New double-sum expansions for certain Mock theta functions

Qiuxia Hu; Bilal Khan; Serkan Araci; Mehmet Acikgoz; Qiuxia Hu; Bilal Khan; Serkan Araci; Mehmet Acikgoz

doi:10.3934/math.2022948

AIMS Mathematics

2022, Volume 7, Issue 9: 17225-17235. doi: 10.3934/math.2022948

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Research article

New double-sum expansions for certain Mock theta functions

1.
Department of Mathematics, Luoyang Normal University, Luoyang, 471934, China
2.
School of Mathematical Sciences and Shanghai Key Laboratory of PMMP, East China Normal University, 500 Dongchuan Road, Shanghai 200241, China
3.
Department of Basic Sciences, Faculty of Engineering, Hasan Kalyoncu University TR-27410 Gaziantep, Turkey
4.
Department of Mathematics, Faculty of Science and Arts, Gaziantep University, TR-27310 Gaziantep, Turkey

Received: 06 June 2022 Revised: 18 July 2022 Accepted: 19 July 2022 Published: 22 July 2022
MSC : Primary 05A30, 30C45; Secondary 11B65, 47B38

The study of expansions of certain mock theta functions in special functions theory has a long and quite significant history. Motivated by recent correlations between $q$ -series and mock theta functions, we establish a new $q$ -series transformation formula and derive the double-sum expansions for mock theta functions. As an application, we state new double-sum representations for certain mock theta functions.

Keywords:

Citation: Qiuxia Hu, Bilal Khan, Serkan Araci, Mehmet Acikgoz. New double-sum expansions for certain Mock theta functions[J]. AIMS Mathematics, 2022, 7(9): 17225-17235. doi: 10.3934/math.2022948

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Abstract

1. Introduction

Throughout this paper, we use the standard $q$ -series notation in ^[5]:

$\begin{equation*} (a;q)_0 = 1, \ \ (a;q)_n = \prod\limits_{k = 1}^n(1-aq^{k-1}), n = 1, 2, 3, \cdots. \end{equation*}$

We also adopt the following compact notation for multiple $q$ -shifted factorials:

$\begin{equation*} (a_1, a_2, \cdots, a_m;q)_n = (a_1;q)_n(a_2;q)_n\cdots(a_m;q)_n, \end{equation*}$

where $n$ is an integer or $\infty$ .

The basic hypergeometric series $_r\phi_s$ is defined by

$\begin{equation*} _r\phi_s{\left[ \begin{array}{ccccc} a_1, & a_2, & \cdots, & a_r & \\ & b_1, & \cdots, & b_s & \end{array} ;q, z\right]} = \sum\limits_{n = 0}^\infty\frac{(a_1;q)_n(a_2;q)_n\cdots(a_r;q)_n}{ (q)_n(b_1;q)_n\cdots(b_s;q)_n} \ \left[(-1)^nq^{\binom{n}{2}}\right]^{1+s-r}z^n. \end{equation*}$

The theory of mock theta functions is a very important research area of the theory of the basic hypergeometric series. Mock theta function was first introduced by Ramanujan in his last letter to Hardy ^[24]. Ramanujan listed $17$ mock theta functions and called them orders of $3$ , $5$ and $7$ . But he did not say explicitly what he meant and did not also give exact definition of order. Until now, exact definition of order has still not been given. The study of mock theta functions has attracted many experts and scholars. Recently, Patkowski ^[21] gave some new expansions for Ramanujan's $10$ th-order, $7$ th-order, and $5$ th-order mock theta functions $\mathcal{F}_{2}(q^4)$ , $\phi(q^4)$ and $\chi_1(q^4)$ by establishing some new Bailey pairs. In ^[19], Lovejoy and Osburn used Bailey pairs and Bailey transformation to obtain many mock theta functions in terms of $q$ -hypergeometric double sums and gave connections to known single-sum mock theta functions. Then, Zhang and Li ^[25] derived some similar nice mock theta double sums by the same method on the previous basis. Patkowski ^[22] obtained double-sum expansions for mock theta functions of Andrew's third-order $\overline{\psi}_{1}(q)$ Ramanujan's $7$ th-order $\phi(q)$ , and $10$ th-order $\mathcal{F}_{2}(q)$ . Some more recent investigations on this subject can be found in ^[9,10,17,18].

