# A note on higher-order Bernoulli polynomials

## Abstract

Let ${\mathbb{P}}_{n}=\left\{p\left(x\right)âˆˆ\mathbf{Q}\left[x\right]|degp\left(x\right)â‰¤n\right\}$ be the $\left(n+1\right)$-dimensional vector space over Q. From the property of the basis ${B}_{0}^{\left(r\right)},{B}_{1}^{\left(r\right)},â€¦,{B}_{n}^{\left(r\right)}$ for the space ${\mathbb{P}}_{n}$, we derive some interesting identities of higher-order Bernoulli polynomials.

## 1 Introduction

Let $\mathbf{N}=\left\{1,2,3,â€¦\right\}$ and ${\mathbf{Z}}_{+}=\mathbf{N}âˆª\left\{0\right\}$. For a fixed $râˆˆ{\mathbf{Z}}_{+}$, the n th Bernoulli polynomials are defined by the generating function to be

${\left(\frac{t}{{e}^{t}âˆ’1}\right)}^{r}{e}^{xt}={e}^{{B}^{\left(r\right)}\left(x\right)t}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{B}_{n}^{\left(r\right)}\left(x\right){t}^{n}}{n!}\phantom{\rule{1em}{0ex}}\left(\text{see [1â€“11]}\right)$
(1)

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

From (1), we note that

$\begin{array}{rl}{B}_{n}^{\left(r\right)}\left(x\right)& =\underset{k=0}{\overset{n}{âˆ‘}}\left(\begin{array}{c}n\\ k\end{array}\right){B}_{k}^{\left(r\right)}{x}^{nâˆ’k}=\underset{k=0}{\overset{n}{âˆ‘}}\left(\begin{array}{c}n\\ k\end{array}\right){B}_{nâˆ’k}^{\left(r\right)}{x}^{k}\\ =\underset{{n}_{1}+â‹¯+{n}_{r}+{n}_{r+1}=n}{âˆ‘}\left(\begin{array}{c}n\\ {n}_{1},â€¦,{n}_{r},{n}_{r+1}\end{array}\right){B}_{{n}_{1}}â‹¯{B}_{{n}_{r}}{x}^{{n}_{r+1}}.\end{array}$
(2)

Thus, by (2) we get the Euler-type sums of products of Bernoulli numbers as follows:

${B}_{n}^{\left(r\right)}=\underset{{n}_{1}+â‹¯+{n}_{r}=n}{âˆ‘}\left(\begin{array}{c}n\\ {n}_{1},â€¦,{n}_{r}\end{array}\right){B}_{{n}_{1}}{B}_{{n}_{2}}â‹¯{B}_{{n}_{r}}\phantom{\rule{1em}{0ex}}\left(\text{see [11â€“17]}\right).$
(3)

By (2) and (3), we see that ${B}_{n}^{\left(r\right)}\left(x\right)$ is a monic polynomial of degree n with coefficients in Q.

From (2), we note that

${\left({B}_{n}^{\left(r\right)}\left(x\right)\right)}^{â€²}=\frac{d}{dx}{B}_{n}^{\left(r\right)}\left(x\right)=n{B}_{nâˆ’1}^{\left(r\right)}\left(x\right)\phantom{\rule{1em}{0ex}}\left(\text{see [11â€“17]}\right)$
(4)

and

${B}_{n}^{\left(r\right)}\left(x+1\right)âˆ’{B}_{n}^{\left(r\right)}\left(x\right)=n{B}_{nâˆ’1}^{\left(râˆ’1\right)}\left(x\right).$
(5)

Let Î© denote the space of real-valued differential functions on $\left(âˆ’\mathrm{âˆž},\mathrm{âˆž}\right)=\mathbf{R}$. Now, we define three linear operators I, â–³, D on Î© as follows:

$If\left(x\right)={âˆ«}_{x}^{x+1}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\phantom{\rule{2em}{0ex}}\mathrm{â–³}f\left(x\right)=f\left(x+1\right)âˆ’f\left(x\right),\phantom{\rule{2em}{0ex}}Df\left(x\right)={f}^{â€²}\left(x\right).$
(6)

Then we see that (i) $DI=ID=\mathrm{â–³}$, (ii) $\mathrm{â–³}I=I\mathrm{â–³}$, (iii) $\mathrm{â–³}D=D\mathrm{â–³}$.

