Some New Transformation Formulas for q-Series through the Bailey Transform

Lijun Hao, Liangliang Xu

School of Science, Zhejiang Sci-Tech University, Hangzhou, China.

DOI: 10.4236/apm.2023.1310045                

Abstract

In the literature, the Bailey transform has many applications in basic hypergeometric series. In this paper, we derive many new transformation formulas for q-series by means of the Bailey transform. Meanwhile, We also obtain some new terminated identities. Furthermore, we establish a companion identity to the Rogers-Ramanujan identity labelled by number (23) on Slater’s list.

Keywords

q-Series, Bailey Transform, Transformation Formulas, Rogers-Ramanujan Identities

1. Introduction

Throughout this paper, $a\mathrm{,}x$ and q are complex number with $\left|q\right|<1$ . Here and in what follows, we adopt the standard q-series notation [1] . For any positive integer n,

${\left(a;q\right)}_0:=1,{\left(a;q\right)}_n:={\displaystyle \underset{k=0}{\overset{n-1}{\prod }}}\left(1-a{q}^k\right),{\left(a;q\right)}_{\infty }:={\displaystyle \underset{k=0}{\overset{\infty }{\prod }}}\left(1-a{q}^k\right),$

${\left(a_1\mathrm{,}{a}_2\mathrm{,}{a}_3\mathrm{,}\cdots \mathrm{,}{a}_m\mathrm{<mo>\; ;\; </mo>}q\right)}_n\mathrm{<mo>\; :\; </mo>}={\left(a_1\mathrm{<mo>\; ;\; </mo>}q\right)}_n{\left(a_2\mathrm{<mo>\; ;\; </mo>}q\right)}_n{\left(a_3\mathrm{<mo>\; ;\; </mo>}q\right)}_n\cdots {\left(a_m\mathrm{<mo>\; ;\; </mo>}q\right)}_n\mathrm{,}$

${\left(a_1\mathrm{,}{a}_2\mathrm{,}{a}_3\mathrm{,}\cdots \mathrm{,}{a}_m\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }\mathrm{<mo>\; :\; </mo>}={\left(a_1\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }{\left(a_2\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }{\left(a_3\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }\cdots {\left(a_m\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }\mathrm{.}$

For convenience, we use ${\left(a\right)}_n$ to denote ${\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_n$ . We will often use basic properties without reference, such as

${\left(a;q\right)}_{n+k}={\left(a;q\right)}_n{\left(a{q}^n;q\right)}_k,{\left(a;q\right)}_{\infty }={\left(a;q\right)}_n{\left(a{q}^n;q\right)}_{\infty }.$

The q-binomial coefficient is given for any nonnegative integers M and N by

${\left[\begin{array}{c}N\\ M\end{array}\right]}_q\mathrm{<mo>\; :\; </mo>}=\left[\begin{array}{c}N\\ M\end{array}\right]=(\begin{array}{ll}\frac{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_N}{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_M{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_{N-M}}\mathrm{,}\hfill & \text{if}N\ge M\text{,}\hfill \\ \mathrm{0,}\hfill & \text{otherwise}\text{.}\hfill \end{array}$

Following Gasper and Rahman [1] , the bilateral basic hypergeometric series is defined by

${}_r{\psi }_s\left[\begin{array}{l}{a}_1,a_2,\cdots ,a_r\hfill \\ b_1,b_2,\cdots ,b_s\hfill \end{array};q,z\right]={\displaystyle \underset{n=-\infty }{\overset{\infty }{\sum }}}\frac{{\left(a_1,a_2,\cdots ,a_r;q\right)}_n}{{\left(b_1,b_2,\cdots ,b_s;q\right)}_n}{\left\{{\left(-1\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\right\}}^{s-r}{z}^n,$

and the unilateral basic hypergeometric series is defined by

${}_r{\varphi }_s\left[\begin{array}{l}{a}_1,a_2,\cdots ,a_r\hfill \\ b_1,b_2,\cdots ,b_s\hfill \end{array};q,z\right]={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(a_1,a_2,\cdots ,a_r;q\right)}_n}{{\left(q,b_1,b_2,\cdots ,b_s;q\right)}_n}{\left\{{\left(-1\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\right\}}^{1+s-r}{z}^n.$

The Ramanujan’s ${}_1{\psi }_1$ sum ( [1] , Appendix (II. 29)), and the sum of a ${}_1{\varphi }_1$ series, ( [1] , Appendix (II.5)) are stated as follows.

