Advances in Difference Equations

, 2013:103 | Cite as

Higher-order Bernoulli, Euler and Hermite polynomials

  • Dae San Kim
  • Taekyun Kim
  • Dmitry V Dolgy
  • Seog-Hoon Rim
Open Access
Review

Abstract

In (Kim and Kim in J. Inequal. Appl. 2013:111, 2013; Kim and Kim in Integral Transforms Spec. Funct., 2013, doi:10.1080/10652469.2012.754756), we have investigated some properties of higher-order Bernoulli and Euler polynomial bases in P n = { p ( x ) Q [ x ] | deg p ( x ) n } Open image in new window. In this paper, we derive some interesting identities of higher-order Bernoulli and Euler polynomials arising from the properties of those bases for P n Open image in new window.

Keywords

Vector Space Ordinary Differential Equation Linear Operator Functional Equation Explicit Expression 

1 Introduction

For r R Open image in new window, let us define the Bernoulli polynomials of order r as follows:
( t e t 1 ) r e x t = n = 0 B n ( r ) ( x ) t n n ! (see [1–18]). Open image in new window
(1)
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. As is well known, the Euler polynomials of order r are defined by the generating function to be
( 2 e t + 1 ) r e x t = n = 0 E n ( r ) ( x ) t n n ! (see [1–10]). Open image in new window
(2)
For λ ( 1 ) C Open image in new window, the Frobenius-Euler polynomials of order r are also given by
( 1 λ e t λ ) r e x t = n = 0 H n ( r ) ( x | λ ) t n n ! (see [1, 7]). Open image in new window
(3)
The Hermite polynomials are defined by the generating function to be:
e 2 x t t 2 = n = 0 H n ( x ) t n n ! (see [8–10, 19]). Open image in new window
(4)
Thus, by (4), we get
H n ( x ) = ( H + 2 x ) n = l = 0 n ( n l ) H n l 2 l x l (see [14]), Open image in new window
(5)
where H n = H n ( 0 ) Open image in new window are called the n th Hermite numbers. Let P n = { p ( x ) Q [ x ] | deg p ( x ) n } Open image in new window. Then P n Open image in new window is an ( n + 1 ) Open image in new window-dimensional vector space over ℚ. In [8, 10], it is called that { E 0 ( r ) ( x ) , E 1 ( r ) ( x ) , , E n ( r ) ( x ) } Open image in new window and { B 0 ( r ) ( x ) , B 1 ( r ) ( x ) , , B n ( r ) ( x ) } Open image in new window are bases for P n Open image in new window. Let Ω denote the space of real-valued differential functions on ( , ) = R Open image in new window. We define four linear operators on Ω as follows:
I ( f ) ( x ) = x x + 1 f ( x ) d x , Δ ( f ) ( x ) = f ( x + 1 ) f ( x ) , Open image in new window
(6)
Δ ˜ ( f ) ( x ) = f ( x + 1 ) + f ( x ) , D ( f ) ( x ) = f ( x ) . Open image in new window
(7)
Thus, by (6) and (7), we get
I n ( f ) ( x ) = k = 0 n ( n k ) ( 1 ) n l f n ( x + l ) (see [8, 10, 12, 13]), Open image in new window
(8)

where f 1 = f , f 2 = f 1 , , f n = f n 1 Open image in new window, n N Open image in new window.

In this paper, we derive some new interesting identities of higher-order Bernoulli, Euler and Hermite polynomials arising from the properties of bases of higher-order Bernoulli and Euler polynomials for P n Open image in new window.

2 Some identities of higher-order Bernoulli and Euler polynomials

First, we introduce the following theorems, which are important in deriving our results in this paper.

Theorem 1 [8]

For r Z + = N { 0 } Open image in new window, let p ( x ) P n Open image in new window. Then we have
p ( x ) = 1 2 r k = 0 n j = 0 r 1 k ! ( r j ) D k p ( j ) E k ( r ) ( x ) . Open image in new window

Theorem 2 [10]

For r Z + Open image in new window, let p ( x ) P n Open image in new window:
  1. (a)
    If r > n Open image in new window, then 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)
    If r n Open image in new window, then
    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 ( k j ) ( D k r p ( j ) ) B k ( r ) ( x ) . Open image in new window
     

Let us take p ( x ) = H n ( x ) P n Open image in new window.

Then, by (5), we get
p ( k ) ( x ) = D k p ( x ) = 2 k n ( n 1 ) ( n k + 1 ) H n k ( x ) = 2 k n ! ( n k ) ! H n k ( x ) . Open image in new window
(9)
From Theorem 1 and (9), we can derive the following equation (10):
H n ( x ) = 1 2 r k = 0 n { j = 0 r 1 k ! ( r j ) 2 k n ! ( n k ) ! H n k ( j ) } E k ( r ) ( x ) = 1 2 r k = 0 n ( n k ) 2 k [ j = 0 r ( r j ) H n k ( j ) ] E k ( r ) ( x ) . Open image in new window
(10)

Therefore, by (10), we obtain the following theorem.

