---
abstract: 'Recently, Kim introduced Carlitz’s type $q$-Euler numbers and polynomials(see [@05]).  In this paper, we construct the degenerate Carlitz’s type $q$-Euler numbers and polynomials which are derived from the fermionic $p$-adic $q$-integral on ${\mathbb{Z}}_p$. Finally, we give some identities and properties of these numbers and polynomials'
address:
- '$^1$ Institute of Natural Sciences, Far Eastern Federal University, Vladivostok, 690950, Russia.' 
- '$^2$ Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.' 
- '$^3$ Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea.' 
- '$^4$ Department of Applied Mathematics, Pukyong National University, Pusan, Republic of Korea'
author:
- 'Dmitry V. Dolgy$^1$'
- Taekyun Kim$^2$
- 'Jin-Woo Park$^{3,*}$'
- 'Jong-Jin Seo$^4$'
title: 'On degenerate $q$-Euler polynomials'
---

[^1]

Introduction
============

Let $p$ be a fixed odd prime number.  Throughout this paper, ${\mathbb{Z}}_p$, ${\mathbb{Q}}_p$ and ${\mathbb{C}}_p$ will, respectively, denote the ring of $p$-adic rational integers, the field of $p$-adic rational numbers and the completion of the algebraic closure of ${\mathbb{Q}}_p$. Let $\nu_p$ be the normalized exponential valuation of ${\mathbb{C}}_p$ with $|p|_p=p^{-\nu_p(p)}=\frac{1}{p}$. Let $q$ be an indeterminate in ${\mathbb{C}}_p$ such that $|q-1|_p<p^{-\frac{1}{p-1}}$ so that $q^x=\exp(x\log q)$. The [*[$q$-analogue of $x$]{}*]{} is defined as $[x]_q=\frac{1-q^x}{1-q}$. Note that $\lim_{q\rightarrow 1}=x$.

As is well known, the [*[Euler polynomials]{}*]{} are defined by the generating function to be $$\label{1}
\frac{2}{e^t+1}e^{xt}=\sum_{n=0} ^{\infty}E_n(x)\frac{t^n}{n!},{\text{ (see [1-11])}}.$$ When $x=0$, $E_n=E_n(0)$ are called the [*[Euler numbers]{}*]{}.  From , we can derive the recurrence relation for the Euler numbers as follows: $$\label{2}
(E+1)^n+E_n=2\delta_{0,n},~(n\geq1),$$ with the usual convention about replacing $E^n$ by $E_n$ (see \[1-12\]). 

In [@05; @06], Kim considered Carlitz’s type $q$-Euler numbers as follows: $$\label{3}
E_{0,q}=1,~q(qE_q+1)^n+E_{n,q}=0,~(n\geq1),$$ with the usual convention about replacing $E_q ^n$ by $E_{n,q}$.

The $q$-Euler polynomials are defined as $$\label{4}
E_{n,q} (x)=\sum_{l=0} ^n \binom{n}{l}q^{lx}E_{l,q}[x]_q ^{n-l},~(n\geq0),{\text{ (see \cite{05,06})}}.$$

Let $C({\mathbb{Z}}_p)$ be the space of all continuous functions on ${\mathbb{Z}}_p$. For $f\in C({\mathbb{Z}}_p)$, the [*[fermionic $p$-adic $q$-integral on ${\mathbb{Z}}_p$]{}*]{} is defined by Kim to be $$\label{5}
I_{-q}(f)={{\int_{{\mathbb{Z}}_p}}}f(x)d\mu_{-q}(x)=\lim_{N\rightarrow \infty}\frac{1}{\left[p^N\right]_{-q}}\sum_{x=0} ^{p^N-1}f(x)(-q)^x,$$ where $[x]_q=\frac{1-(-q)^x}{1+q}$, (see [@05; @06]). 

