New Mock Theta Function Identities via Fractional q-Calculus and Bilateral ₂ψ₂ Series

Abstract

Mock theta functions, introduced by Ramanujan in his last letter to Hardy, play a significant role in q-series theory and have natural connections to fractional q-calculus. In this paper, we study bilateral hypergeometric series of the form ${}_2\psi _2$= ${\sum }_{n=-\infty }^{\infty }{\displaystyle \frac{{(a,b;q)}_n}{{(c,d;q)}_n}}{z}^n,$ where ${(a;q)}_n$ denotes the q-shifted factorial. Using Slater’s three-term transformation formula for bilateral ${}_2\psi _2$ series, we derive new identities for Ramanujan’s mock theta functions of orders 2, 3, 6, and 8. These transformations reveal previously unknown relationships between different q-series representations and extend the classical theory of mock theta functions within the framework of q-special functions.

1. Introduction

We employ the standard q-series notation from [1] throughout this work. Given complex numbers x, q with $\left|q\right|<1$ and integer n, the q-shifted factorial is expressed as

$$\begin{array}{c}\hfill {(x;q)}_{\infty }:=\prod _{i=0}^{\infty }(1-x{q}^i),{(x;q)}_n:={\displaystyle \frac{{(x;q)}_{\infty }}{{(x{q}^n;q)}_{\infty }}}.\end{array}$$

For notational simplicity, we represent products of q-shifted factorials in the condensed form

$${(x_1,x_2,\dots ,x_n)}_k:={(x_1;q)}_k\dots {(x_n;q)}_k,k\in \mathbb{Z}\cup \left\{\infty \right\}.$$

The bilateral hypergeometric series ${}_2\psi _2$ studied in this work have a natural connection to fractional q-calculus. The q-shifted factorial ${(a;q)}_n$, which appears throughout our identities, forms the foundation of q-calculus through its role in q-difference operators. Specifically, the Jackson q-derivative is defined as

$$D_{q}f\left(x\right)={\displaystyle \frac{f\left(x\right)-f\left(qx\right)}{x(1-q)}},$$

and when applied to the power function $x^n$, it produces ${\left[n\right]}_q{x}^{n-1}$, where the q-number ${\left[n\right]}_q=(1-q^n)/(1-q)$ is expressed in terms of q-shifted factorials. Similarly, the Jackson q-integral

$${\int }_0^{x}f\left(t\right)d_{q}t=x(1-q)\sum _{n=0}^{\infty }{q}^{n}f\left(x{q}^n\right)$$

connects directly to the power series expansions we investigate. The bilateral basic hypergeometric series ${}_2\psi _2$ naturally arise as solutions to fractional q-difference equations of the form $D_q^{\alpha }f\left(x\right)=\lambda f\left(x\right)$, where $D_q^{\alpha }$ denotes the fractional q-derivative of order $\alpha$ constructed using iterated q-differences and q-integrals. Our mock theta function identities thus provide explicit solutions to certain classes of such equations. The free parameters $c_1$ and $c_2$ in our transformation formulas can be viewed as integration constants or boundary conditions in this fractional q-calculus framework, establishing a concrete bridge between classical q-series theory and modern fractional calculus. This perspective opens avenues for applying our results to q-analogs of fractional differential equations arising in quantum mechanics, combinatorics, and mathematical physics.

The unilateral basic hypergeometric series ${}_r\varphi _s$ is defined by

$${}_r\varphi _s\left[\begin{array}{cccc}{a}_1,& a_2,& \cdots ,& a_r\\ & b_1,& \cdots ,& b_s\end{array};q,z\right]=\sum _{n=0}^{\infty }{\displaystyle \frac{{(a_1,a_2,\dots ,a_r)}_n}{{(q,b_1,\cdots ,b_s)}_n}}{\left[{(-1)}^n{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}\right]}^{1+s-r}{z}^n,$$

where $q\ne 0$ when $r>s+1$.

The bilateral basic hypergeometric series is defined by

$${}_r\psi _s\left[\begin{array}{ccc}{a}_1,& \cdots ,& a_r\\ b_1,& \cdots ,& b_s\end{array};q,z\right]=\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a_1,\cdots ,a_r)}_n}{{(b_1,\cdots ,b_s)}_n}}{\left[{(-1)}^n{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}\right]}^{s-r}{z}^n.$$

When $b_1=q$ in the bilateral series above, the bilateral q-series can be transformed to the basic hypergeometric series.

Jacobi’s triple product identity is stated as follows:

$$j(y;q):={(q,y,q/y)}_{\infty }=\sum _{n=-\infty }^{\infty }{(-1)}^n{q}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)}{y}^n,$$

and we note that

$$j(y_1,y_2,\dots ,y_n;q):=j(y_1;q)j(y_2;q)\cdots j(y_n;q).$$

In the sequel, we also use the following definition from [2]:

Definition 1. 

If a and m are integers with $m\ge 1$, then

$${\overline{J}}_{a,m}=j(-q^a;q^m),$$

and

$$J_m=J_{m,3m}={(q^m;q^m)}_{\infty }.$$

Mock theta functions are an important component of the theory of basic hypergeometric series. They were first introduced by Ramanujan in 1920 [3]. In his correspondence with Hardy before his death, Ramanujan presented a collection of 17 functions, which he called mock theta functions and grouped into orders 3, 5, and 7. The letter, however, contained no formal definition of these functions or explanation of what the order classification signified. G. H. Hardy [3] (p. 534) defines such functions as follows: a mock $\theta$-function is a function defined by a q-series convergent for $\left|q\right|<1$, for which we can calculate asymptotic formulas when q tends to a rational point $e^{2r\pi i/s}$, with the same degree of precision as those furnished for the ordinary $\theta$-functions by the theory of linear fractional transformations. Later, similar definitions were given by Andrews and Hickerson [4] and Gordon and McIntosh [5,6], respectively. Zwegers [7,8] established the modern understanding of mock theta functions by proving they are holomorphic components of certain harmonic weak Maass forms. Meanwhile, some second-, sixth-, eighth-, and tenth-order mock theta functions have also appeared. For details of mock theta functions, readers may refer to [3,4,5,6,9,10].

In this paper, we use the following mock theta functions defined by Andrews [11]. The second-order mock theta functions:

$$\begin{array}{cc}\hfill & A\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q^2;q^2)}_n}{{(q;q^2)}_{n+1}}}{q}^{n+1}=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{{(n+1)}^2}{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}.\hfill \end{array}$$

(1)

The equivalence between these two series representations is established by the identity (cf. [12] (Equation (1))). To see this, we start from the general term of the first series and apply transformations on the q-shifted factorials:

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(-q^2;q^2)}_n}{{(q;q^2)}_{n+1}}}{q}^{n+1}& ={\displaystyle \frac{{(-q^2;q^2)}_n·q^{n+1}}{(1-q){(q;q^2)}_n(1-q^{2n+1})}}\hfill \\ \hfill & ={\displaystyle \frac{q^{{(n+1)}^2}{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}.\hfill \end{array}$$

This transformation uses the q-factorial identities and yields the second representation after summing over n. For the complete verification, we refer to McIntosh [12].

