q-Binomial Convolution and Transformations of q-Appell Polynomials

Abstract

In this paper, binomial convolution in the frame of quantum calculus is studied for the set ${\mathcal{A}}_q$ of q-Appell sequences. It has been shown that the set ${\mathcal{A}}_q$ of q-Appell sequences forms an Abelian group under the operation of binomial convolution. Several properties for this Abelian group structure ${\mathcal{A}}_q$ have been studied. A new definition of the q-Appell polynomials associated with a random variable is proposed. Scale transformation as well as transformation based on expectation with respect to a random variable is used to present the determinantal form of q-Appell sequences.

1. Quantum Binomial Convolutions and Generating Functions

Let $\mathbb{N}$ denote the set of positive integers and ${\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}.$ Now, onwards $\mu \in {\mathbb{N}}_0,$ $y\in \mathbb{R},$ and $z\in \mathbb{C},$ satisfying $\mid z\mid \le r,$ $r>0.$ Let us denote by ${\mathcal{G}}_q$ the set of all real sequences ${\mathbf{u}}_q={\left(u_{\mu ,q}\right)}_{\mu \ge 0}$ where $u_{0,q}\ne 0$ and:

$$\sum _{\mu =0}^{\infty }\mid u_{\mu ,q}\mid \frac{r^{\mu }}{{\left[\mu \right]}_q!}<\infty .$$

If ${\mathbf{u}}_q\in {\mathcal{G}}_q,$ then its generating function will be denoted by

$$\mathsf{G}({\mathbf{u}}_q,z)=\sum _{\mu =0}^{\infty }{u}_{\mu ,q}\frac{z^{\mu }}{{\left[\mu \right]}_q!}.$$

The q-binomial convolution [8] of ${\mathbf{u}}_q$ and ${\mathbf{v}}_q,$ will be denoted by ${\mathbf{u}}_q{\times }_q{\mathbf{v}}_q={\left({\left(u_q{\times }_q{v}_q\right)}_{\mu }\right)}_{\mu \ge 0}$ for ${\mathbf{u}}_q$ and ${\mathbf{v}}_q\in {\mathcal{G}}_q$ is defined as

$${\left(u_q{\times }_q{v}_q\right)}_{\mu }=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{u}_{k,q}{v}_{\mu -k,q}.$$

(2)

The q-addition is a special case of the q-binomial convolution [29].

The q-multinomial coefficient is given by

$${\left[\begin{array}{c}\mu \\ j_1,\cdots ,j_m\end{array}\right]}_q=\frac{{\left[\mu \right]}_q!}{{\left[j_1\right]}_q!\cdots {\left[j_m\right]}_q!},j_1+\cdots +j_m=\mu ,j_i=0,1,\cdots ,\mu ,i=0,1,\cdots ,mand\mu \in {\mathbb{N}}_0,$$

Proposition 1.

Let that ${\mathit{u}}_q^{\left(k\right)}={\left(u_{\mu ,q}^{\left(k\right)}\right)}_{\mu \ge 0}\in {\mathcal{G}}_q,k=1,2,\cdots ,m,$μ belong to the set of positive integers (see [29]). Then, ${\mathit{u}}_q^{\left(1\right)}{\times }_q\cdots {\times }_q{\mathit{u}}_q^{\left(m\right)}\in {\mathcal{G}}_q$ and:

$${({\mathit{u}}_q^{\left(1\right)}{\times }_q\cdots {\times }_q{\mathit{u}}_q^{\left(m\right)})}_{\mu }=\sum _{j_1+\cdots +j_m=\mu }{\left[\begin{array}{c}\mu \\ j_1,j_2,\cdots ,j_m\end{array}\right]}_q{u}_{j_1,q}^{\left(1\right)}\cdots u_{j_m,q}^{\left(m\right)}.$$

In addition:

$$\mathsf{G}({\mathit{u}}_q^{\left(1\right)}{\times }_q\cdots {\times }_q{\mathit{u}}_q^{\left(m\right)},z)=\mathsf{G}({\mathit{u}}_q^{\left(1\right)},z)\cdots \mathsf{G}({\mathit{u}}_q^{\left(m\right)},z).$$

(3)

Proof. 

Suppose ${\mathbf{u}}_q$ and ${\mathbf{v}}_q\in {\mathcal{G}}_q$ with r and s as their radii, respectively. Let $t=min(r,s)$. Then, from (2), we have:

$$\begin{array}{c}\hfill \mathsf{G}(\mid \left({\mathbf{u}}_q{\times }_q{\mathbf{v}}_q\right)\mid ,t)=\sum _{\mu =0}^{\infty }\mid {\left(u_q{\times }_q{v}_q\right)}_{\mu }\mid \frac{t^{\mu }}{{\left[\mu \right]}_q!}\le \sum _{\mu =0}^{\infty }\frac{t^{\mu }}{{\left[\mu \right]}_q!}\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q\mid u_{k,q}\mid \mid v_{\mu -k,q}\mid \\ \hfill =\sum _{k=0}^{\infty }\mid u_{k,q}\mid \frac{t^k}{{\left[k\right]}_q!}\sum _{\mu =k}^{\infty }\mid v_{\mu -k,q}\mid \frac{t^{\mu -k}}{{[\mu -k]}_q!}=\mathsf{G}(\mid {\mathbf{u}}_q\mid ,t)\mathsf{G}(\mid {\mathbf{v}}_q\mid ,t).\end{array}$$

□

Then $\mathsf{G}({\mathbf{u}}_q{\times }_q{\mathbf{v}}_q,z)=\mathsf{G}({\mathbf{u}}_q,z)\mathsf{G}({\mathbf{v}}_q,z),$ $\mid z\mid \le t.$ Thus, by applying the induction on m, result follows.

Corollary 1.

$({\mathcal{G}}_q,\times )$ is an Abelian group having an identity element as $e_q={\left({\delta }_{\mu 0}\right)}_{\mu \ge 0}$, where ${\delta }_{00}=1$ and ${\delta }_{\mu 0}=0$ for $\mu \in \mathbb{N}.$

Proof. 

