---
abstract: 'In this paper, we give some interesting and new identities of $q$-Bernoulli numbers which are derived from convolutions on the ring of $p$-adic integers.'
address:
- |
    1\
    Department of Applied Mathematics\
    Pukyong National University\
    Busan 608-737, Republic of Korea
- |
    2\
    Department of Mathematics\
    Kwangwoon National University\
    Seoul 139-701, Republic of Korea
- |
    3\
    Division of General Education\
    Kwangwoon National University\
    Seoul 139-701, Republic of Korea
author:
- 'J.J. Seo, T.Kim, S.H.Lee'
title: 'Some identities of $q$-Bernoulli numbers associated $p$-adic convolutions'
---

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**Introduction**
================

Throughout this paper $ \mathbb{Z}_p, \mathbb{Q}_p$ and $\mathbb{C}_p$ will denote the ring of $p$-adic integers, the field of $p$-adic numbers and the completion of algebraic closure of $\mathbb{Q}_p$. The $p$-adic norm is defined by $|p|_p=p^{-v_p(p)} =p^{-1}$. Now, we set $$U_p=\left\{\alpha\in \mathbb{C}_p | \ |\alpha-1|_p <1 \right\}$$ and $$T_p=\left\{w\in \mathbb{C}_p | w^{p^n} =1 \ \ \textnormal{for some}\ \ n\geq 0 \right\},$$ so that $T_p$ is the union of cyclic (multiplicative) group $\mathbb{C}_{p^n}$ of order $p^n$ $(n\geq 0)$ and $T_p\subset U_p$ (see \[12\]). Let $UD(\mathbb{Z}_p)$ and $C(\mathbb{Z}_p)$ be the space of uniformly differentiable and continuous function on $\mathbb{Z}_p$. For $f\in UD(\mathbb{Z}_p)$, the $p$-adic invariant integral on $\mathbb{Z}_p$ is defined as follows : $$I_0(f)=\int_{\mathbb{Z}_p} f(x) d\mu_0 (x) = \lim_{N\rightarrow \infty} {1\over {p^N}} \sum_{x=0}^{p^N-1} f(x), \  (\textnormal{see}\ [11]).$$

For $w \in T_p$, the $p$-adic Fourier transform of $f$ is given by $${\hat f}_w = I_0 (f\phi_w )=\lim_{N\rightarrow \infty} {1\over {p^N}} \sum_{x=0}^{p^N-1} f(x)\phi_w (x),  \  (\textnormal{see}\ [12]),$$ where $\phi_w(x)=w^x.$

Let $f, g\in UD(\mathbb{Z}_p).$ Then *C. F. Woodcock* defined the convolution of $f$ and $g$ as follows : $$f*g=\sum_w {\widehat f}_w {\widehat g}_w \phi_{w^{-1}},   \ \ (\textnormal{see}\ [12]).$$

From (1.3), we note that $$f*g \in UD(\mathbb{Z}_p),\ \ (\widehat{f*g})_w ={\widehat f}_w {\widehat g}_w, \ (\forall w\in T_p),$$ and $(UD(\mathbb{Z}_p), +, *, V)$ is Banach algebra, where $V(f)=\min\left\{\nu(f), R(f) \right\}, \nu(f)=\inf_{x\in\mathbb{Z}_p} \nu_p\left(f(x) \right)$ and $R(f)=\inf \nu_p \left({{f(x)-f(y)}\over {x-y}} \right), x\not=y \in \mathbb{Z}_p$, (see \[12\]).