Similarly, the second-order mock theta function $B\left(q\right)$ admits two equivalent representations, and we also consider the function $\mu \left(q\right)$:

$$\begin{array}{cc}\hfill & B\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^n{(-q;q^2)}_n}{{(q;q^2)}_{n+1}}}=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2+n}{(-q^2;q^2)}_n}{{(q;q^2)}_{n+1}^2}},\hfill \end{array}$$

(2)

$$\begin{array}{cc}\hfill & \mu \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{q}^{n^2}{(q;q^2)}_n}{{(-q^2;q^2)}_n^2}}.\hfill \end{array}$$

(3)

The equivalence of the two series for $B\left(q\right)$ follows by the same transformation technique applied to $A\left(q\right)$. The function $\mu \left(q\right)$ is given in its canonical form. These functions appeared in Ramanujan’s Lost Notebook [13]. In particular, the function $B\left(q\right)$ arises in the modular transformation formulas for $A\left(q\right)$ and $\mu \left(q\right)$. By using the identity (cf. [12] (Equation (1)))

$$\begin{array}{c}\hfill \sum _{n=0}^{\infty }{\displaystyle \frac{{(-q^2/z;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n(n+1)}{z}^n=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-zq;q^2)}_n}{{(q;q^2)}_{n+1}}}{q}^n,\end{array}$$

McIntosh [12] obtained the equalities of the two series in (1) and (2). Then, Srivastava [14] proved these two identities again using the transformation formula of Fine [15]:

$${(1-x)}_2{\varphi }_1\left[\begin{array}{cc}q,& a\\ & b\end{array};q,x\right]=\sum _{n=0}^{\infty }{\displaystyle \frac{{(b/a)}_n{(-ax)}^n{q}^{n(n-1)/2}}{{\left(b\right)}_n{\left(xq\right)}_n}}.$$

The third-order mock theta functions:

$$\begin{array}{cc}\hfill & f\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q,-q;q)}_n}},\varphi \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}},\hfill \\ \hfill & \psi \left(q\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{q^{n^2}}{{(q;q^2)}_n}},\nu \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n(n+1)}}{{(-q;q^2)}_{n+1}}},\hfill \\ \hfill & \omega \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{2n(n+1)}}{{(q;q^2)}_{n+1}^2}},\xi \left(q\right)=1+2\sum _{n=1}^{\infty }{\displaystyle \frac{q^{6n^2-6n+1}}{{(q;q^6)}_n{(q^5;q^6)}_n}}.\hfill \end{array}$$

The sixth-order mock theta functions:

$$\begin{array}{cc}\hfill & {\psi }_6\left(q\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}{q}^{n^2}{(q;q^2)}_{n-1}}{{(-q;q)}_{2n-1}}},\rho \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q)}_n{q}^{n(n+1)/2}}{{(q;q^2)}_{n+1}}},\hfill \\ \hfill & \sigma \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q)}_n{q}^{(n+2)(n+1)/2}}{{(q;q^2)}_{n+1}}},\lambda \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{q}^n{(q;q^2)}_n}{{(-q;q)}_n}},\hfill \\ \hfill & \mu \left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{(q;q^2)}_n}{{(-q;q)}_n}},{\varphi }_6\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-1)}^n{q}^{n^2}{(q;q^2)}_n}{{(-q;q)}_{2n}}},\hfill \\ \hfill & {\psi }_{-}\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q)}_{2n}{q}^{n+1}}{{(q;q^2)}_{n+1}}},{\varphi }_{-}\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n+1}{(-q)}_{2n+1}}{{(q;q^2)}_{n+1}}}.\hfill \end{array}$$

The eighth-order mock theta functions:

$$\begin{array}{cc}\hfill & S_0\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}{(-q;q^2)}_n}{{(-q^2;q^2)}_n}},S_1\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2+2n}{(-q;q^2)}_n}{{(-q^2;q^2)}_n}},\hfill \\ \hfill & T_0\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{(n+1)(n+2)}{(-q^2;q^2)}_n}{{(-q;q^2)}_{n+1}}},T_1\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n(n+1)}{(-q^2;q^2)}_n}{{(-q;q^2)}_{n+1}}},\hfill \\ \hfill & U_0\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}{(-q;q^2)}_n}{{(-q^4;q^4)}_n}},U_1\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{{(n+1)}^2}{(-q;q^2)}_n}{{(-q^2;q^4)}_{n+1}}},\hfill \\ \hfill & V_0\left(q\right)=-1+2\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}{(-q;q^2)}_n}{{(q;q^2)}_n}},V_1\left(q\right)=\sum _{n=0}^{\infty }{\displaystyle \frac{q^{{(n+1)}^2}{(-q;q^2)}_n}{{(q;q^2)}_{n+1}}}.\hfill \end{array}$$

Several mock theta functions arise as special cases corresponding to one “side” ($n\ge 0$ or $n<0$) of bilateral series having the general form

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a,b;q)}_n}{{(c,d;q)}_n}}{z}^n=\sum _{n=0}^{\infty }{\displaystyle \frac{{(a,b;q)}_n}{{(c,d;q)}_n}}{z}^n+\sum _{n=1}^{\infty }{\displaystyle \frac{{(q/c,q/d)}_n}{{(q/a,q/b)}_n}}{(cd/abz)}^n.\end{array}$$

We utilize Bailey’s transformations and summation results for ${}_2\psi _2$ series [16,17], along with other known q-series identities, to obtain new representations for mock theta functions and additional q-series relations. Using this method, McLaughlin [2] obtained all third-order, eight fifth-order, eight sixth-order, and four eighth-order mock theta function identities. However, we find that the author did not obtain the related identities for the second-order mock theta functions. Taking this as motivation, we do further research on the second-order mock theta functions in this paper. Additionally, we find new results for the 3rd, 6th, and 8th order mock theta functions.

To obtain some of the results in this paper, it is necessary to present the following three-term transformation formula for basic hypergeometric series. The case $r=2$ of Slater’s three-term transformation formula (cf. [1] Equation (5.4.3)):

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & {\displaystyle \frac{{(b_1,b_2,q/a_1,q/a_2,dz,q/dz)}_{\infty }}{{(c_1,c_2,q/c_1,q/c_2)}_{\infty }}}{}_2\psi _2\left[\begin{array}{cc}{a}_1,& a_2\\ b_1,& b_2\end{array};q,z\right]\hfill \\ \hfill & ={\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(c_1/a_1,c_1/a_2,q{b}_1/c_1,q{b}_2/c_1,d{c}_{1}z/q,q^2/d{c}_{1}z)}_{\infty }}{{(c_1,q/c_1,c_1/c_2,q{c}_2/c_1)}_{\infty }}}\hfill \\ \hfill & \times {}_2\psi _2\left[\begin{array}{cc}q{a}_1/c_1,& q{a}_2/c_1\\ q{b}_1/c_1,& q{b}_2/c_1\end{array};q,z\right]+\mathrm{idem}(c_1;c_2),\hfill \end{array}\end{array}$$

(4)

where $d=a_1{a}_2/c_1{c}_2$, $|b_1{b}_2/a_1{a}_2|<|z|<1$, and $\mathrm{idem}(c_1;c_2)$ after the expression means that the preceding expression is repeated with $c_1$ and $c_2$ interchanged. In the rest of the paper, we obtain some new identities for these mock theta functions using the bilateral transformation formula given above.