As an example, double-sum expansion for $10$ th-order mock theta function $\mathcal{F}_{2}(q)$ is restated as follows [22,Theorem 2.2,(2.13)]:

$\begin{equation*} \sum\limits_{n\geq 0}\sum\limits_{n\geq j\geq 0}\frac{(-1)^jq^{2n^2+2n+j^2+j}}{ (-q;q)_{2n+1}(q^2;q^2)_{n-j}(q^2;q^2)_j(1-q^{2j+1})} = \mathcal{F}_{2}(q^2), \end{equation*}$

where $\mathcal{F}_{2}(q) = \sum\limits_{n\geq 0}\frac{q^{n^2+n}}{ (q^{n+1};q)_{n+1}}$ .

In this paper, we make use of the following mock theta functions.

The second-order mock theta functions (see ^[20]):

$\begin{align*} &A(q) = \sum\limits_{n = 0}^\infty \frac{q^{n+1}(-q^2;q^2)_n}{(q;q^2)_{n+1} }, \\ &B(q) = \sum\limits_{n = 0}^\infty\frac{q^n(-q;q^2)_n}{(q;q^2)_{n+1}}, \\ &\mu(q) = \sum\limits_{n = 0}^\infty\frac{(-1)^nq^{n^2}(q;q^2)_n}{ (-q^2;q^2)^2_n}. \end{align*}$

The third-order mock theta functions (see ^[7]):

$\begin{align*} &\phi(q) = \sum\limits_{n = 0}^\infty\frac{q^{n^2}}{(-q^2;q^2)_n}, \\ & \psi(q) = \sum\limits_{n = 1}^\infty\frac{q^{n^2}}{(q;q^2)_{n}}, \\ &\nu(q) = \sum\limits_{n = 0}^\infty\frac{q^{n^2+n}}{(-q;q^2)_{n+1}}. \end{align*}$

The sixth-order mock theta functions (see ^[3]):

$\begin{align*} &\psi_6(q) = \sum\limits_{n = 1}^\infty\frac{(-1)^{n-1}q^{n^2}(q;q^2)_{n-1} }{(-q;q)_{2n-1}}, \\ &\phi_6(q) = \sum\limits_{n = 0}^\infty\frac{(-1)^nq^{n^2}(q;q^2)_n}{ (-q;q)_{2n}}, \\ &\rho(q) = \sum\limits_{n = 0}^\infty\frac{q^{n(n+1)/2}(-q;q)_n}{ (q;q^2)_{n+1}}, \\ &\sigma(q) = \sum\limits_{n = 0}^\infty q^{(n+2)(n+1)/2}\frac{(-q;q)_n}{ (q;q^2)_{n+1}}. \end{align*}$

For more details about mock theta functions, the readers can refer to ^{[1,2,3,4,6,7,23,24]}.

Based on the above research results, we continue to do some research on double-sum expansions for mock theta functions in this paper. The rest of this paper is arranged as follows. In Section $2$ , we first provide a new $q$ -series transformation formula in terms of series rearrangement method. And then as applications, some new double sums for certain mock theta functions are given.

2. Main results

In this section, we give some double-sum representations for certain mock theta functions. First, in the following Proposition 1, we establish a new $q$ -series transformation formula by means of $q$ -series rearrangement.

Proposition 1. For $|q| < 1$ and $|\alpha ab| < 1$ , we have that

$\begin{align} \sum\limits_{n = 0}^\infty&\frac{(q^2/a, -\lambda/q;q^2)_n(\alpha ab)^n}{(\alpha q^2, \lambda;q^2)_n} \\ & = \frac{(\alpha a, \alpha b;q^2)_\infty}{(\alpha q^2, \alpha ab/q^2;q^2)_\infty } \sum\limits_{n = 0}^\infty\frac{(1-\alpha q^{4n})(\alpha, q^2/a, q^2/b;q^2)_n(-\alpha ab/q^2)^nq^{n^2-n}}{(1-\alpha)(q^2, \alpha a, \alpha b;q^2)_n} \\ &\times\sum\limits_{k = 0}^n\frac{(q^{-2n}, \alpha q^{2n}, -\lambda/q;q^2)_kq^{4k}}{ (q^2/b, \alpha q^2, \lambda;q^2)_k}. \end{align}$

