\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Identities Involving Two Kinds of \\ \vskip .1in $q$-Euler Polynomials and Numbers } \vskip 1cm \large Abdelmejid Bayad\\ D\'epartement de Math\'ematiques \\ Universit\'e d'Evry Val d'Essonne\\ B\^atiment I.B.G.B.I., 3\`eme \'etage \\ 23 Bd.\ de France\\ 91037 Evry Cedex \\ France \\ \href{mailto:abayad@maths.univ-evry.fr}{\tt abayad@maths.univ-evry.fr} \\ \ \\ Yoshinori Hamahata\\ Faculty of Engineering Science \\ Kansai University\\ 3-3-35 Yamate-cho, Suita-shi \\ Osaka 564-8680 \\ Japan\\ \href{mailto:hamahata@fc.ritsumei.ac.jp}{\tt hamahata@fc.ritsumei.ac.jp}\\ \end{center} \vskip .2 in \begin{abstract} We introduce two kinds of $q$-Euler polynomials and numbers, and investigate many of their interesting properties. In particular, we establish $q$-symmetry properties of these $q$-Euler polynomials, from which we recover the so-called Kaneko-Momiyama identity for the ordinary Euler polynomials, discovered recently by Wu, Sun, and Pan. Indeed, a $q$-symmetry and $q$-recurrence formulas among sum of product of these $q$-analogues Euler numbers and polynomials are obtained. As an application, from these $q$-symmetry formulas we deduce non-linear recurrence formulas for the product of the ordinary Euler numbers and polynomials. \end{abstract} \section{Introduction and preliminaries} \subsection{The ordinary Euler numbers and polynomials: An analytic overview} Let $\mathbb{N}=\{0,1,2,\ldots\}$. The ordinary Euler polynomials $E_n(x)$ are defined by the generating series $$ \frac{2e^{xt}}{e^t+1}= \sum_{n=0}^{\infty}E_n(x) \frac{t^n}{n!}. $$ The first few values are \begin{eqnarray*} E_0(x)=1, E_1(x)=x-\frac{1}{2}, E_2(x)=x^2-x, E_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{4}. \end{eqnarray*} The ordinary Euler numbers $E_n$ ($n=0, 1, 2, \ldots$) are defined by the generating function \begin{eqnarray}\label{Euler_numbers} \frac{2}{e^t+1}= \sum_{n=0}^{\infty}E_n \frac{t^n}{n!}. \end{eqnarray} The $n^{\textrm{th}}$-Euler number and polynomial are connected by the equation: $E_n=E_n(0).$ The first few values are $E_0=1$, $E_1=-1/2$, $E_2=0$, $E_3=1/4$, and it holds that $E_{2k}=0$ ($k=1, 2, 3, \ldots$). \begin{remark}\label{schol1} Note that the Euler numbers $E_n$ which we consider in this paper are different from the Euler numbers defined by M. Abramowitz and I. A. Stegun \cite[Ch.23]{[1]}. \end{remark} From the definition we can easily deduce the following well-known difference formula: \begin{eqnarray}\label{differenceformula} (-1)^nE_n(-x)+E_n(x)=2x^n, (n\in\mathbb{N}). \end{eqnarray} In 2004, K.J. Wu, Z.W. Sun, and H. Pan \cite{WSP} proved the following important formulae: \begin{eqnarray} &&(-1)^m\sum_{k=0}^m {m\choose k}E_{n+k}(x)= (-1)^n\sum_{k=0}^n {n\choose k}E_{m+k}(-x),\label{Sun}\\ &&(-1)^m\sum_{k=0}^{m+1} {m+1\choose k}(n+k+1)E_{n+k}(x)\notag\\ &&\quad + (-1)^n\sum_{k=0}^{n+1} {n+1\choose k}(m+k+1)E_{m+k}(x)=0.