A note on higher-order Bernoulli polynomials

Abstract

Let ${\mathbb{P}}_n=\{p(x)\in \mathbf{Q}[x]|degp(x)\le n\}$ be the $(n+1)$-dimensional vector space over Q. From the property of the basis $B_0^{(r)},B_1^{(r)},\dots ,B_n^{(r)}$ for the space ${\mathbb{P}}_n$, we derive some interesting identities of higher-order Bernoulli polynomials.

1 Introduction

Let $\mathbf{N}=\{1,2,3,\dots \}$ and ${\mathbf{Z}}_{+}=\mathbf{N}\cup \{0\}$. For a fixed $r\in {\mathbf{Z}}_{+}$, the n th Bernoulli polynomials are defined by the generating function to be

$${\left(\frac{t}{e^t-1}\right)}^r{e}^{xt}=e^{B^{(r)}(x)t}=\sum _{n=0}^{\mathrm{\infty }}\frac{B_n^{(r)}(x)t^n}{n!}\left(\text{see [1–11]}\right)$$

(1)

with the usual convention about replacing ${(B^{(r)}(x))}^n$ by $B_n^{(r)}(x)$. In the special case, $x=0$, $B_n^{(r)}(0)=B_n^{(r)}$ are called the n th Bernoulli numbers of order r.

From (1), we note that

$$\begin{array}{rl}{B}_n^{(r)}(x)& =\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)B_k^{(r)}{x}^{n-k}=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)B_{n-k}^{(r)}{x}^k\\ =\sum _{n_1+\cdots +n_r+n_{r+1}=n}\left(\begin{array}{c}n\\ n_1,\dots ,n_r,n_{r+1}\end{array}\right)B_{n_1}\cdots B_{n_r}{x}^{n_{r+1}}.\end{array}$$

(2)

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

$$B_n^{(r)}=\sum _{n_1+\cdots +n_r=n}\left(\begin{array}{c}n\\ n_1,\dots ,n_r\end{array}\right)B_{n_1}{B}_{n_2}\cdots B_{n_r}\left(\text{see [11–17]}\right).$$

(3)

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

From (2), we note that

$${(B_n^{(r)}(x))}^{\prime }=\frac{d}{dx}{B}_n^{(r)}(x)=n{B}_{n-1}^{(r)}(x)\left(\text{see [11–17]}\right)$$

(4)

and

$$B_n^{(r)}(x+1)-B_n^{(r)}(x)=n{B}_{n-1}^{(r-1)}(x).$$

(5)

Let Ω denote the space of real-valued differential functions on $(-\mathrm{\infty },\mathrm{\infty })=\mathbf{R}$. Now, we define three linear operators I, △, D on Ω as follows:

$$If(x)={\int }_x^{x+1}f(t)dt,\mathrm{△}f(x)=f(x+1)-f(x),Df(x)=f^{\prime }(x).$$

(6)

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

Let ${\mathbb{P}}_n=\{p(x)\in \mathbf{Q}(x)|degp(x)\le n\}$ be the $(n+1)$-dimensional vector space over Q. Probably, $\{1,x,\dots ,x^n\}$ is the most natural basis for this space. But $\{B_0^{(r)}(x),B_1^{(r)}(x),B_2^{(r)}(x),\dots ,B_n^{(r)}(x)\}$ is also a good basis for the space ${\mathbb{P}}_n$ for our purpose of arithmetical and combinatorial applications.

Let $p(x)\in {\mathbb{P}}_n$. Then $p(x)$ can be generated by $B_0^{(r)}(x),B_1^{(r)}(x),B_2^{(r)}(x),\dots ,B_n^{(r)}(x)$ as follows:

$$p(x)=\sum _{k=0}^n{a}_k{B}_k^{(r)}(x).$$

In this paper, we develop methods for uniquely determining $a_k$ from the information of $p(x)$. 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 $B_n^{(0)}=x^n$ ($n\in {\mathbf{Z}}_{+}$). Let $p(x)\in {\mathbb{P}}_n$.