(2.1)

Proof. In terms of series rearrangement, the right-hand side of (2.1) equals that

$\begin{align} &\sum\limits_{k = 0}^\infty\sum\limits_{n = k}^\infty\frac{(1-\alpha q^{4n})(\alpha, q^2/a, q^2/b;q^2)_n(-\alpha ab/q^2)^nq^{n^2-n}}{(1-\alpha)(q^2, \alpha a, \alpha b;q^2)_n}\frac{(q^{-2n}, \alpha q^{2n}, -\lambda/q;q^2)_kq^{4k}}{(q^2/b, \alpha q^2, \lambda;q^2)_k}\\ & = \sum\limits_{k = 0}^\infty\sum\limits_{n = 0}^\infty\frac{(1-\alpha q^{4n+4k})(\alpha, q^2/a, q^2/b;q^2)_{n+k}(-\alpha ab/q^2)^{n+k}q^{(n+k)^2-(n+k)}}{(1-\alpha)(q^2, \alpha a, \alpha b;q^2)_{n+k}}\\ &\times\frac{(q^{-2(n+k)}, \alpha q^{2(n+k)}, -\lambda/q;q^2)_kq^{4k}}{(q^2/b, \alpha q^2, \lambda;q^2)_k}\\ & = \sum\limits_{k = 0}^\infty\frac{(\alpha, q^2/a, -\lambda/q;q^2)_k}{(q^2, \alpha a, \alpha b, \alpha q^2, \lambda;q^2)_k} (-\alpha ab)^kq^{k^2+k}\sum\limits_{n = 0}^\infty\frac{(1-\alpha q^{4n+4k})(\alpha q^{2k}, q^{2k+2}/a, q^{2k+2}/b;q^2)_n} {(1-\alpha)(q^{2k+2}, \alpha aq^{2k}, \alpha bq^{2k};q^2)_n}\\ &\times(q^{-2(n+k)}, \alpha q^{2(n+k)};q^2)_k(-\alpha ab)^nq^{n^2+2nk-3n}\\ & = \sum\limits_{k = 0}^\infty\frac{(1-\alpha q^{4k})(q^2/a, -\lambda/q;q^2)_k(\alpha;q^2)_{2k}}{(1-\alpha)(\alpha a, \alpha b, \alpha q^2, \lambda;q^2)_k} (\alpha ab)^k\sum\limits_{n = 0}^\infty\frac{(\alpha q^{4k};q^2)_n(1-\alpha q^{4n+4k})}{(q^2;q^2)_n(1-\alpha q^{4k})}\\ &\times\frac{(q^{2k+2}/a, q^{2k+2}/b;q^2)_n}{(\alpha aq^{2k}, \alpha bq^{2k};q^2)_n}(-\frac{\alpha ab}{q^2})^nq^{n^2-n}. \end{align}$

(2.2)

Let $q\rightarrow q^2, a\rightarrow \alpha q^{4k}, b\rightarrow q^{2k+2}/a, c\rightarrow q^{2k+2}/b, d\rightarrow \infty$ in the following sum of a very-well-poised $_6\phi_5$ series [5,Ⅱ.20]

$\begin{align} &_6\phi_5{\left[\begin{array}{cccccc}a, &qa^{1/2}, &-qa^{1/2}, &b, &c, &d\\ &a^{1/2}, &-a^{1/2}, &aq/b, &aq/c, &aq/d\end{array};q, aq/bcd\right]}\\ & = \frac{(aq, aq/bc, aq/bd, aq/cd;q)_\infty}{(aq/b, aq/c, aq/d, aq/bcd;q)_\infty}. \end{align}$

(2.3)

Then the second term of the right-hand side of (2.2) gives

$\begin{align*} \frac{(\alpha q^{4k+2}, \alpha ab/q^2;q^2)_\infty}{(\alpha aq^{2k}, \alpha bq^{2k};q^2)_\infty} = \frac{(\alpha q^2, \alpha ab/q^2;q^2)_\infty}{(\alpha a, \alpha b;q^2)_\infty}\frac{(\alpha a, \alpha b;q^2)_k}{(\alpha q^2;q^2)_{2k}}. \end{align*}$

