Generalized homogeneous q-difference equations for q-polynomials and their applications to generating functions and fractional q-integrals

Abstract

In this paper, our aim is to build generalized homogeneous q-difference equations for q-polynomials. We also consider their applications to generating functions and fractional q-integrals by using the perspective of q-difference equations. In addition, we also reveal relevant relations of various special cases of our main results involving some known results.

1 Introduction

The objective of this paper is to give an extension of know results on generalized Verma–Jain polynomials [13] and the Hahn polynomials [1, 6, 12, 30]. Here, we will give and prove generating functions for the q-polynomials $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q), \zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$, and several q-identities by using the q-difference equations and the fractional q-integrals. In this article, we begin our investigation by reviewing some definitions as in [33] with $0< q<1$. The basic hypergeometric function ${}_{\mathfrak{r}}\Phi _{\mathfrak{s}}$ is defined in [16, 25] (see also for details [24, Chap. 3] and [32, p. 347, Eq. (272)]):

$${}_{\mathfrak{r}}\mathrm{\Phi }_{\mathfrak{s}}[\begin{array}{c}\begin{array}{r}{a}_1,a_2,\dots ,a_{\mathfrak{r}};\\ b_1,b_2,\dots ,b_{\mathfrak{s}};\end{array}\end{array}q;z]=\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,a_2,\dots ,a_{\mathfrak{r}};q)}_n}{{(q,b_1,b_2,\dots ,b_{\mathfrak{s}};q)}_n}{\left[{(-1)}^n{q}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}\right]}^{1+\mathfrak{s}-\mathfrak{r}}{z}^n.$$

(1.1)

For all z if $\mathfrak{r}\leq \mathfrak{s}$ and for $|z|<1$ if $\mathfrak{r}=\mathfrak{s}+1$, the basic hypergeometric function converges absolutely. (See [31] for some recent applications of the basic hypergeometric function.) For any real or complex parameter x, the q-shifted factorials of ${}_{\mathfrak{r}}\Phi _{\mathfrak{s}}$ are defined, respectively, by

$$\begin{aligned} (x;q)_{0}=1,\qquad (x;q)_{n}=\prod_{k=0}^{n-1} \bigl(1-xq^{k}\bigr), n\geq 1,\quad x\in \mathbb{C} \end{aligned}$$

(1.2)

and

$$\begin{aligned} (x;q)_{\infty }=\prod_{k=0}^{\infty } \bigl(1-xq^{k}\bigr). \end{aligned}$$

(1.3)

For $m\in \{1, 2, 3, \ldots \}$, the product of several q-shifted factorials are given by

$$\begin{aligned} &(x_{1},x_{2},\ldots,x_{m};q)_{n}=(x_{1};q)_{n} (x_{2};q)_{n}\ldots (x_{m};q)_{n},\\ & (x_{1},x_{2},\ldots,x_{m};q)_{\infty }=(x_{1};q)_{\infty }(x_{2};q)_{\infty } \ldots (x_{m};q)_{\infty }. \end{aligned}$$

Taking $x=aq^{-n}, a\neq q^{n}$ in (1.2), we have the following relation:

$$\begin{aligned} \bigl(aq^{-n};q\bigr)_{n}= \frac{(aq^{-n};q)_{\infty }}{(a;q)_{\infty }}=(q/a;q)_{n}(-a)^{n}q^{-n-(\frac{n}{2})}. \end{aligned}$$

(1.4)

The q-binomial coefficient is defined as [16]

$$\begin{aligned} \begin{bmatrix} n \\ k \end{bmatrix}_{q}=\frac{(q^{-n};q)_{k}}{(q;q)_{k}}(-1)^{k}q^{n k-(\frac{k}{2})},\quad 0\leq k\leq n. \end{aligned}$$

(1.5)

Chen et al. [15] introduced the homogeneous q-difference operator $D_{xy}$, Saad and Sukhi [23] introduced the dual homogeneous q-difference operator ${\theta }_{xy}$ as

$$\begin{aligned} D_{xy} \bigl\{ f(x,y)\bigr\} :=\frac{f(x,q^{-1}y)-f( qx, y)}{x-q^{-1}y}, \qquad{ \theta }_{xy} \bigl\{ f(x,y)\bigr\} :=\frac{f(q^{-1}x,y)-f( x,qy)}{q^{-1}x-y}. \end{aligned}$$

(1.6)

Al-Salam and Carlitz [2, Eqs. (1.11) and (1.15)] have introduced the following polynomials:

$$\begin{aligned} \phi _{n}^{(a)}(x|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q}(a;q)_{k}x^{k} \quad \text{and} \quad\psi _{n}^{(a)}(x|q)= \sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q}q^{k(k-n)} \bigl(aq^{1-k};q\bigr)_{k}x^{k}. \end{aligned}$$

(1.7)

Since then, these polynomials are called “Al-Salam–Carlitz polynomials” by many authors. Because of their considerable role in the theories of q-series and q-orthogonal polynomials, many authors investigated an extension of the Al-Salam–Carlitz polynomials (see [7, 12, 28, 35]).

Recently, Cao [7, Eq. (4.7)] has introduced two families of generalized Al-Salam–Carlitz polynomials,

$$\begin{aligned} &\phi _{n}^{(a,b,c)}(x,y|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(a,b;q)_{k}}{(c;q)_{k}}x^{k}y^{n-k}, \end{aligned}$$

(1.8)

$$\begin{aligned} &\psi _{n}^{(a,b,c)}(x,y|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(-1)^{k}q^{\binom{k+1}{2}-nk}(a,b;q)_{k}}{(c;q)_{k}}x^{k}y^{n-k}, \end{aligned}$$

(1.9)

together with the following generating functions [7, Eqs. (4.10) and (4.11)]:

$$\sum _{n=0}^{\mathrm{\infty }}{\varphi }_n^{(a,b,c)}(x,y|q)\frac{t^n}{{(q;q)}_n}=\frac{1}{{(xt;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_1[\begin{array}{c}\begin{array}{r}a,b;\\ c;\end{array}\end{array}q;yt](max\{|yt|,|xt|\}<1),$$

(1.10)

$$\sum _{n=0}^{\mathrm{\infty }}{\psi }_n^{(a,b,c)}(x,y|q)\frac{{(-1)}^n{q}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{t}^n}{{(q;q)}_n}=(xt;q)_{\mathrm{\infty }}{}_2{\mathrm{\Phi }}_1[\begin{array}{c}\begin{array}{c}a,b;\\ c;\end{array}\end{array}q;yt](|xt|<1).$$

(1.11)

Motivated by the work of Cao [7], the authors [12] introduced a new extension of the Al-Salam–Carlitz polynomials $\phi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)$, $\psi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)$,

$$\begin{aligned} &\phi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(a,b,c;q)_{k}}{(d,e;q)_{k}}x^{n-k}y^{k}, \end{aligned}$$

(1.12)

$$\begin{aligned} &\psi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(-1)^{k}q^{k(k-n)}(a,b,c;q)_{k}}{(d,e;q)_{k}}x^{n-k}y^{k}, \end{aligned}$$

(1.13)

and obtained the following results.

Proposition 1

([12, Theorem 4])

Let $f(a,b,c,d,e,x,y)$ be a seven-variable analytic function in a neighborhood of $(a,b,c,d,e,x,y)=(0,0,0,0,0,0,0)\in \mathbb{C}^{7}$.