Let ${\mathbb{P}}_{n}=\left\{p\left(x\right)âˆˆ\mathbf{Q}\left(x\right)|degp\left(x\right)â‰¤n\right\}$ be the $\left(n+1\right)$-dimensional vector space over Q. Probably, $\left\{1,x,â€¦,{x}^{n}\right\}$ is the most natural basis for this space. But $\left\{{B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),{B}_{2}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)\right\}$ is also a good basis for the space ${\mathbb{P}}_{n}$ for our purpose of arithmetical and combinatorial applications.

Let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be generated by ${B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),{B}_{2}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)$ as follows:

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(r\right)}\left(x\right).$

In this paper, we develop methods for uniquely determining ${a}_{k}$ from the information of $p\left(x\right)$. From those methods, we derive some interesting identities of higher-order Bernoulli polynomials.

## 2 Higher-order Bernoulli polynomials

For $r=0$, by (1), we get ${B}_{n}^{\left(0\right)}={x}^{n}$ ($nâˆˆ{\mathbf{Z}}_{+}$). Let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$.

For a fixed $râˆˆ{\mathbf{Z}}_{+}$, $p\left(x\right)$ can be generated by ${B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),{B}_{2}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)$ as follows:

$p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(r\right)}\left(x\right).$
(7)

From (6) and (7), we can derive the following identities:

$\begin{array}{rcl}I{B}_{n}^{\left(r\right)}\left(x\right)& =& {âˆ«}_{x}^{x+1}{B}_{n}^{\left(r\right)}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{n+1}\left({B}_{n+1}^{\left(r\right)}\left(x+1\right)âˆ’{B}_{n+1}^{\left(r\right)}\left(x\right)\right).\end{array}$
(8)

By (5) and (8), we get

$I{B}_{n}^{\left(r\right)}\left(x\right)=\frac{n+1}{n+1}{B}_{n}^{\left(râˆ’1\right)}\left(x\right)={B}_{n}^{\left(râˆ’1\right)}\left(x\right).$
(9)

It is easy to show that

$\mathrm{â–³}{B}_{n}^{\left(r\right)}\left(x\right)={B}_{n}^{\left(r\right)}\left(x+1\right)âˆ’{B}_{n}^{\left(r\right)}\left(x\right)=n{B}_{nâˆ’1}^{\left(râˆ’1\right)}\left(x\right),$
(10)

and

$D{B}_{n}^{\left(r\right)}\left(x\right)=n{B}_{nâˆ’1}^{\left(r\right)}\left(x\right).$
(11)

By (7) and (9), we get

${I}^{r}p\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(0\right)}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{x}^{k}.$
(12)

From (6) and (12), we note that

${D}^{k}{I}^{r}p\left(x\right)=\underset{l=k}{\overset{n}{âˆ‘}}{a}_{l}\frac{l!}{\left(lâˆ’k\right)!}{x}^{lâˆ’k}.$
(13)

Thus, by (13) we get

${D}^{k}{I}^{r}p\left(0\right)=k!{a}_{k}.$
(14)

Hence, from (14) we have

${a}_{k}=\frac{{D}^{k}{I}^{r}p\left(0\right)}{k!}.$
(15)

Case 1. Let $r>n$. Then $r>k$ for all $k=0,1,2,â€¦,n$.

