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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.

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)]):

Φ s r [ a 1 , a 2 , , a r ; b 1 , b 2 , , b s ; q ; z ] = n = 0 ( a 1 , a 2 , , a r ; q ) n ( q , b 1 , b 2 , , b s ; q ) n [ ( 1 ) n q ( n 2 ) ] 1 + s 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)]:

n = 0 ϕ n ( a , b , c ) (x,y|q) t n ( q ; q ) n = 1 ( x t ; q ) 2 Φ 1 [ a , b ; c ; q ; y t ] ( max { | y t | , | x t | } < 1 ) ,
(1.10)
n = 0 ψ n ( a , b , c ) (x,y|q) ( 1 ) n q ( n 2 ) t n ( q ; q ) n = ( x t ; q ) 2 Φ 1 [ a , b ; c ; q ; y t ] ( | x t | < 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. (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. (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. (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. (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. (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. (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. (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.

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. □

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

n = 0 ω n ( a , b , c d , e ) (x,y,z|q) t n ( q ; q ) n = ( y t ; q ) ( x t ; q ) 3 Φ 2 [ a , b , c ; d , e ; q ; z t ] ( max { | x t | , | z t | } < 1 )
(3.1)

and

n = 0 ζ n ( a , b , c d , e ) (x,y,z|q) t n ( q ; q ) n = ( x t ; q ) ( y t ; q ) 3 Φ 3 [ a , b , c ; 0 , d , e ; q ; z t ] ( | y t | < 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. □

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

n = 0 ϕ n ( α ) (z|q) ( λ ; q ) n t n ( q ; q ) n = ( λ t ; q ) ( t ; q ) 2 Φ 1 [ λ , α ; λ t ; q ; z t ] ( max { | t | , | x t | } < 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

n = 0 ϕ n ( α ) ( x | q ) ω n ( a , b , c d , e ) ( u , v , z | q ) t n ( q ; q ) n = ( v t , α x ; q ) ( u t , x ; q ) k = 0 ( u t , α ; q ) k q k ( q / x , v t , q ; q ) k 3 Φ 2 [ a , b , c ; d , e ; q ; z t q k ] ( max { | u t | , | z t | , | x | } < 1 )
(4.3)

and

n = 0 ψ n ( α ) ( x | q ) ζ n ( a , b , c d , e ) ( u , v , z | q ) t n ( q ; q ) n = ( q / x , u x t q ; q ) ( α q , v x t q ; q ) × n = 0 ( 1 ) n q ( n 2 ) ( 1 / ( α x ) , 1 / ( u x t ) ; q ) n ( q / x , 1 / ( v x t ) , q ; q ) n ( α u q v ) n 3 Φ 3 [ a , b , c ; 0 , d , e ; q ; z x t q 1 n ] ( max { | α q | , | v x t | } < 1 ) .
(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

f ( a , b , c , d , e , u , v , 0 ) = n = 0 a n P n ( u , v ) = ( v t , α x ; q ) ( u t , x ; q ) k = 0 ( u t , α ; q ) k q k ( q / x , v t , q ; q ) k = ( v t , α x ; q ) ( u t , x ; q ) 3 Φ 2 [ u t , α , 0 ; q / x , v t ; q ; q ] by (4.5) = ( v t ; q ) ( u t ; q ) 2 Φ 1 [ v / u , α ; v t ; q ; x u t ] = n = 0 ϕ n ( α ) ( x ) P n ( u , v ) t n ( q ; q ) n .