2. Main Results

In this section, we consider the q-series

$$\begin{array}{c}\hfill T(a,b,c;z,q)=1+\sum _{n=1}^{\infty }{\displaystyle \frac{{(a;q^2)}_n}{{(b,c;q^2)}_n}}{z}^n{q}^{n^2},\end{array}$$

and the bilateral series is defined by

$$\begin{array}{c}\hfill T^{*}(a,b,c;z,q)=\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a;q^2)}_n}{{(b,c;q^2)}_n}}{z}^n{q}^{n^2}.\end{array}$$

First, we give the following result to be used later.

Lemma 1. 

If $\left|q\right|<1$, $|bc/azq|<1$, $|b{c}_1/azq|<1$, $|b{c}_2/azq|<1$, then the following identity holds:

$$\begin{array}{cc}\hfill T^{*}(a,b,c;z,q)& =\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a;q^2)}_n}{{(b,c;q^2)}_n}}{z}^n{q}^{n^2}\hfill \\ \hfill & ={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/a,q^{2}b/c_1,q^{2}c/c_1,-az/c_{2}q,-c_2{q}^3/az,c_2,q^2/c_2;q^2)}_{\infty }}{{(c_1/c_2,q^2{c}_2/c_1,b,c,q^2/a,-azq/c_1{c}_2,-c_1{c}_{2}q/az;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a{q}^2/c_1;q^2)}_n}{{(b{q}^2/c_1,c{q}^2/c_1;q^2)}_n}}{\left(z/c_1\right)}^n{q}^{n^2+2n}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(5)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Proof. 

Making the substitutions $q\to q^2$, $a_1\to a$, $b_1\to b$, $b_2\to c$, $z\to -zq/a_2$ in (4), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(b,c,q^2/a,q^2/a_2,-dzq/a_2,-q{a}_2/dz;q^2)}_{\infty }}{{(c_1,c_2,q^2/c_1,q^2/c_2;q^2)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a,a_2;q^2)}_n}{{(b,c;q^2)}_n}}{(-zq/a_2)}^n\hfill \\ \hfill =& {\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/a,c_1/a_2,q^{2}b/c_1,q^{2}c/c_1,-d{c}_{1}z/a_{2}q,-q^3{a}_2/d{c}_{1}z;q^2)}_{\infty }}{{(c_1,q^2/c_1,c_1/c_2,q^2{c}_2/c_1;q^2)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^{2}a/c_1,q^2{a}_2/c_1;q^2)}_n}{{(q^{2}b/c_1,q^{2}c/c_1;q^2)}_n}}{(-zq/a_2)}^n+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Replacing d by $a{a}_2/c_1{c}_2$ and after some simplifications, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(b,c,q^2/a,q^2/a_2,-azq/c_1{c}_2,-c_1{c}_{2}q/az;q^2)}_{\infty }}{{(c_1,c_2,q^2/c_1,q^2/c_2;q^2)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a,a_2;q^2)}_n}{{(b,c;q^2)}_n}}{(-zq/a_2)}^n\hfill \\ \hfill =& {\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/a,c_1/a_2,q^{2}b/c_1,q^{2}c/c_1,-az/c_{2}q,-q^3{c}_2/az;q^2)}_{\infty }}{{(c_1,q^2/c_1,c_1/c_2,q^2{c}_2/c_1;q^2)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^{2}a/c_1,q^2{a}_2/c_1;q^2)}_n}{{(q^{2}b/c_1,q^{2}c/c_1;q^2)}_n}}{(-zq/a_2)}^n+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Setting $a_2\to \infty$ in the above identity, we now examine the limiting behavior of each factor. As $a_2\to \infty$, the factors ${(q^2/a_2;q^2)}_{\infty }$ and ${(c_1/a_2;q^2)}_{\infty }$ both approach 1. The crucial observation is that the product ${(a_2;q^2)}_n·{(-zq/a_2)}^n$ must be evaluated jointly: since ${(a_2;q^2)}_n\sim {(-a_2)}^n{q}^{n(n-1)}$ for large $a_2$, we obtain

$$\begin{array}{c}\hfill {(a_2;q^2)}_n·{(-zq/a_2)}^n\sim {(-a_2)}^n{q}^{n(n-1)}·{\displaystyle \frac{{(-1)}^n{z}^n{q}^n}{a_2^n}}=z^n{q}^{n^2},\end{array}$$

where the powers of $a_2$ cancel. The factors ${(-azq/c_1{c}_2;q^2)}_{\infty }$ and ${(-c_1{c}_{2}q/az;q^2)}_{\infty }$ remain unchanged. Finally, substituting $d=a{a}_2/c_1{c}_2$ and taking the limit simplifies ${(-d{c}_{1}z/a_{2}q;q^2)}_{\infty }$ and ${(-q^3{a}_2/d{c}_{1}z;q^2)}_{\infty }$ to ${(-az/c_{2}q;q^2)}_{\infty }$ and ${(-c_2{q}^3/az;q^2)}_{\infty }$, respectively. After collecting these simplifications, we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(b,c,q^2/a,-azq/c_1{c}_2,-c_1{c}_{2}q/az;q^2)}_{\infty }}{{(c_1,c_2,q^2/c_1,q^2/c_2;q^2)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a;q^2)}_n}{{(b,c;q^2)}_n}}{z}^n{q}^{n^2}\hfill \\ \hfill & ={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/a,q^{2}b/c_1,q^{2}c/c_1,-az/c_{2}q,-c_2{q}^3/az;q^2)}_{\infty }}{{(c_1,q^2/c_1,c_1/c_2,q^2{c}_2/c_1;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^{2}a/c_1;q^2)}_n}{{(q^{2}b/c_1,q^{2}c/c_1;q^2)}_n}}{(z/c_1)}^n{q}^{n^2+2n}+\mathrm{idem}(c_1;c_2),\hfill \end{array}$$

that is,

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a;q^2)}_n}{{(b,c;q^2)}_n}}{z}^n{q}^{n^2}& ={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/a,q^{2}b/c_1,q^{2}c/c_1,-az/c_{2}q,-c_2{q}^3/az,c_2,q^2/c_2;q^2)}_{\infty }}{{(c_1/c_2,q^2{c}_2/c_1,b,c,q^2/a,-azq/c_1{c}_2,-c_1{c}_{2}q/az;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(a{q}^2/c_1;q^2)}_n}{{(b{q}^2/c_1,c{q}^2/c_1;q^2)}_n}}{\left(z/c_1\right)}^n{q}^{n^2+2n}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

This completes the proof of identity (5). □

Theorem 1. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identity for the second-order mock theta function $A\left(q\right)$ holds:

$$\begin{array}{cc}\hfill A\left(q\right)+\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n^2{q}^{2n+1}}{{(-q;q^2)}_{n+1}}}& ={\displaystyle \frac{q^3{J}_4}{c_1{J}_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(q^5/c_1;q^2)}_{\infty }^2{j}^2(c_2;q^2)}{j(c_1/c_2;q^2)j(c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(6)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$. These parameters cannot be zero as they appear in denominators throughout the series expansions and in the arguments of the Jacobi theta functions.

