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# Some identities on Bernoulli and Euler polynomials arising from orthogonality of Legendre polynomials

Journal of Inequalities and Applications20122012:227

https://doi.org/10.1186/1029-242X-2012-227

• Received: 26 June 2012
• Accepted: 25 September 2012
• Published:

## Abstract

The purpose of this paper is to investigate some interesting identities on the Bernoulli and Euler polynomials arising from the orthogonality of Legendre polynomials in the inner product space ${\mathbb{P}}_{n}$.

## Keywords

• Differential Equation
• Generate Function
• Gamma Function
• Product Space
• Legendre Polynomial

## 1 Introduction

As is well known, the Legendre polynomial ${P}_{n}\left(x\right)$ is a solutions of the following differential equation:
$\left(1-{x}^{2}\right){u}^{″}-2x{u}^{\prime }+n\left(n+1\right)u=0\phantom{\rule{1em}{0ex}}\left(\text{see [1–7]}\right),$

where $n=0,1,2,\dots$ .

It is a polynomial of degree n. If n is even or odd, then ${P}_{n}\left(x\right)$ is accordingly even or odd. They are determined up to constant and normalized so that ${P}_{n}\left(1\right)=1$.

Rodrigues’ formula is given by
${P}_{n}\left(x\right)=\frac{1}{{2}^{n}n!}\left\{{\left(\frac{d}{dx}\right)}^{n}{\left({x}^{2}-1\right)}^{n}\right\},\phantom{\rule{1em}{0ex}}n\in {\mathbb{Z}}_{+}.$
(1.1)
Integrating by parts, we can derive
${\int }_{-1}^{1}{P}_{m}\left(x\right){P}_{n}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=\frac{2}{2n+1}{\delta }_{m,n}\phantom{\rule{1em}{0ex}}\left(\text{see [1–7]}\right),$
(1.2)

where ${\delta }_{m,n}$ is the Kronecker symbol.

By (1.1), we get
${P}_{n}\left(x\right)=\frac{1}{{2}^{n}}\sum _{k=0}^{\left[\frac{n}{2}\right]}{\left(-1\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{2n-2k}{n}\right){x}^{n-2k}.$
(1.3)
The generating function is given by
${\left(1-2xt+{t}^{2}\right)}^{-\frac{1}{2}}=\sum _{n=0}^{\mathrm{\infty }}{P}_{n}\left(x\right){t}^{n}.$
(1.4)
The Bernoulli polynomial is defined by a generating function to be
$\frac{t}{{e}^{t}-1}{e}^{xt}={e}^{B\left(x\right)t}=\sum _{n=0}^{\mathrm{\infty }}{B}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [8–13]}\right)$
(1.5)

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 Bernoulli numbers.

From (1.5), we have
${B}_{n}\left(x\right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){B}_{n-l}{x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [10–26]}\right).$
(1.6)
As is well known, the Euler numbers are defined by
${E}_{0}=1,\phantom{\rule{2em}{0ex}}{\left(E+1\right)}^{n}+{E}_{n}=2{\delta }_{0,n}\phantom{\rule{1em}{0ex}}\left(\text{see [10–13]}\right)$
(1.7)

with the usual convention about replacing ${E}^{n}$ by ${E}_{n}$.

The Euler polynomials are defined as
${E}_{n}\left(x\right)=\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n}{l}\right){E}_{n-l}{x}^{l}\phantom{\rule{1em}{0ex}}\left(\text{see [27–31]}\right).$
(1.8)
Let ${\mathbb{P}}_{n}=\left\{p\left(x\right)\in \mathbb{O}\left[x\right]|degp\left(x\right)\le n\right\}$. Then ${\mathbb{P}}_{n}$ is an inner product space with respect to the inner product $〈\cdot ,\cdot 〉$ with
$〈{q}_{1}\left(x\right),{q}_{2}\left(x\right)〉={\int }_{-1}^{1}{q}_{1}\left(x\right){q}_{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,$

where ${q}_{1}\left(x\right),{q}_{2}\left(x\right)\in {\mathbb{P}}_{n}$.

From (1.2), we can show that $\left\{{P}_{0}\left(x\right),{P}_{1}\left(x\right),\dots ,{P}_{n}\left(x\right)\right\}$ is an orthogonal basis for ${\mathbb{P}}_{n}$. In this paper, we derive some interesting identities on the Bernoulli and Euler polynomials from the orthogonality of Legendre polynomials in ${\mathbb{P}}_{n}$.

