A note on higher-order Bernoulli polynomials

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Part of the following topical collections:
  1. Proceedings of the Conference 2012 on Mathematical Inequalities and Nonlinear Functional Analysis with Applications
  2. Proceedings of the Conference 2012 on Mathematical Inequalities and Nonlinear Functional Analysis with Applications

Abstract

Let P n = { p ( x ) Q [ x ] | deg p ( x ) n } Open image in new window be the ( n + 1 ) Open image in new window-dimensional vector space over Q. From the property of the basis B 0 ( r ) , B 1 ( r ) , , B n ( r ) Open image in new window for the space P n Open image in new window, 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 , } Open image in new window and Z + = N { 0 } Open image in new window. For a fixed r Z + Open image in new window, 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] ) Open image in new window
(1)

with the usual convention about replacing ( B ( r ) ( x ) ) n Open image in new window by B n ( r ) ( x ) Open image in new window. In the special case, x = 0 Open image in new window, B n ( r ) ( 0 ) = B n ( r ) Open image in new window 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 . Open image in new window
(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] ) . Open image in new window
(3)

By (2) and (3), we see that B n ( r ) ( x ) Open image in new window 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] ) Open image in new window
(4)
and
B n ( r ) ( x + 1 ) B n ( r ) ( x ) = n B n 1 ( r 1 ) ( x ) . Open image in new window
(5)
Let Ω denote the space of real-valued differential functions on ( , ) = R Open image in new window. 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 ) . Open image in new window
(6)

Then we see that (i) D I = I D = Open image in new window, (ii) I = I Open image in new window, (iii) D = D Open image in new window.

Let P n = { p ( x ) Q ( x ) | deg p ( x ) n } Open image in new window be the ( n + 1 ) Open image in new window-dimensional vector space over Q. Probably, { 1 , x , , x n } Open image in new window 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 ) } Open image in new window is also a good basis for the space P n Open image in new window for our purpose of arithmetical and combinatorial applications.

Let p ( x ) P n Open image in new window. Then p ( x ) Open image in new window can be generated by B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , B 2 ( r ) ( x ) , , B n ( r ) ( x ) Open image in new window as follows:
p ( x ) = k = 0 n a k B k ( r ) ( x ) . Open image in new window

In this paper, we develop methods for uniquely determining a k Open image in new window from the information of  p ( x ) Open image in new window. From those methods, we derive some interesting identities of higher-order Bernoulli polynomials.

2 Higher-order Bernoulli polynomials

For r = 0 Open image in new window, by (1), we get B n ( 0 ) = x n Open image in new window ( n Z + Open image in new window). Let p ( x ) P n Open image in new window.

For a fixed r Z + Open image in new window, p ( x ) Open image in new window can be generated by B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , B 2 ( r ) ( x ) , , B n ( r ) ( x ) Open image in new window as follows:
p ( x ) = k = 0 n a k B k ( r ) ( x ) . Open image in new window
(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 ) ) . Open image in new window
(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 ) . Open image in new window
(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 ) , Open image in new window
(10)
and
D B n ( r ) ( x ) = n B n 1 ( r ) ( x ) . Open image in new window
(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 . Open image in new window
(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 . Open image in new window
(13)
Thus, by (13) we get
D k I r p ( 0 ) = k ! a k . Open image in new window
(14)
Hence, from (14) we have
a k = D k I r p ( 0 ) k ! . Open image in new window
(15)

Case 1. Let r > n Open image in new window. Then r > k Open image in new window for all k = 0 , 1 , 2 , , n Open image in new window.

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 ) . Open image in new window
(16)
Case 2. Assume that r n Open image in new window.
  1. (i)
    For 0 k r Open image in new window, by (15) we get
    a k = 1 k ! j = 0 k ( 1 ) k j ( k j ) I r k p ( j ) . Open image in new window
    (17)
     
  2. (ii)
    For r k n Open image in new window, 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 ) . Open image in new window
    (18)
     

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

Theorem 1
  1. (a)
    For r > n Open image in new window, 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 ) . Open image in new window
     
  2. (b)
    For r n Open image in new window, 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 ) . Open image in new window
     
Let us take p ( x ) = x n P n Open image in new window. Then x n Open image in new window can be expressed as a linear combination of B 0 ( r ) , B 1 ( r ) , , B n ( r ) Open image in new window. For r > n Open image in new window, 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 . Open image in new window
(19)

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

Corollary 2 For n , r Z + Open image in new window with r > n Open image in new window, 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 ) . Open image in new window
Let us assume that r , n Z + Open image in new window with r n Open image in new window. 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 . Open image in new window
(20)