Theorem 3 For n , r Z + Open image in new window, we have
H n ( x ) = 1 2 r k = 0 n ( n k ) 2 k [ j = 0 r ( r j ) H n k ( j ) ] E k ( r ) ( x ) . Open image in new window
We recall an explicit expression for Hermite polynomials as follows:
H n ( x ) = l = 0 [ n 2 ] ( 1 ) l n ! l ! ( n 2 l ) ! ( 2 x ) n 2 l . Open image in new window
(11)
By (11), we get
H n k ( j ) = l = 0 [ n k 2 ] ( 1 ) l ( n k ) ! l ! ( n k 2 l ) ! ( 2 j ) n k 2 l . Open image in new window
(12)

Thus, by Theorem 3 and (12), we obtain the following corollary.

Corollary 4 For n , r Z + Open image in new window, we have
H n ( x ) = 1 2 r k = 0 n { j = 0 r l = 0 [ n k 2 ] ( n k ) ( r j ) 2 k ( 1 ) l ( n k ) ! ( 2 j ) n k 2 l l ! ( n k 2 l ) ! } E k ( r ) ( x ) . Open image in new window

Now, we consider the identities of Hermite polynomials arising from the property of the basis of higher-order Bernoulli polynomials in P n Open image in new window.

For r > k Open image in new window, by (6) and (8), we get
I r k H n ( x ) = l = 0 r k ( r k l ) ( 1 ) r k l H n + r k ( x + l ) 2 r k ( n + 1 ) ( n + r k ) = l = 0 r k ( r k l ) ( 1 ) r k l n ! H n + r k ( x + l ) 2 r k ( n + r k ) ! . Open image in new window
(13)

Therefore, by Theorem 2 and (13), we obtain the following theorem.

Theorem 5 For n , r Z + Open image in new window, with r > n Open image in new window, we have
H n ( x ) = n ! k = 0 n { j = 0 k l = 0 r k ( r k l ) ( k j ) ( 1 ) r j l H n + r k ( j + l ) 2 r k k ! ( n + r k ) ! } B k ( r ) ( x ) . Open image in new window
Let us assume that r , k Z + Open image in new window, with r n Open image in new window. Then, by (b) of Theorem 2, we get
H n ( x ) = n ! k = 0 r 1 { j = 0 k l = 0 r k ( r k l ) ( k j ) ( 1 ) r j l H n + r k ( j + l ) 2 r k k ! ( n + r k ) ! } B k ( r ) ( x ) + n ! k = r n { j = 0 r ( 1 ) r j ( r j ) 2 k r H n + r k ( j ) k ! ( n + r k ) ! } B k ( r ) ( x ) . Open image in new window
(14)

Therefore, by (14), we obtain the following theorem.

Theorem 6 For n , r Z + Open image in new window, with r n Open image in new window, we have
H n ( x ) = n ! k = 0 r 1 { j = 0 k l = 0 r k ( r k l ) ( k j ) ( 1 ) r j l H n + r k ( j + l ) 2 r k k ! ( n + r k ) ! } B k ( r ) ( x ) + n ! k = r n { j = 0 r ( 1 ) r j ( r j ) 2 k r k ! ( n + r k ) ! H n + r k ( j ) } B k ( r ) ( x ) . Open image in new window
Remark From (12), we note that
H n + r k ( j + l ) = m = 0 [ n + r k 2 ] ( 1 ) m ( n + r k ) ! m ! ( n + r k 2 m ) ! ( 2 j + 2 l ) n + r k 2 m Open image in new window
(15)
and
H n + r k ( j ) = m = 0 [ n + r k 2 ] ( 1 ) m ( n + r k ) ! m ! ( n + r k 2 m ) ! ( 2 j ) n + r k 2 m . Open image in new window
(16)

Theorem 7 [10]

For n , r Z + Open image in new window, with r > n Open image in new window and p ( x ) P n Open image in new window, we have
p ( x ) = k = 0 n { j = 0 k l = 0 n ( r k ) ! S 2 ( l + r k , r k ) ( l + r k ) ! k ! ( 1 ) k j ( k j ) p ( l ) ( j ) } B k ( r ) ( x ) , Open image in new window

where S 2 ( l , n ) Open image in new window is the Stirling number of the second kind and p ( l ) ( j ) = D l p ( j ) Open image in new window.

Theorem 8 [10]

For n , r Z + Open image in new window, with r n Open image in new window and p ( x ) P n Open image in new window, we have
p ( x ) = k = 0 r 1 { j = 0 k l = 0 n ( r k ) ! S 2 ( l + r k , r k ) ( l + r k ) ! k ! ( 1 ) k j ( k j ) p ( l ) ( j ) } B k ( r ) ( x ) + k = r n { j = 0 r ( 1 ) r j k ! ( r j ) p ( k r ) ( j ) } B k ( r ) ( x ) . Open image in new window

Let us take p ( x ) = H n ( x ) P n Open image in new window. Then, by Theorem 7 and Theorem 8, we obtain the following corollary.

