q)=∑n=0∞qn(n+1)(-q2;q2)n(q;q2)n+12=∑n=0∞qn(-q;q2)n(q;q2)n+1.
Since the even part of B(q) is the ordinary theta function(1.5)B(q)+B(-q)2=(q4;q4)∞(-q2;q2)∞4,
it follows that the odd part and h1(q) are second-order mock theta functions. Thus 𝒟5(q) is a linear combination of second-order and third-order mock theta function. In some sense, mock theta functions of orders 1, 2, 3, 4, and 6 are all in the same family.The paper is divided as follows.In Section3 we expand 𝒟5(q) as a bilateral q-series and show that it is also a sum of the second-order mock theta function 𝒟5(q) and the third-order mock theta function ω(q). By using Bailey’s transformation we have the interesting result that the bilateral 𝒟5,c(q) is the same as the bilateral ωc(q).In Section4, using bilateral transformation of Slater, we write 𝒟5,c(q) as a bilateral series ψ22 series with a free parameter c.In Section5, a mild generalization 𝒟5,c(z,α) of 𝒟5,c(q) is given and we show that this generalized function is a Fq-function.In Section6 we show that 𝒟5(q), outside the unit circle |q|=1, is a theta function.In Section7 we state a generalized Lambert Series expansion for h1(q) as given in [8].In Section8 we show that h1(q) is a coefficient of z0 of a theta function.In Section9 we prove an identity for h1(q) using h1(q) as a coefficient of z0 of a theta function.In Section10 a double series expansion for h1(q) is obtained by using Bailey pair method.


## 2. Basic Preliminaries

We first introduce some standard notation.Ifq and a are complex numbers with |q|<1 and n is a nonnegative integer, then(2.1)(a)0=(a;q)0=1,(a)n=(a;q)n=∏k=0n-1(1-aqk),(a)∞=(a;q)∞=∏k=0∞(1-aqk),(a1,…,am)n=(a1,…,am;q)n=(a1;q)n,…,(am;q)n.
Ramanujan’s mock theta function of third-order ω(q) and ν(q) is(2.2)ω(q)=∑n=0∞q2n(n+1)(q;q2)n+12,(2.3)ν(q)=∑n=0∞qn(n+1)(-q;q2)n+1,(2.4)φ(q)=∑n=-∞∞qn2=(-q;q2)∞2(q2;q2)∞=(-q;-q)∞(q;-q)∞.
We will use the following notations for θ-functions.Definition 2.1.
If|q|<1 and x≠0, then
(2.5)j(x,q)=(x,qx,q;q)∞.
If  m is a positive integer and a is an integer,
(2.6)Ja,m=j(qa,qm),(2.7)J̅a,m=j(-qa,qm),(2.8)Jm=j(qm,q3m)=(qm;qm)∞,(2.9)j(qx,q)=j(x,q),(2.10)j(x,q)=-xj(x-1,q),(2.11)j(qnx,q)=(-1)nq-n(n-1)/2x-nj(x,q),ifnisaninteger.
By Jacobi’s triple product identity [12, page 282]
(2.12)j(x,q)=∑n=-∞∞(-1)nqn(n-1)/2xn.