(I) $f(a,b,c,d,e,x,y)$ can be expanded in terms of $\phi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)$ if and only if

$$\begin{aligned} & x\bigl\{ f(a,b,c,d,e,x,y)-f(a,b,c,d,e,x,yq) \\ &\qquad{}-(d+e)q^{-1} \bigl[f(a,b,c,d,e,x,yq)-f\bigl(a,b,c,d,e,x,yq^{2}\bigr)\bigr] \\ &\qquad{} +deq^{-2}\bigl[f\bigl(a,b,c,d,e,x,yq^{2}\bigr)-f \bigl(a,b,c,d,e,x,yq^{3}\bigr)\bigr] \bigr\} \\ &\quad=y\bigl\{ \bigl[f(a,b,c,d,e,x,y)-f(a,b,c,d,e,xq,y)\bigr] \\ &\qquad{}-(a+b+c) \bigl[f(a,b,c,d,e,x,yq)-f(a,b,c,d,e,xq,yq)\bigr] \\ &\qquad{} +(ab+ac+bc)\bigl[f\bigl(a,b,c,d,e,x,yq^{2}\bigr)-f \bigl(a,b,c,d,e,xq,yq^{2}\bigr)\bigr] \\ &\qquad{} -abc\bigl[f\bigl(a,b,c,d,e,x,yq^{3}\bigr)-f\bigl(a,b,c,d,e,xq,yq^{3} \bigr)\bigr] \bigr\} . \end{aligned}$$

(1.14)

(II) $f(a,b,c,d,e,x,y)$ can be expanded in terms of $\psi _{n}^{\binom{a,b,c}{d,e}}(x,y|q)$ if and only if

$$\begin{aligned} & x\bigl\{ f(a,b,c,d,e,xq,y)-f(a,b,c,d,e,xq,yq) \\ &\qquad{}-(d+e)q^{-1} \bigl[f(a,b,c,d,e,xq,yq)-f\bigl(a,b,c,d,e,xq,yq^{2}\bigr)\bigr] \\ &\qquad{} +deq^{-2}\bigl[f\bigl(a,b,c,d,e,xq,yq^{2}\bigr)-f \bigl(a,b,c,d,e,xq,yq^{3}\bigr)\bigr] \bigr\} \\ &\quad=y\bigl\{ \bigl[f(a,b,c,d,e,xq,yq)-f(a,b,c,d,e,x,yq)\bigr] \\ &\qquad{}-(a+b+c)\bigl[f \bigl(a,b,c,d,e,xq,yq^{2}\bigr)-f\bigl(a,b,c,d,e,x,yq^{2} \bigr)\bigr] \\ &\qquad{} +(ab+ac+bc)\bigl[f\bigl(a,b,c,d,e,xq,yq^{3}\bigr)-f \bigl(a,b,c,d,e,x,yq^{3}\bigr)\bigr] \\ & \qquad{}-abc\bigl[f\bigl(a,b,c,d,e,xq,yq^{4}\bigr)-f\bigl(a,b,c,d,e,x,yq^{4} \bigr)\bigr] \bigr\} . \end{aligned}$$

(1.15)

Subsequently, Cao et al. [13], gave another extension of Al-Salam–Carlitz polynomials called “generalized Verma–Jain polynomials”,

$$\begin{aligned} &\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(a,b,c;q)_{k}}{(d,e;q)_{k}}P_{n-k}(x,y)z^{k}, \end{aligned}$$

(1.16)

$$\begin{aligned} &\mu _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{(a,b,c;q)_{k}}{(d,e;q)_{k}}P_{n-k}(y,x)z^{k} , \end{aligned}$$

(1.17)

where

$$\begin{aligned} P_{n}(x,y)=(x-y) (x-q y)...\bigl(x-q^{n-1}y \bigr)=(y/x;q)_{n} x^{n} \end{aligned}$$

(1.18)

are the Cauchy polynomials.

Remark 2

Upon setting $(y,z)=(0,y)$, the polynomial (1.16) reduces to (1.12).

Motivated by the recent work of Cao [7], Cao et al. [12, 13] and with the aid of the polynomials (1.17), we introduce the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$.

Definition 3

The q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ are defined by

$$\begin{aligned} \zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)&=\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{q^{\binom{k}{2}}(a,b,c;q)_{k}}{(d,e;q)_{k}}P_{n-k}(y,x) z^{k} . \end{aligned}$$

(1.19)

Remark 4

The q-polynomials (1.19) can be viewed as a general form of the Hahn polynomials.

• (1) Taking $r=s=3, \mathbf{a}=(a,b,c)$ and $\mathbf{b}=(d,e,0)$ in [29, Definition 1], the q-polynomials (1.19) is a special case of the generalized q-hypergeometric polynomials $\Psi _{n}^{(\mathbf{a},\mathbf{b})}(x,y,z|q)$, i.e.,

  $$\begin{aligned} \zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)=(-1)^{n}q^{(\frac{n}{2})} \Psi _{n}^{({ \mathbf{a}},\mathbf{b})}(x,y,z|q). \end{aligned}$$

• (2) Upon setting $(y,z)=(0,y)$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ defined in (1.19) reduce to the polynomials $\psi _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ [12],

  $$\begin{aligned} \zeta _{n}^{\binom{a,b,c}{d,e}}(x,0,y|q)=(-1)^{n}q^{-\binom{n}{2}} \psi _{n}^{\binom{a,b,c }{d,e}}(x,y|q). \end{aligned}$$

• (3) For $b=c=d=e=0$ and $z=-b$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ reduce to the generalized Hahn polynomials $h_{n}(x,y,a,b|q)$ [30],

  $$\begin{aligned} \zeta _{n}^{\binom{a,0,0}{0,0}}(x,y,-b|q) = h_{n}(x,y,a,b|q). \end{aligned}$$

• (4) Setting $a=b=c=d=e=0$ and $z=-z$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ reduce to the trivariate q-polynomials $F_{n}(x,y,z;q)$ [1],

  $$\begin{aligned} \zeta _{n}^{\binom{0,0,0}{0,0}}(x,y,-z|q) = (-1)^{n}q^{\binom{n}{2}}F_{n}(x,y,z;q). \end{aligned}$$

• (5) If we let $a=b=c=d=e=0, y = ax$ and $z=-y$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ reduce to $\psi _{n}^{(a)}(x,y|q)$ [6],

  $$\begin{aligned} \zeta _{n}^{\binom{0,0,0}{0,0}}(x,ax,-y|q) = (-1)^{n}q^{\binom{n}{2}} \psi _{n}^{(a)}(x,y|q). \end{aligned}$$

• (6) For $b=c=d=e=0, x=0, y=x$ and $z=-y$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ reduce to the polynomials $P_{n}(x,y,a)$ [3],

  $$\begin{aligned} \zeta _{n}^{\binom{a,0,0}{0,0}}(0,x,-y|q) = P_{n}(x,y,a). \end{aligned}$$

• (7) Also, $a=b=c=d=e=0, y = ax$ and $z=-1$, the q-polynomials $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ reduce to Hahn polynomials $\psi _{n}^{(a)}(x|q)$ [2],

  $$\begin{aligned} \zeta _{n}^{\binom{0,0,0}{0,0}}(x,ax,-1|q) = (-1)^{n}q^{\binom{n}{2}} \psi _{n}^{(a)}(x|q). \end{aligned}$$

The paper is organized as follows. In Sect. 2, we give and prove our main results to be used in the sequel. In Sect. 3, we obtain generating function for q-polynomials. In Sect. 4, we obtain the Srivastava–Agarwal type generating function for q-hypergeometric polynomials. In Sect. 5, we deduce mixed generating functions for the Rajković–Marinković–Stanković polynomials. In Sect. 6, we derive $U(n+1)$ generalizations of the generating functions for q-hypergeometric polynomials.

2 Proof of main results

In this section, we will give and prove our main results to be used in the sequel.

Theorem 5

Let $f(a,b,c,d,e,x,y,z)$ be an eight-variable analytic function in a neighborhood of $(a,b,c,d,e,x,y,z)=(0,0,0,0,0,0,0,0)\in \mathbb{C}^{8}$.