The right-hand side of (2.2) yields

$\begin{align*} \frac{(\alpha q^2, \alpha ab/q^2;q^2)_\infty}{(\alpha a, \alpha b;q^2)_\infty}&\sum\limits_{k = 0}^\infty\frac{(1-\alpha q^{4k})(q^2/a, q^2/b, -\lambda/q;q^2)_k(\alpha;q^2)_{2k}}{(1-\alpha)(\alpha a, \alpha b, q^2/b, \alpha q^2, \lambda;q^2)_k} (\alpha ab)^k\frac{(\alpha a, \alpha b;q^2)_k}{(\alpha q^2;q^2)_{2k}}\\ & = \frac{(\alpha q^2, \alpha ab/q^2;q^2)_\infty}{(\alpha a, \alpha b;q^2)_\infty} \sum\limits_{k = 0}^\infty\frac{(q^2/a, -\lambda/b;q^2)_k}{(\alpha q^2, \lambda;q^2)_k}(\alpha ab)^k. \end{align*}$

After some simplifications, we derive our desired result. This completes the proof.

Next, as applications of the identity (2.1), in the following theorems we give some new double-sum representations for certain mock theta functions.

Theorem 1. The following double-sum representation for the second-ordermock theta function $A(q)$ is true:

$\begin{align} A(q) = &-q^2\frac{(-q;q^2)^2_\infty}{(q;q^2)^2_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{ (-1)^{n+k}q^{n^2+k^2-2n+3k-2nk}(1-q^{4n+1})(q;q^2)_{n+k}(-q^2;q^2)^2_n}{ (1-q)(q^2;q^2)_{n-k}(-q;q^2)^2_n(-q^2;q^2)_k(q;q^2)_{k+1}}. \end{align}$

(2.4)

Proof. Let $a = b = -1, \alpha = q, \lambda = 0$ in (2.1). Then, we derive that

$\begin{align*} &\sum\limits_{n = 0}^\infty\frac{(-q^2;q^2)_nq^n}{(q^3;q^2)_n}\\ = &\frac{(-q, -q;q^2)_\infty}{(q^3, q^{-1};q^2)_\infty}\sum\limits_{n = 0}^\infty\frac{(1-q^{4n+1})(q, -q^2, -q^2;q^2)_n}{(1-q)(q^2, -q, -q;q^2)_n}(-1)^nq^{n^2-2n}\\ &\times\sum\limits_{k = 0}^n\frac{(q^{-2n}, q^{2n+1};q^2)_kq^{4k}}{(-q^2, q^3;q^2)_k}. \end{align*}$

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

$\begin{align*} \sum\limits_{n = 0}^\infty&\frac{(-q^2;q^2)_nq^{n+1}}{(q;q^2)_{n+1}}\\ = &-\frac{q^2(-q;q^2)^2_\infty}{(1-q)(q;q^2)^2_\infty} \sum\limits_{n = 0}^\infty\frac{1-q^{4n+1}}{1-q} \frac{(q;q^2)_n(-q^2;q^2)^2_n}{(q^2;q^2)_n(-q;q^2)^2_n}(-1)^nq^{n^2-2n}\\ &\times\sum\limits_{k = 0}^n\frac{(q^{-2n}, q^{2n+1};q^2)_kq^{4k}}{(-q^2, q^3;q^2)_k}\\ = &-q^2\frac{(-q;q^2)^2_\infty}{(q;q^2)^2_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{(-1)^{n+k}q^{n^2+k^2-2n+3k-2nk}(1-q^{4n+1})(q;q^2)_{n+k}(-q^2;q^2)^2_n}{(1-q)(q^2;q^2)_{n-k}(-q;q^2)^2_n(-q^2;q^2)_k(q;q^2)_{k+1}}. \end{align*}$

This completes the proof of the identity (2.4).

