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Some identities for the product of two Bernoulli and Euler polynomials

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Abstract

Let 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 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

t e t - 1 e x t = n = 0 B n ( x ) t n n ! , 2 e t + 1 e x t = n = 0 E n ( x ) 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 [117]).

From (1), we note that

B n ( x ) = l = 0 n n l B l x n - l , E n ( x ) = l = 0 n n l E l x n - l .
(2)

For n ≥ 0, we have

d d x B n ( x ) = n B n - 1 ( x ) , d d x E n ( x ) = n E n - 1 ( x ) ,
(3)

(see [7, 8]).

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

B 0 = 1 , B n ( 1 ) - B n = δ 1 , n and E 0 = 1 , E n ( 1 ) + E n = 2 δ 0 , n ,
(4)

where δk, nis the Kronecker symbol (see [117]).

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

0 1 B n ( x ) d x = δ 0 , n n + 1 , 0 1 E n ( x ) d x = - 2 E n + 1 n + 1 .
(5)

It is known [12] that

0 A B m 1 x a 1 B m n x a n d x = a 1 1 - m 1 a n 1 - m n 0 1 B m 1 ( x ) B m n ( x ) d x ,
(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

0 1 B p ( x ) B q ( x ) d x = ( - 1 ) p + 1 B p + q p + q q ,
(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 ( x ) B n ( x ) = r m 2 r n + n 2 r m B 2 r B m + n - 2 r ( x ) m + n - 2 r + ( - 1 ) m + 1 B m + n m + n n , for m + n 2 .
(8)

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

0 1 B m ( x ) B n ( x ) B p ( x ) d x = ( - 1 ) p + 1 p ! r m 2 r n + n 2 r m ( m + n - 2 r - 1 ) ! ( m + n + p - 2 r ) ! B 2 r B m + n + p - 2 r ,
(9)

(see [4]).

In [8], it is known that for n +,

B n ( x ) = k = 0 k 1 n n k B k E n - k ( x )
(10)

and

E n ( x ) = - 2 l = 0 n n l E l + 1 l + 1 B n - l ( x ) .
(11)

Let n = {∑ i a i xi|a i } 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 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

e x t = 1 t t ( e t - 1 ) e t - 1 e x t = 1 t n = 0 ( B n ( x + 1 ) - B n ( x ) ) t n n ! = 1 t n = 1 ( B n ( x + 1 ) - B n ( x ) ) t n n ! = n = 0 B n + 1 ( x + 1 ) - B n + 1 ( x ) n + 1 t n n ! .
(12)

Thus, from (12), we have

x n = 1 n + 1 ( B n + 1 ( x + 1 ) - B n + 1 ( x ) ) = 1 n + 1 l = 0 n n + 1 l B l ( x ) .
(13)

From (13), we note that {B0(x), B1(x), . . . , B n (x)} spans n . For p(x) n , let p ( x ) = k = 0 n a k B k ( x ) and g(x) = p(x + 1) - p(x). Then we have

g ( x ) = k = 0 n a k ( B k ( x + 1 ) - B k ( x ) ) = k = 0 n k a k x k - 1 .
(14)

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

g ( r ) ( x ) = k = r + 1 n k ( k - 1 ) ( k - r ) a k x k - r - 1 ,
(15)

where g ( r ) ( x ) = d r g ( x ) d x r and r = 0, 1, 2, . . . , n. Let us take x = 0 in (15). Then we have

g ( r ) ( 0 ) = ( r + 1 ) ! a r + 1 .
(16)

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

a r = g ( r - 1 ) ( 0 ) r ! = 1 r ! ( p ( r - 1 ) ( 1 ) - p ( r - 1 ) ( 0 ) ) .
(17)

Let 0=p ( x ) = k = 0 n a k B k ( x ) . Then, from (17), we have

a r = 1 r ! g ( r - 1 ) ( 0 ) = 1 r ! ( p ( r - 1 ) ( 1 ) - p ( r - 1 ) ( 0 ) ) = 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 n .