### 2.1. More Definitions

Ifz is a complex number with |z|≠1, then(2.13)ε(z)={1if|z|<1,-1if|z|>1.Ifs is an integer, then(2.14)sg(s)={1ifs≥0,-1ifs<0.Using these definitions,(2.15)11-z=ε(z)∑s=-∞sg(s)=ε(z)∞zs.
We shall use the following theorems.Theorem 2.2 (see [13, Theorem 1.3, page 644]).
Letq be fixed,  0<|q|<1. Let a,b, and m be fixed integers with b≠0 and m≥1. Define
(2.16)F(z)=1j(qazb,qm).
Then F is meromorphic for z≠0, with simple poles at all points z0 such that z0b=qkm-a for some integer k. The residue of F(z) at such a point z0 is
(2.17)(-1)k+1qmk(k-1)/2z0bJm3.Theorem 2.3 (see [13, Theorem 1.8(a), page 647]).
Suppose that(2.18)F(z)=∑rFrzr
for all z≠0 and that F(z) satisfies
(2.19)F(qz)=Cz-nF(z),
where 0<|q|<1 and C≠0. Then
(2.20)F(z)=∑r=0n-1Frzrj(-C-1qrzn,qn).
Truesdell [14] calls the functions which satisfy the difference equation
(2.21)∂∂zF(z,α)=F(z,α+1)
as F-function. He unified the study of these F-functions.
The functions which satisfy theq-analogue of the difference equation
(2.22)Dq,zF(z,α)=F(z,α+1),
where
(2.23)zDq,zF(z,α)=F(z,α)-F(zq,α)
are called Fq-functions.


## 2.1. More Definitions

Ifz is a complex number with |z|≠1, then(2.13)ε(z)={1if|z|<1,-1if|z|>1.Ifs is an integer, then(2.14)sg(s)={1ifs≥0,-1ifs<0.Using these definitions,(2.15)11-z=ε(z)∑s=-∞sg(s)=ε(z)∞zs.
We shall use the following theorems.Theorem 2.2 (see [13, Theorem 1.3, page 644]).
Letq be fixed,  0<|q|<1. Let a,b, and m be fixed integers with b≠0 and m≥1. Define
(2.16)F(z)=1j(qazb,qm).
Then F is meromorphic for z≠0, with simple poles at all points z0 such that z0b=qkm-a for some integer k. The residue of F(z) at such a point z0 is
(2.17)(-1)k+1qmk(k-1)/2z0bJm3.Theorem 2.3 (see [13, Theorem 1.8(a), page 647]).
Suppose that(2.18)F(z)=∑rFrzr
for all z≠0 and that F(z) satisfies
(2.19)F(qz)=Cz-nF(z),
where 0<|q|<1 and C≠0. Then
(2.20)F(z)=∑r=0n-1Frzrj(-C-1qrzn,qn).
Truesdell [14] calls the functions which satisfy the difference equation
(2.21)∂∂zF(z,α)=F(z,α+1)
as F-function. He unified the study of these F-functions.
The functions which satisfy theq-analogue of the difference equation
(2.22)Dq,zF(z,α)=F(z,α+1),
where
(2.23)zDq,zF(z,α)=F(z,α)-F(zq,α)
are called Fq-functions.


## 3. Bilateral𝒟5(q) as a Sum of Two Mock Theta Functions of Different Orders

(i) We shall denote the bilateral of𝒟5(q) by 𝒟5,c(q). We define it as(3.1)(q;q2)∞2𝒟5,c(q)=∑n=-∞∞(q;q2)n2q2n.
Now(3.2)(q;q2)∞2𝒟5,c(q)=∑n=-∞∞(q;q2)n2q2n=∑n=0∞(q;q2)n2q2n+∑n=-1-∞(q;q2)n2q2n=∑n=0∞(q;q2)n2q2n+∑n=0∞q2n2+2n(q;q2)n+12,
and we use (1.2) in the first summation and (2.2) in the second summation, to write(3.3)(q;q2)∞2𝒟5,c(q)=(q;q2)∞2𝒟5(q)+ω(q).Thus𝒟5,c(q) is a sum of a second-order mock theta function and a third-order mock theta function.(ii)  Transformation of Bilateral𝒟5,c(q) into bilateral ωc(q) is as follows.It is very interesting that the bilateral𝒟5,c(q) can be written as bilateral third-order mock theta function ωc(q).We use Bailey’s bilateral transformation [15, 5.20(ii), page 137]:(3.4)ψ22[a,bc,d;q,z]=(az,bz,cq/abz,dq/abz;q)∞(q/a,q/b,c,d;q)∞×ψ22[abz/c,abz/daz,bz;q,cd/abz].
Letting q→q2, and  setting a=b=q, c=d=0, and z=q2 in (3.4), we get(3.5)(q;q2)∞2𝒟5,c(q)=(q3,q3;q2)∞(q,q;q2)∞∑n=-∞∞q2n2+2n(q3;q2)n2=∑n=-∞∞q2n2+2n(q;q2)n+12=ωc(q).