By taking $a = -q, b = -q^{-1}, \alpha = q, \lambda = 0$ in (2.1), we deduce the following identity

Corollary 1. The following double-sum representation for the second-order mock thetafunction $B(q)$ is true:

$\begin{equation} B(q) = -2q\frac{(-q^2;q^2)^2_\infty}{(q;q^2)^2_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{ (-1)^{n+k}q^{n^2+k^2-2n+3k-2nk}(1-q^{4n+1})(q;q^2)_{n+k}(-q;q^2)_n(-q;q^2)_{n+1} }{(1-q)(q^2;q^2)_{n-k}(-q^2, -1;q^2)_n(q^2;q^4)_{k+1}}. \end{equation}$

(2.5)

Taking $\alpha = -1, \lambda = -q^2, b = -q^{-1}, a\rightarrow 0$ in (2.1), we can also attain the identity (2.6) for the second-order mock theta function $\mu(q)$

Corollary 2. The following double-sum representation for the second-order mock thetafunction $\mu(q)$ is true:

$\begin{array}{l} \mu(q) = \frac{(q^{-1};q^2)_\infty}{2(-q^2;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}(1+q^{4n+1})\\ \frac{ (-1)^kq^{2n^2+k^2-3n+3k-2nk}(-1;q^2)_{n+k}(-q^3;q^2)_n(q^2;q^4)_k}{ (q^2;q^2)_{n-k}(q^{-1};q^2)_n(-q^2;q)_{2k}(-q;q)_{2k}}. \end{array}$

(2.6)

For certain third-order mock theta functions, we can also gain the similar conclusions as follows.

Theorem 2. The following double-sum representations for the third-ordermock theta functions $\phi (q), \psi (q), \nu (q)$ hold true:

$\begin{array}{l} \phi (q) = -\frac{(1+q)(-q;q^{2})_{\infty }}{(q;q^{2})_{\infty }}\sum\limits_{n\geq 0}\sum\limits_{n\geq k\geq 0}\frac{{}}{{}}\\ \frac{ (-1)^{n+k}q^{2n^{2}+k^{2}-3n+3k-2nk}(1-q^{4n-1})(q^{-1};q^{2})_{n+k}(-q^{2};q^{2})_{n} }{(1-q)(q^{2};q^{2})_{n-k}(-q^{-1};q^{2})_{n}(-q^{2};q^{2})_{k}^{2}} \end{array}$

(2.7)

$\begin{align} \psi (q) = & \frac{q(-q;q^{2})_{\infty }}{(q;q^{2})_{\infty }}\sum\limits_{n\geq 0}\sum\limits_{n\geq k\geq 0}\frac{ (-1)^{n+k}q^{2n^{2}+k^{2}-n+3k-2nk}(1-q^{4n+1})(q;q^{2})_{n+k}(-q^{2};q^{2})_{n} }{(q^{2};q^{2})_{n-k}(-q;q^{2})_{n}(-q^{2};q^{2})_{k}(q;q^{2})_{k+1}} \end{align}$

(2.8)

$\begin{align} \nu (q) = & \frac{2(-q^{2};q^{2})_{\infty }}{(-q;q^{2})_{\infty }}\sum\limits_{n\geq 0}\sum\limits_{n\geq k\geq 0}\frac{ (-1)^{n+k}q^{2n^{2}+k^{2}-2n+3k-2nk}(1+q^{4n+1})(-q;q^{2})_{n+k}(q^{3};q^{2})_{n} }{(1+q)(q^{2};q^{2})_{n-k}(-1;q^{2})_{n}(q^{6};q^{4})_{k}}. \end{align}$

(2.9)

Proof. For (2.7), taking $\alpha = q^{-1}, \lambda = -q^2, b = -1, a\rightarrow 0$ in (2.1), we attain that

$\begin{align*} &\sum\limits_{n = 0}^\infty\frac{q^{n^2}}{(-q^2;q^2)_n}\\ & = \frac{(-q^{-1};q^2)_\infty}{(q;q^2)_\infty} \sum\limits_{n = 0}^\infty\frac{1-q^{4n-1}}{1-q^{-1}}\frac{(q^{-1}, -q^2;q^2)_n}{(q^2, -q^{-1};q^2)_n}(-1)^nq^{2n^2-3n} \sum\limits_{k = 0}^n\frac{(q^{-2n}, q^{2n-1};q^2)_k}{(-q^2, -q^2;q^2)_k}q^{4k}\\ & = -\frac{(1+q)(-q;q^2)_\infty}{(q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{1-q^{4n-1}}{1-q} \frac{(-1)^{n+k}q^{2n^2+k^2-3n+3k-2nk}(q^{-1};q^2)_{n+k}(-q^2;q^2)_n}{(q^2;q^2)_{n-k}(-q^{-1};q^2)_n(-q^2;q^2)^2_k}. \end{align*}$

This completes the proof of the identity (2.7).

