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Discrete Dynamics in Nature and Society
Volume 2012 (2012), Article ID 584643, 11 pages
http://dx.doi.org/10.1155/2012/584643
Research Article

Some Identities on Bernoulli and Hermite Polynomials Associated with Jacobi Polynomials

1Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
2Department of Mathematics, Sogang University, Seoul, Republic of Korea
3Hanrimwon, Kwangwoon University, Seoul 139-701, Republic of Korea

Received 17 July 2012; Accepted 9 August 2012

Academic Editor: Josef Diblík

Copyright © 2012 Taekyun Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We investigate some identities on the Bernoulli and the Hermite polynomials arising from the orthogonality of Jacobi polynomials in the inner product space Pn.

1. Introduction

For 𝛼,𝛽 with 𝛼>1 and 𝛽>1, the Jacobi polynomials 𝑃𝑛(𝛼,𝛽)(𝑥) are defined as 𝑃𝑛(𝛼,𝛽)(𝑥)=(𝛼+1)𝑛𝑛!2𝐹1𝑛,1+𝛼+𝛽+𝑛;𝛼+1;1𝑥2=(𝛼+1)𝑛𝑛!𝑛𝑘=0(𝑛𝑘)(1+𝛼+𝛽+𝑛)𝑘(𝛼+1)𝑘𝑥12𝑘,(1.1)(see [14]), where (𝛼)𝑛=𝛼(𝛼+1)(𝛼+𝑛1)=Γ(𝛼+𝑛)/Γ(𝛼).

From (1.1), we note that 𝑃𝑛(𝛼,𝛽)Γ(𝑥)=(𝛼+1+𝑛)𝑛!Γ(𝛼+𝛽+𝑛+1)𝑛𝑘=0(𝑛𝑘)Γ(𝛼+𝛽+𝑛+𝑘+1)Γ(𝛼+𝑘+1)𝑥12𝑘.(1.2) By (1.2), we see that 𝑃𝑛(𝛼,𝛽)(𝑥) is polynomial of degree 𝑛 with real coefficients. It is not difficult to show that the leading coefficient of 𝑃𝑛(𝛼,𝛽)(𝑥) is 2𝑛𝑛𝛼+𝛽+2𝑛. From (1.2), we have 𝑃𝑛(𝛼,𝛽)(1)=(𝑛𝛼+𝑛).

By (1.1), we get 𝑑𝑑𝑥𝑘𝑃𝑛(𝛼,𝛽)(𝑥)=2𝑘Γ(𝑛+𝛼+𝛽+𝑘+1)𝑃Γ(𝑛+𝛼+𝛽+1)(𝛼+𝑘,𝛽+𝑘)𝑛𝑘=1(𝑥)2𝑘(𝑛+𝛼+𝛽+𝑘)(𝑛+𝛼+𝛽+𝑘1)(𝑛+𝛼+𝛽+1)𝑃(𝛼+𝑘,𝛽+𝑘)𝑛𝑘(𝑥),(1.3) where 𝑘 is a positive integer (see [14]).

The Rodrigues' formula for 𝑃𝑛(𝛼,𝛽)(𝑥) is given by (1𝑥)𝛼(1+𝑥)𝛽𝑃𝑛(𝛼,𝛽)(𝑥)=(1)𝑛2𝑛𝑑𝑛!𝑑𝑥𝑘(1𝑥)𝑛+𝛼(1+𝑥)𝑛+𝛽.(1.4) It is easy to show that 𝑢=𝑃𝑛(𝛼,𝛽)(𝑥) is a solution of the following differential equation: 1𝑥2𝑢+{𝛽𝛼(𝛼+𝛽+2)𝑥}𝑢+𝑛(𝑛+𝛼+𝛽+1)𝑢=0.(1.5) As is well known, the generating function of 𝑃𝑛(𝛼,𝛽)(𝑥) is given by 𝐹(𝑥,𝑡)=𝑛=0𝑃𝑛(𝛼,𝛽)(𝑥)𝑡𝑛=2𝛼+𝛽𝑅(1𝑡+𝑅)𝛼(1+𝑡+𝑅)𝛽,(1.6) where 𝑅=12𝑥𝑡+𝑡2, (see [14]).

