# 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

${\left(a:q\right)}_{0}=1,\phantom{\rule{2em}{0ex}}{\left(a:q\right)}_{n}=\prod _{i=0}^{n-1}\left(1-a{q}^{i}\right),\phantom{\rule{2em}{0ex}}{\left(a:q\right)}_{\mathrm{\infty }}=\prod _{i=0}^{\mathrm{\infty }}\left(1-a{q}^{i}\right).$
(1.1)

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

${e}_{q}\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{z}^{n}}{{\left[n\right]}_{q}!}=\frac{1}{{\left(\left(1-q\right)z:q\right)}_{\mathrm{\infty }}}\phantom{\rule{1em}{0ex}}\left(\text{see [1–4]}\right),$
(1.2)

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

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

${\int }_{0}^{x}f\left(t\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}t=\left(1-q\right)\sum _{a=0}^{\mathrm{\infty }}f\left({q}^{a}x\right)x{q}^{a}\phantom{\rule{1em}{0ex}}\left(\text{see [1, 2, 5])}.$
(1.3)

The q-difference operator ${D}_{q}$ is defined by

(1.4)

where

$\underset{q\to 1}{lim}{D}_{q}f\left(x\right)=\frac{df\left(x\right)}{dx}\phantom{\rule{1em}{0ex}}\left(\text{see [1, 2, 4, 6]}\right).$

By using an exponential function ${e}_{q}\left(x\right)$, Hegazi and Mansour defined q-Bernoulli polynomials as follows:

$\sum _{n=0}^{\mathrm{\infty }}{B}_{n,q}\left(x\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}=\frac{t}{{e}_{q}\left(t\right)-1}{e}_{q}\left(xt\right)\phantom{\rule{1em}{0ex}}\left(\text{see [1, 2, 4, 7]}\right).$
(1.5)

In the special case, $x=0$, ${B}_{n,q}\left(0\right)={B}_{n,q}$ are called the n th q-Bernoulli numbers.

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

${B}_{n,q}\left(x\right)=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{B}_{n-l,q}{x}^{l}=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{B}_{l,q}{x}^{n-l},$
(1.6)

where

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

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  defined a q-extension of Euler polynomials. In a recent paper (see ), 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 [16, 811]). Let be the complex number field, and let be the set of all formal power series in variable t over with

$\mathcal{F}=\left\{f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{{a}_{k}}{{\left[k\right]}_{q}!}{t}^{k}|{a}_{k}\in \mathbb{C}\right\}.$
(1.7)

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

$〈L+M|p\left(x\right)〉=〈L|p\left(x\right)〉+〈M|p\left(x\right)〉,\phantom{\rule{2em}{0ex}}〈cL|p\left(x\right)〉=c〈L|p\left(x\right)〉,$

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

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

(1.8)

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

$〈{t}^{k}|{x}^{n}〉={\left[n\right]}_{q}!{\delta }_{n,k}\phantom{\rule{1em}{0ex}}\left(n,k\ge 0\right),$
(1.9)

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

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

$f\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}〈f\left(t\right)|{x}^{k}〉\frac{{t}^{k}}{{\left[k\right]}_{q}!},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}〈{t}^{k}|p\left(x\right)〉\frac{{x}^{k}}{{\left[k\right]}_{q}!}\phantom{\rule{1em}{0ex}}\left(\text{see }\right).$
(1.10)

From (1.10), we have

${p}^{\left(k\right)}\left(x\right)={D}_{q}^{k}p\left(x\right)=\sum _{l=k}^{\mathrm{\infty }}\frac{〈{t}^{l}|p\left(x\right)〉}{{\left[l\right]}_{q}!}{\left[l\right]}_{q}\cdots {\left[l-k+1\right]}_{q}{x}^{l-k}.$
(1.11)

By (1.11), we get

${p}^{\left(k\right)}\left(0\right)=〈{t}^{k}|p\left(x\right)〉\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}〈1|{p}^{\left(k\right)}\left(x\right)〉={p}^{\left(k\right)}\left(0\right).$
(1.12)

Thus from (1.12), we note that

${t}^{k}p\left(x\right)={p}^{\left(k\right)}\left(x\right)={D}_{q}^{k}p\left(x\right).$
(1.13)

