Some identities of q-Euler polynomials arising from q-umbral calculus

Abstract

Recently, Araci-Acikgoz-Sen derived some interesting identities on weighted q-Euler polynomials and higher-order q-Euler polynomials from the applications of umbral calculus (see (Araci et al. in J. Number Theory 133(10):3348-3361, 2013)). In this paper, we develop the new method of q-umbral calculus due to Roman, and we study a new q-extension of Euler numbers and polynomials which are derived from q-umbral calculus. Finally, we give some interesting identities on our q-Euler polynomials related to the q-Bernoulli numbers and polynomials of Hegazi and Mansour.

1 Introduction

Throughout this paper we will assume q to be a fixed real number between 0 and 1. We define the q-shifted factorials by

$${(a:q)}_0=1,{(a:q)}_n=\prod _{i=0}^{n-1}(1-a{q}^i),{(a:q)}_{\mathrm{\infty }}=\prod _{i=0}^{\mathrm{\infty }}(1-a{q}^i).$$

(1.1)

If x is a classical object, such as a complex number, its q-version is defined as ${[x]}_q=\frac{1-q^x}{1-q}$. We now introduce the q-extension of exponential function as follows:

$$e_q(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{{[n]}_q!}=\frac{1}{{((1-q)z:q)}_{\mathrm{\infty }}}(\text{see [1–4]}),$$

(1.2)

where $z\in \mathbb{C}$ with $|z|<1$.

The Jackson definite q-integral of the function f is defined by

$${\int }_0^{x}f(t)d_{q}t=(1-q)\sum _{a=0}^{\mathrm{\infty }}f\left(q^{a}x\right)x{q}^a(\text{see [1, 2, 5])}.$$

(1.3)

The q-difference operator $D_q$ is defined by

$$D_{q}f(x)=\frac{d_{q}f(x)}{d_{q}x}=\{\begin{array}{cc}\frac{f(x)-f(qx)}{(1-q)x}\hfill & \text{if }x\ne 0,\hfill \\ \frac{df(x)}{dx}\hfill & \text{if }x=0,\hfill \end{array}$$

(1.4)

where

$$\underset{q\to 1}{lim}{D}_{q}f(x)=\frac{df(x)}{dx}(\text{see [1, 2, 4, 6]}).$$

By using an exponential function $e_q(x)$, Hegazi and Mansour defined q-Bernoulli polynomials as follows:

$$\sum _{n=0}^{\mathrm{\infty }}{B}_{n,q}(x)\frac{t^n}{{[n]}_q!}=\frac{t}{e_q(t)-1}{e}_q(xt)(\text{see [1, 2, 4, 7]}).$$

(1.5)

In the special case, $x=0$, $B_{n,q}(0)=B_{n,q}$ are called the n th q-Bernoulli numbers.

From (1.5), we can easily derive the following equation:

$$B_{n,q}(x)=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{B}_{n-l,q}{x}^l=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{B}_{l,q}{x}^{n-l},$$

(1.6)

where

$${\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q=\frac{{[n]}_q!}{{[n-l]}_q!{[l]}_q!}=\frac{{[n]}_q{[n-1]}_q\cdots {[n-l+1]}_q}{{[l]}_q!}(\text{see [2, 7]}).$$

In the next section, we will consider new q-extensions of Euler numbers and polynomials by using the method of Hegazi and Mansour. More than five decades ago, Carlitz [8] defined a q-extension of Euler polynomials. In a recent paper (see [3]), Kupershmidt constructed reflection symmetries of q-Bernoulli polynomials which differ from Carlitz’s q-Bernoulli numbers and polynomials. By using the method of Kupershmidt, Hegazi and Mansour also introduced a new q-extension of Bernoulli numbers and polynomials (see [1, 3, 4]). From the q-exponential function, Kurt and Cenkci derived some interesting new formulae of q-extension of Genocchi polynomials. Recently, several authors have studied various q-extensions of Bernoulli and Euler polynomials (see [1–6, 8–11]). Let ℂ be the complex number field, and let ℱ be the set of all formal power series in variable t over ℂ with

$$\mathcal{F}=\{f(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{a_k}{{[k]}_q!}{t}^k|a_k\in \mathbb{C}\}.$$

(1.7)

Let $\mathbb{P}=\mathbb{C}[t]$ and let ${\mathbb{P}}^{\ast }$ be the vector space of all linear functionals on ℙ. $〈L|p(x)〉$ denotes the action of linear functional L on the polynomial $p(x)$, and it is well known that the vector space operations on ${\mathbb{P}}^{\ast }$ are defined by

$$〈L+M|p(x)〉=〈L|p(x)〉+〈M|p(x)〉,〈cL|p(x)〉=c〈L|p(x)〉,$$

where c is a complex constant (see [7, 9, 11]).

