Some identities for the product of two Bernoulli and Euler polynomials

Abstract

Let ℙ _n be the space of polynomials of degree less than or equal to n. In this article, using the Bernoulli basis {B₀(x), . . . , B _n (x)} for ℙ _n consisting of Bernoulli polynomials, we investigate some new and interesting identities and formulae for the product of two Bernoulli and Euler polynomials like Carlitz did.

1 Introduction

The Bernoulli and Euler polynomials are defined by means of

$$\frac{t}{e^t-1}{e}^{xt}={\displaystyle \sum _{n=0}^{\infty }}{B}_n\left(x\right)\frac{t^n}{n!},\frac{2}{e^t+1}{e}^{xt}={\displaystyle \sum _{n=0}^{\infty }}{E}_n\left(x\right)\frac{t^n}{n!}.$$

(1)

In the special case, x = 0, B _n (0) = B _n and E _n (0) = E _n are called the n-th Bernoulli and Euler numbers (see [1–17]).

From (1), we note that

$$B_n\left(x\right)={\displaystyle \sum _{l=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)B_l{x}^{n-l},E_n\left(x\right)={\displaystyle \sum _{l=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)E_l{x}^{n-l}.$$

(2)

For n ≥ 0, we have

$$\frac{d}{dx}{B}_n\left(x\right)=n{B}_{n-1}\left(x\right),\frac{d}{dx}{E}_n\left(x\right)=n{E}_{n-1}\left(x\right),$$

(3)

(see [7, 8]).

By (1), we get the following recurrence for the Bernoulli and the Euler numbers:

$$B_0=1,B_n\left(1\right)-B_n={\delta }_{1,n}\text{and}{E}_0=1,E_n\left(1\right)+E_n=2{\delta }_{0,n},$$

(4)

where δ_{k, n}is the Kronecker symbol (see [1–17]).

Thus, from (3) and (4), we have

$$\underset{0}{\overset{1}{\int }}{B}_n\left(x\right)dx=\frac{{\delta }_{0,n}}{n+1},\underset{0}{\overset{1}{\int }}{E}_n\left(x\right)dx=-\frac{2E_{n+1}}{n+1}.$$

(5)

It is known [12] that

$$\underset{0}{\overset{A}{\int }}{B}_{m_1}\left(\frac{x}{a_1}\right)\dots B_{m_n}\left(\frac{x}{a_n}\right)dx=a_1^{1-m_1}\dots a_n^{1-m_n}\underset{0}{\overset{1}{\int }}{B}_{m_1}\left(x\right)\dots B_{m_n}\left(x\right)dx,$$

(6)

where a₁, a₂, . . . , a _n are positive integers that are relatively prime in pairs A = a₁a₂ . . . a _n .

For n = 2, there is the formula

$$\underset{0}{\overset{1}{\int }}{B}_p\left(x\right)B_q\left(x\right)dx={\left(-1\right)}^{p+1}\frac{B_{p+q}}{\left(\begin{array}{c}\hfill p+q\hfill \\ \hfill q\hfill \end{array}\right)},$$

(7)

where p + q ≥ 2 (see [3, 4]). In [3, 4], we can find the following formula for a product of two Bernoulli polynomials:

$$B_m\left(x\right)B_n\left(x\right)={\displaystyle \sum _r}\left[\left(\begin{array}{c}\hfill m\hfill \\ \hfill 2r\hfill \end{array}\right)n+\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2r\hfill \end{array}\right)m\right]\frac{B_{2r}{B}_{m+n-2r}\left(x\right)}{m+n-2r}+{\left(-1\right)}^{m+1}\frac{B_{m+n}}{\left(\begin{array}{c}\hfill m+n\hfill \\ \hfill n\hfill \end{array}\right)},\text{for}m+n\ge 2.$$

(8)

Assume m, n, p ≥ 1. Then, by (7) and (8), we get

$$\underset{0}{\overset{1}{\int }}{B}_m\left(x\right)B_n\left(x\right)B_p\left(x\right)dx={\left(-1\right)}^{p+1}p!{\displaystyle \sum _r}\left[\left(\begin{array}{c}\hfill m\hfill \\ \hfill 2r\hfill \end{array}\right)n+\left(\begin{array}{c}\hfill n\hfill \\ \hfill 2r\hfill \end{array}\right)m\right]\frac{\left(m+n-2r-1\right)!}{\left(m+n+p-2r\right)!}{B}_{2r}{B}_{m+n+p-2r},$$

(9)

(see [4]).

In [8], it is known that for n ∈ ℤ₊,

$$B_n\left(x\right)={\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne 1\end{array}}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_k{E}_{n-k}\left(x\right)$$

(10)

and

$$E_n\left(x\right)=-2{\displaystyle \sum _{l=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)\frac{E_{l+1}}{l+1}{B}_{n-l}\left(x\right).$$

(11)

Let ℙ _n = {∑ _i a _i xⁱ|a _i ∈ ℚ} be the space of polynomials of degree less than or equal to n. In this article, using the Bernoulli basis {B₀(x), . . . , B _n (x)} for ℙ _n consisting of Bernoulli polynomials, we investigate some new and interesting identities and formulae for the product of two Bernoulli and Euler polynomials like Carlitz did.

