# Hermite Polynomials and their Applications Associated with

Authors: Dae San Kim; Taekyun Kim; Seog-Hoon Rim; Sang Hun Lee
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We derive some interesting identities and arithmetic properties of Bernoulli and Euler polynomials from the orthogonality of Hermite polynomials. LetPn={p(x)∈ℚ[x]∣degp(x)≤n} be the (n+1)-dimensional vector space over ℚ. Then we show that {H0(x),H1(x),…,Hn(x)} is a good basis for the space Pn for our purpose of arithmetical and combinatorial applications.


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## Body

## 1. Introduction

As is well known, the Euler polynomials,En(x), are defined by the generating function as follows:
(1.1)2et+1ext=eE(x)t=∑n=0∞En(x)tnn!
(see [1–8]), with the usual convention about replacing En(x) by En(x).In the special case,x=0, En(0)=En is called the nthEuler number. From (1.1) and definition of Euler numbers, we note that
(1.2)En(x)=(E+x)n=∑l=0n(nl)Elxn-l=∑l=0n(nl)En-lxl
with the usual convention about replacing En by En.TheBernoulli numbers are defined as
(1.3)B0=1,(B+1)n-Bn=δ1,n
(see [9–14]), where δk,n is a Kronecker symbol.As is well known,Bernoulli polynomials are also defined by
(1.4)Bn(x)=(B+x)n=∑l=0n(nl)Blxn-l=∑l=0n(nl)Bn-lxl
with the usual convention about replacing Bn by Bn (see [1, 15–18]).TheHermite polynomials are defined by the generating function as follows:
(1.5)e2xt-t2=eH(x)t=∑n=0∞Hn(x)tnn!
(see [5, 19]), with the usual convention about replacing Hn(x) by Hn(x).From (1.5), we can derive the following identities:
(1.6)Hn(x)=(∂∂t)ne2xt-t2|t=0=ex2(∂∂t)ne-(x-t)2|t=0=(-1)nex2(∂∂x)ne-(x-t)2|t=0=(-1)nex2(dndxne-x2).Let us consider two operators as follows:(1.7)f⟼O1f=-(ex2ddxe-x2)f=2xf-dfdx,f⟼O2f=(ex2/2(x-ddx)e-x2/2)f=2xf-dfdx.By (1.7), we get O1=O2. In particular, if we take f=1, then we have
(1.8)-ex2(ddxe-x2)=ex2/2(x-ddx)e-x2/2.We note that(1.9)(-1)nex2(dndxne-x2)=(-ex2ddxe-x2)n.From (1.8), we note that
(1.10)(-1)nex2(dne-x2dxn)=(-ex2de-x2dx)n=(ex2/2(x-ddx)e-x2/2)n=ex2/2(x-ddx)ne-x2/2.Thus, by (1.10), we get
(1.11)Hn(x)=ex2/2(x-ddx)ne-x2/2
(see [5, 19–23]). In the special case, x=0, Hn(0)=Hn are called the Hermite numbers.From (1.5), we can derive the following identities:
(1.12)Hn(x)=(H+2x)n=∑l=0n(nl)Hn-l2lxl
(cf. [5, 19]), with the usual convention about replacing Hn by Hn. It is easy to show that
(1.13)∑n=0∞Hntnn!=e-t2=∑l=0∞(-1)nn!t2n.By comparing coefficients on the both sides of (1.13), we get
(1.14)H2n=(-1)n2n(2n-1)⋯(n+1)=(-1)n(2n)!n!,H2n-1=0,
where n∈ℕ. From (1.12), we have
(1.15)dHn(x)dx=2nHn-1(x)(n∈N).LetPn={p∈ℚ[x]∣degp(x)≤n} be the (n+1)-dimensional vector space over ℚ. Probably, {1,x,x2,…,xn} is the most natural basis for this space. But {H0(x),H1(x),H2(x),…,Hn(x)} is also a good basis for the space Pn, for our purpose of arithmetical and combinatorial applications.Forp(x)∈Pn,
(1.16)p(x)=∑k=0nCkHk(x),
for some uniquely determined bl∈ℚ.The purpose of this paper is to develop methods for computingCk from the information of p(x). By using these methods, we define some interesting identities.