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

$h\left(t\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈h\left(t\right)|{S}_{k}\left(x\right)〉}{{\left[k\right]}_{q}!}g\left(t\right)f{\left(t\right)}^{k},\phantom{\rule{2em}{0ex}}p\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{〈g\left(t\right)f{\left(t\right)}^{k}|p\left(x\right)〉}{{\left[k\right]}_{q}!}{S}_{k}\left(x\right),$
(1.14)

and

(1.15)

where $\overline{f}\left(t\right)$ is the compositional inverse of $f\left(t\right)$ (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 ). 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

$\frac{2}{{e}_{q}\left(t\right)+1}{e}_{q}\left(xt\right)=\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}\left(x\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}.$
(2.1)

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

${E}_{n,q}\left(x\right)=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{E}_{l,q}{x}^{n-l}=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{E}_{n-l,q}{x}^{l}.$
(2.2)

By (2.1), we easily get

${E}_{0,q}=1,\phantom{\rule{2em}{0ex}}{E}_{n,q}\left(1\right)+{E}_{n,q}=2{\delta }_{0,n}.$
(2.3)

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

${E}_{n,q}\left(x\right)\sim \left(\frac{{e}_{q}\left(t\right)+1}{2},t\right)$
(2.4)

and

$\frac{2}{{e}_{q}\left(t\right)+1}{x}^{n}={E}_{n,q}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(2.5)

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

$t{E}_{n,q}\left(x\right)=\frac{2}{{e}_{q}\left(t\right)+1}t{x}^{n}={\left[n\right]}_{q}\frac{2}{{e}_{q}\left(t\right)+1}{x}^{n-1}={\left[n\right]}_{q}{E}_{n-1,q}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(2.6)

Indeed, by (1.9), we get

$\begin{array}{rl}〈\frac{{e}_{q}\left(t\right)+1}{2}{t}^{k}|{E}_{n,q}\left(x\right)〉& =\frac{{\left[k\right]}_{q}!}{2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}〈{e}_{q}\left(t\right)+1|{E}_{n-k,q}\left(x\right)〉\\ =\frac{{\left[k\right]}_{q}!}{2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}\left({E}_{n-k,q}\left(1\right)+{E}_{n-k,q}\right).\end{array}$
(2.7)

From (2.4), we have

$〈\left(\frac{{e}_{q}\left(t\right)+1}{2}\right){t}^{k}|{E}_{n,q}\left(x\right)〉={\left[n\right]}_{q}!{\delta }_{n,k}.$
(2.8)

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

(2.9)

This is equivalent to

(2.10)

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

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

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

From (2.2) we have

$\begin{array}{rl}{\int }_{x}^{x+y}{E}_{n,q}\left(u\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}u& =\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{E}_{n-l,q}\frac{1}{{\left[l+1\right]}_{q}}\left\{{\left(x+y\right)}^{l+1}-{x}^{l+1}\right\}\\ =\frac{1}{{\left[n+1\right]}_{q}}\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n+1}{l+1}\right)}_{q}{E}_{n-l,q}\left\{{\left(x+y\right)}^{l+1}-{x}^{l+1}\right\}\\ =\frac{1}{{\left[n+1\right]}_{q}}\sum _{l=1}^{n+1}{\left(\genfrac{}{}{0}{}{n+1}{l}\right)}_{q}{E}_{n+1-l,q}\left\{{\left(x+y\right)}^{l}-{x}^{l}\right\}\\ =\frac{1}{{\left[n+1\right]}_{q}}\sum _{l=0}^{n+1}{\left(\genfrac{}{}{0}{}{n+1}{l}\right)}_{q}{E}_{n+1-l,q}\left\{{\left(x+y\right)}^{l}-{x}^{l}\right\}\\ =\frac{1}{{\left[n+1\right]}_{q}}\left\{{E}_{n+1,q}\left(x+y\right)-{E}_{n+1,q}\left(x\right)\right\}.\end{array}$
(2.11)