Proof. 

We apply Lemma 1 with the substitutions $b=c=q^3$, $a=-q$, $z=q^2$ in identity (5). First, we verify the convergence conditions of Lemma 1. The constraint $|bc/azq|<1$ becomes

$$\left|{\displaystyle \frac{q^3·q^3}{(-q)·q^2·q}}\right|=\left|{\displaystyle \frac{q^6}{-q^4}}\right|=\left|q^2\right|<1,$$

which is satisfied since $\left|q\right|<1$. Similarly, the conditions $|b{c}_1/azq|<1$ and $|b{c}_2/azq|<1$ follow from $\left|q\right|<1$ for any fixed non-zero $c_1,c_2$. With these substitutions in (5), we obtain for identity (6):

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q^3;q^2)}_n^2}}{q}^{n^2+2n}& ={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(q^5/c_1;q^2)}_{\infty }^2{(q^2/c_2;q^2)}_{\infty }^2{(c_2;q^2)}_{\infty }^2}{{(q^3;q^2)}_{\infty }^2{(c_1/c_2,q^2{c}_2/c_1,-q,q^4/c_1{c}_2,c_1{c}_2/q^2;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Multiplying both sides of the above identity by $q/{(1-q)}^2$, we obtain

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}& ={\displaystyle \frac{q^3}{c_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(q^5/c_1;q^2)}_{\infty }^2{(q^2/c_2;q^2)}_{\infty }^2{(c_2;q^2)}_{\infty }^2}{{(q;q^2)}_{\infty }^2{(c_1/c_2,q^2{c}_2/c_1,-q,q^4/c_1{c}_2,c_1{c}_2/q^2;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2)\hfill \\ \hfill & ={\displaystyle \frac{q^3{J}_4}{c_1{J}_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(q^5/c_1;q^2)}_{\infty }^2{j}^2(c_2;q^2)}{j(c_1/c_2;q^2)j(c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

(7)

To obtain the identity (6), we need to show how the mock theta function $A\left(q\right)$ appears explicitly. We start by separating the bilateral series on the left-hand side of (7) into positive and negative index parts:

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}+\sum _{n=-\infty }^{-1}{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}.\end{array}$$

(8)

For the first sum (with $n\ge 0$), we rewrite the general term by comparing with the definition of $A\left(q\right)$ in (1):

$$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{{(n+1)}^2}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\displaystyle \frac{q^{{(n+1)}^2}{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}\hfill \\ \hfill & =A\left(q\right),\hfill \end{array}$$

(9)

where we used the second representation of $A\left(q\right)$ from Equation (1).

For the second sum (with $n<0$), we make the substitution $n\to -n-1$ and apply the reflection formula for q-shifted factorials

$$\begin{array}{c}\hfill {(a;q)}_{-n}={\displaystyle \frac{{(-q/a)}^n{q}^{\left(\genfrac{}{}{0pt}{}{n}{2}\right)}}{{(q/a;q)}_n}}.\end{array}$$

(10)

This yields:

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{-1}{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q^2)}_{-n-1}}{{(q;q^2)}_{-n}^2}}{q}^{{(-n-1)}^2+2(-n-1)+1}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n^2{q}^{2n+1}}{{(-q;q^2)}_{n+1}}},\hfill \end{array}$$

(11)

after applying the reflection formula and simplifying the exponents.

Combining Equations (9) and (11), we have:

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}^2}}{q}^{n^2+2n+1}=A\left(q\right)+\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n^2{q}^{2n+1}}{{(-q;q^2)}_{n+1}}}.\end{array}$$

(12)

Equating this with the right-hand side of Equation (7), we obtain our desired result (6). □

For the second-order mock theta function $B\left(q\right)$, we can derive some nice results as follows.

Theorem 2. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identities for the second-order mock theta function $B\left(q\right)$ hold:

$$\begin{array}{cc}\hfill B\left(q\right)+{\displaystyle \frac{1}{2}}\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n^2}{{(-q^2;q^2)}_n}}{q}^{2n}& ={\displaystyle \frac{q^2{J}_2^3{j}^2(c_2;q^2){(-c_1/q^2,q^5/c_1,q^5/c_1;q^2)}_{\infty }}{2c_1{J}_1^2{J}_{4}j(c_1/c_2;q^2)j(c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^4/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+3n}}{c_1^n}}+\mathit{idem}(c_1;c_2).\hfill \end{array}$$

(13)

$$\begin{array}{c}\hfill B\left(q\right)+B(-q)={\displaystyle \frac{q}{c_1(1+c_1^2/q)}}{\displaystyle \frac{J_4^2{j}^2(c_2;q)j(-c_1^2/q^2;q^2)}{J_2^{3}j(c_1/c_2;q)j(c_1{c}_2/q;q)}}+\mathit{idem}(c_1;c_2),\end{array}$$

(14)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Proof. 

Following the procedure for the above identities (6), we take $a=-q^2$, $b=c=q^3$, $z=q$ in (5). The convergence condition $|bc/azq|<1$ becomes $|q^6/(-q^2)·q·q|=|q^2|<1$, which holds for $\left|q\right|<1$. Thus we derive identity (13) in Theorem 2. For the identity (14), we give the detailed proof as follows.

Let $b_1=-q^{3/2}$, $b_2=q^{3/2}$, $a_1=i{q}^{1/2}$, $a_2=-i{q}^{1/2}$, $z=q$ in (4). Then, we can rewrite identity (4) as

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{(q^3;q^2)}_{\infty }{(-q;q^2)}_{\infty }{(q^2/c_1{c}_2,c_1{c}_2/q)}_{\infty }}{{(c_1,c_2,q/c_1,q/c_2)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q^3;q^2)}_n}}{q}^n\hfill \\ \hfill =& {\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(-c_1^{2}q,q^5/c_1^2;q^2)}_{\infty }{(q/c_2,c_2)}_{\infty }}{{(c_1,q/c_1,c_1/c_2,q{c}_2/c_1)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1^2;q^2)}_n{q}^n}{{(q^5/c_1^2;q^2)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

(15)

Multiplying both sides of the above equation by ${\displaystyle \frac{1}{1-q}}$ and after some simplifications, we obtain

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n}{{(q;q^2)}_{n+1}}}{q}^n& ={\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(-c_1^{2}q,q^5/c_1^2;q^2)}_{\infty }{j}^2(c_2;q)}{{(q^2;q^4)}_{\infty }j(c_1/c_2;q)j(c_1{c}_2/q;q)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1^2;q^2)}_n{q}^n}{{(q^5/c_1^2;q^2)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

(16)

Recall Ramanujan’s bilateral sum [1]:

$$\begin{array}{c}\hfill {}_1\psi _1\left[\begin{array}{c}a\\ b\end{array};q,z\right]={\displaystyle \frac{{(q,b/a,az,q/az)}_{\infty }}{{(b,q/a,z,b/az)}_{\infty }}}.\end{array}$$

(17)

