Theory and Modern Applications

# Higher-order Bernoulli, Euler and Hermite polynomials

## Abstract

In (Kim and Kim in J. Inequal. Appl. 2013:111, 2013; Kim and Kim in Integral Transforms Spec. Funct., 2013, doi:10.1080/10652469.2012.754756), we have investigated some properties of higher-order Bernoulli and Euler polynomial bases in ${\mathbb{P}}_{n}=\left\{p\left(x\right)âˆˆ\mathbb{Q}\left[x\right]|degp\left(x\right)â‰¤n\right\}$. In this paper, we derive some interesting identities of higher-order Bernoulli and Euler polynomials arising from the properties of those bases for ${\mathbb{P}}_{n}$.

## 1 Introduction

For $râˆˆ\mathbb{R}$, let us define the Bernoulli polynomials of order r as follows:

${\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{r}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [1â€“18]).}$
(1)

In the special case, $x=0$, ${B}_{n}^{\left(r\right)}\left(0\right)={B}_{n}^{\left(r\right)}$ are called the n th Bernoulli numbers of order r. As is well known, the Euler polynomials of order r are defined by the generating function to be

${\left(\frac{2}{{e}^{t}+1}\right)}^{r}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{E}_{n}^{\left(r\right)}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [1â€“10]).}$
(2)

For , the Frobenius-Euler polynomials of order r are also given by

${\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right)}^{r}{e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [1, 7]).}$
(3)

The Hermite polynomials are defined by the generating function to be:

${e}^{2xtâˆ’{t}^{2}}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}\left(x\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [8â€“10, 19]).}$
(4)

Thus, by (4), we get

${H}_{n}\left(x\right)={\left(H+2x\right)}^{n}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{l}\right){H}_{nâˆ’l}{2}^{l}{x}^{l}\phantom{\rule{1em}{0ex}}\text{(see [14]),}$
(5)

where ${H}_{n}={H}_{n}\left(0\right)$ are called the n th Hermite numbers. Let ${\mathbb{P}}_{n}=\left\{p\left(x\right)âˆˆ\mathbb{Q}\left[x\right]|degp\left(x\right)â‰¤n\right\}$. Then ${\mathbb{P}}_{n}$ is an $\left(n+1\right)$-dimensional vector space over â„š. In [8, 10], it is called that $\left\{{E}_{0}^{\left(r\right)}\left(x\right),{E}_{1}^{\left(r\right)}\left(x\right),â€¦,{E}_{n}^{\left(r\right)}\left(x\right)\right\}$ and $\left\{{B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)\right\}$ are bases for ${\mathbb{P}}_{n}$. Let Î© denote the space of real-valued differential functions on $\left(\mathrm{âˆž},âˆ’\mathrm{âˆž}\right)=\mathbb{R}$. We define four linear operators on Î© as follows:

$I\left(f\right)\left(x\right)={âˆ«}_{x}^{x+1}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{2em}{0ex}}\mathrm{Î”}\left(f\right)\left(x\right)=f\left(x+1\right)âˆ’f\left(x\right),$
(6)
$\stackrel{Ëœ}{\mathrm{Î”}}\left(f\right)\left(x\right)=f\left(x+1\right)+f\left(x\right),\phantom{\rule{2em}{0ex}}D\left(f\right)\left(x\right)={f}^{â€²}\left(x\right).$
(7)

Thus, by (6) and (7), we get

${I}^{n}\left(f\right)\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){\left(âˆ’1\right)}^{nâˆ’l}{f}_{n}\left(x+l\right)\phantom{\rule{1em}{0ex}}\text{(see [8, 10, 12, 13]),}$
(8)

where ${f}_{1}^{â€²}=f,{f}_{2}^{â€²}={f}_{1},â€¦,{f}_{n}^{â€²}={f}_{nâˆ’1}$, $nâˆˆ\mathbb{N}$.

In this paper, we derive some new interesting identities of higher-order Bernoulli, Euler and Hermite polynomials arising from the properties of bases of higher-order Bernoulli and Euler polynomials for ${\mathbb{P}}_{n}$.

## 2 Some identities of higher-order Bernoulli and Euler polynomials

First, we introduce the following theorems, which are important in deriving our results in this paper.