2 Bernoulli identities arising from Bernoulli basis polynomials

From (1), we note that

$$\begin{array}{cc}\hfill e^{xt}& =\frac{1}{t}\left.\left(\frac{t\left(e^t-1\right)}{e^t-1}\right)e^{xt}\right)=\frac{1}{t}{\displaystyle \sum _{n=0}^{\infty }}\left(B_n\left(x+1\right)-B_n\left(x\right)\right)\frac{t^n}{n!}\hfill \\ =\frac{1}{t}{\displaystyle \sum _{n=1}^{\infty }}\left(B_n\left(x+1\right)-B_n\left(x\right)\right)\frac{t^n}{n!}\hfill \\ ={\displaystyle \sum _{n=0}^{\infty }}\left(\frac{B_{n+1}\left(x+1\right)-B_{n+1}\left(x\right)}{n+1}\right)\frac{t^n}{n!}.\hfill \end{array}$$

(12)

Thus, from (12), we have

$$x^n=\frac{1}{n+1}\left(B_{n+1}\left(x+1\right)-B_{n+1}\left(x\right)\right)=\frac{1}{n+1}{\displaystyle \sum _{l=0}^n}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)B_l\left(x\right).$$

(13)

From (13), we note that {B₀(x), B₁(x), . . . , B _n (x)} spans ℙ _n . For p(x) ∈ ℙ _n , let $p\left(x\right)={\sum }_{k=0}^n{a}_k{B}_k\left(x\right)$ and g(x) = p(x + 1) - p(x). Then we have

$$g\left(x\right)={\displaystyle \sum _{k=0}^n}{a}_k\left(B_k\left(x+1\right)-B_k\left(x\right)\right)={\displaystyle \sum _{k=0}^n}k{a}_k{x}^{k-1}.$$

(14)

From (14), we can derive the following Equation (15):

$$g^{\left(r\right)}\left(x\right)={\displaystyle \sum _{k=r+1}^n}k\left(k-1\right)\dots \left(k-r\right)a_k{x}^{k-r-1},$$

(15)

where $g^{\left(r\right)}\left(x\right)=\frac{d^{r}g\left(x\right)}{d{x}^r}$ and r = 0, 1, 2, . . . , n. Let us take x = 0 in (15). Then we have

$$g^{\left(r\right)}\left(0\right)=\left(r+1\right)!a_{r+1}.$$

(16)

By (16), we get, for r = 1, 2, . . . , n,

$$a_r=\frac{g^{\left(r-1\right)}\left(0\right)}{r!}=\frac{1}{r!}\left(p^{\left(r-1\right)}\left(1\right)-p^{\left(r-1\right)}\left(0\right)\right).$$

(17)

Let $0=p\left(x\right)={\sum }_{k=0}^n{a}_k{B}_k\left(x\right)$. Then, from (17), we have

$$a_r=\frac{1}{r!}{g}^{\left(r-1\right)}\left(0\right)=\frac{1}{r!}\left(p^{\left(r-1\right)}\left(1\right)-p^{\left(r-1\right)}\left(0\right)\right)=0.$$

(18)

From (18), we note that {B₀(x), B₁(x), . . . , B _n (x)} is a linearly independent set. Therefore, we obtain the following theorem.

Proposition 1 The set of Bernoulli polynomials {B₀(x), B₁(x), . . . , B _n (x)} is a basis for ℙ _n .

Let us consider polynomial p(x) ∈ ℙ _n as a linear combination of Bernoulli basis polynomials with

$$p\left(x\right)=C_0{B}_0\left(x\right)+C_1{B}_1\left(x\right)+\cdots +C_n{B}_n\left(x\right).$$

(19)

We can write (19) as a dot product of two variables:

$$p\left(x\right)=\left(B_0\left(x\right),B_1\left(x\right),\dots ,B_n\left(x\right)\right)\left(\begin{array}{c}\hfill C_0\hfill \\ \hfill C_1\hfill \\ \hfill ⋮\hfill \\ \hfill C_n\hfill \end{array}\right).$$

(20)

From (20), we can derive the following equation:

$$p\left(x\right)=\left(1,x,x^2,\dots ,x^n\right)\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill b_{12}\hfill & \hfill b_{13}\hfill & \hfill \cdots \hfill & \hfill b_{1n+1}\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill b_{23}\hfill & \hfill \cdots \hfill & \hfill b_{2n+1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \cdots \hfill & \hfill b_{3n+1}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill b_{nn+1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill C_0\hfill \\ \hfill C_1\hfill \\ \hfill C_2\hfill \\ \hfill ⋮\hfill \\ \hfill C_n\hfill \end{array}\right),$$

(21)

where b _ij are the coefficients of the power basis that are used to determine the respective Bernoulli polynomials. It is easy to show that

$$B_0\left(x\right)=1,B_1\left(x\right)=x-\frac{1}{2},B_2\left(x\right)=x^2-x+\frac{1}{6},B_3\left(x\right)=x^3-\frac{3}{2}{x}^2+\frac{1}{2}x,\dots .$$

In the quadratic case (n = 2), the matrix representation is

$$p\left(x\right)=\left(1,x,x^2\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill -\frac{1}{2}\hfill & \hfill \frac{1}{6}\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill C_0\hfill \\ \hfill C_1\hfill \\ \hfill C_2\hfill \end{array}\right).$$

(22)

In the cubic case (n = 3), the matrix representation is

$$p\left(x\right)=\left(1,x,x^2,x^3\right)\left(\begin{array}{cccc}\hfill 1\hfill & \hfill -\frac{1}{2}\hfill & \hfill \frac{1}{6}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill & \hfill \frac{1}{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -\frac{3}{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill C_0\hfill \\ \hfill C_1\hfill \\ \hfill C_2\hfill \\ \hfill C_3\hfill \end{array}\right).$$

(23)

In many applications of Bernoulli polynomials, a matrix formulation for the Bernoulli polynomials seems to be useful.

There are many ways of obtaining polynomial identities in general. Here, in Theorems 2-9, we use the Bernoulli basis in order to express certain polynomials as linear combinations of that basis and hence to get some new and interesting polynomial identities.

