# Some Identities on Bernoulli and Hermite Polynomials Associated with Jacobi Polynomials

Authors: Taekyun Kim; Dae San Kim; Dmitry V. Dolgy
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate some identities on the Bernoulli and the Hermite polynomials arising from the orthogonality of Jacobi polynomials in the inner product spacePn.


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## Body

## 1. Introduction

Forα,β∈ℝ with α>-1 and β>-1, the Jacobi polynomials Pn(α,β)(x) are defined as (1.1)Pn(α,β)(x)=(α+1)nn!2F1(-n,1+α+β+n;α+1;1-x2)=(α+1)nn!∑k=0n‍(nk)(1+α+β+n)k(α+1)k(x-12)k,(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)Pn(α,β)(x)=Γ(α+1+n)n!Γ(α+β+n+1)∑k=0n‍(nk)Γ(α+β+n+k+1)Γ(α+k+1)(x-12)k.
By (1.2), we see that Pn(α,β)(x) is polynomial of degree n with real coefficients. It is not difficult to show that the leading coefficient of Pn(α,β)(x) is 2-n(α+β+2nn). From (1.2), we have Pn(α,β)(1)=(α+nn).By (1.1), we get
(1.3)(ddx)kPn(α,β)(x)=2-kΓ(n+α+β+k+1)Γ(n+α+β+1)Pn-k(α+k,β+k)(x)=12k(n+α+β+k)(n+α+β+k-1)⋯(n+α+β+1)Pn-k(α+k,β+k)(x),
where k is a positive integer (see [1–4]).The Rodrigues' formula forPn(α,β)(x) is given by
(1.4)(1-x)α(1+x)βPn(α,β)(x)=(-1)n2nn!(ddx)k{(1-x)n+α(1+x)n+β}.
It is easy to show that u=Pn(α,β)(x) is a solution of the following differential equation:
(1.5)(1-x2)u′′+{β-α-(α+β+2)x}u′+n(n+α+β+1)u=0.
As is well known, the generating function of Pn(α,β)(x) is given by
(1.6)F(x,t)=∑n=0∞‍Pn(α,β)(x)tn=2α+βR(1-t+R)α(1+t+R)β,
where R=1-2xt+t2, (see [1–4]).From (1.3), (1.4), and (1.6), we can derive the following identity:
(1.7)∫-11‍Pm(α,β)(x)Pn(α,β)(x)(1-x)α(1+x)βdx=2α+β+1Γ(n+α+1)Γ(n+β+1)(2n+α+β+1)Γ(n+α+β+1)Γ(n+1)δn,m,
where δn,m is the Kronecker symbol.LetPn={p(x)∈ℝ[x]∣degp(x)≤n}. Then Pn is an inner product space with respect to the inner product 〈q1(x),q2(x)〉=∫-11‍(1-x)α(1+x)βq1(x)q2(x)dx, where q1(x),q2(x)∈Pn. From (1.7), we note that {P0(α,β)(x),P1(α,β)(x),…,Pn(α,β)(x)} is an orthogonal basis for Pn.The so-called Euler polynomialsEn(x) may be defined by means of
(1.8)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!,
(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)tet-1ext=eB(x)t=∑n=0∞‍Bn(x)tnn!,
(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)Bn(x)=∑k=0n‍(nk)Bn-kxk,En(x)=∑k=0n‍(nk)En-kxk.
For n∈ℤ+, we have
(1.11)dBn(x)dx=nBn-1(x),dEn(x)dx=nEn-1(x)
(see [23–29]) By the definition of Bernoulli and Euler polynomials, we get
(1.12)B0=1,Bn(1)-Bn=δ1,n,E0=1,En(1)+En=2δ0,n.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 spacePn.