Replacing q, a, b, z by $q^2$, $-q^3/c_1^2$, $q^5/c_1^2$, q in (17), we obtain

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1^2;q^2)}_n{q}^n}{{(q^5/c_1^2;q^2)}_n}}={\displaystyle \frac{{(q^4;q^4)}_{\infty }{(-q^4/c_1^2,-c_1^2/q^2;q^2)}_{\infty }}{{(q^2;q^4)}_{\infty }{(q^5/c_1^2,-c_1^2/q;q^2)}_{\infty }}}.\end{array}$$

(18)

Combining (16) with (18) and after some computations, we obtain

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}={\displaystyle \frac{q}{c_1(1+c_1^2/q)}}{\displaystyle \frac{J_4^2{j}^2(c_2;q)j(-c_1^2/q^2;q^2)}{J_2^{3}j(c_1/c_2;q)j(c_1{c}_2/q;q)}}+\mathrm{idem}(c_1;c_2).\end{array}$$

(19)

To show how $B\left(q\right)$ and $B(-q)$ appear in identity (14), we separate the bilateral series on the left-hand side of (19). Using the q-shifted factorial reflection formula

$$\begin{array}{c}\hfill {(a;q)}_{-n}={\displaystyle \frac{{(-q/a)}^n{q}^{\left(\genfrac{}{}{0pt}{}{n}{2}\right)}}{{(q/a;q)}_n}},\end{array}$$

(20)

we split the series into positive and negative indices:

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}+\sum _{n=-\infty }^{-1}{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}.\hfill \end{array}$$

(21)

The first sum matches exactly the definition of $B\left(q\right)$ from Equation (2). For the second sum, we substitute $n\to -n-1$ and apply the reflection Formula (20):

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{-1}{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}=\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n{(-1)}^n{q}^n}{{(-q;q^2)}_{n+1}}}=B(-q),\end{array}$$

(22)

where the last equality follows by replacing q with $-q$ in the definition of $B\left(q\right)$. Therefore:

$$\begin{array}{c}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q;q^2)}_n{q}^n}{{(q;q^2)}_{n+1}}}=B\left(q\right)+B(-q).\end{array}$$

(23)

Inserting this into the left side of identity (19), we obtain our desired result (14). This completes the proof of Theorem 2. □

Similar to the proofs of identities in Theorems 1 and 2, we take $a=q$, $b=c=-q^2$, $z=-1$ in (5). The convergence condition $|bc/azq|<1$ becomes $|(-q^2)(-q^2)/\left(q\right)(-1)\left(q\right)|=|q^4/(-q^2)|=|q^2|<1$, which is satisfied for $\left|q\right|<1$. After simplifications, we obtain the following identity for the second-order mock theta function $\mu \left(q\right)$.

Theorem 3. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identity for the second-order mock theta function $\mu \left(q\right)$ holds:

$$\begin{array}{cc}\hfill \mu \left(q\right)+4\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q^2;q^2)}_n^2{q}^{2n+2}}{{(q;q^2)}_{n+1}}}& ={\displaystyle \frac{q^2{J}_2^3{(-q^4/c_1;q^2)}_{\infty }^2{(c_1/q;q^2)}_{\infty }}{c_1{J}_1{J}_4^2}}{\displaystyle \frac{j(1/c_2;q^2)j(c_2;q^2)}{j(c_1/c_2;q^2)j(c_1{c}_2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1;q^2)}_n^2}}{(-1)}^n{q}^{n^2+2n}/c_1^n+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(24)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Remark 1. 

If we define the bilateral series associated with a mock theta function by ${\displaystyle B(M;q):=\sum _{n=-\infty }^{\infty }c(n;q)}$, where ${\displaystyle M\left(q\right):=\sum _{n=0}^{\infty }c(n;q)}$ is a mock theta function, we can regard the left-hand sides of identities (6), (13), and (24) as the bilateral series corresponding to the second-order mock theta functions $A\left(q\right)$, $B\left(q\right)$, and $\mu \left(q\right)$. It is interesting that we find that we can obtain different expansion identities by choosing $c_1$ and $c_2$ suitably in the above four identities.

Next, using the above obtained identities and known results, we obtain the following identities as corollaries. First, combining (6) and (24) with the identity [11] (3.28)

$$\begin{array}{c}\hfill 4A\left(q\right)+\mu (-q)={\displaystyle \frac{{\overline{J}}_{1,2}{J}_2^6}{J_1^3{J}_4^3}},\end{array}$$

(25)

and after some simplifications, we deduce the following transformation formula between the unilateral q-series and the bilateral q-series as Corollary 1.

Corollary 1. 

The following identity holds:

$$\begin{array}{cc}\hfill & {}_3\varphi _2\left[\begin{array}{ccc}q,& q,& q^2\\ & -q^3,& 0\end{array};q^2,q^2\right]+q{}_3\varphi _2\left[\begin{array}{ccc}-q^2,& -q^2,& q^2\\ & -q^3,& 0\end{array};q^2,q^2\right]\hfill \\ \hfill & ={\displaystyle \frac{(1+q)q^2{J}_4}{c_1{J}_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(q^5/c_1;q^2)}_{\infty }^2{j}^2(c_2;q^2)}{j(c_1/c_2;q^2)j(c_1{c}_2/q^2;q^2)}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n^2}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}\hfill \\ \hfill & +{\displaystyle \frac{q(1+q)J_1}{4J_4}}{\displaystyle \frac{{(-q^4/c_1;q^2)}_{\infty }^2{(c_1/q;q^2)}_{\infty }}{c_1}}{\displaystyle \frac{j(1/c_2;q^2)j(c_2;q^2)}{j(c_1/c_2;q^2)j(c_1{c}_2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1;q^2)}_n^2}}{(-1)}^n{q}^{n^2+2n}/c_1^n+\mathit{idem}(c_1;c_2)-{\displaystyle \frac{{\overline{J}}_{1,2}{J}_2^6}{4J_1^3{J}_4^3}},\hfill \end{array}$$

(26)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Recall that [18] (1.5):

$$\begin{array}{c}\hfill B\left(q\right)+B(-q)=2{\displaystyle \frac{J_4^5}{J_2^4}}.\end{array}$$

(27)

Then, plugging (27) into (14), we obtain the following theta identity as Corollary 2.

Corollary 2. 