Let $I_{m,n}={\int }_0^1{B}_m\left(x\right)B_n\left(x\right)dx\text{for}m,n\in {\mathbb{Z}}_{+}$. Then, by integration by parts, we get

$$I_{0,n}=I_{m,0}=0,I_{m,n}={\left(-1\right)}^{m+n}\frac{B_{m+n}}{\left(\begin{array}{c}\hfill m+n\hfill \\ \hfill m\hfill \end{array}\right)},\left(m,n\ge 2\right).$$

(24)

For n ∈ ℤ₊ with n ≥ 2, let us consider the following polynomials in ℙ _n :

$$p\left(x\right)={\displaystyle \sum _{k=0}^n}{B}_k\left(x\right)B_{n-k}\left(x\right)\in {\mathbb{P}}_n.$$

(25)

Then, from (25), we have

$$p^{\left(r\right)}\left(x\right)=\frac{\left(n+1\right)!}{\left(n-r+1\right)!}{\displaystyle \sum _{k=r}^n}{B}_{k-r}\left(x\right)B_{n-k}\left(x\right),$$

(26)

where r = 0, 1, 2, . . . n.

By Proposition 1, we see that p(x) can be written as

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

(27)

From (25) and (27), we note that

$$a_0={\displaystyle \underset{0}{\overset{1}{\int }}}p\left(t\right)dt={\displaystyle \sum _{k=0}^n}{I}_{k,n-k}=B_n{\displaystyle \sum _{k=1}^{n-1}}\frac{{\left(-1\right)}^{k-1}}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}=B_n\frac{\left(1+{\left(-1\right)}^n\right)}{n+2}=\frac{2}{n+2}{B}_n.$$

By (18) and (26), we get

$$\begin{array}{cc}\hfill a_{r+1}& =\frac{1}{\left(r+1\right)!}\left(p^{\left(r\right)}\left(1\right)-p^{\left(r\right)}\left(0\right)\right)\hfill \\ =\frac{\left(n+1\right)!}{\left(r+1\right)!\left(n-r+1\right)!}{\displaystyle \sum _{k=r}^n}\left(B_{k-r}\left(1\right)B_{n-k}\left(1\right)-B_{k-r}{B}_{n-k}\right)\hfill \\ =\frac{1}{n+2}\left(\begin{array}{c}\hfill n+r\hfill \\ \hfill r+1\hfill \end{array}\right){\displaystyle \sum _{k=r}^n}\left\{\left({\delta }_{1,k-r}+B_{k-r}\right)\left({\delta }_{1,n-k}+B_{n-k}\right)-B_{k-r}{B}_{n-k}\right\}\hfill \\ =\frac{1}{n+2}\left(\begin{array}{c}n+2\hfill \\ r+1\hfill \end{array}\right)\left(B_{n-r-1}+B_{n-r-1}+{\delta }_{r,n-2}\right)\hfill \\ =\left\{\begin{array}{cc}\frac{2}{n+2}\left(\begin{array}{c}n+2\\ r+1\end{array}\right)B_{n-r-1}\hfill & \text{if }r\ne n-2.\hfill \\ 0\hfill & \text{if }r=n-2.\hfill \end{array}\right.\hfill \end{array}$$

(28)

Therefore, by (25), (27) and (28), we obtain the following theorem.

Theorem 2 For n ∈ ℤ₊ with n ≥ 2, we have

$${\displaystyle \sum _{k=0}^n}{B}_k\left(x\right)B_{n-k}\left(x\right)=\frac{2}{n+2}{\displaystyle \sum _{k=0}^{n-2}}\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)B_{n-k}{B}_k\left(x\right)+\left(n+1\right)B_n\left(x\right).$$

For n ∈ ℤ₊ with n ≥ 2, let us take polynomial p(x) in ℙ _n as follows:

$$p\left(x\right)={\displaystyle \sum _{k=0}^n}\frac{1}{k!\left(n-k\right)!}{B}_k\left(x\right)B_{n-k}\left(x\right)\in {\mathbb{P}}_n.$$

(29)

From Proposition 1, we note that p(x) is given by means of Bernoulli basis polynomials:

$$p\left(x\right)={\displaystyle \sum _{k=0}^n}{a}_k{B}_k\left(x\right)\in {\mathbb{P}}_n.$$

(30)

By (24), (29) and (30), we get

$$\begin{array}{cc}\hfill a_0& ={\displaystyle \underset{0}{\overset{1}{\int }}}p\left(t\right)dt={\displaystyle \sum _{k=0}^n}\frac{1}{k!\left(n-k\right)!}{I}_{k,n-k}=\frac{2I_{0,n}}{n!}+{\displaystyle \sum _{k=1}^{n-1}}\frac{{\left(-1\right)}^{k-1}}{k!\left(n-k\right)!\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{B}_n\hfill \\ =\frac{B_n}{n!}{\displaystyle \sum _{k=1}^{n-1}}{\left(-1\right)}^{k-1}=\frac{B_n}{n!}\frac{\left(1+{\left(-1\right)}^n\right)}{2}=\frac{B_n}{n!}.\hfill \end{array}$$

(31)

From (29), we have that for r = 0, 1, 2, . . . , n,

$$p^{\left(r\right)}\left(x\right)=2^r{\displaystyle \sum _{k=r}^n}\frac{B_{k-r}\left(x\right)B_{n-k}\left(x\right)}{\left(k-r\right)!\left(n-k\right)!}.$$

(32)

