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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:

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.

Keywords

  • Linear Combination
  • Vector Space
  • Linear Operator
  • Dimensional Vector
  • Good Basis

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, Seoul, 121-742, Republic of Korea
(2)
Department of Mathematics, Kwangwoon University, Seoul, 139-701, Republic of Korea

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Copyright

© Kim and Kim; licensee Springer. 2013

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