Thus, by (2.11), we get

$\begin{array}{rl}〈\frac{{e}_{q}\left(t\right)-1}{t}|{E}_{n,q}\left(x\right)〉& =\frac{1}{{\left[n+1\right]}_{q}}〈\frac{{e}_{q}\left(t\right)-1}{t}|t{E}_{n+1,q}\left(x\right)〉\\ =\frac{1}{{\left[n+1\right]}_{q}}〈{e}_{q}\left(t\right)-1|{E}_{n+1,q}\left(x\right)〉\\ =\frac{1}{{\left[n+1\right]}_{q}}\left\{{E}_{n+1,q}\left(1\right)-{E}_{n+1,q}\right\}\\ ={\int }_{0}^{1}{E}_{n,q}\left(u\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}u.\end{array}$
(2.12)

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

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

$〈\frac{{e}_{q}\left(t\right)-1}{t}|{E}_{n,q}\left(x\right)〉={\int }_{0}^{1}{E}_{n,q}\left(u\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}u.$

Let

${\mathbb{P}}_{n}=\left\{p\left(x\right)\in \mathbb{C}\left[x\right]|degp\left(x\right)\le n\right\}.$
(2.13)

For $p\left(x\right)\in {\mathbb{P}}_{n}$, let us assume that

$p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}{E}_{k,q}\left(x\right).$
(2.14)

Then, by (2.4), we get

$〈\left(\frac{{e}_{q}\left(t\right)+1}{2}\right){t}^{k}|{E}_{n,q}\left(x\right)〉={\left[n\right]}_{q}!{\delta }_{n,k}.$
(2.15)

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

$\begin{array}{rl}〈\left(\frac{{e}_{q}\left(t\right)+1}{2}\right){t}^{k}|p\left(x\right)〉& =\sum _{l=0}^{n}{b}_{l,q}〈\left(\frac{{e}_{q}\left(t\right)+1}{2}\right){t}^{k}|{E}_{l,q}\left(x\right)〉\\ =\sum _{l=0}^{n}{b}_{l,q}{\left[l\right]}_{q}!{\delta }_{l,k}={\left[k\right]}_{q}!{b}_{k,q}.\end{array}$
(2.16)

Thus, by (2.16), we get

$\begin{array}{rl}{b}_{k,q}& =\frac{1}{{\left[k\right]}_{q}!}〈\left(\frac{{e}_{q}\left(t\right)+1}{2}\right){t}^{k}|p\left(x\right)〉=\frac{1}{2{\left[k\right]}_{q}!}〈\left({e}_{q}\left(t\right)+1\right){t}^{k}|p\left(x\right)〉\\ =\frac{1}{2{\left[k\right]}_{q}!}〈{e}_{q}\left(t\right)+1|{p}^{\left(k\right)}\left(x\right)〉=\frac{1}{2{\left[k\right]}_{q}!}\left\{{p}^{\left(k\right)}\left(1\right)+{p}^{\left(k\right)}\left(0\right)\right\},\end{array}$
(2.17)

where ${p}^{\left(k\right)}\left(x\right)={D}_{q}^{k}p\left(x\right)$.

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

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

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

where ${p}^{\left(k\right)}\left(x\right)={D}_{q}^{k}p\left(x\right)$.

From (1.5), we note that

${B}_{n,q}\left(x\right)\sim \left(\frac{{e}_{q}\left(t\right)-1}{t},t\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(2.18)

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

${B}_{n,q}\left(x\right)=p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}{E}_{k,q}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\ge 0\right),$
(2.19)

where

$\begin{array}{rl}{b}_{k,q}& =\frac{1}{2{\left[k\right]}_{q}!}〈\left({e}_{q}\left(t\right)+1\right){t}^{k}|{B}_{n,q}\left(x\right)〉\\ =\frac{{\left[n\right]}_{q}{\left[n-1\right]}_{q}\cdots {\left[n-k+1\right]}_{q}}{2{\left[k\right]}_{q}!}〈{e}_{q}\left(t\right)+1|{B}_{n-k,q}\left(x\right)〉\\ =\frac{1}{2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}〈{e}_{q}\left(t\right)+1|{B}_{n-k,q}\left(x\right)〉=\frac{1}{2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}\left\{{B}_{n-k,q}\left(1\right)+{B}_{n-k,q}\right\}.\end{array}$
(2.20)