\label{Kaneko-Momiyama-Sun} \end{eqnarray} The last identity (\ref{Kaneko-Momiyama-Sun}) is an Euler polynomial version of Kaneko-Momiyama relations among Bernoulli numbers. See M. Kaneko \cite{Kaneko}, H. Momiyama \cite{Momiyama}, I. M. Gessel \cite{Gessel2} and Wu-Sun-Pan \cite{WSP} for details. The identity (\ref{Sun}) can be viewed as an integral version of the Kaneko-Momiyama type identity for the Euler polynomials. In this present paper, we introduce and investigate two kinds of $q$-Euler polynomials and numbers. For instance, we find $q$-analogues for the identities (\ref{Sun}) and (\ref{Kaneko-Momiyama-Sun}). On the other hand, we also establish a relation between sums of products of our $q$-Euler polynomials. \subsection{$q$-shifted factorials} Let $a\in\mathbb{C}$. The {\it $q$-shifted factorials} are defined by $$ (a,q)_0=1,\quad (a,q)_n= \prod_{k=0}^{n-1}(1-aq^k)\quad (n=1,2,\ldots ). $$ If $|q|<1$, then we define $$ (a,q)_{\infty}=\lim_{n\to\infty}(a,q)_n =\prod_{k=0}^{\infty}(1-aq^k). $$ We also denote \begin{eqnarray*} \left[x\right]_q &=& \frac{1-q^x}{1-q}, \quad x\in\mathbb{C},\\ \left[n\right]_{q}! &=& \frac{(q,q)_n}{(1-q)^n},\quad n\in\mathbb{N},\\ \begin{bmatrix}n\\ k\end{bmatrix}_q &=& \frac{[n]_q!}{[k]_q![n-k]_q!} ,\quad k,n\in\mathbb{N},\\ \left[\begin{array}{c}n\\ i_1, \ldots ,i_m\end{array} \right]_q &=& \frac{[n]_q!}{[i_1]_q!\cdots [i_m]_q!} ,\quad n, i_1,\ldots , i_m\in \mathbb{N}, \textrm{ with } i_1+\cdots + i_m=n . \end{eqnarray*} \subsection{$q$-Exponential functions} The {\it $q$-exponential functions} are given by \begin{eqnarray}\label{exponential functions} e_q(z):=\sum_{n=0}^{\infty}\frac{z^n}{[n]_q!},\textrm{ and }\quad e_{q^{-1}}(z):=\sum_{n=0}^{\infty} \frac{z^n}{[n]_{q^{-1}}!}. \end{eqnarray} It is easy to see that $[n]_{q^{-1}}!= q^{-{n\choose 2}}[n]_q!$. Hence $$ e_{q^{-1}}(z)=\sum_{n=0}^{\infty} \frac{q^{n\choose 2}z^n}{[n]_{q}!}. $$ Recently both $q$-exponential functions are intensively studied in $q$-calulus and and quatum theory. See I. M. Gessel \cite{Gessel1}, W. P. Johnson \cite{Johnson} for related topics. As is well-known, these functions are related to the infinite product $ \left(z,q\right)_{\infty}$ by $$ e_q(z)=\frac{1}{\left( (1-q)z,q\right)_{\infty}}, \qquad e_{q^{-1}}(z)=\left( -(1-q)z,q\right)_{\infty}. $$ This yields $ e_q(z)e_{q^{-1}}(-z)=1. $ \subsection{$q$-Euler polynomials and numbers} \begin{definition} We define two kinds of $q$-Euler polynomials $E_n(x,q)$ and $F_n(x,q^{-1})$ ($n=0, 1, 2, \ldots$) by \begin{eqnarray*} \frac{2e_q(xt)}{e_q(t)+1} &=& \sum_{n=0}^{\infty}E_n(x,q)\frac{t^n}{[n]_q!},\\ \frac{2e_q(xt)}{e_{q^{-1}}(t)+1} &=& \sum_{n=0}^{\infty}F_n(x,q^{-1})\frac{t^n}{[n]_{q}!}. \end{eqnarray*} We call $E_n(x,q)$ (resp. $F_n(x,q^{-1})$) the {\it first} (resp. {\it second}) {\it $q$-Euler polynomials}. In particular, we call $E_n(0,q)$ (resp. $F_n(0,q^{-1})$) the {\it first} (resp. {\it second}) {\it $q$-Euler numbers}. \end{definition} \begin{example} \begin{eqnarray*} E_0(x,q)&=&1, E_1(x,q)=x-\frac{1}{2},\\ E_2(x,q)&=&x^2-\frac{[2]_q}{2}x-\frac{1}{2} +\frac{[2]_q}{4},\\ E_3(x,q)&=&x^3-\frac{[3]_q}{2}x^2 +\left(\frac{[3]_q[2]_q}{4}-\frac{[3]_q}{2}\right) x-\frac{1}{2}-\frac{[3]_q[2]_q}{8}+\frac{[3]_q}{2}. \end{eqnarray*} \begin{eqnarray*} F_0(x,q^{-1})&=&1, F_1(x,q^{-1})=x-\frac{1}{2},\\ F_2(x,q^{-1})&=&x^2-\frac{[2]_q}{2}x-\frac{q}{2} +\frac{[2]_q}{4},\\ F_3(x,q^{-1})&=&x^3-\frac{[3]_q}{2}x^2 +\left(\frac{[3]_q[2]_q}{4}-\frac{[3]_q}{2}q\right) x-\frac{q^3}{2}-\frac{[3]_q[2]_q}{8}+\frac{[3]_q}{2}q. \end{eqnarray*} \end{example} \begin{remark}\ \begin{enumerate} \item The reason for introducing both kinds of $q$-analogue Bernoulli polynomials $E_n(x,q)$ and $F_n(x,q^{-1})$ is that they are needed in the $q$-analogues of symmetry, difference, recurrence and complementary argument formulas. \item The case $q=1$ corresponds to the ordinary Euler polynomials and numbers. \item In the literature there are many $q$-analogues of the Euler numbers and polynomials. The $q$-analogues which we consider here are closely related to $q$-calculus, $q$-Jackson integral and hypergeometric series. \item Various $q$-analoques of the Euler numbers and polynomials are studied by many mathematicians. For more details for example you can refer to T. Kim \cite{Kim1},C. S. Ryoo \cite{Ryoo}, Y. Simsek \cite{Simsek} and others. \item It seems to be difficult to clarify the connections between all the $q$-analogues of the Euler numbers and polynomials. \end{enumerate} \end{remark} \section{$q$-Recurrence, $q$-Addition, $q$-Derivative and $q$-integral formulae} In this section, we establish a series of formulas involving the polynomials $E_n(x,q)$ and $F_n(x,q^{-1})$, like $q$-addition, $q$-derivative, $q$-integral and $q$-recurrence formulas. \subsection{$q$-Recurrence formulae} \begin{proposition}[$q$-Recurrence formula] For any $n\geq 1$, we have \begin{eqnarray*} E_n(x,q)&=&x^n-\frac{1}{2} \sum_{i=0}^{n-1}\begin{bmatrix}n\\ i\end{bmatrix}_q E_i(x,q),\\ F_n(x,q^{-1})&=&x^n-\frac{1}{2} \sum_{i=0}^{n-1}\begin{bmatrix}n\\ i\end{bmatrix}_q F_i(x,q^{-1})q^{\binom{n-i}{2}}. \end{eqnarray*} \end{proposition} \begin{proof} As for the first identity, we make use of $$ \left(\sum_{n=0}^{\infty}E_n(x,q)\frac{t^n}{[n]_q!} \right)(e_q(t)+1)=2e_q(xt). $$ We deduce from this identity $$ \sum_{n=0}^{\infty}\left( \sum_{i=0}^n\begin{bmatrix}n\\ i\end{bmatrix}_q E_i(x,q)+E_n(x,q)\right) \frac{t^n}{[n]_q!}= \sum_{n=0}2x^n \frac{t^n}{[n]_q!}, $$ which yields the result. We get the second result in the similar way. \end{proof} \par As $q\to 1$, one has a recurrence formula for the ordinary Euler polynomials: \begin{eqnarray*} E_n(x)=x^n-\frac{1}{2} \sum_{i=0}^{n-1}\binom{n}{i} E_i(x)\qquad (n\geq 1), \end{eqnarray*} then for the Euler numbers $E_{2n+1}$ we recover the well-known recurrence formula \begin{eqnarray}\label{Euler_recurrence_1} E_{2n+1}(x)=-\frac{1}{2} \sum_{k=0}^{2n}\binom{2n+1}{k} E_k\qquad (n\geq 0). \end{eqnarray} \subsection{$q$-Derivative and $q$-integral} The {\it $q$-derivative} of a function $f$ is given by $$ D_qf(x):= \frac{f(x)-f(qx)}{(1-q)x}\qquad (x\ne 0, q\ne 1), $$ where $x$ and $qx$ should be in the domain of $f$. If $f$ is differentiable on an open set $I$, then for all $x\in I$, $$ \lim_{q\to 1}D_qf(x)=f'(x). $$ Besides, for all $n\in\mathbb{N}$, \begin{eqnarray*} D_q(x^n)&=& [n]_qx^{n-1}, D_q(x,q)_n= -[n]_q(xq,q)_{n-1},\\ D_{q^{-1}}(x,q)_n&=& -[n]_q(x,q)_{n-1}, D_q\left( \frac{x^n}{[n]_q!