1. Introduction
For α,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 for Pn(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.
Let Pn={p(x)∈ℝ[x]∣deg p(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 polynomials En(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 space Pn.
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).
For q(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.
For q(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 take q(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.
For n∈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 take q(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.
For n∈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).
Let q(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).
Let Hn(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.
For n∈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 Theorem 2.3, 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).