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

p ( x ) = C 0 B 0 ( x ) + C 1 B 1 ( x ) + + C n B n ( x ) .
(19)

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

p ( x ) = ( B 0 ( x ) , B 1 ( x ) , , B n ( x ) ) C 0 C 1 C n .
(20)

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

p ( x ) = ( 1 , x , x 2 , , x n ) 1 b 12 b 13 b 1 n + 1 0 1 b 23 b 2 n + 1 0 0 1 b 3 n + 1 0 0 0 b n n + 1 0 0 0 1 C 0 C 1 C 2 C n ,
(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 ( x ) = 1 , B 1 ( x ) = x - 1 2 , B 2 ( x ) = x 2 - x + 1 6 , B 3 ( x ) = x 3 - 3 2 x 2 + 1 2 x , .

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

p ( x ) = ( 1 , x , x 2 ) 1 - 1 2 1 6 0 1 - 1 0 0 1 C 0 C 1 C 2 .
(22)

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

p ( x ) = ( 1 , x , x 2 , x 3 ) 1 - 1 2 1 6 0 0 1 - 1 1 2 0 0 1 - 3 2 0 0 0 1 C 0 C 1 C 2 C 3 .
(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 = 0 1 B m ( x ) B n ( x ) dx for m,n + . Then, by integration by parts, we get

I 0 , n = I m , 0 = 0 , I m , n = ( - 1 ) m + n B m + n m + n m , ( m , n 2 ) .
(24)

For n + with n ≥ 2, let us consider the following polynomials in n :

p ( x ) = k = 0 n B k ( x ) B n - k ( x ) n .
(25)

Then, from (25), we have

p ( r ) ( x ) = ( n + 1 ) ! ( n - r + 1 ) ! k = r n B k - r ( x ) B n - k ( x ) ,
(26)

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

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

p ( x ) = k = 0 n a k B k ( x ) .
(27)

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

a 0 = 0 1 p ( t ) d t = k = 0 n I k , n - k = B n k = 1 n - 1 ( - 1 ) k - 1 n k = B n ( 1 + ( - 1 ) n ) n + 2 = 2 n + 2 B n .

By (18) and (26), we get

a r + 1 = 1 ( r + 1 ) ! ( p ( r ) ( 1 ) - p ( r ) ( 0 ) ) = ( n + 1 ) ! ( r + 1 ) ! ( n - r + 1 ) ! k = r n ( B k - r ( 1 ) B n - k ( 1 ) - B k - r B n - k ) = 1 n + 2 n + r r + 1 k = r n { ( δ 1 , k - r + B k - r ) ( δ 1 , n - k + B n - k ) - B k - r B n - k } = 1 n + 2 n + 2 r + 1 ( B n - r - 1 + B n - r - 1 + δ r , n - 2 ) = 2 n + 2 n + 2 r + 1 B n - r - 1 if  r n - 2 . 0 if  r = n - 2 .
(28)

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

Theorem 2 For n + with n ≥ 2, we have

k = 0 n B k ( x ) B n - k ( x ) = 2 n + 2 k = 0 n - 2 n + 2 k B n - k B k ( x ) + ( n + 1 ) B n ( x ) .

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

p ( x ) = k = 0 n 1 k ! ( n - k ) ! B k ( x ) B n - k ( x ) n .
(29)

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

p ( x ) = k = 0 n a k B k ( x ) n .
(30)

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

a 0 = 0 1 p ( t ) d t = k = 0 n 1 k ! ( n - k ) ! I k , n - k = 2 I 0 , n n ! + k = 1 n - 1 ( - 1 ) k - 1 k ! ( n - k ) ! n k B n = B n n ! k = 1 n - 1 ( - 1 ) k - 1 = B n n ! ( 1 + ( - 1 ) n ) 2 = B n n ! .
(31)

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

p ( r ) ( x ) = 2 r k = r n B k - r ( x ) B n - k ( x ) ( k - r ) ! ( n - k ) ! .
(32)

By (18), we get

a r + 1 = 1 ( r + 1 ) ! ( p ( r ) ( 1 ) - p ( r ) ( 0 ) ) = 2 r ( r + 1 ) ! k = r n 1 ( k - r ) ! ( n - k ) ! ( B k - r ( 1 ) B n - k ( 1 ) - B k - r B n - k ) = 2 r ( r + 1 ) ! 2 B n - r - 1 ( n - 1 - r ) ! + k = r n δ 1 , k - r δ 1 , n - k = 2 r + 1 n ! n r + 1 B n - r - 1 if  r n - 2 , 0 if  r = n - 2 .
(33)

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

Theorem 3 For n + with n ≥ 2, we have

k = 0 n n k B k ( x ) B n - k ( x ) = k = 0 k n - 1 n 2 k n k B n - k B k ( x ) .