## 2 Some identities on the Bernoulli and Euler polynomials

For $q\left(x\right)\in {\mathbb{P}}_{n}$, let
$q\left(x\right)=\sum _{k=0}^{n}{C}_{k}{P}_{k}\left(x\right).$
(2.1)
Then, from (1.2), we have
$\begin{array}{rcl}〈q\left(x\right),{P}_{k}\left(x\right)〉& =& {C}_{k}〈{P}_{k}\left(x\right),{P}_{k}\left(x\right)〉\\ =& {C}_{k}{\int }_{-1}^{1}{\left\{{P}_{k}\left(x\right)\right\}}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{2}{2k+1}{C}_{k}.\end{array}$
(2.2)
By (2.2), we get
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{2}〈q\left(x\right),{P}_{k}\left(x\right)〉=\frac{2k+1}{2}{\int }_{-1}^{1}{P}_{k}\left(x\right)q\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \left(\frac{2k+1}{2}\right)\frac{1}{{2}^{k}k!}{\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right)q\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \left(\frac{2k+1}{{2}^{k+1}k!}\right){\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right)q\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.3)

Therefore, by (2.1) and (2.3), we obtain the following proposition.

Proposition 2.1 For$q\left(x\right)\in {\mathbb{P}}_{n}$, let
$q\left(x\right)=\sum _{k=0}^{n}{C}_{k}{P}_{k}\left(x\right).$
Then
${C}_{k}=\frac{2k+1}{{2}^{k+1}k!}{\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right)q\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$

Let us assume that $q\left(x\right)={x}^{n}\in {\mathbb{P}}_{n}$.

From Proposition 2.1, we have
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{{2}^{k+1}k!}{\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right){x}^{n}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{2k+1}{{2}^{k+1}}{\left(-1\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right){\int }_{-1}^{1}{\left({x}^{2}-1\right)}^{k}{x}^{n-k}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(1+{\left(-1\right)}^{n-k}\right){\int }_{0}^{1}{\left(1-{x}^{2}\right)}^{k}{x}^{n-k}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.4)
For $n-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)$, by (2.4), we get
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right){\int }_{0}^{1}{\left(1-y\right)}^{k}{y}^{\frac{n-k-1}{2}}\phantom{\rule{0.2em}{0ex}}dy\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)B\left(k+1,\frac{n-k+1}{2}\right)\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{\mathrm{\Gamma }\left(k+1\right)\mathrm{\Gamma }\left(\frac{n-k+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n+k+1}{2}+1\right)}\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{k!\mathrm{\Gamma }\left(\frac{n-k+1}{2}\right)}{\left(\frac{n+k+1}{2}\right)\left(\frac{n+k-1}{2}\right)\cdots \left(\frac{n-k+1}{2}\right)\mathrm{\Gamma }\left(\frac{n-k+1}{2}\right)}\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)k!{2}^{k+1}\frac{\left(n-k\right)!\left(n+k+2\right)\left(n+k\right)\cdots \left(n-k+2\right)}{\left(n+k+2\right)!}\\ =& \frac{\left(2k+1\right){2}^{k+1}}{\left(n+k+2\right)!}×\frac{n!\left(\frac{n+k+2}{2}\right)!}{\left(\frac{n-k}{2}\right)!}.\end{array}$
(2.5)
Here the beta function $B\left(x,y\right)$ is defined by
$B\left(x,y\right)={\int }_{0}^{1}{t}^{x-1}{\left(1-t\right)}^{y-1}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(Re\left(x\right),Re\left(y\right)>0\right),$
and it is well known that
$B\left(x,y\right)=\frac{\mathrm{\Gamma }\left(x\right)\mathrm{\Gamma }\left(y\right)}{\mathrm{\Gamma }\left(x+y\right)},$

where $\mathrm{\Gamma }\left(s\right)={\int }_{0}^{\mathrm{\infty }}{t}^{s-1}{e}^{-t}\phantom{\rule{0.2em}{0ex}}dt$ ($Re\left(s\right)>0$) is the gamma function.

By Proposition 2.1 and (2.5), we get
${x}^{n}=\sum _{0\le k\le n,n-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{\left(2k+1\right)n!{2}^{k+1}\left(\frac{n+k+2}{2}\right)!}{\left(n+k+2\right)!\left(\frac{n-k}{2}\right)!}{P}_{k}\left(x\right).$
(2.6)
From (1.5), we can easily derive the following equation (2.7):
${x}^{n}=\frac{1}{n+1}\sum _{l=0}^{n}\left(\genfrac{}{}{0}{}{n+1}{l}\right){B}_{l}\left(x\right)\phantom{\rule{1em}{0ex}}\left(n\in {\mathbb{Z}}_{+}\right).$
(2.7)

Therefore, by (2.6) and (2.7), we obtain the following Proposition 2.2.

