Open Access

A note on higher-order Bernoulli polynomials

Journal of Inequalities and Applications20132013:111

https://doi.org/10.1186/1029-242X-2013-111

Received: 23 September 2012

Accepted: 28 February 2013

Published: 19 March 2013

Abstract

Let P n = { p ( x ) Q [ x ] | deg p ( x ) n } be the ( n + 1 ) -dimensional vector space over Q. From the property of the basis B 0 ( r ) , B 1 ( r ) , , B n ( r ) for the space P n , we derive some interesting identities of higher-order Bernoulli polynomials.

1 Introduction

Let N = { 1 , 2 , 3 , } and Z + = N { 0 } . For a fixed r Z + , the n th Bernoulli polynomials are defined by the generating function to be
( t e t 1 ) r e x t = e B ( r ) ( x ) t = n = 0 B n ( r ) ( x ) t n n ! ( see [1–11] )
(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
B n ( r ) ( x ) = k = 0 n ( n k ) B k ( r ) x n k = k = 0 n ( n k ) B n k ( r ) x k = n 1 + + n r + n r + 1 = n ( n n 1 , , n r , n r + 1 ) B n 1 B n r x n r + 1 .
(2)
Thus, by (2) we get the Euler-type sums of products of Bernoulli numbers as follows:
B n ( r ) = n 1 + + n r = n ( n n 1 , , n r ) B n 1 B n 2 B n r ( see [11–17] ) .
(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 ) ) = d d x B n ( r ) ( x ) = n B n 1 ( r ) ( x ) ( see [11–17] )
(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 ( , ) = R . Now, we define three linear operators I, , D on Ω as follows:
I f ( x ) = x x + 1 f ( t ) d t , f ( x ) = f ( x + 1 ) f ( x ) , D f ( x ) = f ( x ) .
(6)

Then we see that (i) D I = I D = , (ii) I = I , (iii) D = D .

Let P n = { p ( x ) Q ( x ) | deg p ( x ) n } be the ( n + 1 ) -dimensional vector space over Q. Probably, { 1 , x , , x n } is the most natural basis for this space. But { B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , B 2 ( r ) ( x ) , , B n ( r ) ( x ) } is also a good basis for the space P n for our purpose of arithmetical and combinatorial applications.

Let p ( x ) P n . Then p ( x ) can be generated by B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , B 2 ( r ) ( x ) , , B n ( r ) ( x ) as follows:
p ( x ) = 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 Z + ). Let p ( x ) P n .

For a fixed r Z + , p ( x ) can be generated by B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , B 2 ( r ) ( x ) , , B n ( r ) ( x ) as follows:
p ( x ) = k = 0 n a k B k ( r ) ( x ) .
(7)
From (6) and (7), we can derive the following identities:
I B n ( r ) ( x ) = x x + 1 B n ( r ) ( x ) d x = 1 n + 1 ( B n + 1 ( r ) ( x + 1 ) B n + 1 ( r ) ( x ) ) .
(8)
By (5) and (8), we get
I B n ( r ) ( x ) = n + 1 n + 1 B n ( r 1 ) ( x ) = B n ( r 1 ) ( x ) .
(9)
It is easy to show that
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 ) = k = 0 n a k B k ( 0 ) ( x ) = k = 0 n a k x k .
(12)
From (6) and (12), we note that
D k I r p ( x ) = l = k n a l 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 = D k I r p ( 0 ) k ! .
(15)

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

By (15), we get
a k = 1 k ! D k I k I r k p ( 0 ) = 1 k ! ( D I ) k I r k p ( 0 ) = 1 k ! k I r k p ( 0 ) = 1 k ! j = 0 k ( k j ) ( 1 ) k j I r k p ( j ) .
(16)
Case 2. Assume that r n .
  1. (i)
    For 0 k r , by (15) we get
    a k = 1 k ! j = 0 k ( 1 ) k j ( k j ) I r k p ( j ) .
    (17)
     
  2. (ii)
    For r k n , by (15) we see that
    a k = 1 k ! D k r D r I r p ( 0 ) = 1 k ! D k r ( D I ) r p ( 0 ) = 1 k ! D k r r p ( 0 ) = 1 k ! r D k r p ( 0 ) = 1 k ! j = 0 r ( r j ) ( 1 ) r j D k r p ( j ) .
    (18)
     

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

Theorem 1
  1. (a)
    For r > n , we have
    p ( x ) = k = 0 n ( j = 0 k 1 k ! ( 1 ) k j ( k j ) I r k p ( j ) ) B k ( r ) ( x ) .
     
