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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 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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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© 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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