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\begin{center}
\vskip 1cm{\LARGE\bf 
A New Approach to Multivariate $q$-Euler  \\
\vskip .1in
Polynomials Using the Umbral Calculus 
}
\vskip 1cm
\normalsize
Serkan Araci \\
Atat\"{u}rk Street \\
31290 Hatay \\
Turkey\\
\href{mailto:mtsrkn@hotmail.com}{\tt mtsrkn@hotmail.com} \\

\ \\

Xiangxing Kong\\
Department of Mathematics and Statistics\\
Central South University\\
Changsha 410075\\
China \\
\href{mailto:xiangxingkong@gmail.com}{\tt xiangxingkong@gmail.com}\\

\ \\

Mehmet Acikgoz\\
Department of Mathematics \\
University of Gaziantep\\
27310 Gaziantep \\
Turkey \\
\href{mailto:acikgoz@gantep.edu.tr}{\tt acikgoz@gantep.edu.tr}\\

\ \\

Erdo\u{g}an \c{S}en\\
Department of Mathematics \\
Namik Kemal University \\
59030 Tekirda\u{g}\\
Turkey \\
\href{mailto:erdogan.math@gmail.com}{\tt erdogan.math@gmail.com}\\
\end{center}

\vskip .2 in

\begin{abstract}
We derive numerous identities for multivariate $q$-Euler
polynomials by using the umbral calculus.
\end{abstract}

\section{Preliminaries}

Throughout this paper, we use the following notation, where $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ denotes the set of complex numbers, $\mathcal{F}$ denotes the set of all
formal power series in the variable $t$ over $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $\mathcal{F}=\left\{ f(t) =\sum_{k=0}^{\infty }a_{k}\frac{%
t^{k}}{k!}\mid a_{k}\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\right\} $, $\mathcal{P}=%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}[x] $ and $\mathcal{P}^{\ast}$ denotes the vector space of all
linear functional on $\mathcal{P}$, $\left\langle L\mid 
p(x)
\right\rangle$ denotes the action of the linear functional $L$ on the
polynomial $p(x)$, and it is well-known that the vector space
operation on $\mathcal{P}^{\ast}$ is defined by 
\begin{eqnarray*}
\left\langle L+M\mid p(x) \right\rangle &amp;=&amp;\left\langle L\mid
p(x) \right\rangle +\left\langle M\mid p(x)
\right\rangle , \\
\left\langle cL\mid p(x) \right\rangle &amp;=&amp;c\left\langle L\mid
p(x) \right\rangle ,
\end{eqnarray*}%
where $c$ is some constant in $\mathbb{C}$ (for details, see \cite{Kim2,Kim11,Dere,Roman}).

The formal power series are known by the rule%
\begin{equation*}
f(t) =\sum_{k=0}^{\infty }a_{k}\frac{t^{k}}{k!}\in \mathcal{F}
\end{equation*}%
which defines a linear functional on $\mathcal{P}$ as $\left\langle 
f(t) \mid x^{n}\right\rangle =a_{n}$ for all $n\geq 0$ (for details, see \cite{Kim2,Kim11,Dere,Roman}]). 
Additionally,%
\begin{equation}
\left\langle t^{k}\mid x^{n}\right\rangle =n!\delta_{n,k},
\label{euation2}
\end{equation}%
where $\delta_{n,k}$ is the Kronecker symbol. When we take $f_{L}(t) 
=\sum_{k=0}^{\infty }\left\langle L\mid x^{k}\right\rangle \frac{%
t^{k}}{k!},$ then we obtain $\left\langle f_{L}(t) \mid
x^{n}\right\rangle =\left\langle L\mid x^{n}\right\rangle $ and so as linear
functionals $L=f_{L}(t)$ (see \cite{Kim2,Kim11,Dere,Roman}). Additionally, the
map $L\rightarrow f_{L}(t)$ is a vector space isomorphism from $%
\mathcal{P}^{\ast}$ onto $\mathcal{F}$. Henceforth, $\mathcal{F}$ 
denotes both the algebra of the formal power series in $t$ and the vector
space of all linear functionals on $\mathcal{P}$, and so an element 
$f(t)$ of $\mathcal{F}$ can be thought of as both a formal power series
and a linear functional.  The algebra $\mathcal{F}$ is called the
umbral algebra (see \cite{Kim2,Kim11,Dere,Roman}).