Theorem 1 [8]

For $râˆˆ{\mathbb{Z}}_{+}=\mathbb{N}âˆª\left\{0\right\}$, let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then we have

$p\left(x\right)=\frac{1}{{2}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\underset{j=0}{\overset{r}{âˆ‘}}\frac{1}{k!}\left(\genfrac{}{}{0}{}{r}{j}\right){D}^{k}p\left(j\right){E}_{k}^{\left(r\right)}\left(x\right).$

Theorem 2 [10]

For $râˆˆ{\mathbb{Z}}_{+}$, let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$:

1. (a)

If $r>n$, then we have

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\underset{j=0}{\overset{k}{âˆ‘}}\frac{1}{k!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right)\left({I}^{râˆ’k}p\left(j\right)\right){B}_{k}^{\left(r\right)}\left(x\right).$
2. (b)

If $râ‰¤n$, then

$\begin{array}{rl}p\left(x\right)=& \underset{k=0}{\overset{râˆ’1}{âˆ‘}}\underset{j=0}{\overset{k}{âˆ‘}}\frac{1}{k!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right)\left({I}^{râˆ’k}p\left(j\right)\right){B}_{k}^{\left(r\right)}\left(x\right)\\ +\underset{k=r}{\overset{n}{âˆ‘}}\underset{j=0}{\overset{r}{âˆ‘}}\frac{1}{k!}{\left(âˆ’1\right)}^{râˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right)\left({D}^{kâˆ’r}p\left(j\right)\right){B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Let us take $p\left(x\right)={H}_{n}\left(x\right)âˆˆ{\mathbb{P}}_{n}$.

Then, by (5), we get

$\begin{array}{rl}{p}^{\left(k\right)}\left(x\right)& ={D}^{k}p\left(x\right)={2}^{k}n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+1\right){H}_{nâˆ’k}\left(x\right)\\ ={2}^{k}\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\left(x\right).\end{array}$
(9)

From Theorem 1 and (9), we can derive the following equation (10):

$\begin{array}{rl}{H}_{n}\left(x\right)& =\frac{1}{{2}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{1}{k!}\left(\genfrac{}{}{0}{}{r}{j}\right){2}^{k}\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\left(j\right)\right\}{E}_{k}^{\left(r\right)}\left(x\right)\\ =\frac{1}{{2}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left[\underset{j=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{j}\right){H}_{nâˆ’k}\left(j\right)\right]{E}_{k}^{\left(r\right)}\left(x\right).\end{array}$
(10)

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

Theorem 3 For $n,râˆˆ{\mathbb{Z}}_{+}$, we have

${H}_{n}\left(x\right)=\frac{1}{{2}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left[\underset{j=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{j}\right){H}_{nâˆ’k}\left(j\right)\right]{E}_{k}^{\left(r\right)}\left(x\right).$

We recall an explicit expression for Hermite polynomials as follows:

${H}_{n}\left(x\right)=\underset{l=0}{\overset{\left[\frac{n}{2}\right]}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{l}n!}{l!\left(nâˆ’2l\right)!}{\left(2x\right)}^{nâˆ’2l}.$
(11)

By (11), we get

${H}_{nâˆ’k}\left(j\right)=\underset{l=0}{\overset{\left[\frac{nâˆ’k}{2}\right]}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{l}\left(nâˆ’k\right)!}{l!\left(nâˆ’kâˆ’2l\right)!}{\left(2j\right)}^{nâˆ’kâˆ’2l}.$
(12)

Thus, by Theorem 3 and (12), we obtain the following corollary.

Corollary 4 For $n,râˆˆ{\mathbb{Z}}_{+}$, we have

${H}_{n}\left(x\right)=\frac{1}{{2}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\underset{l=0}{\overset{\left[\frac{nâˆ’k}{2}\right]}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{r}{j}\right){2}^{k}{\left(âˆ’1\right)}^{l}\left(nâˆ’k\right)!{\left(2j\right)}^{nâˆ’kâˆ’2l}}{l!\left(nâˆ’kâˆ’2l\right)!}\right\}{E}_{k}^{\left(r\right)}\left(x\right).$

Now, we consider the identities of Hermite polynomials arising from the property of the basis of higher-order Bernoulli polynomials in ${\mathbb{P}}_{n}$.