(I) If $f(a,b,c,d,e,x,y,z)$ can be expanded in terms of $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ if and only if

$$\begin{aligned} & \bigl(x-q^{-1}y\bigr) \bigl\{ \bigl[f(a,b,c,d,e,x,y,z)-f(a,b,c,d,e,x,y,qz) \bigr] \\ &\qquad{} -(d+e)q^{-1}\bigl[f(a,b,c,d,e,x,y,qz)-f\bigl(a,b,c,d,e,x,y,q^{2}z \bigr)\bigr] \\ &\qquad{}+deq^{-2}\bigl[f\bigl(a,b,c,d,e,x,y,q^{2}z\bigr)-f \bigl(a,b,c,d,e,x,y,q^{3}z\bigr)\bigr] \bigr\} \\ &\quad=z \bigl\{ \bigl[f\bigl(a,b,c,d,e,x,q^{-1}y,z\bigr)-f(a,b,c,,d,e,qx,y,z) \bigr] \\ &\qquad{} -(a+b+c) \bigl[f\bigl(a,b,c,d,e,x,q^{-1}y,qz \bigr)-f(a,b,c,d,e,qx,y,qz) \bigr] \\ &\qquad{}+(ab+ac+bc) \bigl[f\bigl(a,b,c,d,e,x,q^{-1}y,q^{2}z \bigr)-f\bigl(a,b,c,d,e,qx,y,q^{2}z\bigr)\bigr] \\ &\qquad{}-abc\bigl[f\bigl(a,b,c,d,e,x,q^{-1}y,q^{3}z\bigr)-f \bigl(a,b,c,d,e,qx,y,q^{3}z\bigr) \bigr] \bigr\} . \end{aligned}$$

(2.1)

(II) If $f(a,b,c,d,e,x,y,z)$ can be expanded in terms of $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$ if and only if

$$\begin{aligned} & \bigl(q^{-1}x-y\bigr) \bigl\{ \bigl[f(a,b,c,d,e,x,y,z)-f(a,b,c,d,e,x,y,qz) \bigr] \\ &\qquad{}-(d+e)q^{-1}\bigl[f(a,b,c,d,e,x,y,qz)-f\bigl(a,b,c,d,e,x,y,q^{2}z \bigr)\bigr] \\ &\qquad{}+deq^{-2}\bigl[f\bigl(a,b,c,d,e,x,y,q^{2}z\bigr)-f \bigl(a,b,c,d,e,x,y,q^{3}z\bigr)\bigr] \bigr\} \\ &\quad=z \bigl\{ \bigl[f(a,b,c,d,e, x,qy,qz)-f\bigl(a,b,c,d,e,q^{-1}x,y,qz \bigr) \bigr] \\ &\qquad{}-(a+b+c) \bigl[f\bigl(a,b,c,d,e, x,qy,q^{2}z\bigr) \\ &\qquad{}-f\bigl(a,b,c,d,e,q^{-1}x,y,q^{2}z\bigr) \bigr] \\ &\qquad{}+(ab+ac+bc) \bigl[f\bigl(a,b,c,d,e, x,qy,q^{3}z\bigr)-f \bigl(a,b,c,d,e,q^{-1}x, y,q^{3}z\bigr)\bigr] \\ &\qquad{}-abc\bigl[f\bigl(a,b,c,d,e, x,qy,q^{4}z\bigr)-f \bigl(a,b,c,d,e,q^{-1}x, y,q^{4}z\bigr) \bigr] \bigr\} . \end{aligned}$$

(2.2)

Remark 6

For $(y,z)=(0,y)$, Eq. (2.1) reduces to (1.14).

To prove Theorem 5, we need the following lemmas.

Lemma 7

([17, Hartogs theorem])

If a complex-valued function is holomorphic (analytic) in each variable separately in an open domain $D \in \mathbb{C}^{n}$, then it is holomorphic (analytic) in D.

Lemma 8

([20, Proposition 1])

If $f(x_{1},x_{2},\ldots,x_{k})$ is analytic at the origin $(0,0,\ldots,0)\in \mathbb{C}^{k}$, then f can be expanded in an absolutely convergent power series,

$$\begin{aligned} f(x_{1},x_{2},\ldots,x_{k})=\sum _{n_{1},n_{2},\ldots,n_{k}=0}^{\infty }\alpha _{n_{1},n_{2},\ldots,n_{k}}x_{1}^{n_{1}}x_{2}^{n_{2}}...x_{k}^{n_{k}}. \end{aligned}$$

Proof of Theorem 5

(I) From Lemmas 7 and 8, we assume that there exists a sequence $\{A_{n}\}$ such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }A_{n}(a,b,c,d,e,x,y)z^{n}. \end{aligned}$$

(2.3)

First, substituting (2.3) into Eq. (2.1), we have

$$\begin{aligned} &\bigl(x-q^{-1}y\bigr)\sum_{n=0}^{\infty } \bigl[1-q^{n}-(d+e)q^{n-1}+(d+e)q^{2n-1}+deq^{2n-2}-deq^{3n-2} \bigr]\\ &\qquad{}\times A_{n}(a,b,c,d,e,x,y)z^{n} \\ &\quad=\sum_{n=0}^{\infty }\bigl[1-(a+b+c)q^{n}+(ab+bc+ac)q^{2n}-abcq^{3n} \bigr]\\ &\qquad{}\times \bigl[A_{n}\bigl(a,b,c,d,e,x,q^{-1}y \bigr)-A_{n}(a,b,c,d,e,qx,y)\bigr]z^{n+1}, \end{aligned}$$

which is equal to

$$\begin{aligned} & \bigl(x-q^{-1}y\bigr)\sum_{n=0}^{\infty } \bigl(1-q^{n}\bigr) \bigl(1-dq^{n-1}\bigr) \bigl(1-eq^{n-1} \bigr)A_{n}(a,b,c,d,e,x,y)z^{n} \\ &\quad=\sum_{n=0}^{\infty }\bigl(1-aq^{n} \bigr) \bigl(1-bq^{n}\bigr) \bigl(1-cq^{n}\bigr) \\ &\qquad{}\times \bigl[A_{n}\bigl(a,b,c,d,e,x,q^{-1}y\bigr)-A_{n}(a,b,c,d,e,qx,y) \bigr]z^{n+1}. \end{aligned}$$

(2.4)

Comparing coefficients of $z^{n} n\geq 1$, on both sides of Eq. (2.4), we readily find that

$$\begin{aligned} &\bigl(x-q^{-1}y\bigr) \bigl(1-q^{n}\bigr) \bigl(1-dq^{n-1}\bigr) \bigl(1-eq^{n-1}\bigr)A_{n}(a,b,c,d,e,x,y) \\ & \quad=\bigl(1-aq^{n-1}\bigr) \bigl(1-bq^{n-1}\bigr) \bigl(1-cq^{n-1}\bigr) \\ &\qquad{}\times\bigl[A_{n-1}\bigl(a,b,c,d,e,x,q^{-1}y \bigr)-A_{n-1}(a,b,c,d,e,qx,y) \bigr], \end{aligned}$$

which is equivalent to

$$\begin{aligned} A_{n}(a,b,c,d,e,x,y)= \frac{(1-aq^{n-1})(1-bq^{n-1})(1-cq^{n-1})}{(1-q^{n})(1-dq^{n-1})(1-eq^{n-1})}D_{xy}{A_{n-1}(a,b,c,d,e,x,y)}. \end{aligned}$$

By iteration, we obtain

$$\begin{aligned} A_{n}(a,b,c,d,e,x,y)=\frac{(a,b,c;q)_{n}}{(q,d,e;q)_{n}}D_{xy}^{n} \bigl\{ A_{0}(a,b,c,d,e,x,y) \bigr\} . \end{aligned}$$

(2.5)

Taking $f(a,b,c,d,e,x,y,0)=A_{0}(a,b,c,d,e,x,y)=\sum_{n=0}^{\infty }\beta _{n}P_{n}(x,y)$ yields

$$\begin{aligned} A_{k}(a,b,c,d,e,x,y)=\frac{(a,b,c;q)_{k}}{(q,d,e;q)_{k}}\cdot \sum _{n=0}^{\infty }\beta _{n} \frac{(q;q)_{n}}{(q;q)_{n-k}}P_{n-k}(x,y), \end{aligned}$$

(2.6)

and we have

$$\begin{aligned} f(a,b,c,d,e,x,y,z)&=\sum_{k=0}^{\infty } \frac{(a,b,c;q)_{k}}{(q,d,e;q)_{k}}\sum_{n=0}^{\infty }\beta _{n} \frac{(q;q)_{n}}{(q;q)_{n-k}}P_{n-k}(x,y)z^{k} \\ &=\sum_{n=0}^{\infty }\beta _{n}\sum _{k=0}^{\infty } \begin{bmatrix} n \\ k \end{bmatrix}_{q}\frac{(a,b,c;q)_{k}}{(d,e;q)_{k}}P_{n-k}(x,y)z^{k} \\ &=\sum_{n=0}^{\infty }\beta _{n}\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q). \end{aligned}$$

Second, if $f(a,b,c,d,e,x,y,z)$ can be expanded in terms of $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$, we can verify that it satisfies (2.1).

