1. Introduction
For
α,
β ∈
ℝ with
α > −1 and
β > −1, the Jacobi polynomials
are defined as
(1.1)
(see [
1–
4]), where (
α)
n =
α(
α + 1)⋯(
α +
n − 1) = Γ(
α +
n)/Γ(
α).
From (
1.1), we note that
(1.2)
By (
1.2), we see that
is polynomial of degree
n 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
(1.3)
where
k is a positive integer (see [
1–
4]).
The Rodrigues′ formula for
is given by
(1.4)
It is easy to show that
is a solution of the following differential equation:
(1.5)
As is well known, the generating function of
is given by
(1.6)
where
, (see [
1–
4]).
From (
1.3), (
1.4), and (
1.6), we can derive the following identity:
(1.7)
where
δn,m is the Kronecker symbol.
Let Pn = {p(x) ∈ ℝ[x]∣deg p(x) ≤ n}. Then Pn is an inner product space with respect to the inner product , where q1(x), q2(x) ∈ Pn. From (1.7), we note that is an orthogonal basis for Pn.
The so-called Euler polynomials
En(
x) may be defined by means of
(1.8)
(see [
5–
22]), 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
(1.9)
(see [
11–
21]), with the usual convention about replacing
Bn(
x) by
Bn(
x).
From (
1.8) and (
1.9), we note that
(1.10)
For
n ∈
ℤ+, we have
(1.11)
(see [
23–
29]) By the definition of Bernoulli and Euler polynomials, we get
(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
(2.1)
By (
2.1), we have
(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
.
For
q(
x) ∈
Pn, let
(2.3)
From (
1.7), we note that
(2.4)
Thus, by (
2.4), we get
(2.5)
Therefore, by (
1.7), (
2.3), and (
2.5), we obtain the following proposition.
Proposition 2.1. For q(x) ∈ Pn(n ∈ ℕ), one has
(2.6)
where
(2.7)
Let us take
q(
x) =
xn ∈
Pn. First, we consider the following integral:
(2.8)
From (
2.5) and (16), we have
(2.9)
By Proposition
2.1, we get
(2.10)
From (
1.9), we have
(2.11)
By (
2.11), we get
(2.12)
Therefore, by (
2.10) and (
2.12), we obtain the following theorem.
Theorem 2.2. For n ∈ ℤ+, one has
(2.13)
Let us take
q(
x) =
Bn(
x) ∈
Pn. Then we evaluate the following integral:
(2.14)
Finding (
2.5) and (21), we have
(2.15)
Theorem 2.3. For n ∈ ℤ+, one has
(2.16)
Let
. From Proposition
2.1, we firstly evaluate the following integral:
(2.17)
By (
2.1) and (
2.17), we get
(2.18)
It is easy to show that
(2.19)
From (
2.5), (
2.18), and (
2.19), we can derive the following equation:
(2.20)
Therefore, by Proposition
2.1, we obtain the following theorem.
Theorem 2.4. For (n ∈ ℤ+), one has
(2.21)
Let
Hn(
x) be the Hermite polynomial with
(2.22)
where
(2.23)
Integrating by parts, one has
(2.24)
By (
2.23) and (29), we get
(2.25)
Therefore, by (
2.22) and (
2.25), we obtain the following theorem.
Theorem 2.5. For n ∈ ℤ+, one has
(2.26)
where
Hn is the
nth Hermite number.
Remark 2.6. By the same method as Theorem 2.3, we get
(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.