Closure: for ${\mathbf{u}}_q,{\mathbf{v}}_q\in {\mathcal{G}}_q,$ then ${\mathbf{u}}_q{\times }_q{\mathbf{v}}_q={\left({\left(u_q{\times }_q{v}_q\right)}_{\mu }\right)}_{\mu \ge 0}\in {\mathcal{G}}_q$ as ${\left(u_q{\times }_q{v}_q\right)}_{\mu }={\displaystyle \sum _{k=0}^{\mu }}{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{u}_{k,q}{v}_{\mu -k,q}\in \mathbb{R}.$

Associativity: let ${\mathbf{u}}_q,{\mathbf{v}}_q,{\mathbf{w}}_q\in {\mathcal{G}}_q$ be any elements:

$\left({\mathbf{u}}_q{\times }_q{\mathbf{v}}_q\right){\times }_q{\mathbf{w}}_q={\left({\left(\left(u_q{\times }_q{v}_q\right){\times }_q{w}_q\right)}_{\mu }\right)}_{\mu \ge 0}$ and ${\mathbf{u}}_q{\times }_q\left({\mathbf{v}}_q{\times }_q{\mathbf{w}}_q\right)={\left((u_q{\times }_q{\left(v_q{\times }_q{w}_q\right)}_{\mu }\right)}_{\mu \ge 0}$

$$\begin{array}{cc}\hfill {\left(u_q{\times }_q\left(v_q{\times }_q{w}_q\right)\right)}_{\mu }& =\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{u}_{s,q}{\left(v_q{\times }_q{w}_q\right)}_{\mu -s}\hfill \\ \hfill & =\sum _{s=0}^{\mu }\sum _{r=0}^{\mu -s}{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{\left[\begin{array}{c}\mu -s\\ r\end{array}\right]}_q{u}_{s,q}{v}_{r,q}{w}_{\mu -s-r,q}\hfill \\ \hfill & =\sum _{s=0}^{\mu }\sum _{r=s}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{\left[\begin{array}{c}\mu -s\\ r-s\end{array}\right]}_q{u}_{s,q}{v}_{r-s,q}{w}_{\mu -r,q}\hfill \\ \hfill & =\sum _{r=0}^{\mu }\sum _{s=0}^r{\left[\begin{array}{c}\mu \\ r\end{array}\right]}_q{\left[\begin{array}{c}r\\ s\end{array}\right]}_q{u}_{s,q}{v}_{r-s,q}{w}_{\mu -r,q}\hfill \\ \hfill & =\sum _{r=0}^{\mu }{\left[\begin{array}{c}\mu \\ r\end{array}\right]}_q{\left(u_q{\times }_q{v}_q\right)}_r{w}_{\mu -r,q}\hfill \\ \hfill & ={\left(\left(u_q{\times }_q{v}_q\right){\times }_q{w}_q\right)}_{\mu }\hfill \end{array}$$

Existence of identity: it is easy to see that ${\mathbf{u}}_q{\times }_q{e}_q=e_q{\times }_q{\mathbf{u}}_q={\mathbf{u}}_q$ for all ${\mathbf{u}}_q\in {\mathcal{G}}_q$ where $e_q={\left({\delta }_{\mu 0}\right)}_{\mu \ge 0},$ Existence of inverse: let ${\mathbf{u}}_q\in {\mathcal{G}}_q.$ Since $\mathsf{G}({\mathbf{u}}_q,0)=u_{0,q}\ne 0,$ then $\mid \mathsf{G}({\mathbf{u}}_q,z)\mid >0,\mid z\mid <\lambda ,$ for some $\lambda >0.$ This implies that $\frac{1}{\mathsf{G}({\mathbf{u}}_q,z)}$ is a well-defined function that can be represented via power series due to analyticity as

$$\frac{1}{\mathsf{G}({\mathbf{u}}_q,z)}=\sum _{\mu =0}^{\infty }{v}_{\mu ,q}\frac{z^{\mu }}{{\left[\mu \right]}_q!}=:\mathsf{G}({\mathbf{v}}_q,z),\mid z\mid \le \rho ,$$

(4)

for some real sequence ${\mathbf{v}}_q={\left(v_{\mu ,q}\right)}_{\mu \ge 0}$ and some $\rho >0.$ Here, one can observe that $v_{0,q}=\frac{1}{u_{0,q}}\ne 0$ by (4), and that ${\mathbf{v}}_q\in {\mathcal{G}}_q.$ Again, it can be observed from (3) and (4), that ${\mathbf{v}}_q$ is the inverse of ${\mathbf{u}}_q.$ Thus, ${\mathbf{v}}_q$ is the unique solution to the systems of equations:

$${\left(u_q{\times }_q{v}_q\right)}_{\mu }=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{u}_{k,q}{v}_{\mu -k,q}={\delta }_{\mu 0}.$$

(5)

Commutative: it is easy to see that ${\mathbf{u}}_q{\times }_q{\mathbf{v}}_q={\mathbf{v}}_q{\times }_q{\mathbf{u}}_q$ for all ${\mathbf{u}}_q,{\mathbf{v}}_q\in {\mathcal{G}}_q.$

The proof is complete. □

Let ${\mathbf{A}}_q\left(y\right)={\left(A_{\mu ,q}\left(y\right)\right)}_{\mu \ge 0}$ be a sequence of polynomials such that ${\mathbf{A}}_q\left(0\right)={\left(A_{\mu ,q}\left(0\right)\right)}_{\mu \ge 0}$$\in {\mathcal{G}}_q$. Recall that ${\mathbf{A}}_q\left(y\right)$ is called a q-Appell sequence if one of the following equivalent conditions is satisfied:

$${\mathsf{D}}_{q,y}\left(A_{\mu ,q}\left(y\right)\right)={\left[\mu \right]}_q{A}_{\mu -1,q}\left(y\right),\mu \in \mathbb{N},$$

(6)

$$A_{\mu ,q}\left(y\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{A}_{k,q}\left(0\right)y^{\mu -k},$$

