Skip to main content

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 .

References

  1. Abramowitz M, Stegun IA: Handbook of Mathematical Functions. Natl. Bur. of Standards, Washington; 1964.

    MATH  Google Scholar 

  2. Araci S, Acikgoz M: A note on the Frobenius-Euler numbers and polynomials associated with Bernstein polynomials. Adv. Stud. Contemp. Math - Jang’jun Math. Soc. 2012, 22(3):399–406.

    MATH  MathSciNet  Google Scholar 

  3. Bayad A, Kim T: Identities involving values of Bernstein, q -Bernoulli, and q -Euler polynomials. Russ. J. Math. Phys. 2011, 18(2):133–143. 10.1134/S1061920811020014

    MATH  MathSciNet  Article  Google Scholar 

  4. Gould HW: Explicit formulas for Bernoulli numbers. Am. Math. Mon. 1972, 79: 44–51. 10.2307/2978125

    MATH  Article  Google Scholar 

  5. Kim T: Symmetry p -adic invariant integral on Z p for Bernoulli and Euler polynomials. J. Differ. Equ. Appl. 2008, 14(12):1267–1277. 10.1080/10236190801943220

    MATH  Article  Google Scholar 

  6. Kim DS, Kim T, Kim YH, Dolgy DV: A note on Eulerian polynomials associated with Bernoulli and Euler numbers and polynomials. Adv. Stud. Contemp. Math - Jang’jun Math. Soc. 2012, 22(3):379–389.

    MATH  MathSciNet  Google Scholar 

  7. Kim DS, Kim T: Euler basis, identities, and their applications. Int. J. Math. Math. Sci. 2012., 2012: Article ID 343981 10.1155/2012/343981

    Google Scholar 

  8. Kim DS, Kim T: Bernoulli basis and the product of several Bernoulli polynomials. Int. J. Math. Math. Sci. 2012., 2012: Article ID 463659 10.1155/2012/463659

    Google Scholar 

  9. Kim T: Sums of products of q -Bernoulli numbers. Arch. Math. 2001, 76(3):190–195. 10.1007/s000130050559

    MATH  MathSciNet  Article  Google Scholar 

  10. Kim T, Adiga C: Sums of products of generalized Bernoulli numbers. Int. Math. J. 2004, 5(1):1–7.

    MathSciNet  Google Scholar 

  11. Kim T: Symmetry of power sum polynomials and multivariate fermionic p -adic invariant integral on Z p . Russ. J. Math. Phys. 2009, 16(1):93–96. 10.1134/S1061920809010063

    MATH  MathSciNet  Article  Google Scholar 

  12. Kim T: Identities involving Frobenius-Euler polynomials arising from non-linear differential equations. J. Number Theory 2012, 132(12):2854–2865. 10.1016/j.jnt.2012.05.033

    MATH  MathSciNet  Article  Google Scholar 

  13. Kim T: Some formulae for the q -Bernstein polynomials and q -deformed binomial distributions. J. Comput. Anal. Appl. 2012, 14(5):917–933.

    MATH  MathSciNet  Google Scholar 

  14. Petojević A: New sums of products of Bernoulli numbers. Integral Transforms Spec. Funct. 2008, 19(1–2):105–114.

    MATH  MathSciNet  Article  Google Scholar 

  15. Rim S-H, Jeong J: On the modified q -Euler numbers of higher order with weight. Adv. Stud. Contemp. Math - Jang’jun Math. Soc. 2012, 22(1):93–98.

    MATH  MathSciNet  Google Scholar 

  16. Ryoo CS: Some relations between twisted q -Euler numbers and Bernstein polynomials. Adv. Stud. Contemp. Math - Jang’jun Math. Soc. 2011, 21(2):217–223.

    MATH  MathSciNet  Google Scholar 

  17. Kim T: An identity of the symmetry for the Frobenius-Euler polynomials associated with the fermionic p -adic invariant q -integrals on Z p . Rocky Mt. J. Math. 2011, 41(1):239–247. 10.1216/RMJ-2011-41-1-239

    MATH  Article  Google Scholar 

Download references

Acknowledgements

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

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Taekyun Kim.

Additional information

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.

Rights and permissions

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

Reprints and Permissions

About this article

Cite this article

Kim, D.S., Kim, T. A note on higher-order Bernoulli polynomials. J Inequal Appl 2013, 111 (2013). https://doi.org/10.1186/1029-242X-2013-111

Download citation

  • Received:

  • Accepted:

  • Published:

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

Keywords

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