From , we have $$qI_{-q}(f_1)+I_{-q}(f)=[2]_qf(0),{\text{ where }}f_1(x)=f(x+1).$$ The $q$-Euler polynomials can be represented by the fermionic $p$-adic $q$-integral on ${\mathbb{Z}}_p$ as follows: $$\label{6}
\sum_{n=0} ^{\infty}E_{n,q}(x)\frac{t^n}{n!}={{\int_{{\mathbb{Z}}_p}}}e^{[x+y]_qt}d\mu_{-q}(y),{\text{ (see \cite{05,06})}}.$$ Thus, by , we get $$\label{7}
{{\int_{{\mathbb{Z}}_p}}}[x+y]_q ^nd\mu_{-q}(y)=E_{n,q}(x),~(n\geq 0).$$

In this paper, we consider the degenerate Carlitz $q$-Euler polynomials and numbers which are derived from the fermionic $p$-adic $q$-integral on ${\mathbb{Z}}_p$ and we investigate some properties and identities of those polynomials. 

Degenerate Carlitz $q$-Bernoulli numbers and polynomials
========================================================

In this section, we assume that $\lambda,t\in{\mathbb{C}}_p$ with $|\lambda t|_p<p^{-\frac{1}{p-1}}$ so that $(1+\lambda t)^{\frac{x}{\lambda}}=\exp\left(\frac{x}{\lambda}\log(1+\lambda t)\right)$.

It is well known that $$\label{8}
\lim_{\lambda\rightarrow 0}(1+\lambda t)^{\frac{1}{\lambda}}=\lim_{\lambda\rightarrow0}\sum_{n=0} ^{\infty}\binom{\frac{1}{\lambda}}{n}\lambda^nt^n=\sum_{n=0} ^{\infty}\frac{t^n}{n!}=e^{xt}.$$ From and , we consider the [*[degenerate $q$-Euler polynomials]{}*]{} which are given by the generating function to be

$$\label{9}
{{\int_{{\mathbb{Z}}_p}}}(1+\lambda t)^{\frac{1}{\lambda}[x+y]_q}d\mu_{-q}(y)=\sum_{n=0} ^{\infty} E_{n,q}(x|\lambda)\frac{t^n}{n!}.$$

Note that $$\label{10}
\begin{split}
\sum_{n=0} ^{\infty}\lim_{\lambda\rightarrow0}E_{n,q} (x|\lambda)\frac{t^n}{n!}=&\lim_{\lambda\rightarrow0}{{\int_{{\mathbb{Z}}_p}}}\left(1+\lambda t\right)^{\frac{[x+y]_q}{\lambda}}d\mu_{-q}(y)\\
=&{{\int_{{\mathbb{Z}}_p}}}e^{[x+y]_qt}d\mu_{-q  }(y)=\sum_{n=0} ^{\infty}E_{n,q}(x)\frac{t^n}{n!}. 
\end{split}$$ Thus, by , we get $$\label{11}
\lim_{\lambda\rightarrow0}E_{n,q}(x|\lambda)=E_{n,q}(x),~(n\geq0).$$ When $x=0$, $E_{n,q}(\lambda)=E_{n,q}(0|\lambda)$ are called [*[degenerate $q$-Euler numbers]{}*]{}. 

From , we have $$\label{12}
\begin{split}
\sum_{n=0} ^{\infty}E_{n,q}(x|\lambda)\frac{t^n}{n!}=&\sum_{m=0} ^{\infty}\lambda^{-m}{{\int_{{\mathbb{Z}}_p}}}[x+y]_q ^md\mu_{-q}(y)\frac{\left(\log(1+t)\right)^m}{m!}\\
=&\sum_{m=0} ^{\infty}\lambda^{-m}E_{m,q}(x)\sum_{n=m} ^{\infty}S_1(n,m)\frac{\lambda^nt^n}{n!}\\
=&\sum_{n=0} ^{\infty}\left(\sum_{m=0} ^n\lambda^{n-m}E_{m,q}(x)S_1(n,m)\right)\frac{t^n}{n!}. 
\end{split}$$ By comparing the coefficients on the both sides of , we obtain the following theorem. 

\[thm1\] For $n \geq 0$, we have $$E_{n,q}(x|\lambda)=\sum_{m=0} ^n \lambda^{n-m}E_{m,q}(x)S_1(n,m),.$$ where $S_1(n,m)$ is the Stirling number of the first kind. 