For a fixed $r\in {\mathbf{Z}}_{+}$, $p(x)$ can be generated by $B_0^{(r)}(x),B_1^{(r)}(x),B_2^{(r)}(x),\dots ,B_n^{(r)}(x)$ as follows:

$$p(x)=\sum _{k=0}^n{a}_k{B}_k^{(r)}(x).$$

(7)

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

$$\begin{array}{rcl}I{B}_n^{(r)}(x)& =& {\int }_x^{x+1}{B}_n^{(r)}(x)dx\\ =& \frac{1}{n+1}(B_{n+1}^{(r)}(x+1)-B_{n+1}^{(r)}(x)).\end{array}$$

(8)

By (5) and (8), we get

$$I{B}_n^{(r)}(x)=\frac{n+1}{n+1}{B}_n^{(r-1)}(x)=B_n^{(r-1)}(x).$$

(9)

It is easy to show that

$$\mathrm{△}{B}_n^{(r)}(x)=B_n^{(r)}(x+1)-B_n^{(r)}(x)=n{B}_{n-1}^{(r-1)}(x),$$

(10)

and

$$D{B}_n^{(r)}(x)=n{B}_{n-1}^{(r)}(x).$$

(11)

By (7) and (9), we get

$$I^{r}p(x)=\sum _{k=0}^n{a}_k{B}_k^{(0)}(x)=\sum _{k=0}^n{a}_k{x}^k.$$

(12)

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

$$D^k{I}^{r}p(x)=\sum _{l=k}^n{a}_l\frac{l!}{(l-k)!}{x}^{l-k}.$$

(13)

Thus, by (13) we get

$$D^k{I}^{r}p(0)=k!a_k.$$

(14)

Hence, from (14) we have

$$a_k=\frac{D^k{I}^{r}p(0)}{k!}.$$

(15)

Case 1. Let $r>n$. Then $r>k$ for all $k=0,1,2,\dots ,n$.

By (15), we get

$$\begin{array}{rl}{a}_k& =\frac{1}{k!}{D}^k{I}^k{I}^{r-k}p(0)=\frac{1}{k!}{(DI)}^k{I}^{r-k}p(0)\\ =\frac{1}{k!}{\mathrm{△}}^k{I}^{r-k}p(0)=\frac{1}{k!}\sum _{j=0}^k\left(\begin{array}{c}k\\ j\end{array}\right){(-1)}^{k-j}{I}^{r-k}p(j).\end{array}$$

(16)

Case 2. Assume that $r\le n$.

• (i) For $0\le k\le r$, by (15) we get

  $$a_k=\frac{1}{k!}\sum _{j=0}^k{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)I^{r-k}p(j).$$

  (17)

• (ii) For $r\le k\le n$, by (15) we see that

  $$\begin{array}{rl}{a}_k& =\frac{1}{k!}{D}^{k-r}{D}^r{I}^{r}p(0)=\frac{1}{k!}{D}^{k-r}{(DI)}^{r}p(0)=\frac{1}{k!}{D}^{k-r}{\mathrm{△}}^{r}p(0)\\ =\frac{1}{k!}{\mathrm{△}}^r{D}^{k-r}p(0)=\frac{1}{k!}\sum _{j=0}^r\left(\begin{array}{c}r\\ j\end{array}\right){(-1)}^{r-j}{D}^{k-r}p(j).\end{array}$$

  (18)

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

Theorem 1

• (a) For $r>n$, we have

  $$p(x)=\sum _{k=0}^n(\sum _{j=0}^k\frac{1}{k!}{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)I^{r-k}p(j))B_k^{(r)}(x).$$

• (b) For $r\le n$, we have

  $$\begin{array}{rcl}p(x)& =& \sum _{k=0}^{r-1}(\sum _{j=0}^k\frac{1}{k!}{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)I^{r-k}p(j))B_k^{(r)}(x)\\ +\sum _{k=r}^n(\sum _{j=0}^r\frac{1}{k!}{(-1)}^{r-j}{D}^{k-r}p(j))B_k^{(r)}(x).\end{array}$$

Let us take $p(x)=x^n\in {\mathbb{P}}_n$. Then $x^n$ can be expressed as a linear combination of $B_0^{(r)},B_1^{(r)},\dots ,B_n^{(r)}$. For $r>n$, we have

$$I^{r-k}{x}^n=\frac{n!}{(n+r-k)!}\sum _{l=0}^{r-k}{(-1)}^{r-k-l}\left(\begin{array}{c}r-k\\ l\end{array}\right){(x+l)}^{n+r-k}.$$

(19)

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

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

$$x^n=\sum _{k=0}^n\left\{\sum _{j=0}^k\sum _{l=0}^{r-k}{(-1)}^{r-j-l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}r-k\\ l\end{array}\right)}{k!(n+r-k)!}{(j+l)}^{n+r-k}\right\}{B}_k^{(r)}(x).$$

