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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 $r≤n$.

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

(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, 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 $a 0 , a 1 , a 2 ,…, a n$.

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## Acknowledgements

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

## Author information

Correspondence to Taekyun Kim.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

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

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