In almost the same way, we assume that there exists a sequence $\{B_{n}\}$ such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }B_{n}(a,b,c,d,e,x,y)z^{n}. \end{aligned}$$

(2.7)

Now, substituting Eq. (2.7) into Eq. (2.2), we have

$$\begin{aligned} &\bigl(q^{-1}x- y\bigr)\sum_{n=0}^{\infty } \bigl[1-q^{n}-(d+e)q^{n-1}+(d+e)q^{2n-1}+deq^{2n-2}-deq^{3n-2} \bigr] \\ &\qquad{}\times B_{n}(a,b,c,d,e,x,y)z^{n} \\ &\quad=\sum_{n=0}^{\infty }\bigl[q^{n}-(a+b+c)q^{2n}+(ab+bc+ac)q^{3n}-abcq^{4n} \bigr] \\ &\qquad{}\times\bigl[B_{n}(a,b,c,d,e,x,qy)-B_{n}\bigl(a,b,c,d,e,q^{-1}x,y \bigr)\bigr]z^{n+1}, \end{aligned}$$

which is equal to

$$\begin{aligned} &\bigl(q^{-1}x- y\bigr)\sum _{n=0}^{\infty }\bigl(1-q^{n}\bigr) \bigl(1-dq^{n-1}\bigr) \bigl(1-eq^{n-1}\bigr)B_{n}(a,b,c,d,e,x,y)z^{n} \\ &\quad=\sum_{n=0}^{\infty }q^{n} \bigl(1-aq^{n}\bigr) \bigl(1-bq^{n}\bigr) \bigl(1-cq^{n}\bigr) \\ &\qquad{}\times\bigl[B_{n}(a,b,c,d,e,x,qy)-B_{n} \bigl(a,b,c,d,e,q^{-1}x,y\bigr)\bigr]z^{n+1}. \end{aligned}$$

(2.8)

Comparing coefficients of $z^{n} n\geq 1$, on both sides of Eq. (2.8), we readily find that

$$\begin{aligned} &\bigl(q^{-1}x- y\bigr) \bigl(1-q^{n}\bigr) \bigl(1-dq^{n-1}\bigr) \bigl(1-eq^{n-1}\bigr)B_{n}(a,b,c,d,e,x,y) \\ & \quad=q^{n-1}\bigl(1-aq^{n-1}\bigr) \bigl(1-bq^{n-1}\bigr) \bigl(1-cq^{n-1}\bigr) \\ &\qquad{}\times\bigl[B_{n-1}(a,b,c,d,e,x,qy)-B_{n-1} \bigl(a,b,c,d,e,q^{-1}x,y\bigr) \bigr], \end{aligned}$$

which is equivalent to

$$\begin{aligned} B_{n}(a,b,c,d,e,x,y)=-q^{n-1} \frac{(1-aq^{n-1})(1-bq^{n-1})(1-cq^{n-1})}{(1-q^{n})(1-dq^{n-1})(1-eq^{n-1})} \theta _{xy}{B_{n-1}(a,b,c,d,e,x,y)}. \end{aligned}$$

By iteration, we obtain

$$\begin{aligned} B_{n}(a,b,c,d,e,x,y)= \frac{(-1)^{n}q^{\binom{n}{2}}(a,b,c;q)_{n}}{(q,d,e;q)_{n}}\theta _{xy}^{n} \bigl\{ B_{0}(a,b,c,d,e,x,y)\bigr\} . \end{aligned}$$

(2.9)

Upon setting $f(a,b,c,d,e,x,y,0)=B_{0}(a,b,c,d,e,x,y)=\sum_{n=0}^{\infty }\beta _{n}P_{n}(y,x)$,

$$\begin{aligned} B_{k}(a,b,c,d,e,x,y)= \frac{q^{\binom{k}{2}}(a,b,c;q)_{k}}{(q,d,e;q)_{k}}\cdot \sum _{n=0}^{\infty }\beta _{n} \frac{(q;q)_{n}}{(q;q)_{n-k}}P_{n-k}(y,x), \end{aligned}$$

(2.10)

we have

$$\begin{aligned} f(a,b,c,d,e,x,y,z)&=\sum_{k=0}^{\infty } \frac{q^{\binom{k}{2}}(a,b,c;q)_{k}}{(q,d,e;q)_{k}}\sum_{n=0}^{\infty }\beta _{n}\frac{(q;q)_{n}}{(q;q)_{n-k}}P_{n-k}(x,y)z^{k} \\ &=\sum_{n=0}^{\infty }\beta _{n}\sum _{k=0}^{\infty } \begin{bmatrix} n \\ k \end{bmatrix}_{q}\frac{q^{\binom{k}{2}}(a,b,c;q)_{k}}{(d,e;q)_{k}}P_{n-k}(x,y)z^{k} \\ &=\sum_{n=0}^{\infty }\beta _{n}\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q). \end{aligned}$$

Finally, if $f(a,b,c,d,e,x,y,z)$ can be written in terms of $\zeta _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q)$, we can verify that $f(a,b,c,d,e,x,y,z)$ satisfies (2.2). The proof of Theorem 5 is complete. □

3 Generating function for new generalized q-polynomials

In this section, our aim is to give and prove the generating functions for q-polynomials by means of the q-difference equations.

Theorem 9

It is asserted that

$$\sum _{n=0}^{\mathrm{\infty }}{\omega }_n^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(x,y,z|q)\frac{t^n}{{(q;q)}_n}=\frac{{(yt;q)}_{\mathrm{\infty }}}{{(xt;q)}_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e;\end{array}\end{array}q;zt](max\{|xt|,|zt|\}<1)$$

(3.1)

and

$$\sum _{n=0}^{\mathrm{\infty }}{\zeta }_n^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(x,y,z|q)\frac{t^n}{{(q;q)}_n}=\frac{{(xt;q)}_{\mathrm{\infty }}}{{(yt;q)}_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_3[\begin{array}{c}\begin{array}{r}a,b,c;\\ 0,d,e;\end{array}\end{array}q;-zt](|yt|<1).$$

(3.2)

Remark 10

Equations (3.1) and (3.2) reduce to Eqs. (1.10) and (1.11), respectively, when $c=e=y=0$ in Theorem 9.

Proof of Theorem 9

We denote the right-hand side of Eq. (3.1) by $f(a,b,c,d,e,x,y,z)$. One can verify that $f(a,b,c,d,e,x,y,z)$ satisfies (2.1). So, there exists a sequence $\{\beta _{n}\}$, such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }\beta _{n}\omega _{n}^{\binom{a,b,c }{d,e}}(x,y,z|q). \end{aligned}$$

(3.3)

Upon setting $z=0$ in Eq. (3.3) and then using the obvious fact $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,0|q)=P_{n}(x,y)$, we have

$$\begin{aligned} f(a,b,c,d,e,x,y,0)=\sum_{n=0}^{\infty }\beta _{n} P_{n}(x,y)= \frac{(yt;q)_{\infty }}{(xt;q)_{\infty }}=\sum _{n=0}^{\infty }P_{n}(x,y) \frac{t^{n}}{(q;q)_{n}}. \end{aligned}$$

So, the function $f(a,b,c,d,e,x,y,z)$ is equivalent to

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty } \frac{t^{n}}{(q;q)_{n}}\omega _{n}^{\binom{a,b,c }{d,e}}(x,y,z|q), \end{aligned}$$

which equals the right-hand side of Eq. (3.1). Similarly, we prove Eq. (3.2).