} \right) = \frac{x^{n-1}}{[n-1]_q!}. \end{eqnarray*} From the last identity, for instance, we have $D_qe_q(x)=e_q(x)$.\\ Our $q$-Euler polynomials form ``$q$-Appell sequences'': \begin{proposition}[$q$-Derivative formula] For any $n\geq 0$, we have \begin{eqnarray*} D_qE_{n+1}(x,q)&=&[n+1]_qE_n(x,q),\\ D_qF_{n+1}(x,q^{-1})&=&[n+1]_qF_n(x,q^{-1}). \end{eqnarray*} \end{proposition} \begin{proof} Since $$ \sum_{n=0}^{\infty}D_qE_n(x,q) \frac{t^n}{[n]_q!}= \frac{2te_q(xt)}{e_q(t)+1} =\sum_{n=1}^{\infty}[n]_qE_{n-1}(x,q) \frac{t^n}{[n]_q!}, $$ we have the first identity. The second identity can be obtained similarly. \end{proof} \par As $q\to 1$, we have the identities of Appell sequences of the ordinary Euler polynomials: \begin{eqnarray*} \frac{d}{dx}E_{n+1}(x)=(n+1)E_n(x). \end{eqnarray*} \par For the product of two functions $f$ and $g$, the following formula holds: \begin{eqnarray*} D_q(f\cdot g)(x)&=& g(x)D_q(x)+f(qx)D_qg(x)\\ &=& f(x)D_qg(x)+g(qx)D_qf(x). \end{eqnarray*} We next treat the composition of $f(x)$ and $g(x)$. When $g(x)=-x$, the following chain rule for the $q$-derivative is valid: $$ D_q(f\circ g)(x)=D_qf(g(x))D_qg(x), $$ which will be used in the proofs of Theorems \ref{3.5} and \ref{3.9}. However, in general, the rule above does not hold. If we modify the definition of the composition of two functions, then a new chain rule for the $q$-derivative is gained. We refer to I. M. Gessel \cite{Gessel1} for this topic. The $q$-{\it Jackson integral} from $0$ to $a$ is defined by \begin{eqnarray*} \int_0^af(x)d_qx &:=& (1-q)a\sum_{n=0}^{\infty}f(aq^n)q^n \end{eqnarray*} provided the infinite sums converge absolutely. The $q$-Jackson integral in the generic interval $[a,b]$ is given by $$ \int_a^bf(x)d_qx= \int_0^bf(x)d_qx- \int_0^af(x)d_qx. $$ For any function $f$ we have $$ D_q\int_0^xf(t)d_qt= f(x). $$ \begin{proposition}[$q$-Integral formula] For any $n\geq 0$, \begin{eqnarray*} \int_a^xE_n(t,q)d_qt&=& \frac{E_{n+1}(x,q)-E_{n+1}(a,q)} {[n+1]_q},\\ \int_a^xF_n(t,q^{-1})d_qt&=& \frac{F_{n+1}(x,q^{-1})-F_{n+1}(a,q^{-1})} {[n+1]_q}. \end{eqnarray*} \end{proposition} This result follows from $q$-derivative formula. As $q\to 1$, we have integral formula for the classical Euler polynomials: \begin{equation*} \int_a^xE_n(t)dt= \frac{E_{n+1}(x)-E_{n+1}(a)} {n+1}. \end{equation*} \subsection{$q$-Binomial formula} Let $q\in\mathbb{C}$, and take two $q$-commuting variables $x$ and $y$ which satisfy the relation $$ xy=q^{-1}yx. $$ Let $\mathbb{C}_q[x,y]$ be the complex associative algebra with $1$ generated by $x$ and $y$. Then the following identity is valid in the algebra $\mathbb{C}_q[x,y]$: $$ (x+y)^n= \sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q x^ky^{n-k},\qquad n\in\mathbb{N}, $$ or alternatively, $$ (x+y)^n= \sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}} y^kx^{n-k},\qquad n\in\mathbb{N}. $$ For details, we refer to Andrews-Askey-Roy \cite{AAR}, Gasper-Rahman \cite{Gasper-Rahman}. \subsection{$q$-Exponential identity} Let $x, y$ be the $q$-commuting variables satisfying the relation $xy=q^{-1}yx$. Let $\mathbb{C}_q[[x,y]]$ be the complex associative algebra with $1$ of formal power series $$ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} a_{m,n}x^my^n $$ with arbitrary complex coefficients $a_{m,n}$. One knows in Andrews-Askey-Roy \cite{AAR}, Gasper-Rahman \cite{Gasper-Rahman} that in $\mathbb{C}_q[[x,y]]$, we have the following identity $$ e_q(x+y)=e_q(x)e_q(y). $$ \begin{proposition}[$q$-Addition formula] Let $x, y$ be the $q$-commuting variables satisfying the relation $xy=q^{-1}yx$. For any $n\geq 0$, we have \begin{eqnarray*} E_n(x+y,q)&=& \sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q E_k(x,q)y^{n-k},\\ F_n(x+y,q^{-1})&=& \sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q F_k(x,q^{-1})y^{n-k}. \end{eqnarray*} Particularly, it follows that \begin{eqnarray*} E_n(y,q)&=&\sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q E_k(0,q)y^{n-k},\\ F_n(y,q^{-1})&=&\sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q F_k(0,q^{-1})y^{n-k}. \end{eqnarray*} \end{proposition} \begin{proof} The first identity follows from $$ \frac{2e_q((x+y)t)}{e_q(t)+1}= \frac{2e_q(xt)}{e_q(t)+1}\cdot e_q(yt). $$ One can easily prove the remaining identities. \end{proof} \par As $q\to 1$, we have the classical formula: \begin{equation*} E_{n+1}(x+y)= \sum_{k=0}^n \binom{n}{k} E_k(x)y^{n-k}. \end{equation*} Particularly, it holds that \begin{equation*} E_n(y)=\sum_{k=0}^n \binom{n}{k} E_ky^{n-k}. \end{equation*} At the end of this section, we give a list of limit of $q$-analogues. \begin{eqnarray*} \lim_{q\to 1}e_q(z) &=& \lim_{q\to 1}e_{q^{-1}}(z) =e^z,\\ \lim_{q\to 1}[n]_q &=& n,\\ \lim_{q\to 1}[n]_q! &=& n!,\\ \lim_{q\to 1} \begin{bmatrix}n\\ k\end{bmatrix}_q &=& {n\choose k},\\ \lim_{q\to 1}\left[ \begin{array}{c}n\\ i_1,\ldots ,i_m\end{array} \right]_q &=& \left( \begin{array}{c}n\\ i_1,\ldots ,i_m\end{array} \right) := \frac{n!}{i_1!\cdots i_m!},\\ \lim_{q\to 1}E_n(x,q) &=& \lim_{q\to 1}F_n(x,q^{-1})= E_n(x). \end{eqnarray*} \section{$q$-symmetry and $q$-analogues to Kaneko-Momiyama identities} \subsection{$q$-symmetry fo Sums of products} \begin{theorem}[Sums of products]\label{3.1} Let $m$ be a given positive integer. Then for any $n\geq 0$, we have \begin{multline*} (-1)^n \sum_{i_1+\cdots +i_m=n} \left[\begin{array}{c}n\\ i_1,\ldots ,i_m\end{array}\right]_q F_{i_1}(-x,q^{-1})\cdots F_{i_m}(-x,q^{-1})\\ =\sum_{j=0}^m(-1)^j2^{m-j} \binom{m}{j} \sum_{k_1+\cdots k_m=n} \left[\begin{array}{c}n\\ k_1,\ldots ,k_m\end{array}\right]_q E_{k_1}(x,q)\cdots E_{k_j}(x,q) x^{n-(k_1+\cdots +k_j)}. \end{multline*} \end{theorem} \begin{remark}\label{3.1bis} The above theorem implies the following results. \begin{enumerate} \item If $m=1$, then \begin{eqnarray}\label{E-F} (-1)^nF_n(-x,q^{-1})+E_n(x,q)=2x^n. \end{eqnarray} This relation can be viewed as a $q$-difference formula. If $q\to 1$ we recover the usual difference formula for the Euler polynomials (\ref{differenceformula}). \item If $m=2$, then \begin{eqnarray*} &&(-1)^n\sum_{i=0}^n \begin{bmatrix}n\\ i\end{bmatrix}_q F_i(-x,q^{-1})F_{n-i}(-x,q^{-1})\\ &&\hspace{2cm}=\sum_{i=0}^n \begin{bmatrix}n\\ i\end{bmatrix}_q E_i(x,q)E_{n-i}(x,q) -4\sum_{i=0}^n \begin{bmatrix}n\\ i\end{bmatrix}_q E_i(x,q)x^{n-i} +4x^n\sum_{i=0}^n \begin{bmatrix}n\\ i\end{bmatrix}_q. \end{eqnarray*} \item It should be noted that Simsek \cite{Simsek} found formulae for sums of products of another kind of $q$-Euler polynomials. \end{enumerate} \end{remark} \begin{proof} In view of $e_q(t)e_{q^{-1}}(-t)=1$, we have $$ \frac{1}{e_{q^{-1}}(-t)+1}=1- \frac{1}{e_q(t)+1}. $$ Hence for $m\geq 1$, $$ \left(\frac{2e_q((-x)(-t))}{e_{q^{-1}}(-t)+1} \right)^m =\left(2e_q(xt)- \frac{2e_q(xt)}{e_q(t)+1}\right)^m. $$ The left-hand side of the identity is $$ \sum_{n=0}^{\infty}(-1)^n \sum_{i_1+\cdots +i_m=n} \left[\begin{array}{c}n\\ i_1,\ldots ,i_m\end{array}\right]_q F_{i_1}(-x,q^{-1})\cdots F_{i_m}(-x,q^{-1}) \frac{t^n}{[n]_q!}. $$ The right-hand side becomes \begin{eqnarray*} &&\sum_{j=0}^m(-1)^j \binom{m}{j} \left(\frac{2e_q(xt)}{e_q(t)+1}\right)^j 2^{m-j}e_q(xt)^{m-j}\\ &&= \sum_{j=0}^m(-1)^j 2^{m-j}\binom{m}{j} \sum_{n=0}^{\infty} \sum_{k_1+\cdots +k_m=n} \left[\begin{array}{c}n\\ k_1,\ldots ,k_m\end{array}\right]_q E_{k_1}(x,q)\cdots E_{k_j}(x,q)x^{k_{j+1}}\cdots x^{k_m} \frac{t^n}{[n]_q!}. \end{eqnarray*} \end{proof} As $q\to 1$ in the formula of Theorem \ref{3.1}, we have \begin{theorem}\label{3.2} Let $m$ be a given positive integer. Then for any $n\geq 0$, \begin{multline*} (-1)^n \sum_{i_1+\cdots +i_m=n} \left(\begin{array}{c}n\\ i_1,\ldots ,i_m\end{array}\right) E_{i_1}(-x)\cdots E_{i_m}(-x)\\ =\sum_{j=0}^m(-1)^j2^{m-j} \binom{m}{j} \sum_{k_1+\cdots k_m=n} \left(\begin{array}{c}n\\ k_1,\ldots ,k_m\end{array}\right) E_{k_1}(x)\cdots E_{k_j}(x) x^{n-(k_1+\cdots +k_j)}. \end{multline*} \end{theorem} \begin{corollary}\label{3.2bis} Especially in the cases $m=1, 2$, the following results hold: \par {\rm (1)} For any $n\geq 0$, we have $(-1)^nE_n(-x)+E_n(x)=2x^n$. \par {\rm (2)} For any $k\geq 1$, $E_{2k}=0$. \par {\rm (3)} For any $n\geq 0$, we get the {\bf non-linear recurrence} formulae $$ (-1)^n\sum_{i=0}^n\binom{n}{i}E_i(-x)E_{n-i}(-x) =\sum_{i=0}^n\binom{n}{i} E_i(x)E_{n-i}(x)-4\sum_{i=0}^n\binom{n}{i} E_i(x)x^{n-i}+2^{n+2}x^n. $$ \par {\rm (4)} If $n$ is an odd positive integer, then we obtain the well-known {\bf Euler non-linear recurrence} formula $$ \sum_{i=0}^n\binom{n}{i}E_iE_{n-i}=2E_n. $$ \end{corollary} \subsection{$q$-Symmetry} \begin{theorem}[$q$-Symmetry 1]\label{3.3} For any $m, n\in\mathbb{N}$, we have \begin{equation}\label{Momiyama} (-1)^m\sum_{k=0}^m \begin{bmatrix}m\\ k\end{bmatrix}_q E_{n+k}(x,q) q^{-kn+mn} = (-1)^n\sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}} F_{m+k}(-x,q^{-1}) q^{\binom{n}{2}-\binom{k}{2}}. \end{equation} \end{theorem} This identity (\ref{Momiyama}) can be viewed as a $q$-analogue to the polynomial version of the integral Kaneko-Momiyama's formulae on Euler polynomials (\ref{Sun}).\\ \begin{proof} Let $x, y$ be two $q$-commuting variables with $xy=q^{-1}yx$. We compute the generating functions \begin{eqnarray*} L(w,x,y) &=& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^m\sum_{k=0}^m \begin{bmatrix}m\\ k\end{bmatrix}_q E_{n+k}(w,q)q^{-kn+mn} \frac{x^m}{[m]_q!