## 2. Properties of Hermite Polynomials

From (1.5) and (1.13), we note that
(2.1)1=(∑m=0∞Hmtmm!)(∑l=0∞t2ll!)=(∑m=0∞H2mt2m(2m)!)(∑l=0∞(2l)(2l-1)⋯(l+1)(2l)!t2l)=∑n=0∞(∑l=0n(2l)(2l-1)⋯(l+1)(2l)!(2n-2l)!H2n-2l(2n)!)t2n(2n)!=∑n=0∞(∑l=0nl!(2ll)(2n2l)H2n-2l)t2n(2n)!.Thus, by (2.1), we obtain the following recurrence formula.Proposition 2.1.
Forn∈ℤ+=ℕ∪{0}, one has
(2.2)∑l=0nl!(2ll)(2n2l)H2n-2l={1,ifn=00,ifn≠0.By, (1.5), we get
(2.3)∑n=0∞Hn(-x)tnn!=e2t(-x)-t2=e2x(-t)-(-t)2=∑n=0∞Hn(x)(-1)ntnn!.From (2.3), we can derive the following reflection symmetric identity of Hn(x):
(2.4)Hn(-x)=(-1)nHn(x).By (1.5), we easily see that
(2.5)∂∂t(e2xt-t2)=(2x-2t)e2xt-t2.Thus, by (1.5) and (2.5), we get
(2.6)∂∂t(∑n=0∞Hn(x)tnn!)=(2x-2t)(∑n=0∞Hn(x)tnn!).(2.7)LHS of(2.5)=∑n=1∞Hn(x)tn-1(n-1)!=∑n=0∞Hn+1(x)tnn!,(2.8)RHSof(2.5)=∑n=0∞(2xHn(x)tnn!)-∑n=0∞2Hn(x)tn+1n!=∑n=0∞(2xHn(x)tnn!)-∑n=1∞2Hn-1(x)tn(n-1)!=∑n=0∞(2xHn(x))tnn!-∑n=1∞2nHn-1(x)tnn!.Thus, by (2.6) and (2.7), we get
(2.9)Hn+1(x)=2xHn(x)-2nHn-1(x),(n∈N).From (1.15) and (2.9), we note that
(2.10)Hn+1(x)-2xHn(x)+Hn′(x)=0.Differentiating on both sides, we have(2.11)2(n+1)Hn(x)-2Hn(x)-2xHn′(x)+Hn′(x)=0.Thus, we have(2.12)Hn′′(x)-2xHn′(x)+2nHn(x)=0.From (2.12), we note that Hn(x) is a solution of the following second-order linear differential equation:
(2.13)u′′-2xu′+2nu=0.From (1.5), we note that
(2.14)∑m=0∞‍Hn(x)tnn!=e2tx-t2=(∑l=0∞‍(2x)ll!tl)(∑k=0∞‍(-1)kk!t2k)=∑n=0∞‍(∑k=0[n/2]‍(-1)kn!(2x)n-2kk!(n-2k)!)tnn!.Thus, by (2.14), we get
(2.15)Hn(x)=∑k=0[n/2](-1)kn!k!(n-2k)!(2x)n-2k={∑l=0n/2(-1)n/2-ln!22l(n/2-l)!(2l)!x2l,ifn≡0(mod 2),∑l=0(n-1)/2(-1)(n-1)/2-ln!22l+1((n-1)/2-l)!(2l+1)!x2l+1,ifn≡1(mod 2).