## 2. Bernoulli, Euler and Jacobi Polynomials

From (1.4), we have
(2.1)Pn(α,β)(x)=∑k=0n‍(n+αn-k)(n+βk)(x-12)k(x+12)n-k.
By (2.1), we have
(2.2)∑n=0∞‍Pn(α,β)(x)tn=12πi∮‍(1+((x+1)/2)z)n+α(1+((x-1)/2)z)n+βzn+1dz,
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 Pn(α,β)(-x)=(-1)nPn(β,α)(x).Forq(x)∈Pn, let
(2.3)q(x)=∑k=0n‍CkPk(α,β)(x),(Ck∈ℝ).
From (1.7), we note that
(2.4)〈q(x),Pk(α,β)(x)〉=Ck〈Pk(α,β)(x),Pk(α,β)(x)〉=Ck∫-11‍(1-x)α(1+x)β(Pk(α,β)(x))2dx=Ck2α+β+1Γ(k+α+1)Γ(k+β+1)(2k+α+β+1)Γ(α+β+k+1)k!.
Thus, by (2.4), we get
(2.5)Ck=(2k+α+β+1)Γ(α+β+k+1)k!2α+β+1Γ(k+α+1)Γ(k+β+1)∫-11‍(1-x)α(1+x)βPk(α,β)(x)q(x)dx.
Therefore, by (1.7), (2.3), and (2.5), we obtain the following proposition.Proposition 2.1.
Forq(x)∈Pn(n∈ℕ), one has
(2.6)q(x)=∑k=0n‍CkPk(α,β)(x),
where
(2.7)Ck=(-1)k(2k+α+β+1)Γ(k+α+β+1)2α+β+1+kΓ(α+k+1)Γ(β+k+1)∫-11‍(dkdxk(1-x)k+α(1+x)k+β)q(x)dx.Let us takeq(x)=xn∈Pn. First, we consider the following integral:
(2.8)∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}q(x)dx=∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}xndx=(-n)∫-11‍(ddx)k-1{(1-x)k+α(1+x)k+β}xn-1dx=⋯=(-1)kn!(n-k)!∫-11‍(1-x)k+α(1+x)k+βxn-kdx=(-1)kn!22k+α+β+1(n-k)!∫01‍yk+β(1-y)k+α(2y-1)n-kdy=(-1)kn!(n-k)!22k+α+β+1∑l=0n-k‍(n-kl)2l(-1)n-k-lB(k+l+β+1,k+α+1)=(-1)kn!(n-k)!22k+α+β+1∑l=0n-k‍(n-kl)2l(-1)n-k-lΓ(k+l+β+1)Γ(k+α+1)Γ(2k+α+β+l+2).
From (2.5) and (16), we have
(2.9)Ck=(-1)k(2k+α+β+1)Γ(k+α+β+1)2α+β+1+kΓ(α+k+1)Γ(β+k+1)×∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}xndx=(-1)k(2k+α+β+1)Γ(k+α+β+1)2α+β+1+kΓ(α+k+1)Γ(β+k+1)·(-1)kn!22k+α+β+1(n-k)!×∑l=0n-k‍(n-kl)2l(-1)n-k-lΓ(k+l+β+1)Γ(k+α+1)Γ(2k+α+β+l+2)=(2k+α+β+1)Γ(k+α+β+1)n!2kΓ(β+k+1)(n-k)!∑l=0n-k‍(-1)n-k-l(n-kl)2lΓ(k+l+β+1)Γ(2k+α+β+l+2).
By Proposition 2.1, we get
(2.10)xn=n!∑k=0n‍∑l=0n-k‍((2k+α+β+1)Γ(k+α+β+1)Γ(k+β+1)(n-k)!2k)×((-1)n-k-l(n-kl)2lΓ(k+l+β+1)Γ(2k+α+β+l+2))Pk(α,β)(x).
From (1.9), we have
(2.11)ext=1ttet-1ext(et-1)=∑n=0∞‍(Bn+1(x+1)-Bn+1(x)n+1)tnn!.
By (2.11), we get
(2.12)xn=Bn+1(x+1)-Bn+1(x)n+1,(n∈ℤ+).