}\frac{y^n}{[n]_q!},\\ R(w,x,y) &=& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^n\sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_q F_{m+k}(-w,q^{-1})q^{\binom{n}{2}-\binom{k}{2}} \frac{x^m}{[m]_{q}!}\frac{y^n}{[n]_{q}!}, \end{eqnarray*} where $w$ is a commuting variable with $x$ and $y$. \begin{eqnarray*} L(w,x,y) &=& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^m\sum_{k=0}^m \begin{bmatrix}m\\ k\end{bmatrix}_q E_{n+k}(w,q)q^{-kn} \frac{y^n}{[n]_q!}\frac{x^m}{[m]_q!}\\ &=& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^m\sum_{k=0}^m E_{n+k}(w,q)q^{-kn} \frac{y^n}{[n]_q!} \frac{x^k}{[k]_q!} \frac{x^{m-k}}{[m-k]_q!}\\ &=& \sum_{j=0}^{\infty} \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} E_{n+k}(w,q) \frac{(-x)^k}{[k]_q!} \frac{y^n}{[n]_q!} \frac{(-x)^{j}}{[j]_q!}\\ &=& \left( \sum_{i=0}^{\infty}E_i(x,q) \sum_{k=0}^i \frac{(-x)^k}{[k]_q!} \frac{y^{i-k}}{[i-k]_q!} \right) e_q(-x)\\ &=& \frac{2e_q(w(y-x))}{e_q(y-x)+1} e_q(-x). \end{eqnarray*} \begin{eqnarray*} R(w,x,y)&=& \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} (-1)^n\sum_{k=0}^n F_{m+k}(-w,q^{-1}) \frac{x^m}{[m]_q!}\frac{y^k}{[k]_q!} \frac{q^{\binom{n-k}{2}}y^{n-k}}{[n-k]_q!}\\ &=& \sum_{j=0}^{\infty}\sum_{k=0}^{\infty} \sum_{m=0}^{\infty}(-1)^{j+k} F_{m+k}(-w,q^{-1}) \frac{x^m}{[m]_{q}!} \frac{y^k}{[k]_{q}!} \frac{y^j}{[j]_{q^{-1}}!}\\ &=& \left( \sum_{m=0}^{\infty}\sum_{k=0}^{\infty} F_{m+k}(-w,q^{-1}) \frac{x^m}{[m]_{q}!} \frac{(-y)^k}{[k]_{q}!} \right) e_q(-y)\\ &=& \frac{2e_q(-w(y-x))}{e_{q^{-1}}(x-y)+1} e_{q^{-1}}(-y). \end{eqnarray*} Hence it follows that \begin{eqnarray*} R(w,x,y)e_q(y) &=& \frac{2e_q(w(y-x))}{e_{q^{-1}}(x-y)+1}\\ &=& \frac{2e_q(w(y-x))}{e_q(y-x)+1} e_q(y-x)\\ &=&L(w,x,y)e_q(y), \end{eqnarray*} which provides $R(w,x,y)=L(w,x,y)$. Therefore we can complete the proof. \end{proof} As $q\to 1$ in (\ref{Momiyama}) of Theorem \ref{3.3}, we have a symmetric relation for the ordinary Euler polynomials: \begin{theorem}\label{3.4} For any $m, n\in\mathbb{N}$, we have \begin{equation*} (-1)^m\sum_{k=0}^m \begin{pmatrix}m\\ k\end{pmatrix} E_{n+k}(x)= (-1)^n\sum_{k=0}^n \begin{pmatrix}n\\ k\end{pmatrix} E_{m+k}(-x). \end{equation*} \end{theorem} \begin{theorem}[$q$-Symmetry 2]\label{3.5} For any $m, n\in\mathbb{N}$, we have \begin{multline}\label{(2)} (-1)^m\sum_{k=0}^{m+1} \begin{bmatrix}m+1\\ k\end{bmatrix}_q [n+k+1]_q E_{n+k}(x,q) q^{-k(n+1)-{n\choose 2}+mn+1}\\ + (-1)^n\sum_{k=0}^{n+1} \begin{bmatrix}n+1\\ k\end{bmatrix}_{q^{-1}} [m+k+1]_{q^{-1}} F_{m+k}(-x,q^{-1}) q^{k(m+1)+{m\choose 2}}=0. \end{multline} \end{theorem} \begin{proof} Applying $q$-derivative formula to the identity (\ref{Momiyama}) in Theorem \ref{3.3} replaced $m$, $n$ by $m+1$, $n+1$, respectively, we have the result. \end{proof} As $q\to 1$ in (\ref{(2)}) of Theorem \ref{3.5}, we have another symmetric formula for the ordinary Euler polynomials: \begin{theorem}\label{3.6} For any $m, n\in\mathbb{N}$, we have \begin{equation}\label{(3)} (-1)^m\sum_{k=0}^{m+1} \begin{pmatrix}m+1\\ k\end{pmatrix} (n+k+1)E_{n+k}(x) + (-1)^n\sum_{k=0}^{n+1} \begin{pmatrix}n+1\\ k\end{pmatrix} (m+k+1)E_{m+k}(-x)=0. \end{equation} \end{theorem} This can be regarded as an Euler polynomial version of Kaneko-Momiyama formulae for Bernoulli numbers. To be precise, put $m=n$ and $x=0$ in (\ref{(3)}). Then we have an analogue of Kaneko's formula: \begin{theorem}\label{3.7} For any $n\in\mathbb{N}$, $$ \sum_{k=0}^{n+1}\binom{n+1}{k} (n+k+1)E_{n+k}=0. $$ \end{theorem} Then we obtain the nice formula \begin{eqnarray}\label{Euler_recurrence_2} E_{2n+1}=-\frac1{n+1}\sum_{k=0}^{n} {n+1\choose k}(n+k+1)E_{n+k}. \end{eqnarray} \begin{remark} The formula (\ref{Euler_recurrence_2}) has a strong resemblance to the usual recurrence (\ref{Euler_recurrence_1}), ($0\leq k\leq ~2n$). Using the formula (\ref{Euler_recurrence_1}) and according to the fact that $E_k=0$ for $k$ even positive integers, we need the $n$ first terms with odd indexes $k$ to compute $E_{2n+1}$. But the recurrence formula (\ref{Euler_recurrence_2}) needs only half the number of those terms ($E_k$ with $n\leq k\leq 2n$ with $k$ odd) to calculate $E_{2n+1}$. \end{remark} Put $x=0$ in (\ref{(3)}). Then we have an analogue of Momiyama's formula: \begin{theorem}\label{3.8} For any $m, n\in\mathbb{N}$, we have \begin{equation*} (-1)^m\sum_{k=0}^{m+1} \begin{pmatrix}m+1\\ k\end{pmatrix} (n+k+1)E_{n+k} + (-1)^n\sum_{k=0}^{n+1} \begin{pmatrix}n+1\\ k\end{pmatrix} (m+k+1)E_{m+k}=0. \end{equation*} \end{theorem} \par Using $q$-integral formula to (\ref{Momiyama}) in Theorem \ref{3.3}, we have \begin{theorem}[$q$-Symmetry 3]\label{3.9} For any $m, n\in\mathbb{N}$ and $a, b\in\mathbb{R}$, \begin{multline*} (-1)^m\sum_{k=0}^m \begin{bmatrix}m\\ k\end{bmatrix}_q \frac{E_{n+k+1}(a,q)-E_{n+k+1}(b,q)}{[n+k+1]_q} q^{-kn+mn}\\ + (-1)^n\sum_{k=0}^n \begin{bmatrix}n\\ k\end{bmatrix}_{q^{-1}} \frac{F_{m+k+1}(-a,q^{-1})-F_{m+k+1}(-b,q^{-1})}{[m+k+1]_q} q^{\binom{n}{2}-\binom{k}{2}}=0. \end{multline*} \end{theorem} As $q\to 1$, we get \begin{theorem}\label{3.10} For any $m, n\in\mathbb{N}$ and $a, b\in\mathbb{R}$, \begin{equation*} (-1)^m\sum_{k=0}^m \binom{m}{k} \frac{E_{n+k+1}(a)-E_{n+k+1}(b)}{n+k+1} + (-1)^n\sum_{k=0}^n \binom{n}{k} \frac{E_{m+k+1}(-a)-E_{m+k+1}(-b)}{m+k+1} =0. \end{equation*} \end{theorem} \section{Acknowledgement} The second named author was supported by Grant-in-Aid for Scientific Research (No.\ 20540026), Japan Society for the Promotion of Science. \begin{thebibliography}{12} \bibitem{[1]} M. 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Pan, \newblock Some identities for Bernoulli and Euler polynomials. \newblock {\it Fibonacci Quart.} {\bf 42} (2004), 295--299. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B68; Secondary 05A30. \noindent \emph{Keywords: } $q$-analogue, Euler number, Euler polynomial, $q$-symmetry, Keneko-Momiyama identity. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received February 15 2012; revised version received March 31 2012. Published in {\it Journal of Integer Sequences}, April 9 2012. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}