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

$\begin{array}{rl}t& =\left(\sum _{l=0}^{\mathrm{\infty }}{B}_{l,q}\frac{{t}^{l}}{{\left[l\right]}_{q}!}\right)\left({e}_{q}\left(t\right)-1\right)\\ =\sum _{n=0}^{\mathrm{\infty }}\left(\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{B}_{l,q}\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}-\sum _{n=0}^{\mathrm{\infty }}{B}_{n,q}\frac{{t}^{n}}{{\left[n\right]}_{q}!}\\ =\sum _{n=0}^{\mathrm{\infty }}\left({B}_{n,q}\left(1\right)-{B}_{n,q}\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}.\end{array}$
(2.21)

Thus, by (2.21), we get

(2.22)

For example, ${B}_{0,q}=1$, ${B}_{1,q}=-\frac{1}{{\left[2\right]}_{q}}$, ${B}_{2,q}=\frac{{q}^{2}}{{\left[3\right]}_{q}{\left[2\right]}_{q}},\dots$ .

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

$\begin{array}{rl}{B}_{n,q}\left(x\right)& ={b}_{n,q}{E}_{n,q}\left(x\right)+{b}_{n-1,q}{E}_{n-1,q}\left(x\right)+\sum _{k=0}^{n-2}{b}_{k,q}{E}_{k,q}\left(x\right)\\ ={E}_{n,q}\left(x\right)+\frac{{\left[n\right]}_{q}}{2}\left(1-\frac{2}{{\left[2\right]}_{q}}\right){E}_{n-1,q}\left(x\right)+\sum _{k=0}^{n-2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}{B}_{n-k,q}{E}_{k,q}\left(x\right)\\ ={E}_{n,q}\left(x\right)-\frac{{\left[n\right]}_{q}\left(1-q\right)}{2{\left[2\right]}_{q}}{E}_{n-1,q}\left(x\right)+\sum _{k=0}^{n-2}{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}{B}_{n-k,q}{E}_{k,q}\left(x\right).\end{array}$
(2.23)

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

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

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

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

$\begin{array}{c}{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r}{e}_{q}\left(xt\right)=\underset{\underset{r-\text{times}}{⏟}}{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)×\cdots ×\left(\frac{2}{{e}_{q}\left(t\right)+1}\right){e}_{q}\left(xt\right)}\\ =\sum _{n=0}^{\infty }{E}_{n,q}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}.\end{array}$
(2.24)

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

Let

${g}^{r}\left(t\right)={\left(\frac{{e}_{q}\left(t\right)+1}{2}\right)}^{r}\phantom{\rule{1em}{0ex}}\left(r\in {\mathbb{Z}}_{\ge 0}\right).$
(2.25)

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

$\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}=\frac{1}{{g}^{r}\left(t\right)}{e}_{q}\left(xt\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{g}^{r}\left(t\right)}{x}^{n}\frac{{t}^{n}}{{\left[n\right]}_{q}!}.$
(2.26)

By (2.26), we get

${E}_{n,q}^{\left(r\right)}\left(x\right)=\frac{1}{{g}^{r}\left(t\right)}{x}^{n},$
(2.27)

and

$t{E}_{n,q}^{\left(r\right)}\left(x\right)=\frac{1}{{g}^{r}\left(t\right)}t{x}^{n}={\left[n\right]}_{q}\frac{1}{{g}^{r}\left(t\right)}{x}^{n-1}={\left[n\right]}_{q}{E}_{n-1,q}^{\left(r\right)}\left(x\right).$
(2.28)

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

${E}_{n,q}^{\left(r\right)}\left(x\right)\sim \left({\left(\frac{{e}_{q}\left(t\right)+1}{2}\right)}^{r},t\right).$
(2.29)

By (1.9) and (2.24), we get

$〈{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r}{e}_{q}\left(yt\right)|{x}^{n}〉={E}_{n,q}^{\left(r\right)}\left(y\right)=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{E}_{n-l,q}^{\left(r\right)}{y}^{l}.$
(2.30)