By (15), we get

$\begin{array}{rl}{a}_{k}& =\frac{1}{k!}{D}^{k}{I}^{k}{I}^{râˆ’k}p\left(0\right)=\frac{1}{k!}{\left(DI\right)}^{k}{I}^{râˆ’k}p\left(0\right)\\ =\frac{1}{k!}{\mathrm{â–³}}^{k}{I}^{râˆ’k}p\left(0\right)=\frac{1}{k!}\underset{j=0}{\overset{k}{âˆ‘}}\left(\begin{array}{c}k\\ j\end{array}\right){\left(âˆ’1\right)}^{kâˆ’j}{I}^{râˆ’k}p\left(j\right).\end{array}$
(16)

Case 2. Assume that $râ‰¤n$.

1. (i)

For $0â‰¤kâ‰¤r$, by (15) we get

${a}_{k}=\frac{1}{k!}\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\left(\begin{array}{c}k\\ j\end{array}\right){I}^{râˆ’k}p\left(j\right).$
(17)
2. (ii)

For $râ‰¤kâ‰¤n$, by (15) we see that

$\begin{array}{rl}{a}_{k}& =\frac{1}{k!}{D}^{kâˆ’r}{D}^{r}{I}^{r}p\left(0\right)=\frac{1}{k!}{D}^{kâˆ’r}{\left(DI\right)}^{r}p\left(0\right)=\frac{1}{k!}{D}^{kâˆ’r}{\mathrm{â–³}}^{r}p\left(0\right)\\ =\frac{1}{k!}{\mathrm{â–³}}^{r}{D}^{kâˆ’r}p\left(0\right)=\frac{1}{k!}\underset{j=0}{\overset{r}{âˆ‘}}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’1\right)}^{râˆ’j}{D}^{kâˆ’r}p\left(j\right).\end{array}$
(18)

Therefore, by (7), (16), (17) and (18), we obtain the following theorem.

Theorem 1

1. (a)

For $r>n$, we have

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

For $râ‰¤n$, we have

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

Let us take $p\left(x\right)={x}^{n}âˆˆ{\mathbb{P}}_{n}$. Then ${x}^{n}$ can be expressed as a linear combination of ${B}_{0}^{\left(r\right)},{B}_{1}^{\left(r\right)},â€¦,{B}_{n}^{\left(r\right)}$. For $r>n$, we have

${I}^{râˆ’k}{x}^{n}=\frac{n!}{\left(n+râˆ’k\right)!}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’kâˆ’l}\left(\begin{array}{c}râˆ’k\\ l\end{array}\right){\left(x+l\right)}^{n+râˆ’k}.$
(19)

Therefore, by Theorem 1 and (19), we obtain the following corollary.

Corollary 2 For $n,râˆˆ{\mathbf{Z}}_{+}$ with $r>n$, we have

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

Let us assume that $r,nâˆˆ{\mathbf{Z}}_{+}$ with $râ‰¤n$. Observe that

${D}^{kâˆ’r}{x}^{n}=n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+r+1\right){x}^{nâˆ’k+r}=\frac{n!}{\left(nâˆ’k+r\right)!}{x}^{nâˆ’k+r}.$
(20)

Thus, by Theorem 1 and (20), we obtain the following corollary.

Corollary 3 For $n,râˆˆ{\mathbf{Z}}_{+}$ with $râ©½n$, we have

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

Let us take $p\left(x\right)={B}_{n}^{\left(s\right)}\left(x\right)âˆˆ{\mathbb{P}}_{n}$ ($sâˆˆ{\mathbf{Z}}_{+}$). Then $p\left(x\right)$ can be generated by $\left\{{B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)\right\}$ as follows:

${B}_{n}^{\left(s\right)}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(r\right)}\left(x\right).$
(21)

For $r>n$, we have

${I}^{râˆ’k}{B}_{n}^{\left(s\right)}\left(x\right)={B}_{n}^{\left(sâˆ’r+k\right)}\left(x\right).$
(22)

Thus, by Theorem 1 and (22), we obtain the following theorem.