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

g ( a , b , c , d , e , u , v , 0 ) = n = 0 b n P n ( v , u ) = ( q / x , u x t q ; q ) ( α q , v x t q ; q ) n = 0 ( 1 ) n q ( n 2 ) ( 1 / ( α x ) , 1 / ( u x t ) ; q ) n ( q / x , 1 / ( v x t ) , q ; q ) n ( α u q v ) n = ( q / x , u x t q ; q ) ( α q , v x t q ; q ) 2 Φ 2 [ 1 / ( α x ) , 1 / ( u x t ) ; q / x , 1 / ( v x t ) ; q ; α u q v ] by (4.6) = ( u x t q ; q ) ( v x t q ; q ) 2 Φ 2 [ u / v , 1 / ( α x ) ; 1 / ( v x t ) ; q ; α q ] = n = 0 ψ n ( α ) ( x | q ) P n ( v , u ) ( q t ) n ( q ; q ) n .

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

n = 0 ω n + s ( a , b , c d , e ) ( x , y , z | q ) t n ( q ; q ) n = ( y t ; q ) t s ( x t ; q ) k = 0 s ( q s , x t ; q ) k q k ( q ; y t ; q ) k 3 Φ 2 [ a , b , c ; d , e ; q ; z t q k ] ( max { | x t | , | z t | } < 1 )
(4.9)

and

n = 0 ζ n + s ( a , b , c d , e ) ( x , y , z | q ) t n ( q ; q ) n = ( x t ; q ) t s ( y t ; q ) k = 0 s ( q s , y t ; q ) k q k ( q ; x t ; q ) k 3 Φ 3 [ a , b , c ; d , e , 0 ; q ; z t q k ] ( | y t | < 1 ) .
(4.10)

Corollary 16

([35, Eq. (2.1)])

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

n = 0 Ω n + s (x;a,b|q) z n ( q ; q ) n = ( b , a x z , b z q s ; q ) ( z , x z , b q s ; q ) 3 Φ 2 [ q s , a , x ; a x z , q 1 s / b ; q ; q x 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 k = 0 s ( q s ; q ) k q k ( q ; q ) k ( y t q k ; q ) ( x t q k ; q ) 3 Φ 2 [ a , b , c ; d , e ; q ; z t 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

f ( a , b , c , d , e , x , y , 0 ) = n = 0 β n P n ( x , y ) = ( y t ; q ) t s ( x t ; q ) 2 Φ 1 [ q s , x t ; y t ; q ; q ] by (4.12) = ( y t q s ; q ) P s ( x , y ) ( x t ; q ) = n = 0 P n + s ( x , y ) t n ( q ; q ) n = n = s P n ( x , y ) t n s ( q ; q ) n s .

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. □

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

I q , a α { ( b x z , t x ; q ) ( x s , x z ; q ) } = ( 1 q ) α ( a b z , a t ; q ) ( a s , a z ; q ) k = 0 x α + k ( a / x ; q ) α + k a k ( q ; q ) α + k 3 Φ 2 [ q k , a s , a z ; a t , a b z ; q ; q ] .
(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

n = 0 P n ( α , a , x | q ) ω n ( a 1 , b 1 , c 1 d 1 , e 1 ) ( s , t , r | q ) w n ( q ; q ) n = ( 1 q ) α ( a w t ; q ) ( a w s ; q ) n = 0 ( a 1 , b 1 , c 1 , a w t ; q ) n ( r / s ) n ( q , d 1 , e 1 , t / s ; q ) n k = 0 n ( q n , a w s ; q ) k q k ( q , a w t ; q ) k × m = 0 x α + m ( a / x ; q ) α + m a m ( q ; q ) α + m 3 Φ 2 [ q m , a w s q k , a w t q n ; a w t , a w t q k ; q ; q ]
(5.8)

and

n = 0 P n ( α , a , x | q ) ζ n ( a 1 , b 1 , c 1 d 1 , e 1 ) ( s , t , r | q ) w n ( q ; q ) n = ( 1 q ) α ( a w s ; q ) ( a w t ; q ) n = 0 q ( n 2 ) ( a 1 , b 1 , c 1 , a w s ; q ) n ( r / t ) n ( q , d 1 , e 1 , s / t ; q ) n k = 0 n ( q n , a w t ; q ) k q k ( q , a w s ; q ) k × m = 0 x α + m ( a / x ; q ) α + m a m ( q ; q ) α + m 3 Φ 2 [ q m , a w t q k , a w s q n ; a w s , a w s q k ; q ; q ] .
(5.9)

