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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,

1

Taekyun Kim,

2

Seog-Hoon Rim,

3

and Sang Hun Lee

4

1Department 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.

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The Bernoulli numbers are defined as

B01, B1nBn 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

xd

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

xd

dx

e−x2/2. 1.8

We note that

−1nex2 dn

dxne−x2

−ex2 d dxe−x2

n

. 1.9

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From1.8, we note that

−1nex2

dne−x2 dxn

−ex2de−x2 dx

n

ex2/2

xd

dx

e−x2/2 n

ex2/2

xd dx

n e−x2/2.

1.10

Thus, by1.10, we get

Hnx ex2/2

xd 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|degpxn}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.

ForpxPn,

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.

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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

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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

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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 > mornmwithnm≡1mod 2, m!−1n

π

2m−nm−n/2! ifnmwithnm≡0mod 2.

3.4 From 3.2, we note thatH0x, H1x, . . . , Hnx are orthogonal basis for the spacePn {px∈Qx|degpxn}with respect to the inner product

px, qx

−∞e−x2pxqxdx. 3.5

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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!nk/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

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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

nk l

Bn−k−l

−∞e−x2xldx −1kn!

n−k!

0≤l≤n−k l≡0mod 2

n−k!Bn−k−l

l!nkl! × l!π 2ll/2!

−1kn!

π

0≤l≤n−k l≡0mod 2

Bn−k−l

n−kl!2ll/2!.

3.13

Thus, from3.11and3.13, we have Ck n!

2kk!

0≤l≤n−k l≡0mod 2

Bn−k−l

n−kl!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!nkl!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!nkl!l/2!Hkx. 3.16

Now, we considerpx Hnx. From3.6, we note thatpxcan be rewritten as

Hnx n

k0

CkHkx, 3.17

(9)

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

nk l

2lHn−k−l

−∞e−x2xldx

−1k2kn!

n−k!

n−k

l≡0mod 2l0

2ln−k!

l!nkl!Hn−k−l l!π 2ll/2!

−1k2kn!π

n−k

l≡0mod 2l0

Hn−k−l

n−kl!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−kl!l/2!

⎟⎟

n!

k!

0≤l≤n−k l≡0mod 2

Hn−k−l

n−kl!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!nkl!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!nkl!l/2!Hkx n!Hnx

n! . 3.22

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Thus, we have, for 0≤knk,

0≤l≤n−k l≡0mod 2

Hn−k−l

n−kl!l/2! 0. 3.23

Letl, k∈Zwithkl. Then we easily see that

−∞

dke−x2 dxk

Blxdx −1kl!

π

0jl−k j0mod 2

Bl−k−j

lkj

!2j j/2

!, 3.24

−∞

dke−x2 dxk

Elxdx −1kl!

π

0≤j≤l−k j≡0mod 2

El−k−j lkj

!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

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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≤kn−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×

0jl−k j≡0mod 2

Bl−k−j

lkj

!2j j/2

!

n1−1kn!

π

0jn−k j≡0mod 2

Bn−k−j

nkj

!2j j/2

!

⎫⎪

⎪⎬

⎪⎪

⎭ 2

n2

n−2

lk

0jl−k j≡0mod 2

n2 l

Bn−lBl−k−jl!

2kjk!

lkj

! j/2

!

n1!

0jn−k j≡0mod 2

Bn−k−j k!

nkj

! j/2

!2kj.

3.32 Therefore, by3.27and3.32, we obtain the following theorem.

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Theorem 3.5. Forn∈Z, one has

n k0

BkxBn−kx

n−2

k0

⎧⎪

⎪⎨

⎪⎪

⎩ 2 n2

n−2 lk

0jn−k j≡0mod 2

n2 l

l!Bn−lBl−k−j 2kjk!

lkj

! j/2

!

n1!

0≤jn−k j≡0mod 2

Bn−k−j

2kjk!

nkj

! 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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