For $f(t)={\sum }_{k=0}^{\mathrm{\infty }}\frac{a_k}{{[k]}_q!}{t}^k\in \mathcal{F}$, we define the linear functional on ℙ by setting

$$〈f(t)|x^n〉=a_n\text{for all }n\ge 0.$$

(1.8)

From (1.7) and (1.8), we note that

$$〈t^k|x^n〉={[n]}_q!{\delta }_{n,k}(n,k\ge 0),$$

(1.9)

where ${\delta }_{n,k}$ is the Kronecker symbol.

Let us assume that $f_L(t)={\sum }_{k=0}^{\mathrm{\infty }}〈L|x^n〉\frac{t^k}{k!}$. Then by (1.9) we easily see that $〈f_L(t)|x^n〉=〈L|x^n〉$. That is, $f_L(t)=L$. Additionally, the map $L\mapsto f_L(t)$ is a vector space isomorphism from ${\mathbb{P}}^{\ast }$ onto ℱ. Henceforth ℱ denotes both the algebra of formal power series in t and the vector space of all linear functionals on ℙ, and so an element $f(t)$ of ℱ will be thought of as a formal power series and a linear functional. We call it the q-umbral algebra. The q-umbral calculus is the study of q-umbral algebra. By (1.2) and (1.3), we easily see that $〈e_q(yt)|x^n〉=y^n$ and so $〈e_q(yt)|p(x)〉=p(y)$ for $p(x)\in \mathbb{P}$. The order $o(f(t))$ of the power series $f(t)\ne 0$ is the smallest integer for which $a_k$ does not vanish. If $o(f(t))=0$, then $f(t)$ is called an invertible series. If $o(f(t))=1$, then$f(t)$ is called a delta series (see [7, 9, 11, 12]). For $f(t),g(t)\in \mathcal{F}$, we have $〈f(t)g(t)|p(x)〉=〈f(t)|g(t)p(x)〉=〈g(t)|f(t)p(x)〉$. Let $f(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$. Then we have

$$f(t)=\sum _{k=0}^{\mathrm{\infty }}〈f(t)|x^k〉\frac{t^k}{{[k]}_q!},p(x)=\sum _{k=0}^{\mathrm{\infty }}〈t^k|p(x)〉\frac{x^k}{{[k]}_q!}(\text{see [12]}).$$

(1.10)

From (1.10), we have

$$p^{(k)}(x)=D_q^{k}p(x)=\sum _{l=k}^{\mathrm{\infty }}\frac{〈t^l|p(x)〉}{{[l]}_q!}{[l]}_q\cdots {[l-k+1]}_q{x}^{l-k}.$$

(1.11)

By (1.11), we get

$$p^{(k)}(0)=〈t^k|p(x)〉\text{and}〈1|p^{(k)}(x)〉=p^{(k)}(0).$$

(1.12)

Thus from (1.12), we note that

$$t^{k}p(x)=p^{(k)}(x)=D_q^{k}p(x).$$

(1.13)

Let $f(t),g(t)\in \mathcal{F}$ with $o(f(t))=1$ and $o(g(t))=0$. Then there exists a unique sequence $S_n(x)$ ($deg{S}_n(x)=n$) of polynomials such that $〈g(t)f{(t)}^k|S_n(x)〉={[n]}_q!{\delta }_{n,k}$ ($n,k\ge 0$). The sequence $S_n(x)$ is called the q-Sheffer sequence for $(g(t),f(t))$ which is denoted by $S_n(x)\sim (g(t),f(t))$. Let $S_n(x)\sim (g(t),f(t))$. For $h(t)\in \mathcal{F}$ and $p(x)\in \mathbb{P}$, we have

$$h(t)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈h(t)|S_k(x)〉}{{[k]}_q!}g(t)f{(t)}^k,p(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈g(t)f{(t)}^k|p(x)〉}{{[k]}_q!}{S}_k(x),$$

(1.14)

and

$$\frac{1}{g(\overline{f}(t)}{e}_q(y\overline{f}(t))=\sum _{k=0}^{\mathrm{\infty }}\frac{S_k(y)}{{[k]}_q!}{t}^k\text{for all }y\in \mathbb{C},$$

(1.15)

where $\overline{f}(t)$ is the compositional inverse of $f(t)$ (see [7, 12]).