Replacing $t$ by $\frac{1}{\lambda}\left(e^{\lambda t}-1\right)$ in , we get $$\label{13}
\begin{split}
\sum_{m=0} ^{\infty}E_{m,q}(x|\lambda)\lambda^{-m}\frac{\left(e^{\lambda t}-1\right)^m}{m!}=&{{\int_{{\mathbb{Z}}_p}}}e^{[x+y]_qt}d\mu_{-q}(y)\\
=&\sum_{n=0} ^{\infty}E_{n,q}(x)\frac{t^n}{n!}. 
\end{split}$$ On the other hand, $$\label{14}
\begin{split}
\sum_{m=0} ^{\infty}E_{m,q}(x|\lambda)\lambda^{-m}\frac{1}{m!}\left(e^{\lambda t}-1\right)^m=&\sum_{m=0} ^{\infty}E_{m,q}(x|\lambda)\lambda^{-m}\sum_{n=m} ^{\infty}S_2(n,m)\frac{\lambda^nt^n}{n!}\\
=&\sum_{n=0} ^{\infty}\left(\sum_{m=0} ^nE_{m,q}(x|\lambda)\lambda^{n-m}S_2(n,m)\right)\frac{t^n}{n!}. 
\end{split}$$ Therefore, by and , we obtain the following theorem. 

\[thm2\] For $n\geq 0$, we have $$E_{n,q}(x)=\sum_{m=0} ^n E_{m,q}(x|\lambda)\lambda^{n-m}S_2(n,m).$$ where $S_2(n,m)$ is the Stirling number of the second kind. 

We observe that $$\label{15}
(1+\lambda t)^{\frac{[x+y]_q}{\lambda}}=\sum_{n=0} ^{\infty}\left(\frac{[x+y]_q}{\lambda}\right)_n\frac{\lambda^nt^n}{n!}$$ where $(x)_n=x(x-1)\cdots(x-n+1)$.

Now we define the [*[degenerate falling factorial]{}*]{} as follows: $$\label{16}
(x)_{n,\lambda}=x(x-\lambda)(x-2\lambda)\cdots(x-(n-1)\lambda).$$ Note that $(x)_{n,1}=x(x-1)\cdots(x-n+1)=(x)_n$, $n\geq0$. Therefore, by and , we obtain the following theorem. 

\[thm3\] For $n \geq 0$, we have $${{\int_{{\mathbb{Z}}_p}}}\left([x+y]_q\right)_{n,\lambda}d\mu_{-q}(y)=E_{n,q}(x|\lambda),$$ where $\left([x+y]_q\right)_{n,\lambda}=[x+y]_q\left([x+y]_q-\lambda\right)\cdots\left([x+y]_q-(n-1)\lambda\right)$.

From , we can easily derive the following equation: $$\label{17}
\begin{split}
{{\int_{{\mathbb{Z}}_p}}}[x+y]_q ^nd\mu_{-q}(y)=&\frac{[2]_q}{(1-q)^n}\sum_{l=0} ^n \binom{n}{l}(-1)^lq^{lx}\frac{1}{1+q^{l+1}}\\
=&[2]_q\sum_{m=0} ^{\infty}(-1)^m[x+m+1]_q ^n. 
\end{split}$$ Thus, by , we get the generating function of the $q$-Euler polynomials as follows: $$\label{18}
\sum_{n=0} ^{\infty}E_{n,q}(x)\frac{t^n}{n!}=[2]_q\sum_{m=0} ^{\infty}(1-)^me^{[x+m+1]_q}.$$ From and , we can derive the generating function of degenerate $q$-Euler polynomials which is given by $$\label{19}
\sum_{n=0} ^{\infty}E_{n,q}(x|\lambda)\frac{t^n}{n!}=[2]_q\sum_{m=0} ^{\infty}(-1)^m\left(1+\lambda t\right)^{\frac{[x+m+1]_q}{\lambda}}.$$ By , we easily get $$\label{20}
\sum_{n=0} ^{\infty}E_{n,q}(x|\lambda)\frac{t^n}{n!}=\sum_{n=0} ^{\infty}\left([2]_q\sum_{m=0} ^{\infty}(-1)^m\left([x+m+1]_q\right)_{n,\lambda}\right)\frac{t^n}{n!}.$$ Therefore, by , we obtain the following theorem. 