From (1.3), (1.4), and (1.6), we can derive the following identity: 11𝑃𝑚(𝛼,𝛽)(𝑥)𝑃𝑛(𝛼,𝛽)(𝑥)(1𝑥)𝛼(1+𝑥)𝛽=2𝑑𝑥𝛼+𝛽+1Γ(𝑛+𝛼+1)Γ(𝑛+𝛽+1)𝛿(2𝑛+𝛼+𝛽+1)Γ(𝑛+𝛼+𝛽+1)Γ(𝑛+1)𝑛,𝑚,(1.7) where 𝛿𝑛,𝑚 is the Kronecker symbol.

Let 𝐏𝑛={𝑝(𝑥)[𝑥]deg𝑝(𝑥)𝑛}. Then 𝐏𝑛 is an inner product space with respect to the inner product 𝑞1(𝑥),𝑞2(𝑥)=11(1𝑥)𝛼(1+𝑥)𝛽𝑞1(𝑥)𝑞2(𝑥)𝑑𝑥, where 𝑞1(𝑥),𝑞2(𝑥)𝐏𝑛. From (1.7), we note that {𝑃0(𝛼,𝛽)(𝑥),𝑃1(𝛼,𝛽)(𝑥),,𝑃𝑛(𝛼,𝛽)(𝑥)} is an orthogonal basis for 𝐏𝑛.

The so-called Euler polynomials 𝐸𝑛(𝑥) may be defined by means of 2𝑒𝑡𝑒+1𝑥𝑡=𝑒𝐸(𝑥)𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛,𝑛!(1.8) (see [522]), with the usual convention about replacing 𝐸𝑛(𝑥) by 𝐸𝑛(𝑥). In the special case, 𝑥=0, 𝐸𝑛(0)=𝐸𝑛 are called the Euler numbers.

The Bernoulli polynomials are also defined by the generating function to be 𝑡𝑒𝑡𝑒1𝑥𝑡=𝑒𝐵(𝑥)𝑡=𝑛=0𝐵𝑛(𝑡𝑥)𝑛,𝑛!(1.9) (see [1121]), with the usual convention about replacing 𝐵𝑛(𝑥) by 𝐵𝑛(𝑥).

From (1.8) and (1.9), we note that 𝐵𝑛(𝑥)=𝑛𝑘=0𝑛𝑘𝐵𝑛𝑘𝑥𝑘,𝐸𝑛(𝑥)=𝑛𝑘=0𝑛𝑘𝐸𝑛𝑘𝑥𝑘.(1.10) For 𝑛+, we have 𝑑𝐵𝑛(𝑥)𝑑𝑥=𝑛𝐵𝑛1(𝑥),𝑑𝐸𝑛(𝑥)𝑑𝑥=𝑛𝐸𝑛1(𝑥)(1.11) (see [2329]) By the definition of Bernoulli and Euler polynomials, we get 𝐵0=1,𝐵𝑛(1)𝐵𝑛=𝛿1,𝑛,𝐸0=1,𝐸𝑛(1)+𝐸𝑛=2𝛿0,𝑛.(1.12)

In this paper we give some interesting identities on the Bernoulli and the Hermite polynomials arising from the orthogonality of Jacobi polynomials in the inner product space 𝐏𝑛.

2. Bernoulli, Euler and Jacobi Polynomials

From (1.4), we have 𝑃𝑛(𝛼,𝛽)(𝑥)=𝑛𝑘=0𝑘𝑛+𝛼𝑛𝑘𝑛+𝛽𝑥12𝑘𝑥+12𝑛𝑘.(2.1) By (2.1), we have 𝑛=0𝑃𝑛(𝛼,𝛽)(𝑥)𝑡𝑛=12𝜋𝑖(1+((𝑥+1)/2)𝑧)𝑛+𝛼(1+((𝑥1)/2)𝑧)𝑛+𝛽𝑧𝑛+1𝑑𝑧,(2.2) where we assume 𝑥±1 and circle around 0 is taken so small that 2(𝑥±1)1 lie neither on it nor in its interior. It is not so difficult to show that 𝑃𝑛(𝛼,𝛽)(𝑥)=(1)𝑛𝑃𝑛(𝛽,𝛼)(𝑥).

