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# Some identities of Genocchi polynomials arising from Genocchi basis

Journal of Inequalities and Applications20132013:43

https://doi.org/10.1186/1029-242X-2013-43

• Received: 21 December 2012
• Accepted: 13 January 2013
• Published:

## Abstract

In this paper, we give some interesting identities which are derived from the basis of Genocchi. From our methods which are treated in this paper, we can derive some new identities associated with Bernoulli and Euler polynomials.

MSC:11B68, 11S80.

## Keywords

• Vector Space
• Dimensional Vector
• Good Basis
• Natural Basis
• Euler Number

## 1 Introduction

As is well known, the Genocchi polynomials are defined by the generating function to be
$\frac{2t}{{e}^{t}+1}{e}^{xt}={e}^{G\left(x\right)t}=\sum _{n=0}^{\mathrm{\infty }}{G}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1–9]}\right)$
(1)

with the usual convention about replacing ${G}^{n}\left(x\right)$ by ${G}_{n}\left(x\right)$.

In the special case $x=0$, ${G}_{n}\left(0\right)={G}_{n}$ are called the n th Genocchi numbers. From (1), we note that
${G}_{0}=0,\phantom{\rule{2em}{0ex}}{G}_{n}\left(1\right)+{G}_{n}=2{\delta }_{n,1}\phantom{\rule{1em}{0ex}}\left(\text{see [10–16]}\right),$
(2)
where ${\delta }_{n,k}$ is the Kronecker symbol.
${G}_{n}\left(x\right)={\left(G+x\right)}^{n}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){G}_{l}{x}^{n-l}\phantom{\rule{1em}{0ex}}\left(\text{see [6–8, 17]}\right).$
(3)
Thus, by (2) and (3), we see that
$\frac{d}{dx}{G}_{n}\left(x\right)=n{G}_{n-1}\left(x\right),\phantom{\rule{2em}{0ex}}deg{G}_{n}\left(x\right)=n-1.$
(4)
The n th Bernoulli polynomials are also defined by the generating function to be
$\frac{t}{{e}^{t}-1}{e}^{xt}={e}^{B\left(x\right)t}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [14–16]}\right)$
(5)

with the usual convention about replacing ${B}^{n}\left(x\right)$ by ${B}_{n}\left(x\right)$.

In the special case $x=0$, ${B}_{n}\left(0\right)={B}_{n}$ are called the n th Bernoulli numbers. By (5), we get
${B}_{0}=1,\phantom{\rule{2em}{0ex}}{B}_{n}\left(1\right)-{B}_{n}={\delta }_{1,n}\phantom{\rule{1em}{0ex}}\left(\text{see [8, 9, 17]}\right)$
(6)
and
${B}_{n}\left(x\right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{l}{x}^{n-l}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{x}^{l}.$
(7)
The Euler numbers are defined by
${E}_{0}=1,\phantom{\rule{2em}{0ex}}{\left(E+1\right)}^{n}+{E}_{n}=2{\delta }_{0,n}.$
(8)
The Euler polynomials are defined by
${E}_{n}\left(x\right)={\left(E+x\right)}^{n}=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){E}_{n-l}{x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [7–13, 17]}\right).$
(9)

Let ${\mathbb{P}}_{n}=\left\{p\left(x\right)\in \mathbb{Q}\left[x\right]\mid degp\left(x\right)\le n\right\}$ be the $\left(n+1\right)$-dimensional vector space over . Probably, $\left\{1,x,\dots ,{x}^{n}\right\}$ is the most natural basis for ${\mathbb{P}}_{n}$. But $\left\{{G}_{1}\left(x\right),{G}_{2}\left(x\right),\dots ,{G}_{n+1}\left(x\right)\right\}$ is also a good basis for the space ${\mathbb{P}}_{n}$ for our purpose of arithmetical applications of Genocchi polynomials. Let $p\left(x\right)\in {\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be expressed by $p\left(x\right)={\sum }_{1\le k\le n+1}{b}_{k}{G}_{k}\left(x\right)$.