For (2.8), set $\alpha = q, \lambda = 0, b = -1, a\rightarrow 0$ in (2.1). We deduce that

$\begin{align*} &\sum\limits_{n = 0}^\infty\frac{q^{n^2+2n}}{(q^3;q^2)_n}\nonumber\\ = &\frac{(-q;q^2)_\infty}{(q^3;q^2)_\infty} \sum\limits_{n = 0}^\infty\frac{(1-q^{4n+1})(q, -q^2;q^2)_n}{(1-q)(q^2, -q;q^2)_n}(-1)^nq^{2n^2-n} \sum\limits_{k = 0}^n\frac{(q^{-2n}, q^{2n+1};q^2)_k}{(-q^2, q^3;q^2)_k}q^{4k}\nonumber\\ & = \frac{(-q;q^2)_\infty}{(q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}(1-q^{4n+1})\frac{(q;q^2)_{n+k}(-q^2;q^2)_n}{(q^2;q^2)_{n-k}(-q;q^2)_n}\nonumber\\ &\times\frac{(-1)^{n+k}q^{2n^2+k^2-n+3k-2nk}}{(-q^2, q^3;q^2)_k}. \end{align*}$

Multiplying both sides of the above identity by $\frac{q}{1-q}$ , we get the following double-sum representation for the third-order mock theta function $\psi(q)$ :

$\begin{align*} \psi(q) = \sum\limits_{n = 0}^\infty\frac{q^{n^2+2n+1}}{(q;q^2)_{n+1}}& = \frac{q(-q;q^2)_\infty}{(q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}(1-q^{4n+1})\frac{(q;q^2)_{n+k}(-q^2;q^2)_n}{(q^2;q^2)_{n-k}(-q;q^2)_n}\nonumber\\ &\times\frac{(-1)^{n+k}q^{2n^2+k^2-n+3k-2nk}}{(-q^2;q^2)_k(q;q^2)_{k+1}}. \end{align*}$

This completes the proof of the identity (2.8).

For (2.9), taking $\alpha = -q, \lambda = 0, b = q^{-1}, a\rightarrow 0$ in (2.1), we obtain that

$\begin{align} \sum\limits_{n = 0}^\infty\frac{q^{n^2+n}}{(-q^3;q^2)_n}& = \frac{(-1;q^2)_\infty}{(-q^3;q^2)_\infty} \sum\limits_{n = 0}^\infty\frac{(1+q^{4n+1})(-q, q^3;q^2)_n(-1)^nq^{2n^2-2n}}{(1+q)(q^2, -1;q^2)_n}\\ &\times\sum\limits_{k = 0}^n\frac{(q^{-2n}, -q^{2n+1};q^2)_kq^{4k}}{(q^3, -q^3;q^2)_k}. \end{align}$

(2.10)

Multiplying both sides of the identity (2.10) by $\frac{1}{1+q}$ and after applications as seen above, we get our desired result (2.7). Thus we complete the proof of Theorem 2 by obtaining the Eq (2.9).

Theorem 3. The following double-sum representations for the third-order mock thetafunctions $\psi_6(q), \phi_6(q), \rho(q), \sigma(q)$ hold true:

$\begin{align} \psi_6(q)& = \frac{q(q;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{ (-1)^kq^{2n^2+k^2-n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^4;q^4)_n(q^2;q^4)_k } {(q^2;q^2)_{n-k}(q;q)_{2n}(-q;q)_{2k}(-q;q)_{2k+1}} \end{align}$

(2.11)

$\begin{array}{l}\phi_6(q) = \frac{(q^{-1};q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\\ \frac{ (-1)^kq^{2n^2+k^2-3n+3k-2nk}(1+q^{4n-1})(-q^{-1};q^2)_{n+k}(-q^2;q^2)_n(q^2;q^4)_k }{(1+q^{-1})(q^2;q^2)_{n-k}(q^{-1};q^2)_n(-q;q)^2_{2k}} \end{array}$