Therefore, by (2.10) and (2.12), we obtain the following theorem.Theorem 2.2.
Forn∈ℤ+, one has
(2.13)1(n+1)!{Bn+1(x+1)-Bn+1(x)}=∑k=0n‍(∑l=0n-k‍(-1)n-k-l2k+l(2k+α+β+1)(n-kl)Γ(k+β+1)Γ(2k+α+β+l+2)(n-k)!×Γ(k+α+β+1)Γ(k+l+β+1)∑l=kn‍)Pk(α,β)(x).Let us takeq(x)=Bn(x)∈Pn. Then we evaluate the following integral:
(2.14)∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Bn(x)dx=∑l=kn‍(nl)Bn-l(-1)kl!(l-k)!22k+α+β+1∫01‍yk+β(1-y)k+α(2y-1)l-kdy=∑l=kn‍(nl)Bn-l(-1)kl!(l-k)!22k+α+β+1∑m=0l-k‍(l-km)2m(-1)l-k-m×Γ(k+m+β+1)Γ(k+α+1)Γ(2k+α+β+m+2)=∑l=kn‍∑m=0l-k‍(nl)Bn-l(-1)l-ml!22k+α+β+1(l-km)2mΓ(k+m+β+1)Γ(k+α+1)(l-k)!Γ(2k+α+β+m+2).
Finding (2.5) and (21), we have
(2.15)Ck=(-1)k(2k+α+β+1)Γ(α+β+k+1)2α+β+k+1Γ(α+k+1)Γ(β+k+1)×∫-11‍(dkdxk(1-x)k+α(1+x)k+β)Bn(x)dx=∑l=kn‍∑m=0l-k‍2k+m(nl)Bn-l(-1)l-m-kl!(2k+α+β+1)(l-km)Γ(β+k+1)(l-k)!Γ(2k+α+β+m+2)×Γ(k+m+β+1)Γ(k+α+β+1).Theorem 2.3.
Forn∈ℤ+, one has
(2.16)Bn(x)=∑k=0n‍(∑l=kn‍∑m=0l-k‍2k+m(nl)Bn-l(-1)l-m-kl!(2k+α+β+1)(l-km)Γ(β+k+1)(l-k)!Γ(2k+α+β+m+2)×Γ(k+m+β+1)Γ(k+α+β+1)∑l=kn‍)Pk(α,β)(x).Letq(x)=Pn(α,β)(x)∈Pn. From Proposition 2.1, we firstly evaluate the following integral:
(2.17)∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Pn(α,β)(x)dx=(-1)k12kΓ(n+α+β+k+1)Γ(n+α+β+1)∫-11‍(1-x)k+α(1+x)k+βPn-k(α+k,β+k)(x)dx.By (2.1) and (2.17), we get
(2.18)∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Pn(α,β)(x)dx=(-1)k2kΓ(n+α+β+k+1)Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)×∫-11‍(1-x)k+α(1+x)k+β(x-12)l(x+12)n-k-ldx=(-1)k2kΓ(n+α+β+k+1)Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(-1)l22k+α+β+1×∫01‍(1-y)k+α+lyn+β-ldy=(-1)k2α+β+k+1Γ(n+α+β+k+1)Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(-1)l×B(k+α+l+1,n+β-l+1)=(-1)k2α+β+k+1Γ(n+α+β+k+1)Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(-1)l×Γ(α+k+l+1)Γ(n+β-l+1)Γ(α+β+k+n+2)=(-1)k2α+β+k+11Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(-1)l×Γ(α+k+l+1)Γ(n+β-l+1)(α+β+k+n+1).
It is easy to show that