For 𝑞(𝑥)𝐏𝑛, let 𝑞(𝑥)=𝑛𝑘=0𝐶𝑘𝑃𝑘(𝛼,𝛽)𝐶(𝑥),𝑘.(2.3) From (1.7), we note that 𝑞(𝑥),𝑃𝑘(𝛼,𝛽)(𝑥)=𝐶𝑘𝑃𝑘(𝛼,𝛽)(𝑥),𝑃𝑘(𝛼,𝛽)(𝑥)=𝐶𝑘11(1𝑥)𝛼(1+𝑥)𝛽𝑃𝑘(𝛼,𝛽)(𝑥)2𝑑𝑥=𝐶𝑘2𝛼+𝛽+1Γ(𝑘+𝛼+1)Γ(𝑘+𝛽+1).(2𝑘+𝛼+𝛽+1)Γ(𝛼+𝛽+𝑘+1)𝑘!(2.4) Thus, by (2.4), we get 𝐶𝑘=(2𝑘+𝛼+𝛽+1)Γ(𝛼+𝛽+𝑘+1)𝑘!2𝛼+𝛽+1Γ(𝑘+𝛼+1)Γ(𝑘+𝛽+1)11(1𝑥)𝛼(1+𝑥)𝛽𝑃𝑘(𝛼,𝛽)(𝑥)𝑞(𝑥)𝑑𝑥.(2.5) Therefore, by (1.7), (2.3), and (2.5), we obtain the following proposition.

Proposition 2.1. For 𝑞(𝑥)𝐏𝑛(𝑛), one has 𝑞(𝑥)=𝑛𝑘=0𝐶𝑘𝑃𝑘(𝛼,𝛽)(𝑥),(2.6) where 𝐶𝑘=(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)2𝛼+𝛽+1+𝑘Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)11𝑑𝑘𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑞(𝑥)𝑑𝑥.(2.7)

Let us take 𝑞(𝑥)=𝑥𝑛𝐏𝑛. First, we consider the following integral: 11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽=𝑞(𝑥)𝑑𝑥11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥𝑛𝑑𝑥=(𝑛)11𝑑𝑑𝑥𝑘1(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥𝑛1𝑑𝑥==(1)𝑘𝑛!(𝑛𝑘)!11(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥𝑛𝑘=𝑑𝑥(1)𝑘𝑛!22𝑘+𝛼+𝛽+1(𝑛𝑘)!10𝑦𝑘+𝛽(1𝑦)𝑘+𝛼(2𝑦1)𝑛𝑘=(𝑑𝑦1)𝑘𝑛!2(𝑛𝑘)!2𝑘+𝛼+𝛽+1𝑛𝑘𝑙=0𝑙2𝑛𝑘𝑙(1)𝑛𝑘𝑙=(B(𝑘+𝑙+𝛽+1,𝑘+𝛼+1)1)𝑘𝑛!2(𝑛𝑘)!2𝑘+𝛼+𝛽+1𝑛𝑘𝑙=0𝑙2𝑛𝑘𝑙(1)𝑛𝑘𝑙Γ(𝑘+𝑙+𝛽+1)Γ(𝑘+𝛼+1).Γ(2𝑘+𝛼+𝛽+𝑙+2)(2.8) From (2.5) and (16), we have 𝐶𝑘=(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)2𝛼+𝛽+1+𝑘×Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥𝑛=𝑑𝑥(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)2𝛼+𝛽+1+𝑘Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)(1)𝑘𝑛!22𝑘+𝛼+𝛽+1×(𝑛𝑘)!𝑛𝑘𝑙=0𝑙2𝑛𝑘𝑙(1)𝑛𝑘𝑙Γ(𝑘+𝑙+𝛽+1)Γ(𝑘+𝛼+1)=Γ(2𝑘+𝛼+𝛽+𝑙+2)(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)𝑛!2𝑘Γ(𝛽+𝑘+1)(𝑛𝑘)!𝑛𝑘𝑙=0(1)𝑛𝑘𝑙𝑙𝑛𝑘2𝑙Γ(𝑘+𝑙+𝛽+1).Γ(2𝑘+𝛼+𝛽+𝑙+2)(2.9) By Proposition 2.1, we get 𝑥𝑛=𝑛!𝑛𝑘=0𝑛𝑘𝑙=0(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)Γ2(𝑘+𝛽+1)(𝑛𝑘)!𝑘×(1)𝑛𝑘𝑙𝑙𝑛𝑘2𝑙Γ(𝑘+𝑙+𝛽+1)𝑃Γ(2𝑘+𝛼+𝛽+𝑙+2)𝑘(𝛼,𝛽)(𝑥).(2.10) From (1.9), we have 𝑒𝑥𝑡=1𝑡𝑡𝑒𝑡𝑒1𝑥𝑡𝑒𝑡=1𝑛=0𝐵𝑛+1(𝑥+1)𝐵𝑛+1(𝑥)𝑡𝑛+1𝑛.𝑛!(2.11) By (2.11), we get 𝑥𝑛=𝐵𝑛+1(𝑥+1)𝐵𝑛+1(𝑥),𝑛+1𝑛+.(2.12) Therefore, by (2.10) and (2.12), we obtain the following theorem.