Theorem 4 For $r,n,sâˆˆ{\mathbf{Z}}_{+}$ with $r>n$, we have

${B}_{n}^{\left(s\right)}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right){B}_{n}^{\left(s+kâˆ’r\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).$

In particular, for $r=s$, we have

$\begin{array}{rcl}{B}_{n}^{\left(r\right)}\left(x\right)& =& 0{B}_{0}^{\left(r\right)}\left(x\right)+0{B}_{1}^{\left(r\right)}\left(x\right)+â‹¯+0{B}_{nâˆ’1}^{\left(r\right)}\left(x\right)+1{B}_{n}^{\left(r\right)}\left(x\right)\\ =& \underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right){B}_{n}^{\left(k\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$
(23)

By comparing coefficients on the both sides of (23), we get

(24)

Therefore, by (24), we obtain the following corollary.

Corollary 5

1. (a)

For $n,kâˆˆ{\mathbf{Z}}_{+}$ with $0â‰¤kâ‰¤nâˆ’1$, we have

$\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\left(\begin{array}{c}k\\ j\end{array}\right){B}_{n}^{\left(k\right)}\left(j\right)=0.$
2. (b)

In particular, $k=n$, we get

$\underset{j=0}{\overset{n}{âˆ‘}}{\left(âˆ’1\right)}^{nâˆ’j}\left(\begin{array}{c}n\\ j\end{array}\right){B}_{n}^{\left(n\right)}\left(j\right)=n!.$

Let us assume that $râ‰¤n$ in (21). Then we have

${D}^{kâˆ’r}{B}_{n}^{\left(s\right)}\left(x\right)=n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+r+1\right){B}_{n+râˆ’k}^{\left(s\right)}\left(x\right)=\frac{n!}{\left(nâˆ’k+r!\right)}{B}_{n+râˆ’k}^{\left(s\right)}\left(x\right).$
(25)

Therefore, by Theorem 1, (21) and (25), we obtain the following theorem.

Theorem 6 For $r,nâˆˆ{\mathbf{Z}}_{+}$ with $râ‰¤n$, we have

$\begin{array}{rcl}{B}_{n}^{\left(s\right)}\left(x\right)& =& \underset{k=0}{\overset{nâˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right){B}_{n}^{\left(s+kâˆ’r\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’j}\frac{n!\left(\begin{array}{c}r\\ j\end{array}\right)}{k!\left(n+râˆ’k\right)!}{B}_{n+râˆ’k}^{\left(s\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Let $p\left(x\right)={E}_{n}^{\left(s\right)}\left(x\right)$ ($sâˆˆ{\mathbf{Z}}_{+}$) be Euler polynomials of order s. Then ${E}_{n}^{\left(s\right)}$ can be expressed as a linear combination of ${B}_{0}^{\left(r\right)}\left(x\right),{B}_{1}^{\left(r\right)}\left(x\right),â€¦,{B}_{n}^{\left(r\right)}\left(x\right)$.

Assume that $r,nâˆˆ{\mathbf{Z}}_{+}$ with $r>n$.

By (6), we get

$\begin{array}{rl}{I}^{râˆ’k}{E}_{n}^{\left(s\right)}\left(x\right)& =\frac{1}{\left(n+1\right)â‹¯\left(n+râˆ’k\right)}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’kâˆ’l}\left(\begin{array}{c}râˆ’k\\ l\end{array}\right){E}_{n+râˆ’k}^{\left(s\right)}\left(x+l\right)\\ =\frac{n!}{\left(n+râˆ’k\right)!}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’kâˆ’l}\left(\begin{array}{c}râˆ’k\\ l\end{array}\right){E}_{n+râˆ’k}^{\left(s\right)}\left(x+l\right).\end{array}$
(26)

Therefore, by Theorem 1 and (26), we obtain the following theorem.

