Volume 2012, Article ID 974632,13pages doi:10.1155/2012/974632
Research Article
Hermite Polynomials and their Applications Associated with Bernoulli and Euler Numbers
Dae San Kim,
1Taekyun Kim,
2Seog-Hoon Rim,
3and Sang Hun Lee
41Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
2Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
3Department of Mathematics Education, Kyungpook National University, Taegu 702-701, Republic of Korea
4Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea
Correspondence should be addressed to Taekyun Kim,[email protected] Received 7 May 2012; Accepted 15 May 2012
Academic Editor: Garyfalos Papaschinopoulos
Copyrightq2012 Dae San 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.
We derive some interesting identities and arithmetic properties of Bernoulli and Euler polynomials from the orthogonality of Hermite polynomials. LetPn {px ∈ Qx | degpx ≤ n}be the n1-dimensional vector space overQ. Then we show that{H0x, H1x, . . . , Hnx}is a good basis for the spacePnfor our purpose of arithmetical and combinatorial applications.
1. Introduction
As is well known, the Euler polynomials,Enx, are defined by the generating function as follows:
2
et1exteExt∞
n0
Enxtn
n! 1.1
see1–8, with the usual convention about replacingEnxbyEnx.
In the special case,x 0,En0 En is called the nth Euler number. From1.1and definition of Euler numbers, we note that
Enx Exnn
l0
n l
Elxn−ln
l0
n l
En−lxl 1.2
with the usual convention about replacingEnbyEn.
The Bernoulli numbers are defined as
B01, B1n−Bn δ1,n 1.3
see9–14, whereδk,nis a Kronecker symbol.
As is well known, Bernoulli polynomials are also defined by
Bnx Bxnn
l0
n l
Blxn−ln
l0
n l
Bn−lxl 1.4
with the usual convention about replacingBnbyBnsee1,15–18.
The Hermite polynomials are defined by the generating function as follows:
e2xt−t2eHxt∞
n0
Hnxtn
n! 1.5
see5,19, with the usual convention about replacingHnxbyHnx.
From1.5, we can derive the following identities:
Hnx ∂
∂t n
e2xt−t2
t0 ex2 ∂
∂t n
e−x−t2 t0
−1nex2 ∂
∂x n
e−x−t2
t0 −1nex2 dn
dxne−x2
.
1.6
Let us consider two operators as follows:
f −→O1f −
ex2 d dxe−x2
f2xf− df dx, f−→O2f
ex2/2
x− d
dx
e−x2/2
f 2xf− df dx.
1.7
By1.7, we getO1O2. In particular, if we takef 1, then we have
−ex2 d dxe−x2
ex2/2
x− d
dx
e−x2/2. 1.8
We note that
−1nex2 dn
dxne−x2
−ex2 d dxe−x2
n
. 1.9
From1.8, we note that
−1nex2
dne−x2 dxn
−ex2de−x2 dx
n
ex2/2
x− d
dx
e−x2/2 n
ex2/2
x− d dx
n e−x2/2.
1.10
Thus, by1.10, we get
Hnx ex2/2
x− d dx
n
e−x2/2 1.11
see5,19–23. In the special case,x0,Hn0 Hnare called the Hermite numbers.
From1.5, we can derive the following identities:
Hnx H2xnn
l0
n l
Hn−l2lxl 1.12
cf.5,19, with the usual convention about replacingHnbyHn. It is easy to show that ∞
n0
Hntn
n! e−t2∞
l0
−1n
n! t2n. 1.13
By comparing coefficients on the both sides of1.13, we get
H2n −1n2n2n−1· · ·n1 −1n2n!
n! , H2n−10, 1.14 wheren∈N. From1.12, we have
dHnx
dx 2nHn−1x n∈N. 1.15
LetPn{p∈Qx|degpx≤n}be then1-dimensional vector space overQ. Probably, {1, x, x2, . . . , xn}is the most natural basis for this space. But{H0x, H1x, H2x, . . . , Hnx}
is also a good basis for the space Pn, for our purpose of arithmetical and combinatorial applications.