Theorem 2.2. For 𝑛+, one has 1𝐵(𝑛+1)!𝑛+1(𝑥+1)𝐵𝑛+1=(𝑥)𝑛𝑘=0𝑛𝑘𝑙=0(1)𝑛𝑘𝑙2𝑘+𝑙(2𝑘+𝛼+𝛽+1)𝑙𝑛𝑘Γ𝑃(𝑘+𝛽+1)Γ(2𝑘+𝛼+𝛽+𝑙+2)(𝑛𝑘)!×Γ(𝑘+𝛼+𝛽+1)Γ(𝑘+𝑙+𝛽+1)𝑘(𝛼,𝛽)(𝑥).(2.13)

Let us take 𝑞(𝑥)=𝐵𝑛(𝑥)𝐏𝑛. Then we evaluate the following integral: 11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝐵𝑛=(𝑥)𝑑𝑥𝑛𝑙=𝑘𝑛𝑙𝐵𝑛𝑙(1)𝑘𝑙!2(𝑙𝑘)!2𝑘+𝛼+𝛽+110𝑦𝑘+𝛽(1𝑦)𝑘+𝛼(2𝑦1)𝑙𝑘=𝑑𝑦𝑛𝑙=𝑘𝑛𝑙𝐵𝑛𝑙(1)𝑘𝑙!(2𝑙𝑘)!2𝑘+𝛼+𝛽+1𝑙𝑘𝑚=0𝑚2𝑙𝑘𝑚(1)𝑙𝑘𝑚×Γ(𝑘+𝑚+𝛽+1)Γ(𝑘+𝛼+1)=Γ(2𝑘+𝛼+𝛽+𝑚+2)𝑛𝑙=𝑘𝑙𝑘𝑚=0(𝑛𝑙)𝐵𝑛𝑙(1)𝑙𝑚𝑙!22𝑘+𝛼+𝛽+1𝑚𝑙𝑘2𝑚Γ(𝑘+𝑚+𝛽+1)Γ(𝑘+𝛼+1).(𝑙𝑘)!Γ(2𝑘+𝛼+𝛽+𝑚+2)(2.14) Finding (2.5) and (21), we have 𝐶𝑘=(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝛼+𝛽+𝑘+1)2𝛼+𝛽+𝑘+1×Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)11𝑑𝑘𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝐵𝑛=(𝑥)𝑑𝑥𝑛𝑙=𝑘𝑙𝑘𝑚=02𝑘+𝑚(𝑛𝑙)𝐵𝑛𝑙(1)𝑙𝑚𝑘𝑙!(2𝑘+𝛼+𝛽+1)𝑚𝑙𝑘Γ(𝛽+𝑘+1)(𝑙𝑘)!Γ(2𝑘+𝛼+𝛽+𝑚+2)×Γ(𝑘+𝑚+𝛽+1)Γ(𝑘+𝛼+𝛽+1).(2.15)

Theorem 2.3. For 𝑛+, one has 𝐵𝑛(𝑥)=𝑛𝑘=0𝑛𝑙=𝑘𝑙𝑘𝑚=02𝑘+𝑚(𝑛𝑙)𝐵𝑛𝑙(1)𝑙𝑚𝑘𝑙!(2𝑘+𝛼+𝛽+1)𝑚𝑙𝑘Γ𝑃(𝛽+𝑘+1)(𝑙𝑘)!Γ(2𝑘+𝛼+𝛽+𝑚+2)×Γ(𝑘+𝑚+𝛽+1)Γ(𝑘+𝛼+𝛽+1)𝑘(𝛼,𝛽)(𝑥).(2.16)