We have

$${\displaystyle \frac{q}{c_1(1+c_1^2/q)}}{\displaystyle \frac{J_4^2{j}^2(c_2;q)j(-c_1^2/q^2;q^2)}{J_2^{3}j(c_1/c_2;q)j(c_1{c}_2/q;q)}}+\mathit{idem}(c_1;c_2)=2{\displaystyle \frac{J_4^5}{J_2^4}},$$

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Theorem 4. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identities for the third-order mock theta functions hold:

$$\begin{array}{cc}\hfill f\left(q\right)=-4\sum _{n=0}^{\infty }{(-q;q)}_n^2{q}^{n+1}+& {\displaystyle \frac{qj(c_2,1/c_2,-c_1/q,-c_1/q;q)}{c_1{J}_2^2{(-c_1/q;q)}_{\infty }^{2}j(c_1/c_2,c_1{c}_2;q)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^2/c_1,-q^2/c_1;q)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill \omega \left(q\right)=-\sum _{n=0}^{\infty }{(q;q^2)}_n^2{q}^{2n}& +{\displaystyle \frac{q^2{j}^2(c_1/q^3,c_2;q^2)}{c_1{J}_1^2{(c_1/q^3;q^2)}_{\infty }^{2}j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{2n^2+6n}/c_1^{2n}}{{(q^5/c_1;q^2)}_n^2}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill \varphi \left(q\right)=-2\sum _{n=0}^{\infty }{(-q^2;q^2)}_n{q}^{n+1}+& {\displaystyle \frac{qj(-c_1^2/q^2;q^2)j(c_2,1/c_2;q)}{c_1{J}_4{(-c_1^2/q^2;q^2)}_{\infty }j(c_1{c}_2,c_1/c_2;q)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^4/c_1^2;q^2)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill \psi \left(q\right)=-\sum _{n=0}^{\infty }{(-1)}^n{q}^n{(q;q^2)}_n+& {\displaystyle \frac{q^{2}j(c_1^2/q^3;q^2)j(c_2/q,c_2;q)}{c_1{J}_1{(c_1^2/q^3;q^2)}_{\infty }j(c_1/c_2,c_1{c}_2/q^2;q)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+4n}/c_1^{2n}}{{(q^5/c_1^2;q^2)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill \nu \left(q\right)=-\sum _{n=0}^{\infty }{(-q;q^2)}_n{q}^n+& {\displaystyle \frac{q{J}_1{J}_{4}j{(-c_1^2/q;q^2)}_{\infty }{j}^2(c_2;q)}{c_1{J}_2^2{(-c_1^2/q;q^2)}_{\infty }j(c_1/c_2,c_1{c}_2/q;q)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+3n}/c_1^{2n}}{{(-q^3/c_1^2;q^2)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill \xi \left(q\right)=1-2\sum _{n=0}^{\infty }{(q,q^5;q^6)}_n{q}^{6n+1}+& {\displaystyle \frac{2q^7{(q^7/c_1,q^{11}/c_1;q^6)}_{\infty }}{c_1{(q,q^5;q^6)}_{\infty }}}{\displaystyle \frac{j(c_2,c_2{q}^{12};q^6)}{j(c_1/c_2,1/c_1{c}_2;q^6)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{6n^2+6n}/c_1^{2n}}{{(q^7/c_1,q^{11}/c_1;q^6)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(33)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Proof. 

We verify the convergence conditions for each identity by checking $|bc/azq|<1$ with the respective parameter substitutions.

For (28), we replace $q^2$, b, c, z by q, $-q$, $-q$, $-q^{1/2}/a$ in (5). The convergence condition $|bc/azq|<1$ becomes $|(-q)(-q)/\left(a\right)(-q^{1/2}/a)\left(q^{1/2}\right)|=|q^2/q|=|q|<1$, which is satisfied. Letting $a\to \infty$, we obtain

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q;q)}_n^2}}=& {\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(-q^2/c_1,-q^2/c_1,1/c_2,c_{2}q,c_2,q/c_2;q)}_{\infty }}{{(c_1/c_2,q{c}_2/c_1,-q,-q,q/c_1{c}_2,c_1{c}_2;q)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^2/c_1,-q^2/c_1;q)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

(34)

Using (20) on the terms of negative index in the series on the left-hand side and Jacobi’s triple product on the right-hand side, and after some simplifications, we derive our desired result (28).

For identity (29), we apply (5) with $b=c=q^3$ and replace z by $-q^3/a$. The convergence condition is satisfied for $\left|q\right|<1$. Letting $a\to \infty$ and dividing through by ${(1-q)}^2$ yields (29).

Replace $q^2$, b, c, z by q, $iq$, $-iq$, $-q^{1/2}/a$ in (5), and then let $a\to \infty$. Then, we have

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}=& {\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(-q^4/c_1^2;q^2)}_{\infty }{(1/c_2,c_{2}q,c_2,q/c_2)}_{\infty }}{{(-q^2;q^2)}_{\infty }{(c_1/c_2,q{c}_2/c_1,q/c_1{c}_2,c_1{c}_2)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^4/c_1^2;q^2)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Multiplying the numerator and denominator by ${(q;q)}_{\infty }^2$, ${(q^2;q^2)}_{\infty }$, ${(-c_1^2/q^2;q^2)}_{\infty }$ on the right-hand side of the above identity, we obtain

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}=& {\displaystyle \frac{q}{c_1}}{\displaystyle \frac{{(q^2,-c_1^2/q^2,-q^3/c_1^2;q^2)}_{\infty }{(q,q,1/c_2,c_{2}q,c_2,q/c_2)}_{\infty }}{{(q^2,-c_1^2/q^2;q^2)}_{\infty }{(-q^2;q^2)}_{\infty }{(q,q,c_1/c_2,q{c}_2/c_1,q/c_1{c}_2,c_1{c}_2)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^4/c_1^2;q^2)}_n}}+\mathrm{idem}(c_1;c_2)\hfill \\ \hfill =& {\displaystyle \frac{qj(-c_1^2/q^2;q^2)j(c_2,1/c_2;q)}{c_1{J}_4{(-c_1^2/q^2;q^2)}_{\infty }j(c_1{c}_2,c_1/c_2;q)}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2+2n}/c_1^{2n}}{{(-q^4/c_1^2;q^2)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

On the right-hand side of the above identity, we use Jacobi’s triple product. Using (20) on the left-hand side again, we obtain

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}& =\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}+\sum _{n=-\infty }^{-1}{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\displaystyle \frac{q^{n^2}}{{(-q^2;q^2)}_n}}+2\sum _{n=0}^{\infty }{(-q^2;q^2)}_n{q}^{n+1}.\hfill \end{array}$$

(35)

This completes the proof of identity (30).

For (31), in (5) again replace $q^2$ with q, and then replace b, c, z with $q^{3/2}$, $-q^{3/2}$, $-q^{5/2}/a$, respectively. The convergence condition $|bc/azq|<1$ becomes $|\left(q^{3/2}\right)(-q^{3/2})/\left(a\right)(q^{5/2}/a)\left(q^{1/2}\right)|=|q^3/q^3|=1$ in the limit as $a\to \infty$, which is satisfied. And then let $a\to \infty$, and multiply through by $q/(1-q)$.

A similar application of (5): again replace $q^2$ with q, and then replace b, c, z with $i{q}^{3/2}$, $-i{q}^{3/2}$, $-q^{3/2}/a$, respectively. The convergence condition $|bc/azq|<1$ becomes $|\left(i{q}^{3/2}\right)(-i{q}^{3/2})/\left(a\right)(-q^{3/2}/a)\left(q^{1/2}\right)|=|q^3/q^2|=|q|<1$, which is satisfied. let $a\to \infty$, and divide through by $1+q$, leading to identity (32).