Recently, Araci-Acikgoz-Sen derived some new interesting properties on the new family of q-Euler numbers and polynomials from some applications of umbral algebra (see [9]). The properties of q-Euler and q-Bernoulli polynomials seem to be of interest and worthwhile in the areas of both number theory and mathematical physics. In this paper, we develop the new method of q-umbral calculus due to Roman and study a new q-extension of Euler numbers and polynomials which are derived from q-umbral calculus. Finally, we give new explicit formulas on q-Euler polynomials related to Hegazi-Mansour’s q-Bernoulli polynomials.

2 q-Euler numbers and polynomials

We consider the new q-extension of Euler polynomials which are generated by the generating function to be

$$\frac{2}{e_q(t)+1}{e}_q(xt)=\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}(x)\frac{t^n}{{[n]}_q!}.$$

(2.1)

In the special case, $x=0$, $E_{n,q}(0)=E_{n,q}$ are called the n th q-Euler numbers. From (2.1), we note that

$$E_{n,q}(x)=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{E}_{l,q}{x}^{n-l}=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{E}_{n-l,q}{x}^l.$$

(2.2)

By (2.1), we easily get

$$E_{0,q}=1,E_{n,q}(1)+E_{n,q}=2{\delta }_{0,n}.$$

(2.3)

For example, $E_{0,q}=1$, $E_{1,q}=-\frac{1}{2}$, $E_{2,q}=\frac{q-1}{4}$, $E_{3,q}=\frac{q+q^2-1}{4}+\frac{(1-q){[3]}_q}{8},\dots$ . From (1.15) and (2.1), we have

$$E_{n,q}(x)\sim (\frac{e_q(t)+1}{2},t)$$

(2.4)

and

$$\frac{2}{e_q(t)+1}{x}^n=E_{n,q}(x)(n\ge 0).$$

(2.5)

Thus, by (1.13) and (2.5), we get

$$t{E}_{n,q}(x)=\frac{2}{e_q(t)+1}t{x}^n={[n]}_q\frac{2}{e_q(t)+1}{x}^{n-1}={[n]}_q{E}_{n-1,q}(x)(n\ge 0).$$

(2.6)

Indeed, by (1.9), we get

$$\begin{array}{rl}〈\frac{e_q(t)+1}{2}{t}^k|E_{n,q}(x)〉& =\frac{{[k]}_q!}{2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q〈e_q(t)+1|E_{n-k,q}(x)〉\\ =\frac{{[k]}_q!}{2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q(E_{n-k,q}(1)+E_{n-k,q}).\end{array}$$

(2.7)

From (2.4), we have

$$〈\left(\frac{e_q(t)+1}{2}\right)t^k|E_{n,q}(x)〉={[n]}_q!{\delta }_{n,k}.$$

(2.8)

Thus, by (2.7) and (2.8), we get

$$0=E_{n-k,q}(1)+E_{n-k,q}=\sum _{l=0}^{n-k}{\left(\genfrac{}{}{0ex}{}{n-k}{l}\right)}_q{E}_{l,q}+E_{n-k,q}(n,k\in {\mathbb{Z}}_{\ge 0}\text{ with }n>k).$$

(2.9)

This is equivalent to

$$-2E_{n-k,q}=\sum _{l=0}^{n-k-1}{\left(\genfrac{}{}{0ex}{}{n-k}{l}\right)}_q{E}_{l,q},\text{where }n,k\in {\mathbb{Z}}_{\ge 0}\text{ with }n>k.$$

(2.10)

Therefore, by (2.10), we obtain the following lemma.