Let 𝑞(𝑥)=𝑃𝑛(𝛼,𝛽)(𝑥)𝐏𝑛. From Proposition 2.1, we firstly evaluate the following integral: 11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑃𝑛(𝛼,𝛽)(𝑥)𝑑𝑥=(1)𝑘12𝑘Γ(𝑛+𝛼+𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)11(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑃(𝛼+𝑘,𝛽+𝑘)𝑛𝑘(𝑥)𝑑𝑥.(2.17)

By (2.1) and (2.17), we get 11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑃𝑛(𝛼,𝛽)=(𝑥)𝑑𝑥(1)𝑘2𝑘Γ(𝑛+𝛼+𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙×𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽11(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥12𝑙𝑥+12𝑛𝑘𝑙=𝑑𝑥(1)𝑘2𝑘Γ(𝑛+𝛼+𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽(1)𝑙22𝑘+𝛼+𝛽+1×10(1𝑦)𝑘+𝛼+𝑙𝑦𝑛+𝛽𝑙𝑑𝑦=(1)𝑘2𝛼+𝛽+𝑘+1Γ(𝑛+𝛼+𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽(1)𝑙×𝐵(𝑘+𝛼+𝑙+1,𝑛+𝛽𝑙+1)=(1)𝑘2𝛼+𝛽+𝑘+1Γ(𝑛+𝛼+𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽(1)𝑙×Γ(𝛼+𝑘+𝑙+1)Γ(𝑛+𝛽𝑙+1)Γ(𝛼+𝛽+𝑘+𝑛+2)=(1)𝑘2𝛼+𝛽+𝑘+11Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽(1)𝑙×Γ(𝛼+𝑘+𝑙+1)Γ(𝑛+𝛽𝑙+1).(𝛼+𝛽+𝑘+𝑛+1)(2.18) It is easy to show that Γ(𝑛+𝛽𝑙+1)=Γ(𝛽+𝑘+1)(𝑛+𝛽𝑙)𝛽Γ(𝛽)=(𝛽+𝑘)𝛽Γ(𝛽)=(𝑛+𝛽𝑙)(𝛽+𝑘+1)𝑛+𝛽𝑙𝑛𝑘𝑙(𝑛𝑘𝑙)!.(2.19) From (2.5), (2.18), and (2.19), we can derive the following equation: 𝐶𝑘=(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)2𝛼+𝛽+𝑘+1×Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑃𝑛(𝛼,𝛽)=(𝑥)𝑑𝑥(2𝑘+𝛼+𝛽+1)Γ(𝛼+𝛽+𝑘+1)Γ(𝛽+𝑘+1)Γ(𝑛+𝛼+𝛽+1)𝑛𝑘𝑙=0𝑙𝑙𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽𝛼+𝑘+𝑙×𝑙!(1)𝑙Γ(𝑛+𝛽𝑙+1)(𝛼+𝛽+𝑘+𝑛+1)=(2𝑘+𝛼+𝛽+1)Γ(𝛼+𝛽+𝑘+1)𝑛𝑘𝑙=0𝑙×𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽𝑛+𝛽𝑙𝑛𝑘𝑙(𝑛𝑘𝑙)!𝑙!𝛼+𝛽+𝑘+𝑛+1(1)𝑙.(2.20) Therefore, by Proposition 2.1, we obtain the following theorem.

Theorem 2.4. For (𝑛+), one has Γ(𝑛+𝛼+𝛽+1)𝑃𝑛(𝛼,𝛽)(𝑥)=Γ(𝛼+𝛽+1)𝑛𝑘=0𝑛𝑘𝑙=0𝑘×𝑙𝑙(2𝑘+𝛼+𝛽+1)𝛼+𝛽+𝑘𝑛+𝛼𝑛𝑘𝑙𝑛+𝛽𝛼+𝑘+𝑙𝑛+𝛽𝑙𝑛𝑘𝑙(1)𝑙(𝑛𝑘𝑙)!𝑘!𝑙!𝑃𝛼+𝛽+𝑛+𝑘+1𝑘(𝛼,𝛽)(𝑥).(2.21)