For (33), replace q, b, c, z with $q^3$, q, $q^5$, $-1/a{q}^3$ in (5). The convergence condition $|bc/azq|<1$ becomes $|\left(q\right)\left(q^5\right)/\left(a\right)(-1/a{q}^3)\left(q^3\right)|=|q^6|<1$, which is satisfied. And then let $a\to \infty$. Thus, we have

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{6n^2-6n}}{{(q,q^5;q^6)}_n}}=& {\displaystyle \frac{q^6}{c_1}}{\displaystyle \frac{{(q^7/c_1,q^{11}/c_1,1/q^6{c}_2,c_2{q}^{12},c_2,q^6/c_2;q^6)}_{\infty }}{{(c_1/c_2,q^6{c}_2/c_1,1/c_1{c}_2,q^6{c}_1{c}_2;q^6)}_{\infty }{(q,q^5;q^6)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{6n^2+6n}/c_1^{2n}}{{(q^7/c_1,q^{11}/c_1;q^6)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Multiplying both sides by $2q$ and breaking the left side into the positive terms and negative terms, we obtain

$$\begin{array}{cc}\hfill 2\sum _{n=1}^{\infty }& {\displaystyle \frac{q^{6n^2-6n+1}}{{(q,q^5;q^6)}_n}}+2\sum _{n=-\infty }^0{\displaystyle \frac{q^{6n^2-6n+1}}{{(q,q^5;q^6)}_n}}\hfill \\ \hfill =& 2\sum _{n=1}^{\infty }{\displaystyle \frac{q^{6n^2-6n+1}}{{(q,q^5;q^6)}_n}}+2\sum _{n=0}^{\infty }{(q,q^5;q^6)}_n{q}^{6n+1}\hfill \\ \hfill =& 2{\displaystyle \frac{q^7}{c_1}}{\displaystyle \frac{{(q^7/c_1,q^{11}/c_1,1/q^6{c}_2,c_2{q}^{12},c_2,q^6/c_2;q^6)}_{\infty }}{{(c_1/c_2,q^6{c}_2/c_1,1/c_1{c}_2,q^6{c}_1{c}_2;q^6)}_{\infty }{(q,q^5;q^6)}_{\infty }}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{6n^2+6n}/c_1^{2n}}{{(q^7/c_1,q^{11}/c_1;q^6)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Finally, adding 1 to both sides of the above identity and after some simplifications, we derive

$$\begin{array}{cc}\hfill 1+2\sum _{n=1}^{\infty }{\displaystyle \frac{q^{6n^2-6n+1}}{{(q,q^5;q^6)}_n}}=& 1-2\sum _{n=0}^{\infty }{(q,q^5;q^6)}_n{q}^{6n+1}\hfill \\ \hfill +& 2{\displaystyle \frac{q^7{(q^{11}/c_1;q^6)}_{\infty }}{c_1{(c_1/q;q^6)}_{\infty }}}{\displaystyle \frac{j(c_1/q,c_2,c_2{q}^{12};q^6)}{j(q,c_1/c_2,1/c_1{c}_2;q^6)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{q^{6n^2+6n}/c_1^{2n}}{{(q^7/c_1,q^{11}/c_1;q^6)}_n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

This completes the proof of identity (33). □

Theorem 5. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identities for the sixth-order mock theta functions hold:

$$\begin{array}{cc}\hfill {\psi }_6\left(q\right)+2{\psi }_{-}\left(q\right)& ={\displaystyle \frac{q^3{(c_1/q;q^2)}_{\infty }{(-q^4/c_1;q)}_{\infty }{j}^2(c_2;q^2)}{c_{1}j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1,-q^5/c_1;q^2)}_n}}{\displaystyle \frac{{(-1)}^n{q}^{n^2+4n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill \rho \left(q\right)+{\displaystyle \frac{1}{2}}\lambda \left(q\right)=& {\displaystyle \frac{q{(-c_1/q;q)}_{\infty }{(q^5/c_1^2;q^2)}_{\infty }{j}^2(c_2;q)}{2c_{1}j(c_1/c_2,c_1{c}_2/q;q)}}\hfill \\ \hfill \times & \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^2/c_1;q)}_n}{{(q^5/c_1^2;q^2)}_n}}{q}^{{\displaystyle \frac{n^2+3n}{2}}}/c_1^n+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill \sigma \left(q\right)+{\displaystyle \frac{1}{2}}\mu \left(q\right)=& {\displaystyle \frac{q^2{(-c_1/q;q)}_{\infty }{(q^5/c_1^2;q^2)}_{\infty }j(c_2/q,c_2;q)}{2c_{1}j(c_1/c_2,c_1{c}_2/q^2;q)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^2/c_1;q)}_n{q}^{\frac{n^2+5n}{2}}/c_1^n}{{(q^5/c_1^2;q^2)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill {\varphi }_6\left(q\right)+2{\varphi }_{-}\left(q\right)=& {\displaystyle \frac{q^2{(c_1/q;q^2)}_{\infty }{(-q^3/c_1;q)}_{\infty }j(1/c_2,c_2;q^2)}{c_{1}j(c_1/c_2,c_1{c}_2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n{(-1)}^n{q}^{n^2+2n}/c_1^n}{{(-q^3/c_1,-q^4/c_1;q^2)}_n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(39)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Proof. 

We verify the convergence conditions for each identity.

For identity (36), we replace a, b, c, z with q, $-q^2$, $-q^3$, $-q^2$ in (5). The convergence condition $|bc/azq|<1$ becomes $|(-q^2)(-q^3)/\left(q\right)(-q^2)\left(q\right)|=|q^5/q^4|=|q|<1$, which is satisfied. We obtain

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n{(-1)}^n{q}^{n^2+2n}}{{(-q^2;q)}_{2n}}}+\sum _{n=-\infty }^{-1}{\displaystyle \frac{{(q;q^2)}_n{(-1)}^n{q}^{n^2+2n}}{{(-q^2;q)}_{2n}}}\hfill \\ \hfill & ={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(c_1/q,-q^4/c_1,-q^5/c_1,q^2/c_2,c_2,c_2,q^2/c_2;q^2)}_{\infty }}{{(c_1/c_2,q^2{c}_2/c_1,-q^2,-q^3,q,q^4/c_1{c}_2,c_1{c}_2/q^2;q^2)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1,-q^5/c_1;q^2)}_n}}{\displaystyle \frac{{(-1)}^n{q}^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Making the substitution $n\to -n-1$ in the second term of the left-hand side and after some simplifications, we obtain

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n{(-1)}^n{q}^{n^2+2n}}{{(-q^2;q)}_{2n}}}+2\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q)}_{2n}{q}^n(1+q)}{{(q;q^2)}_{n+1}}}\hfill \\ \hfill & ={\displaystyle \frac{q^2}{c_1{(-q^2,-q^3;q^2)}_{\infty }{(q;q^2)}_{\infty }}}{\displaystyle \frac{j^2(c_2;q^2){(c_1/q;q^2)}_{\infty }{(-q^4/c_1)}_{\infty }}{j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1,-q^5/c_1;q^2)}_n}}{\displaystyle \frac{{(-1)}^n{q}^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Multiplying both sides of the above identity by ${\displaystyle \frac{q}{1+q}}$, we obtain

$$\begin{array}{cc}\hfill & \sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_n{(-1)}^n{q}^{{(n+1)}^2}}{{(-q;q)}_{2n+1}}}+2\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q;q)}_{2n}{q}^{n+1}}{{(q;q^2)}_{n+1}}}\hfill \\ \hfill & ={\displaystyle \frac{q^3}{c_1{(-q)}_{\infty }{(q;q^2)}_{\infty }}}{\displaystyle \frac{j^2(c_2;q^2){(c_1/q;q^2)}_{\infty }{(-q^4/c_1)}_{\infty }}{j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1,-q^5/c_1;q^2)}_n}}{\displaystyle \frac{{(-1)}^n{q}^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