For $r>k$, by (6) and (8), we get

$\begin{array}{rl}{I}^{râˆ’k}{H}_{n}\left(x\right)& =\underset{l=0}{\overset{râˆ’k}{âˆ‘}}\left(\genfrac{}{}{0}{}{râˆ’k}{l}\right){\left(âˆ’1\right)}^{râˆ’kâˆ’l}\frac{{H}_{n+râˆ’k}\left(x+l\right)}{{2}^{râˆ’k}\left(n+1\right)â‹¯\left(n+râˆ’k\right)}\\ =\underset{l=0}{\overset{râˆ’k}{âˆ‘}}\left(\genfrac{}{}{0}{}{râˆ’k}{l}\right){\left(âˆ’1\right)}^{râˆ’kâˆ’l}\frac{n!{H}_{n+râˆ’k}\left(x+l\right)}{{2}^{râˆ’k}\left(n+râˆ’k\right)!}.\end{array}$
(13)

Therefore, by Theorem 2 and (13), we obtain the following theorem.

Theorem 5 For $n,râˆˆ{\mathbb{Z}}_{+}$, with $r>n$, we have

${H}_{n}\left(x\right)=n!\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{râˆ’k}{l}\right)\left(\genfrac{}{}{0}{}{k}{j}\right){\left(âˆ’1\right)}^{râˆ’jâˆ’l}{H}_{n+râˆ’k}\left(j+l\right)}{{2}^{râˆ’k}k!\left(n+râˆ’k\right)!}\right\}{B}_{k}^{\left(r\right)}\left(x\right).$

Let us assume that $r,kâˆˆ{\mathbb{Z}}_{+}$, with $râ‰¤n$. Then, by (b) of Theorem 2, we get

$\begin{array}{rl}{H}_{n}\left(x\right)=& n!\underset{k=0}{\overset{râˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{râˆ’k}{l}\right)\left(\genfrac{}{}{0}{}{k}{j}\right){\left(âˆ’1\right)}^{râˆ’jâˆ’l}{H}_{n+râˆ’k}\left(j+l\right)}{{2}^{râˆ’k}k!\left(n+râˆ’k\right)!}\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +n!\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{râˆ’j}\left(\genfrac{}{}{0}{}{r}{j}\right){2}^{kâˆ’r}{H}_{n+râˆ’k}\left(j\right)}{k!\left(n+râˆ’k\right)!}\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$
(14)

Therefore, by (14), we obtain the following theorem.

Theorem 6 For $n,râˆˆ{\mathbb{Z}}_{+}$, with $râ‰¤n$, we have

$\begin{array}{rl}{H}_{n}\left(x\right)=& n!\underset{k=0}{\overset{râˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{râˆ’k}{l}\right)\left(\genfrac{}{}{0}{}{k}{j}\right){\left(âˆ’1\right)}^{râˆ’jâˆ’l}{H}_{n+râˆ’k}\left(j+l\right)}{{2}^{râˆ’k}k!\left(n+râˆ’k\right)!}\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +n!\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{râˆ’j}\left(\genfrac{}{}{0}{}{r}{j}\right){2}^{kâˆ’r}}{k!\left(n+râˆ’k\right)!}{H}_{n+râˆ’k}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Remark From (12), we note that

${H}_{n+râˆ’k}\left(j+l\right)=\underset{m=0}{\overset{\left[\frac{n+râˆ’k}{2}\right]}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(n+râˆ’k\right)!}{m!\left(n+râˆ’kâˆ’2m\right)!}{\left(2j+2l\right)}^{n+râˆ’kâˆ’2m}$
(15)

and

${H}_{n+râˆ’k}\left(j\right)=\underset{m=0}{\overset{\left[\frac{n+râˆ’k}{2}\right]}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{m}\left(n+râˆ’k\right)!}{m!\left(n+râˆ’kâˆ’2m\right)!}{\left(2j\right)}^{n+râˆ’kâˆ’2m}.$
(16)

Theorem 7 [10]

For $n,râˆˆ{\mathbb{Z}}_{+}$, with $r>n$ and $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$, we have

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{n}{âˆ‘}}\frac{\left(râˆ’k\right)!{S}_{2}\left(l+râˆ’k,râˆ’k\right)}{\left(l+râˆ’k\right)!k!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right){p}^{\left(l\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right),$

where ${S}_{2}\left(l,n\right)$ is the Stirling number of the second kind and ${p}^{\left(l\right)}\left(j\right)={D}^{l}p\left(j\right)$.