\[thm4\] For $n \geq 0$, we have $$E_{n,q}(x|\lambda)=[2]_q\sum_{m=0} ^{\infty}(-1)^m\left([x+m+1]_q\right)_{n,\lambda}.$$

We observe that $$\label{21}
\begin{split}
(1+\lambda t)^{\frac{[x+y]_q}{\lambda}}=&(1+\lambda t)^{\frac{[x]_q}{\lambda}}(1+\lambda t)^{\frac{q^x[y]_q}{\lambda}}\\
=&\left(\sum_{m=0} ^{\infty}\left([x]_q\right)_{m,\lambda}\frac{t^m}{m!}\right)\left(\sum_{l=0} ^{\infty}\lambda^{-l}q^{lx}\frac{[y]_q ^l \left(\log(1+\lambda t)\right)^l}{l!}\right)\\
=&\sum_{n=0} ^{\infty}\left(\sum_{k=0} ^n \sum_{l=0} ^k\left([x]_q\right)_{n-k,\lambda}\lambda^{k-l}q^{lx}[y]_q ^lS_1(k,l)\binom{n}{k}\right)\frac{t^n}{n!}. 
\end{split}$$ From and , we can derive the following theorem. 

\[thm5\] For $n \geq 0$, we have $$E_{n,q}(x|\lambda)=\sum_{k=0} ^n \sum_{l=0} ^k\binom{n}{k}\left([x]_q\right)_{n-k,\lambda}\lambda^{k-l}q^{lx}S_1(k,l)E_{l,q}.$$

It is not difficult to show that $$\label{22}
q{{\int_{{\mathbb{Z}}_p}}}\left([x+y+1]_q\right)_{n,\lambda}d\mu_{-q}(y)+{{\int_{{\mathbb{Z}}_p}}}\left([x+y]_q\right)_{n,\lambda}d\mu_{-q}(y)=[2]_q\left([x]_q\right)_{n,\lambda}.$$ Thus, by Theorem \[thm5\] and , we obtain the following theorem. 

\[thm6\] For $n \geq 0$, we have $$qE_{n,q}(x+1|\lambda)+E_{n,q}(x|\lambda)=[2]_q\left([x]_q\right)_{n,\lambda}.$$

Let $r\in{\mathbb{N}}$. Now, we recall the Carlitz’s $q$-Euler polynomials of order $r$ which are given by the generating function to be $$\label{23}
{{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}e^{[x_1+\cdots+x_r+x]_q}d\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r)\\
=\sum_{n=0} ^{\infty}E_{n,q} ^{(r)}(x)\frac{t^n}{n!}.$$ Thus, by , we get $$\label{24}
{{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}[x_1+\cdots+x_r+x]_q ^nd\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r)=E_{n,q} ^{(r)}(x),~(n\geq0),{\text{ (see \cite{05})}}.$$