Also, the evaluation functional for $y$ in $\mathbb{C}$ is defined 
to be power series $e^{yt}$. We can write that $\left\langle
e^{yt}\mid x^{n}\right\rangle =y^{n}$ and so $\left\langle e^{yt}\mid
p(x) \right\rangle =p(y)$ (see \cite{Kim2,Kim11,Dere,Roman}). We
note that for all $f(t)$ in $\mathcal{F}$%
\begin{equation}
f(t) =\sum_{k=0}^{\infty }\left\langle f(t) \mid
x^{k}\right\rangle \frac{t^{k}}{k!}  \label{euation3}
\end{equation}%
and for all polynomials $p(x)$, 
\begin{equation}
p(x) =\sum_{k=0}^{\infty }\left\langle t^{k}\mid p(x)
\right\rangle \frac{x^{k}}{k!},  \label{euation4}
\end{equation}%
(for details, see \cite{Kim2,Kim11,Dere,Roman}). The order $o(f(t))$ 
of the power series $f(t) \neq 0$ is the smallest
integer $k$ for which $a_{k}$ does not vanish. It is considered 
$o(f(t)) =\infty$ if $f(t) =0$. We see that $%
o(f(t) g(t)) =o(f(t)) + o(g(t)) $
and $o(f(t) +g(t)) \geq
\min \left\{ o(f(t)),o(g(t)) \right\}$. The series $f(t)$
has a multiplicative inverse,
denoted by $f(t)^{-1}$ or $\frac{1}{f(t)}$,
if and only if $o(f(t)) =0$.
Such series is called an invertible series.
A series $f(t)$ for
which $o(f(t)) =1$ is called a delta series (see
\cite{Kim2,Kim11,Dere,Roman}). For $f(t) ,g(t) \in \mathcal{F}$,
we have $\left\langle f(t) g(t) \mid p(x) \right\rangle =\left\langle f(t) \mid g(t)
p(x) \right\rangle $.

A delta series $f(t)$ has a compositional inverse $\overline{f}(t) $ 
such that $f\left( \overline{f}(t)\right) =%
\overline{f}\left(f(t)\right) =t$.

For $f(t), g(t) \in \mathcal{F}$ , we have $%
\left\langle f(t) g(t) \mid p(x)
\right\rangle =\left\langle f(t) \mid g(t) p(x) \right\rangle $. By (\ref{euation4}), we have%
\begin{equation}
p^{\left( k\right) }\left( x\right) =\frac{d^{k}p(x) }{dx^{k}}%
=\sum_{l=k}^{\infty }\frac{\left\langle t^{l}\mid p(x)
\right\rangle }{l!}l\left( l-1\right) \cdots \left(l-k+1\right) x^{l-k}%
\text{.}  \label{euation5}
\end{equation}

Thus, we see that%
\begin{equation}
p^{\left( k\right) }\left( 0\right) =\left\langle t^{k}\mid p(x)
\right\rangle =\left\langle 1\mid p^{\left( k\right) }\left( x\right)
\right\rangle \text{.}  \label{euation6}
\end{equation}

By (\ref{euation5}), we get%
\begin{equation}
t^{k}p(x) =p^{\left( k\right) }\left( x\right) =\frac{%
d^{k}p(x) }{dx^{k}}\text{.}  \label{euation7}
\end{equation}

So, we have%
\begin{equation}
e^{yt}p(x) =p\left( x+y\right) \text{.}  \label{euation8}
\end{equation}

Let $f(t)$ be a delta series and let $g(t) $ be an
invertible series. Then there exists a unique sequence $S_{n}(x) 
$ of polynomials, with $\deg S_{n}(x) =n$, such that $%
\left\langle g(t) f(t) ^{k}\mid S_{n}(x)
\right\rangle =n!\delta _{n,k}$ for all $n,k\geq 0$. The sequence $%
S_{n}(x)$ is called the Sheffer sequence for $\left(g(t), f(t)\right) $ 
or that $S_{n}(t)$ is
Sheffer for $\left(g(t), f(t)\right)$.

The Sheffer sequence for $\left(1,f(t)\right)$ is called the
associated sequence for $f(t)$; we also say
$S_{n}(x)$ is
associated with $f(t)$. The Sheffer sequence for 
$\left(g(t), t\right)$ is called the Appell sequence for $g(t)$; 
we also say
$S_{n}(x) $ is Appell for $g(t)$.