By (18), we get

$$\begin{array}{cc}\hfill a_{r+1}& =\frac{1}{\left(r+1\right)!}\left(p^{\left(r\right)}\left(1\right)-p^{\left(r\right)}\left(0\right)\right)\hfill \\ =\frac{2^r}{\left(r+1\right)!}{\displaystyle \sum _{k=r}^n}\frac{1}{\left(k-r\right)!\left(n-k\right)!}\left(B_{k-r}\left(1\right)B_{n-k}\left(1\right)-B_{k-r}{B}_{n-k}\right)\hfill \\ =\frac{2^r}{\left(r+1\right)!}\left(\frac{2B_{n-r-1}}{\left(n-1-r\right)!}+{\displaystyle \sum _{k=r}^n}{\delta }_{1,k-r}{\delta }_{1,n-k}\right)\hfill \\ =\left\{\begin{array}{cc}\hfill \frac{2^{r+1}}{n!}\left(\begin{array}{c}\hfill n\hfill \\ \hfill r+1\hfill \end{array}\right)B_{n-r-1}\hfill & \hfill \text{if }r\ne n-2,\hfill \\ \hfill 0\hfill & \hfill \text{if }r=n-2.\hfill \end{array}\right.\hfill \end{array}$$

(33)

Therefore, from (29), (30) and (33), we obtain the following theorem.

Theorem 3 For n ∈ ℤ₊ with n ≥ 2, we have

$${\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_k\left(x\right)B_{n-k}\left(x\right)={\displaystyle \sum _{\begin{array}{c}k=0\\ k\ne n-1\end{array}}^n}{2}^k\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_{n-k}{B}_k\left(x\right).$$

Let n ∈ ℤ₊ with n ≥ 2. Then we consider polynomial p(x) in ℙ _n with

$$p\left(x\right)={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{B}_k\left(x\right)B_{n-k}\left(x\right).$$

By Proposition 1, we see that p(x) is written as

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

(34)

From (34), we have

$$\begin{array}{cc}\hfill a_0& ={\displaystyle \underset{0}{\overset{1}{\int }}}p\left(t\right)dt={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{\displaystyle \underset{0}{\overset{1}{\int }}}{B}_k\left(t\right)B_{n-k}\left(t\right)dt\hfill \\ ={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}\frac{{\left(-1\right)}^{k-1}}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{B}_n=\left(\frac{1+{\left(-1\right)}^n}{n^2}\right)B_n=\frac{2B_n}{n^2}.\hfill \end{array}$$

It is easy to show that for r = 1, 2 , . . . , n - 1,

$$p^{\left(r\right)}\left(x\right)=2C_r{B}_{n-r}\left(x\right)+\left(n-1\right)\cdots \left(n-r\right){\displaystyle \sum _{k=r+1}^{n-1}}\frac{B_{k-r}\left(x\right)B_{n-k}\left(x\right)}{\left(k-r\right)\left(n-k\right)},$$

(35)

where $C_r=\frac{1}{n-r}{\sum }_{j=1}^r\left(n-1\right)...\left(n-j+1\right)\left(n-j-1\right)...\left(n-r\right).$

By (17), we get

$$\begin{array}{cc}\hfill a_{r+1}& =\frac{1}{\left(r+1\right)!}\left(p^{\left(r\right)}\left(1\right)-p^{\left(r\right)}\left(0\right)\right)\hfill \\ =\frac{1}{\left(r+1\right)!}\left\{2C_r\left(B_{n-r}\left(1\right)-B_{n-r}\right)\right.\hfill \\ \left.+\left(n-1\right)\dots \left(n-r\right){\displaystyle \sum _{k=r+1}^{n-1}}\frac{B_{k-r}\left(1\right)B_{n-k}\left(1\right)-B_{k-r}{B}_{n-k}}{\left(k-r\right)\left(n-k\right)}\right\}\hfill \\ =\frac{2C_r}{\left(r+1\right)!}{\delta }_{r,n-1}+\frac{1}{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill r+1\hfill \end{array}\right){\displaystyle \sum _{k=r+1}^{n-1}}\frac{B_{k-r}{\delta }_{1,n-k}+{\delta }_{1,k-r}{B}_{n-k}+{\delta }_{1,k-r}{\delta }_{1,n-k}}{\left(k-r\right)\left(n-k\right)}\hfill \\ =\left\{\begin{array}{cc}\frac{2}{n\left(n-r-1\right)}\left(\begin{array}{c}\hfill n\hfill \\ \hfill r+1\hfill \end{array}\right)B_{n-r-1}\hfill & \text{if}0\le r\le n-3,\hfill \\ 0\hfill & \text{if}r=n-2,\hfill \\ \frac{2}{n!}{C}_{n-1}\hfill & \text{if}r=n-1.\hfill \end{array}\right.\hfill \end{array}$$

(36)

From the definition of C _r , we have

$$\frac{2}{n!}{C}_{n-1}=\frac{2}{n!}{\displaystyle \sum _{i=1}^{n-1}}\frac{\left(n-1\right)!}{n-i}=\frac{2}{n}{\displaystyle \sum _{i=1}^{n-1}}\frac{1}{i}=\frac{2}{n}{H}_{n-1},$$

(37)

where $H_n={\sum }_{i=1}^n\frac{1}{i}.$

Therefore, by (34), (36) and (37), we obtain the following theorem.