The proof of Theorem 9 is complete. □

4 Srivastava–Agarwal type generating function for q-hypergeometric polynomials

We recall that the following Srivastava–Agarwal type generating functions for the Al-Salam–Carlitz polynomials. See also [10, 19] for some recent work on generating functions.

Proposition 11

([27, Eq. (3.20)] and [5, Eq. (5.4)])

We have

$$\sum _{n=0}^{\mathrm{\infty }}{\varphi }_n^{(\alpha )}(z|q)\frac{{(\lambda ;q)}_n{t}^n}{{(q;q)}_n}=\frac{{(\lambda t;q)}_{\mathrm{\infty }}}{{(t;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_1[\begin{array}{c}\begin{array}{r}\lambda ,\alpha ;\\ \lambda t;\end{array}\end{array}q;zt](max\{|t|,|xt|\}<1)$$

(4.1)

and

$$\begin{aligned} & \sum_{n=0}^{\infty }\psi _{n}^{(\alpha )}(x|q) (1/\lambda;q)_{n} \frac{ (\lambda tq)^{n}}{(q;q)_{n}} = \frac{(xtq; q)_{\infty }}{(\lambda xtq;q)_{\infty }} {}_{2}\Phi _{1}\left [ \textstyle\begin{array}{rr} 1/ \lambda,1/ (\alpha x); \\ 1/(\lambda xt); \end{array}\displaystyle q; \alpha q \right ] \\ &\quad\bigl(\max \bigl\{ \vert \lambda x tq \vert , \vert \alpha q \vert \bigr\} < 1 \bigr). \end{aligned}$$

(4.2)

In this section, we state and prove the Srivastava–Agarwal type bilinear generating functions for q-hypergeometric polynomials by the method of homogeneous q-difference equations.

Theorem 12

It is asserted that

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\varphi }_n^{(\alpha )}(x|q){\omega }_n^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(u,v,z|q)\frac{t^n}{{(q;q)}_n}\\ & =\frac{{(vt,\alpha x;q)}_{\mathrm{\infty }}}{{(ut,x;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(ut,\alpha ;q)}_k{q}^k}{{(q/x,vt,q;q)}_k}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e;\end{array}\end{array}q;zt{q}^k]\\ & (max\{|ut|,|zt|,|x|\}<1)\end{array}$$

(4.3)

and

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\psi }_n^{(\alpha )}(x|q){\zeta }_n^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(u,v,z|q)\frac{t^n}{{(q;q)}_n}\\ & =\frac{{(q/x,uxtq;q)}_{\mathrm{\infty }}}{{(\alpha q,vxtq;q)}_{\mathrm{\infty }}}\\ & \times \sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{q}^{(\frac{n}{2})}{(1/(\alpha x),1/(uxt);q)}_n}{{(q/x,1/(vxt),q;q)}_n}\left(\frac{\alpha uq}{v}\right)^n{}_3{\mathrm{\Phi }}_3[\begin{array}{c}\begin{array}{r}a,b,c;\\ 0,d,e;\end{array}\end{array}q;-zxt{q}^{1-n}]\\ & (max\{|\alpha q|,|vxt|\}<1).\end{array}$$

(4.4)

Remark 13

For $c=e=0, b=d$ and $y=0,x=1$ in Theorem 12, Eq. (4.3) reduces to (4.1)

To prove Theorem 12, the following proposition is necessary.

Proposition 14

([4, Theorem 5.2] and [16, Eq. (III.4)])

We have

$$\begin{aligned} {}_{2}\Phi _{1}\left [ \textstyle\begin{array}{rr} a, b; \\ c; \end{array}\displaystyle q; z \right ] = \frac{ (abz/c;q)_{\infty }}{ (az/c;q)_{\infty }} {}_{3}\Phi _{2}\left [ \textstyle\begin{array}{r} b, c/a,0; \\ qc/(az),c; \end{array}\displaystyle q; q \right ] \end{aligned}$$

(4.5)

and

$$\begin{aligned} {}_{2}\Phi _{1}\left [ \textstyle\begin{array}{rr} a, b; \\ c; \end{array}\displaystyle q; z \right ]= \frac{ (bz; q)_{\infty }}{ (z;q)_{\infty }} {}_{2} \Phi _{2}\left [ \textstyle\begin{array}{rr} b,c/a; \\ bz,c; \end{array}\displaystyle q; az \right ]. \end{aligned}$$

(4.6)

Proof of Theorem 12

If we use $f(a,b,c,d,e,x,y,z)$ to denote the right-hand side of (4.3), we calculate that $f(a,b,c,d,e,x,y,z)$ satisfies (2.1). Thus, there exists a sequence $\{a_{n}\}$ independent of $x, y$ and z such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }a_{n} {}\omega _{n}^{\binom{a,b,c }{d,e}}(u,v,z|q). \end{aligned}$$

(4.7)

Letting $z=0$ in Eq. (4.7) and utilizing the obvious fact $\omega _{n}^{\binom{a,b,c}{d,e}}(u,v,0|q)=P_{n}(u,v)$, we have

$$\begin{array}{rl}f(a,b,c,d,e,u,v,0)& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{P}_n(u,v)=\frac{{(vt,\alpha x;q)}_{\mathrm{\infty }}}{{(ut,x;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{{(ut,\alpha ;q)}_k{q}^k}{{(q/x,vt,q;q)}_k}\\ & =\frac{{(vt,\alpha x;q)}_{\mathrm{\infty }}}{{(ut,x;q)}_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}ut,\alpha ,0;\\ q/x,vt;\end{array}\end{array}q;q]\text{by (4.5)}\\ & =\frac{{(vt;q)}_{\mathrm{\infty }}}{{(ut;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_1[\begin{array}{c}\begin{array}{r}v/u,\alpha ;\\ vt;\end{array}\end{array}q;xut]\\ & =\sum _{n=0}^{\mathrm{\infty }}{\varphi }_n^{(\alpha )}(x)P_n(u,v)\frac{t^n}{{(q;q)}_n}.\end{array}$$

Hence

$$\begin{aligned} f(a,b,c,d,e,u,v,z)=\sum_{n=0}^{\infty }\phi _{n}^{(\alpha )}(x)\omega _{n}^{\binom{a,b,c }{d,e}}(u,v,z|q) \frac{ t^{n}}{(q;q)_{n}}, \end{aligned}$$

which is equal to the left-hand side of (4.3).

Similarly, if we use $f(a,b,c,d,e,x,y,z)$ to denote the right-hand side of (4.4), we test that $g(a,b,c,d,e,x,y,z)$ satisfies (2.2). Thus, there exists a sequence $\{b_{n}\}$ independent of $x, y$ and z such that

$$\begin{aligned} g(a,b,c,d,e,u,v,z)=\sum_{n=0}^{\infty }b_{n} {}\zeta _{n}^{\binom{a,b,c }{d,e}}(u,v,z|q). \end{aligned}$$

(4.8)

Setting $z=0$ in Eq. (4.8), using the obvious fact $\zeta _{n}^{\binom{a,b,c}{d,e}}(u,v,0|q)=P_{n}(v,u)$, we have

$$\begin{array}{rl}& g(a,b,c,d,e,u,v,0)\\ & =\sum _{n=0}^{\mathrm{\infty }}{b}_n{P}_n(v,u)=\frac{{(q/x,uxtq;q)}_{\mathrm{\infty }}}{{(\alpha q,vxtq;q)}_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{q}^{(\frac{n}{2})}{(1/(\alpha x),1/(uxt);q)}_n}{{(q/x,1/(vxt),q;q)}_n}{\left(\frac{\alpha uq}{v}\right)}^n\\ & =\frac{{(q/x,uxtq;q)}_{\mathrm{\infty }}}{{(\alpha q,vxtq;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}1/(\alpha x),1/(uxt);\\ q/x,1/(vxt);\end{array}\end{array}q;\frac{\alpha uq}{v}]\text{by (4.6)}\\ & =\frac{{(uxtq;q)}_{\mathrm{\infty }}}{{(vxtq;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}u/v,1/(\alpha x);\\ 1/(vxt);\end{array}\end{array}q;\alpha q]\\ & =\sum _{n=0}^{\mathrm{\infty }}{\psi }_n^{(\alpha )}(x|q)P_n(v,u)\frac{{(qt)}^n}{{(q;q)}_n}.\end{array}$$