Let $p(x) \in \mathcal{P}$. Then we have%
\begin{eqnarray}
\left\langle \frac{e^{yt}-1}{t}\mid p(x) \right\rangle
&amp;=&amp;\int_{0}^{y}p(u) du,  \notag \\
\left\langle f(t) \mid xp(x) \right\rangle
&amp;=&amp;\left\langle \partial _{t}f(t) \mid p(x)
\right\rangle =\left\langle f{\acute{}}(t) \mid p(x)\right\rangle ,  \label{euation9} \\
\left\langle e^{yt}-1\mid p(x) \right\rangle &amp;=&amp;p(y)
-p(0) ,\text{(see \cite{Kim2,Kim11,Dere,Roman}).}  \notag
\end{eqnarray}

Let $S_{n}\left( x\right) $ be Sheffer for $\left( g\left( t\right) ,f\left(
t\right) \right) $. Then the following results are known in \cite{Roman}:%
\begin{eqnarray}
h\left( t\right) &amp;=&amp;\sum_{k=0}^{\infty }\frac{\left\langle h\left( t\right)
\mid S_{k}\left( x\right) \right\rangle }{k!}g\left( t\right) f\left(
t\right) ^{k},\text{ }h\left( t\right) \in \mathcal{F}  \notag \\
p(x) &amp;=&amp;\sum_{k=0}^{\infty }\frac{\left\langle g\left( t\right)
f(t) ^{k}\mid p(x) \right\rangle }{k!}S_{k}\left(
x\right) ,\text{ }p(x) \in \mathcal{P},  \notag \\
\frac{1}{g\left( \overline{f}\left( t\right) \right) }e^{y\overline{f}\left(
t\right) } &amp;=&amp;\sum_{k=0}^{\infty }S_{k}\left( y\right) \frac{t^{k}}{k!},%
\text{ for all }y\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,  \label{euation10} \\
f(t) S_{n}\left( x\right) &amp;=&amp;nS_{n-1}(x) \text{.} 
\notag
\end{eqnarray}

Let $a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{r}$ be positive integers. Kim and
Rim \cite{Kim1} defined the generating function for multivariate $q$-Euler
polynomials as follows:%
\begin{gather}
F_{q}(t,x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r})
=\sum_{n=0}^{\infty }E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \frac{t^{n}}{n!}  \label{euation11} \\
=\frac{2^{r}}{\left( q^{b_{1}}e^{a_{1}t}+1\right) \cdots \left(
q^{b_{r}}e^{a_{r}t}+1\right) }e^{xt}\text{.}  \notag
\end{gather}

Note that%
\begin{equation*}
E_{0,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =\frac{%
2^{r}}{\left[ 2\right] _{q^{b_{1}}}\left[ 2\right] _{q^{b_{2}}}\cdots \left[
2\right] _{q^{b_{r}}}},
\end{equation*}%
where $\left[ x\right] _{q}$ is $q$-extension of $x$ defined by%
\begin{equation*}
\left[ x\right] _{q}=\frac{q^{x}-1}{q-1}=1+q+q^{2}+\cdots +q^{x-1}.
\end{equation*}

We assume that $q\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $\left\vert q\right\vert &lt;1$. Also, we note that $%
\lim_{q\rightarrow 1}\left[ x\right] _{q}=x$ (see \cite{Kim1}--\cite{Roman}). In the special
case, $x=0$, $E_{n,q}\left( 0\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) :=E_{n,q}\left( a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) 
$ are called multivariate $q$-Euler numbers. By (\ref{euation11}), we
obtain the following:%
\begin{equation}
E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
=\sum_{k=0}^{n}\binom{n}{k}x^{k}E_{n-k,q}\left( a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \text{.}  \label{euation12}
\end{equation}

Kim and Kim \cite{Kim2} studied some interesting identities for
Frobenius-Euler polynomials arising from umbral calculus. They derived not
only new but also fascinating identities in modern classical umbral calculus.

By the same motivation, we also get numerous identities for multivariate $q$%
-Euler polynomials by utilizing from the umbral calculus.