Lemma 2.1 For $n\ge 1$, we have

$$-2E_{n,q}=\sum _{l=0}^{n-1}{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{E}_{l,q}.$$

From (2.2) we have

$$\begin{array}{rl}{\int }_x^{x+y}{E}_{n,q}(u)d_{q}u& =\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{E}_{n-l,q}\frac{1}{{[l+1]}_q}\{{(x+y)}^{l+1}-x^{l+1}\}\\ =\frac{1}{{[n+1]}_q}\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n+1}{l+1}\right)}_q{E}_{n-l,q}\{{(x+y)}^{l+1}-x^{l+1}\}\\ =\frac{1}{{[n+1]}_q}\sum _{l=1}^{n+1}{\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)}_q{E}_{n+1-l,q}\{{(x+y)}^l-x^l\}\\ =\frac{1}{{[n+1]}_q}\sum _{l=0}^{n+1}{\left(\genfrac{}{}{0ex}{}{n+1}{l}\right)}_q{E}_{n+1-l,q}\{{(x+y)}^l-x^l\}\\ =\frac{1}{{[n+1]}_q}\{E_{n+1,q}(x+y)-E_{n+1,q}(x)\}.\end{array}$$

(2.11)

Thus, by (2.11), we get

$$\begin{array}{rl}〈\frac{e_q(t)-1}{t}|E_{n,q}(x)〉& =\frac{1}{{[n+1]}_q}〈\frac{e_q(t)-1}{t}|t{E}_{n+1,q}(x)〉\\ =\frac{1}{{[n+1]}_q}〈e_q(t)-1|E_{n+1,q}(x)〉\\ =\frac{1}{{[n+1]}_q}\{E_{n+1,q}(1)-E_{n+1,q}\}\\ ={\int }_0^1{E}_{n,q}(u)d_{q}u.\end{array}$$

(2.12)

Therefore, by (2.12), we obtain the following theorem.

Theorem 2.2 For $n\ge 0$, we have

$$〈\frac{e_q(t)-1}{t}|E_{n,q}(x)〉={\int }_0^1{E}_{n,q}(u)d_{q}u.$$

Let

$${\mathbb{P}}_n=\{p(x)\in \mathbb{C}[x]|degp(x)\le n\}.$$

(2.13)

For $p(x)\in {\mathbb{P}}_n$, let us assume that

$$p(x)=\sum _{k=0}^n{b}_{k,q}{E}_{k,q}(x).$$

(2.14)

Then, by (2.4), we get

$$〈\left(\frac{e_q(t)+1}{2}\right)t^k|E_{n,q}(x)〉={[n]}_q!{\delta }_{n,k}.$$

(2.15)

From (2.14) and (2.15), we can derive the following equation:

$$\begin{array}{rl}〈\left(\frac{e_q(t)+1}{2}\right)t^k|p(x)〉& =\sum _{l=0}^n{b}_{l,q}〈\left(\frac{e_q(t)+1}{2}\right)t^k|E_{l,q}(x)〉\\ =\sum _{l=0}^n{b}_{l,q}{[l]}_q!{\delta }_{l,k}={[k]}_q!b_{k,q}.\end{array}$$

(2.16)

Thus, by (2.16), we get

$$\begin{array}{rl}{b}_{k,q}& =\frac{1}{{[k]}_q!}〈\left(\frac{e_q(t)+1}{2}\right)t^k|p(x)〉=\frac{1}{2{[k]}_q!}〈(e_q(t)+1)t^k|p(x)〉\\ =\frac{1}{2{[k]}_q!}〈e_q(t)+1|p^{(k)}(x)〉=\frac{1}{2{[k]}_q!}\{p^{(k)}(1)+p^{(k)}(0)\},\end{array}$$

(2.17)

where $p^{(k)}(x)=D_q^{k}p(x)$.

Therefore, by (2.14) and (2.17), we obtain the following theorem.

Theorem 2.3 For $p(x)\in {\mathbb{P}}_n$, let $p(x)={\sum }_{k=0}^n{b}_{k,q}{E}_{k,q}(x)$. Then we have

$$\begin{array}{rl}{b}_{k,q}& =\frac{1}{2{[k]}_q!}〈(e_q(t)+1)t^k|p(x)〉\\ =\frac{1}{2{[k]}_q!}\{p^{(k)}(1)+p^{(k)}(0)\},\end{array}$$

where $p^{(k)}(x)=D_q^{k}p(x)$.

