Some identities on the twisted q-Bernoulli numbers and polynomials with weight α

Abstract

In this paper, we consider the twisted q-Bernoulli numbers and polynomials with weight α by using the bosonic q-integral on . From the construction of the twisted q-Bernoulli numbers with weight α, we derive some identities and relations.

MSC: 11B68, 11S40, 11S80.

1 Introduction

Let p be a fixed prime number. Throughout this paper, , , and will, respectively, denote the ring of p-adic rational integers, the field of p-adic rational numbers, and the p-adic completion of the algebraic closure of , respectively. Let ${\nu }_p$ be the normalized exponential valuation of with ${|p|}_p=p^{{\nu }_p(p)}=p^{-1}$. When one talks of q-extension, q is variously considered as an indeterminate, a complex number , or a p-adic number . If , one normally assume $|q|<1$. If , then we assume that ${|1-q|}_p<p^{-\frac{1}{1-p}}$, so that $q^x=exp(xlogq)$ for ${|x|}_p\le 1$. The q-number is defined by ${[x]}_q=\frac{1-q^x}{1-q}$ (see [1–25]). Note that ${lim}_{q\to 1}{[x]}_q=x$.

We say that f is uniformly differentiable function at a point , and denote this property by , if the difference quotient $F_f(x,y)=\frac{f(x)-f(y)}{x-y}$ has a limit $f^{\prime }(a)$ as $(x,y)\to (a,a)$. For , the p-adic q-integral on , which is called the bosonic q-integral, defined by Kim as follows:

(1)

From (1), we note that

(2)

where ${\parallel f\parallel }_1=sup\{{|f(0)|}_p,{sup}_{x\ne y}|\frac{f(x)-f(y)}{x-y}{|}_p\}$.

In [3], Carlitz defined q-Bernoulli numbers which are called the Carlitz’s q-Bernoulli numbers, by

$${\beta }_{0,q}=1\text{and}q{(q{\beta }_q+1)}^n-{\beta }_{n,q}=\{\begin{array}{cc}1\hfill & \text{if }m=1,\hfill \\ 0\hfill & \text{if }m>1,\hfill \end{array}$$

(3)

with the usual convention about replacing ${({\beta }_q^{(h)})}^n$ by ${\beta }_{n,q}^{(h)}$. In [3], Carlitz also considered the expansion of q-Bernoulli numbers as follows:

$${\beta }_{0,q}^{(h)}=\frac{h}{{[h]}_q}\text{and}{q}^h{(q{\beta }_q^{(h)}+1)}^n-{\beta }_{n,q}^{(h)}=\{\begin{array}{cc}1\hfill & \text{if }n=1,\hfill \\ 0\hfill & \text{if }n>1,\hfill \end{array}$$

(4)

with the usual convention about replacing ${({\beta }_q^{(h)})}^n$ by ${\beta }_{n,q}^{(h)}$. In [15], for and , q-Bernoulli numbers with weight α is defined by Kim as follows:

$${\beta }_{0,q}^{(\alpha )}\text{and}q{(q^{\alpha }{\tilde{\beta }}_q^{(\alpha )}+1)}^n-{\tilde{\beta }}_{n,q}^{(\alpha )}=\{\begin{array}{cc}\frac{\alpha }{{[\alpha ]}_q}\hfill & \text{if }n=1,\hfill \\ 0\hfill & \text{if }n>1,\hfill \end{array}$$

(5)

with the usual convention about replacing ${({\beta }_q^{(\alpha )})}^n$ by ${\beta }_{n,q}^{(\alpha )}$.

Let $C_{p^n}=\{\xi |{\xi }^{p^n}=1\}$ be the cyclic group of order $p^n$, and let $T_p={lim}_{n\to \mathrm{\infty }}{C}_{p^n}={\bigcup }_{n\ge 0}{C}_{p^n}$ (see [8, 13, 20–24]). Note that $T_p$ is a locally constant space. For $\xi \in T_p$, the twisted Bernoulli numbers are defined by

$$\frac{t}{\xi e^t-1}=e^{B_{\xi }t}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n,\xi }\frac{t^n}{n!},$$

(6)

with the usual convention about replacing ${(B_{\xi })}^n$ by $B_{n,q}$ (see [15, 19–24]).