(7)

or:

$$\mathsf{G}\left({\mathbf{A}}_q\left(y\right),z\right)=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),z\right)e_q^{yz}.$$

(8)

The set of all q-Appell sequence will be denoted by ${\mathcal{A}}_q.$ Let ${\mathbf{I}}_q\left(y\right)={\left(y^{\mu }\right)}_{\mu \ge 0}$ be the unit q-Appell sequence. Using (2), the condition (7) can be expressed as

$${\mathbf{A}}_q\left(y\right)={\mathbf{A}}_q\left(0\right){\times }_q{\mathbf{I}}_q\left(y\right).$$

(9)

From Proposition (1), ${\mathbf{A}}_q\left(y\right)\in {\mathcal{G}}_q,$ for any $y\in \mathbb{R}.$ From the binomial identity, ${\mathbf{I}}_q\left(y{\oplus }_{q}x\right)={\mathbf{I}}_q\left(y\right){\times }_q{\mathbf{I}}_q\left(x\right),$ for $x\in \mathbb{R}.$ Thus, from Equation (9) and Corollary (1):

$${\mathbf{A}}_q\left(y{\oplus }_{q}x\right)={\mathbf{A}}_q\left(y\right){\times }_q{\mathbf{I}}_q\left(x\right)={\mathbf{A}}_q\left(x\right){\times }_q{\mathbf{I}}_q\left(y\right),x\in \mathbb{R}.$$

(10)

2. The Abelian Group Structure of ${\mathcal{A}}_{\mathit{q}}$

Let ${\mathbf{A}}_q\left(y\right),{\mathbf{C}}_q\left(y\right)\in {\mathcal{A}}_q.$ The q-binomial convolution of ${\mathbf{A}}_q\left(y\right)$ and ${\mathbf{C}}_q\left(y\right),$ denoted by $\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)={\left({\left(A_q{\times }_q{C}_q\right)}_{\mu }\left(y\right)\right)}_{\mu \ge 0}$ and is defined as

$$\left(A_q{\times }_q{C}_q\right)\left(y\right)={\mathbf{A}}_q\left(y\right){\times }_q{\mathbf{C}}_q\left(0\right)={\mathbf{A}}_q\left(0\right){\times }_q{\mathbf{C}}_q\left(y\right)={\mathbf{A}}_q\left(0\right){\times }_q{\mathbf{C}}_q\left(0\right){\times }_q{\mathbf{I}}_q\left(y\right),$$

(11)

The last two equalities of (11) can be obtained using (9) and Corollary (1). Equivalently:

$$\begin{array}{cc}\hfill {\left(A_q{\times }_q{C}_q\right)}_{\mu }\left(y\right)& =\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{A}_{k,q}\left(0\right)C_{\mu -k,q}\left(y\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{C}_{k,q}\left(0\right)A_{\mu -k,q}\left(y\right)\hfill \\ \hfill & =\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{y}^{\mu -k}\sum _{j=0}^k{\left[\begin{array}{c}k\\ j\end{array}\right]}_q{A}_{j,q}\left(0\right)C_{k-j,q}\left(0\right)\hfill \\ \hfill & =\sum _{j_1+j_2+j_3=\mu }{\left[\begin{array}{c}\mu \\ j_1,j_2,j_3\end{array}\right]}_q{A}_{j_1,q}\left(0\right)C_{j_2,q}\left(0\right)y^{j_3}.\hfill \end{array}$$

(12)

Theorem 1.

Let ${\mathit{A}}_q\left(y\right),{\mathit{C}}_q\left(y\right)\in {\mathcal{A}}_q.$ Then, $\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right)$ is an q-Appell sequences characterized by its generating function:

$$\mathsf{G}\left(\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right),z\right)=\mathsf{G}\left({\mathit{A}}_q\left(0\right),z\right)\mathsf{G}\left({\mathit{C}}_q\left(0\right),z\right)e_q^{yz}.$$

(13)

As a consequence, $({\mathcal{A}}_q,{\times }_q)$ is an Abelian group with identity element ${\mathit{I}}_q\left(y\right).$ In addition, we have:

$$\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y{\oplus }_{q}x\right)={\mathit{A}}_q\left(y\right){\times }_q{\mathit{C}}_q\left(x\right),x\in \mathbb{R}.$$

(14)

In general, for any ${\mathit{A}}_q^{\left(i\right)}\left(y\right)\in {\mathcal{A}}_q$ and $y_i\in \mathbb{R},i=1,\cdots ,m,$ with $y_1{\oplus }_q\cdots {\oplus }_q{y}_m=y$:

$${\mathit{A}}_q^{\left(1\right)}\left(y_1\right){\times }_q\cdots {\times }_q{\mathit{A}}_q^{\left(m\right)}\left(y_m\right)=\left({\mathit{A}}_q^{\left(1\right)}{\times }_q\cdots {\times }_q{\mathit{A}}_q^{\left(m\right)}\right)\left(y\right).$$

(15)

Proof. 

By (12), ${\left(A_q{\times }_q{C}_q\right)}_0\left(0\right)=A_{0,q}\left(0\right)C_{0,q}\left(0\right)\ne 0.$ Using (11) and proposition 1, we have:

$$\mathsf{G}\left(\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right),z\right)=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),z\right)\mathsf{G}\left({\mathbf{C}}_q\left(0\right),z\right)e_q^{yz}=\mathsf{G}\left(\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(0\right),z\right)e_q^{yz}.$$

Thus, the first statement in Theorem 1 is evident from (8). Similarly, from (13) and Proposition 1, Formula (14) can be obtained. Now, we will show that ${\mathbf{C}}_q\left(y\right)\in {\mathcal{A}}_q$ will be the inverse of ${\mathbf{A}}_q\left(y\right)\in {\mathcal{A}}_q.$ Similar to the method used in Corollary 1, let ${\mathbf{C}}_q\left(0\right)={\left(C_{\mu ,q}\left(0\right)\right)}_{\mu \ge 0}\in {\mathcal{G}}_q$ be the real sequence having a generating function as

$$\mathsf{G}\left({\mathbf{C}}_q\left(0\right),z\right)=\frac{1}{\mathsf{G}({\mathbf{A}}_q\left(0\right),z)}.$$