## 4. Another Bilateral Transformation

Slater [15, (5.4.3), page 129] gave the following transformation formula, and we have taken r=2:(4.1)(b1,b2,q/a1,q/a2,dz,q/dz;q)∞(c1,c2,q/c1,q/c2;q)∞ψ22[a1,a2b1,b2;q,z]=qc1(c1/a1,c1/a2,qb1/c1,qb2/c1,dc1z/q,q2/dc1z;q)∞(c1,q/c1,c1/c2,qc2/c1;q)∞ψ22[qa1/c1,qa2/c1qb1/c1,qb2/c1;q,z]+idem(c1;c2),
where   d=a1a2/c1c2,    |b1b2/a1a2|<|z|<1,  and idem(c1;c2) after the expression means that the preceding expression is repeated with c1 and c2 interchanged.In the transformation it is interesting that thec’s are absent in the ψ22 series on the left side of (4.1). This gives us the freedom to choose the c’s  in a convenient way.Lettingq→q2 and  setting, a1=a2=q,  b1=b2=0, and  z=q2 in (4.1), so  d=q2/c1c2  and 0<|z|<1,   to get(4.2)(q;q2)∞4(q4/c1c2;q2)∞(c1c2/q2;q2)∞(c2;q2)∞(q2/c2;q2)∞𝒟5,c(q)=q2c1(c1/q;q2)∞2(q2/c2;q2)∞(c2;q2)∞(c1/c2;q2)∞(q2c2/c1;q2)∞∑n=-∞∞(q3/c1;q2)n2q2n+idem(c1;c2).
By choosing c1 suitably we can have different expansion identities. Moreover (4.2) can be seen as a generalization of (3.3).


## 5. Mild Generalization of𝒟5,c(q)

We define the bilateral generalized function𝒟5,c(z,α) as(5.1)(q;q2)∞2𝒟5,c(z,α)=1(z)∞∑n=-∞∞(q;q2)n2(z)nqnα+n.
For α=1, z=0, 𝒟5,c(z,α) reduce to 𝒟5,c(q).Now(5.2)Dq,z[𝒟5,c(z,α)]=1z[𝒟5,c(z,α)-𝒟5,c(zq,α)]=1z(q;q2)∞2[1(z)∞∑n=-∞∞(q;q2)n2(z)nqnα+n-1(zq)∞∑n=-∞∞(q;q2)n2(zq)nqnα+n]=1z(q;q2)∞21(z)∞∑n=-∞∞(q;q2)n2(z)nqnα+n(1-(1-zqn))=1(q;q2)∞2(z)∞∑n=-∞∞(q;q2)n2(z)nqnα+2n=𝒟5,c(z,α+1).
So 𝒟5,c(z,α+1) is an Fq-function.BeingFq-function it has unified properties of Fq-functions. For example, one has the following.(i) The inverse operatorDq,x-1 of q-differentiation is related to q-integration as(5.3)Dq,x-1ϕ(x)=(1-q)-1∫ϕ(x)dqx.
See Jackson  [16].(ii)Dq,znFq(z,α)=Fq(z,α+n), where n is a nonnegative integer.


## 6. Behaviour of𝒟5(q) outside the Unit Circle

By definition (1.1)(6.1)𝒟5(q)=∑n=0∞(-q;q)n(q;q2)n+1qn.
Replacing q by 1/q and writing 𝒟5*(q) for 𝒟5(1/q) [10],(6.2)𝒟5*(q)=∑n=0∞(-1)nq(n2+n)/2(-q;q)n(q;q2)n+1=1-q2+q6-q12+q20-q30+⋯=∑n=0∞(-1)nqn2+n,
which is a θ-function.