  2. (b)
    For r n , we have
    p ( x ) = k = 0 r 1 ( j = 0 k 1 k ! ( 1 ) k j ( k j ) I r k p ( j ) ) B k ( r ) ( x ) + k = r n ( j = 0 r 1 k ! ( 1 ) r j D k r p ( j ) ) B k ( r ) ( x ) .
     
Let us take p ( x ) = x n P n . Then x n can be expressed as a linear combination of B 0 ( r ) , B 1 ( r ) , , B n ( r ) . For r > n , we have
I r k x n = n ! ( n + r k ) ! l = 0 r k ( 1 ) r k l ( r k l ) ( x + l ) n + r k .
(19)

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

Corollary 2 For n , r Z + with r > n , we have
x n = k = 0 n { j = 0 k l = 0 r k ( 1 ) r j l n ! ( k j ) ( r k l ) k ! ( n + r k ) ! ( j + l ) n + r k } B k ( r ) ( x ) .
Let us assume that r , n Z + with r n . Observe that
D k r x n = n ( n 1 ) ( n k + r + 1 ) x n k + r = 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 Z + with r n , we have
x n = k = 0 r 1 { j = 0 k l = 0 r k ( 1 ) r j l n ! ( k j ) ( r k l ) k ! ( n + r k ) ! ( j + l ) n + r k } B k ( r ) ( x ) + k = r n { j = 0 r ( 1 ) r j n ! ( r j ) k ! ( n + r k ) ! j n + r k } B k ( r ) ( x ) .
Let us take p ( x ) = B n ( s ) ( x ) P n ( s Z + ). Then p ( x ) can be generated by { B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , , B n ( r ) ( x ) } as follows:
B n ( s ) ( x ) = 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 Z + with r > n , we have
B n ( s ) ( x ) = k = 0 n { j = 0 k ( 1 ) k j 1 k ! ( k j ) B n ( s + k r ) ( j ) } B k ( r ) ( x ) .
In particular, for r = s , we have
B n ( r ) ( x ) = 0 B 0 ( r ) ( x ) + 0 B 1 ( r ) ( x ) + + 0 B n 1 ( r ) ( x ) + 1 B n ( r ) ( x ) = k = 0 n { j = 0 k ( 1 ) k j 1 k ! ( k j ) B n ( k ) ( j ) } B k ( r ) ( x ) .
(23)
By comparing coefficients on the both sides of (23), we get
j = 0 k ( 1 ) k j 1 k ! ( k j ) B n ( k ) ( j ) = δ k n , for  0 k n .
(24)

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

Corollary 5
  1. (a)
    For n , k Z + with 0 k n 1 , we have
    j = 0 k ( 1 ) k j ( k j ) B n ( k ) ( j ) = 0 .
     
  2. (b)
    In particular, k = n , we get
    j = 0 n ( 1 ) n j ( n j ) B n ( n ) ( j ) = n ! .
     
Let us assume that r n in (21). Then we have
D k r B n ( s ) ( x ) = n ( n 1 ) ( n k + r + 1 ) B n + r k ( s ) ( x ) = 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 Z + with r n , we have
B n ( s ) ( x ) = k = 0 n 1 { j = 0 k ( 1 ) k j 1 k ! ( k j ) B n ( s + k r ) ( j ) } B k ( r ) ( x ) + k = r n { j = 0 r ( 1 ) r j n ! ( r j ) k ! ( n + r k ) ! B n + r k ( s ) ( j ) } B k ( r ) ( x ) .

Let p ( x ) = E n ( s ) ( x ) ( s 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 ) , , B n ( r ) ( x ) .

Assume that r , n Z + with r > n .

By (6), we get
I r k E n ( s ) ( x ) = 1 ( n + 1 ) ( n + r k ) l = 0 r k ( 1 ) r k l ( r k l ) E n + r k ( s ) ( x + l ) = n ! ( n + r k ) ! l = 0 r k ( 1 ) r k l ( r k l ) E n + r k ( s ) ( x + l ) .
(26)