Hence

$$\begin{aligned} g(a,b,c,d,e,u,v,z)=\sum_{n=0}^{\infty }\psi _{n}^{(\alpha )}(x)\zeta _{n}^{\binom{a,b,c }{d,e}}(u,v,z|q) \frac{ (qt)^{n}}{(q;q)_{n}}, \end{aligned}$$

which is equal to the left-hand side of (2.2). This completes the proof of Theorem 12. □

Theorem 15

For $s\in \mathbb{N}$, we have

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\omega }_{n+s}^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(x,y,z|q)\frac{t^n}{{(q;q)}_n}=\frac{{(yt;q)}_{\mathrm{\infty }}}{t^s{(xt;q)}_{\mathrm{\infty }}}\sum _{k=0}^s\frac{{(q^{-s},xt;q)}_k{q}^k}{{(q;yt;q)}_k}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e;\end{array}\end{array}q;zt{q}^k]\\ & (max\{|xt|,|zt|\}<1)\end{array}$$

(4.9)

and

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\zeta }_{n+s}^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(x,y,z|q)\frac{t^n}{{(q;q)}_n}=\frac{{(xt;q)}_{\mathrm{\infty }}}{t^s{(yt;q)}_{\mathrm{\infty }}}\sum _{k=0}^s\frac{{(q^{-s},yt;q)}_k{q}^k}{{(q;xt;q)}_k}{}_3{\mathrm{\Phi }}_3[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e,0;\end{array}\end{array}q;-zt{q}^k]\\ & (|yt|<1).\end{array}$$

(4.10)

Corollary 16

([35, Eq. (2.1)])

For $s\in \mathbb{N}$ and $\max \{|z|,|xz|,|b|\}<1$, we have

$$\sum _{n=0}^{\mathrm{\infty }}{\mathrm{\Omega }}_{n+s}(x;a,b|q)\frac{z^n}{{(q;q)}_n}=\frac{{(b,axz,bz{q}^s;q)}_{\mathrm{\infty }}}{{(z,xz,b{q}^s;q)}_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{q}^{-s},a,x;\\ axz,q^{1-s}/b;\end{array}\end{array}q;\frac{qx}{b}].$$

(4.11)

Remark 17

For $c=e=0, b=d$ and $x=1$ in Theorem 15, Eq. (4.9) reduces to (4.11).

To prove Theorem 15, we need the following lemma.

Lemma 18

([16, Eq. (II.6)])

The q-Chu–Vandermonde formula is given by

$$\begin{aligned} {}_{2}\Phi _{1}\left [ \textstyle\begin{array}{rr} q^{-n}, a; \\ c; \end{array}\displaystyle q; q \right ]=\frac{(c/a;q)_{n}}{(c;q)_{n}} a^{n} \quad\bigl(n\in \mathbb{N}_{0}:=\mathbb{N}\cup \{0\}\bigr). \end{aligned}$$

(4.12)

Proof of Theorem 15

If we denote the right-hand side of Eq. (4.9) equivalently by

$$f(a,b,c,d,e,x,y,z)=t^{-s}\sum _{k=0}^s\frac{{(q^{-s};q)}_k{q}^k}{{(q;q)}_k}\frac{{(yt{q}^k;q)}_{\mathrm{\infty }}}{{(xt{q}^k;q)}_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e;\end{array}\end{array}q;zt{q}^k],$$

we test that $f(a,b,c,d,e,x,y,z)$ satisfies Eq. (2.1). Thus, there exists a sequence $\{\alpha _{n}\}$ independent of $x, y$ and z such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }\alpha _{n} \omega _{n}^{\binom{a,b,c }{d,e}}(x,y,z|q). \end{aligned}$$

We set $z=0$ in the above equation, using the notable fact $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,0|q)=P_{n}(x,y)$, we have

$$\begin{array}{rl}f(a,b,c,d,e,x,y,0)=& \sum _{n=0}^{\mathrm{\infty }}{\beta }_n{P}_n(x,y)=\frac{{(yt;q)}_{\mathrm{\infty }}}{t^s{(xt;q)}_{\mathrm{\infty }}}{}_2{\mathrm{\Phi }}_1[\begin{array}{c}\begin{array}{r}{q}^{-s},xt;\\ yt;\end{array}\end{array}q;q]\text{by (4.12)}\\ =& \frac{{(yt{q}^s;q)}_{\mathrm{\infty }}{P}_s(x,y)}{{(xt;q)}_{\mathrm{\infty }}}=\sum _{n=0}^{\mathrm{\infty }}{P}_{n+s}(x,y)\frac{t^n}{{(q;q)}_n}=\sum _{n=s}^{\mathrm{\infty }}{P}_n(x,y)\frac{t^{n-s}}{{(q;q)}_{n-s}}.\end{array}$$

We immediately conclude that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q) \frac{t^{n-s}}{(q;q)_{n-s}}=\sum _{n=0}^{\infty }\omega _{n+s}^{\binom{a,b,c }{d,e}}(x,y,z|q) \frac{t^{n}}{(q;q)_{n}}, \end{aligned}$$

which is equal to the left-hand side of (4.9).

Similarly, we get (4.10). This completes the proof of Theorem 15. □

5 Some new mixed generating functions for the Rajković–Marinković–Stanković polynomials

In this section, we give and prove the mixed generating functions for the Rajković–Marinković–Stanković polynomials.

Let a and b be two real numbers, the Thomae–Jackson q-integral is defined as [16, 18, 34]

$$\begin{aligned} \int _{a}^{b}f(x)\,\mathrm {d}_{q}x=(1-q) \sum_{n=0}^{\infty } \bigl[bf\bigl(bq^{n} \bigr)-af\bigl(aq^{n}\bigr) \bigr]q^{n}. \end{aligned}$$

(5.1)

Assume that $\alpha \in \mathbb{R}^{+}$ and $0< a< x<1$, the generalized Riemann–Liouville fractional q-integral operator is defined by [22] (see [11])

$$\begin{aligned} \bigl(I_{q,a}^{\alpha }f \bigr) (x)= \frac{x^{\alpha -1}}{\Gamma _{q}(\alpha )} \int _{a}^{x} (qt/x;q )_{\alpha -1}f(t)\,\mathrm {d}_{q}t. \end{aligned}$$

(5.2)

Due to the q-integral (5.1), we rewrite fractional q-integral (5.2) equivalently as follows (see [9, 11, 14]):

$$\begin{aligned} \bigl(I_{q,a}^{\alpha }f \bigr) (x)&= \frac{x^{\alpha -1}(1-q)}{\Gamma _{q}(\alpha )} \sum_{n=0}^{\infty } \bigl[x \bigl(q^{n+1};q \bigr)_{\alpha -1}f \bigl(xq^{n} \bigr) -a \bigl(aq^{n+1}/x;q \bigr)_{\alpha -1}f \bigl(aq^{n} \bigr) \bigr]q^{n}. \end{aligned}$$

(5.3)

Recall that the Rajković–Marinković–Stanković polynomials are defined [22] (see [8, 11]) by

$$\begin{aligned} \mathcal{P}_{n}(\alpha,a,x|q)=I^{\alpha }_{q,a} \bigl\{ x^{n} \bigr\} = \sum_{k=0}^{\infty } \begin{bmatrix} n \\ k \end{bmatrix}_{q} \frac{[k]_{q}! a^{n-k}}{\Gamma _{q}(\alpha +k+1)} x^{\alpha +k}(a/x;q)_{ \alpha +k}, \end{aligned}$$

(5.4)

where $\alpha \in \mathbb{R}^{\ast }$ and $0< a< x<1$.

We have the following lemmas.