\section{On the multivariate $q$-Euler polynomials arising from umbral
calculus}

Assume that $S_{n}\left( x\right) $ is an Appell sequence for $g\left(
t\right) $. By (\ref{euation10}), we have 
\begin{equation}
\frac{1}{g(t)}x^{n}=S_{n}(x) \text{ if and only if }%
x^{n}=g(t) S_{n}(x) \text{, }\left( n\geq 0\right) .
\label{euation13}
\end{equation}

Let us take 
\begin{equation*}
g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =\frac{\left(
q^{b_{1}}e^{a_{1}t}+1\right) \cdots \left( q^{b_{r}}e^{a_{r}t}+1\right) }{%
2^{r}}\in \mathcal{F}\text{.}
\end{equation*}

Then we readily see that $g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) $ is an invertible series. By (\ref{euation13}), we have 
\begin{equation}
\sum_{n=0}^{\infty }E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \frac{t^{n}}{n!}=\frac{1}{g\left( t\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) }e^{xt}\text{.}  \label{euation14}
\end{equation}

By (\ref{euation14}), we obtain the following%
\begin{equation}
\frac{1}{g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) }%
x^{n}=E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) 
\text{.}  \label{euation15}
\end{equation}

Also, by (\ref{euation7}), we have%
\begin{gather}
tE_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =E%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
\label{euation16} \\
=nE_{n-1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) \text{%
.}  \notag
\end{gather}

By (\ref{euation15}) and (\ref{euation16}), we have the following
proposition.

\begin{proposition}
\label{Proposition1}\textrm{For }$n\geq 0$\textrm{, }$E_{n,q}\left( x\mid
a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) $\textrm{\ is an Appell
sequence for} 
\begin{equation*}
g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =\frac{\left(
q^{b_{1}}e^{a_{1}t}+1\right) \cdots \left( q^{b_{r}}e^{a_{r}t}+1\right) }{%
2^{r}}\text{.}
\end{equation*}
\end{proposition}

By (\ref{euation11}), we see that%
\begin{align}
\sum_{n=1}^{\infty }E_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \frac{t^{n}}{n!}&amp; =\frac{xge^{xt}-g%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
e^{xt}}{g^{2}}  \label{euation17} \\
&amp; =\sum_{n=0}^{\infty }\left( x\frac{1}{g}x^{n}-\frac{g%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
}{g}\frac{1}{g}x^{n}\right) \frac{t^{n}}{n!}  \notag
\end{align}%
where we used $g:=g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) $. Because of (\ref{euation15}) and (\ref{euation17}), we
discover the following:%
\begin{gather}
E_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
\label{euation18} \\
=xE_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) -\frac{g%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
}{g}E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) \text{%
.}  \notag
\end{gather}

Therefore, we deduce the following theorem.

\begin{theorem}
\label{Theorem1}\textrm{Let }$g:=g\left( t\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) =\frac{\left( q^{b_{1}}e^{a_{1}t}+1\right)
\cdots \left( q^{b_{r}}e^{a_{r}t}+1\right) }{2^{r}}\in \mathcal{F}$\textrm{.
Then we have for }$n\geq 0:$%
\begin{equation}
E_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =\left(
x-\frac{g%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
}{g}\right) E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \text{.}  \label{euation19}
\end{equation}
\end{theorem}

From (\ref{euation11}), we derive that%
\begin{gather}
\sum_{n=0}^{\infty }\left( q^{b_{r}}E_{n,q}\left( x+a_{r}\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) +E_{n,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \right) \frac{t^{n}}{n!}
\label{euation20} \\
=2\sum_{n=0}^{\infty }E_{n,q}\left( x\mid a_{1},\ldots ,a_{r-1};b_{1},\ldots
,b_{r-1}\right) \frac{t^{n}}{n!}\text{.}  \notag
\end{gather}

By comparing the coefficients in the both sides of $\frac{t^{n}}{n!}$ on the
above, we obtain the following%
\begin{gather}
2E_{n,q}\left( x\mid a_{1},\ldots ,a_{r-1};b_{1},\ldots ,b_{r-1}\right)
=q^{b_{r}}E_{n,q}\left( x+a_{r}\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right)  \label{euation21} \\
+E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) \text{.}
\notag
\end{gather}

From Theorem \ref{Theorem1}, we get the following equation%
\begin{gather}
gE_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
\label{euation22} \\
=gxE_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) -g%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) .  \notag
\end{gather}

By using (\ref{euation21}) and (\ref{euation22}), we arrive at the desired
theorem.