Theorem 4 For n ∈ ℤ₊ with n ≥ 2, we have

$${\displaystyle \sum _{k=1}^{n-1}}\frac{B_k\left(x\right)B_{n-k}\left(x\right)}{k\left(n-k\right)}=\frac{2}{n}{\displaystyle \sum _{k=0}^{n-2}}\frac{1}{n-k}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_{n-k}{B}_k\left(x\right)+\frac{2}{n}{H}_{n-1}{B}_n\left(x\right).$$

Let $J_{m,n}={\int }_0^1{E}_m\left(t\right)E_n\left(t\right)dt$, for m, n ∈ ℤ₊. Then we see that

$$J_{m,n}=\frac{2{\left(-1\right)}^{m-1}}{\left(n+m+1\right)\left(\begin{array}{c}\hfill n+m\hfill \\ \hfill m\hfill \end{array}\right)}{E}_{n+m+1},\left(\text{see}\left[3,4,7,8\right]\right).$$

(38)

Let us take polynomials p(x) in ℙ _n with $p\left(x\right)={\sum }_{k=0}^n{E}_k\left(x\right)E_{n-k}\left(x\right)$. Then, by Proposition 1, p(x) is written as $p\left(x\right)={\sum }_{k=0}^n{a}_k{B}_k\left(x\right)$.

It is not difficult to show that

$$a_0=\underset{0}{\overset{1}{\int }}p\left(t\right)dt={\displaystyle \sum _{k=0}^n}{J}_{k,n-k}=\frac{2E_{n+1}}{n+1}{\displaystyle \sum _{k=0}^n}\frac{{\left(-1\right)}^{k-1}}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}=-2E_{n+1}\left(\frac{1+{\left(-1\right)}^n}{n+2}\right)=\frac{-4E_{n+1}}{n+2}$$

and

$$p^{\left(r\right)}\left(x\right)=\frac{\left(n+1\right)!}{\left(n+1-r\right)!}{\displaystyle \sum _{k=r}^n}{E}_{k-r}\left(x\right)E_{n-k}\left(x\right),\left(r=0,1,2,\dots ,n\right).$$

(39)

By (17) and (39), we get

$$\begin{array}{cc}\hfill a_k& =\frac{1}{k!}\left(p^{\left(k-1\right)}\left(1\right)-p^{\left(k-1\right)}\left(0\right)\right)\hfill \\ =\frac{\left(n+1\right)!}{k!\left(n-k+2\right)!}{\displaystyle \sum _{l=k-1}^n}\left(E_{l-k+1}\left(1\right)E_{n-l}\left(1\right)-E_{l-k+1}{E}_{n-l}\right)\hfill \\ =\frac{\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)}{n+2}{\displaystyle \sum _{l=k-1}^n}\left\{\left(-E_{l-k+1}+2{\delta }_{0,l-k+1}\right)\left(-E_{n-l}+2{\delta }_{0,n-l}\right)-E_{l-k+1}{E}_{n-l}\right\}\hfill \\ =-\frac{4\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)}{n+2}{E}_{n-k+1},\hfill \end{array}$$

(40)

where k = 0, 1, 2, . . . , n. Therefore, by (40), we obtain the following theorem.

Theorem 5 For n ∈ ℤ₊, we have

$${\displaystyle \sum _{k=0}^n}{E}_k\left(x\right)E_{n-k}\left(x\right)=-\frac{4}{n+2}{\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)E_{n-k+1}{B}_k\left(x\right).$$

Let us take the polynomial p(x) in ℙ _n as follows:

$$p\left(x\right)={\displaystyle \sum _{k=0}^n}\frac{1}{k!\left(n-k\right)!}{E}_k\left(x\right)E_{n-k}\left(x\right).$$

(41)

Then, by (41), we get

$$p^{\left(r\right)}\left(x\right)=2^r{\displaystyle \sum _{k=r}^n}\frac{E_{k-r}\left(x\right)E_{n-k}\left(x\right)}{\left(k-r\right)!\left(n-k\right)!},$$

(42)

where r = 0, 1, 2, . . . , n.

By Proposition 1, we see that p(x) can be written as

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

(43)

From (41), (42) and (43), we have

$$\begin{array}{cc}\hfill a_0& ={\displaystyle \underset{0}{\overset{1}{\int }}}p\left(t\right)dt={\displaystyle \sum _{k=0}^n}\frac{1}{k!\left(n-k\right)!}{J}_{k,n-k}\hfill \\ =\frac{2E_{n+1}}{\left(n+1\right)!}{\displaystyle \sum _{k=0}^n}{\left(-1\right)}^{k-1}=-\frac{2E_{n+1}}{\left(n+1\right)!}\left(\frac{1+{\left(-1\right)}^n}{2}\right)=\frac{-2E_{n+1}}{\left(n+1\right)!}\hfill \end{array}$$

(44)

and

$$\begin{array}{cc}\hfill a_r& =\frac{1}{r!}\left(p^{\left(r-1\right)}\left(1\right)-p^{\left(r-1\right)}\left(0\right)\right)\hfill \\ =\frac{2^{r-1}}{r!}{\displaystyle \sum _{k=r-1}^n}\frac{E_{k-r+1}\left(1\right)E_{n-k}\left(1\right)-E_{k-r+1}{E}_{n-k}}{\left(k-r+1\right)!\left(n-k\right)!}\hfill \\ =\frac{2^{r-1}}{r!}\left(-\frac{2E_{n-r+1}}{\left(n-r+1\right)!}-\frac{2E_{n-r+1}}{\left(n-r+1\right)!}+4{\delta }_{n+1,r}\right)\hfill \\ =-\frac{2^{r+1}}{\left(n+1\right)!}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill r\hfill \end{array}\right)E_{n-r+1},\hfill \end{array}$$

(45)

where r = 1, 2, . . . , n.

Therefore, by (41), (43) and (45), we obtain the following theorem.