(16)

Then, the q-Appell sequences ${\mathbf{C}}_q\left(y\right)={\mathbf{C}}_q\left(0\right){\times }_q{\mathbf{I}}_q\left(y\right)$ will be inverse of ${\mathbf{A}}_q\left(y\right).$ Equivalently, ${\mathbf{C}}_q\left(0\right)$ will be the unique solution to the systems of equations:

$${\left(A_q{\times }_q{C}_q\right)}_{\mu }\left(y\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{C}_{k,q}\left(0\right)A_{\mu -k,q}\left(y\right)={\mathbf{I}}_{\mu ,q}\left(y\right)=y^{\mu },$$

(17)

which completes the proof. □

Note that Theorem 1 is equivalent to q-Appell polynomials determinantal approach, now we state the following:

Corollary 2.

(Determinantal form). For the real sequence ${\mathit{A}}_q\left(y\right)\in {\mathcal{A}}_q$ and ${\mathit{C}}_q\left(0\right)\in {\mathcal{G}}_q$ whose generating function is represented in (16). Then, $A_{0,q}\left(y\right)=\frac{1}{C_{0,q}\left(0\right)}$ and we have for $\mu \in \mathbb{N}$:

$$\begin{array}{ccc}\hfill A_{\mu ,q}\left(y\right)& =& \frac{{(-1)}^{\mu }}{{\left(C_{0,q}\left(0\right)\right)}^{\mu +1}}\hfill \\ & \times & \left|\begin{array}{cccccc}1& y& y^2& \cdots & y^{\mu -1}& y^{\mu }\\ C_{0,q}\left(0\right)& C_{1,q}\left(0\right)& C_{2,q}\left(0\right)& \cdots & C_{\mu -1,q}\left(0\right)& C_{\mu ,q}\left(0\right)\\ 0& C_{0,q}\left(0\right)& {\left[\begin{array}{c}2\\ 1\end{array}\right]}_q{C}_{1,q}\left(0\right)& \cdots & {\left[\begin{array}{c}\mu -1\\ 1\end{array}\right]}_q{C}_{\mu -2,q}\left(0\right)& {\left[\begin{array}{c}\mu \\ 1\end{array}\right]}_q{C}_{\mu -1,q}\left(0\right)\\ 0& 0& C_{0,q}\left(0\right)& \cdots & {\left[\begin{array}{c}\mu -1\\ 2\end{array}\right]}_q{C}_{\mu -3,q}\left(0\right)& {\left[\begin{array}{c}\mu \\ 2\end{array}\right]}_q{C}_{\mu -2,q}\left(0\right)\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & C_{0,q}\left(0\right)& {\left[\begin{array}{c}\mu \\ \mu -1\end{array}\right]}_q{C}_{1,q}\left(0\right)\end{array}\right|\hfill \end{array}$$

Proof. 

It suffices to put formula (17) in a determinantal form. □

Corollary 2 has applications in Corollary 3.

3. Scale Transformations

Now, we will study scale transformations. For $\alpha \in \mathbb{R}$ and ${\mathbf{A}}_q\left(y\right)\in {\mathcal{A}}_q,$ $T_{\alpha }{A}_q\left(y\right)={\left(T_{\alpha }{A}_{\mu ,q}\left(y\right)\right)}_{\mu \ge 0}$ is defined as

$$\left\{\begin{array}{c}{T}_{\alpha }{A}_{\mu ,q}\left(y\right)={\alpha }^{\mu }{A}_{\mu ,q}(y/\alpha )={\displaystyle \sum _{k=0}^{\mu }}{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{\alpha }^k{A}_{k,q}\left(0\right)y^{\mu -k},a\ne 0\hfill \\ T_0{A}_{\mu ,q}\left(y\right)=A_{0,q}\left(0\right)y^{\mu }\hfill \end{array}\right.$$

(18)

where the last equality of the first equation of (18) is by using (7). These transformations will be characterized next.

Proposition 2.

Let $\alpha ,\beta \in \mathbb{R},$ then $T_{\alpha }{\mathit{A}}_q\left(y\right)$ is a q-Appell sequence if ${\mathit{A}}_q\left(y\right),{\mathit{C}}_q\left(y\right)\in {\mathcal{A}}_q$ and characterized by generating function:

$$\mathsf{G}\left(T_{\alpha }{\mathit{A}}_q\left(y\right),z\right)=\mathsf{G}\left({\mathit{A}}_q\left(0\right),\alpha z\right)e_q^{yz}$$

(19)

In addition

$$\mathsf{G}\left(\left(T_{\alpha }{\mathbf{A}}_q{\times }_q{T}_{\beta }{\mathbf{C}}_q\right)\left(y\right),z\right)=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),\alpha z\right)\mathsf{G}\left({\mathbf{C}}_q\left(0\right),\beta z\right)e_q^{yz}.$$

(20)

As a consequence, the map $T_{\alpha }:{\mathcal{A}}_q⟶{\mathcal{A}}_q$ is an isomorphism, whenever $a\ne 0.$

Proof. 

By (18), $T_{\alpha }{A}_{0,q}\left(0\right)=A_{0,q}\left(0\right)\ne 0.$ Again by (18) and Proposition 1, we have:

$$\mathsf{G}\left(T_{\alpha }{\mathbf{A}}_q\left(y\right),z\right)=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),\alpha z\right)e_q^{yz}=\mathsf{G}\left(T_{\alpha }{\mathbf{A}}_q\left(0\right),z\right)e_q^{yz}$$

since $T_{\alpha }{A}_{0,q}\left(0\right)={\alpha }^{\mu }{A}_{\mu ,q}\left(0\right).$ Hence, the first statement in Proposition 2 follows from (8). On the other hand, we have from (13) and (19):

$$\mathsf{G}\left(\left(T_{\alpha }{\mathbf{A}}_q{\times }_q{T}_{\beta }{\mathbf{C}}_q\right)\left(y\right),z\right)=\mathsf{G}\left(T_{\alpha }{\mathbf{A}}_q\left(0\right),z\right)\mathsf{G}\left(T_{\beta }{\mathbf{C}}_q\left(0\right),z\right)e_q^{yz}=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),\alpha z\right)\mathsf{G}\left({\mathbf{C}}_q\left(0\right),\beta z\right)e_q^{yz},$$

thus showing (20). Moreover, by (13) and (19), we have

$$T_{\alpha }\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=\left(T_{\alpha }{\mathbf{A}}_q{\times }_q{T}_{\alpha }{\mathbf{C}}_q\right)\left(y\right)$$

(21)

as both sides of (21) have the same generating function.