\begin{theorem}
\textrm{For }$n\geq 0$\textrm{, we have}%
\begin{gather}
2E_{n,q}\left( x\mid a_{1},\ldots ,a_{r-1};b_{1},\ldots ,b_{r-1}\right)
=q^{b_{r}}E_{n,q}\left( x+a_{r}\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right)  \label{euation23} \\
+E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) \text{.}
\notag
\end{gather}
\end{theorem}

Now, we consider that%
\begin{eqnarray*}
&amp;&amp;\int_{x}^{x+y}E_{n,q}\left( u\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) du \\
&amp;=&amp;\frac{1}{n+1}\left( E_{n+1,q}\left( x+y\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) -E_{n+1,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \right) \\
&amp;=&amp;\frac{1}{n+1}\sum_{j=1}^{\infty }\binom{n+1}{j}E_{n+1-j,q}\left( x\mid
a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) y^{j} \\
&amp;=&amp;\sum_{j=1}^{\infty }\frac{n\left( n-1\right) \left( n-2\right) \cdots
\left( n-j+2\right)}{j!}E_{n+1-j,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) y^{j} \\
&amp;=&amp;\frac{1}{t}\left( \sum_{j=0}^{\infty }\frac{y^{j}t^{j}}{j!}-1\right)
E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) \\
&amp;=&amp;\frac{e^{yt}-1}{t}E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \text{.}
\end{eqnarray*}

Therefore, we discover the following theorem:

\begin{theorem}
\textrm{For }$n\geq 0$\textrm{, we have}%
\begin{equation}
\int_{x}^{x+y}E_{n,q}\left( u\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) du=\frac{e^{yt}-1}{t}E_{n,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \text{.}  \label{euation24}
\end{equation}
\end{theorem}

By (\ref{euation16}) and Proposition \ref{Proposition1}, we have%
\begin{equation}
t\left\{ \frac{1}{n+1}E_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \right\} =E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \text{.}  \label{euation25}
\end{equation}

Thanks to (\ref{euation25}), we readily derive the following:%
\begin{eqnarray}
&amp;&amp;\left\langle e^{yt}-1\mid \frac{E_{n+1,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) }{n+1}\right\rangle  \label{euation26} \\
&amp;=&amp;\left\langle \frac{e^{yt}-1}{t}\mid t\left\{ \frac{E_{n+1,q}\left( x\mid
a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)}{n+1}\right\} \right\rangle 
\notag \\
&amp;=&amp;\left\langle \frac{e^{yt}-1}{t}\mid E_{n,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \right\rangle .  \notag
\end{eqnarray}

On account of (\ref{euation9}) and (\ref{euation25}), we get%
\begin{gather*}
\left\langle \frac{e^{yt}-1}{t}\mid E_{n,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \right\rangle =\left\langle e^{yt}-1\mid 
\frac{E_{n+1,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) }{%
n+1}\right\rangle \\
=\frac{1}{n+1}\left\{ E_{n+1,q}\left( y\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) -E_{n+1,q}\left( a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \right\} \\
=\int_{0}^{y}E_{n,q}\left( u\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) du.
\end{gather*}

Consequently, we obtain the following theorem.

\begin{theorem}
\textrm{For }$n\geq 0$\textrm{, we have}%
\begin{equation}
\left\langle \frac{e^{yt}-1}{t}\mid E_{n,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \right\rangle =\int_{0}^{y}E_{n,q}\left(
u\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) du\text{.}
\label{euation27}
\end{equation}
\end{theorem}

Assume that 
\begin{equation*}
\mathcal{P}\left( q\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
=\left\{ p(x) \in Q\left( q\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \left[ x\right] \mid \deg p(x) \leq n\right\}
\end{equation*}%
is a vector space over $Q\left( q\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) $ which are the space of all polynomials including
coefficients $q,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{r}$.

For $p(x) \in \mathcal{P}\left( q\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right)$, let us consider 
\begin{equation}
p(x) =\sum_{k=0}^{n}b_{k}E_{k,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \text{.}  \label{euation28}
\end{equation}

By Proposition \ref{Proposition1}, $E_{n,q}\left(u\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) $ is an Appell sequence for 
\begin{equation*}
g:=g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right) =\frac{%
\left( q^{b_{1}}e^{a_{1}t}+1\right) \cdots \left(
q^{b_{r}}e^{a_{r}t}+1\right) }{2^{r}}\text{.}
\end{equation*}