Proof of Theorem 22

The LHS of Eq. (5.8) is equal to

I q , a α { n = 0 ω n ( a 1 , b 1 , c 1 d 1 , e 1 ) ( s , t , r | q ) ( x w ) n ( q ; q ) n } = I q , a α { ( x w t ; q ) x w s ; q 3 Φ 2 [ a 1 , b 1 , c 1 ; d 1 , e 1 ; q ; r x w ] } = I q , a α { ( x w t ; q ) x w s ; q n = 0 ( a 1 , b 1 , c 1 ; q ) n ( q , d 1 , e 1 ; q ) n ( r w ) n x n } = I q , a α { ( x w t ; q ) x w s ; q n = 0 ( a 1 , b 1 , c 1 ; q ) n ( r w ) n ( q , d 1 , e 1 ; q ) n ( x w t ; q ) n ( t / s ; q ) n ( w s ) n k = 0 n ( q n , x w s ; q ) k ( q , x w t ; q ) k q k } = I q , a α { n = 0 ( a 1 , b 1 , c 1 ; q ) n ( q , d 1 , e 1 ; q ) n ( r / s ) n ( x w t ; q ) n ( t / s ; q ) n k = 0 n ( x w t q k ; q ) ( q n ; q ) k ( x w s q k ; q ) ( q ; q ) k q k } = n = 0 ( a 1 , b 1 , c 1 ; q ) n ( q , d 1 , e 1 ; q ) n ( r / s ) n ( t / s ; q ) n k = 0 n ( q n ; q ) k q k ( q ; q ) k I q , a α { ( x w t , x w t q k ; q ) ( x w t q n , x w s q k ; q ) } = ( 1 q ) α ( a w t ; q ) ( a w s ; q ) n = 0 ( a 1 , b 1 , c 1 , a w t ; q ) n ( r / s ) n ( q , d 1 , e 1 , t / s ; q ) n k = 0 n ( q n , a w s ; q ) k q k ( q , a w t ; q ) k × m = 0 x α + m ( a / x ; q ) α + m a m ( q ; q ) α + m 3 Φ 2 [ q m , a w s q k , a w t q n ; a w t , a w t q k ; q ; q ] ,

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).

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:

y k 0 k = 1 , 2 , , n { 1 r < s n [ 1 x r x s q y r y s ) 1 x r x s ] r , s = 1 n ( q x r x s ; q ) y r 1 i = 1 n ( x i ) n y i ( y 1 + + y n ) ( 1 ) ( n 1 ) ( y 1 + + y n ) × q y 2 + 2 y 3 + + ( n 1 ) y n + ( n 1 ) [ ( y 1 2 ) + + ( y n 2 ) ] e 2 ( y 1 , , y n ) ω s + y 1 + + y n ( a , b , c d , e ) ( x , y , z | q ) t y 1 + + y n } = ( y t ; q ) t s ( x t ; q ) k = 0 s ( q s , x t ; q ) k q k ( q , y t ; q ) k 3 Φ 2 [ a , b , c ; d , e ; q ; z t q k ] ,
(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]).

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This work was supported by the Zhejiang Provincial Natural Science Foundation of China (No. LY21A010019).

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Cao, J., Zhou, HL. & Arjika, S. Generalized homogeneous q-difference equations for q-polynomials and their applications to generating functions and fractional q-integrals. Adv Differ Equ 2021, 329 (2021). https://doi.org/10.1186/s13662-021-03484-9

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MSC

  • 05A30
  • 11B65
  • 33D15
  • 33D45
  • 33D60
  • 39A13
  • 39B32

Keywords

  • Homogeneous q-difference equations
  • Verma–Jain polynomials
  • Srivastava–Agarwal type generating functions
  • Fractional q-integrals
  • \(U(n+1)\) type generating function