# Some Identities on Laguerre Polynomials in Connection with Bernoulli and Euler Numbers

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

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## Abstract

We study some interesting identities and properties of Laguerre polynomials in connection with Bernoulli and Euler numbers. These identities are derived from the orthogonality of Laguerre polynomials with respect to inner product〈f,g〉=∫0∞e−x2f(x)g(x)dx.


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## Body

## 1. Introduction/Preliminaries

As is well known, Laguerre polynomials are defined by the generating function as(1.1)exp⁡(-xt/(1-t))1-t=∑n=0∞‍Ln(x)tn
(see [1, 2]). By (1.1), we get
(1.2)∑n=0∞‍Ln(x)tn=exp⁡(-xt/(1-t))1-t=∑r=0∞‍(-1)rxrtrr!(1-t)-(r+1)=∑n=0∞‍∑r=0∞‍(-1)rxr(r+s)!r!r!s!tr+s=∑n=0∞‍(∑r=0n‍(-1)r(nr)r!xr)tn.
Thus, from (1.2), we have
(1.3)Ln(x)=∑r=0n‍(-1)r(nr)r!xr.
By (1.3), we see that Ln(x) is a polynomial of degree n with rational coefficients and the leading coefficient (-1)n/n!. It is well known that Rodrigues' formula is given by
(1.4)Ln(x)=1n!ex(dndxne-xxn)
(see [1–27]). From (1.1), we can derive the following of Laguerre polynomials:
(1.5)L0(x)=1,L1(x)=-x+1,(n+1)Ln+1(x)=(2n+1-x)Ln(x)-nLn-1(x),(n≥1),(1.6)Ln′(x)=Ln-1′(x)-Ln-1(x)=0,(n≥1),(1.7)xLn′(x)=nLn(x)-nLn-1(x)=0,(n≥1).
By (1.7), we easily see that u=Ln(x) is a solution of the following differential equation of order 2:
(1.8)xu′′(x)+(1-x)u′(x)+nu(x)=0.
The Bernoulli numbers, Bn, are defined by the generating function as
(1.9)tet-1=eBt=∑n=0∞‍Bntnn!
(see [1–28, 28]), with the usual convention about replacing Bn by Bn.It is well known that Bernoulli polynomials of degreen are given by
(1.10)Bn(x)=(B+x)n=∑l=0n‍(nl)Bn-lxl
(see [2, 26]). Thus, from (1.10), we have
(1.11)Bn′(x)=dBn(x)dx=nBn-1(x)
(see [3–12]). From (1.9) and (1.10), we can derive the following recurrence relation:
(1.12)B0=1,(B+1)n-Bn=δ1,n
where δn,k is Kronecker's symbol.The Euler polynomialsEn(x) are also defined by the generating function as
(1.13)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!
(see [27, 28]), with the usual convention about replacing En(x) by En(x).In this special case,x=0, En(0)=En are called the nth Euler numbers. From (1.13), we note that the recurrence formula of En is given by
(1.14)E0=1,(E+1)n+En=2δ0,n
(see [24]). Finally, we introduce Hermite polynomials, which are defined by
(1.15)e2xt-t2=eH(x)t=∑n=0∞‍Hn(x)tnn!
(see [29]). In the special case, x=0, Hn(0)=Hn is called the n-th Hermite number. By (1.15), we get
(1.16)Hn(x)=(H+2x)n=∑l=0n‍(nl)Hn-l2lxl
(see [29]). It is not difficult to show that
(1.17)∫0∞‍e-xLm(x)Ln(x)dx=δm,n,(m,n∈Z+=N∪{0}).
In the present paper, we investigate some interesting identities and properties of Laguerre polynomials in connection with Bernoulli, Euler, and Hermite polynomials. These identities and properties are derived from (1.17).