In this paper, we develop some new methods to obtain some new identities and properties of Genocchi polynomials which are derived from the Genocchi basis.

## 2 Genocchi basis and some identities of Genocchi polynomials

Let us take $p\left(x\right)\in {\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be expressed as a -linear combination of ${G}_{1}\left(x\right),{G}_{2}\left(x\right),\dots ,{G}_{n+1}\left(x\right)$ as follows:
$p\left(x\right)=\sum _{1\le k\le n+1}{b}_{k}{G}_{k}\left(x\right)={b}_{1}{G}_{1}\left(x\right)+{b}_{2}{G}_{2}\left(x\right)+\cdots +{b}_{n+1}{G}_{n+1}\left(x\right).$
(10)
Now, let us define the operator $\stackrel{˜}{\mathrm{△}}$ by
$\stackrel{˜}{\mathrm{△}}p\left(x\right)=p\left(x+1\right)+p\left(x\right).$
(11)
Then, by (10) and (11), we set
$g\left(x\right)=\stackrel{˜}{\mathrm{△}}p\left(x\right)=\sum _{1\le k\le n+1}{b}_{k}\left({G}_{k}\left(x+1\right)+{G}_{k}\left(x\right)\right).$
(12)
From (1), we note that
$\sum _{n=0}^{\mathrm{\infty }}\left\{{G}_{n}\left(x+1\right)+{G}_{n}\left(x\right)\right\}\frac{{t}^{n}}{n!}=\frac{2t}{{e}^{t}+1}{e}^{\left(x+1\right)t}+\frac{2t}{{e}^{t}+1}{e}^{xt}.$
(13)
By (2), (3) and (13), we get
$\frac{{G}_{n+1}\left(x+1\right)+{G}_{n+1}\left(x\right)}{n+1}=2{x}^{n}.$
(14)
From (12) and (14), we get
$g\left(x\right)=\stackrel{˜}{\mathrm{△}}p\left(x\right)=2\sum _{1\le k\le n+1}k{b}_{k}{x}^{k-1}.$
(15)
For $r\in \mathbb{N}$, let us take the r th derivative of $g\left(x\right)$ in (15) as follows:
${g}^{\left(r\right)}\left(x\right)=\frac{{d}^{r}g\left(x\right)}{d{x}^{r}}=2\sum _{1\le k\le n+1}k\left(k-1\right)\cdots \left(k-1-r+1\right){b}_{k}{x}^{k-1-r}.$
(16)
Thus, by (16), we get
${g}^{\left(r\right)}\left(0\right)=\frac{{d}^{r}g\left(x\right)}{d{x}^{r}}{|}_{x=0}=2\left(r+1\right)!{b}_{r+1}.$
(17)
From (11) and (17), we have
${b}_{r+1}=\frac{1}{2\left(r+1\right)!}\left\{{p}^{\left(r\right)}\left(1\right)+{p}^{\left(r\right)}\left(0\right)\right\},$
(18)

where ${p}^{\left(r\right)}\left(a\right)=\frac{{d}^{r}g\left(x\right)}{d{x}^{r}}{|}_{x=a}$.

Therefore, by (10) and (18), we obtain the following theorem.

Theorem 1 For $n\in \mathbb{N}$, let $p\left(x\right)\in {\mathbb{P}}_{n}$ with $p\left(x\right)={\sum }_{1\le k\le n+1}{b}_{k}{G}_{k}\left(x\right)$.