From (1.5), we note that

$$B_{n,q}(x)\sim (\frac{e_q(t)-1}{t},t)(n\ge 0).$$

(2.18)

Let us take $p(x)=B_{n,q}(x)\in {\mathbb{P}}_n$. Then $B_{n,q}(x)$ can be represented as a linear combination of $\{E_{0,q}(x),E_{1,q}(x),\dots ,E_{n,q}(x)\}$ as follows:

$$B_{n,q}(x)=p(x)=\sum _{k=0}^n{b}_{k,q}{E}_{k,q}(x)(n\ge 0),$$

(2.19)

where

$$\begin{array}{rl}{b}_{k,q}& =\frac{1}{2{[k]}_q!}〈(e_q(t)+1)t^k|B_{n,q}(x)〉\\ =\frac{{[n]}_q{[n-1]}_q\cdots {[n-k+1]}_q}{2{[k]}_q!}〈e_q(t)+1|B_{n-k,q}(x)〉\\ =\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q〈e_q(t)+1|B_{n-k,q}(x)〉=\frac{1}{2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q\{B_{n-k,q}(1)+B_{n-k,q}\}.\end{array}$$

(2.20)

From (1.5), we can derive the following recurrence relation for the q-Bernoulli numbers:

$$\begin{array}{rl}t& =\left(\sum _{l=0}^{\mathrm{\infty }}{B}_{l,q}\frac{t^l}{{[l]}_q!}\right)(e_q(t)-1)\\ =\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{B}_{l,q}\right)\frac{t^n}{{[n]}_q!}-\sum _{n=0}^{\mathrm{\infty }}{B}_{n,q}\frac{t^n}{{[n]}_q!}\\ =\sum _{n=0}^{\mathrm{\infty }}(B_{n,q}(1)-B_{n,q})\frac{t^n}{{[n]}_q!}.\end{array}$$

(2.21)

Thus, by (2.21), we get

$$B_{0,q}=1,B_{n,q}(1)-B_{n,q}=\{\begin{array}{cc}1\hfill & \text{if }n=1,\hfill \\ 0\hfill & \text{if }n>1.\hfill \end{array}$$

(2.22)

For example, $B_{0,q}=1$, $B_{1,q}=-\frac{1}{{[2]}_q}$, $B_{2,q}=\frac{q^2}{{[3]}_q{[2]}_q},\dots$ .

By (2.19), (2.20) and (2.22), we get

$$\begin{array}{rl}{B}_{n,q}(x)& =b_{n,q}{E}_{n,q}(x)+b_{n-1,q}{E}_{n-1,q}(x)+\sum _{k=0}^{n-2}{b}_{k,q}{E}_{k,q}(x)\\ =E_{n,q}(x)+\frac{{[n]}_q}{2}(1-\frac{2}{{[2]}_q})E_{n-1,q}(x)+\sum _{k=0}^{n-2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q{B}_{n-k,q}{E}_{k,q}(x)\\ =E_{n,q}(x)-\frac{{[n]}_q(1-q)}{2{[2]}_q}{E}_{n-1,q}(x)+\sum _{k=0}^{n-2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q{B}_{n-k,q}{E}_{k,q}(x).\end{array}$$

(2.23)

Therefore, by (2.23), we obtain the following theorem.

Theorem 2.4 For $n\ge 2$, we have

$$B_{n,q}(x)=E_{n,q}(x)+\frac{{[n]}_q(q-1)}{2{[2]}_q}{E}_{n-1,q}(x)+\sum _{k=0}^{n-2}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q{B}_{n-k,q}{E}_{k,q}(x).$$

For $r\in {\mathbb{Z}}_{\ge 0}$, the q-Euler polynomials, $E_{n,q}^{(r)}(x)$, of order r are defined by the generating function to be

$$\begin{array}{c}{\left(\frac{2}{e_q\left(t\right)+1}\right)}^r{e}_q\left(xt\right)=\underset{\underset{r-\text{times}}{⏟}}{\left(\frac{2}{e_q\left(t\right)+1}\right)\times \cdots \times \left(\frac{2}{e_q\left(t\right)+1}\right)e_q\left(xt\right)}\\ =\sum _{n=0}^{\infty }{E}_{n,q}^{(r)}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$

(2.24)

In the special case, $x=0$, $E_{n,q}^{(r)}(0)=E_{n,q}^{(r)}$ are called the n th q-Euler numbers of order r.