Theorem 7 For $r,nâˆˆ{\mathbf{Z}}_{+}$ with $r>n$, we have

${E}_{n}^{\left(s\right)}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’jâˆ’l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}râˆ’k\\ l\end{array}\right)}{k!\left(n+râˆ’k\right)!}{E}_{n+râˆ’k}^{\left(s\right)}\left(j+l\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).$

For $r,nâˆˆ{\mathbf{Z}}_{+}$ with $râ‰¤n$, we have

${D}^{kâˆ’r}{E}_{n}^{\left(s\right)}\left(x\right)=n\left(nâˆ’1\right)â‹¯\left(nâˆ’k+r+1\right){E}_{nâˆ’k+r}^{\left(s\right)}\left(x\right).$
(27)

By Theorem 1 and (27), we obtain the following theorem.

Theorem 8 For $r,nâˆˆ{\mathbf{Z}}_{+}$ with $râ‰¤n$, we have

$\begin{array}{rcl}{E}_{n}^{\left(s\right)}\left(x\right)& =& \underset{k=0}{\overset{râˆ’1}{âˆ‘}}\left\{\underset{j=0}{\overset{k}{âˆ‘}}\underset{l=0}{\overset{râˆ’k}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’jâˆ’l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}râˆ’k\\ l\end{array}\right)}{k!\left(n+râˆ’k\right)!}{E}_{n+râˆ’k}^{\left(s\right)}\left(j+l\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right)\\ +\underset{k=r}{\overset{n}{âˆ‘}}\left\{\underset{j=0}{\overset{r}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’j}\frac{n!\left(\begin{array}{c}r\\ k\end{array}\right)}{k!\left(n+râˆ’k\right)!}{E}_{n+râˆ’k}^{\left(s\right)}\left(j\right)\right\}{B}_{k}^{\left(r\right)}\left(x\right).\end{array}$

Remarks (a) For $râ‰¤0$, by (40) we get

${I}^{r}{x}^{n}=\frac{n!}{\left(n+r\right)!}\underset{j=0}{\overset{r}{âˆ‘}}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’1\right)}^{râˆ’j}{\left(x+j\right)}^{n+r}=\frac{1}{\left(\begin{array}{c}n+r\\ r\end{array}\right)}\frac{1}{r!}\underset{j=0}{\overset{r}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’j}\left(\begin{array}{c}r\\ j\end{array}\right){\left(x+j\right)}^{n+r}.$

Thus, for $x=0$, we have

${I}^{r}{x}^{n}{|}_{x=0}=\frac{n!}{\left(n+r\right)!}\underset{j=0}{\overset{r}{âˆ‘}}\left(\begin{array}{c}r\\ j\end{array}\right){\left(âˆ’1\right)}^{râˆ’j}{j}^{n+r}=\frac{1}{\left(\begin{array}{c}n+r\\ r\end{array}\right)}\frac{1}{r!}{\mathrm{â–³}}^{r}{0}^{n+r}=\frac{S\left(n+r,r\right)}{\left(\begin{array}{c}n+r\\ r\end{array}\right)},$
(28)

where $S\left(n,r\right)$ is the Stirling number of the second kind.

1. (b)

Assume

$\underset{k=0}{\overset{n}{âˆ‘}}{\mathrm{Î±}}_{k}{x}^{k}=\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(r\right)}\left(x\right)\phantom{\rule{1em}{0ex}}\left(râ‰¥0\right).$
(29)

Applying ${I}^{t}$ on both sides ($tâ‰¥0$), we get

$\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(râˆ’t\right)}\left(x\right)=\underset{k=0}{\overset{n}{âˆ‘}}{\mathrm{Î±}}_{k}{I}^{t}{x}^{k}=\underset{k=0}{\overset{n}{âˆ‘}}\frac{{\mathrm{Î±}}_{k}}{\left(\begin{array}{c}\mathrm{Î±}+t\\ t\end{array}\right)}\frac{1}{t!}\underset{j=0}{\overset{t}{âˆ‘}}{\left(âˆ’1\right)}^{\left(tâˆ’j\right)}\left(\begin{array}{c}t\\ j\end{array}\right){\left(x+j\right)}^{k+t}.$
(30)