Lemma 19

([8, Lemma 10])

For $\alpha \in \mathbb{R}^{+}, 0< a< x<1$, we have

$$\begin{aligned} \sum_{n=0}^{\infty } \mathcal{P}_{n}(\alpha,a,x|q) \frac{w^{n}}{(q;q)_{n}}=\frac{(1-q)^{\alpha }}{(aw;q)_{\infty }}\sum _{k=0}^{\infty }\frac{x^{\alpha +k}(a/x;q)_{\alpha +k}w^{k}}{(q;q)_{\alpha +k}}. \end{aligned}$$

(5.5)

Lemma 20

([8, Theorem 3])

For $\alpha \in \mathbb{R}^{+}, 0< a< x<1$ and if $\max \{ |at|,|az| \} <1$, we have

$$\begin{array}{rl}& I_{q,a}^{\alpha }\left\{\frac{{(bxz,tx;q)}_{\mathrm{\infty }}}{{(xs,xz;q)}_{\mathrm{\infty }}}\right\}\\ & =\frac{{(1-q)}^{\alpha }{(abz,at;q)}_{\mathrm{\infty }}}{{(as,az;q)}_{\mathrm{\infty }}}\sum _{k=0}^{\mathrm{\infty }}\frac{x^{\alpha +k}{(a/x;q)}_{\alpha +k}}{a^k{(q;q)}_{\alpha +k}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{q}^{-k},as,az;\\ at,abz;\end{array}\end{array}q;q].\end{array}$$

(5.6)

Remark 21

Upon taking $z=0$ in (5.6) and by the means of the q-Chu–Vandermonde formula (4.12), we obtain

$$\begin{aligned} I^{\alpha }_{q,a} \biggl\{ \frac{(xt;q)_{\infty }}{(xs;q)_{\infty }} \biggr\} = \frac{(1-q)^{\alpha }(at;q)_{\infty }}{(as;q)_{\infty }}\sum_{k=0}^{\infty } \frac{x^{\alpha +k}(a/x;q)_{\alpha +k}}{(q;q)_{\alpha +k}} \frac{(t/s;q)_{k} s^{k}}{(at;q)_{k}},\quad \vert as \vert < 1. \end{aligned}$$

(5.7)

Using Lemma 20 and the theory of q-difference equations, we are able to deduce the following new mixed generating functions for the Rajković–Marinković–Stanković polynomials.

Theorem 22

For $\alpha \in \mathbb{R}^{+},0< a< x<1$, and $\max \{ |aws|,|awt| \} <1$, we have

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_n(\alpha ,a,x|q){\omega }_n^{\left(\genfrac{}{}{0ex}{}{a_1,b_1,c_1}{d_1,e_1}\right)}(s,t,r|q)\frac{w^n}{{(q;q)}_n}\\ & =\frac{{(1-q)}^{\alpha }{(awt;q)}_{\mathrm{\infty }}}{{(aws;q)}_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1,awt;q)}_n{(r/s)}^n}{{(q,d_1,e_1,t/s;q)}_n}\sum _{k=0}^n\frac{{(q^{-n},aws;q)}_k{q}^k}{{(q,awt;q)}_k}\\ & \times \sum _{m=0}^{\mathrm{\infty }}\frac{x^{\alpha +m}{(a/x;q)}_{\alpha +m}}{a^m{(q;q)}_{\alpha +m}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{q}^{-m},aws{q}^k,awt{q}^n;\\ awt,awt{q}^k;\end{array}\end{array}q;q]\end{array}$$

(5.8)

and

$$\begin{array}{rl}& \sum _{n=0}^{\mathrm{\infty }}{\mathcal{P}}_n(\alpha ,a,x|q){\zeta }_n^{\left(\genfrac{}{}{0ex}{}{a_1,b_1,c_1}{d_1,e_1}\right)}(s,t,r|q)\frac{w^n}{{(q;q)}_n}\\ & =\frac{{(1-q)}^{\alpha }{(aws;q)}_{\mathrm{\infty }}}{{(awt;q)}_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{q^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{(a_1,b_1,c_1,aws;q)}_n{(r/t)}^n}{{(q,d_1,e_1,s/t;q)}_n}\sum _{k=0}^n\frac{{(q^{-n},awt;q)}_k{q}^k}{{(q,aws;q)}_k}\\ & \times \sum _{m=0}^{\mathrm{\infty }}\frac{x^{\alpha +m}{(a/x;q)}_{\alpha +m}}{a^m{(q;q)}_{\alpha +m}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{q}^{-m},awt{q}^k,aws{q}^n;\\ aws,aws{q}^k;\end{array}\end{array}q;q].\end{array}$$

(5.9)

Proof of Theorem 22

The LHS of Eq. (5.8) is equal to

$$\begin{array}{rl}& I_{q,a}^{\alpha }\{\sum _{n=0}^{\mathrm{\infty }}{\omega }_n^{\left(\genfrac{}{}{0ex}{}{a_1,b_1,c_1}{d_1,e_1}\right)}(s,t,r|q)\frac{{(xw)}^n}{{(q;q)}_n}\}\\ & =I_{q,a}^{\alpha }\left\{\frac{{(xwt;q)}_{\mathrm{\infty }}}{xws;q_{\mathrm{\infty }}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{a}_1,b_1,c_1;\\ d_1,e_1;\end{array}\end{array}q;rxw]\right\}\\ & =I_{q,a}^{\alpha }\left\{\frac{{(xwt;q)}_{\mathrm{\infty }}}{xws;q_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1;q)}_n}{{(q,d_1,e_1;q)}_n}{(rw)}^n{x}^n\right\}\\ & =I_{q,a}^{\alpha }\left\{\frac{{(xwt;q)}_{\mathrm{\infty }}}{xws;q_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1;q)}_n{(rw)}^n}{{(q,d_1,e_1;q)}_n}\frac{{(xwt;q)}_n}{{(t/s;q)}_n{(ws)}^n}\sum _{k=0}^n\frac{{(q^{-n},xws;q)}_k}{{(q,xwt;q)}_k}{q}^k\right\}\\ & =I_{q,a}^{\alpha }\left\{\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1;q)}_n}{{(q,d_1,e_1;q)}_n}{(r/s)}^n\frac{{(xwt;q)}_n}{{(t/s;q)}_n}\sum _{k=0}^n\frac{{(xwt{q}^k;q)}_{\mathrm{\infty }}{(q^{-n};q)}_k}{{(xws{q}^k;q)}_{\mathrm{\infty }}{(q;q)}_k}{q}^k\right\}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1;q)}_n}{{(q,d_1,e_1;q)}_n}\frac{{(r/s)}^n}{{(t/s;q)}_n}\sum _{k=0}^n\frac{{(q^{-n};q)}_k{q}^k}{{(q;q)}_k}{I}_{q,a}^{\alpha }\left\{\frac{{(xwt,xwt{q}^k;q)}_{\mathrm{\infty }}}{{(xwt{q}^n,xws{q}^k;q)}_{\mathrm{\infty }}}\right\}\\ & =\frac{{(1-q)}^{\alpha }{(awt;q)}_{\mathrm{\infty }}}{{(aws;q)}_{\mathrm{\infty }}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(a_1,b_1,c_1,awt;q)}_n{(r/s)}^n}{{(q,d_1,e_1,t/s;q)}_n}\sum _{k=0}^n\frac{{(q^{-n},aws;q)}_k{q}^k}{{(q,awt;q)}_k}\\ & \times \sum _{m=0}^{\mathrm{\infty }}\frac{x^{\alpha +m}{(a/x;q)}_{\alpha +m}}{a^m{(q;q)}_{\alpha +m}}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}{q}^{-m},aws{q}^k,awt{q}^n;\\ awt,awt{q}^k;\end{array}\end{array}q;q],\end{array}$$

which equals the RHS of Eq. (5.8) after using (5.6). Similarly, we get (5.9). This completes the proof of Theorem 22. □

Remark 23

For $(t,r)=(0,0)$ in Theorem 22, we get (5.5).