From , we have $$\label{25}
\begin{split}
E_{n,q} ^{(r)}(x)=&\frac{[2]_q ^r}{(1-q)^n}\sum_{l=0} ^n \binom{n}{l}(-1)^lq^{lx}\left(\frac{1}{1+q^{l+1}}\right)^r\\
=&[2]_q ^r\sum_{m=0} ^{\infty}(-1)^m\binom{r+m-1}{m}q^m[m+x]_q ^n. 
\end{split}$$ Thus, by , we get the generating function of Carlitz’s $q$-Euler polynomials of order $r$ which given by $$\label{26}
\begin{split}
\sum_{n=0} ^{\infty}E_{n,q} ^{(r)}(x)\frac{t^n}{n!}=[2]_q ^r\sum_{m=0} ^{\infty}(-1)^m\binom{r+m-1}{m}q^me^{[m+x]_q t}. 
\end{split}$$ We consider [*[degenerate $q$-Euler polynomials of order $r$]{}*]{} which are given by the generating function to be $$\label{27}
\begin{split}
\sum_{n=0} ^{\infty}E_{n,q} ^{(r)}(x|\lambda)\frac{t^n}{n!}=[2]_q ^r\sum_{m=0} ^{\infty}(-1)^m\binom{r+m-1}{m}q^m(1+\lambda t)^{[m+x]_q t}. 
\end{split}$$ Thus, by , we have $$\label{28}
\begin{split}
&\sum_{n=0} ^{\infty}E_{n,q} ^{(r)}(x|\lambda)\frac{t^n}{n!}\\
=&[2]_q ^r \sum_{n=0} ^{\infty}\left(\sum_{m=0} ^{\infty}(-1)^m\binom{r+m-1}{m}q^m\left([m+x]_q\right)_{n,\lambda}\right)\frac{t^n}{n!}. 
\end{split}$$ By comparing the coefficients on the both sides of , we obtain the following theorem. 

\[thm7\] For $n\geq 0$, we have $$E_{n,q} ^{(r)}(x|\lambda)=[2]_q ^r\sum_{m=0} ^{\infty}(-1)^m\binom{r+m-1}{m}q^m\left([m+x]_q\right)_{n,\lambda}.$$

From and Theorem \[thm7\], we note that $$\label{29}
\sum_{n=0} ^{\infty}E_{n,q} ^{(r)}(x|\lambda)\frac{t^n}{n!}={{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}(1+\lambda t)^{\frac{[x_1+\cdots+x_r+x]_q}{\lambda}}d\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r).$$ Thus, by , we get $$\label{30}
\begin{split}
E_{n,q} ^{(r)}(x|\lambda)
=&\sum_{m=0} ^{\infty}{{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}\left([x_1+\cdots+x_r+x]_q\right)_{n,\lambda} d\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r)\\
=&\sum_{m=0} ^nS_1(n,m)\lambda^{n-m}{{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}[x_1+\cdots+x_r+x]_q ^md\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r). 
\end{split}$$ Therefore, by and , we obtain the following theorem. 

\[thm8\] For $n \geq 0$, we have $$E_{n,q} ^{(r)}(x|\lambda)={{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}\left([x_1+\cdots+x_r+x]_q\right)_{n,\lambda} d\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r).$$ Furthermore, $$E_{n,q} ^{(r)}(x|\lambda)=\sum_{m=0} ^n S_1(n,m)\lambda^{n-m}E_{m,q} ^{(r)}(x).$$

Replacing $t$ by $\frac{1}{\lambda}\left(e^{\lambda t}-1\right)$ in , we get $$\label{31}
\begin{split}
&{{\int_{{\mathbb{Z}}_p}}}\cdots{{\int_{{\mathbb{Z}}_p}}}e^{[x_1+\cdots+x_r+x]_q t}d\mu_{-q}(x_1)\cdots d\mu_{-q}(x_r)\\
=&\sum_{m=0} ^{\infty}E_{m,q} ^{(r)}(x|\lambda)\frac{1}{m!}\lambda^{-m}\left(e^{\lambda t}-1\right)^m\\
=&\sum_{m=0} ^{\infty}E_{m,q} ^{(r)}(x|\lambda)\lambda^{n-m}S_2(n,m)\frac{t^n}{n!}\\
=&\sum_{n=0} ^{\infty}\left(\sum_{m=0} ^n \lambda^{n-m}E_{m,q} ^{(r)}(x|\lambda)S_2(n,m)\right)\frac{t^n}{n!}. 
\end{split}$$ Therefore, by and , we obtain the following theorem. 

For $n\geq 0$, we have $$E_{n,q} ^{(r)}(x)=\sum_{m=0} ^nE_{m,q} ^{(r)}(x|\lambda)\lambda^{n-m}S_2(n,m).$$

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[^1]: $*$ corresponding author