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

Theorem 7 For r , n Z + with r > n , we have
E n ( s ) ( x ) = k = 0 n { j = 0 k l = 0 r k ( 1 ) r j l n ! ( k j ) ( r k l ) k ! ( n + r k ) ! E n + r k ( s ) ( j + l ) } B k ( r ) ( x ) .
For r , n Z + with r n , we have
D k r E n ( s ) ( x ) = n ( n 1 ) ( 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 Z + with r n , we have
E n ( s ) ( x ) = k = 0 r 1 { j = 0 k l = 0 r k ( 1 ) r j l n ! ( k j ) ( r k l ) k ! ( n + r k ) ! E n + r k ( s ) ( j + l ) } B k ( r ) ( x ) + k = r n { j = 0 r ( 1 ) r j n ! ( r k ) k ! ( n + r k ) ! E n + r k ( s ) ( j ) } B k ( r ) ( x ) .
Remarks (a) For r 0 , by (40) we get
I r x n = n ! ( n + r ) ! j = 0 r ( r j ) ( 1 ) r j ( x + j ) n + r = 1 ( n + r r ) 1 r ! j = 0 r ( 1 ) r j ( r j ) ( x + j ) n + r .
Thus, for x = 0 , we have
I r x n | x = 0 = n ! ( n + r ) ! j = 0 r ( r j ) ( 1 ) r j j n + r = 1 ( n + r r ) 1 r ! r 0 n + r = S ( n + r , r ) ( n + r r ) ,
(28)
where S ( n , r ) is the Stirling number of the second kind.
  1. (b)
    Assume
    k = 0 n α k x k = k = 0 n a k B k ( r ) ( x ) ( r 0 ) .
    (29)
     
Applying I t on both sides ( t 0 ), we get
k = 0 n a k B k ( r t ) ( x ) = k = 0 n α k I t x k = k = 0 n α k ( α + t t ) 1 t ! j = 0 t ( 1 ) ( t j ) ( t j ) ( x + j ) k + t .
(30)
From (28) and (30), we have
k = 0 n a k B k ( r t ) = k = 0 n α k ( α + t t ) S ( k + t , t ) .
Remark Let us define two operators d, d ˜ as follows:
d = e D = n = 0 ( 1 ) n n ! D n , d ˜ = e D = n = 0 D n n ! .
(31)
From (31), we note that
d ˜ x n = l = 0 n ( n l ) x n l = ( x + 1 ) n , d x n = l = 0 n ( n l ) ( 1 ) l x n l = ( x 1 ) n .
(32)
Thus, by (31) and (32), we get
d ˜ B n ( r ) ( x ) = B n ( r ) ( x + 1 ) , d B n ( r ) ( x ) = B n ( r ) ( x 1 ) ,
(33)
and
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 , , r n Z + , { B 0 ( r 0 ) ( x ) , B 1 ( r 1 ) ( x ) , , B 0 ( r n ) ( x ) } forms a basis for P n . Let r = max { r i | i = 0 , 1 , 2 , , n } . Let p ( x ) P n . Then p ( x ) can be expressed as a linear combination of B 0 ( r 0 ) ( x ) , B 1 ( r 1 ) ( x ) , , B n ( r n ) ( x ) as follows:
p ( x ) = a 0 B 0 ( r 0 ) ( x ) + a 1 B 1 ( r 1 ) ( x ) + + a n B n ( r n ) ( x ) = l = 0 n a l B l ( r l ) ( x ) .
(35)
Thus, by (6) and (35), we get
I r p ( x ) = l = 0 n a l I r B l ( r l ) ( x ) = l = 0 n a l I r r l I r l B l ( r l ) ( x ) = l = 0 n a l I r r l B l ( 0 ) ( x ) = l = 0 n a l I r r l x l .
(36)
Now, for each k = 0 , 1 , 2 , , n , by (36) we get
D k I r p ( x ) = l = 0 n a l D k I r r l x l = l = 0 n a l I r r l ( D k x l ) = l = k n a l I r r l ( l ! ( l k ) ! x l k ) = l = k n a l l ! ( l k ) ! I r r l x l k .
(37)
Let us take x = 0 in (37). Then, by (28) and (37), we get
D k I r p ( 0 ) = l = k n l ! a l ( l k ) ! × S ( l k + r r l , r r l ) ( l k + r r l r r l ) = l = k n a l ( r r l ) ! l ! ( l k ) ! S ( l k + r r l , r r l ) .
(38)
Case 1. For r > n , we have
D k I r p ( 0 ) = D k I k I r k p ( 0 ) = ( D I ) k I r k p ( 0 ) = k I r k p ( 0 ) = j = 0 k ( 1 ) k j ( k j ) I r k p ( j ) .
(39)
Case 2. Let r n .
  1. (i)
    For 0 k < r , we have
    D k I r p ( 0 ) = j = 0 k ( 1 ) k j ( k j ) I r k p ( j ) .
    (40)
     
  2. (ii)
    For r k n , we have
    D k I r p ( 0 ) = D k r D r I r p ( 0 ) = D k r ( D I ) r p ( 0 ) = D k r r p ( 0 ) = r D k r p ( 0 ) = j = 0 r ( 1 ) r j ( r j ) D k r p ( j ) .
    (41)
     

Thus, by (38), (39), (40) and (41), we can determine a 0 , a 1 , a 2 , , a n .

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Sogang University
(2)
Department of Mathematics, Kwangwoon University

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© Kim and Kim; licensee Springer. 2013

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