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

p ( x ) = k = 1 n - 1 1 k ( n - k ) B k ( x ) B n - k ( x ) .

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

p ( x ) = k = 0 n a k B k ( x ) .
(34)

From (34), we have

a 0 = 0 1 p ( t ) d t = k = 1 n - 1 1 k ( n - k ) 0 1 B k ( t ) B n - k ( t ) d t = k = 1 n - 1 1 k ( n - k ) ( - 1 ) k - 1 n k B n = 1 + ( - 1 ) n n 2 B n = 2 B n n 2 .

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

p ( r ) ( x ) = 2 C r B n - r ( x ) + ( n - 1 ) ( n - r ) k = r + 1 n - 1 B k - r ( x ) B n - k ( x ) ( k - r ) ( n - k ) ,
(35)

where C r = 1 n - r j = 1 r ( n - 1 ) ... ( n - j + 1 ) ( n - j - 1 ) ... ( n - r ) .

By (17), we get

a r + 1 = 1 ( r + 1 ) ! ( p ( r ) ( 1 ) - p ( r ) ( 0 ) ) = 1 ( r + 1 ) ! 2 C r ( B n - r ( 1 ) - B n - r ) + ( n - 1 ) ( n - r ) k = r + 1 n - 1 B k - r ( 1 ) B n - k ( 1 ) - B k - r B n - k ( k - r ) ( n - k ) = 2 C r ( r + 1 ) ! δ r , n - 1 + 1 n n r + 1 k = r + 1 n - 1 B k - r δ 1 , n - k + δ 1 , k - r B n - k + δ 1 , k - r δ 1 , n - k ( k - r ) ( n - k ) = 2 n ( n - r - 1 ) n r + 1 B n - r - 1 if 0 r n - 3 , 0 if r = n - 2 , 2 n ! C n - 1 if r = n - 1 .
(36)

From the definition of C r , we have

2 n ! C n - 1 = 2 n ! i = 1 n - 1 ( n - 1 ) ! n - i = 2 n i = 1 n - 1 1 i = 2 n H n - 1 ,
(37)

where H n = i = 1 n 1 i .

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

Theorem 4 For n + with n ≥ 2, we have

k = 1 n - 1 B k ( x ) B n - k ( x ) k ( n - k ) = 2 n k = 0 n - 2 1 n - k n k B n - k B k ( x ) + 2 n H n - 1 B n ( x ) .

Let J m , n = 0 1 E m ( t ) E n ( t ) dt, for m, n +. Then we see that

J m , n = 2 ( - 1 ) m - 1 ( n + m + 1 ) n + m m E n + m + 1 , ( see [ 3 , 4 , 7 , 8 ] ) .
(38)

Let us take polynomials p(x) in n with p ( x ) = k = 0 n E k ( x ) E n - k ( x ) . Then, by Proposition 1, p(x) is written as p ( x ) = k = 0 n a k B k ( x ) .

It is not difficult to show that

a 0 = 0 1 p ( t ) d t = k = 0 n J k , n - k = 2 E n + 1 n + 1 k = 0 n ( - 1 ) k - 1 n k = - 2 E n + 1 1 + ( - 1 ) n n + 2 = - 4 E n + 1 n + 2

and

p ( r ) ( x ) = ( n + 1 ) ! ( n + 1 - r ) ! k = r n E k - r ( x ) E n - k ( x ) , ( r = 0 , 1 , 2 , , n ) .
(39)

By (17) and (39), we get

a k = 1 k ! ( p ( k - 1 ) ( 1 ) - p ( k - 1 ) ( 0 ) ) = ( n + 1 ) ! k ! ( n - k + 2 ) ! l = k - 1 n ( E l - k + 1 ( 1 ) E n - l ( 1 ) - E l - k + 1 E n - l ) = n + 2 k n + 2 l = k - 1 n { ( - E l - k + 1 + 2 δ 0 , l - k + 1 ) ( - E n - l + 2 δ 0 , n - l ) - E l - k + 1 E n - l } = - 4 n + 2 k n + 2 E n - k + 1 ,
(40)

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

Theorem 5 For n +, we have

k = 0 n E k ( x ) E n - k ( x ) = - 4 n + 2 k = 0 n n + 2 k E n - k + 1 B k ( x ) .