Let $int$ $\mathbb{Z}_p=\left\{f\in UD(\mathbb{Z}_p) | f^{\prime}=0 \right\}$. Then $int$ $\mathbb{Z}_p$ is $*$-ideal of $UD(\mathbb{Z}_p)$. Differentiation induced a linear isometry $$UD(\mathbb{Z}_p)/ {int } \\ \mathbb{Z}_p  \longrightarrow C(\mathbb{Z}_p) \ \ \textnormal{by}\ \ (f*g)^{\prime}=-f^{\prime} \circledast g^{\prime},$$ where $f, g \in UD(\mathbb{Z}_p) $ (see \[12\]). By (1.5), we get $$(f\circledast g)(n)=\sum_{i=0}^n f(i)g(n-i),$$ where $f, g \in C(\mathbb{Z}_p).$

For $f, g \in UD(\mathbb{Z}_p)$ and $z \in \mathbb{Z}_p$, it is known that $$f*g(z)=I_0^{(x)} (f(x)g(z-x))-f \circledast g^{\prime}(z),$$ where $I_0^{(x)}(f) = \lim_{N\rightarrow \infty}  \frac{1}{p^N}\sum_{x=0}^{p^N -1} f(x)$.

When one talks of $q$-extensions, $q$ is variously considered as an indeterminate, a complex number $q\in \mathbb{C}$, or a $p$-adic number $q\in \mathbb{C}_p$. If $q\in \mathbb{C}_p$, then we normally assume $|1-q|_p < p^{- 1\over {p-1}}$ so that $q^x=\exp(\log q) $ for $|x|_p \leq 1$. We use the notation $\left[x\right]_q=\left[x:q\right]={{1-q^x}\over{1-q}}$. Thus, $\lim_{q\rightarrow 1}\left[x\right]_q =x$.

As is well known, the usual *$Bernoulli \ numbers$* are defined by $$B_0=1, \ \ (B+1)^n -B_n =
\left\{
\begin{array}{cc}
1 , & \text{if }n=1 \\
0, & \text{if\ }n> 1\text{,}\end{array}\right.$$

with the usual convention about replacing $\beta^i$ by $\beta_i$ (see \[1-10\]). In \[1.3\], $Carlitz$ defined the $q$-extension of $Bernoulli\ numbers$ as follows:

$$\beta_{0,q} =1, \ \ q (q\beta_q+1)^n -\beta_{n,q}=
\left\{
\begin{array}{cc}
1 , & \text{if }n=1 \\
0, & \text{if\ }n> 1\text{,}\end{array}\right.$$

with the usual convention about replacing $\beta_q^i$ by $\beta_{i,q}$. By (1.9), we easily see that $$\beta_{0,q}=1,\ \beta_{1,q}={-1\over {[2]_q}},\ \beta_{2,q}={q\over {[2]_q [3]_q}},\ \beta_{3,q}={{(1-q)}\over {[3]_q [4]_q}}, \cdots.$$

In this paper, we consider the modified $q\emph{-Bernoulli\ numbers},$ which are slightly different $\emph{Carlitz's}$ $q$*-Bernoulli numbers*, as follows : $$\widetilde{\beta}_{0,q} =1, \ \left(q\widetilde{\beta}_q+1 \right)^n -\widetilde{\beta}_{n,q}=
\left\{
\begin{array}{cc}
\frac{logq}{q-1} , & \text{if }n=1 \\
0, & \text{if\ }n> 1\text{,}\end{array}\right.$$ with the usual convention about replacing $\widetilde{\beta}_q^i$ by $\widetilde{\beta}_{i,q}.$

The purpose of our paper is to give some interesting and new identities of the modified $q-Bernoulli\ numbers$ $\widetilde{\beta}_{n,q}$ which are derived from convolutions on the ring of $p$-adic integers.