Thus, we have

$\begin{array}{rl}〈{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r}|{x}^{n}〉& =\sum _{m=0}^{\mathrm{\infty }}\left(\sum _{{i}_{1}+\cdots +{i}_{r}=m}\frac{{E}_{{i}_{1},q}\cdots {E}_{{i}_{r},q}}{{\left[{i}_{1}\right]}_{q}!\cdots {\left[{i}_{r}\right]}_{q}!}\right)〈{t}^{m}|{x}^{n}〉\\ =\sum _{{i}_{1}+\cdots +{i}_{r}=n}\frac{{\left[n\right]}_{q}!}{{\left[{i}_{1}\right]}_{q}!\cdots {\left[{i}_{r}\right]}_{q}!}{E}_{{i}_{1},q}\cdots {E}_{{i}_{r},q}\\ =\sum _{{i}_{1}+\cdots +{i}_{r}=n}{\left(\genfrac{}{}{0}{}{n}{{i}_{1},\dots ,{i}_{r}}\right)}_{q}{E}_{{i}_{1},q}\cdots {E}_{{i}_{r},q},\end{array}$
(2.31)

where ${\left(\genfrac{}{}{0}{}{n}{{i}_{1},\dots ,{i}_{r}}\right)}_{q}=\frac{{\left[n\right]}_{q}!}{{\left[{i}_{1}\right]}_{q}!\cdots {\left[{i}_{r}\right]}_{q}!}$.

By (2.30), we easily get

$〈{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r}|{x}^{n}〉={E}_{n,q}^{\left(r\right)}.$
(2.32)

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

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

${E}_{n,q}^{\left(r\right)}=\sum _{{i}_{1}+\cdots +{i}_{r}=n}{\left(\genfrac{}{}{0}{}{n}{{i}_{1},\dots ,{i}_{r}}\right)}_{q}{E}_{{i}_{1},q}\cdots {E}_{{i}_{r},q},$

where ${\left(\genfrac{}{}{0}{}{n}{{i}_{1},\dots ,{i}_{r}}\right)}_{q}=\frac{{\left[n\right]}_{q}!}{{\left[{i}_{1}\right]}_{q}!\cdots {\left[{i}_{r}\right]}_{q}!}$.

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

${E}_{n,q}^{\left(r\right)}\left(x\right)=p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}{E}_{k,q}\left(x\right),$
(2.33)

where

$\begin{array}{rl}{b}_{k,q}& =\frac{1}{2{\left[k\right]}_{q}!}〈\left({e}_{q}\left(t\right)+1\right){t}^{k}|p\left(x\right)〉=\frac{1}{2{\left[k\right]}_{q}!}〈\left({e}_{q}\left(t\right)+1\right)|{t}^{k}p\left(x\right)〉\\ =\frac{{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}}{2}〈\left({e}_{q}\left(t\right)+1\right)|{E}_{n-k,q}^{\left(r\right)}\left(x\right)〉=\frac{{\left(\genfrac{}{}{0}{}{n}{k}\right)}_{q}}{2}\left\{{E}_{n-k,q}^{\left(r\right)}\left(1\right)+{E}_{n-k,q}^{\left(r\right)}\right\}.\end{array}$
(2.34)

From (2.24), we have

$\begin{array}{rl}\sum _{k=0}^{\mathrm{\infty }}\left\{{E}_{n,q}^{\left(r\right)}\left(1\right)+{E}_{n,q}^{\left(r\right)}\right\}\frac{{t}^{n}}{{\left[n\right]}_{q}!}& ={\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r}\left({e}_{q}\left(t\right)+1\right)\\ =2{\left(\frac{2}{{e}_{q}\left(t\right)+1}\right)}^{r-1}=2\sum _{n=0}^{\mathrm{\infty }}{E}_{n,q}^{\left(r-1\right)}\frac{{t}^{n}}{{\left[n\right]}_{q}!}.\end{array}$
(2.35)

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

${E}_{n,q}^{\left(r\right)}\left(1\right)+{E}_{n,q}^{\left(r\right)}=2{E}_{n,q}^{\left(r-1\right)}\phantom{\rule{1em}{0ex}}\left(n\ge 0\right).$
(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

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

Let us assume that

$p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}^{r}{E}_{k,q}^{\left(r\right)}\left(x\right)\in {\mathbb{P}}_{n}.$
(2.37)

By (2.29) and (2.37), we get

$\begin{array}{rl}〈{\left(\frac{{e}_{q}\left(t\right)+1}{2}\right)}^{r}{t}^{k}|p\left(x\right)〉& =\sum _{l=0}^{n}{b}_{l,q}^{r}〈{\left(\frac{{e}_{q}\left(t\right)+1}{2}\right)}^{r}{t}^{k}|{E}_{l,q}^{\left(r\right)}\left(x\right)〉\\ =\sum _{l=0}^{n}{b}_{l,q}^{r}{\left[l\right]}_{q}!{\delta }_{l,k}={\left[k\right]}_{q}!{b}_{k,q}^{r}.\end{array}$
(2.38)