${}_1{\psi }_1\left[\begin{array}{l}a\hfill \\ b\hfill \end{array};q,z\right]=\frac{{\left(q,b/a,az,q/az;q\right)}_{\infty }}{{\left(b,q/a,z,b/az;q\right)}_{\infty }},$ (1.1)

${}_1{\varphi }_1\left[\begin{array}{l}a\hfill \\ c\hfill \end{array}\mathrm{<mo>\; ;\; </mo>}q\mathrm{,}\frac{c}{a}\right]=\frac{{\left(c/a\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}{{\left(c\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}\mathrm{.}$ (1.2)

Among the other formulas needed for this paper, we have separately documented the q-binomial formula and its consequence ( [1] (1.3.2)) in the sequel,

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(a\mathrm{<mo>\; ;\; </mo>}q\right)}_n}{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_n}{x}^n=\frac{{\left(ax\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}{{\left(x\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }}\mathrm{.}$ (1.3)

Setting $q\to q^2\mathrm{,}a\to \infty \mathrm{,}x\to xq/a$ in (1.3), we have

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{q}^{n^2}{x}^n}{{\left(q^2;q^2\right)}_n}={\left(xq;q^2\right)}_{\infty }.$ (1.4)

Furthermore, Euler’s formulas (cf. [1] , Corollary 2.2)

${\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{x}^n}{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_n}={\left(-x\mathrm{,}q\right)}_{\infty }\text{and}{\displaystyle \underset{n\ge 0}{\sum }}\frac{x^n}{{\left(q\mathrm{<mo>\; ;\; </mo>}q\right)}_n}=\frac{1}{{\left(x\mathrm{,}q\right)}_{\infty }}\mathrm{.}$ (1.5)

Cauchy’s identity (cf. [2] , Theorem 3.3)

${\left(x;q\right)}_n={\displaystyle \underset{k=0}{\overset{n}{\sum }}}{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}\left[\begin{array}{c}n\\ k\end{array}\right]x^k,$ (1.6)

and the formula ( [1] , Exercise 1.16) which is a special case of the Bailey-Daum sum ( [1] , (1.8.1))

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(x;q\right)}_n{q}^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}}{{\left(q;q\right)}_n}={\left(-q;q\right)}_{\infty }{\left(xq;q^2\right)}_{\infty }.$ (1.7)

Besides, if $a\ne \mathrm{0,1}-b{q}^n\ne 0$ , the following identity was given by Ramanujan [3] ,

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-b/a;q\right)}_n{a}^n{q}^{n\left(n+1\right)/2}}{{\left(q;q\right)}_n{\left(bq;q\right)}_n}=\frac{{\left(-aq;q\right)}_{\infty }}{{\left(bq;q\right)}_{\infty }}.$ (1.8)

A pair of sequences $\left({\alpha }_n\mathrm{,}{\beta }_n\right)$ is called a Bailey pair relative to a if

${\beta }_n={\displaystyle \underset{r=0}{\overset{n}{\sum }}}\frac{{\alpha }_r}{{\left(q;q\right)}_{n-r}{\left(aq;q\right)}_{n+r}}.$ (1.9)

And a conjugate Bailey pair relative to a is a pair of sequences $\left({\delta }_n\mathrm{,}{\gamma }_n\right)$ satisfying

${\gamma }_n={\displaystyle \underset{k=n}{\overset{\infty }{\sum }}}\frac{{\delta }_k}{{\left(q;q\right)}_{k-n}{\left(aq;q\right)}_{k+n}}.$ (1.10)

In fact, the Bailey pair and the conjugate Bailey pair are the special cases of the following Bailey transform.