In the view point of (6), we will try to study the twisted q-Bernoulli numbers with weight α. By using the p-adic q-integral on , we give some identities and relations on the twisted q-Bernoulli numbers and polynomials with weight α.

2 The twisted q-Bernoulli numbers and polynomials with weight α

Let $f_n(x)=f(x+n)$. In [[15], Theorem 1], Kim proved the following integral equation:

$$q^n{I}_q(f_n)-I_q(f)=(q-1)\sum _{l=0}^{n-1}{q}^{l}f(l)+\frac{q-1}{logq}\sum _{l=0}^{n-1}{q}^l{f}^{\prime }(l).$$

(7)

In particular, when $n=1$, we have

$$q{I}_q(f_1)-I_q(f)=(q-1)f(0)+\frac{q-1}{logq}{f}^{\prime }(0).$$

(8)

Let and . The n th q-Bernoulli polynomials with weight α are defined by

(9)

Then, by using the bosonic p-adic q-integral on , we evaluate the above equation (9) as follows:

(10)

In the special case $x=0$, ${\tilde{\beta }}_{n,\xi ,q}^{\alpha }(0)={\tilde{\beta }}_{n,\xi ,q}^{\alpha }$ are called the n th twisted q-Bernoulli numbers with weight α.

From (10), we note that

(11)

From (11), when $x=0$, we get

$${\tilde{\beta }}_{n,\xi ,q}^{\alpha }=\frac{-n\alpha }{{[\alpha ]}_q}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^{m\alpha +m}{[m]}_{q^{\alpha }}^{n-1}+(1-q)\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^m{[m]}_{q^{\alpha }}^n.$$

(12)

Therefore, by (10) and (12), we obtain the following theorem.

Theorem 1 Let and . Then we

When $x=0$, we can obtain some identity on the twisted q-Bernoulli numbers with weight α as follows.

Corollary 2 For and . Then we have

$$-\frac{{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}}{n}=\frac{\alpha }{{[\alpha ]}_q}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^{m\alpha +m}{[m]}_{q^{\alpha }}^{n-1}+\frac{1-q}{n}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^m{[m]}_{q^{\alpha }}^n.$$

For $\alpha =1$, we note that ${\tilde{\beta }}_{n,\xi ,q}^{(1)}(x)$ are the twisted Carlitz’s q-Bernoulli polynomials and ${\tilde{\beta }}_{n,\xi ,q}^{(1)}$ are the twisted Carlitz’s q-Bernoulli numbers. By Corollary 2, we easily get

(13)

From (13), we have

$$\sum _{n=0}^{\mathrm{\infty }}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}\frac{t^n}{n!}=-t\frac{\alpha }{{[\alpha ]}_q}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^{(m+1)\alpha +m}{e}^{{[m]}_{q^{\alpha }}t}+(1-q)\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^m{e}^{{[m]}_{q^{\alpha }}t}.$$

(14)

Therefore, we obtain the following corollary.

Corollary 3 Let and $F_{\xi ,q}^{(\alpha )}(t)={\sum }_{n=0}^{\mathrm{\infty }}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}\frac{t^n}{n!}$. Then we have

$$F_{\xi ,q}^{(\alpha )}(t)=-t\frac{\alpha }{{[\alpha ]}_q}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^{(m+1)\alpha +m}{e}^{{[m]}_{q^{\alpha }}t}+(1-q)\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^m{e}^{{[m]}_{q^{\alpha }}t}.$$

Let $F_{\xi ,q}^{(\alpha )}(t,x)={\sum }_{n=0}^{\mathrm{\infty }}{\sum }_{n=0}^{\mathrm{\infty }}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x)\frac{t^n}{n!}$. From Theorem 1, we have

(15)

By simple calculation, we easily get

(16)

Therefore, by (12), (15), and (16), we obtain the following theorem.