Let us assume that $r,n\in {\mathbf{Z}}_{+}$ with $r\le n$. Observe that

$$D^{k-r}{x}^n=n(n-1)\cdots (n-k+r+1)x^{n-k+r}=\frac{n!}{(n-k+r)!}{x}^{n-k+r}.$$

(20)

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

Corollary 3 For $n,r\in {\mathbf{Z}}_{+}$ with $r⩽n$, we have

$$\begin{array}{rcl}{x}^n& =& \sum _{k=0}^{r-1}\left\{\sum _{j=0}^k\sum _{l=0}^{r-k}{(-1)}^{r-j-l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}r-k\\ l\end{array}\right)}{k!(n+r-k)!}{(j+l)}^{n+r-k}\right\}{B}_k^{(r)}(x)\\ +\sum _{k=r}^n\left\{\sum _{j=0}^r{(-1)}^{r-j}\frac{n!\left(\begin{array}{c}r\\ j\end{array}\right)}{k!(n+r-k)!}{j}^{n+r-k}\right\}{B}_k^{(r)}(x).\end{array}$$

Let us take $p(x)=B_n^{(s)}(x)\in {\mathbb{P}}_n$ ($s\in {\mathbf{Z}}_{+}$). Then $p(x)$ can be generated by $\{B_0^{(r)}(x),B_1^{(r)}(x),\dots ,B_n^{(r)}(x)\}$ as follows:

$$B_n^{(s)}(x)=\sum _{k=0}^n{a}_k{B}_k^{(r)}(x).$$

(21)

For $r>n$, we have

$$I^{r-k}{B}_n^{(s)}(x)=B_n^{(s-r+k)}(x).$$

(22)

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

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

$$B_n^{(s)}(x)=\sum _{k=0}^n\{\sum _{j=0}^k{(-1)}^{k-j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right)B_n^{(s+k-r)}(j)\}{B}_k^{(r)}(x).$$

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

$$\begin{array}{rcl}{B}_n^{(r)}(x)& =& 0B_0^{(r)}(x)+0B_1^{(r)}(x)+\cdots +0B_{n-1}^{(r)}(x)+1B_n^{(r)}(x)\\ =& \sum _{k=0}^n\{\sum _{j=0}^k{(-1)}^{k-j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right)B_n^{(k)}(j)\}{B}_k^{(r)}(x).\end{array}$$

(23)

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

$$\sum _{j=0}^k{(-1)}^{k-j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right)B_n^{(k)}(j)={\delta }_{kn},\text{for }0\le k\le n.$$

(24)

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

Corollary 5

• (a) For $n,k\in {\mathbf{Z}}_{+}$ with $0\le k\le n-1$, we have

  $$\sum _{j=0}^k{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)B_n^{(k)}(j)=0.$$

• (b) In particular, $k=n$, we get

  $$\sum _{j=0}^n{(-1)}^{n-j}\left(\begin{array}{c}n\\ j\end{array}\right)B_n^{(n)}(j)=n!.$$

Let us assume that $r\le n$ in (21). Then we have

$$D^{k-r}{B}_n^{(s)}(x)=n(n-1)\cdots (n-k+r+1)B_{n+r-k}^{(s)}(x)=\frac{n!}{(n-k+r!)}{B}_{n+r-k}^{(s)}(x).$$

(25)

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

Theorem 6 For $r,n\in {\mathbf{Z}}_{+}$ with $r\le n$, we have

$$\begin{array}{rcl}{B}_n^{(s)}(x)& =& \sum _{k=0}^{n-1}\{\sum _{j=0}^k{(-1)}^{k-j}\frac{1}{k!}\left(\begin{array}{c}k\\ j\end{array}\right)B_n^{(s+k-r)}(j)\}{B}_k^{(r)}(x)\\ +\sum _{k=r}^n\{\sum _{j=0}^r{(-1)}^{r-j}\frac{n!\left(\begin{array}{c}r\\ j\end{array}\right)}{k!(n+r-k)!}{B}_{n+r-k}^{(s)}(j)\}{B}_k^{(r)}(x).\end{array}$$

Let $p(x)=E_n^{(s)}(x)$ ($s\in {\mathbf{Z}}_{+}$) be Euler polynomials of order s. Then $E_n^{(s)}$ can be expressed as a linear combination of $B_0^{(r)}(x),B_1^{(r)}(x),\dots ,B_n^{(r)}(x)$.