From (28) and (30), we have

$\underset{k=0}{\overset{n}{âˆ‘}}{a}_{k}{B}_{k}^{\left(râˆ’t\right)}=\underset{k=0}{\overset{n}{âˆ‘}}\frac{{\mathrm{Î±}}_{k}}{\left(\begin{array}{c}\mathrm{Î±}+t\\ t\end{array}\right)}S\left(k+t,t\right).$

Remark Let us define two operators d, $\stackrel{Ëœ}{d}$ as follows:

$d={e}^{âˆ’D}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(âˆ’1\right)}^{n}}{n!}{D}^{n},\phantom{\rule{2em}{0ex}}\stackrel{Ëœ}{d}={e}^{D}=\underset{n=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{D}^{n}}{n!}.$
(31)

From (31), we note that

$\begin{array}{r}\stackrel{Ëœ}{d}{x}^{n}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\begin{array}{c}n\\ l\end{array}\right){x}^{nâˆ’l}={\left(x+1\right)}^{n},\\ d{x}^{n}=\underset{l=0}{\overset{n}{âˆ‘}}\left(\begin{array}{c}n\\ l\end{array}\right){\left(âˆ’1\right)}^{l}{x}^{nâˆ’l}={\left(xâˆ’1\right)}^{n}.\end{array}$
(32)

Thus, by (31) and (32), we get

$\stackrel{Ëœ}{d}{B}_{n}^{\left(r\right)}\left(x\right)={B}_{n}^{\left(r\right)}\left(x+1\right),\phantom{\rule{2em}{0ex}}d{B}_{n}^{\left(r\right)}\left(x\right)={B}_{n}^{\left(r\right)}\left(xâˆ’1\right),$
(33)

and

$\stackrel{Ëœ}{d}{E}_{n}^{\left(r\right)}\left(x\right)={E}_{n}^{\left(r\right)}\left(x+1\right),\phantom{\rule{2em}{0ex}}d{E}_{n}^{\left(r\right)}\left(x\right)={E}_{n}^{\left(r\right)}\left(xâˆ’1\right).$
(34)

## 3 Further remarks

For any ${r}_{0},{r}_{1},â€¦,{r}_{n}âˆˆ{\mathbf{Z}}_{+}$, $\left\{{B}_{0}^{\left({r}_{0}\right)}\left(x\right),{B}_{1}^{\left({r}_{1}\right)}\left(x\right),â€¦,{B}_{0}^{\left({r}_{n}\right)}\left(x\right)\right\}$ forms a basis for ${\mathbb{P}}_{n}$. Let $r=max\left\{{r}_{i}|i=0,1,2,â€¦,n\right\}$. Let $p\left(x\right)âˆˆ{\mathbb{P}}_{n}$. Then $p\left(x\right)$ can be expressed as a linear combination of ${B}_{0}^{\left({r}_{0}\right)}\left(x\right),{B}_{1}^{\left({r}_{1}\right)}\left(x\right),â€¦,{B}_{n}^{\left({r}_{n}\right)}\left(x\right)$ as follows:

$p\left(x\right)={a}_{0}{B}_{0}^{\left({r}_{0}\right)}\left(x\right)+{a}_{1}{B}_{1}^{\left({r}_{1}\right)}\left(x\right)+â‹¯+{a}_{n}{B}_{n}^{\left({r}_{n}\right)}\left(x\right)=\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{B}_{l}^{\left({r}_{l}\right)}\left(x\right).$
(35)