Theorem 6 For n ∈ ℤ₊, we have

$${\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)E_k\left(x\right)E_{n-k}\left(x\right)=-\frac{2}{n+1}{\displaystyle \sum _{k=0}^n}{2}^k\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)E_{n-k+1}{B}_k\left(x\right).$$

Let us take

$$p\left(x\right)={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{E}_k\left(x\right)E_{n-k}\left(x\right)$$

in ℙ _n . Then, by Proposition 1, p(x) is given by means of basis polynomials:

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

(46)

It is easy to show that

$$\begin{array}{cc}\hfill a_0& ={\displaystyle \underset{0}{\overset{1}{\int }}}p\left(t\right)dt={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{J}_{k,n-k}\hfill \\ =\frac{2E_{n+1}}{n+1}{\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}\frac{{\left(-1\right)}^{k-1}}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}=\frac{2\left(1+{\left(-1\right)}^n\right)}{n^2\left(n+1\right)}{E}_{n+1}=\frac{4E_{n+1}}{n^2\left(n+1\right)}\hfill \end{array}$$

and

$$p^{\left(k\right)}\left(x\right)=2C_k{E}_{n-k}\left(x\right)+\left(n-1\right)\dots \left(n-k\right){\displaystyle \sum _{l=k+1}^{n-1}}\frac{E_{l-k}\left(x\right)E_{n-l}\left(x\right)}{\left(l-k\right)\left(n-l\right)},\left(k=1,2,\dots ,n-1\right)$$

where $C_k=\frac{1}{\left(n-k\right)}{\sum }_{j=1}^k\left(n-1\right)\dots \left(n-j+1\right)\left(n-j-1\right)\dots \left(n-k\right)$.

By the same method, we get

$$\begin{array}{cc}\hfill a_k& =\frac{1}{k!}\left(p^{\left(k-1\right)}\left(1\right)-p^{\left(k-1\right)}\left(0\right)\right)\hfill \\ =\frac{1}{k!}\left\{2C_{k-1}\left(E_{n-k+1}\left(1\right)-E_{n-k+1}\right)\right.\hfill \\ \left.+\left(n-1\right)\dots \left(n-k+1\right){\displaystyle \sum _{l=k}^{n-1}}\frac{E_{l-k+1}\left(1\right)E_{n-l}\left(1\right)-E_{l-k+1}{E}_{n-l}}{\left(l-k+1\right)\left(n-l\right)}\right\}\hfill \\ =-\frac{4C_{k-1}}{k!}{E}_{n-k+1}.\hfill \end{array}$$

From the construction of C _k , we note that

$$\begin{array}{cc}\hfill \frac{C_{k-1}}{k!}& =\frac{1}{k!\left(n-k+1\right)}{\displaystyle \sum _{j=1}^{k-1}}\left(n-1\right)\dots \left(n-j+1\right)\left(n-j-1\right)\dots \left(n-k+1\right)\hfill \\ =\frac{1}{k!\left(n-k+1\right)}{\displaystyle \sum _{j=1}^{k-1}}\frac{\left(n-1\right)!}{\left(n-k\right)!\left(n-j\right)}=\frac{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{n\left(n-k+1\right)}{\displaystyle \sum _{j=1}^{k-1}}\frac{1}{n-j}\hfill \\ =\frac{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{n\left(n-k+1\right)}\left({\displaystyle \sum _{j=1}^{n-1}}\frac{1}{j}-{\displaystyle \sum _{j=1}^{n-k}}\frac{1}{j}\right)=\frac{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{n\left(n-k+1\right)}\left(H_{n-1}-H_{n-k}\right).\hfill \end{array}$$

Therefore, by the same method, we obtain the following theorem.

Theorem 7 For n ∈ ℤ₊ with n ≥ 2, we have

$${\displaystyle \sum _{k=1}^{n-1}}\frac{E_k\left(x\right)E_{n-k}\left(x\right)}{k\left(n-k\right)}=\frac{4E_{n+1}}{n^2\left(n+1\right)}-\frac{4}{n}{\displaystyle \sum _{k=1}^n}\frac{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{n-k+1}\left(H_{n-1}-H_{n-k}\right)E_{n-k+1}{B}_k\left(x\right).$$

Let

$$T_{m,n}=\underset{0}{\overset{1}{\int }}{B}_m\left(t\right)E_n\left(t\right)dt,\text{for}m,n\in {\mathbb{Z}}_{+}.$$

(47)

From (47), we have that

$$T_{m,0}=\underset{0}{\overset{1}{\int }}{B}_m\left(t\right)dt=\frac{{\delta }_{0,m}}{m+1}\text{and}{T}_{0,n}=\underset{0}{\overset{1}{\int }}{E}_n\left(t\right)dt=-\frac{2E_{n+1}}{n+1}.$$

For m, n ∈ ℕ, we have

$$T_{m,n}=\frac{2{\left(-1\right)}^m}{\left(m+n+1\right)\left(\begin{array}{c}\hfill m+n\hfill \\ \hfill m\hfill \end{array}\right)}{\displaystyle \sum _{l=m+1}^{m+n}}{\left(-1\right)}^l\left(\begin{array}{c}\hfill m+n+1\hfill \\ \hfill l\hfill \end{array}\right)B_l{E}_{n+m+1-l}.$$

(48)

Let us consider the following polynomial in ℙ _n :

$$p\left(x\right)={\displaystyle \sum _{k=0}^n}{B}_k\left(x\right)E_{n-k}\left(x\right).$$

(49)

For n ∈ ℕ with n ≥ 2, by Proposition 1, p(x) is given by

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

(50)