Theorem 4 For and , we have

$$\begin{array}{rcl}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x)& =& \frac{1-q}{{(1-q^{\alpha })}^n}\sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right){(-1)}^l{q}^{\alpha lx}\left(\frac{\alpha l+1}{1-\xi q^{\alpha l+1}}\right)\\ =& -n\frac{\alpha }{{[\alpha ]}_q}\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^{(m+1)\alpha +m}{[m+x]}_{q^{\alpha }}^{n-1}\\ +(1-q)\sum _{m=0}^{\mathrm{\infty }}{\xi }^m{q}^m{[m+x]}_{q^{\alpha }}^n.\end{array}$$

(17)

Moreover, ${\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x)={\sum }_{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right){[x]}_{q^{\alpha }}^{n-l}{q}^{\alpha lx}{\tilde{\beta }}_{l,\xi ,q}^{(\alpha )}$.

By (7), we see that

(18)

Therefore, we obtain the following theorem.

Theorem 5 For , , and , we have

If we take $f(x)={\xi }^x{e}^{{[x]}_{q^{\alpha }}t}$, by (8), then we have

(19)

Therefore, by Theorem 5, we obtain the following theorem.

Theorem 6 For and , we have

$$\xi q{\tilde{\beta }}_{0,\xi ,q}^{(\alpha )}(1)-{\tilde{\beta }}_{0,\xi ,q}^{(\alpha )}=\{\begin{array}{cc}q-1\hfill & \mathit{\text{if}}n=0,\hfill \\ \frac{\alpha }{{[\alpha ]}_q}\hfill & \mathit{\text{if}}n=1,\hfill \\ 0\hfill & \mathit{\text{if}}n>1.\hfill \end{array}$$

Remark that when $n=0$, we have ${\tilde{\beta }}_{0,\xi ,q}^{(\alpha )}=\frac{q-1}{\xi q-1}$. By (16) and Theorem 6, we get

$$\begin{array}{rcl}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x)& =& \sum _{l=0}^n\left(\genfrac{}{}{0ex}{}{n}{l}\right){[x]}_{q^{\alpha }}^{n-l}{q}^{\alpha lx}{\tilde{\beta }}_{l,\xi ,q}^{(\alpha )}\\ =& {({[x]}_{q^{\alpha }}+q^{\alpha x}{\tilde{\beta }}_{\xi ,q}^{(\alpha )})}^n,\end{array}$$

(20)

with the usual convention about replacing ${({\tilde{\beta }}_{\xi ,q}^{(\alpha )})}^n$ by ${\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}$. By (20) and Theorem 6, we obtain the following theorem.

Theorem 7 For and , we have

$$\xi q{(q^{\alpha }{\tilde{\beta }}_{\xi ,q}^{(\alpha )}+1)}^n-{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}=\{\begin{array}{cc}q-1\hfill & \mathit{\text{if}}n=0,\hfill \\ \frac{\alpha }{{[\alpha ]}_q}\hfill & \mathit{\text{if}}n=1,\hfill \\ 0\hfill & \mathit{\text{if}}n>1,\hfill \end{array}$$

with the usual convention about replacing ${({\tilde{\beta }}_{\xi ,q}^{(\alpha )})}^n$ by ${\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}$.

From (8), we can easily derive the following equation (21). For , we get

(21)

Therefore, by (21), we obtain the following theorem.

Theorem 8 For , , and , we have

$${\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x)=\frac{{[d]}_{q^{\alpha }}^n}{{[d]}_q}\sum _{a=0}^{d-1}{q}^a{\tilde{\beta }}_{n,\xi ,q^d}^{(\alpha )}\left(\frac{x+a}{d}\right).$$

From (9), we note that

(22)

Therefore, by (22), we obtain the following theorem.

Theorem 9 For and , we have

$${\tilde{\beta }}_{n,{\xi }^{-1},q^{-1}}^{(\alpha )}(1-x)={(-1)}^n{q}^{\alpha n}{\tilde{\beta }}_{n,\xi ,q}^{(\alpha )}(x).$$

It is easy to show that

(23)

By (22) and (23), we obtain the following corollary.

Corollary 10 For and , we have