Thus, by (6) and (35), we get

$\begin{array}{rl}{I}^{r}p\left(x\right)& =\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{I}^{r}{B}_{l}^{\left({r}_{l}\right)}\left(x\right)\\ =\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{I}^{râˆ’{r}_{l}}{I}^{{r}_{l}}{B}_{l}^{\left({r}_{l}\right)}\left(x\right)=\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{I}^{râˆ’{r}_{l}}{B}_{l}^{\left(0\right)}\left(x\right)\\ =\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{I}^{râˆ’{r}_{l}}{x}^{l}.\end{array}$
(36)

Now, for each $k=0,1,2,â€¦,n$, by (36) we get

$\begin{array}{rl}{D}^{k}{I}^{r}p\left(x\right)& =\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{D}^{k}{I}^{râˆ’{r}_{l}}{x}^{l}=\underset{l=0}{\overset{n}{âˆ‘}}{a}_{l}{I}^{râˆ’{r}_{l}}\left({D}^{k}{x}^{l}\right)\\ =\underset{l=k}{\overset{n}{âˆ‘}}{a}_{l}{I}^{râˆ’{r}_{l}}\left(\frac{l!}{\left(lâˆ’k\right)!}{x}^{lâˆ’k}\right)=\underset{l=k}{\overset{n}{âˆ‘}}\frac{{a}_{l}l!}{\left(lâˆ’k\right)!}{I}^{râˆ’{r}_{l}}{x}^{lâˆ’k}.\end{array}$
(37)

Let us take $x=0$ in (37). Then, by (28) and (37), we get

$\begin{array}{rl}{D}^{k}{I}^{r}p\left(0\right)& =\underset{l=k}{\overset{n}{âˆ‘}}\frac{l!{a}_{l}}{\left(lâˆ’k\right)!}Ã—\frac{S\left(lâˆ’k+râˆ’{r}_{l},râˆ’{r}_{l}\right)}{\left(\begin{array}{c}lâˆ’k+râˆ’{r}_{l}\\ râˆ’{r}_{l}\end{array}\right)}\\ =\underset{l=k}{\overset{n}{âˆ‘}}\frac{{a}_{l}\left(râˆ’{r}_{l}\right)!l!}{\left(lâˆ’k\right)!}S\left(lâˆ’k+râˆ’{r}_{l},râˆ’{r}_{l}\right).\end{array}$
(38)

Case 1. For $r>n$, we have

$\begin{array}{rl}{D}^{k}{I}^{r}p\left(0\right)& ={D}^{k}{I}^{k}{I}^{râˆ’k}p\left(0\right)={\left(DI\right)}^{k}{I}^{râˆ’k}p\left(0\right)={\mathrm{â–³}}^{k}{I}^{râˆ’k}p\left(0\right)\\ =\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\left(\begin{array}{c}k\\ j\end{array}\right){I}^{râˆ’k}p\left(j\right).\end{array}$
(39)

Case 2. Let $râ©½n$.

1. (i)

For $0â©½k, we have

${D}^{k}{I}^{r}p\left(0\right)=\underset{j=0}{\overset{k}{âˆ‘}}{\left(âˆ’1\right)}^{kâˆ’j}\left(\begin{array}{c}k\\ j\end{array}\right){I}^{râˆ’k}p\left(j\right).$
(40)
2. (ii)

For $râ©½kâ©½n$, we have

$\begin{array}{rl}{D}^{k}{I}^{r}p\left(0\right)& ={D}^{kâˆ’r}{D}^{r}{I}^{r}p\left(0\right)={D}^{kâˆ’r}{\left(DI\right)}^{r}p\left(0\right)={D}^{kâˆ’r}{\mathrm{â–³}}^{r}p\left(0\right)\\ ={\mathrm{â–³}}^{r}{D}^{kâˆ’r}p\left(0\right)=\underset{j=0}{\overset{r}{âˆ‘}}{\left(âˆ’1\right)}^{râˆ’j}\left(\begin{array}{c}r\\ j\end{array}\right){D}^{kâˆ’r}p\left(j\right).\end{array}$
(41)

Thus, by (38), (39), (40) and (41), we can determine ${a}_{0},{a}_{1},{a}_{2},â€¦,{a}_{n}$.