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

p ( x ) = k = 0 n 1 k ! ( n - k ) ! E k ( x ) E n - k ( x ) .
(41)

Then, by (41), we get

p ( r ) ( x ) = 2 r k = r n E k - r ( x ) E n - k ( x ) ( k - r ) ! ( n - k ) ! ,
(42)

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

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

p ( x ) = k = 0 n a k B k ( x ) .
(43)

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

a 0 = 0 1 p ( t ) d t = k = 0 n 1 k ! ( n - k ) ! J k , n - k = 2 E n + 1 ( n + 1 ) ! k = 0 n ( - 1 ) k - 1 = - 2 E n + 1 ( n + 1 ) ! 1 + ( - 1 ) n 2 = - 2 E n + 1 ( n + 1 ) !
(44)

and

a r = 1 r ! ( p ( r - 1 ) ( 1 ) - p ( r - 1 ) ( 0 ) ) = 2 r - 1 r ! k = r - 1 n E k - r + 1 ( 1 ) E n - k ( 1 ) - E k - r + 1 E n - k ( k - r + 1 ) ! ( n - k ) ! = 2 r - 1 r ! - 2 E n - r + 1 ( n - r + 1 ) ! - 2 E n - r + 1 ( n - r + 1 ) ! + 4 δ n + 1 , r = - 2 r + 1 ( n + 1 ) ! n + 1 r E n - r + 1 ,
(45)

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

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

Theorem 6 For n +, we have

k = 0 n n k E k ( x ) E n - k ( x ) = - 2 n + 1 k = 0 n 2 k n + 1 k E n - k + 1 B k ( x ) .

Let us take

p ( x ) = k = 1 n - 1 1 k ( n - k ) E k ( x ) E n - k ( x )

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

p ( x ) = k = 0 n a k B k ( x ) .
(46)

It is easy to show that

a 0 = 0 1 p ( t ) d t = k = 1 n - 1 1 k ( n - k ) J k , n - k = 2 E n + 1 n + 1 k = 1 n - 1 1 k ( n - k ) ( - 1 ) k - 1 n k = 2 ( 1 + ( - 1 ) n ) n 2 ( n + 1 ) E n + 1 = 4 E n + 1 n 2 ( n + 1 )

and

p ( k ) ( x ) = 2 C k E n - k ( x ) + ( n - 1 ) ( n - k ) l = k + 1 n - 1 E l - k ( x ) E n - l ( x ) ( l - k ) ( n - l ) , ( k = 1 , 2 , , n - 1 )

where C k = 1 ( n - k ) j = 1 k ( n - 1 ) ( n - j + 1 ) ( n - j - 1 ) ( n - k ) .

By the same method, we get

a k = 1 k ! ( p ( k - 1 ) ( 1 ) - p ( k - 1 ) ( 0 ) ) = 1 k ! 2 C k - 1 ( E n - k + 1 ( 1 ) - E n - k + 1 ) + ( n - 1 ) ( n - k + 1 ) l = k n - 1 E l - k + 1 ( 1 ) E n - l ( 1 ) - E l - k + 1 E n - l ( l - k + 1 ) ( n - l ) = - 4 C k - 1 k ! E n - k + 1 .

From the construction of C k , we note that

C k - 1 k ! = 1 k ! ( n - k + 1 ) j = 1 k - 1 ( n - 1 ) ( n - j + 1 ) ( n - j - 1 ) ( n - k + 1 ) = 1 k ! ( n - k + 1 ) j = 1 k - 1 ( n - 1 ) ! ( n - k ) ! ( n - j ) = n k n ( n - k + 1 ) j = 1 k - 1 1 n - j = n k n ( n - k + 1 ) j = 1 n - 1 1 j - j = 1 n - k 1 j = n k n ( n - k + 1 ) ( H n - 1 - H n - k ) .