From (2.38), we have

$\begin{array}{rl}{b}_{k,q}^{r}& =\frac{1}{{\left[k\right]}_{q}!}〈{\left(\frac{{e}_{q}\left(t\right)+1}{2}\right)}^{r}{t}^{k}|p\left(x\right)〉=\frac{1}{{2}^{r}{\left[k\right]}_{q}!}〈{\left({e}_{q}\left(t\right)+1\right)}^{r}|{t}^{k}p\left(x\right)〉\\ =\frac{1}{{2}^{r}{\left[k\right]}_{q}!}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right)\sum _{m\ge 0}\left(\sum _{{i}_{1}+\cdots +{i}_{l}=m}{\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}\right)\frac{1}{{\left[m\right]}_{q}!}〈1|{t}^{m+k}p\left(x\right)〉\\ =\frac{1}{{2}^{r}{\left[k\right]}_{q}!}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right)\sum _{m\ge 0}\sum _{{i}_{1}+\cdots +{i}_{l}=m}{\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}\frac{1}{{\left[m\right]}_{q}!}{p}^{\left(m+k\right)}\left(0\right).\end{array}$
(2.39)

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

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

Then we have

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

where ${p}^{\left(k\right)}\left(x\right)={D}_{q}^{k}p\left(x\right)$.

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

${E}_{n,q}\left(x\right)=p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}^{r}{E}_{k,q}^{\left(r\right)}\left(x\right),$
(2.40)

where

$\begin{array}{rl}{b}_{k,q}=& \frac{1}{{2}^{r}{\left[k\right]}_{q}!}\sum _{m=0}^{n-k}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right)\sum _{{i}_{1}+\cdots +{i}_{l}=m}{\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}\\ ×\frac{1}{{\left[m\right]}_{q}!}{\left[n\right]}_{q}\cdots {\left[n-m-k+1\right]}_{q}{E}_{n-m-k,q}\\ =& \frac{1}{{2}^{r}}\sum _{m=0}^{n-k}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right)\sum _{{i}_{1}+\cdots +{i}_{l}=m}{\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}\\ ×\frac{{\left[m+k\right]}_{q}!}{{\left[m\right]}_{q}!{\left[k\right]}_{q}!}\frac{{\left[n\right]}_{q}\cdots {\left[n-m-k+1\right]}_{q}}{{\left[m+k\right]}_{q}!}{E}_{n-m-k,q}\\ =& \frac{1}{{2}^{r}}\sum _{m=0}^{n-k}\sum _{l=0}^{r}\sum _{{i}_{1}+\cdots +{i}_{l}=m}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}{\left(\genfrac{}{}{0}{}{m+k}{m}\right)}_{q}{\left(\genfrac{}{}{0}{}{n}{m+k}\right)}_{q}{E}_{n-m-k,q}.\end{array}$
(2.41)

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

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

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

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

$\begin{array}{c}{\left(\frac{2}{{e}_{q}\left(t\right)-1}\right)}^{r}{e}_{q}\left(xt\right)=\underset{\underset{r-\text{times}}{⏟}}{\left(\frac{2}{{e}_{q}\left(t\right)-1}\right)×\cdots ×\left(\frac{2}{{e}_{q}\left(t\right)-1}\right){e}_{q}\left(xt\right)}\\ =\sum _{n=0}^{\infty }{B}_{n,q}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{{\left[n\right]}_{q}!}.\end{array}$
(2.42)

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

${B}_{n,q}^{\left(r\right)}\left(x\right)=\sum _{l=0}^{n}{\left(\genfrac{}{}{0}{}{n}{l}\right)}_{q}{B}_{l,q}^{\left(r\right)}{x}^{n-l}\in {\mathbb{P}}_{n}.$
(2.43)