From (49) and (50), we note that

$$\begin{array}{cc}\hfill a_0& =\underset{0}{\overset{1}{\int }}p\left(t\right)dt=T_{0,n}+{\displaystyle \sum _{k=1}^{n-1}}{T}_{k,n-k}+T_{n,0}\hfill \\ =-\frac{2E_{n+1}}{n+1}+\frac{2}{n+1}{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \sum _{l=k+1}^n}{\left(-1\right)}^{k+l}\frac{\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{B}_l{E}_{n+1-l}.\hfill \end{array}$$

(51)

For k = 0, 1, 2, . . . , n, we have

$$\begin{array}{cc}\hfill p^{\left(k\right)}\left(x\right)& =\left(n+1\right)n\dots \left(n+2-k\right){\displaystyle \sum _{l=k}^n}{B}_{l-k}\left(x\right)E_{n-l}\left(x\right)\hfill \\ =\frac{\left(n+1\right)!}{\left(n-k+1\right)!}{\displaystyle \sum _{l=k}^n}{B}_{l-k}\left(x\right)E_{n-l}\left(x\right).\hfill \end{array}$$

(52)

By (17), we get

$$\begin{array}{cc}\hfill a_k& =\frac{1}{k!}\left(p^{\left(k-1\right)}\left(1\right)-p^{\left(k-1\right)}\left(0\right)\right)\hfill \\ =\frac{\left(n+1\right)!}{k!\left(n-k+2\right)!}{\displaystyle \sum _{l=k-1}^n}\left(B_{l-k+1}\left(1\right)E_{n-l}\left(1\right)-B_{l-k+1}{E}_{n-l}\right)\hfill \\ =\frac{\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)}{n+2}{\displaystyle \sum _{l=k-1}^n}\left\{\left(B_{l-k+1}+{\delta }_{1,l-k+1}\right)\left(-E_{n-l}+2{\delta }_{0,n-l}\right)-B_{l-k+1}{E}_{n-l}\right\}\hfill \\ =\frac{\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)}{n+2}\left(-2{\displaystyle \sum _{l=k-1}^n}{B}_{l-k+1}{E}_{n-l}-E_{n-k}+2B_{n-k+1}+2{\delta }_{n,k}\right).\hfill \end{array}$$

(53)

Therefore, by (49), (50) and (53), we obtain the following theorem.

Theorem 8 For n ∈ ℤ₊ with n ≥ 2, we have

$$\begin{array}{c}{\displaystyle \sum _{k=0}^n}{B}_k\left(x\right)E_{n-k}\left(x\right)\hfill \\ =-\frac{2E_{n+1}}{n+1}+\frac{2}{n+1}{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \sum _{l=k+1}^n}{\left(-1\right)}^{k+l}\frac{\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)}{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{B}_l{E}_{n+1-l}+\left(n+1\right)B_n\left(x\right)\hfill \\ +\frac{1}{n+2}{\displaystyle \sum _{k=1}^{n-2}}\left(\begin{array}{c}\hfill n+2\hfill \\ \hfill k\hfill \end{array}\right)\left(-2{\displaystyle \sum _{l=k-1}^n}{B}_{l-k+1}{E}_{n-l}-E_{n-k}+2B_{n-k+1}\right)B_k\left(x\right).\hfill \end{array}$$

For n ∈ ℕ with n ≥ 2, let us take $p\left(x\right)={\sum }_{k=0}^n\frac{B_k\left(x\right)E_{n-k}\left(x\right)}{k!\left(n-k\right)!}$in ℙ _n . Then we have

$$p^{\left(k\right)}\left(x\right)=2^k{\displaystyle \sum _{l=k}^n}\frac{1}{\left(l-k\right)!\left(n-l\right)!}{B}_{l-k}\left(x\right)E_{n-l}\left(x\right).$$

(54)

From Proposition 1, we note that p(x) can be written as

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

(55)

Thus, by (55), we get

$$\begin{array}{cc}\hfill a_0& =\underset{0}{\overset{1}{\int }}p\left(t\right)dt={\displaystyle \sum _{k=0}^n}\frac{1}{k!\left(n-k\right)!}{T}_{k,n-k}\hfill \\ =\frac{T_{0,n}}{n!}+{\displaystyle \sum _{k=1}^{n-1}}\frac{T_{k,n-k}}{k!\left(n-k\right)!}+\frac{T_{n,0}}{n!}\hfill \\ =-\frac{2E_{n+1}}{\left(n+1\right)!}+\frac{2}{\left(n+1\right)!}{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \sum _{l=k+1}^n}{\left(-1\right)}^{k+l}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)B_l{E}_{n+1-l}.\hfill \end{array}$$

(56)

From (17), we note that

$$\begin{array}{cc}\hfill a_k& =\frac{1}{k!}\left(p^{\left(k-1\right)}\left(1\right)-p^{\left(k-1\right)}\left(0\right)\right)\hfill \\ =\frac{2^{k-1}}{k!}{\displaystyle \sum _{l=k-1}^n}\frac{B_{l-k+1}\left(1\right)E_{n-l}\left(1\right)-B_{l-k+1}{E}_{n-l}}{\left(l-k+1\right)!\left(n-l\right)!}\hfill \\ =\frac{2^{k-1}}{k!}\left({\displaystyle \sum _{l=k-1}^n}\frac{-2B_{l-k+1}{E}_{n-l}}{\left(l-k+1\right)!\left(n-l\right)!}-\frac{E_{n-k}}{\left(n-k\right)!}+\frac{2B_{n-k+1}}{\left(n-k+1\right)!}+2{\delta }_{n,k}\right).\hfill \end{array}$$

(57)

Therefore, by (54), (55) and (57), we obtain the following theorem.