Forpx∈Pn,
px n
k0
CkHkx, 1.16
for some uniquely determinedbl∈Q.
The purpose of this paper is to develop methods for computing Ck from the information ofpx. By using these methods, we define some interesting identities.
2. Properties of Hermite Polynomials
From1.5and1.13, we note that
1 ∞
m0
Hmtm m!
∞
l0
t2l l!
∞
m0
H2m t2m 2m!
∞
l0
2l2l−1· · ·l1 2l! t2l
∞
n0
n
l0
2l2l−1· · ·l1
2l!2n−2l! H2n−2l2n!
t2n 2n!
∞
n0
n
l0
l!
2l l
2n 2l
H2n−2l
t2n 2n!.
2.1
Thus, by2.1, we obtain the following recurrence formula.
Proposition 2.1. Forn∈ZN∪ {0}, one has
n l0
l!
2l l
2n 2l
H2n−2l
1, if n0
0, if n /0 . 2.2
By,1.5, we get
∞ n0
Hn−xtn
n! e2t−x−t2e2x−t−−t2 ∞
n0
Hnx−1ntn
n!. 2.3
From2.3, we can derive the following reflection symmetric identity ofHnx:
Hn−x −1nHnx. 2.4
By1.5, we easily see that
∂
∂t e2xt−t2
2x−2te2xt−t2. 2.5
Thus, by1.5and2.5, we get
∂
∂t ∞
n0
Hnxtn n!
2x−2t ∞
n0
Hnxtn n!
. 2.6
LHS of2.5 ∞
n1
Hnx tn−1
n−1! ∞
n0
Hn1xtn
n!, 2.7
RHS of2.5 ∞
n0
2xHnxtn n!
−∞
n0
2Hnxtn1 n!
∞
n0
2xHnxtn n!
−∞
n1
2Hn−1x tn n−1!
∞
n0
2xHnxtn n!−∞
n1
2nHn−1xtn n!.
2.8
Thus, by2.6and2.7, we get
Hn1x 2xHnx−2nHn−1x, n∈N. 2.9 From1.15and2.9, we note that
Hn1x−2xHnx Hnx 0. 2.10 Differentiating on both sides, we have
2n1Hnx−2Hnx−2xHnx Hnx 0. 2.11
Thus, we have
Hnx−2xHnx 2nHnx 0. 2.12 From2.12, we note thatHnxis a solution of the following second-order linear differential equation:
u−2xu2nu0. 2.13
From1.5, we note that ∞ m0
Hnxtn
n! e2tx−t2 ∞
l0
2xl l! tl
∞
k0
−1k k! t2k
∞
n0
n/2
k0
−1kn!2xn−2k k!n−2k!
tn n!.
2.14
Thus, by2.14, we get
Hnx n/2
k0
−1kn!
k!n−2k!2xn−2k
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
n/2
l0
−1n/2−ln!22l
n/2−l!2l!x2l, ifn≡0mod 2,
n−1/2
l0
−1n−1/2−ln!22l1
n−1/2−l!2l1!x2l1, ifn≡1mod 2.
2.15
3. Main Results
By1.6, we easily get ∞
−∞e−x2HnxHmxdx −1n ∞
−∞
dn dxne−x2
Hmxdx. 3.1
From3.1, we note that ∞
−∞e−x2HnxHmxdx2nn!√
πδm,n. 3.2
It is easy to show that ∞
−∞e−x2xldx
⎧⎪
⎪⎨
⎪⎪
⎩
0 ifl≡1mod 2, l!√
π
2ll/2! ifl≡0mod 2, 3.3 wherel∈Z N∪ {0}. By3.3, we get
∞
−∞
dne−x2 dxn
xmdx
⎧⎪
⎪⎨
⎪⎪
⎩
0 ifn > morn≤mwithn−m≡1mod 2, m!−1n√
π
2m−nm−n/2! ifn≤mwithn−m≡0mod 2.