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

${B}_{n,q}^{\left(r\right)}\left(x\right)=p\left(x\right)=\sum _{k=0}^{n}{b}_{k,q}^{r}{E}_{k,q}^{\left(r\right)}\left(x\right),$
(2.44)

where

$\begin{array}{rl}{b}_{k,q}^{r}=& \frac{1}{{2}^{r}{\left[k\right]}_{q}!}〈{\left({e}_{q}\left(t\right)+1\right)}^{r}{t}^{k}|{B}_{n,q}^{\left(r\right)}\left(x\right)\right)〉\\ =& \frac{1}{{2}^{r}{\left[k\right]}_{q}!}\sum _{m=0}^{n-k}\sum _{l=0}^{r}\left(\genfrac{}{}{0}{}{r}{l}\right)\sum _{{i}_{1}+\cdots +{i}_{l}=m}{\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}\frac{{\left[n\right]}_{q}\cdots {\left[n-m-k+1\right]}_{q}}{{\left[m\right]}_{q}!}{B}_{n-m-k,q}^{\left(r\right)}\\ =& \frac{1}{{2}^{r}}\sum _{m=0}^{n-k}\sum _{l=0}^{r}\sum _{{i}_{1}+\cdots +{i}_{l}=m}\left(\genfrac{}{}{0}{}{r}{l}\right){\left(\genfrac{}{}{0}{}{m}{{i}_{1},\dots ,{i}_{l}}\right)}_{q}{\left(\genfrac{}{}{0}{}{m+k}{m}\right)}_{q}{\left(\genfrac{}{}{0}{}{n}{m+k}\right)}_{q}{B}_{n-m-k,q}^{\left(r\right)}.\end{array}$
(2.45)

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

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

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

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

## References

1. Hegazi AS, Mansour M: A note on q -Bernoulli numbers and polynomials. J. Nonlinear Math. Phys. 2006,13(1):9-18. 10.2991/jnmp.2006.13.1.2

2. Kim T: q -Generalized Euler numbers and polynomials. Russ. J. Math. Phys. 2006, 13: 293-298. 10.1134/S1061920806030058

3. Kupershmidt BO: Reflection symmetries of q -Bernoulli polynomials. J. Nonlinear Math. Phys. 2005, 12: 412-422. (suppl. 1) 10.2991/jnmp.2005.12.s1.34

4. Kurt V: Some relation between the Bernstein polynomials and second kind Bernoulli polynomials. Adv. Stud. Contemp. Math. 2013,23(1):43-48.

5. Mahmudov NI, Keleshteri ME: On a class of generalized q -Bernoulli and q -Euler polynomials. Adv. Differ. Equ. 2013., 2013: Article ID 115

6. Mahmudov NI: On a class of q -Bernoulli and q -Euler polynomials. Adv. Differ. Equ. 2013., 2013: Article ID 108

7. Roman S: The Umbral Calculus. Dover, New York; 2005.

8. Carlitz L: q -Bernoulli numbers and polynomials. Duke Math. J. 1948, 15: 987-1000. 10.1215/S0012-7094-48-01588-9

9. Araci S, Acikgoz M, Sen E: On the extended Kim’s p -adic q -deformed fermionic integrals in the p -adic integer ring. J. Number Theory 2013,133(10):3348-3361. 10.1016/j.jnt.2013.04.007

10. Ismail MEH, Rahman M: Inverse operators, q -fractional integrals, and q -Bernoulli polynomials. J. Approx. Theory 2002,114(2):269-307. 10.1006/jath.2001.3644

11. Kim DS, Kim T, Lee S-H, Rim S-H: Some identities of Bernoulli, Euler and Abel polynomials arising from umbral calculus. Adv. Differ. Equ. 2013., 2013: Article ID 15

12. Roman S: More on the umbral calculus, with emphasis on the q -umbral calculus. J. Math. Anal. Appl. 1985, 107: 222-254. 10.1016/0022-247X(85)90367-1

13. Aral A, Gupta V, Agarwal RP: Applications of q-Calculus in Operator Theory. Springer, New York; 2013. xii+262 pp.

## Acknowledgements

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MOE) (No. 2012R1A1A2003786). The authors express their sincere gratitude to the referees for their valuable suggestions and comments.

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Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T. Some identities of q-Euler polynomials arising from q-umbral calculus. J Inequal Appl 2014, 1 (2014). https://doi.org/10.1186/1029-242X-2014-1

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• DOI: https://doi.org/10.1186/1029-242X-2014-1

### Keywords

• Formal Power Series
• Euler Number
• Euler Polynomial
• Vector Space Isomorphism
• Interesting Identity 