Theorem 9 For n ∈ ℕ with n ≥ 2, we have

$$\begin{array}{c}{\displaystyle \sum _{k=0}^n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_k\left(x\right)E_{n-k}\left(x\right)\hfill \\ =-\frac{2E_{n+1}}{n+1}+\frac{2}{n+1}{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \sum _{l=k+1}^n}{\left(-1\right)}^{k+l}\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)B_l{E}_{n+1-l}\hfill \\ +{\displaystyle \sum _{k=1}^{n-2}}\left(-\frac{2^k\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)}{n+1}{\displaystyle \sum _{l=k-1}^n}\left(\begin{array}{c}\hfill n-k+1\hfill \\ \hfill n-l\hfill \end{array}\right)B_{l-k+1}{E}_{n-l}-2^{k-1}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)E_{n-k}\right.\hfill \\ \left.+\frac{2^k\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill k\hfill \end{array}\right)}{n+1}{B}_{n-k+1}\right)B_k\left(x\right)+2^n{B}_n\left(x\right).\hfill \end{array}$$

For n ∈ ℕ with n ≥ 2, let us consider the polynomial $p\left(x\right)={\sum }_{k=1}^{n-1}\frac{B_k\left(x\right)E_{n-k}\left(x\right)}{k\left(n-k\right)}$ in ℙ _n .

From Proposition 1, we note that p(x) can be written as $p\left(x\right)={\sum }_{k=0}^n{a}_k{B}_k\left(x\right)$. Then the k-th derivative of p(x) is given by

$$p^{\left(k\right)}\left(x\right)=C_k\left(B_{n-k}\left(x\right)+E_{n-k}\left(x\right)\right)+\left(n-1\right)\dots \left(n-k\right){\displaystyle \sum _{l=k+1}^n}\frac{B_{l-k}\left(x\right)E_{n-l}\left(x\right)}{\left(l-k\right)\left(n-l\right)},$$

(58)

where k = 1, 2, . . . , n - 1 and

$$C_k=\frac{1}{n-k}{\displaystyle \sum _{j=1}^k}\left(n-1\right)\left(n-2\right)\dots \left(n-j+1\right)\left(n-j-1\right)\dots \left(n-k\right).$$

In addition,

$$p^{\left(n\right)}\left(x\right)={\left(p^{\left(n-1\right)}\left(x\right)\right)}^{\prime }={\left(C_{n-1}\left(B_1\left(x\right)+E_1\left(x\right)\right)\right)}^{\prime }=2C_{n-1}=2\left(n-1\right)!H_{n-1}.$$

From (17), we note that

$$\begin{array}{c}{a}_k=\frac{1}{k!}\left(p^{\left(k-1\right)}\left(1\right)-p^{\left(k-1\right)}\left(0\right)\right)\hfill \\ =\frac{C_{k-1}}{k!}\left\{\left(B_{n-k+1}\left(1\right)-B_{n-k+1}\right)+\left(E_{n-k+1}\left(1\right)-E_{n-k+1}\right)\right\}\hfill \\ +\frac{\left(n-1\right)\dots \left(n-k+1\right)}{k!}{\displaystyle \sum _{l=k}^{n-1}}\frac{1}{\left(l-k+1\right)\left(n-l\right)}\left(B_{l-k+1}\left(1\right)E_{n-l}\left(1\right)-B_{l-k+1}{E}_{n-l}\right)\hfill \\ =\frac{C_{k-1}}{k!}\left(-2E_{n-k+1}+{\delta }_{1,n-k+1}\right)+\frac{\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)}{n}\left({\displaystyle \sum _{l=k}^{n-1}}\frac{-2B_{l-k+1}{E}_{n-l}}{\left(l-k+1\right)\left(n-l\right)}-\frac{E_{n-k}}{n-k}\right).\hfill \end{array}$$

(59)

It is easy to show that

$$\begin{array}{cc}\hfill a_0& =\underset{0}{\overset{1}{\int }}p\left(t\right)dt={\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{T}_{k,n-k}\hfill \\ =\frac{2}{\left(n+1\right)n\left(n-1\right)}{\displaystyle \sum _{k=0}^{n-2}}\frac{{\left(-1\right)}^{k+1}}{\left(\begin{array}{c}\hfill n-2\hfill \\ \hfill k\hfill \end{array}\right)}{\displaystyle \sum _{l=k+2}^n}{\left(-1\right)}^l\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)B_l{E}_{n+1-l}.\hfill \end{array}$$

(60)

Therefore, from (59) and (60), we have

$$\begin{array}{c}{\displaystyle \sum _{k=1}^{n-1}}\frac{1}{k\left(n-k\right)}{B}_k\left(x\right)E_{n-k}\left(x\right)\hfill \\ =\frac{2}{n\left(n^2-1\right)}{\displaystyle \sum _{k=0}^{n-2}}{\displaystyle \sum _{l=k+2}^n}{\left(-1\right)}^{k+l+1}\frac{\left(\begin{array}{c}\hfill n+1\hfill \\ \hfill l\hfill \end{array}\right)}{\left(\begin{array}{c}\hfill n-2\hfill \\ \hfill k\hfill \end{array}\right)}{B}_l{E}_{n+1-l}\hfill \\ +{\displaystyle \sum _{k=1}^{n-2}}\left\{\frac{-2}{n\left(n-k+1\right)}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)\left(H_{n-1}-H_{n-k}\right)E_{n-k+1}\right.\hfill \\ \left.+\frac{1}{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)\left(-2{\displaystyle \sum _{l=k}^{n-1}}\frac{B_{l-k+1}{E}_{n-l}}{\left(l-k+1\right)\left(n-l\right)}-\frac{E_{n-k}}{n-k}\right)\right\}{B}_k\left(x\right)+\frac{2}{n}{H}_{n-1}{B}_n\left(x\right).\hfill \end{array}$$