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Higher-order Bernoulli, Euler and Hermite polynomials

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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]|degp(x)n}. 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 .

1 Introduction

For rR, 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]).
(1)

In the special case, x=0, B n ( r ) (0)= B n ( r ) 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]).
(2)

For λ(1)C, 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]).
(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]).
(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]),
(5)

where H n = H n (0) are called the n th Hermite numbers. Let P n ={p(x)Q[x]|degp(x)n}. Then P n is an (n+1)-dimensional vector space over . In [8, 10], it is called that { E 0 ( r ) (x), E 1 ( r ) (x),, E n ( r ) (x)} and { B 0 ( r ) (x), B 1 ( r ) (x),, B n ( r ) (x)} are bases for P n . Let Ω denote the space of real-valued differential functions on (,)=R. We define four linear operators on Ω as follows:

I(f)(x)= x x + 1 f(x)dx,Δ(f)(x)=f(x+1)f(x),
(6)
Δ ˜ (f)(x)=f(x+1)+f(x),D(f)(x)= f (x).
(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]),
(8)

where f 1 =f, f 2 = f 1 ,, f n = f n 1 , nN.

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 .

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}, let p(x) P n . 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).

Theorem 2 [10]

For r Z + , let p(x) P n :

  1. (a)

    If r>n, 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).
  2. (b)

    If rn, 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 ) .

Let us take p(x)= H n (x) P n .

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 ) .
(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 ) .
(10)

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

Theorem 3 For n,r Z + , 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).

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 .
(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 .
(12)

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

Corollary 4 For n,r Z + , 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).

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

For r>k, 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 ) ! .
(13)

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

Theorem 5 For n,r Z + , with r>n, 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).

Let us assume that r,k Z + , with rn. 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 ) .
(14)

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

Theorem 6 For n,r Z + , with rn, 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 ) .

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
(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 .
(16)

Theorem 7 [10]

For n,r Z + , with r>n and p(x) P n , 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),

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

Theorem 8 [10]

For n,r Z + , with rn and p(x) P n , 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 ) .

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

Corollary 9 For n,r Z + :

  1. (a)

    For r>n, 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).
  2. (b)

    For rn, 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 ) .

Theorem 10 [9]

For p(x) P n , 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).

Let us take p(x)= H n (x) P n . 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 ) .
(17)

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

Corollary 11 For n Z + , 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).

For r=1, 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]).
(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 ) ) .
(19)

For n Z + , let p(x) P n . Then we note that

(1λ)p(x)= k = 0 n 1 k ! { p ( k ) ( 1 ) λ p ( k ) ( 0 ) } H k (x|λ)(see [9]).
(20)

Let us take p(x)= H n (x). 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]).
(21)

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

Theorem 12 For n Z + , we have

(1λ) H n (x)= k = 0 n ( n k ) 2 k ( H n k ( 1 ) λ H n k ) H k (x|λ).

Let us take d d λ 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 ) ) .
(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 ) ) .
(23)

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

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Keywords

  • Vector Space
  • Ordinary Differential Equation
  • Linear Operator
  • Functional Equation
  • Explicit Expression