Let

$$g^r(t)={\left(\frac{e_q(t)+1}{2}\right)}^r(r\in {\mathbb{Z}}_{\ge 0}).$$

(2.25)

Then $g^r(t)$ is an invertible series. From (2.24) and (2.25), we have

$$\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}^{(r)}(x)\frac{t^n}{{[n]}_q!}=\frac{1}{g^r(t)}{e}_q(xt)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{g^r(t)}{x}^n\frac{t^n}{{[n]}_q!}.$$

(2.26)

By (2.26), we get

$$E_{n,q}^{(r)}(x)=\frac{1}{g^r(t)}{x}^n,$$

(2.27)

and

$$t{E}_{n,q}^{(r)}(x)=\frac{1}{g^r(t)}t{x}^n={[n]}_q\frac{1}{g^r(t)}{x}^{n-1}={[n]}_q{E}_{n-1,q}^{(r)}(x).$$

(2.28)

Thus, by (2.26), (2.27) and (2.28), we see that

$$E_{n,q}^{(r)}(x)\sim ({\left(\frac{e_q(t)+1}{2}\right)}^r,t).$$

(2.29)

By (1.9) and (2.24), we get

$$〈{\left(\frac{2}{e_q(t)+1}\right)}^r{e}_q(yt)|x^n〉=E_{n,q}^{(r)}(y)=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{E}_{n-l,q}^{(r)}{y}^l.$$

(2.30)

Thus, we have

$$\begin{array}{rl}〈{\left(\frac{2}{e_q(t)+1}\right)}^r|x^n〉& =\sum _{m=0}^{\mathrm{\infty }}\left(\sum _{i_1+\cdots +i_r=m}\frac{E_{i_1,q}\cdots E_{i_r,q}}{{[i_1]}_q!\cdots {[i_r]}_q!}\right)〈t^m|x^n〉\\ =\sum _{i_1+\cdots +i_r=n}\frac{{[n]}_q!}{{[i_1]}_q!\cdots {[i_r]}_q!}{E}_{i_1,q}\cdots E_{i_r,q}\\ =\sum _{i_1+\cdots +i_r=n}{\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)}_q{E}_{i_1,q}\cdots E_{i_r,q},\end{array}$$

(2.31)

where ${\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)}_q=\frac{{[n]}_q!}{{[i_1]}_q!\cdots {[i_r]}_q!}$.

By (2.30), we easily get

$$〈{\left(\frac{2}{e_q(t)+1}\right)}^r|x^n〉=E_{n,q}^{(r)}.$$

(2.32)

Therefore, by (2.31) and (2.32), we obtain the following theorem.

Theorem 2.5 For $n\ge 0$, we have

$$E_{n,q}^{(r)}=\sum _{i_1+\cdots +i_r=n}{\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)}_q{E}_{i_1,q}\cdots E_{i_r,q},$$

where ${\left(\genfrac{}{}{0ex}{}{n}{i_1,\dots ,i_r}\right)}_q=\frac{{[n]}_q!}{{[i_1]}_q!\cdots {[i_r]}_q!}$.

Let us take $p(x)=E_{n,q}^{(r)}(x)\in {\mathbb{P}}_n$. Then, by Theorem 2.3, we get

$$E_{n,q}^{(r)}(x)=p(x)=\sum _{k=0}^n{b}_{k,q}{E}_{k,q}(x),$$

(2.33)

where

$$\begin{array}{rl}{b}_{k,q}& =\frac{1}{2{[k]}_q!}〈(e_q(t)+1)t^k|p(x)〉=\frac{1}{2{[k]}_q!}〈(e_q(t)+1)|t^{k}p(x)〉\\ =\frac{{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q}{2}〈(e_q(t)+1)|E_{n-k,q}^{(r)}(x)〉=\frac{{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q}{2}\{E_{n-k,q}^{(r)}(1)+E_{n-k,q}^{(r)}\}.\end{array}$$

(2.34)

From (2.24), we have

$$\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}\{E_{n,q}^{(r)}(1)+E_{n,q}^{(r)}\}\frac{t^n}{{[n]}_q!}& ={\left(\frac{2}{e_q(t)+1}\right)}^r(e_q(t)+1)\\ =2{\left(\frac{2}{e_q(t)+1}\right)}^{r-1}=2\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}^{(r-1)}\frac{t^n}{{[n]}_q!}.\end{array}$$

(2.35)

By comparing the coefficients on the both sides of (2.35), we get

$$E_{n,q}^{(r)}(1)+E_{n,q}^{(r)}=2E_{n,q}^{(r-1)}(n\ge 0).$$

(2.36)

Therefore, by (2.33), (2.34) and (2.36), we obtain the following theorem.