Then we have
${b}_{k}=\frac{1}{2k!}\left({p}^{\left(k-1\right)}\left(1\right)+{p}^{\left(k-1\right)}\left(0\right)\right).$
Let us assume that $p\left(x\right)={B}_{n}\left(x\right)$. Then by Theorem 1, we get
${B}_{n}\left(x\right)=\sum _{1\le k\le n+1}{b}_{k}{G}_{k}\left(x\right),$
(19)
where
${b}_{k}=\frac{1}{2k!}\left\{{p}^{\left(k-1\right)}\left(1\right)+{p}^{\left(k-1\right)}\left(0\right)\right\}=\frac{1}{2k!}{\left(n\right)}_{k-1}\left\{{B}_{n-k+1}\left(1\right)+{B}_{n-k+1}\right\}.$
(20)
From (6) and (20), we have
${b}_{k}=\frac{1}{2\left(n+1\right)}\left(\genfrac{}{}{0}{}{n+1}{k}\right)\left\{{\delta }_{n,k}+2{B}_{n-k+1}\right\}.$
(21)
By (19) and (21), we get
$\begin{array}{rl}{B}_{n}\left(x\right)& =\frac{1}{n+1}\sum _{1\le k\le n-1}\left(\genfrac{}{}{0}{}{n+1}{k}\right){B}_{n-k+1}{G}_{k}\left(x\right)+\frac{1}{2}\left(1+2{B}_{1}\right){G}_{n}\left(x\right)+\frac{1}{2\left(n+1\right)}2{G}_{n+1}\left(x\right)\\ =\frac{1}{n+1}\sum _{1\le k\le n-1}\left(\genfrac{}{}{0}{}{n+1}{k}\right){B}_{n-k+1}{G}_{k}\left(x\right)+\frac{1}{n+1}{G}_{n+1}\left(x\right).\end{array}$
(22)