3.4 From 3.2, we note thatH0x, H1x, . . . , Hnx are orthogonal basis for the spacePn {px∈Qx|degpx≤n}with respect to the inner product
px, qx
∞
−∞e−x2pxqxdx. 3.5
Forpx∈Pn, the polynomialpxis given by px ∞
k0
CkHkx, 3.6
where
Ck 1 2kk!√
π
px, Hkx
−1k 2kk!√
π ∞
−∞
dke−x2 dxk
pxdx.
3.7
Let us takepx xn∈Pn. Forn≡0mod 2, we computeCkin3.6as follows
Ck −1k 2kk!√
π ∞
−∞
dke−x2 dxk
xndx
⎧⎪
⎪⎨
⎪⎪
⎩
−1k 2kk!√
π × −1kn!√ π
2n−kn−k/2! ifk≡0mod 2, 0 ifk≡1mod 2.
3.8
Letn≡1mod 2. Then we have
Ck −1k 2kk!√
π ∞
−∞
dke−x2 dxk
xndx
⎧⎪
⎨
⎪⎩
n!
2nk!n−k/2! ifk≡1mod 2, 0 ifk≡0mod 2.
3.9
Therefore, by3.6,3.8, and3.9, we obtain the following proposition.
Proposition 3.1. One has
x2n 2n!
22n n k0
1
2k!n−k!H2kx,
x2n1 2n1!
22n1 n k0
1
2k1!n−k!H2k1x.
3.10
Let us takepx Bnx. From3.4,Pxcan be rewritten by Bnx n
k0
CkHkx, 3.11
where
Ck −1k 2kk!√
π ∞
−∞
dke−x2 dxk
Bnxdx. 3.12
By integrating by parts, we get ∞
−∞
dke−x2 dxk
Bnx −n−n−1· · ·−n−k1 ∞
−∞e−x2Bn−kxdx −1k n!
n−k!
n−k
l0
n−k l
Bn−k−l
∞
−∞e−x2xldx −1kn!
n−k!
0≤l≤n−k l≡0mod 2
n−k!Bn−k−l
l!n−k−l! × l!√ π 2ll/2!
−1kn!√
π
0≤l≤n−k l≡0mod 2
Bn−k−l
n−k−l!2ll/2!.
3.13
Thus, from3.11and3.13, we have Ck n!
2kk!
0≤l≤n−k l≡0mod 2
Bn−k−l
n−k−l!2ll/2!. 3.14
Therefore, by3.11and3.14, we obtain the following theorem.
Theorem 3.2. Forn∈Z, one has
Bnx n!
n k0
0≤l≤n−k l≡0mod 2
Bn−k−l
2klk!n−k−l!l/2!Hkx. 3.15
Remark 3.3. Let us takepx Enx. Then, by the same method, we obtain the following identity:
Enx n!
n k0
0≤l≤n−k l≡0mod 2
En−k−l
2klk!n−k−l!l/2!Hkx. 3.16
Now, we considerpx Hnx. From3.6, we note thatpxcan be rewritten as
Hnx n
k0
CkHkx, 3.17
where
Ck −1k 2kk!√
π ∞
−∞
dke−x2 dxk
Hnxdx. 3.18
By integrating by parts, we get ∞
−∞
dke−x2 dxk
Hnxdx −2n· · ·−2n−k1 ∞
−∞e−x2Hn−kxdx −1k2kn!
n−k!
n−k
l0
n−k l
2lHn−k−l
∞
−∞e−x2xldx
−1k2kn!
n−k!
n−k
l≡0mod 2l0
2ln−k!
l!n−k−l!Hn−k−l l!√ π 2ll/2!
−1k2kn!√ π
n−k
l≡0mod 2l0
Hn−k−l
n−k−l!l/2!.
3.19
From3.17and3.19, we note that
Ck
−1k 2kk!√
π
×
⎛
⎜⎜
⎝−1k2kn!√
π
0≤l≤n−k l≡0mod 2
Hn−k−l
n−k−l!l/2!
⎞
⎟⎟
⎠
n!
k!
0≤l≤n−k l≡0mod 2
Hn−k−l
n−k−l!l/2!.
3.20
Therefore, by3.17and3.20, we obtain the following theorem.