Theorem 2.6 For $n\in {\mathbb{Z}}_{\ge 0}$, $r\in {\mathbb{Z}}_{>0}$, we have

$$E_{n,q}^{(r)}(x)=\sum _{k=0}^{\mathrm{\infty }}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}_q{E}_{n-k,q}^{(r-1)}{E}_{k,q}(x).$$

Let us assume that

$$p(x)=\sum _{k=0}^n{b}_{k,q}^r{E}_{k,q}^{(r)}(x)\in {\mathbb{P}}_n.$$

(2.37)

By (2.29) and (2.37), we get

$$\begin{array}{rl}〈{\left(\frac{e_q(t)+1}{2}\right)}^r{t}^k|p(x)〉& =\sum _{l=0}^n{b}_{l,q}^r〈{\left(\frac{e_q(t)+1}{2}\right)}^r{t}^k|E_{l,q}^{(r)}(x)〉\\ =\sum _{l=0}^n{b}_{l,q}^r{[l]}_q!{\delta }_{l,k}={[k]}_q!b_{k,q}^r.\end{array}$$

(2.38)

From (2.38), we have

$$\begin{array}{rl}{b}_{k,q}^r& =\frac{1}{{[k]}_q!}〈{\left(\frac{e_q(t)+1}{2}\right)}^r{t}^k|p(x)〉=\frac{1}{2^r{[k]}_q!}〈{(e_q(t)+1)}^r|t^{k}p(x)〉\\ =\frac{1}{2^r{[k]}_q!}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{m\ge 0}\left(\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\right)\frac{1}{{[m]}_q!}〈1|t^{m+k}p(x)〉\\ =\frac{1}{2^r{[k]}_q!}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{m\ge 0}\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\frac{1}{{[m]}_q!}{p}^{(m+k)}(0).\end{array}$$

(2.39)

Therefore by (2.37) and (2.39), we obtain the following theorem.

Theorem 2.7 For $n\ge 0$, let $p(x)={\sum }_{k=0}^n{b}_{k,q}^r{E}_{k,q}^{(r)}(x)\in {\mathbb{P}}_n$.

Then we have

$$\begin{array}{rl}{b}_{k,q}^r& =\frac{1}{2^r{[k]}_q!}=〈{(e_q(t)+1)}^r{t}^k|p(x)〉\\ =\frac{1}{2^r{[k]}_q!}\sum _{m\ge 0}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\frac{1}{{[m]}_q!}{p}^{(m+k)}(0),\end{array}$$

where $p^{(k)}(x)=D_q^{k}p(x)$.

Let us take $p(x)=E_{n,q}(x)\in {\mathbb{P}}_n$. Then, by Theorem 2.7, we get

$$E_{n,q}(x)=p(x)=\sum _{k=0}^n{b}_{k,q}^r{E}_{k,q}^{(r)}(x),$$

(2.40)

where

$$\begin{array}{rl}{b}_{k,q}=& \frac{1}{2^r{[k]}_q!}\sum _{m=0}^{n-k}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\\ \times \frac{1}{{[m]}_q!}{[n]}_q\cdots {[n-m-k+1]}_q{E}_{n-m-k,q}\\ =& \frac{1}{2^r}\sum _{m=0}^{n-k}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\\ \times \frac{{[m+k]}_q!}{{[m]}_q!{[k]}_q!}\frac{{[n]}_q\cdots {[n-m-k+1]}_q}{{[m+k]}_q!}{E}_{n-m-k,q}\\ =& \frac{1}{2^r}\sum _{m=0}^{n-k}\sum _{l=0}^r\sum _{i_1+\cdots +i_l=m}\left(\genfrac{}{}{0ex}{}{r}{l}\right){\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q{\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)}_q{\left(\genfrac{}{}{0ex}{}{n}{m+k}\right)}_q{E}_{n-m-k,q}.\end{array}$$

(2.41)

Therefore, by (2.40) and (2.41), we obtain the following theorem.