Lemma 1.1 ( [4] ). Let n be a nonnegative integer and ${\left\{A_n\right\}}_{n=0}^{\infty },{\left\{B_n\right\}}_{n=0}^{\infty },{\left\{C_n\right\}}_{n=0}^{\infty },{\left\{D_n\right\}}_{n=0}^{\infty }$ be sequences of complex numbers. Assuming convergence of the series, if

$B_n={\displaystyle \underset{j=0}{\overset{n}{\sum }}}{A}_j{U}_{n-j}{V}_{n+j}\text{and}{C}_n={\displaystyle \underset{j=n}{\overset{\infty }{\sum }}}{D}_j{U}_{j-n}{V}_{j+n}\mathrm{,}$ (1.11)

then

${\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{A}_n{C}_n={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{B}_n{D}_n.$

This terminology was first proposed by Slater [5] . Bailey transform is widely used in mathematics for a long time, especially, in the area of basic hypergeometric series. For example, Andrews [6] , Kim and Lovejoy [7] , and Lovejoy [8] established multiple sums Rogers-Ramanujan type identities and partial theta identities. Andrews and Warnaar [9] applied the Bailey transform to give another proof of false theta functions. Bailey [4] [10] , Bressoud [2] , and Slater [11] [12] used this transform to derive a number of identities of Rogers-Ramanujan type identities. Ji and Zhao [13] established the Hecke-Rogers identities for the universal mock theta functions by means of the Bailey transform.

Notice that if we take

$U_n=\frac{1}{{\left(q;q\right)}_n},V_n=\frac{1}{{\left(aq;q\right)}_n}$

in lemma 1.1, the pair of sequences $\left(A_n\mathrm{,}{B}_n\right)$ is a Bailey pair relative to a, and the pair of sequences $\left(C_n\mathrm{,}{D}_n\right)$ is a conjugate Bailey pair relative to a.

The main motivation for this work came from some Bailey’s transform of M. E. Bachraoui which appear in [14] , specifically the ( $V_n$ ) is the constant sequences with value 1. Applying the ( [14] , Theorem 3, 4, 6) and choosing appropriate $D_n$ , we establish some new transformation formulas for q-series.

2. Main Results

Theorem 2.1. We have

${\left(a{q}^{\frac{1}{2}};q\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(a{q}^{\frac{1}{2}}\right)}^n\left(1-q^n\right){\left(x;q\right)}_n}{{\left(q;q\right)}_n^2}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(ax\right)}^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2}.$ (2.1)

Corollary 2.2. There holds

$\frac{x^n\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n}={\displaystyle \underset{k=0}{\overset{n}{\sum }}}\frac{{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k+1\\ 2\end{array}\right)-nk}\left(1-q^k\right){\left(x;q\right)}_k}{{\left(q;q\right)}_k}\left[\begin{array}{c}n\\ k\end{array}\right],$ (2.2)

${\left(q;q\right)}_{\infty }{\displaystyle \underset{n\ge 1}{\sum }}\frac{q^n{\left(-q;q\right)}_{n-1}}{{\left(q;q\right)}_n{\left(q;q\right)}_{n-1}}={\displaystyle \underset{n\ge 1}{\sum }}\frac{q^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}{\left(-q;q\right)}_{n-1}}{{\left(q;q\right)}_n{\left(q;q\right)}_{n-1}},$ (2.3)

${\left(-1;q\right)}_{\infty }{\displaystyle \underset{n\ge 1}{\sum }}\frac{{\left(-1\right)}^n{\left(-q;q\right)}_{n-1}}{{\left(q;q\right)}_n{\left(q;q\right)}_{n-1}}={\displaystyle \underset{n\ge 1}{\sum }}\frac{{\left(-1\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{\left(-q;q\right)}_{n-1}}{{\left(q;q\right)}_n{\left(q;q\right)}_{n-1}}.$ (2.4)