On the other hand, if $T_{\alpha }{\mathbf{A}}_q\left(y\right)=T_{\alpha }{\mathbf{C}}_q\left(y\right),$ then ${\mathbf{A}}_q\left(0\right)={\mathbf{C}}_q\left(0\right),$ as follows from (19) and we have ${\mathbf{A}}_q\left(y\right)={\mathbf{C}}_q\left(y\right).$ By (19), $T_{\alpha }\left(T_{{\alpha }^{-1}}{\mathbf{A}}_q\right)\left(y\right)={\mathbf{A}}_q\left(y\right),$ thus it shows that $T_{\alpha }$ is an isomorphism and thus the proof is completed. □

The order m generalized q-Bernoulli polynomials can be expressed in terms of the q-Bernoulli polynomials ${\mathbf{B}}_q\left(y\right)$ as

$${\mathbf{B}}_q({\alpha }_1,\cdots ,{\alpha }_m;y)=\left(T_{{\alpha }_1}{\mathbf{B}}_q{\times }_q\cdots {\times }_q{T}_{{\alpha }_m}{\mathbf{B}}_q\right)\left(y\right).$$

(22)

Relation (22) can be obtained using above the table and Proposition 2 as follows:

$$\begin{array}{cc}\hfill \mathsf{G}\left({\mathbf{B}}_q({\alpha }_1,\cdots ,{\alpha }_m;y),z\right)& =e_q^{yz}\prod _{i=1}^m\frac{{\alpha }_{i}z}{e_q^{{\alpha }_{i}z}-1}\hfill \\ \hfill & =e_q^{yz}\prod _{i=1}^m\mathsf{G}\left({\mathbf{B}}_q\left(0\right),{\alpha }_{i}z\right)\hfill \\ \hfill & =\mathsf{G}\left((T_{{\alpha }_1}{\mathbf{B}}_q{\times }_q{T}_{{\alpha }_2}{\mathbf{B}}_q{\times }_q\cdots {\times }_q{T}_{{\alpha }_m}{\mathbf{B}}_q)\left(y\right),z\right).\hfill \end{array}$$

Similarly, by using the result of relation (20) in Proposition 2, the order m generalized q-Euler polynomials can be expressed by means of the type q-Euler polynomials $\mathbf{E}\left(y\right)$ as

$${\mathbf{E}}_q({\alpha }_1,\cdots ,{\alpha }_m;y)=\left(T_{{\alpha }_1}{\mathbf{E}}_q{\times }_q\cdots {\times }_q{T}_{{\alpha }_m}{\mathbf{E}}_q\right)\left(y\right).$$

(23)

Finally, we have the relating q-Bernoulli and q-Euler polynomials:

$$\left({\mathbf{B}}_q{\times }_q{\mathbf{E}}_q\right)\left(y\right)=T_2{\mathbf{B}}_q\left(y\right).$$

(24)

From Table 1:

$$\begin{array}{cc}\hfill \sum _{\mu =0}^{\infty }{B}_{\mu ,q}\left(\frac{y}{2}\right)\frac{z^{\mu }}{{\left[\mu \right]}_q!}& =\frac{z{e}_q^{yz/2}}{e_q^z-1}=\frac{z/2}{e_q^{z/2}-1}\frac{2e_q^{yz/2}}{e_q^{z/2}+1}\hfill \\ \hfill & =\sum _{\mu =0}^{\infty }{B}_{\mu ,q}\left(0\right)\frac{{(z/2)}^{\mu }}{{\left[\mu \right]}_q!}\sum _{s=0}^{\infty }{E}_{s,q}\left(y\right)\frac{{(z/2)}^s}{{\left[s\right]}_q!}\hfill \\ \hfill & =\sum _{\mu =0}^{\infty }\left(2^{-\mu }\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{B}_{\mu -s,q}\left(0\right)E_{s,q}\left(y\right)\right)\frac{z^{\mu }}{{\left[\mu \right]}_q!}.\hfill \end{array}$$

Table 1. Generating the function of some q-Appell type polynomials.

Notation	Polynomials	Generating Functions
$B_q\left(y\right)$	q-Bernoulli polynomials [26]	$\frac{z{e}_q^{yz}}{e_q^z-1}$
$B_q({\alpha }_1,\cdots ,{\alpha }_m;y)$	Order m generalized q-Bernoulli polynomials [29] (p. 117, 4.125)	$e_q^{yz}{\prod }_{i=1}^m\frac{{\alpha }_{i}z}{e_q^{{\alpha }_{i}z}-1}$
$E_q\left(y\right)$	q-Euler polynomials [26]	$\frac{2e_q^{yz}}{e_q^z+1}$
$E_q({\alpha }_1,\cdots ,{\alpha }_m;y)$	Order m generalized q-Euler polynomials [29] (p. 129, 4.197)	$e_q^{yz}{\prod }_{i=1}^m\frac{2}{e_q^{{\alpha }_{i}z}+1}$
$G_q\left(y\right)$	q-Genocchi polynomials [26]	$\frac{2z{e}_q^{yz}}{e_q^z+1}$

Then, we have:

$$B_{\mu ,q}\left(\frac{y}{2}\right)=2^{-\mu }\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{B}_{s,q}\left(0\right)E_{\mu -s,q}\left(y\right).$$