Theorem 2.8 For $n,r\ge 0$, we have

$$\begin{array}{rcl}{E}_{n,q}(x)& =& \frac{1}{2^r}\sum _{k=0}^n\{\sum _{m=0}^{n-k}\sum _{l=0}^r\sum _{i_1+\cdots +i_l=m}\left(\genfrac{}{}{0ex}{}{r}{l}\right){\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q{\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)}_q{\left(\genfrac{}{}{0ex}{}{n}{m+k}\right)}_q\\ \times E_{n-m-k,q}\}{E}_{k,q}^{(r)}(x).\end{array}$$

For $r\in {\mathbb{Z}}_{\ge 0}$, let us consider q-Bernoulli polynomials of order r which are defined by the generating function to be

$$\begin{array}{c}{\left(\frac{2}{e_q\left(t\right)-1}\right)}^r{e}_q\left(xt\right)=\underset{\underset{r-\text{times}}{⏟}}{\left(\frac{2}{e_q\left(t\right)-1}\right)\times \cdots \times \left(\frac{2}{e_q\left(t\right)-1}\right)e_q\left(xt\right)}\\ =\sum _{n=0}^{\infty }{B}_{n,q}^{(r)}\left(x\right)\frac{t^n}{{\left[n\right]}_q!}.\end{array}$$

(2.42)

In the special case, $x=0$, $B_{n,q}^{(r)}(0)=B_{n,q}^{(r)}$ are called the n th q-Bernoulli numbers of order r. By (2.42), we easily get

$$B_{n,q}^{(r)}(x)=\sum _{l=0}^n{\left(\genfrac{}{}{0ex}{}{n}{l}\right)}_q{B}_{l,q}^{(r)}{x}^{n-l}\in {\mathbb{P}}_n.$$

(2.43)

Let us take $p(x)=B_{n,q}^{(r)}(x)\in {\mathbb{P}}_n$. Then, by Theorem 2.7, we get

$$B_{n,q}^{(r)}(x)=p(x)=\sum _{k=0}^n{b}_{k,q}^r{E}_{k,q}^{(r)}(x),$$

(2.44)

where

$$\begin{array}{rl}{b}_{k,q}^r=& \frac{1}{2^r{[k]}_q!}〈{(e_q(t)+1)}^r{t}^k|B_{n,q}^{(r)}(x))〉\\ =& \frac{1}{2^r{[k]}_q!}\sum _{m=0}^{n-k}\sum _{l=0}^r\left(\genfrac{}{}{0ex}{}{r}{l}\right)\sum _{i_1+\cdots +i_l=m}{\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q\frac{{[n]}_q\cdots {[n-m-k+1]}_q}{{[m]}_q!}{B}_{n-m-k,q}^{(r)}\\ =& \frac{1}{2^r}\sum _{m=0}^{n-k}\sum _{l=0}^r\sum _{i_1+\cdots +i_l=m}\left(\genfrac{}{}{0ex}{}{r}{l}\right){\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q{\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)}_q{\left(\genfrac{}{}{0ex}{}{n}{m+k}\right)}_q{B}_{n-m-k,q}^{(r)}.\end{array}$$

(2.45)

Therefore, by (2.44) and (2.45), we obtain the following theorem.

Theorem 2.9 For $n,r\ge 0$, we have

$$\begin{array}{rcl}{B}_{n,q}^{(r)}(x)& =& \frac{1}{2^r}\sum _{k=0}^n\{\sum _{m=0}^{n-k}\sum _{l=0}^r\sum _{i_1+\cdots +i_l=m}\left(\genfrac{}{}{0ex}{}{r}{l}\right){\left(\genfrac{}{}{0ex}{}{m}{i_1,\dots ,i_l}\right)}_q{\left(\genfrac{}{}{0ex}{}{m+k}{m}\right)}_q{\left(\genfrac{}{}{0ex}{}{n}{m+k}\right)}_q\\ \times B_{n-m-k,q}^{(r)}\}{E}_{k,q}^{(r)}(x).\end{array}$$

Remark Recently, Aral, Gupta and Agarwal introduced many interesting properties and applications of q-calculus which are related to this paper (see [13]).