Theorem 2.3. We have

$\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n\left(1-q^n\right){\left(x,a;q\right)}_n}{{\left(q;q\right)}_n^2}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(cx/a\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\left(1-q^n\right){\left(x^{-1},a;q\right)}_n}{{\left(q;q\right)}_n^2{\left(c;q\right)}_n}.$ (2.5)

Corollary 2.4. There holds

$\frac{1}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{c}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\left(1-q^n\right){\left(x;q\right)}_n}{{\left(q;q\right)}_n^2}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(cx\right)}^n{q}^{n\left(n-1\right)}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2{\left(c;q\right)}_n},$ (2.6)

${\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\left(1-q^n\right)}{{\left(q;q\right)}_n^2{\left(-q^n;q\right)}_{\infty }}={\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{n\left(n-1\right)}\left(1-q^n\right)}{{\left(q;q\right)}_n^2}.$ (2.7)

Theorem 2.5. We have

${\left(a{q}^{\frac{1}{2}};q\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(a{q}^{\frac{1}{2}}\right)}^n{\left(x;q\right)}_n}{{\left(q;q\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(ax\right)}^n{q}^{\frac{n^2}{2}}}{{\left(q;q\right)}_n}.$ (2.8)

Theorem 2.6. There holds

$\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n{\left(x,a;q\right)}_n}{{\left(q;q\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(cx/a\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{\left(a;q\right)}_n}{{\left(q,c;q\right)}_n}.$ (2.9)

Corollary 2.7. We have

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n{\left(a{q}^k;q\right)}_n}{{\left(q;q\right)}_n}=\frac{{\left(c{q}^k;q\right)}_{\infty }}{{\left(c/a;q\right)}_{\infty }},$ (2.10)

$\frac{1}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{c}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{\left(x;q\right)}_n}{{\left(q;q\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(cx\right)}^n{q}^{n\left(n-1\right)}}{{\left(c,q;q\right)}_n}.$ (2.11)

Remark. Setting $c=-q,x=q$ in (2.11), we derive

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

Furthermore, we have

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

Combining the above two identities, we arrive at

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{q}^{n^2+n}}{{\left(q^2;q^2\right)}_n}={\left(q^2;q^2\right)}_{\infty }.$

The above identity is a companion to number (23) on Slater’s list [12] as follows.

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{q}^{n^2}}{{\left(q^2;q^2\right)}_n}={\left(q;q^2\right)}_{\infty }.$

Theorem 2.8. We have

${\left(-q^2;q^2\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{n^2}{\left(x;q^2\right)}_n{\left(x{q}^{2n+2};q^4\right)}_{\infty }}{{\left(q^2;q^2\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}{\left(x;q^2\right)}_n}{{\left(q;q\right)}_n}.$ (2.12)

Corollary 2.9. There holds

${\left(-q^2;q^2\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}{q}^{n^2}{\left(q^{2n+4};q^4\right)}_{\infty }={\displaystyle \underset{n\ge 0}{\sum }}{q}^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}{\left(-q;q\right)}_n,$ (2.13)

${\left(-q^2;q^2\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{n^2}{\left(q;q^2\right)}_n{\left(q^{2n+3};q^4\right)}_{\infty }}{{\left(q^2;q^2\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}{\left(q;q^2\right)}_n}{{\left(q;q\right)}_n}.$ (2.14)

Theorem 2.10. We have

$\frac{{\left(-a{q}^3;q^2\right)}_{\infty }}{{\left(b{q}^2;q^2\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{a^n{q}^{n^2+n}{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(q^2;q^2\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}\frac{a^n{q}^{\left(\begin{array}{c}n+2\\ 2\end{array}\right)-1}{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(q;q\right)}_n{\left(b{q}^2;q^2\right)}_n}.$ (2.15)