Since

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{(-q)}_{\infty }{(q;q^2)}_{\infty }}}=1,\end{array}$$

(40)

and after some simplifications, we obtain

$$\begin{array}{cc}\hfill {\psi }_6\left(q\right)+2{\psi }_{-}\left(q\right)& ={\displaystyle \frac{q^3}{c_1}}{\displaystyle \frac{j^2(c_2;q^2){(c_1/q;q^2)}_{\infty }{(-q^4/c_1)}_{\infty }}{j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(q^3/c_1;q^2)}_n}{{(-q^4/c_1,-q^5/c_1;q^2)}_n}}{\displaystyle \frac{{(-1)}^n{q}^{n^2+4n}}{c_1^n}}+\mathrm{idem}(c_1;c_2).\hfill \end{array}$$

This completes the proof of (36).

The proofs of (37)–(39) follow similarly by appropriate substitutions in (5), making the substitution $n\to -n-1$ in the negative index terms, using (20), and applying Jacobi’s triple product and (40) where needed. □

Theorem 6. 

Suppose $\left|q\right|<1$ and the convergence conditions from Lemma 1 are satisfied. Then the following identities for the eighth-order mock theta functions hold:

$$\begin{array}{cc}\hfill & S_0\left(q\right)+2T_0\left(q\right)={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(-c_1/q,-q^4/c_1;q^2)}_{\infty }j(c_2,1/c_2;q^2)}{{(-q)}_{\infty }j(c_1/c_2,c_1{c}_2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(-q^4/c_1;q^2)}_n}}{\displaystyle \frac{q^{n^2+2n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill & S_1\left(q\right)+2T_1\left(q\right)={\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(-c_1/q,-q^4/c_1;q^2)}_{\infty }{j}^2(c_2;q^2)}{{(-q)}_{\infty }j(c_1/c_2,c_1{c}_2/q^2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(-q^4/c_1;q^2)}_n}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill & U_0\left(q\right)=-2\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q^4;q^4)}_n{q}^{2n+2}}{{(-q;q^2)}_{n+1}}}+{\displaystyle \frac{q^2}{c_1}}{\displaystyle \frac{{(-c_1/q;q^2)}_{\infty }{(-q^8/c_1^2;q^4)}_{\infty }j(c_2,1/c_2;q^2)}{{(-q^4;q^4)}_{\infty }{(-q;q^2)}_{\infty }j(c_1/c_2,c_1{c}_2;q^2)}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(-q^8/c_1^2;q^4)}_n}}{\displaystyle \frac{q^{n^2+2n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill & U_1\left(q\right)=-\sum _{n=0}^{\infty }{\displaystyle \frac{{(-q^2;q^4)}_n{q}^{2n+1}}{{(-q;q^2)}_{n+1}}}+{\displaystyle \frac{q^3{J}_1{J}_8{(-c_1/q;q^2)}_{\infty }{(-q^{10}/c_1^2;q^4)}_{\infty }}{c_1{J}_2{J}_{4}j(c_1/c_2;q^2)}}\hfill \\ \hfill & \times {\displaystyle \frac{j^2(c_2;q^2)}{j(c_1{c}_2/q^2;q^2)}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(-q^{10}/c_1^2;q^4)}_n}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & V_0\left(q\right)=-1-2\sum _{n=0}^{\infty }{\displaystyle \frac{{(q;q^2)}_{n+1}{(-1)}^{n+1}{q}^{{(n+1)}^2}}{{(-q;q^2)}_{n+1}}}+{\displaystyle \frac{2q^{2}j(c_2,1/c_2;q^2)}{c_{1}j(c_1/c_2,c_1{c}_2;q^2)}}\hfill \\ \hfill & \times {\displaystyle \frac{{(-c_1/q,q^3/c_1;q^2)}_{\infty }}{{(q^2;q^4)}_{\infty }}}\sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^3/c_1;q^2)}_n}}{\displaystyle \frac{q^{n^2+2n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(45)

$$\begin{array}{cc}\hfill & V_1\left(q\right)-V_1(-q)={\displaystyle \frac{q^3{j}^2(c_2;q^2){(-c_1/q,q^5/c_1;q^2)}_{\infty }}{c_{1}j(c_1/c_2,c_1{c}_2/q^2;q^2){(q^2;q^4)}_{\infty }}}\hfill \\ \hfill & \times \sum _{n=-\infty }^{\infty }{\displaystyle \frac{{(-q^3/c_1;q^2)}_n}{{(q^5/c_1;q^2)}_n}}{\displaystyle \frac{q^{n^2+4n}}{c_1^n}}+\mathit{idem}(c_1;c_2),\hfill \end{array}$$

(46)

where $c_1$ and $c_2$ are arbitrary non-zero complex numbers satisfying $c_1\ne c_2$.

Proof. 

The proofs follow similarly by appropriate substitutions in (5), making the substitution $n\to -n-1$ in the negative index terms, using (20), and applying Jacobi’s triple product and (40) where needed. □

3. Conclusions

We have successfully applied Slater’s three-term transformation formula for bilateral ${}_2\psi _2$ series to derive new identities for mock theta functions of orders 2, 3, 6, and 8. The most significant contribution is the development of bilateral series representations for second-order mock theta functions, which were notably absent from McLaughlin’s earlier investigation. Our results demonstrate that the bilateral series approach offers a systematic and powerful method for studying mock theta functions across different orders.

The identities derived in this work have immediate applications in several areas. First, in partition theory, mock theta functions enumerate restricted partitions, and our bilateral representations provide new generating functions for partition identities with specific congruence conditions. For instance, the parameters $c_1$ and $c_2$ in our formulas can be specialized to obtain partition statistics for various arithmetic progressions. Second, following Zwegers’ groundbreaking work [7], our identities contribute to the understanding of harmonic Maass forms, where mock theta functions appear as holomorphic projection components. The bilateral series representations we obtain facilitate the computation of Fourier coefficients and period integrals of these forms. Third, in the framework of fractional q-calculus, the bilateral ${}_2\psi _2$ series serve as explicit solutions to specific fractional q-difference equations arising in quantum calculus and mathematical physics.

Natural extensions of this research include several promising directions. First, developing analogous identities for mock theta functions of orders 5, 7, and 10 using higher-order bilateral transformations and generalizations of Slater’s formula would complete the catalog of classical mock theta functions. Second, investigating the geometric and number-theoretic interpretation of the free parameters $c_1$ and $c_2$ may reveal new families of mock theta identities and shed light on the modular properties of these functions. Third, applying our bilateral series approach to prove new congruences for partition functions and establishing connections with recent developments in quantum modular forms would extend the impact of these results. Finally, exploring computational aspects of our identities, particularly convergence rates for different parameter choices, could lead to efficient algorithms for numerical evaluation of mock theta functions and related special functions.