Corollary 9 For n , r Z + Open image in new window:
  1. (a)
    For r > n Open image in new window, we have
    H n ( x ) = n ! k = 0 n { j = 0 k l = 0 n ( r k ) ! S 2 ( l + r k , r k ) ( l + r k ) ! k ! ( n l ) ! ( 1 ) k j ( k j ) 2 l H n l ( j ) } B k ( r ) ( x ) . Open image in new window
     
  2. (b)
    For r n Open image in new window, we have
    H n ( x ) = n ! k = 0 r 1 { j = 0 k l = 0 n ( r k ) ! S 2 ( l + r k , r k ) ( l + r k ) ! k ! ( n l ) ! ( 1 ) k j ( k j ) 2 l H n l ( j ) } B k ( r ) ( x ) + n ! k = r n { j = 0 r ( 1 ) r j ( r j ) 2 k r H n k + r ( j ) k ! ( n k + r ) ! } B k ( r ) ( x ) . Open image in new window
     

Theorem 10 [9]

For p ( x ) P n Open image in new window, we have
p ( x ) = 1 ( 1 λ ) r k = 0 n { j = 0 r 1 k ! ( r j ) ( λ ) r j p ( k ) ( j ) } H k ( r ) ( x | λ ) . Open image in new window
Let us take p ( x ) = H n ( x ) P n Open image in new window. Then
H n ( x ) = 1 ( 1 λ ) r k = 0 n { j = 0 r 1 k ! ( r j ) ( λ ) r j 2 k n ! ( n k ) ! H n k ( j ) } H k ( r ) ( x | λ ) = 1 ( 1 λ ) r k = 0 n ( n k ) 2 k { j = 0 r ( r j ) ( λ ) r j H n k ( j ) } H k ( r ) ( x | λ ) = 1 ( 1 λ ) r k = 0 n { j = 0 r l = 0 [ n k 2 ] ( n k ) 2 k ( r j ) ( λ ) r j ( 1 ) l ( n k ) ! ( 2 j ) n k 2 l l ! ( n k 2 l ) ! } H k ( r ) ( x | λ ) . Open image in new window
(17)

Therefore, by (17), we obtain the following corollary.

Corollary 11 For n Z + Open image in new window, we have
H n ( x ) = 1 ( 1 λ ) r k = 0 n { j = 0 r l = 0 [ n k 2 ] ( n k ) 2 k ( r j ) ( λ ) r j ( 1 ) l ( n k ) ! ( 2 j ) n k 2 l l ! ( n k 2 l ) ! } H k ( r ) ( x | λ ) . Open image in new window
For r = 1 Open image in new window, the Frobenius-Euler polynomials are defined by the generating function to be
( 1 λ e t λ ) e x t = n = 0 H n ( x | λ ) t n n ! (see [9]). Open image in new window
(18)
Thus, by (18), we get
d d λ H n ( x | λ ) = 1 1 λ ( H n ( 2 ) ( x | λ ) H n ( x | λ ) ) . Open image in new window
(19)
For n Z + Open image in new window, let p ( x ) P n Open image in new window. Then we note that
( 1 λ ) p ( x ) = k = 0 n 1 k ! { p ( k ) ( 1 ) λ p ( k ) ( 0 ) } H k ( x | λ ) (see [9]). Open image in new window
(20)
Let us take p ( x ) = H n ( x ) Open image in new window. Then, by (20), we get
( 1 λ ) H n ( x ) = k = 0 n 1 k ! { 2 k n ! ( n k ) ! H n k ( 1 ) λ 2 k n ! ( n k ) ! H n k } H k ( x | λ ) = k = 0 n ( n k ) 2 k ( H n k ( 1 ) λ H n k ) H k ( x | λ ) (see [9]). Open image in new window
(21)

Therefore, by (21), we obtain the following theorem.

Theorem 12 For n Z + Open image in new window, we have
( 1 λ ) H n ( x ) = k = 0 n ( n k ) 2 k ( H n k ( 1 ) λ H n k ) H k ( x | λ ) . Open image in new window

Let us take d d λ Open image in new window on the both sides of Theorem 12.

Then, we have
H n ( x ) = k = 0 n ( n k ) 2 k H n k H k ( x | λ ) + k = 0 n ( n k ) 2 k ( H n k ( 1 ) λ H n k ) ( d d λ H k ( x | λ ) ) . Open image in new window
(22)
By (22), we get
H n ( x ) = k = 0 n ( n k ) 2 k H n k H k ( x | λ ) + k = 0 n ( n k ) ( λ H n k H n k ( 1 ) ) 2 k ( d d λ H k ( x | λ ) ) . Open image in new window
(23)

Notes

Acknowledgements

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

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Copyright information

© Kim et al.; 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

  • Dae San Kim
    • 1
  • Taekyun Kim
    • 2
  • Dmitry V Dolgy
    • 3
  • Seog-Hoon Rim
    • 4
  1. 1.Department of MathematicsSogang UniversitySeoulS. Korea
  2. 2.Department of MathematicsKwangwoon UniversitySeoulS. Korea
  3. 3.Hanrimwon, Kwangwoon UniversitySeoulS. Korea
  4. 4.Department of Mathematics EducationKyungpook National UniversityTaeguS. Korea

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