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

Corollary 3 For n , r Z + Open image in new window with r n Open image in new window, 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 ) . Open image in new window
Let us take p ( x ) = B n ( s ) ( x ) P n Open image in new window ( s Z + Open image in new window). Then p ( x ) Open image in new window can be generated by { B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , , B n ( r ) ( x ) } Open image in new window as follows:
B n ( s ) ( x ) = k = 0 n a k B k ( r ) ( x ) . Open image in new window
(21)
For r > n Open image in new window, we have
I r k B n ( s ) ( x ) = B n ( s r + k ) ( x ) . Open image in new window
(22)

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

Theorem 4 For r , n , s Z + Open image in new window with r > n Open image in new window, 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 ) . Open image in new window
In particular, for r = s Open image in new window, 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 ) . Open image in new window
(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 . Open image in new window
(24)

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

Corollary 5
  1. (a)
    For n , k Z + Open image in new window with 0 k n 1 Open image in new window, we have
    j = 0 k ( 1 ) k j ( k j ) B n ( k ) ( j ) = 0 . Open image in new window
     
  2. (b)
    In particular, k = n Open image in new window, we get
    j = 0 n ( 1 ) n j ( n j ) B n ( n ) ( j ) = n ! . Open image in new window
     
Let us assume that r n Open image in new window 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 ) . Open image in new window
(25)

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

Theorem 6 For r , n Z + Open image in new window with r n Open image in new window, 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 ) . Open image in new window

Let p ( x ) = E n ( s ) ( x ) Open image in new window ( s Z + Open image in new window) be Euler polynomials of order s. Then E n ( s ) Open image in new window can be expressed as a linear combination of B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , , B n ( r ) ( x ) Open image in new window.

Assume that r , n Z + Open image in new window with r > n Open image in new window.

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 ) . Open image in new window
(26)

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

Theorem 7 For r , n Z + Open image in new window with r > n Open image in new window, 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 ) . Open image in new window
For r , n Z + Open image in new window with r n Open image in new window, we have
D k r E n ( s ) ( x ) = n ( n 1 ) ( n k + r + 1 ) E n k + r ( s ) ( x ) . Open image in new window
(27)

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

Theorem 8 For r , n Z + Open image in new window with r n Open image in new window, 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 ) . Open image in new window
Remarks (a) For r 0 Open image in new window, 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 . Open image in new window
Thus, for x = 0 Open image in new window, 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 ) , Open image in new window
(28)
where S ( n , r ) Open image in new window 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 ) . Open image in new window
    (29)
     
Applying I t Open image in new window on both sides ( t 0 Open image in new window), 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 . Open image in new window
(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 ) . Open image in new window
Remark Let us define two operators d, d ˜ Open image in new window as follows:
d = e D = n = 0 ( 1 ) n n ! D n , d ˜ = e D = n = 0 D n n ! . Open image in new window
(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 . Open image in new window
(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 ) , Open image in new window
(33)
and
d ˜ E n ( r ) ( x ) = E n ( r ) ( x + 1 ) , d E n ( r ) ( x ) = E n ( r ) ( x 1 ) . Open image in new window
(34)

3 Further remarks

For any r 0 , r 1 , , r n Z + Open image in new window, { B 0 ( r 0 ) ( x ) , B 1 ( r 1 ) ( x ) , , B 0 ( r n ) ( x ) } Open image in new window forms a basis for P n Open image in new window. Let r = max { r i | i = 0 , 1 , 2 , , n } Open image in new window. Let p ( x ) P n Open image in new window. Then p ( x ) Open image in new window can be expressed as a linear combination of B 0 ( r 0 ) ( x ) , B 1 ( r 1 ) ( x ) , , B n ( r n ) ( x ) Open image in new window 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 ) . Open image in new window
(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 . Open image in new window
(36)
Now, for each k = 0 , 1 , 2 , , n Open image in new window, 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 . Open image in new window
(37)
Let us take x = 0 Open image in new window 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 ) . Open image in new window
(38)
Case 1. For r > n Open image in new window, 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 ) . Open image in new window
(39)
Case 2. Let r n Open image in new window.
  1. (i)
    For 0 k < r Open image in new window, we have
    D k I r p ( 0 ) = j = 0 k ( 1 ) k j ( k j ) I r k p ( j ) . Open image in new window
    (40)
     
  2. (ii)
    For r k n Open image in new window, 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 ) . Open image in new window
    (41)
     

Thus, by (38), (39), (40) and (41), we can determine a 0 , a 1 , a 2 , , a n Open image in new window.

Notes

Acknowledgements

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

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© Kim and Kim; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsSogang UniversitySeoulRepublic of Korea
  2. 2.Department of MathematicsKwangwoon UniversitySeoulRepublic of Korea

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