Similarly we can get:

$$B_{\mu ,q}\left(\frac{y}{2}\right)=2^{-\mu }\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{E}_{s,q}\left(0\right)B_{\mu -s,q}\left(y\right).$$

Then:

$$\begin{array}{cc}\hfill {\left(B_q{\times }_q{E}_q\right)}_{\mu }\left(y\right)& =\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{B}_{s,q}\left(0\right)E_{\mu -s,q}\left(y\right)\hfill \\ \hfill & =\sum _{s=0}^{\mu }{\left[\begin{array}{c}\mu \\ s\end{array}\right]}_q{E}_{s,q}\left(0\right)B_{\mu -s,q}\left(y\right)\hfill \\ \hfill & =2^{-\mu }{B}_{\mu ,q}\left(\frac{y}{2}\right)\hfill \\ \hfill & =T_2{B}_{\mu ,q}\left(y\right).\hfill \end{array}$$

4. Transformations Based on Expectations

Let X be a random variables such that:

$$E{e}_q^{r\mid X\mid }<\infty ,\mathrm{for}r>0.$$

Here, we consider expectations and transformations of q-Appell sequences ${\mathbf{A}}_q\left(y\right)$ by replacing x by X in (10) similar to a classical analogue [7,30]. These transformations are the result due to a probabilistic approach to q-Appell polynomials. For ${\mathbf{A}}_q\left(y\right)\in {\mathcal{A}}_q$ and a random variable X, we define $R_X{\mathbf{A}}_q\left(y\right)={\left(R_X{A}_{\mu ,q}\left(y\right)\right)}_{\mu \ge 0}$ as

$$R_X{A}_{\mu ,q}\left(y\right)=E{A}_{\mu ,q}\left(y{\oplus }_{q}X\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{A}_{k,q}\left(0\right)E{\left(y{\oplus }_{q}X\right)}^{\mu -k}$$

$$=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_{q}E{X}^k{A}_{\mu -k,q}\left(y\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_{q}E{A}_{k,q}\left(X\right)y^{\mu -k},$$

(25)

follow from (7) and (10). Notice that whenever $X=0,$ $R_X{\mathbf{A}}_q\left(y\right)={\mathbf{A}}_q\left(y\right).$ In addition:

$$R_X{\mathbf{I}}_q\left(y\right)={\left(E{\left(y{\oplus }_{q}X\right)}^{\mu }\right)}_{\mu \ge 0}.$$

(26)

Identity ${\mathbf{I}}_q\left(y\right)$ transformation plays an important role as which is evident from equality in (25), we obtain:

$$R_X{\mathbf{A}}_q\left(y\right)=\left({\mathbf{A}}_q{\times }_q{R}_X{\mathbf{I}}_q\right)\left(y\right).$$

(27)

Now, we will study some results following the characterization of the transformations based on expectations and will present some of their properties.

Proposition 3.

Let Y and X denote two independent random variables and ${\mathit{A}}_q\left(y\right),{\mathit{C}}_q\left(y\right)\in {\mathcal{A}}_q.$ Then, $R_X{\mathit{A}}_q\left(y\right)$ is a q-Appell sequence represented in terms of generating a function as

$$\mathsf{G}\left(R_X{\mathit{A}}_q\left(y\right),z\right)=\mathsf{G}\left({\mathit{A}}_q\left(0\right),z\right)E{e}_q^{z\left(y{\oplus }_{q}X\right)}.$$

(28)

Moreover, the following properties are true:

(a) 

$R_X\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right)=\left(R_X{\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right)=\left({\mathit{A}}_q{\times }_q{R}_X{\mathit{C}}_q\right)\left(y\right).$

(b) 

$\left(R_Y{\mathit{A}}_q{\times }_q{R}_X{\mathit{C}}_q\right)\left(y\right)=R_{Y{\oplus }_{q}X}\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right).$

(c) 

$R_Y{R}_X{\mathit{A}}_q\left(y\right)=R_{Y{\oplus }_{q}X}{\mathit{A}}_q\left(y\right).$

Proof. 

Using (25), $R_X{\mathbf{A}}_{0,q}\left(0\right)={\mathbf{A}}_{0,q}\left(0\right)\ne 0.$ By interchanging sum with expectation, from (26) we have:

$$\mathsf{G}\left(R_X{\mathbf{I}}_q\left(y\right),z\right)=\sum _{\mu =0}^{\infty }E{\left(y{\oplus }_{q}X\right)}^{\mu }\frac{z^{\mu }}{{\left[\mu \right]}_q!}=E{e}_q^{z\left(y{\oplus }_{q}X\right)}.$$

(29)

□

Thus, from (13), (27)–(29) can be obtained. Formula (28) implies that:

$$\mathsf{G}\left(R_X{\mathbf{A}}_q\left(y\right),z\right)=\mathsf{G}\left(R_X{\mathbf{A}}_q\left(0\right),z\right)e_q^{yz}.$$

Combining it with (8), will prove the first statement in Proposition 3. From Proposition 1 and (27), we obtain:

$$R_X\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q{\times }_q{R}_X{\mathbf{I}}_q\right)\left(y\right)=\left(R_X{\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=\left({\mathbf{A}}_q{\times }_q{R}_X{\mathbf{C}}_q\right)\left(y\right),$$

hence, we obtain (a). Notice that:

$$\left(R_Y{\mathbf{I}}_q{\times }_q{R}_X{\mathbf{I}}_q\right)\left(y\right)=R_{Y{\oplus }_{q}X}{\mathbf{I}}_q\left(y\right),$$

as it is clear from (13), (29), and the independence between the variables Y and X. From (27) and Proposition 1, we obtain:

$$\left(R_Y{\mathbf{A}}_q{\times }_q{R}_X{\mathbf{C}}_q\right)\left(y\right)=\left({\mathbf{A}}_q{\times }_q{R}_Y{\mathbf{I}}_q{\times }_q{\mathbf{C}}_q{\times }_q{R}_X{\mathbf{I}}_q\right)\left(y\right)=R_{Y{\oplus }_{q}X}\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right),$$

which justifies (b). Similarly, one can prove (c).