## 3. Main Results

By (1.6), we easily get
(3.1)∫-∞∞e-x2Hn(x)Hm(x)dx=(-1)n∫-∞∞(dndxne-x2)Hm(x)dx.From (3.1), we note that
(3.2)∫-∞∞e-x2Hn(x)Hm(x)dx=2nn!πδm,n.It is easy to show that(3.3)∫-∞∞e-x2xldx={0ifl≡1(mod 2),l!π2l(l/2)!ifl≡0(mod 2),
where l∈ℤ+=ℕ∪{0}. By (3.3), we get
(3.4)∫-∞∞(dne-x2dxn)xmdx={0ifn>morn≤mwithn-m≡1(mod 2),m!(-1)nπ2m-n((m-n)/2)!ifn≤mwithn-m≡0(mod 2).From (3.2), we note that H0(x),H1(x),…,Hn(x) are orthogonal basis for the space ℙn={p(x)∈ℚ[x]∣degp(x)≤n} with respect to the inner product
(3.5)〈p(x),q(x)〉=∫-∞∞e-x2p(x)q(x)dx.Forp(x)∈ℙn, the polynomial p(x) is given by
(3.6)p(x)=∑k=0∞CkHk(x),
where
(3.7)Ck=12kk!π〈p(x),Hk(x)〉=(-1)k2kk!π∫-∞∞(dke-x2dxk)p(x)dx.Let us takep(x)=xn∈ℙn. For n≡0(mod 2), we compute Ck in (3.6) as follows
(3.8)Ck=(-1)k2kk!π∫-∞∞(dke-x2dxk)xndx={(-1)k2kk!π×(-1)kn!π2n-k((n-k)/2)!ifk≡0(mod 2),0ifk≡1(mod 2).Letn≡1(mod 2). Then we have
(3.9)Ck=(-1)k2kk!π∫-∞∞(dke-x2dxk)xndx={n!2nk!((n-k)/2)!ifk≡1(mod 2),0ifk≡0(mod 2).Therefore, by (3.6), (3.8), and (3.9), we obtain the following proposition.Proposition 3.1.
One has(3.10)x2n=(2n)!22n∑k=0n1(2k)!(n-k)!H2k(x),x2n+1=(2n+1)!22n+1∑k=0n1(2k+1)!(n-k)!H2k+1(x).Let us takep(x)=Bn(x). From (3.4), P(x) can be rewritten by
(3.11)Bn(x)=∑k=0nCkHk(x),
where
(3.12)Ck=(-1)k2kk!π∫-∞∞(dke-x2dxk)Bn(x)dx.By integrating by parts, we get(3.13)∫-∞∞(dke-x2dxk)Bn(x)=(-n)(-(n-1))⋯(-(n-k+1))∫-∞∞e-x2Bn-k(x)dx=(-1)kn!(n-k)!∑l=0n-k(n-kl)Bn-k-l∫-∞∞e-x2xldx=(-1)kn!(n-k)!∑0≤l≤n-kl≡0(mod 2)(n-k)!Bn-k-ll!(n-k-l)!×l!π2l(l/2)!=(-1)kn!π∑0≤l≤n-kl≡0(mod 2)Bn-k-l(n-k-l)!2l(l/2)!.Thus, from (3.11) and (3.13), we have
(3.14)Ck=n!2kk!∑0≤l≤n-kl≡0(mod 2)Bn-k-l(n-k-l)!2l(l/2)!.Therefore, by (3.11) and (3.14), we obtain the following theorem.Theorem 3.2.
Forn∈ℤ+, one has
(3.15)Bn(x)=n!∑k=0n∑0≤l≤n-kl≡0(mod2)Bn-k-l2k+lk!(n-k-l)!(l/2)!Hk(x).Remark 3.3.
Let us takep(x)=En(x). Then, by the same method, we obtain the following identity:
(3.16)En(x)=n!∑k=0n∑0≤l≤n-kl≡0(mod 2)En-k-l2k+lk!(n-k-l)!(l/2)!Hk(x).Now, we considerp(x)=Hn(x). From (3.6), we note that p(x) can be rewritten as
(3.17)Hn(x)=∑k=0nCkHk(x),
where
(3.18)Ck=(-1)k2kk!π∫-∞∞(dke-x2dxk)Hn(x)dx.By integrating by parts, we get(3.19)∫-∞∞(dke-x2dxk)Hn(x)dx=(-2n)⋯(-2(n-k+1))∫-∞∞e-x2Hn-k(x)dx=(-1)k2kn!(n-k)!∑l=0n-k(n-kl)2lHn-k-l∫-∞∞e-x2xldx=(-1)k2kn!(n-k)!∑l=0l≡0(mod 2)n-k2l(n-k)!l!(n-k-l)!Hn-k-ll!π2l(l/2)!=(-1)k2kn!π∑l=0l≡0(mod 2)n-kHn-k-l(n-k-l)!(l/2)!.From (3.17) and (3.19), we note that