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

Theorem 2 For $n\in \mathbb{N}$, we have
${B}_{n}\left(x\right)=\frac{1}{n+1}\sum _{1\le k\le n-1}\left(\genfrac{}{}{0}{}{n+1}{k}\right){B}_{n-k+1}{G}_{k}\left(x\right)+\frac{1}{n+1}{G}_{n+1}\left(x\right).$
In particular, if we take $p\left(x\right)={E}_{n}\left(x\right)\in {\mathbb{P}}_{n}$, then we have
${E}_{n}\left(x\right)=\sum _{1\le k\le n+1}{b}_{k}{G}_{k}\left(x\right),$
(23)
where
${b}_{k}=\frac{1}{2k!}\left\{{p}^{\left(k-1\right)}\left(1\right)+{p}^{\left(k-1\right)}\left(0\right)\right\}=\frac{1}{2k!}{\left(n\right)}_{k-1}\left\{{E}_{n-k+1}\left(1\right)+{E}_{n-k+1}\right\}.$
(24)
By (8) and (24), we get
$\begin{array}{rl}{b}_{k}& =\frac{1}{2\left(n+1\right)}\left(\genfrac{}{}{0}{}{n+1}{k}\right)\left\{2{\delta }_{n-k+1,0}-{E}_{n-k+1}+{E}_{n-k+1}\right\}\\ =\frac{1}{n+1}\left(\genfrac{}{}{0}{}{n+1}{k}\right){\delta }_{n+1,k}.\end{array}$
(25)
From (23) and (25), we have
${E}_{n}\left(x\right)=\frac{1}{n+1}{G}_{n+1}\left(x\right).$
Let us take $p\left(x\right)\in {\mathbb{P}}_{n}$ with
$p\left(x\right)=\sum _{0\le k\le n}{B}_{k}\left(x\right){B}_{n-k}\left(x\right).$
(26)
Continuing this process, we get
$\begin{array}{rl}\frac{{d}^{k}p\left(x\right)}{d{x}^{k}}& ={p}^{\left(k\right)}\left(x\right)=\left(n+1\right)n\cdots \left(n+1-k+1\right)\sum _{l=k}^{n}{B}_{l-k}\left(x\right){B}_{n-l}\left(x\right)\\ =\frac{\left(n+1\right)!}{\left(n+1-k\right)!}\sum _{l=k}^{n}{B}_{l-k}\left(x\right){B}_{n-l}\left(x\right).\end{array}$
(27)
From (27), we have
${p}^{\left(k-1\right)}\left(1\right)=\frac{\left(n+1\right)!}{\left(n+2-k\right)!}\sum _{l=k-1}^{n}{B}_{l+1-k}\left(1\right){B}_{n-l}\left(1\right).$
(28)
By (6), we get
$\begin{array}{rl}{B}_{l+1-k}\left(1\right){B}_{n-l}\left(1\right)& =\left({\delta }_{l+1-k,1}+{B}_{l+1-k}\right)\left({\delta }_{n-l,1}+{B}_{n-l}\right)\\ =\left\{{\delta }_{k,n-1}+{B}_{n-k}+{B}_{n-k}+{B}_{l+1-k}{B}_{n-l}\right\}.\end{array}$
(29)
From (28) and (29), we have
${p}^{\left(k-1\right)}\left(1\right)=\frac{\left(n+1\right)!}{\left(n+2-k\right)!}\left\{{\delta }_{k,n-1}+2{B}_{n-k}+\sum _{k-1\le l\le n}{B}_{l+1-k}{B}_{n-l}\right\}.$
(30)
By Theorem 1, $p\left(x\right)={\sum }_{0\le k\le n}{B}_{k}\left(x\right){B}_{n-k}\left(x\right)$ can be expressed by
$p\left(x\right)=\sum _{1\le k\le n+1}{b}_{k}\left(x\right){G}_{k}\left(x\right),$
(31)
where
$\begin{array}{rl}{b}_{k}& =\frac{1}{2k!}\left\{{p}^{\left(k-1\right)}\left(1\right)+{p}^{\left(k-1\right)}\left(0\right)\right\}\\ =\frac{\left(n+1\right)!}{2k!\left(n+2-k\right)!}\left\{{\delta }_{k,n-1}+2{B}_{n-k}+2\sum _{l=k-1}{B}_{l+1-k}{B}_{n-l}\right\}.\end{array}$
(32)
Thus, by (31) and (32), we get
$\begin{array}{rcl}p\left(x\right)& =& \frac{n\left(n+1\right)}{12}{G}_{n-1}\left(x\right)+\sum _{1\le k\le n+1}\frac{1}{k}\left(\genfrac{}{}{0}{}{n+1}{k-1}\right){B}_{n-k}{G}_{k}\left(x\right)\\ +\sum _{1\le k\le n+1}\frac{1}{k}\left(\genfrac{}{}{0}{}{n+1}{k-1}\right)\sum _{l=k-1}^{n}{B}_{l+1-k}{B}_{n-l}{G}_{k}\left(x\right).\end{array}$
(33)

Therefore, by (31) and (33), we obtain the following theorem.

Theorem 3 For $n\in \mathbb{N}$, we have
$\begin{array}{rcl}\sum _{k=0}^{n}{B}_{k}\left(x\right){B}_{n-k}\left(x\right)& =& \frac{n\left(n+1\right)}{12}{G}_{n-1}\left(x\right)+\sum _{1\le k\le n+1}\frac{1}{k}\left(\genfrac{}{}{0}{}{n+1}{k-1}\right){B}_{n-k}{G}_{k}\left(x\right)\\ +\sum _{1\le k\le n+1}\left(\sum _{k-1\le l\le n}\frac{1}{k}\left(\genfrac{}{}{0}{}{n+1}{k-1}\right){B}_{l+1-k}{B}_{n-l}\right){G}_{k}\left(x\right).\end{array}$

## Declarations

### Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to express their gratitude for the valuable comments and suggestions of referees. This research was supported by Kwangwoon University in 2013.

## Authors’ Affiliations

(1)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, South Korea
(2)
Department of Mathematics Education, Kyungpook National University, Daegu, 702-701, South Korea
(3)
Hanrimwon, Kwangwoon University, Seoul, 139-701, South Korea
(4)
Division of General Education, Kwangwoon University, Seoul, 139-701, South Korea

## References 