3. Proofs of Theorem 2.1 and Corollary 2.2

Proof of Theorem 2.1. Setting

$\begin{array}{l}{A}_n=\frac{{\left(-1\right)}^n{q}^{\frac{n}{2}}\left(1-q^n\right){\left(x;q\right)}_n}{{\left(q;q\right)}_n^2},U_n=\frac{q^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},\\ B_n=\frac{x^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2},D_n={\left(-1\right)}^n{a}^n\end{array}$

and $V_n=1$ in Lemma 1.1, we obtain

$\begin{array}{c}{C}_n={\displaystyle \underset{k=n}{\overset{\infty }{\sum }}}{D}_k{U}_{k-n}{V}_{k+n}={\displaystyle \underset{k\ge 0}{\sum }}{D}_{n+k}{U}_k={\displaystyle \underset{k\ge 0}{\sum }}{\left(-a\right)}^{n+k}\frac{q^{\frac{k^2}{2}}}{{\left(q;q\right)}_k}\\ ={\left(-a\right)}^n{\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(-1\right)}^k{a}^k{q}^{\frac{k^2}{2}}}{{\left(q;q\right)}_k}={\left(-1\right)}^n{a}^n{\left(a{q}^{\frac{1}{2}};q\right)}_{\infty },\end{array}$ (3.1)

where the last step follows by (1.4).

Thus,

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n={\left(a{q}^{\frac{1}{2}};q\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(a{q}^{\frac{1}{2}}\right)}^n\left(1-q^n\right){\left(x;q\right)}_n}{{\left(q;q\right)}_n^2}\\ ={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(ax\right)}^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2}.\end{array}$

This completes the proof.

Proof of Corollary 2.2. Based on (1.4), we can rewrite (2.1) as follows

${\displaystyle \underset{m\ge 0}{\sum }}\frac{{\left(-1\right)}^m{q}^{\frac{m^2}{2}}{a}^m}{{\left(q;q\right)}_m}{\displaystyle \underset{k\ge 0}{\sum }}\frac{q^{\frac{k}{2}}\left(1-q^k\right){\left(x;q\right)}_k{a}^k}{{\left(q;q\right)}_k^2}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{x}^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n{a}^n}{{\left(q;q\right)}_n^2}.$

Equating terms of the corresponding powers of $a^n$ , we achieve

${\displaystyle \underset{k=0}{\overset{n}{\sum }}}\frac{{\left(-1\right)}^{n-k}{q}^{\frac{{\left(n-k\right)}^2}{2}}}{{\left(q;q\right)}_{n-k}}\cdot \frac{q^{\frac{k}{2}}\left(1-q^k\right){\left(x;q\right)}_k}{{\left(q;q\right)}_k^2}=\frac{{\left(-1\right)}^n{x}^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2},$

which is (2.2) by some basic simplifications. (2.3) (2.4) follow from (2.1) upon letting $a=q^{\frac{1}{2}},x=-1$ ( $a=-q^{-\frac{1}{2}},x=-1$ ), respectively.

4. Proofs of Theorem 2.3 and Corollary 2.4

Proof of Theorem 2.3. We now apply Lemma 1.1 with

$\begin{array}{l}{A}_n=\frac{{\left(-1\right)}^n{q}^{\frac{n}{2}}\left(1-q^n\right){\left(x;q\right)}_n}{{\left(q;q\right)}_n^2},\\ U_n=\frac{q^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},\\ B_n=\frac{x^n{q}^{\frac{n^2}{2}}\left(1-q^n\right){\left(x^{-1};q\right)}_n}{{\left(q;q\right)}_n^2},\\ D_n={\left(-1\right)}^n{\left(c/a\right)}^n{q}^{-\frac{n}{2}}\frac{{\left(a;q\right)}_n}{{\left(c;q\right)}_n},\end{array}$

