# Higher-order Bernoulli, Euler and Hermite polynomials

## 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 $r∈R$, 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

$H n (x)= ( H + 2 x ) n = ∑ l = 0 n ( n l ) H n − l 2 l x l (see ),$
(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 $f 1 ′ =f, f 2 ′ = f 1 ,…, f n ′ = f n − 1$, $n∈N$.

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 

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 

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 $r≤n$, 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 $r≤n$. 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 $r≤n$, 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 

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 

For $n,r∈ Z +$, with $r≤n$ 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 $r≤n$, 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 

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

$( 1 − λ e t − λ ) e x t = ∑ n = 0 ∞ H n (x|λ) t n n ! (see ).$
(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

$(1−λ)p(x)= ∑ k = 0 n 1 k ! { p ( k ) ( 1 ) − λ p ( k ) ( 0 ) } H k (x|λ)(see ).$
(20)

Let us take $p(x)= H n (x)$. Then, by (20), we get

$( 1 − λ ) H n ( x ) = ∑ k = 0 n 1 k ! { 2 k n ! ( n − k ) ! H n − k ( 1 ) − λ 2 k n ! ( n − k ) ! H n − k } H k ( x | λ ) = ∑ k = 0 n ( n k ) 2 k ( H n − k ( 1 ) − λ H n − k ) H k ( x | λ ) (see ).$
(21)

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

Theorem 12 For $n∈ Z +$, we have

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

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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The authors declare that they have no competing interests.

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

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Kim, D.S., Kim, T., Dolgy, D.V. et al. Higher-order Bernoulli, Euler and Hermite polynomials. Adv Differ Equ 2013, 103 (2013). https://doi.org/10.1186/1687-1847-2013-103 