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

Journal of Inequalities and Applications20122012:176

https://doi.org/10.1186/1029-242X-2012-176

• Received: 24 April 2012
• Accepted: 1 August 2012
• Published:

## 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.

## Keywords

• Bernoulli numbers and polynomials
• bosonic q-integral
• twisted q-Bernoulli numbers and polynomials with weight α

## 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 ${|p|}_{p}={p}^{{\nu }_{p}\left(p\right)}={p}^{-1}$. 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 ${q}^{x}=exp\left(xlogq\right)$ for ${|x|}_{p}\le 1$. The q-number is defined by ${\left[x\right]}_{q}=\frac{1-{q}^{x}}{1-q}$ (see ). Note that ${lim}_{q\to 1}{\left[x\right]}_{q}=x$.

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

where ${\parallel f\parallel }_{1}=sup\left\{{|f\left(0\right)|}_{p},{sup}_{x\ne y}|\frac{f\left(x\right)-f\left(y\right)}{x-y}{|}_{p}\right\}$.

In , Carlitz defined q-Bernoulli numbers which are called the Carlitz’s q-Bernoulli numbers, by
(3)
with the usual convention about replacing ${\left({\beta }_{q}^{\left(h\right)}\right)}^{n}$ by ${\beta }_{n,q}^{\left(h\right)}$. In , Carlitz also considered the expansion of q-Bernoulli numbers as follows:
(4)
with the usual convention about replacing ${\left({\beta }_{q}^{\left(h\right)}\right)}^{n}$ by ${\beta }_{n,q}^{\left(h\right)}$. In , for and , q-Bernoulli numbers with weight α is defined by Kim as follows:
(5)

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

Let ${C}_{{p}^{n}}=\left\{\xi |{\xi }^{{p}^{n}}=1\right\}$ 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, 2024]). 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 ${\left({B}_{\xi }\right)}^{n}$ by ${B}_{n,q}$ (see [15, 1924]).

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 ${f}_{n}\left(x\right)=f\left(x+n\right)$. In [, Theorem 1], Kim proved the following integral equation:
${q}^{n}{I}_{q}\left({f}_{n}\right)-{I}_{q}\left(f\right)=\left(q-1\right)\sum _{l=0}^{n-1}{q}^{l}f\left(l\right)+\frac{q-1}{logq}\sum _{l=0}^{n-1}{q}^{l}{f}^{\prime }\left(l\right).$
(7)
In particular, when $n=1$, we have
$q{I}_{q}\left({f}_{1}\right)-{I}_{q}\left(f\right)=\left(q-1\right)f\left(0\right)+\frac{q-1}{logq}{f}^{\prime }\left(0\right).$
(8)
Let and . The n th q-Bernoulli polynomials with weight α are defined by
Then, by using the bosonic p-adic q-integral on , we evaluate the above equation (9) as follows:

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

From (11), when $x=0$, we get
${\stackrel{˜}{\beta }}_{n,\xi ,q}^{\alpha }=\frac{-n\alpha }{{\left[\alpha \right]}_{q}}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m\alpha +m}{\left[m\right]}_{{q}^{\alpha }}^{n-1}+\left(1-q\right)\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m}{\left[m\right]}_{{q}^{\alpha }}^{n}.$
(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
$-\frac{{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}}{n}=\frac{\alpha }{{\left[\alpha \right]}_{q}}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m\alpha +m}{\left[m\right]}_{{q}^{\alpha }}^{n-1}+\frac{1-q}{n}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m}{\left[m\right]}_{{q}^{\alpha }}^{n}.$
For $\alpha =1$, we note that ${\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(1\right)}\left(x\right)$ are the twisted Carlitz’s q-Bernoulli polynomials and ${\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(1\right)}$ are the twisted Carlitz’s q-Bernoulli numbers. By Corollary 2, we easily get
From (13), we have
$\sum _{n=0}^{\mathrm{\infty }}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\frac{{t}^{n}}{n!}=-t\frac{\alpha }{{\left[\alpha \right]}_{q}}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{\left(m+1\right)\alpha +m}{e}^{{\left[m\right]}_{{q}^{\alpha }}t}+\left(1-q\right)\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m}{e}^{{\left[m\right]}_{{q}^{\alpha }}t}.$
(14)

Therefore, we obtain the following corollary.