Theorem 8 [10]

For $n,râˆˆ{\mathbb{Z}}_{+}$, with $râ‰¤n$ and $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$, we have

$\begin{array}{rl}p\left(x\right)=& \underset{k=0}{\overset{râˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{n}{âˆ‘}}\frac{\left(râˆ’k\right)!{S}_{2}\left(l+râˆ’k,râˆ’k\right)}{\left(l+râˆ’k\right)!k!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right){p}^{\left(l\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{râˆ’j}}{k!}\left(\genfrac{}{}{0}{}{r}{j}\right){p}^{\left(kâˆ’r\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Let us take $p\left(x\right)={H}_{n}\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then, by Theorem 7 and Theorem 8, we obtain the following corollary.

Corollary 9 For $n,râˆˆ{\mathbb{Z}}_{+}$:

1. (a)

For $r>n$, we have

${H}_{n}\left(x\right)=n!\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{n}{âˆ‘}}\frac{\left(râˆ’k\right)!{S}_{2}\left(l+râˆ’k,râˆ’k\right)}{\left(l+râˆ’k\right)!k!\left(nâˆ’l\right)!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right){2}^{l}{H}_{nâˆ’l}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).$
2. (b)

For $râ‰¤n$, we have

$\begin{array}{rl}{H}_{n}\left(x\right)=& n!\underset{k=0}{\overset{râˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{n}{âˆ‘}}\frac{\left(râˆ’k\right)!{S}_{2}\left(l+râˆ’k,râˆ’k\right)}{\left(l+râˆ’k\right)!k!\left(nâˆ’l\right)!}{\left(âˆ’1\right)}^{kâˆ’j}\left(\genfrac{}{}{0}{}{k}{j}\right){2}^{l}{H}_{nâˆ’l}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +n!\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{râˆ’j}\left(\genfrac{}{}{0}{}{r}{j}\right){2}^{kâˆ’r}{H}_{nâˆ’k+r}\left(j\right)}{k!\left(nâˆ’k+r\right)!}\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Theorem 10 [9]

For $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$, we have

$p\left(x\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{1}{k!}\left(\genfrac{}{}{0}{}{r}{j}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{p}^{\left(k\right)}\left(j\right)\right\}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$

Let us take $p\left(x\right)={H}_{n}\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then

$\begin{array}{rl}{H}_{n}\left(x\right)& =\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\frac{1}{k!}\left(\genfrac{}{}{0}{}{r}{j}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{2}^{k}\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\left(j\right)\right\}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\\ =\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\left(\genfrac{}{}{0}{}{r}{j}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{H}_{nâˆ’k}\left(j\right)\right\}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right)\\ =\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\underset{l=0}{\overset{\left[\frac{nâˆ’k}{2}\right]}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left(\genfrac{}{}{0}{}{r}{j}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{\left(âˆ’1\right)}^{l}\left(nâˆ’k\right)!{\left(2j\right)}^{nâˆ’kâˆ’2l}}{l!\left(nâˆ’kâˆ’2l\right)!}\right\}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).\end{array}$
(17)

Therefore, by (17), we obtain the following corollary.