and $V_n=1$ . We compute

$\begin{array}{c}{C}_n={\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(a;q\right)}_{n+k}}{{\left(c;q\right)}_{n+k}}{\left(-1\right)}^{n+k}{\left(c/a\right)}^{n+k}{q}^{-\frac{n+k}{2}}\cdot \frac{q^{\frac{k^2}{2}}}{{\left(q;q\right)}_k}\\ ={\left(-1\right)}^n{\left(c/a\right)}^n{q}^{-\frac{n}{2}}\frac{{\left(a;q\right)}_n}{{\left(c;q\right)}_n}{\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(a{q}^n;q\right)}_k}{{\left(c{q}^n,q;q\right)}_k}{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{\left(c/a\right)}^k\\ ={\left(-1\right)}^n{\left(c/a\right)}^n{q}^{-\frac{n}{2}}{\left(a;q\right)}_n\cdot \frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }},\end{array}$ (4.1)

where the last step follows by (1.2). Thus, we arrive at

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n=\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n\left(1-q^n\right){\left(x,a;q\right)}_n}{{\left(q;q\right)}_n^2}\\ ={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(cx/a\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}\left(1-q^n\right){\left(x^{-1},a;q\right)}_n}{{\left(q;q\right)}_n^2{\left(c;q\right)}_n},\end{array}$

which completes the proof.

Proof of Corollary 2.4. (2.6) follows from (2.5) upon letting $a\to \infty$ and (2.7) follows from (2.6) by letting $x=c=-1$ , which proves the desired formula.

5. Proof of Theorem 2.5

Proof of Theorem 2.5. Setting

$A_n=\frac{{\left(-1\right)}^n{q}^{\frac{n}{2}}{\left(x;q\right)}_n}{{\left(q;q\right)}_n},U_n=\frac{q^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},B_n=\frac{x^n{q}^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},D_n={\left(-1\right)}^n{a}^n,$

and $V_n$ is the constant sequences with value 1 in Lemma 1.1. Due to (3.1), we have

$C_n={\left(-1\right)}^n{a}^n{\left(a{q}^{\frac{1}{2}}\mathrm{<mo>\; ;\; </mo>}q\right)}_{\infty }\mathrm{.}$

Thus,

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n={\left(a{q}^{\frac{1}{2}};q\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(a{q}^{\frac{1}{2}}\right)}^n{\left(x;q\right)}_n}{{\left(q;q\right)}_n}\\ ={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{\left(ax\right)}^n{q}^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},\end{array}$

which is (2.8).

Notice that we use (1.4) to express the left-hand side of (2.8)

${\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(-1\right)}^k{q}^{\frac{k^2}{2}}{a}^k}{{\left(q;q\right)}_k}{\displaystyle \underset{m\ge 0}{\sum }}\frac{q^{\frac{m}{2}}{\left(x;q\right)}_m{a}^m}{{\left(q;q\right)}_m}={\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^n{x}^n{q}^{\frac{n^2}{2}}{a}^n}{{\left(q;q\right)}_n}.$ (5.1)

Now we equate the terms corresponding to $a^n$ in (5.1) to obtain

${\displaystyle \underset{k=0}{\overset{n}{\sum }}}\frac{{\left(-1\right)}^{n-k}{q}^{\frac{{(n-k)}^2}{2}}}{{\left(q;q\right)}_{n-k}}\cdot \frac{q^{\frac{k}{2}}{\left(x;q\right)}_k}{{\left(q;q\right)}_k}=\frac{{\left(-1\right)}^n{x}^n{q}^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},$

which gives the following identity after straightforward simplifications.

$x^n={\displaystyle \underset{k=0}{\overset{n}{\sum }}}{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k+1\\ 2\end{array}\right)-nk}{\left(x;q\right)}_k\left[\begin{array}{c}n\\ k\end{array}\right].$

The above identity appears in ( [4] , p. 157).