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

Theorem 7 For n + with n ≥ 2, we have

k = 1 n - 1 E k ( x ) E n - k ( x ) k ( n - k ) = 4 E n + 1 n 2 ( n + 1 ) - 4 n k = 1 n n k n - k + 1 ( H n - 1 - H n - k ) E n - k + 1 B k ( x ) .

Let

T m , n = 0 1 B m ( t ) E n ( t ) d t , for m , n + .
(47)

From (47), we have that

T m , 0 = 0 1 B m ( t ) d t = δ 0 , m m + 1 and T 0 , n = 0 1 E n ( t ) d t = - 2 E n + 1 n + 1 .

For m, n , we have

T m , n = 2 ( - 1 ) m ( m + n + 1 ) m + n m l = m + 1 m + n ( - 1 ) l m + n + 1 l B l E n + m + 1 - l .
(48)

Let us consider the following polynomial in n :

p ( x ) = k = 0 n B k ( x ) E n - k ( x ) .
(49)

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

p ( x ) = k = 0 n a k B k ( x ) .
(50)

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

a 0 = 0 1 p ( t ) d t = T 0 , n + k = 1 n - 1 T k , n - k + T n , 0 = - 2 E n + 1 n + 1 + 2 n + 1 k = 1 n - 1 l = k + 1 n ( - 1 ) k + l n + 1 l n k B l E n + 1 - l .
(51)

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

p ( k ) ( x ) = ( n + 1 ) n ( n + 2 - k ) l = k n B l - k ( x ) E n - l ( x ) = ( n + 1 ) ! ( n - k + 1 ) ! l = k n B l - k ( x ) E n - l ( x ) .
(52)

By (17), we get

a k = 1 k ! ( p ( k - 1 ) ( 1 ) - p ( k - 1 ) ( 0 ) ) = ( n + 1 ) ! k ! ( n - k + 2 ) ! l = k - 1 n ( B l - k + 1 ( 1 ) E n - l ( 1 ) - B l - k + 1 E n - l ) = n + 2 k n + 2 l = k - 1 n { ( B l - k + 1 + δ 1 , l - k + 1 ) ( - E n - l + 2 δ 0 , n - l ) - B l - k + 1 E n - l } = n + 2 k n + 2 - 2 l = k - 1 n B l - k + 1 E n - l - E n - k + 2 B n - k + 1 + 2 δ n , k .
(53)

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

Theorem 8 For n + with n ≥ 2, we have

k = 0 n B k ( x ) E n - k ( x ) = - 2 E n + 1 n + 1 + 2 n + 1 k = 1 n - 1 l = k + 1 n ( - 1 ) k + l n + 1 l n k B l E n + 1 - l + ( n + 1 ) B n ( x ) + 1 n + 2 k = 1 n - 2 n + 2 k - 2 l = k - 1 n B l - k + 1 E n - l - E n - k + 2 B n - k + 1 B k ( x ) .

For n with n ≥ 2, let us take p ( x ) = k = 0 n B k ( x ) E n - k ( x ) k ! ( n - k ) ! in n . Then we have

p ( k ) ( x ) = 2 k l = k n 1 ( l - k ) ! ( n - l ) ! B l - k ( x ) E n - l ( x ) .
(54)

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

p ( x ) = k = 0 n a k B k ( x ) .
(55)

Thus, by (55), we get

a 0 = 0 1 p ( t ) d t = k = 0 n 1 k ! ( n - k ) ! T k , n - k = T 0 , n n ! + k = 1 n - 1 T k , n - k k ! ( n - k ) ! + T n , 0 n ! = - 2 E n + 1 ( n + 1 ) ! + 2 ( n + 1 ) ! k = 1 n - 1 l = k + 1 n ( - 1 ) k + l n + 1 l B l E n + 1 - l .
(56)