(2.12)

$\begin{array}{l} \rho(q^2) = \frac{2(-q^2;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \\ \frac{ (-1)^{n+k}q^{2n^2+k^2-2n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^3;q^2)_n(q^4;q^4)_k }{(q^2;q^2)_{n-k}(-1;q^2)_n(q^2;q)_{2k}(q^2;q^4)_{k+1}} \end{array}$

(2.13)

$\begin{array}{l} \sigma(q^2) = \frac{q^2(-q^2;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \\ \frac{ (-1)^{n+k}q^{2n^2+k^2+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^2;q^4)_n(q^4;q^4)_k }{(q^2;q^2)_{n-k}(-q;q)_{2n}(q^2;q^4)_{k+1}(q;q)_{2k} }. \end{array}$

(2.14)

Proof. For (2.11), set $\alpha = -q, \lambda = -q^2, b = -1, a\rightarrow 0$ in (2.1). Then we obtain

$\begin{align*} \sum\limits_{n = 0}^\infty\frac{(q;q^2)_n(-1)^nq^{n^2+2n}}{(-q^3, -q^2;q^2)_n} & = \frac{(q;q^2)_\infty}{(-q^3;q^2)_\infty}\sum\limits_{n = 0}^\infty\frac{(1+q^{4n+1})(-q, -q^2;q^2)_nq^{2n^2-n}}{(1+q)(q^2, q;q^2)_n}\nonumber\\ &\times\sum\limits_{k = 0}^n\frac{(q^{-2n}, -q^{2n+1}, q;q^2)_kq^{4k}}{(-q^2, -q^3, -q^2;q^2)_k}. \end{align*}$

After suitable simplifications, we derive

$\begin{align*} &\sum\limits_{n = 0}^\infty\frac{(q;q^2)_n(-1)^nq^{n^2+2n}}{(-q^2;q)_{2n}}\\ & = \frac{(q;q^2)_\infty}{(-q^3;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{(-1)^kq^{2n^2+k^2-n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^4;q^4)_n(q^2;q^4)_k} {(q^2;q^2)_{n-k}(q;q)_{2n}(-q;q)_{2k}(-q;q)_{2k+1}}. \end{align*}$

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

$\begin{align*} &\sum\limits_{n = 0}^\infty\frac{(q;q^2)_n(-1)^nq^{n^2+2n+1}}{(-q;q)_{2n+1}}\\ & = \frac{q(q;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0}\frac{(-1)^kq^{2n^2+k^2-n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^4;q^4)_n(q^2;q^4)_k} {(q^2;q^2)_{n-k}(q;q)_{2n}(-q;q)_{2k}(-q;q)_{2k+1}}. \end{align*}$

This completes the proof of (2.11).

For (2.12), set $\alpha = -q^{-1}, \lambda = -q^2, b = -1, a\rightarrow 0$ in (2.1). After some simplifications, we attain the identity (2.12). Therefore, we omit the proof.

For (2.13), taking $\alpha = -q, \lambda = q^3, b = q^{-1}, a\rightarrow 0$ in (2.1) and after some simplifications, we get that

$\begin{align*} &\sum\limits_{n = 0}^\infty \frac{q^{n^2+n}(-q^2;q^2)_n}{(q^6;q^4)_n}\\ & = \frac{(-1;q^2)_\infty}{(-q^3;q^2)_\infty} \sum\limits_{n = 0}^\infty \frac{(1+q^{4n+1})(-q, q^3;q^2)_nq^{2n^2-2n}}{(1+q)(q^2, -1;q^2)_n} \sum\limits_{k = 0}^n \frac{(q^{-2n}, -q^{2n+1}, -q^2;q^2)_kq^{4k}}{(q^3, -q^3, q^3;q^2)_k}\\ & = \frac{2(-q^2;q^2)_\infty}{(-q^3;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \frac{(-1)^{n+k}q^{2n^2+k^2-2n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^3;q^2)_n(-q^2;q^2)_k}{(1+q)(q^2;q^2)_{n-k}(-1;q^2)_n(q^3;q^2)_k(q^6;q^4)_k}. \end{align*}$