6. Proofs of Theorem 2.6 and Corollary 2.7

Proof of Theorem 2.6. We apply Lemma 1.1 with

$\begin{array}{l}{A}_n=\frac{{\left(-1\right)}^n{q}^{\frac{n}{2}}{\left(x;q\right)}_n}{{\left(q;q\right)}_n},U_n=\frac{q^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},B_n=\frac{x^n{q}^{\frac{n^2}{2}}}{{\left(q;q\right)}_n},\\ D_n={\left(-1\right)}^n{\left(c/a\right)}^n{q}^{-\frac{n}{2}}\frac{{\left(a;q\right)}_n}{{\left(c;q\right)}_n},\end{array}$

and $V_n$ is the constant sequences with value 1. Then due to (4.1), we get

$C_n={\left(-1\right)}^n{\left(c/a\right)}^n{q}^{-\frac{n}{2}}{\left(a;q\right)}_n\cdot \frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}.$

Thus,

${\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n=\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n{\left(a,x;q\right)}_n}{{\left(q;q\right)}_n}={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n=\frac{{\left(-1\right)}^n{\left(cx/a\right)}^n{q}^{\left(\begin{array}{c}n\\ 2\end{array}\right)}{\left(a;q\right)}_n}{{\left(q,c;q\right)}_n},$

which yields the desired formula.

Proof of Corollary 2.7. To prove (2.10), we first use (1.6) to express the left-hand side of (2.9) as powers series in x. Then

L. H. S. of (2.9) $=\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(c/a\right)}^n{\left(a;q\right)}_n}{{\left(q;q\right)}_n}{\displaystyle \underset{k=0}{\overset{n}{\sum }}}{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}\left[\begin{array}{c}n\\ k\end{array}\right]x^k$

$\begin{array}{l}=\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{x}^k}{{\left(q;q\right)}_k}{\displaystyle \underset{n=k}{\overset{\infty }{\sum }}}\frac{{\left(c/a\right)}^n{\left(a;q\right)}_n}{{\left(q;q\right)}_{n-k}}\\ =\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{k\ge 0}{\sum }}\frac{{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{x}^k}{{\left(q;q\right)}_k}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(c/a\right)}^{n+k}{\left(a;q\right)}_{n+k}}{{\left(q;q\right)}_n}\\ =\frac{{\left(c/a;q\right)}_{\infty }}{{\left(c;q\right)}_{\infty }}{\displaystyle \underset{k\ge 0}{\sum }}\left({\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{\left(c/a\right)}^{n+k}{\left(a;q\right)}_{n+k}}{{\left(q;q\right)}_k{\left(q;q\right)}_n}\right)x^k.\end{array}$

Now equate the terms corresponding to $x^k$ in (2.9) to obtain

${\displaystyle \underset{n\ge 0}{\sum }}\frac{{\left(-1\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{\left(c/a\right)}^{n+k}{\left(a;q\right)}_{n+k}}{{\left(q;q\right)}_k{\left(q;q\right)}_n}=\frac{{\left(c;q\right)}_{\infty }}{{\left(c/a;q\right)}_{\infty }}\cdot \frac{{\left(-1\right)}^k{\left(c/a\right)}^k{q}^{\left(\begin{array}{c}k\\ 2\end{array}\right)}{\left(a;q\right)}_k}{{\left(c,q;q\right)}_k},$

which by some basic calculations down to (2.10). (2.11) follows easily from (2.9) upon letting $a\to \infty$ .

7. Proofs of Theorem 2.8 and Corollary 2.9

Proof of Theorem 2.8. Let us use Lemma 1.1 with a Bailey transform as follows:

$A_n=\frac{q^{n^2+n}}{{\left(q^2;q^2\right)}_n},U_n=\frac{q^{n^2+2n}}{{\left(q^2;q^2\right)}_n},B_n=\frac{q^{\left(\begin{array}{c}n+2\\ 2\end{array}\right)-1}}{{\left(q;q\right)}_n},D_n=q^{-n}{\left(x;q^2\right)}_n,V_n=1.$