It is to note that the map $R_X:{\mathcal{A}}_q⟶{\mathcal{A}}_q$ is not a homomorphism, which follows from Proposition 3(a).

5. The Subset ${\mathcal{R}}_{\mathit{q}}$

Let us consider the exponential moments (finite) $M_X\left(t\right):=E\left(e_q^{tX}\right)$ for a random variable X s.t. $\mid t\mid <\rho$ similar to its classical case defined in [30]. Then, $E\left(X^{\mu }\right)<\infty$ for all $\mu =1,2,\cdots ,$ and:

$$E\left(e_q^{tX}\right)=\sum _{\mu =0}^{\infty }\frac{t^{\mu }}{{\left[\mu \right]}_q!}E\left(X^{\mu }\right),\mid t\mid <\rho .$$

For $t\in \mathbb{C},$ the right hand side of (5) will be a complex analytic function $z\mapsto \phi \left(z\right),\mid z\mid <\rho ,z\in \mathbb{C}.$ As $M_X\left(0\right)=1,$ $\mid \phi \left(z\right)\mid >0$ for $\mid z\mid <\rho$ due to the continuity of $\phi .$ Thus, $z\mapsto \frac{1}{\phi \left(z\right)}$ is a well-defined analytic function and can be expressed in the form of power series:

$$\frac{1}{\phi \left(z\right)}=\sum _{\mu =0}^{\infty }{c}_{\mu ,q}{z}^{\mu },\mid z\mid <\lambda ,$$

where $\lambda >0$ denotes the radius of convergence. For $z=t\in \mathbb{R}$ such that $\mid z\mid <\lambda$ we have:

$$\frac{1}{E\left(e_q^{tX}\right)}=\sum _{\mu =0}^{\infty }{c}_{\mu ,q}{t}^{\mu }=\sum _{\mu =0}^{\infty }{\widehat{c}}_{\mu ,q}\frac{t^{\mu }}{{\left[\mu \right]}_q!},$$

where ${\widehat{c}}_{\mu ,q}=c_{\mu ,q}{\left[\mu \right]}_q!.$ For $y\in \mathbb{R}$ and $\mid t\mid <\lambda$ it holds:

$$\begin{array}{cc}\hfill \frac{e_q^{ty}}{E\left(e_q^{tX}\right)}& =\sum _{s=0}^{\infty }\frac{t^s}{{\left[s\right]}_q!}{y}^s\sum _{m=0}^{\infty }\frac{t^m}{{\left[m\right]}_q!}{\widehat{c}}_{m,q}\hfill \\ \hfill & =\sum _{s=0}^{\infty }\sum _{m=0}^{\infty }\frac{t^{s+m}}{{\left[s\right]}_q!{\left[m\right]}_q!}{\widehat{c}}_{m,q}{y}^s\hfill \\ \hfill & =\sum _{\mu =0}^{\infty }\frac{t^{\mu }}{{\left[\mu \right]}_q!}\sum _{m=0}^{\infty }{\left[\begin{array}{c}\mu \\ m\end{array}\right]}_q{\widehat{c}}_m{y}^{\mu -m}.\hfill \end{array}$$

(30)

As both series are absolutely convergent in the neighborhood of the origin, the sum will therefore not depend on the order of summation. Motivated by (30), we now present the definition of q-Appell polynomials related to a random variable $X.$

Definition 1.

Let X be a random variable having some exponential moments. The polynomials $A_{\mu ,q}\left(y\right),\mu =0,1,2,\cdots ,$ satisfying:

$$\sum _{\mu =0}^{\infty }\frac{t^{\mu }}{{\left[\mu \right]}_q!}{A}_{\mu ,q}\left(y\right)=\frac{e_q^{ty}}{E\left(e_q^{tX}\right)}forally\in \mathbb{R}$$

(31)

where $A_{\mu ,q}\left(y\right)$ are called as q-Appell polynomials of order μ associated with random variable $X.$

Now, we denote by ${\mathcal{R}}_q$ the set of Appell sequences ${\mathbf{A}}_q\left(y\right)$ such that:

$$\mathsf{G}\left({\mathbf{A}}_q\left(y\right),z\right)=\frac{e_q^{yz}}{E{e}_q^{zX}},$$

(32)

for a random variable $X.$ For another random variable Y satisfying (32), due to the uniqueness theorem for characteristic functions, Y and X will follow same law. Thus, ${\mathbf{A}}_q\left(y\right)$ has associated the random variable $X.$ Notice that for $X=0,$ ${\mathbf{I}}_q\left(y\right)\in {\mathcal{R}}_q.$ Then, we present a preposition for the construction of other q-Appell polynomials.

Proposition 4.

For associated independent random variable Y and X and let: ${\mathit{A}}_q\left(y\right),{\mathit{C}}_q\left(y\right)\in {\mathcal{R}}_q$, respectively, where $\alpha ,\beta \in \mathbb{R}.$ Then, $\left(T_{\alpha }{\mathit{A}}_q{\times }_q{T}_{\beta }{\mathit{C}}_q\right)\left(y\right)$ belong to a ${\mathcal{R}}_q$ with the associated random variable $\alpha Y{\oplus }_q\beta X.$

In particular, if ${\mathit{A}}_q\left(y\right),{\mathit{C}}_q\left(y\right)\in {\mathcal{R}}_q,$ then $\left({\mathit{A}}_q{\times }_q{\mathit{C}}_q\right)\left(y\right)$ belongs to ${\mathcal{R}}_q$ with associated random variable $Y{\oplus }_{q}X.$

Proof. 