## 7. Lambert Series Expansion forh1(q)

For the double series expansion, we first require the generalized Lambert series expansion forh1(q).By Entry 12.4.5, of Ramanujan’s Lost Notebook [17, page 277], Hikami [10] noted that(7.1)𝒟5(q)=2h1(q)-(-q;q)∞2ω(q),
where(7.2)h1(q)=∑n=0∞(-q;q)2n(q;q2)n+12qn.
There is a slight misprint in the definition h1(q) in Hikami’s paper [10] which has been corrected and Gordon and McIntosh have also pointed out in their survey [8].In [8] the Lambert series expansion for h1(q) is(7.3)h1(q)=∑n=0∞(-q;q)2n(q;q2)n+12qn=1θ4(0,q)∑n=0∞(-1)nqn(n+2)1-q2n+1=12θ4(0,q)∑n=-∞∞(-1)nqn(n+2)1-q2n+1.


## 8.h1(q)  as a Coefficient of z0 of a θ-Function

In the following theorem of Hickerson [13, Theorem 1.4, page 645],(8.1)∑r=-∞∞xr1-qry=J13j(xy,q)j(x,q)j(y,q)
let q→q2,  and then put y=q, to get(8.2)∑r=-∞∞xr1-q2r+1=J23j(qx,q2)j(x,q2)j(q,q2).
For  |q|<1, and z≠0 and not an integral power of q, let(8.3)A(z)=12θ4(0,q)J23j(qz,q2)j(z,q2)j(q,q2)j(zq,q2).Theorem 8.1.
Letq be fixed with 0<|q|<1.  Then h1(q) is the coefficient of z0 in the Laurent series expansion of A(z) in the annulus |q|<|z|<1.Proof.
By (7.3)
(8.4)2θ4(0,q)h1(q)=∑n=-∞∞(-1)nqn(n+2)1-q2n+1=coefficientofz0in∑n=-∞∞zn1-q2n+1∑s=-∞∞(-1)sqs2+s(zq)-s=coefficientofz0inJ23j(qz,q2)j(z,q2)j(q,q2)j(zq,q2)
dividing by 2θ4(0,q) gives the theorem.