From (17), we note that

a k = 1 k ! ( p ( k - 1 ) ( 1 ) - p ( k - 1 ) ( 0 ) ) = 2 k - 1 k ! l = k - 1 n B l - k + 1 ( 1 ) E n - l ( 1 ) - B l - k + 1 E n - l ( l - k + 1 ) ! ( n - l ) ! = 2 k - 1 k ! l = k - 1 n - 2 B l - k + 1 E n - l ( l - k + 1 ) ! ( n - l ) ! - E n - k ( n - k ) ! + 2 B n - k + 1 ( n - k + 1 ) ! + 2 δ n , k .
(57)

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

Theorem 9 For n with n ≥ 2, we have

k = 0 n n k B k ( x ) E n - k ( x ) = - 2 E n + 1 n + 1 + 2 n + 1 k = 1 n - 1 l = k + 1 n ( - 1 ) k + l n + 1 l B l E n + 1 - l + k = 1 n - 2 - 2 k n + 1 k n + 1 l = k - 1 n n - k + 1 n - l B l - k + 1 E n - l - 2 k - 1 n k E n - k + 2 k n + 1 k n + 1 B n - k + 1 B k ( x ) + 2 n B n ( x ) .

For n with n ≥ 2, let us consider the polynomial p ( x ) = k = 1 n - 1 B k ( x ) E n - k ( x ) k ( n - k ) in n .

From Proposition 1, we note that p(x) can be written as p ( x ) = k = 0 n a k B k ( x ) . Then the k-th derivative of p(x) is given by

p ( k ) ( x ) = C k ( B n - k ( x ) + E n - k ( x ) ) + ( n - 1 ) ( n - k ) l = k + 1 n B l - k ( x ) E n - l ( x ) ( l - k ) ( n - l ) ,
(58)

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

C k = 1 n - k j = 1 k ( n - 1 ) ( n - 2 ) ( n - j + 1 ) ( n - j - 1 ) ( n - k ) .

In addition,

p ( n ) ( x ) = ( p ( n - 1 ) ( x ) ) = C n - 1 ( B 1 ( x ) + E 1 ( x ) ) = 2 C n - 1 = 2 ( n - 1 ) ! H n - 1 .

From (17), we note that

a k = 1 k ! ( p ( k - 1 ) ( 1 ) - p ( k - 1 ) ( 0 ) ) = C k - 1 k ! { ( B n - k + 1 ( 1 ) - B n - k + 1 ) + ( E n - k + 1 ( 1 ) - E n - k + 1 ) } + ( n - 1 ) ( n - k + 1 ) k ! l = k n - 1 1 ( l - k + 1 ) ( n - l ) ( B l - k + 1 ( 1 ) E n - l ( 1 ) - B l - k + 1 E n - l ) = C k - 1 k ! ( - 2 E n - k + 1 + δ 1 , n - k + 1 ) + n k n l = k n - 1 - 2 B l - k + 1 E n - l ( l - k + 1 ) ( n - l ) - E n - k n - k .
(59)

It is easy to show that

a 0 = 0 1 p ( t ) d t = k = 1 n - 1 1 k ( n - k ) T k , n - k = 2 ( n + 1 ) n ( n - 1 ) k = 0 n - 2 ( - 1 ) k + 1 n - 2 k l = k + 2 n ( - 1 ) l n + 1 l B l E n + 1 - l .
(60)

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

k = 1 n - 1 1 k ( n - k ) B k ( x ) E n - k ( x ) = 2 n ( n 2 - 1 ) k = 0 n - 2 l = k + 2 n ( - 1 ) k + l + 1 n + 1 l n - 2 k B l E n + 1 - l + k = 1 n - 2 - 2 n ( n - k + 1 ) n k ( H n - 1 - H n - k ) E n - k + 1 + 1 n n k - 2 l = k n - 1 B l - k + 1 E n - l ( l - k + 1 ) ( n - l ) - E n - k n - k B k ( x ) + 2 n H n - 1 B n ( x ) .

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Acknowledgements

The authors would like to express their deep gratitudes to the referees for their valuable suggestions and comments.

Author information

Correspondence to Taekyun Kim.

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The authors declare that they have no competing interests.

Authors' contributions

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

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Keywords

  • Differential Equation
  • Positive Integer
  • Linear Combination
  • Partial Differential Equation
  • Ordinary Differential Equation