Proposition 2.2 For$n\in {\mathbb{Z}}_{+}$, we have
$\sum _{l=0}^{n}\frac{{B}_{l}\left(x\right)}{\left(n+1-l\right)!l!}=\sum _{0\le k\le n,n-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{\left(2k+1\right){2}^{k+1}\left(\frac{n+k+2}{2}\right)!}{\left(n+k+2\right)!\left(\frac{n-k}{2}\right)!}{P}_{k}\left(x\right).$
Let us take $q\left(x\right)={B}_{n}\left(x\right)\in {\mathbb{P}}_{n}$. By Proposition 2.1, we get
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{{2}^{k+1}k!}{\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right){B}_{n}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}\left(2k+1\right)}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right){\int }_{-1}^{1}{\left({x}^{2}-1\right)}^{k}{B}_{n-k}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{{\left(-1\right)}^{k}\left(2k+1\right)}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\sum _{l=0}^{n-k}\left(\genfrac{}{}{0}{}{n-k}{l}\right){B}_{n-k-l}{\int }_{-1}^{1}{\left({x}^{2}-1\right)}^{k}{x}^{l}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\sum _{l=0}^{n-k}\left(\genfrac{}{}{0}{}{n-k}{l}\right){B}_{n-k-l}\left(1+{\left(-1\right)}^{l}\right){\int }_{0}^{1}{\left(1-{x}^{2}\right)}^{k}{x}^{l}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.8)
For $l\in {\mathbb{Z}}_{+}$ with $l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)$, we have
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\sum _{0\le l\le n-k,l\phantom{\rule{0.25em}{0ex}}\mathrm{is}\phantom{\rule{0.25em}{0ex}}\mathrm{even}}\left(\genfrac{}{}{0}{}{n-k}{l}\right){B}_{n-k-l}{\int }_{0}^{1}{\left(1-y\right)}^{k}{y}^{\frac{l-1}{2}}\phantom{\rule{0.2em}{0ex}}dy\\ =& \frac{2k+1}{{2}^{k+1}}\left(\genfrac{}{}{0}{}{n}{k}\right)\sum _{0\le l\le n-k,l\phantom{\rule{0.25em}{0ex}}\mathrm{is}\phantom{\rule{0.25em}{0ex}}\mathrm{even}}\left(\genfrac{}{}{0}{}{n-k}{l}\right){B}_{n-k-l}\frac{\mathrm{\Gamma }\left(k+1\right)\mathrm{\Gamma }\left(\frac{l+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{2k+l+1}{2}+1\right)}\\ =& \left(2k+1\right){2}^{k+1}n!\sum _{0\le l\le n-k,l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{{B}_{n-k-l}}{\left(n-k-l\right)!}×\frac{\left(\frac{2k+l+2}{2}\right)!}{\left(2k+l+2\right)!\left(\frac{l}{2}\right)!}.\end{array}$
(2.9)
In , we showed that
${B}_{n}\left(x\right)=\sum _{k=0}^{n-2}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{n-k}{E}_{k}\left(x\right)+{E}_{n}\left(x\right)=\sum _{k=0,k\ne n-1}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{n-k}{E}_{k}\left(x\right).$
(2.10)

Therefore, by Proposition 2.1, (2.9) and (2.10), we obtain the following theorem.

Theorem 2.3 For$n\in {\mathbb{Z}}_{+}$, we have
$\frac{1}{n!}\sum _{k=0,k\ne n-1}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right){B}_{n-k}{E}_{k}\left(x\right)=\sum _{k=0}^{n}\left(\sum _{0\le l\le n-k,l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{\left(2k+1\right){2}^{k+1}\left(\frac{l+2k+2}{2}\right)!{B}_{n-k-l}}{\left(n-k-l\right)!\left(l+2k+2\right)!\left(\frac{l}{2}\right)!}\right){P}_{k}\left(x\right).$
By the same method of Theorem 2.3, we easily see that
$\frac{{E}_{n}\left(x\right)}{n!}=\sum _{k=0}^{n}\left(\sum _{0\le l\le n-k,l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{\left(2k+1\right){2}^{k+1}\left(\frac{l+2k+2}{2}\right)!{B}_{n-k-l}}{\left(n-k-l\right)!\left(l+2k+2\right)!\left(\frac{l}{2}\right)!}\right){P}_{k}\left(x\right).$
(2.11)
Let us take $q\left(x\right)={\sum }_{k=0}^{n}{B}_{k}\left(x\right){B}_{n-k}\left(x\right)\in {\mathbb{P}}_{n}$. Then we see that

The equation (2.12) was proved in .