## 9. An Identity forh1(q)

Theorem 9.1.
If0<|q|<1 and z  is neither zero nor an integral power of q, then
(9.1)A(z,q)=j(z,q2)h1(q)-(-12)∑r=-∞∞(-1)rqr2+3r-1zr+21-q2r+2z-12∑r=-∞∞(-1)rqr2+3r+1z-r-11-q2r+2z-1.
Define
(9.2)L(z)=-12∑r=-∞∞(-1)rqr2+3r-1zr+21-q2r+2z(9.3)M(z)=12∑r=-∞∞(-1)rqr2+3r+1z-r-11-q2r+2z-1,(9.4)F(z)=A(z)+L(z)+M(z).
The scheme will be first to show that F(z) satisfies the functional relation:
(9.5)F(q2z)=-z-1F(z).
One considers the poles of L(z) and M(z) and shows that the residue of F(z) at these poles is zero. So F(z) is analytic at these points. One then shows that the coefficients of z0 in L(z) and M(z) are zero and equating the coefficient of  z0 in (9.4) one has the theorem.Proof.
We show that(9.6)F(q2z)=-z-1F(z).
We shall show that each of  A(z), L(z), and M(z) satisfies the functional equation:
(9.7)A(z)=12(-q;q)∞(q;q)∞J23j(qz,q2)j(z,q2)j(q,q2)j(zq,q2),
and so
(9.8)A(q2z)=12(-q;q)∞(q;q)∞J23j(q3z,q2)j(q2z,q2)j(q,q2)j(zq,q2).
We employ (2.11) on the right-hand side to get
(9.9)A(q2z)=12(-q;q)∞(q;q)∞J23(-1)z-1q-1j(zq,q2)(-1)qz-1(-1)z-1j(z,q2)j(q,q2)j(zq,q2)A(q2z)=-z-1A(z).
We now take L(z):
(9.10)L(q2z)=12∑r=-∞∞(-1)rqr2+3r-1(zq2)r+21-q2r+2(zq2).
Writing r-1 for ron the right-hand side we have
(9.11)L(q2z)=-z-1L(z).
Similarly only writing r+1 for r we have
(9.12)M(q2z)=-z-1M(z).
Hence the functional equation (9.4) is proved.
ObviouslyL(z) and M(z) are meromorphic for z≠0. L(z) has simple poles at z=q2k-2 and M(z) has simple poles at z=q2k+2. Hence F(z) is meromorphic for z≠0 with, at most, simple poles at z=q2k±2.
Takingr=0 in (9.2), we calculate the residue of L(z) at the point z=1/q2:
(9.13)ResidueofL(z)=limz→1/q212(z-1q2)z2q-1(z-(1/q2))q2=12q-5.
For the residue of A(z) at z=1/q2, take b=1, k=-1, m=2, a=0 in (2.16) to get
(9.14)ResidueofA(z)=12(-q;q)∞(q;q)∞J23j(1/q,q2)j(1/q3,q2)j(q,q2)1J23=12(-q;q)∞(q;q)∞j(q,q2)j(q3,q2)q4j(q,q2)=12(-q;q)∞(q;q)∞1q4(q3;q2)∞(1q;q2)∞(q2;q2)∞=12(-q;q)∞(q;q)∞1q4(1-(1/q))(1-q)(q;q2)∞(q;q2)∞(q2;q2)∞=-12(-q;q)∞(q;q)∞1q5(q;q)∞(-q;q)∞=-12q-5.
So the residue of F(z) at z=1/q2 is -(1/2)q5+0+(1/2)q5=0.
Now we calculate the residue atz=q2:
(9.15)ResidueofM(z)=limz→q212(z-q2)qz-1(1-q2z-1)=limz→q212(z-q2)q(z-q2)=q2,
and for the residue of A(z) at z=q2, taking b=1, k=1, m=2, and a=0 in (2.16), so
(9.16)ResidueofA(z)=12(-q;q)∞(q;q)∞J23j(q3,q2)j(q,q2)j(q,q2)q2J23=12(-q;q)∞(q;q)∞j(q3,q2)q2=12(-q;q)∞(q;q)∞(q3;q2)∞(1q;q2)∞(q2;q2)∞q2=12(-q;q)∞(q;q)∞(1-(1/q))(1-q)(q;q2)∞(q;q2)∞(q2;q2)∞q2=-12q.
Hence the residue of F(z) at z=q2 is 0+(1/2)q-(1/2)q=0. Hence F(z) is analytic at z=q2.
SinceF(z) satisfies (9.4), so F(z) is analytic at all points of the form z=q2k±2 and hence for all z≠0.
We now apply (2.20) with n=1 and c=-1 and q replaced by q2 to get
(9.17)F(z)=F0j(z,q2),
where F0 is the coefficient of z0 in the Laurent expansion of  F(z),z≠0.
Now for|q|<|z|<1, by Theorem 8.1, the coefficient of z0 in A(z) is h1(q).
For suchz, |q2r+2z|<1 if and only if  r≥0.
That is,
(9.18)ε(q2r+2z)=sg(r).
Hence by (2.15)
(9.19)11-q2r+2z=sg(r)∑r=-∞sg(r)=sg(s)∞q(2r+2)szs.
So
(9.20)L(z)=-12∑sg(r)=sg(s)∞sg(r)(-1)rqr2+3r-1+(2r+2)szr+2+s.
If  sg(r)=sg(s), then r+s+2 is either ≥1 or ≤-1; so coefficient of z0 in L(z) is 0. Similarly the coefficient of z0 in M(z) is 0 and so the coefficient of z0 in F(z) is h1(q).
Hence by (9.17), we have
(9.21)F(z)=h1(q)j(z,q2),
which gives the theorem.