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

By (6), we get

$$\begin{array}{rl}{I}^{r-k}{E}_n^{(s)}(x)& =\frac{1}{(n+1)\cdots (n+r-k)}\sum _{l=0}^{r-k}{(-1)}^{r-k-l}\left(\begin{array}{c}r-k\\ l\end{array}\right)E_{n+r-k}^{(s)}(x+l)\\ =\frac{n!}{(n+r-k)!}\sum _{l=0}^{r-k}{(-1)}^{r-k-l}\left(\begin{array}{c}r-k\\ l\end{array}\right)E_{n+r-k}^{(s)}(x+l).\end{array}$$

(26)

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

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

$$E_n^{(s)}(x)=\sum _{k=0}^n\{\sum _{j=0}^k\sum _{l=0}^{r-k}{(-1)}^{r-j-l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}r-k\\ l\end{array}\right)}{k!(n+r-k)!}{E}_{n+r-k}^{(s)}(j+l)\}{B}_k^{(r)}(x).$$

For $r,n\in {\mathbf{Z}}_{+}$ with $r\le n$, we have

$$D^{k-r}{E}_n^{(s)}(x)=n(n-1)\cdots (n-k+r+1)E_{n-k+r}^{(s)}(x).$$

(27)

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

Theorem 8 For $r,n\in {\mathbf{Z}}_{+}$ with $r\le n$, we have

$$\begin{array}{rcl}{E}_n^{(s)}(x)& =& \sum _{k=0}^{r-1}\{\sum _{j=0}^k\sum _{l=0}^{r-k}{(-1)}^{r-j-l}\frac{n!\left(\begin{array}{c}k\\ j\end{array}\right)\left(\begin{array}{c}r-k\\ l\end{array}\right)}{k!(n+r-k)!}{E}_{n+r-k}^{(s)}(j+l)\}{B}_k^{(r)}(x)\\ +\sum _{k=r}^n\{\sum _{j=0}^r{(-1)}^{r-j}\frac{n!\left(\begin{array}{c}r\\ k\end{array}\right)}{k!(n+r-k)!}{E}_{n+r-k}^{(s)}(j)\}{B}_k^{(r)}(x).\end{array}$$

Remarks (a) For $r\le 0$, by (40) we get

$$I^r{x}^n=\frac{n!}{(n+r)!}\sum _{j=0}^r\left(\begin{array}{c}r\\ j\end{array}\right){(-1)}^{r-j}{(x+j)}^{n+r}=\frac{1}{\left(\begin{array}{c}n+r\\ r\end{array}\right)}\frac{1}{r!}\sum _{j=0}^r{(-1)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right){(x+j)}^{n+r}.$$

Thus, for $x=0$, we have

$$I^r{x}^n{|}_{x=0}=\frac{n!}{(n+r)!}\sum _{j=0}^r\left(\begin{array}{c}r\\ j\end{array}\right){(-1)}^{r-j}{j}^{n+r}=\frac{1}{\left(\begin{array}{c}n+r\\ r\end{array}\right)}\frac{1}{r!}{\mathrm{△}}^r{0}^{n+r}=\frac{S(n+r,r)}{\left(\begin{array}{c}n+r\\ r\end{array}\right)},$$

(28)

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

• (b) Assume

  $$\sum _{k=0}^n{\alpha }_k{x}^k=\sum _{k=0}^n{a}_k{B}_k^{(r)}(x)(r\ge 0).$$

  (29)

Applying $I^t$ on both sides ($t\ge 0$), we get

$$\sum _{k=0}^n{a}_k{B}_k^{(r-t)}(x)=\sum _{k=0}^n{\alpha }_k{I}^t{x}^k=\sum _{k=0}^n\frac{{\alpha }_k}{\left(\begin{array}{c}\alpha +t\\ t\end{array}\right)}\frac{1}{t!}\sum _{j=0}^t{(-1)}^{(t-j)}\left(\begin{array}{c}t\\ j\end{array}\right){(x+j)}^{k+t}.$$

(30)

From (28) and (30), we have

$$\sum _{k=0}^n{a}_k{B}_k^{(r-t)}=\sum _{k=0}^n\frac{{\alpha }_k}{\left(\begin{array}{c}\alpha +t\\ t\end{array}\right)}S(k+t,t).$$

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

$$d=e^{-D}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{D}^n,\tilde{d}=e^D=\sum _{n=0}^{\mathrm{\infty }}\frac{D^n}{n!}.$$

(31)

From (31), we note that

$$\begin{array}{r}\tilde{d}{x}^n=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right)x^{n-l}={(x+1)}^n,\\ d{x}^n=\sum _{l=0}^n\left(\begin{array}{c}n\\ l\end{array}\right){(-1)}^l{x}^{n-l}={(x-1)}^n.\end{array}$$