By (2.12) and Proposition 2.2, we have
$\begin{array}{rcl}{C}_{k}& =& \frac{2k+1}{{2}^{k+1}k!}\left\{\left(n+1\right)\sum _{l=0}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{l}\right)}{n-l+1}\left(\sum _{m=l}^{n}{B}_{m-l}{B}_{n-m}+{B}_{n-1-l}\right)\\ ×{\int }_{-1}^{1}{E}_{l}\left(x\right)\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right)\phantom{\rule{0.2em}{0ex}}dx\\ +\frac{\left({n}^{2}-1\right)n}{12}{\int }_{-1}^{1}\left(\frac{{d}^{k}}{d{x}^{k}}{\left({x}^{2}-1\right)}^{k}\right){E}_{n-2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}.\end{array}$
(2.13)
Therefore, by (2.13), (2.14), (2.15) and (2.16), we get
$\begin{array}{rcl}{C}_{k}& =& \left(2k+1\right){2}^{k+1}\left\{\left(n+1\right)\sum _{l=k}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{k}\right)}{n-l+1}\left(\sum _{m=l}^{n}{B}_{m-l}{B}_{n-m}+{B}_{n-1-l}\right)\\ ×\sum _{k\le j\le l,j-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\left(\genfrac{}{}{0}{}{l}{j}\right){E}_{l-j}\frac{j!}{\left(j+k+2\right)!}×\frac{\left(\frac{j+k+2}{2}\right)!}{\left(\frac{j-k}{2}\right)!}\\ +\frac{\left({n}^{2}-1\right)n}{12}\sum _{k\le j\le n-2,j-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\left(\genfrac{}{}{0}{}{n-2}{j}\right){E}_{n-2-j}\frac{j!}{\left(j+k+2\right)!}×\frac{\left(\frac{j+k+2}{2}\right)!}{\left(\frac{j-k}{2}\right)!}\right\}.\end{array}$
(2.17)

Therefore, by Proposition 2.1 and (2.17), we obtain the following theorem.

Theorem 2.4 For$n\in {\mathbb{Z}}_{+}$, we have
$\begin{array}{c}\sum _{k=0}^{n}{B}_{k}\left(x\right){B}_{n-k}\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{k=0}^{n}\left(2k+1\right){2}^{k+1}\left\{\left(n+1\right)\sum _{l=k}^{n}\frac{\left(\genfrac{}{}{0}{}{n}{k}\right)}{n-l+1}\left(\sum _{m=l}^{n}{B}_{m-l}{B}_{n-m}+{B}_{n-1-l}\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\sum _{k\le j\le l,j-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\left(\genfrac{}{}{0}{}{l}{j}\right){E}_{l-j}\frac{j!}{\left(j+k+2\right)!}×\frac{\left(\frac{j+k+2}{2}\right)!}{\left(\frac{j-k}{2}\right)!}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{\left({n}^{2}-1\right)n}{12}\sum _{k\le j\le n-2,j-k\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\left(\genfrac{}{}{0}{}{n-2}{j}\right){E}_{n-2-j}\frac{j!}{\left(j+k+2\right)!}×\frac{\left(\frac{j+k+2}{2}\right)!}{\left(\frac{j-k}{2}\right)!}\right\}{P}_{k}\left(x\right).\hfill \end{array}$
Remark 2.5 The extended Laguerre polynomials are given by
${L}_{n}^{\alpha }\left(x\right)=\sum _{r=0}^{n}\frac{{\left(-1\right)}^{r}}{r!}\left(\genfrac{}{}{0}{}{n+\alpha }{n-r}\right){x}^{r}\phantom{\rule{1em}{0ex}}\left(\alpha >-1\right).$
By the same method, we get
${L}_{n}^{\alpha }\left(x\right)=\sum _{k=0}^{n}\sum _{0\le l\le n-k,l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{{\left(-1\right)}^{k+l}\left(2k+1\right){2}^{k+1}\left(\genfrac{}{}{0}{}{n+\alpha }{n-k-l}\right)\left(\frac{l+2k+2}{2}\right)!}{\left(l+2k+2\right)!\left(\frac{l}{2}\right)!}{P}_{k}\left(x\right)$
and
${H}_{n}\left(x\right)=\sum _{k=0}^{n}\sum _{0\le l\le n-k,l\equiv 0\phantom{\rule{0.25em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.25em}{0ex}}2\right)}\frac{\left(2k+1\right){2}^{2k+l+1}n!\left(\frac{l+2k+2}{2}\right)!{H}_{n-k-l}}{\left(n-k-l\right)!\left(l+2k+2\right)!\left(\frac{l}{2}\right)!}{P}_{k}\left(x\right),$

where ${H}_{n}\left(x\right)$ is the Hermite polynomial of degree n (see ).

## Declarations

### Acknowledgements

The authors would like to express their sincere gratitude to referee for his/her valuable comments and information. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science, and Technology 2012R1A1A2003786.

## Authors’ Affiliations

(1)
Department of Mathematics, Sogang University, Seoul, 121-742, Republic of Korea
(2)
Department of Mathematics Education, Kyungpook National University, Taegu, 702-701, Republic of Korea
(3)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea

## References

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