6 The $U(n+1)$ generalizations of generating functions for q-hypergeometric polynomials

Lemma 24

([21, Theorem 5.42])

Let $b,z$ and $x_{1},\ldots,x_{n}$ be indeterminate, and let $n\geq 1$. Suppose that none of the denominators in the following identity vanishes, $0<|q|<1$ and $|z|<|x_{1},\ldots,x_{n}||x_{m}|^{-n}|q|^{(n-1)/2}$, for $m=1,2,\ldots,n$. Then we have

$$\begin{aligned} &\mathop{\sum_{y_{k} \geq 0}}_{ k=1,2,\ldots,n} \Biggl\{ \prod _{1\leq r< s\leq n} \biggl[ \frac{1-\frac{x_{r}}{x_{s}}q^{y_{r}-y_{s}})}{1-\frac{x_{r}}{x_{s}}} \biggr]\prod _{r,s=1}^{n} \biggl(q\frac{x_{r}}{x_{s}};q \biggr)^{-1}_{y_{r}} \prod_{i=1}^{n}(x_{i})^{ny_{i}-(y_{1}+\cdots+y_{n})}(-1)^{(n-1)(y_{1}+\cdots+y_{n})} \\ &\qquad{} \times q^{y_{2}+2y_{3}+\cdots+(n-1)y_{n}+(n-1) [\binom{y_{1}}{2}+\cdots+\binom{y_{n}}{2} ]-e_{2}(y_{1},\ldots,y_{n})}(b;q)_{y_{1}+\cdots+y_{n}}z^{y_{1}+\cdots+y_{n}} \Biggr\} \\ & \quad= \frac{(bz;q)_{\infty }}{(z;q)_{\infty }}, \end{aligned}$$

(6.1)

where $e_{2}(y_{1},\ldots,y_{n})$ is the second elementary symmetric function of $\{y_{1},\ldots,y_{n}\}$.

In this part, using the method of homogeneous q-difference equations, we derive the following $U(n+1)$ type generating functions for q-hypergeometric polynomials.

Theorem 25

Let $b, z, x_{1},\ldots x_{n}, n\geq 1$ be indeterminate. Suppose that none of the denominators in the following identity vanishes, and that $0<|q|<1$, and $|z|<|x_{1},\ldots,x_{n}|\times |x_{m}|^{-n}|q|^{(n-1)/2}$, for $m=1,2,\ldots,n$. Then we have the following:

$$\begin{array}{rl}& \underset{k=1,2,\dots ,n}{\sum _{y_k\ge 0}}\{\prod _{1\le r<s\le n}\left[\frac{1-\frac{x_r}{x_s}{q}^{y_r-y_s})}{1-\frac{x_r}{x_s}}\right]\prod _{r,s=1}^n{(q\frac{x_r}{x_s};q)}_{y_r}^{-1}\prod _{i=1}^n{(x_i)}^{n{y}_i-(y_1+\cdots +y_n)}{(-1)}^{(n-1)(y_1+\cdots +y_n)}\\ & \times q^{y_2+2y_3+\cdots +(n-1)y_n+(n-1)[\left(\genfrac{}{}{0ex}{}{y_1}{2}\right)+\cdots +\left(\genfrac{}{}{0ex}{}{y_n}{2}\right)]-e_2(y_1,\dots ,y_n)}{\omega }_{s+y_1+\cdots +y_n}^{\left(\genfrac{}{}{0ex}{}{a,b,c}{d,e}\right)}(x,y,z|q)t^{y_1+\cdots +y_n}\}\\ & =\frac{{(yt;q)}_{\mathrm{\infty }}}{t^s{(xt;q)}_{\mathrm{\infty }}}\sum _{k=0}^s\frac{{(q^{-s},xt;q)}_k{q}^k}{{(q,yt;q)}_k}{}_3{\mathrm{\Phi }}_2[\begin{array}{c}\begin{array}{r}a,b,c;\\ d,e;\end{array}\end{array}q;zt{q}^k],\end{array}$$

(6.2)

where $a=q^{-M}$ and $|xt|<1$.

Remark 26

Setting $n=1$ in Theorem 25, the assertion (6.2) reduces to (4.9).

Proof of Theorem 25

Upon taking $(b,z)=(yq^{s}/x,xt)$ in Eq. (6.1), we obtain

$$\begin{aligned} &\mathop{\sum_{y_{k} \geq 0}}_{ k=1,2,\ldots,n} \Biggl\{ \prod _{1\leq r< s\leq n} \biggl[ \frac{1-\frac{x_{r}}{x_{s}}q^{y_{r}-y_{s}})}{1-\frac{x_{r}}{x_{s}}} \biggr]\prod _{r,s=1}^{n} \biggl(q\frac{x_{r}}{x_{s}};q \biggr)^{-1}_{y_{r}} \prod_{i=1}^{n}(x_{i})^{ny_{i}-(y_{1}+\cdots+y_{n})}(-1)^{(n-1)(y_{1}+\cdots+y_{n})} \\ &\qquad{} \times q^{y_{2}+2y_{3}+\cdots+(n-1)y_{n}+(n-1) [\binom{y_{1}}{2}+\cdots+\binom{y_{n}}{2} ]-e_{2}(y_{1},\ldots,y_{n})}P_{s+y_{1}+\cdots+y_{n}}\bigl(x,yq^{s} \bigr)t^{y_{1}+\cdots+y_{n}} \Biggr\} \\ &\quad = \frac{(ytq^{s};q)_{\infty }}{(xt;q)_{\infty }}. \end{aligned}$$

(6.3)

If we use $f(a,b,c,d,e,x,y,z)$ to denote the left-hand side of Eq. (6.2), we can verify that $f(a,b,c,d,e,x,y,z)$ satisfies Eq. (2.1). There exists a sequence $\{\beta _{n}\}$ such that

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }\beta _{n}{}\omega _{n}^{\binom{a,b,c }{d,e}}(x,y,z|q). \end{aligned}$$

(6.4)

Setting $z=0$ in Eq. (6.4) and then, using the obvious fact $\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,0|q)=P_{n}(x,y)$, we have

$$\begin{aligned} &f(a,b,c,d,e,x,y,0)\\ &\quad=\sum_{n=0}^{\infty }\beta _{n} P_{n}(x,y) \\ &\quad=\mathop{\sum_{y_{k} \geq 0}}_{ k=1,2,\ldots,n} \Biggl\{ \prod _{1\leq r< s\leq n} \biggl[ \frac{1-\frac{x_{r}}{x_{s}}q^{y_{r}-y_{s}})}{1-\frac{x_{r}}{x_{s}}} \biggr]\prod _{r,s=1}^{n} \biggl(q\frac{x_{r}}{x_{s}};q \biggr)^{-1}_{y_{r}} \prod_{i=1}^{n}(x_{i})^{ny_{i}-(y_{1}+\cdots+y_{n})}(-1)^{(n-1)(y_{1}+\cdots+y_{n})} \\ & \qquad{}\times q^{y_{2}+2y_{3}+\cdots+(n-1)y_{n}+(n-1) [\binom{y_{1} }{2}+\cdots+\binom{y_{n}}{2} ]-e_{2}(y_{1},\ldots,y_{n})}P_{s+y_{1}+\cdots+y_{n}}\bigl(x,yq^{s} \bigr)t^{y_{1}+\cdots+y_{n}} \Biggr\} \\ &\quad=\frac{P_{s}(x,y)(ytq^{s};q)_{\infty }}{(xt;q)_{\infty }} \\ &\quad=\sum_{n=0}^{\infty }P_{s+n}(x,y) \frac{t^{n}}{(q;q)_{n}}. \end{aligned}$$

Hence

$$\begin{aligned} f(a,b,c,d,e,x,y,z)=\sum_{n=0}^{\infty }\omega _{n}^{\binom{a,b,c}{d,e}}(x,y,z|q) \frac{t^{n-s}}{(q;q)_{n-s}}, \end{aligned}$$

which is equal to the right-hand side of (6.2) by (4.9). The proof of Theorem 25 is complete. □

We remark in passing that, in a recently-published survey-cum-expository review article, the so-called $(p,q)$-calculus was exposed to be a rather trivial and inconsequential variation of the classical q-calculus, the additional parameter p being redundant or superfluous (see, for details, [26, p. 340]).