From Equations (13), (19), and (32) and using the property that Y and X are independent:

$$\begin{array}{cc}\hfill \mathsf{G}\left(\left(T_{\alpha }{\mathbf{A}}_q{\times }_q{T}_{\beta }{\mathbf{C}}_q\right)\left(y\right),z\right)& =\mathsf{G}\left(T_{\alpha }{\mathbf{A}}_q\left(0\right),z\right)\mathsf{G}\left(T_{\beta }{\mathbf{C}}_q\left(0\right),z\right)e_q^{yz}\hfill \\ \hfill & =\mathsf{G}\left({\mathbf{A}}_q\left(0\right),\alpha z\right)\mathsf{G}\left({\mathbf{C}}_q\left(0\right),\beta z\right)e_q^{yz}\hfill \\ \hfill & =\frac{e_q^{yz}}{E{e}_q^{z\left(\alpha Y{\oplus }_q\beta X\right)}}\hfill \end{array}$$

where we get the required result. □

The q-Appell sequence in ${\mathcal{R}}_q$ is characterized as follows.

Theorem 2.

The following statements are equivalent:

(a) 

${\mathit{A}}_q\left(y\right)\in {\mathcal{R}}_q$ with the associated random variable $X.$

(b) 

$\mathsf{G}\left({\mathit{A}}_q\left(0\right),z\right)={\left(E{e}_q^{zX}\right)}^{-1}.$

(c) 

$R_X{\mathit{A}}_q\left(y\right)={\mathit{I}}_q\left(y\right).$

(d) 

The inverse element of ${\mathit{A}}_q\left(y\right)$ is $R_X{\mathit{I}}_q\left(y\right).$

Proof. 

From (32), equivalence of (a) and (b) can be obtained. Similarly, (27) gives equivalence between (c) and (d). If (b) is true, then from (28), we have:

$$\mathsf{G}\left(R_X{\mathbf{A}}_q\left(y\right),z\right)=\mathsf{G}\left({\mathbf{A}}_q\left(0\right),z\right)E{e}_q^{z\left(y{\oplus }_{q}X\right)}=e_q^{yz}=\mathsf{G}\left({\mathbf{I}}_q\left(y\right),z\right),$$

which together with (8), shows (c). Finally, if (d) holds, we see from (13) and (28):

$$e_q^{yz}=\mathsf{G}\left(\left(R_X{\mathbf{I}}_q{\times }_q\mathbf{A}\right)\left(y\right),z\right)=\mathsf{G}\left(R_X{\mathbf{I}}_q\left(0\right),z\right)\mathsf{G}\left({\mathbf{A}}_q\left(0\right),z\right)e_q^{yz}=E{e}_q^{zX}\mathsf{G}\left({\mathbf{A}}_q\left(0\right),z\right)e_q^{yz}$$

thus showing (b) which completes the proof. □

For any q-Appell sequence, its determinantal form ${\mathbf{A}}_q\left(y\right)\in {\mathcal{R}}_q$ can be expressed in terms of the moments of its associated random variable $X.$

Corollary 3.

Let ${\mathbf{A}}_q\left(y\right)\in {\mathcal{R}}_q$ with associated random variable $X.$ Denote $R_X{I}_{\mu ,q}\left(0\right)=E{X}^{\mu }=:T_{\mu ,q}.$ Then, ${\mathbf{A}}_{0,q}\left(y\right)=\frac{1}{T_{0,q}}=1$ and:

$$A_{\mu ,q}\left(y\right)={(-1)}^{\mu }\left|\begin{array}{cccccc}1& y& y^2& \cdots & y^{\mu -1}& y^{\mu }\\ 1& T_{1,q}& T_{2,q}& \cdots & T_{\mu -1,q}& T_{\mu ,q}\\ 0& 1& {\left[\begin{array}{c}2\\ 1\end{array}\right]}_q{T}_{1,q}& \cdots & {\left[\begin{array}{c}\mu -1\\ 1\end{array}\right]}_q{T}_{\mu -2,q}& {\left[\begin{array}{c}\mu \\ 1\end{array}\right]}_q{T}_{\mu -1,q}\\ 0& 0& 1& \cdots & {\left[\begin{array}{c}\mu -1\\ 2\end{array}\right]}_q{T}_{\mu -3,q}& {\left[\begin{array}{c}\mu \\ 2\end{array}\right]}_q{T}_{\mu -2,q}\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& {\left[\begin{array}{c}\mu \\ \mu -1\end{array}\right]}_q{T}_{1,q}\end{array}\right|,\mu \in \mathbb{N}$$

(33)

Proof. 

From Corollary 2 and Theorem 2(d) proof follows. □

Another important result from Theorem 2 is as follows:

Corollary 4.

Let ${\mathit{A}}_q\left(y\right)\in {\mathcal{R}}_q$ with associated random variable $X.$ For any ${\mathit{C}}_q\left(y\right)\in {\mathcal{A}}_q,$ we have:

$${\mathit{C}}_q\left(y\right)=E\left({\mathit{C}}_q{\times }_q{\mathit{A}}_q\right)\left(y{\oplus }_{q}X\right).$$

(34)

As a consequence, we have for any $x\in \mathbb{R}$:

$$C_{\mu ,q}\left(y{\oplus }_{q}x\right)=E{\left(C_q{\times }_q{A}_q\right)}_{\mu }\left(y{\oplus }_{q}x{\oplus }_{q}X\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_{q}E{C}_{k,q}\left(x{\oplus }_{q}X\right)A_{\mu -k,q}\left(y\right).$$

(35)

In particular:

$$y^{\mu }=E{A}_{\mu ,q}\left(y{\oplus }_{q}X\right)=\sum _{k=0}^{\mu }{\left[\begin{array}{c}\mu \\ k\end{array}\right]}_q{T}_{k,q}{A}_{\mu -k,q}\left(y\right).$$

(36)

Proof. 

By Proposition 3(a) and Theorem 2(c), we have:

$${\mathbf{C}}_q\left(y\right)=\left({\mathbf{I}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=\left(R_X{\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=R_X\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y\right)=E\left({\mathbf{A}}_q{\times }_q{\mathbf{C}}_q\right)\left(y{\oplus }_{q}X\right),$$

which shows (34). Formula (35) follows by replacing y by $y{\oplus }_{q}x$ in (34) and then applying (14). Identity (36) follows by setting ${\mathbf{C}}_q\left(y\right)={\mathbf{I}}_q\left(y\right)$ and $x=0$ in (35). Thus, the proof is completed. □