Thus we have%
\begin{equation}
\left\langle g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
t^{k}\mid E_{n,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
\right\rangle =n!\delta _{n,k}\text{.}  \label{euation29}
\end{equation}

From (\ref{euation28}) and (\ref{euation29}), we compute%
\begin{gather}
\left\langle g\left( t\mid a_{1},\ldots ,a_{r};b_{1},\ldots ,b_{r}\right)
t^{k}\mid p(x) \right\rangle =\sum_{l=0}^{n}b_{l}\left\langle
gt^{k}\mid E_{l,q}\left( x\mid a_{1},\ldots ,a_{r};b_{1},\ldots
,b_{r}\right) \right\rangle  \label{euation30} \\
=\sum_{l=0}^{n}b_{l}l!\delta _{l,k}=k!b_{k}\text{.}  \notag
\end{gather}

Thus, by (\ref{euation30}), we derive%
\begin{eqnarray}
b_{k} &amp;=&amp;\frac{1}{k!}\left\langle gt^{k}\mid p(x) \right\rangle
\label{euation31} \\
&amp;=&amp;\frac{1}{2^{r}k!}\left\langle \left( q^{b_{1}}e^{a_{1}t}+1\right) \cdots
\left( q^{b_{r}}e^{a_{r}t}+1\right) \mid p^{\left( k\right) }\left( x\right)
\right\rangle \text{.}  \notag
\end{eqnarray}

It is not difficult to show the following%
\begin{equation}
\left( q^{b_{1}}e^{a_{1}t}+1\right) \cdots \left(
q^{b_{r}}e^{a_{r}t}+1\right) =\sum_{\underset{k_{1}+k_{2}+\textcolor[rgb]{0,0,0}{\ldots} +k_{r}=1}{%
k_{1},\ldots ,k_{r}\geq 0}}q^{\sum_{l=1}^{r}b_{l}k_{l}}e^{t%
\sum_{j=1}^{r}a_{j}k_{j}}\text{.}  \label{euation32}
\end{equation}

Via the results (\ref{euation31}) and (\ref{euation32}), we easily see that%
\begin{eqnarray*}
b_{k} &amp;=&amp;\frac{1}{2^{r}k!}\sum_{\underset{k_{1}+k_{2}+\ldots +k_{r}=1}{%
k_{1},\ldots ,k_{r}\geq 0}}q^{\sum_{l=1}^{r}b_{l}k_{l}}\left\langle
e^{t\sum_{j=1}^{r}a_{j}k_{j}}\mid p^{\left( k\right) }\left( x\right)
\right\rangle \\
&amp;=&amp;\frac{1}{2^{r}k!}\sum_{\underset{k_{1}+k_{2}+\ldots +k_{r}=1}{%
k_{1},\ldots ,k_{r}\geq 0}}q^{\sum_{l=1}^{r}b_{l}k_{l}}p^{\left( k\right)
}\left( \sum_{j=1}^{r}a_{j}k_{j}\right) \text{.}
\end{eqnarray*}

As a result, we state the following theorem.

\begin{theorem}
\textrm{For }$p(x) \in \mathcal{P}\left( q\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) $\textrm{, when we consider }%
\begin{equation*}
p(x) =\sum_{k=0}^{n}b_{k}E_{k,q}\left( x\mid a_{1},\ldots
,a_{r};b_{1},\ldots ,b_{r}\right) \text{,}
\end{equation*}%
\textrm{we obtain}%
\begin{equation*}
b_{k}=\frac{1}{2^{r}k!}\sum_{\underset{k_{1}+k_{2}+\ldots +k_{r}=1}{%
k_{1},\ldots ,k_{r}\geq 0}}q^{\sum_{l=1}^{r}b_{l}k_{l}}p^{\left( k\right)
}\left( \sum_{j=1}^{r}a_{j}k_{j}\right) \text{.}
\end{equation*}
\end{theorem}

\section{Acknowledgements}

The authors would like to thank 
the anonymous referee for several helpful
comments.

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\bigskip \hrule\bigskip

\noindent 2010 \textit{Mathematics Subject Classification}: Primary 11S80; Secondary
11B68.

\noindent \emph{Keywords: } Appell sequence, Sheffer sequence,
multivariate $q$-Euler polynomial, formal power series.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 23 2013;
revised versions received June 5 2013; December 7 2013.
Published in {\it Journal of Integer Sequences}, December 8 2013.

\bigskip
\hrule
\bigskip

\noindent
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