(2.19)Γ(n+β-l+1)Γ(β+k+1)=(n+β-l)⋯βΓ(β)(β+k)⋯βΓ(β)=(n+β-l)⋯(β+k+1)=(n+β-ln-k-l)(n-k-l)!.
From (2.5), (2.18), and (2.19), we can derive the following equation:
(2.20)Ck=(-1)k(2k+α+β+1)Γ(k+α+β+1)2α+β+k+1Γ(α+k+1)Γ(β+k+1)×∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Pn(α,β)(x)dx=(2k+α+β+1)Γ(α+β+k+1)Γ(β+k+1)Γ(n+α+β+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(α+k+ll)×l!(-1)lΓ(n+β-l+1)(α+β+k+n+1)=(2k+α+β+1)Γ(α+β+k+1)∑l=0n-k‍(n+αn-k-l)(n+βl)(n+β-ln-k-l)×(n-k-l)!l!α+β+k+n+1(-1)l.
Therefore, by Proposition 2.1, we obtain the following theorem.Theorem 2.4.
For(n∈ℤ+), one has
(2.21)Γ(n+α+β+1)Pn(α,β)(x)Γ(α+β+1)=∑k=0n‍{∑l=0n-k‍(2k+α+β+1)(α+β+kk)(n+αn-k-l)×(n+βl)(α+k+ll)(n+β-ln-k-l)(-1)l(n-k-l)!k!l!α+β+n+k+1}Pk(α,β)(x).LetHn(x) be the Hermite polynomial with
(2.22)Hn(x)=q(x)=∑k=0n‍CkPk(α,β)(x),
where
(2.23)Ck=(-1)k(2k+α+β+1)Γ(k+α+β+1)2α+β+k+1Γ(α+k+1)Γ(β+k+1)×∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Hn(x)dx.
Integrating by parts, one has
(2.24)∫-11‍(ddx)k{(1-x)k+α(1+x)k+β}Hn(x)dx=2k(-1)kn!(n-k)!∫-11‍(1-x)k+α(1+x)k+βHn-k(x)dx=2k(-1)kn!(n-k)!∑l=0n-k‍(n-kl)Hn-k-l2l∫-11‍(1-x)k+α(1+x)k+βxldx=2k(-1)kn!(n-k)!∑l=0n-k‍(n-kl)Hn-k-l22k+α+β+l+1∑m=0l‍(lm)(-1)l-m2m×∫01‍(1-y)k+αyk+β+mdy=2k(-1)kn!(n-k)!∑l=0n-k‍∑m=0l‍(n-kl)(lm)Hn-k-l(-1)l-m22k+α+β+m+l+1×Γ(k+α+1)Γ(β+k+m+1)Γ(2k+α+β+m+2).
By (2.23) and (29), we get
(2.25)Ck=∑l=0n-k‍∑m=0l‍(n-kl)(lm)Hn-k-l(-1)l-m(2k+α+β+1)(α+β+kk)k!(α+β+1)(2k+α+β+m+1m+2k)(m+2k)!(n-k)!×22k+m+ln!(β+k+mm)m!.
Therefore, by (2.22) and (2.25), we obtain the following theorem.Theorem 2.5.
Forn∈ℤ+, one has
(2.26)(α+β+1)Hn(x)n!=∑k=0n‍{∑l=0n-k‍∑m=0l‍(n-kl)(lm)Hn-k-l(-1)l-m(2k+α+β+1)(2k+α+β+m+1m+2k)(m+2k)!(n-k)!×(α+β+kk)k!22k+m+l(β+k+mm)m!∑l=0n-k‍}Pk(α,β)(x),
where Hn is the nth Hermite number.Remark 2.6.
By the same method as Theorem2.3, we get
(2.27)12n!{En(x+1)+En(x)}=∑k=0n‍(∑l=0n-k‍2k+l(2k+α+β+1)(n-kl)(-1)n-k-lΓ(k+β+1)Γ(2k+α+β+l+2)(n-k)!×Γ(k+α+β+1)Γ(k+l+β+1)∑l=0n-k‍)Pk(α,β)(x).



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