## 10. Double Series Expansion

Now we derive the double series expansion forh1(q). We shall use the Bailey pair method, as used by Andrews [18] for fifth and seventh-order mock theta functions and by Andrews and Hickerson [5] for sixth-order mock theta functions.We define Bailey pair.Two sequences{αn} and {βn}, n≥0, form a Bailey pair relative to a number a if(10.1)βn=∑r=0nαr(q)n-r(aq)n+r,
for all n≥0.Corollary 10.1 (see [5, Corollary. 2.1, page 70]).
If{αn}  and {βn} form a Bailey pair relative to a,  then
(10.2)∑n=0∞(ρ1)n(ρ2)n(aq/ρ1ρ2)nαn(aq/ρ1)n(aq/ρ2)n=(aq)∞(aq/ρ1ρ2)∞(aq/ρ1)∞(aq/ρ2)∞∑n=0∞(ρ1)n(ρ2)n(aqρ1ρ2)nβn
provided that both sums converge absolutely.We state the theorem of Andrews and Hickerson [5, Theorem 2.3, pages 72-73].Leta, b, c, and q be complex numbers with a≠1, b≠0, c≠0, q≠0, and none a/b, a/c, qb, qc of the form q-k  with k≥0.  For n≥0, define(10.3)An′(a,b,c,q)=qn2(bc)n(1-aq2n)(a/b)n(a/c)n(1-a)(qb)n(qc)n×∑j=0n(-1)j(1-aq2j-1)(a)j-1(b)j(c)jq(2j)(bc)j(q)j(a/b)j(a/c)j,Bn′(a,b,c,q)=1(qb)n(qc)n.
Then the sequences {An′(a,b,c,q)} and {Bn′(a,b,c,q)} form a Bailey pair relative to a.Lettingq→q2 and then taking a=q2, b=c=q, in (10.3), we get(10.4)An′(q2,q,q2)=q2n2+2n(1-q4n+2)(q;q2)n2(1-q2)(q3;q2)n2×∑j=0n(-1)j(1-q4j)(q2;q2)j-1(q;q2)j2qj2+j(q2;q2)j(q;q2)j2=(1-q)2(1+q2n+1)q2n2+2n(1-q2)(1-q2n+1)∑j=0n(-1)jq-j2-j(1+q2j)=(1-q)2(1+q2n+1)q2n2+2n(1-q2)(1-q2n+1)[1+∑j=-nn(-1)jq-j2-j],Bn′(q2,q,q,q2)=1(q3;q2)n2.
Now letting q→q2 and then setting ρ1=-q, ρ2=-q2, a=q2 in (10.2) we get(10.5)∑n=0∞qn(-q;q2)n(-q3;q2)nαn=(q;q2)∞(q4;q2)∞(-q2;q2)∞(-q3;q2)∞∑n=0∞(-q;q2)n(-q2;q2)nqnβn.
Taking An′ and  Bn′ for αn′ and βn′, respectively, in (10.5) and using the definition of h1(q), we get(10.6)(q;q2)∞(q2;q2)∞(-q;q2)∞(-q2;q2)∞h1(q)=∑n=0∞q2n2+3n1-q2n+1[1+∑j=-nn(-1)jq-j2-j]
or(10.7)h1(q)=(-q;q2)∞(-q2;q2)∞(q;q2)∞(q2;q2)∞∑n=0∞q2n2+3n1-q2n+1[1+∑j=-nn(-1)jq-j2-j],
which is the double series expansion for h1(q).This double series expansion can be used to get more properties of𝒟5(q).



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*Source: 978425-2010-04-22.xml*