Some Identities on Bernoulli and Hermite Polynomials Associated with Jacobi Polynomials
Taekyun Kim,1Dae San Kim,2and Dmitry V. Dolgy3
Academic Editor: Josef Diblík
Received17 Jul 2012
Accepted09 Aug 2012
Published10 Sept 2012
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 with and , the Jacobi polynomials are defined as (see [1–4]), where .
From (1.1), we note that
By (1.2), we see that is polynomial of degree with real coefficients. It is not difficult to show that the leading coefficient of is . From (1.2), we have .
By (1.1), we get
where is a positive integer (see [1–4]).
The Rodrigues' formula for is given by
It is easy to show that is a solution of the following differential equation:
As is well known, the generating function of is given by
where , (see [1–4]).
From (1.3), (1.4), and (1.6), we can derive the following identity:
where is the Kronecker symbol.
Let . Then is an inner product space with respect to the inner product , where . From (1.7), we note that is an orthogonal basis for .
The so-called Euler polynomials may be defined by means of
(see [5–22]), with the usual convention about replacing by . In the special case, , are called the Euler numbers.
The Bernoulli polynomials are also defined by the generating function to be
(see [11–21]), with the usual convention about replacing by .
From (1.8) and (1.9), we note that
For , we have
(see [23–29]) By the definition of Bernoulli and Euler polynomials, we get
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 .
2. Bernoulli, Euler and Jacobi Polynomials
From (1.4), we have
By (2.1), we have
where we assume and circle around is taken so small that lie neither on it nor in its interior. It is not so difficult to show that .
For , let
From (1.7), we note that
Thus, by (2.4), we get
Therefore, by (1.7), (2.3), and (2.5), we obtain the following proposition.
Proposition 2.1. For , one has
where
Let us take . First, we consider the following integral:
From (2.5) and (16), we have
By Proposition 2.1, we get
From (1.9), we have
By (2.11), we get
Therefore, by (2.10) and (2.12), we obtain the following theorem.
Theorem 2.2. For , one has
Let us take . Then we evaluate the following integral:
Finding (2.5) and (21), we have
Theorem 2.3. For , one has
Let . From Proposition 2.1, we firstly evaluate the following integral:
By (2.1) and (2.17), we get
It is easy to show that
From (2.5), (2.18), and (2.19), we can derive the following equation:
Therefore, by Proposition 2.1, we obtain the following theorem.
Theorem 2.4. For , one has
Let be the Hermite polynomial with
where
Integrating by parts, one has
By (2.23) and (29), we get
Therefore, by (2.22) and (2.25), we obtain the following theorem.
Theorem 2.5. For , one has
where is the th Hermite number.
Remark 2.6. By the same method as Theorem 2.3, we get
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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