**Some identities of $q$-Bernoulli numbers**
============================================

Let us consider the following $q$-extension of $Bernoulli$ $numbers$ : $$\widetilde{\beta}_{0,q}=1, \ \ \left(q\widetilde{\beta}_q+1\right)^n-\widetilde{\beta}_{n,q} =
\left\{
\begin{array}{cc}
\frac{logq}{q-1} , & \text{if }n=1 \\
0, & \text{if\ }n> 1\text{,}\end{array}\right.$$

Then, by (1.1), we easily see that

$$\begin{split}
\widetilde{\beta}_{n,q}=&\int_{\mathbb{Z}_p} [x]_q^n d\mu_0(x)\\
                       =&{{\log q}\over{(1-q)^{n+1}}}\sum_{l=0}^n \binom{n}{l} (-1)^{l-1} {l \over{[l]_q}}, \ \ n\in \mathbb{Z}_+ = \mathbb{N}\cup \left\{0\right\}.
\end{split}$$

In the equation (1.6) and (1.7), if we take $p$-adic integral on $\mathbb{Z}_p$ with respect to variable $\mathbb{Z}$, then we have $$I_0^{(z)}(f*g)=I_0^{(z)}\left( I_0^{(x)}(f(x)g(z-x))\right)-I_0^{(z)} \left(f\circledast g^{\prime} (z) \right).$$

By (1.4) and (2.3), we get

$$I_0^{(z)}\left(f\circledast g^{\prime}(z)\right)=I_0^{(z)}\left( I_0^{(x)}(f(x)g(z-x))\right)-I_0^{(z)} (f) I_0^{(z)}(g).$$

Let us take $f(x)=[x]_{q^{-1}}^m,$ $g[x]=[x]_q^n$. Then $$q^{\prime}(x)=n{{\log q}\over{q-1}} [x]_q^{n-1}, \ \ (m, n \in \mathbb{N}).$$

Now, we set $$A_{m,n}^q =I_0^{(z)}\left([z]_{q^{-1}}^m \circledast[z]_q^{n-1} \right), \ \ m, n \in \mathbb{N}.$$

From (2.4) and (2.5), we can derive

$$n{{\log q}\over{q-1}}A_{m,n-1}^q = \int_{\mathbb{Z}_p}\int_{\mathbb{Z}_p} [x]_{q^{-1}}^m [z-x]_q^{n} d\mu_0(x) d\mu_0(z)-\widetilde{\beta}_{m,q^{-1}} \widetilde{\beta}_{n,q}.$$

Note that $$[z-x]_{q^{-1}}^n=\left([z]_q-q^{-1}q^z [x]_{q^{-1}} \right)^n =\sum_{l=0}^n  \binom{n}{l}   [z]_q^{n-l}(-1)^lq^{-l}q^{lz} [x]_{q^{-1}}^l.$$

By (2.6) and (2.7), we get

$$\begin{split}
n{{\log q}\over{q-1}}A_{m,n}^q =&\sum_{l=1}^n\binom{n}{l} (-1)^l q^{-l}   \int_{\mathbb{Z}_p} [x]_{q-1}^{m+l} d\mu_0(x)\int_{\mathbb{Z}_p} [z]_q^{n-l} q^{lz} d\mu_0 (z) \\
                               =&\sum_{l=1}^n \binom{n}{l}(-1)^k q^{-l} \widetilde{\beta}_{m+l,q^{-1}} \sum_{k=0}^l \binom{l}{k}(q-1)^k \int_{\mathbb{Z}_p} [z]_q^{n+k-l}d\mu_0 (z)\\
                               =&\sum_{l=1}^n\sum_{k=0}^{l} \binom{n}{l}\binom{l}{k} (-1)^l q^{-l} \widetilde{\beta}_{m+l,q^{-1}} (q-1)^k \widetilde{\beta}_{m+k-l,q}.
\end{split}$$

Thus, from (2.8), we have

$$A_{m,n}^q={{q-1}\over{\log q}}{1\over n}\sum_{l=1}^n \sum_{k=0}^l \binom{n}{l} \binom{l}{k}(-1)^lq^{-l}(q-1)^k \widetilde{\beta}_{m+l,q^{-1}}\widetilde{\beta}_{n+k-l,q}.$$

By (2.5), we easily see that

$$\begin{split}
A_{m,n}^q=&I_0^{(z)} \left([z]_{q^{-1}}^m\circledast [z]_q^{n-1} \right)\\
         =&I_0^{(z)} \left([z]_q^{n-1}\circledast [z]_{q^{-1}}^m \right)\\
         =&A_{n-1,m+1}^q.
\end{split}$$