Multiplying both sides of the above identity by $\frac{1}{1-q^2}$ and through a mass of complex computations, we find the following double-sum representation for the sixth-order mock theta function $\rho(q)$ :

$\begin{align*} &\sum\limits_{n = 0}^\infty \frac{q^{n^2+n}(-q^2;q^2)_n}{(q^2;q^4)_{n+1}}\\ & = \frac{2(-q^2;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \frac{(-1)^{n+k}q^{2n^2+k^2-2n+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^3;q^2)_n(q^4;q^4)_k}{(q^2;q^2)_{n-k}(-1;q^2)_n(q^2;q)_{2k}(q^2;q^4)_{k+1}}. \end{align*}$

This completes the proof of the identity (2.13). For (2.14), take $\alpha = -q, \lambda = q^3, b = q, a\rightarrow 0$ in (2.1). Then we have that

$\begin{align*} &\sum\limits_{n = 0}^\infty \frac{q^{n^2+3n}(-q^2;q^2)_n}{(q^6;q^4)_n}\\ & = \frac{(-q^2;q^2)_\infty}{(-q^3;q^2)_\infty}\sum\limits_{n = 0}^\infty\frac{(1+q^{4n+1})(-q, q;q^2)_n(-1)^nq^{2n^2}}{(1+q)(q^2, -q^2;q^2)_n} \sum\limits_{k = 0}^n\frac{(q^{-2n}, -q^{2n+1}, -q^2;q^2)_k}{(q, -q^3, q^3;q^2)_k}q^{4k}\\ & = \frac{(-q^2;q^2)_\infty}{(-q^3;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \frac{(1+q^{4n+1})(-q;q^2)_{n+k}(q^2;q^4)_n(q^4;q^4)_k(-1)^{n+k}q^{2n^2+k^2+3k-2nk}}{(1+q)(q^2;q^2)_{n-k}(-q;q)_{2n}(q^6;q^4)_k(q;q)_{2k}}. \end{align*}$

In order to get our desired result, we multiply both sides of the above identity by $\frac{q^2}{1-q^2}$ . Thus, we have that

$\begin{align*} &\sum\limits_{n = 0}^\infty \frac{q^{n^2+3n+2}(-q^2;q^2)_n}{(q^2;q^4)_{n+1}}\\ & = \frac{q^2(-q^2;q^2)_\infty}{(-q;q^2)_\infty} \sum\limits_{n\geq 0}\sum\limits_{n \geq k\geq 0} \frac{(-1)^{n+k}q^{2n^2+k^2+3k-2nk}(1+q^{4n+1})(-q;q^2)_{n+k}(q^2;q^4)_n(q^4;q^4)_k}{(q^2;q^2)_{n-k}(-q;q)_{2n}(q^2;q^4)_{k+1}(q;q)_{2k}}. \end{align*}$

3. Conclusions

Basic (or $q$ -) polynomials and (or $q$ -) hypergeometric functions are particularly applicable in many diverse areas of mathematics, physics and other sciences. Here in our present investigation, we have motivated by the work of Patkowski ^[22] and have studied the double-sum expansions for mock theta functions. Then in terms of series rearrangement method, we have established a new $q$ -series transformation formula. As an applications, we have derived some new double-sum representations for certain mock theta functions.

Studies of the special functions and $q$ -polynomials are widely using in different branches of mathematics. For example in ^[11], by make use of certain $q$ -Chebyshev Polynomials, certain subclasses of analytic and bi-univalent functions have been defined systematically. Just like the class defined in ^[11], one can define a similar class by taking this newly established $q$ -series transformation formula instead of $q$ -Chebyshev Polynomials. These kind of investigations can also be found in ^{[8,12,13,14,15,16]}.

Conflict of interest

All authors declare no conflicts of interest in this paper.

Acknowledgments

The first-named author (Q. Hu) is supported by the Natural Science Foundation of Henan Province (No. 212300410211), the National Natural Science Foundation of China (Grant No. 12101287) and the National Project Cultivation Foundation of Luoyang Normal University (No. 2020-PYJJ-011).

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