(32)

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

$$\tilde{d}{B}_n^{(r)}(x)=B_n^{(r)}(x+1),d{B}_n^{(r)}(x)=B_n^{(r)}(x-1),$$

(33)

and

$$\tilde{d}{E}_n^{(r)}(x)=E_n^{(r)}(x+1),d{E}_n^{(r)}(x)=E_n^{(r)}(x-1).$$

(34)

3 Further remarks

For any $r_0,r_1,\dots ,r_n\in {\mathbf{Z}}_{+}$, $\{B_0^{(r_0)}(x),B_1^{(r_1)}(x),\dots ,B_0^{(r_n)}(x)\}$ forms a basis for ${\mathbb{P}}_n$. Let $r=max\{r_i|i=0,1,2,\dots ,n\}$. Let $p(x)\in {\mathbb{P}}_n$. Then $p(x)$ can be expressed as a linear combination of $B_0^{(r_0)}(x),B_1^{(r_1)}(x),\dots ,B_n^{(r_n)}(x)$ as follows:

$$p(x)=a_0{B}_0^{(r_0)}(x)+a_1{B}_1^{(r_1)}(x)+\cdots +a_n{B}_n^{(r_n)}(x)=\sum _{l=0}^n{a}_l{B}_l^{(r_l)}(x).$$

(35)

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

$$\begin{array}{rl}{I}^{r}p(x)& =\sum _{l=0}^n{a}_l{I}^r{B}_l^{(r_l)}(x)\\ =\sum _{l=0}^n{a}_l{I}^{r-r_l}{I}^{r_l}{B}_l^{(r_l)}(x)=\sum _{l=0}^n{a}_l{I}^{r-r_l}{B}_l^{(0)}(x)\\ =\sum _{l=0}^n{a}_l{I}^{r-r_l}{x}^l.\end{array}$$

(36)

Now, for each $k=0,1,2,\dots ,n$, by (36) we get

$$\begin{array}{rl}{D}^k{I}^{r}p(x)& =\sum _{l=0}^n{a}_l{D}^k{I}^{r-r_l}{x}^l=\sum _{l=0}^n{a}_l{I}^{r-r_l}\left(D^k{x}^l\right)\\ =\sum _{l=k}^n{a}_l{I}^{r-r_l}\left(\frac{l!}{(l-k)!}{x}^{l-k}\right)=\sum _{l=k}^n\frac{a_{l}l!}{(l-k)!}{I}^{r-r_l}{x}^{l-k}.\end{array}$$

(37)

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

$$\begin{array}{rl}{D}^k{I}^{r}p(0)& =\sum _{l=k}^n\frac{l!a_l}{(l-k)!}\times \frac{S(l-k+r-r_l,r-r_l)}{\left(\begin{array}{c}l-k+r-r_l\\ r-r_l\end{array}\right)}\\ =\sum _{l=k}^n\frac{a_l(r-r_l)!l!}{(l-k)!}S(l-k+r-r_l,r-r_l).\end{array}$$

(38)

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

$$\begin{array}{rl}{D}^k{I}^{r}p(0)& =D^k{I}^k{I}^{r-k}p(0)={(DI)}^k{I}^{r-k}p(0)={\mathrm{△}}^k{I}^{r-k}p(0)\\ =\sum _{j=0}^k{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)I^{r-k}p(j).\end{array}$$

(39)

Case 2. Let $r⩽n$.

• (i) For $0⩽k<r$, we have

  $$D^k{I}^{r}p(0)=\sum _{j=0}^k{(-1)}^{k-j}\left(\begin{array}{c}k\\ j\end{array}\right)I^{r-k}p(j).$$

  (40)

• (ii) For $r⩽k⩽n$, we have

  $$\begin{array}{rl}{D}^k{I}^{r}p(0)& =D^{k-r}{D}^r{I}^{r}p(0)=D^{k-r}{(DI)}^{r}p(0)=D^{k-r}{\mathrm{△}}^{r}p(0)\\ ={\mathrm{△}}^r{D}^{k-r}p(0)=\sum _{j=0}^r{(-1)}^{r-j}\left(\begin{array}{c}r\\ j\end{array}\right)D^{k-r}p(j).\end{array}$$

  (41)

Thus, by (38), (39), (40) and (41), we can determine $a_0,a_1,a_2,\dots ,a_n$.