Let 𝐻𝑛(𝑥) be the Hermite polynomial with 𝐻𝑛(𝑥)=𝑞(𝑥)=𝑛𝑘=0𝐶𝑘𝑃𝑘(𝛼,𝛽)(𝑥),(2.22) where 𝐶𝑘=(1)𝑘(2𝑘+𝛼+𝛽+1)Γ(𝑘+𝛼+𝛽+1)2𝛼+𝛽+𝑘+1×Γ(𝛼+𝑘+1)Γ(𝛽+𝑘+1)11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝐻𝑛(𝑥)𝑑𝑥.(2.23) Integrating by parts, one has 11𝑑𝑑𝑥𝑘(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝐻𝑛=2(𝑥)𝑑𝑥𝑘(1)𝑘𝑛!(𝑛𝑘)!11(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝐻𝑛𝑘=2(𝑥)𝑑𝑥𝑘(1)𝑘𝑛!(𝑛𝑘)!𝑛𝑘𝑙=0𝑙𝐻𝑛𝑘𝑛𝑘𝑙2𝑙11(1𝑥)𝑘+𝛼(1+𝑥)𝑘+𝛽𝑥𝑙=2𝑑𝑥𝑘(1)𝑘𝑛!(𝑛𝑘)!𝑛𝑘𝑙=0𝑙𝐻𝑛𝑘𝑛𝑘𝑙2𝑙2𝑘+𝛼+𝛽+𝑙+1𝑚=0𝑙𝑚(1)𝑙𝑚2𝑚×10(1𝑦)𝑘+𝛼𝑦𝑘+𝛽+𝑚=2𝑑𝑦𝑘(1)𝑘𝑛!(𝑛𝑘)!𝑛𝑘𝑙𝑙=0𝑚=0𝑙𝑙𝑚𝐻𝑛𝑘𝑛𝑘𝑙(1)𝑙𝑚22𝑘+𝛼+𝛽+𝑚+𝑙+1×Γ(𝑘+𝛼+1)Γ(𝛽+𝑘+𝑚+1)Γ.(2𝑘+𝛼+𝛽+𝑚+2)(2.24) By (2.23) and (29), we get 𝐶𝑘=𝑛𝑘𝑙𝑙=0𝑚=0𝑙𝑛𝑘𝑙𝑚𝐻𝑛𝑘𝑙(1)𝑙𝑚(2𝑘+𝛼+𝛽+1)𝑘𝛼+𝛽+𝑘𝑘!(𝛼+𝛽+1)2𝑘+𝛼+𝛽+𝑚+1𝑚+2𝑘(𝑚+2𝑘)!(𝑛𝑘)!×22𝑘+𝑚+𝑙𝑚𝑛!𝛽+𝑘+𝑚𝑚!.(2.25) Therefore, by (2.22) and (2.25), we obtain the following theorem.

Theorem 2.5. For 𝑛+, one has (𝛼+𝛽+1)𝐻𝑛(𝑥)=𝑛!𝑛𝑘=0𝑛𝑘𝑙𝑙=0𝑚=0𝑙𝑛𝑘𝑙𝑚𝐻𝑛𝑘𝑙(1)𝑙𝑚(2𝑘+𝛼+𝛽+1)2𝑘+𝛼+𝛽+𝑚+1𝑚+2𝑘(×𝑘𝑚+2𝑘)!(𝑛𝑘)!𝛼+𝛽+𝑘𝑘!22𝑘+𝑚+𝑙𝑚𝑃𝛽+𝑘+𝑚𝑚!𝑘(𝛼,𝛽)(𝑥),(2.26) where 𝐻𝑛 is the 𝑛th Hermite number.

Remark 2.6. By the same method as Theorem 2.3, we get 1𝐸2𝑛!𝑛(𝑥+1)+𝐸𝑛=(𝑥)𝑛𝑘=0𝑛𝑘𝑙=02𝑘+𝑙(2𝑘+𝛼+𝛽+1)𝑙𝑛𝑘(1)𝑛𝑘𝑙Γ𝑃(𝑘+𝛽+1)Γ(2𝑘+𝛼+𝛽+𝑙+2)(𝑛𝑘)!×Γ(𝑘+𝛼+𝛽+1)Γ(𝑘+𝑙+𝛽+1)𝑘(𝛼,𝛽)(𝑥).(2.27)

Acknowledgments

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology 2012R1A1A2003786.

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