From (2.5), we have

$$\begin{split}
\nu_p(A_{m,n}^q)=&\nu_p \left(I_0\left([z]_{q^{-1}}^m\circledast [z]_q^{n-1} \right) \right)\\
                \geq &\nu \left([z]_{q^{-1}}^m\circledast [z]_q^{n-1} \right)-1 \\
                \geq &\nu ([z]_{q^{-1}}^m)+\nu([z]_q^{n-1})-1-1\\
                \geq &-2.
\end{split}$$

Therefore, by (2.9), (2.10) and (2.11), we obtain the following theorem.

\[thm 2.1\] For $m\in \mathbb{Z}_+$ and $n \in \mathbb{N}$, we have $$A_{m,n}^q={{q-1}\over{\log q}} {1\over n} \sum_{l=1}^n\sum_{k=0}^l  \binom{n}{l}\binom{l}{k} (-1)^lq^{-l}(q-1)^k \widetilde{\beta}_{m+l,q^{-1}}\widetilde{\beta}_{n+k-l,q} .$$ Furthermore, $$A_{m,n}^q=A_{n-1,m+1}^q, \ \ \nu_p(A_{m,n}^q)\geq -2.$$

In particular, if we take $m=0$, then by (2.10), we get

$$A_{0,n}^q=A_{n-1,1}^q \ \ \textnormal{($q$-analogue of \textit{Euler identity})}.$$

From (2.2), we note that

$$\begin{split}
\widetilde{\beta}_{n,q}=&{{\log q}\over{(1-q)^{n+1}}}\sum_{l=0}^n\binom{n}{l} (-1)^{l-1} l {{1-q}\over{1-q^l}}\\
                       =&n{{\log q}\over{(1-q)^n}}\sum_{l=1}^n\binom{n-1}{l-1} (-1)^{l-1} \sum_{m=0}^\infty q^{lm}\\
                       =&n{{\log q}\over{1-q}}{1\over{(1-q)^{n-1}}} \sum_{l=0}^{n-1} \binom{n-1}{l}(-1)^{l} \sum_{m=0}^\infty q^{(l+1)m}\\
                       =&n{{\log q}\over{1-q}} \sum_{m=0}^{\infty} q^m [m]_q^{n-1},
\end{split}$$

where $n \in \mathbb{N}$. Thus, by (2.13), we obtain the following theorem.

\[thm 2.2\] For $n\in \mathbb{N}$, we have $$- {{\widetilde{\beta}_{n,q}}\over{n}} ={{\log q}\over{q-1}}\sum_{m=1}^\infty q^m [m]_q^{n-1}.$$

Let $F_q(t)$ be the generating function for $\widetilde{\beta}_{n,q}$ with $F_q(t)=\sum_{k=0}^\infty {1\over{k!}}\widetilde{\beta}_{k,q}t^k.$ Then, by (2.13), we get

$$\begin{split}
F_q(t)=&\sum_{k=0}^\infty {\widetilde{\beta}_{k,q}\over{k!}} t^k \\
      =&{{\log q}\over{1-q}}\sum_{m=0}^{\infty} q^m \left\{1+\sum_{k=1}^\infty {k\over{k!}} [m]_q^{k-1}t^k \right\}\\
      =&{{\log q}\over {(1-q)^2}}+t{{\log q}\over{1-q}}\sum_{m=0}^\infty q^m e^{[m]_q^t}.
\end{split}$$

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

\[thm 2.3\] Let $F_q(t)=\sum_{k=0}^\infty \widetilde{\beta}_{k,q} {t^k \over {k!}}$. Then we have $$F_q(t)={{\log q}\over{(1-q)^2}} +t{{\log q}\over{1-q}}\sum_{m=0}^\infty q^m e^{[m]_q^t}.$$

[0]{}

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