Theorem 3.4. Forn∈Z, one has
Hnx n!
n k0
0≤l≤n−k l≡0mod 2
Hn−k−l
k!n−k−l!l/2!Hkx. 3.21
From Theorem3.4, we note that
Hnx n!
n−1
k0
0≤l≤n−k l≡0mod 2
Hn−k−l
k!n−k−l!l/2!Hkx n!Hnx
n! . 3.22
Thus, we have, for 0≤k≤n−k,
0≤l≤n−k l≡0mod 2
Hn−k−l
n−k−l!l/2! 0. 3.23
Letl, k∈Zwithk≤l. Then we easily see that ∞
−∞
dke−x2 dxk
Blxdx −1kl!√
π
0≤j≤l−k j≡0mod 2
Bl−k−j
l−k−j
!2j j/2
!, 3.24
∞
−∞
dke−x2 dxk
Elxdx −1kl!√
π
0≤j≤l−k j≡0mod 2
El−k−j l−k−j
!2j j/2
!. 3.25
Let us consider the following polynomial of degreeninPn: px n
k0
BkxBn−kx. 3.26
From3.6, we note thatpxcan be rewritten as px n
k0
CkHkx, 3.27
where
Ck −1k 2kk!√
π ∞
−∞
dke−x2 dxk
pxdx. 3.28
In15, it is known that
px n
k0
BkxBn−kx
2 n2
n−2
l0
n2 l
Bn−lBlx n1Bnx.
3.29
From3.23and3.29, we have the following:
Ck −1k 2kk!√
π 2
n2
n−2
l0
n2 l
∞
−∞
dke−x2 dxk
Blxdx n1 ∞
−∞
dke−x2 dxk
Bnxdx
, 3.30
By3.24and3.30, we get
Cn −1n 2nn!√
π
×n1 ∞
−∞
dne−x2 dxn
Bnxdx
−1n 2nn!√
π
×
n1−1nn!√ πB0
0!200!
n1 2n , Cn−1
−1n−1 2n−1n−1!√
π
×
n1 ∞
−∞
dn−1e−x2 dxn−1
Bnxdx
−1n−1 2n−1n−1!√
π
×
⎛
⎜⎜
⎝n1−1n−1n!√ π
1 j≡0mod 2j0
B1−j 1−j
!2j j/2
!
⎞
⎟⎟
⎠
−1n−1 2n−1n−1!√
π
× n1−1n−1n!√ πB1
−nn1 2n .
3.31
For 0≤k≤n−2, we have
Ck
−1k 2kk!√
π 2
n2 n−2
lk
n2 l
Bn−l
∞
−∞
dke−x2 dxk
Blxdx n1 ∞
−∞
dke−x2 dxk
Bnxdx
−1k 2kk!√
π
⎧⎪
⎪⎨
⎪⎪
⎩ 2 n2
n−2
lk
n2 l
Bn−l−1kl!√
π×
0≤j≤l−k j≡0mod 2
Bl−k−j
l−k−j
!2j j/2
!
n1−1kn!√
π
0≤j≤n−k j≡0mod 2
Bn−k−j
n−k−j
!2j j/2
!
⎫⎪
⎪⎬
⎪⎪
⎭ 2
n2
n−2
lk
0≤j≤l−k j≡0mod 2
n2 l
Bn−lBl−k−jl!
2kjk!
l−k−j
! j/2
!
n1!
0≤j≤n−k j≡0mod 2
Bn−k−j k!
n−k−j
! j/2
!2kj.
3.32 Therefore, by3.27and3.32, we obtain the following theorem.
Theorem 3.5. Forn∈Z, one has
n k0
BkxBn−kx
n−2
k0
⎧⎪
⎪⎨
⎪⎪
⎩ 2 n2
n−2 lk
0≤j≤n−k j≡0mod 2
n2 l
l!Bn−lBl−k−j 2kjk!
l−k−j
! j/2
!
n1!
0≤j≤n−k j≡0mod 2
Bn−k−j
2kjk!
n−k−j
! j/2
!
⎫⎪
⎪⎬
⎪⎪
⎭Hkx
−nn1
2n Hn−1x n1 2n Hnx.
3.33
Acknowledgment
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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