(3.20)Ck=((-1)k2kk!π)×((-1)k2kn!π∑0≤l≤n-kl≡0(mod 2)Hn-k-l(n-k-l)!(l/2)!)=n!k!∑0≤l≤n-kl≡0(mod 2)Hn-k-l(n-k-l)!(l/2)!.Therefore, by (3.17) and (3.20), we obtain the following theorem.Theorem 3.4.
Forn∈ℤ+, one has
(3.21)Hn(x)=n!∑k=0n∑0≤l≤n-kl≡0(mod 2)Hn-k-lk!(n-k-l)!(l/2)!Hk(x).From Theorem3.4, we note that
(3.22)Hn(x)=n!∑k=0n-1∑0≤l≤n-kl≡0(mod 2)Hn-k-lk!(n-k-l)!(l/2)!Hk(x)+n!Hn(x)n!.Thus, we have, for0≤k≤n-k,
(3.23)∑0≤l≤n-kl≡0(mod 2)Hn-k-l(n-k-l)!(l/2)!=0.Letl,k∈ℤ+ with k≤l. Then we easily see that
(3.24)∫-∞∞(dke-x2dxk)Bl(x)dx=(-1)kl!π∑0≤j≤l-kj≡0(mod 2)Bl-k-j(l-k-j)!2j(j/2)!,(3.25)∫-∞∞(dke-x2dxk)El(x)dx=(-1)kl!π∑0≤j≤l-kj≡0(mod 2)El-k-j(l-k-j)!2j(j/2)!.Let us consider the following polynomial of degreen in ℙn:
(3.26)p(x)=∑k=0nBk(x)Bn-k(x).From (3.6), we note that p(x) can be rewritten as
(3.27)p(x)=∑k=0nCkHk(x),
where
(3.28)Ck=(-1)k2kk!π∫-∞∞(dke-x2dxk)p(x)dx.In [15], it is known that
(3.29)p(x)=∑k=0nBk(x)Bn-k(x)=2n+2∑l=0n-2(n+2l)Bn-lBl(x)+(n+1)Bn(x).From (3.23) and (3.29), we have the following:
(3.30)Ck=(-1)k2kk!π{2n+2∑l=0n-2(n+2l)∫-∞∞(dke-x2dxk)Bl(x)dx+(n+1)∫-∞∞(dke-x2dxk)Bn(x)dx},By (3.24) and (3.30), we get
(3.31)Cn=((-1)n2nn!π)×(n+1)∫-∞∞(dne-x2dxn)Bn(x)dx=((-1)n2nn!π)×((n+1)(-1)nn!πB00!200!)=n+12n,Cn-1=((-1)n-12n-1(n-1)!π)×((n+1)∫-∞∞(dn-1e-x2dxn-1)Bn(x)dx)=((-1)n-12n-1(n-1)!π)×((n+1)(-1)n-1n!π∑j=0j≡0(mod 2)1B1-j(1-j)!2j(j/2)!)=((-1)n-12n-1(n-1)!π)×((n+1)(-1)n-1n!πB1)=-n(n+1)2n.For0≤k≤n-2, we have
(3.32)Ck=(-1)k2kk!π{2n+2∑l=kn-2(n+2l)Bn-l∫-∞∞(dke-x2dxk)Bl(x)dx+(n+1)∫-∞∞(dke-x2dxk)Bn(x)dx}=(-1)k2kk!π{2n+2∑l=kn-2(n+2l)Bn-l(-1)kl!π×∑0≤j≤l-kj≡0(mod 2)Bl-k-j(l-k-j)!2j(j/2)!+(n+1)(-1)kn!π∑0≤j≤n-kj≡0(mod 2)Bn-k-j(n-k-j)!2j(j/2)!}=2n+2∑l=kn-2∑0≤j≤l-kj≡0(mod 2)(n+2l)Bn-lBl-k-jl!2k+jk!(l-k-j)!(j/2)!+(n+1)!∑0≤j≤n-kj≡0(mod 2)Bn-k-jk!(n-k-j)!(j/2)!2k+j.Therefore, by (3.27) and (3.32), we obtain the following theorem.Theorem 3.5.
Forn∈ℤ+, one has
(3.33)∑k=0nBk(x)Bn-k(x)=∑k=0n-2{2n+2∑l=kn-2‍∑0≤j≤n-kj≡0(mod 2)‍(n+2l)l!Bn-lBl-k-j2k+jk!(l-k-j)!(j/2)!+(n+1)!∑0≤j≤n-kj≡0(mod 2)‍Bn-k-j2k+jk!(n-k-j)!(j/2)!}Hk(x)-n(n+1)2nHn-1(x)+n+12nHn(x).



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