Corollary 11 For $nâˆˆ{\mathbb{Z}}_{+}$, we have

${H}_{n}\left(x\right)=\frac{1}{{\left(1âˆ’\mathrm{Î»}\right)}^{r}}\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}\underset{l=0}{\overset{\left[\frac{nâˆ’k}{2}\right]}{âˆ‘}}\frac{\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left(\genfrac{}{}{0}{}{r}{j}\right){\left(âˆ’\mathrm{Î»}\right)}^{râˆ’j}{\left(âˆ’1\right)}^{l}\left(nâˆ’k\right)!{\left(2j\right)}^{nâˆ’kâˆ’2l}}{l!\left(nâˆ’kâˆ’2l\right)!}\right\}{H}_{k}^{\left(r\right)}\left(x|\mathrm{Î»}\right).$

For $r=1$, the Frobenius-Euler polynomials are defined by the generating function to be

$\left(\frac{1âˆ’\mathrm{Î»}}{{e}^{t}âˆ’\mathrm{Î»}}\right){e}^{xt}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}{H}_{n}\left(x|\mathrm{Î»}\right)\frac{{t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\text{(see [9]).}$
(18)

Thus, by (18), we get

$\frac{d}{d\mathrm{Î»}}{H}_{n}\left(x|\mathrm{Î»}\right)=\frac{1}{1âˆ’\mathrm{Î»}}\left({H}_{n}^{\left(2\right)}\left(x|\mathrm{Î»}\right)âˆ’{H}_{n}\left(x|\mathrm{Î»}\right)\right).$
(19)

For $nâˆˆ{\mathbb{Z}}_{+}$, let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then we note that

$\left(1âˆ’\mathrm{Î»}\right)p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\frac{1}{k!}\left\{{p}^{\left(k\right)}\left(1\right)âˆ’\mathrm{Î»}{p}^{\left(k\right)}\left(0\right)\right\}{H}_{k}\left(x|\mathrm{Î»}\right)\phantom{\rule{1em}{0ex}}\text{(see [9]).}$
(20)

Let us take $p\left(x\right)={H}_{n}\left(x\right)$. Then, by (20), we get

$\begin{array}{rl}\left(1âˆ’\mathrm{Î»}\right){H}_{n}\left(x\right)& =\underset{k=0}{\overset{n}{âˆ‘}}\frac{1}{k!}\left\{{2}^{k}\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\left(1\right)âˆ’\mathrm{Î»}{2}^{k}\frac{n!}{\left(nâˆ’k\right)!}{H}_{nâˆ’k}\right\}{H}_{k}\left(x|\mathrm{Î»}\right)\\ =\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left({H}_{nâˆ’k}\left(1\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\right){H}_{k}\left(x|\mathrm{Î»}\right)\phantom{\rule{1em}{0ex}}\text{(see [9]).}\end{array}$
(21)

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

Theorem 12 For $nâˆˆ{\mathbb{Z}}_{+}$, we have

$\left(1âˆ’\mathrm{Î»}\right){H}_{n}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left({H}_{nâˆ’k}\left(1\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\right){H}_{k}\left(x|\mathrm{Î»}\right).$

Let us take $\frac{d}{d\mathrm{Î»}}$ on the both sides of Theorem 12.

Then, we have

$\begin{array}{rl}âˆ’{H}_{n}\left(x\right)=& âˆ’\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}{H}_{nâˆ’k}{H}_{k}\left(x|\mathrm{Î»}\right)\\ +\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}\left({H}_{nâˆ’k}\left(1\right)âˆ’\mathrm{Î»}{H}_{nâˆ’k}\right)\left(\frac{d}{d\mathrm{Î»}}{H}_{k}\left(x|\mathrm{Î»}\right)\right).\end{array}$
(22)

By (22), we get

$\begin{array}{rl}{H}_{n}\left(x\right)=& \underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right){2}^{k}{H}_{nâˆ’k}{H}_{k}\left(x|\mathrm{Î»}\right)\\ +\underset{k=0}{\overset{n}{âˆ‘}}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\mathrm{Î»}{H}_{nâˆ’k}âˆ’{H}_{nâˆ’k}\left(1\right)\right){2}^{k}\left(\frac{d}{d\mathrm{Î»}}{H}_{k}\left(x|\mathrm{Î»}\right)\right).\end{array}$
(23)

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## Acknowledgements

This paper is supported in part by the Research Grant of Kwangwoon University in 2013.

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Correspondence to Taekyun Kim.

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The authors declare that they have no competing interests.

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All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Kim, D.S., Kim, T., Dolgy, D.V. et al. Higher-order Bernoulli, Euler and Hermite polynomials. Adv Differ Equ 2013, 103 (2013). https://doi.org/10.1186/1687-1847-2013-103