Corollary 3 Let and ${F}_{\xi ,q}^{\left(\alpha \right)}\left(t\right)={\sum }_{n=0}^{\mathrm{\infty }}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\frac{{t}^{n}}{n!}$. Then we have
${F}_{\xi ,q}^{\left(\alpha \right)}\left(t\right)=-t\frac{\alpha }{{\left[\alpha \right]}_{q}}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{\left(m+1\right)\alpha +m}{e}^{{\left[m\right]}_{{q}^{\alpha }}t}+\left(1-q\right)\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m}{e}^{{\left[m\right]}_{{q}^{\alpha }}t}.$
Let ${F}_{\xi ,q}^{\left(\alpha \right)}\left(t,x\right)={\sum }_{n=0}^{\mathrm{\infty }}{\sum }_{n=0}^{\mathrm{\infty }}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\left(x\right)\frac{{t}^{n}}{n!}$. From Theorem 1, we have

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

Theorem 4 For and , we have
$\begin{array}{rcl}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\left(x\right)& =& \frac{1-q}{{\left(1-{q}^{\alpha }\right)}^{n}}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){\left(-1\right)}^{l}{q}^{\alpha lx}\left(\frac{\alpha l+1}{1-\xi {q}^{\alpha l+1}}\right)\\ =& -n\frac{\alpha }{{\left[\alpha \right]}_{q}}\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{\left(m+1\right)\alpha +m}{\left[m+x\right]}_{{q}^{\alpha }}^{n-1}\\ +\left(1-q\right)\sum _{m=0}^{\mathrm{\infty }}{\xi }^{m}{q}^{m}{\left[m+x\right]}_{{q}^{\alpha }}^{n}.\end{array}$
(17)

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

Therefore, we obtain the following theorem.

Theorem 5 For , , and , we have
If we take $f\left(x\right)={\xi }^{x}{e}^{{\left[x\right]}_{{q}^{\alpha }}t}$, by (8), then we have

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

Theorem 6 For and , we have
$\xi q{\stackrel{˜}{\beta }}_{0,\xi ,q}^{\left(\alpha \right)}\left(1\right)-{\stackrel{˜}{\beta }}_{0,\xi ,q}^{\left(\alpha \right)}=\left\{\begin{array}{cc}q-1\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n=0,\hfill \\ \frac{\alpha }{{\left[\alpha \right]}_{q}}\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n=1,\hfill \\ 0\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n>1.\hfill \end{array}$
Remark that when $n=0$, we have ${\stackrel{˜}{\beta }}_{0,\xi ,q}^{\left(\alpha \right)}=\frac{q-1}{\xi q-1}$. By (16) and Theorem 6, we get
$\begin{array}{rcl}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\left(x\right)& =& \sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){\left[x\right]}_{{q}^{\alpha }}^{n-l}{q}^{\alpha lx}{\stackrel{˜}{\beta }}_{l,\xi ,q}^{\left(\alpha \right)}\\ =& {\left({\left[x\right]}_{{q}^{\alpha }}+{q}^{\alpha x}{\stackrel{˜}{\beta }}_{\xi ,q}^{\left(\alpha \right)}\right)}^{n},\end{array}$
(20)

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

Theorem 7 For and , we have
$\xi q{\left({q}^{\alpha }{\stackrel{˜}{\beta }}_{\xi ,q}^{\left(\alpha \right)}+1\right)}^{n}-{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}=\left\{\begin{array}{cc}q-1\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n=0,\hfill \\ \frac{\alpha }{{\left[\alpha \right]}_{q}}\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n=1,\hfill \\ 0\hfill & \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}n>1,\hfill \end{array}$

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

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

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

Theorem 8 For , , and , we have
${\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\left(x\right)=\frac{{\left[d\right]}_{{q}^{\alpha }}^{n}}{{\left[d\right]}_{q}}\sum _{a=0}^{d-1}{q}^{a}{\stackrel{˜}{\beta }}_{n,\xi ,{q}^{d}}^{\left(\alpha \right)}\left(\frac{x+a}{d}\right).$

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

Theorem 9 For and , we have
${\stackrel{˜}{\beta }}_{n,{\xi }^{-1},{q}^{-1}}^{\left(\alpha \right)}\left(1-x\right)={\left(-1\right)}^{n}{q}^{\alpha n}{\stackrel{˜}{\beta }}_{n,\xi ,q}^{\left(\alpha \right)}\left(x\right).$

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

Corollary 10 For and , we have

## Declarations

### Acknowledgement

This paper was supported by Kon-Kuk University in 2012.

## Authors’ Affiliations

(1)
Department of Mathematics and Computer Science, Kon-Kuk University, Chungju, 138-701, Korea

## References

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