## References

1. Abramowitz M, Stegun IA: Handbook of Mathematical Functions. Natl. Bur. of Standards, Washington; 1964.

2. Araci S, Acikgoz M: A note on the Frobenius-Euler numbers and polynomials associated with Bernstein polynomials. Adv. Stud. Contemp. Math - Jangâ€™jun Math. Soc. 2012, 22(3):399â€“406.

3. Bayad A, Kim T: Identities involving values of Bernstein, q -Bernoulli, and q -Euler polynomials. Russ. J. Math. Phys. 2011, 18(2):133â€“143. 10.1134/S1061920811020014

4. Gould HW: Explicit formulas for Bernoulli numbers. Am. Math. Mon. 1972, 79: 44â€“51. 10.2307/2978125

5. Kim T: Symmetry p -adic invariant integral on ${\mathbb{Z}}_{p}$ for Bernoulli and Euler polynomials. J. Differ. Equ. Appl. 2008, 14(12):1267â€“1277. 10.1080/10236190801943220

6. Kim DS, Kim T, Kim YH, Dolgy DV: A note on Eulerian polynomials associated with Bernoulli and Euler numbers and polynomials. Adv. Stud. Contemp. Math - Jangâ€™jun Math. Soc. 2012, 22(3):379â€“389.

7. Kim DS, Kim T: Euler basis, identities, and their applications. Int. J. Math. Math. Sci. 2012., 2012: Article ID 343981 10.1155/2012/343981

8. Kim DS, Kim T: Bernoulli basis and the product of several Bernoulli polynomials. Int. J. Math. Math. Sci. 2012., 2012: Article ID 463659 10.1155/2012/463659

9. Kim T: Sums of products of q -Bernoulli numbers. Arch. Math. 2001, 76(3):190â€“195. 10.1007/s000130050559

10. Kim T, Adiga C: Sums of products of generalized Bernoulli numbers. Int. Math. J. 2004, 5(1):1â€“7.

11. Kim T: Symmetry of power sum polynomials and multivariate fermionic p -adic invariant integral on ${\mathbb{Z}}_{p}$ . Russ. J. Math. Phys. 2009, 16(1):93â€“96. 10.1134/S1061920809010063

12. Kim T: Identities involving Frobenius-Euler polynomials arising from non-linear differential equations. J. Number Theory 2012, 132(12):2854â€“2865. 10.1016/j.jnt.2012.05.033

13. Kim T: Some formulae for the q -Bernstein polynomials and q -deformed binomial distributions. J. Comput. Anal. Appl. 2012, 14(5):917â€“933.

14. PetojeviÄ‡ A: New sums of products of Bernoulli numbers. Integral Transforms Spec. Funct. 2008, 19(1â€“2):105â€“114.

15. Rim S-H, Jeong J: On the modified q -Euler numbers of higher order with weight. Adv. Stud. Contemp. Math - Jangâ€™jun Math. Soc. 2012, 22(1):93â€“98.

16. Ryoo CS: Some relations between twisted q -Euler numbers and Bernstein polynomials. Adv. Stud. Contemp. Math - Jangâ€™jun Math. Soc. 2011, 21(2):217â€“223.

17. Kim T: An identity of the symmetry for the Frobenius-Euler polynomials associated with the fermionic p -adic invariant q -integrals on ${\mathbb{Z}}_{p}$ . Rocky Mt. J. Math. 2011, 41(1):239â€“247. 10.1216/RMJ-2011-41-1-239

## Acknowledgements

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

## Author information

Authors

### Corresponding author

Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

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

## Rights and permissions

Reprints and permissions

Kim, D.S., Kim, T. A note on higher-order Bernoulli polynomials. J Inequal Appl 2013, 111 (2013). https://doi.org/10.1186/1029-242X-2013-111