We compute

$\begin{array}{c}{C}_n={\displaystyle \underset{k\ge 0}{\sum }}{q}^{-n-k}{\left(x;q^2\right)}_{n+k}\cdot \frac{q^{k^2+2k}}{{\left(q^2;q^2\right)}_k}\\ =q^{-n}{\left(x;q^2\right)}_n{\displaystyle \underset{k\ge 0}{\sum }}\frac{q^{k\left(k+1\right)}}{{\left(x{q}^{2n};q^2\right)}_k}{\left(q^2;q^2\right)}_k\\ =q^{-n}{\left(x;q^2\right)}_n{\left(-q^2;q^2\right)}_{\infty }{\left(x{q}^{2n+2};q^4\right)}_{\infty },\end{array}$

where in the last step, we used the (1.7) with $q\to q^2\mathrm{,}x\to x{q}^{2n}$ . Then by virtue of Lemma 1.1, we get

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n={\left(-q^2;q^2\right)}_{\infty }{\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{n^2}{\left(x;q^2\right)}_n{\left(x{q}^{2n+2};q^4\right)}_{\infty }}{{\left(q^2;q^2\right)}_n}\\ ={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n={\displaystyle \underset{n\ge 0}{\sum }}\frac{q^{\left(\begin{array}{c}n+1\\ 2\end{array}\right)}{\left(x;q^2\right)}_n}{{\left(q;q\right)}_n},\end{array}$

which completes the proof.

Proof of Corollary 2.9. (2.13) and (2.14) follow from (2.12) by letting, respectively, $x=q^2$ and $x=q$ .

8. Proof of Theorem 2.10

Proof of Theorem 2.10. Let us use Lemma 1.1 with a Bailey transform as follows:

$A_n=\frac{q^{n^2+n}}{{\left(q^2;q^2\right)}_n},U_n=\frac{q^{n^2+2n}}{{\left(q^2;q^2\right)}_n},B_n=\frac{q^{\left(\begin{array}{c}n+2\\ 2\end{array}\right)-1}}{{\left(q;q\right)}_n},D_n=\frac{a^n{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(b{q}^2;q^2\right)}_n},V_n=1.$

We compute

$\begin{array}{c}{C}_n={\displaystyle \underset{k\ge 0}{\sum }}\frac{a^{n+k}{\left(-\frac{b}{aq};q^2\right)}_{n+k}}{{\left(b{q}^2;q^2\right)}_{n+k}}\cdot \frac{q^{k^2+2k}}{{\left(q^2;q^2\right)}_k}\\ =\frac{a^n{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(b{q}^2;q^2\right)}_n}{\displaystyle \underset{k\ge 0}{\sum }}\frac{a^k{q}^{k^2+2k}{\left(-b{q}^{2n+1}/a;q^2\right)}_k}{{\left(q^2;q^2\right)}_k{\left(b{q}^{2n+2};q^2\right)}_k}\\ =a^n{\left(-\frac{b}{aq};q^2\right)}_n\frac{{\left(-a{q}^3;q^2\right)}_{\infty }}{{\left(b{q}^2;q^2\right)}_{\infty }},\end{array}$

where in the last step, we used (1.8) by letting $a=aq,b=b{q}^{2n}$ . Then by virtue of Lemma 1.1, we get

$\begin{array}{c}{\displaystyle \underset{n\ge 0}{\sum }}{A}_n{C}_n=\frac{{\left(-a{q}^3;q^2\right)}_{\infty }}{{\left(b{q}^2;q^2\right)}_{\infty }}{\displaystyle \underset{n\ge 0}{\sum }}\frac{a^n{q}^{n^2+n}{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(q^2;q^2\right)}_n}\\ ={\displaystyle \underset{n\ge 0}{\sum }}{B}_n{D}_n={\displaystyle \underset{n\ge 0}{\sum }}\frac{a^n{q}^{\left(\begin{array}{c}n+2\\ 2\end{array}\right)-1}{\left(-\frac{b}{aq};q^2\right)}_n}{{\left(q;q\right)}_n{\left(b{q}^2;q^2\right)}_n},\end{array}$

which completes the proof.

9. Conclusion

By choosing some sequences, we can derive many identities from the Bailey transform. Furthermore, we should study the generalized Bailey transform [14] deeply to establish the multiple parameterized identities. On the other hand, we can also study the mock theta functions or the Rogers-Ramanujan identities through the Bailey transform.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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