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A note on higher-order Bernoulli polynomials

Abstract

Let P n ={p(x)Q[x]|degp(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:

If(x)= x x + 1 f(t)dt,f(x)=f(x+1)f(x),Df(x)= f (x).
(6)

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

Let P n ={p(x)Q(x)|degp(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 rn.

  1. (i)

    For 0kr, 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 rkn, 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 rn, 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 rn. Observe that

D k r x n =n(n1)(nk+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 rn, 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 0kn.
(24)

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

Corollary 5

  1. (a)

    For n,k Z + with 0kn1, 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 rn in (21). Then we have

D k r B n ( s ) (x)=n(n1)(nk+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 rn, 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 rn, we have

D k r E n ( s ) (x)=n(n1)(nk+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 rn, 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 r0, 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)(r0).
    (29)

Applying I t on both sides (t0), 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 ) (x1),
(33)

and

d ˜ E n ( r ) (x)= E n ( r ) (x+1),d E n ( r ) (x)= E n ( r ) (x1).
(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 rn.

  1. (i)

    For 0k<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 rkn, 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 .

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Acknowledgements

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

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Correspondence to Taekyun Kim.

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The authors declare that they have no competing interests.

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

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Keywords

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