## 2. Some Formulae on Laguerre Polynomials in Connection with Bernoulli, Euler, and Hermite Polynomials

Let(2.1)Pn={p(x)∈Q[x]∣deg⁡p(x)≤n}.
Then Pn is an inner product space with the inner product
(2.2)〈p1(x),p2(x)〉=∫0∞‍e-xp1(x)p2(x)dx,(p1(x),p2(x)∈Pn).
By (1.17), (2.1), and (2.2), we see that L0(x),L1(x),…,Ln(x) are orthogonal basis for Pn.Forp(x)∈Pn, it is given by
(2.3)p(x)=∑k=0n‍CkLk(x),
where
(2.4)Ck=〈p(x),Lk(x)〉=∫0∞‍e-xLk(x)p(x)dx=1k!∫0∞‍(dkdxke-xxk)p(x)dx.
Let us take p(x)=xn∈Pn. From (2.3) and (2.4), we note that
(2.5)Ck=1k!∫0∞‍(dkdxke-xxk)xndx=-nk!∫0∞‍(dk-1dxk-1e-xxk)xn-1dx=(-n)(-(n-1))k!∫0∞‍(dk-2dxk-2e-xxk)xn-2dx=⋯=(-1)kn(n-1)⋯(n-k+1)k!∫0∞‍e-xxndx=(-1)k(nk)n!.
Therefore, by (2.3), (2.4), and (2.5), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.6)xn=n!∑k=0n‍(-1)k(nk)Lk(x).Let us considerp(x)=Bn(x)∈Pn. Then, by (2.3) and (2.4), we get
(2.7)Ck=1k!∫0∞‍(dkdxke-xxk)Bn(x)dx=-nk!∫0∞‍(dk-1dxk-1e-xxk)Bn-1(x)dx=(-n)(-(n-1))k!∫0∞‍(dk-2dxk-2e-xxk)Bn-2(x)dx=⋯=(-1)kn(n-1)⋯(n-k+1)k!∑l=0n-k‍(n-kl)Bn-k-l∫0∞‍e-xxk+ldx=(-1)k(nk)∑l=0n-k‍(n-kl)Bn-k-l(k+l)!=n!(-1)k∑l=0n-k‍(k+lk)Bn-k-l(n-k-l)!.
Therefore, by (2.3), (2.4), and (2.7), we obtain the following theorem.Theorem 2.2.
Forn∈ℤ+, one has
(2.8)Bn(x)=n!∑k=0n‍∑l=0n-k‍(-1)k(k+lk)Bn-k-l(n-k-l)!Lk(x).Let us takep(x)=En(x)∈Pn. By the same method, we easily see that
(2.9)En(x)=n!∑k=0n‍∑l=0n-k‍(-1)k(k+lk)En-k-l(n-k-l)!Lk(x).Forp(x)=Hn(x)∈Pn, we have
(2.10)Hn(x)=∑k=0n‍CkLk(x),
where
(2.11)Ck=1k!∫0∞‍(dkdxke-xxk)Hn(x)dx=-2nk!∫0∞‍(dk-1dxk-1e-xxk)Hn-1(x)dx=(-2n)(-2(n-1))k!∫0∞‍(dk-2dxk-2e-xxk)Hn-2(x)dx=⋯=(-2n)(-2(n-1))⋯(-2(n-k+1))k!∫0∞‍e-xxkHn-k(x)dx=(-1)k2kn(n-1)⋯(n-k+1)k!∑l=0n-k‍(n-kl)Hn-k-l2l∫0∞‍e-xxk+ldx=(-1)k(nk)∑l=0n-k‍(n-kl)2k+lHn-k-l(k+l)!=n!(-1)k∑l=0n-k‍2k+l(k+lk)Hn-k-l(n-k-l)!.
Therefore, by (2.10) and (2.11), we obtain the following theorem.Theorem 2.3.
Forn∈ℤ+, one has
(2.12)Hn(x)=n!∑k=0n‍∑l=0n-k‍(-1)k2k+l(k+lk)Hn-k-l(n-k-l)!Lk(x).Letp(x)=∑k=0n‍Bk(x)Bn-k(x)∈Pn. Then we have
(2.13)p(x)=∑k=0n‍Bk(x)Bn-k(x)=∑k=0n‍CkLk(x),
where
(2.14)Ck=1k!∫0∞‍(dkdxke-xxk)p(x)dx.
In [15], it is known that
(2.15)∑k=0n‍Bk(x)Bn-k(x)=2n+2∑l=0n-2‍(n+2l)Bn-lBl(x)+(n+1)Bn(x).
By (2.14) and (2.15), we get
(2.16)Ck=1k!{2n+2∑l=0n-2‍(n+2l)Bn-l∫0∞‍(dkdxke-xxk)Bl(x)dx1k!2222+(n+1)∫0∞‍(dkdxke-xxk)Bn(x)dx}.
From (2.16), we can derive the following equations ((2.17)-(2.18)):
(2.17)Cn=(-1)n(n+1)!,Cn-1=n(n+1)!(-1)n-1-12(-1)n-1(n+1)!.
For 0≤k≤n-2, we have
(2.18)Ck=2n+2∑l=kn-2‍∑m=0l-k‍(n+2l)(k+mk)l!(-1)kBn-lBl-k-m(l-k-m)!+(-1)k(n+1)!∑m=0n-k‍(k+mk)Bn-k-m(n-k-m)!.
Therefore, by (2.13), (2.17), and (2.18), we obtain the following theorem.Theorem 2.4.
Forn∈ℤ+, one has
(2.19)∑k=0n‍Bk(x)Bn-k(x)=∑k=0n-2‍{2n+2∑l=kn-2‍∑m=0l-k‍(-1)kl!(n+2l)(k+mk)Bn-lBl-k-m(l-k-m)!∑k=0n-2‍22+(-1)k(n+1)!∑m=0n-k‍(k+mk)Bn-k-m(n-k-m)!}Lk(x)+(n(n+1)!(-1)n-1-12(-1)n-1(n+1)!)Ln-1(x)+(-1)n(n+1)!Ln(x).Let us takep(x)=∑k=0n‍Ek(x)En-k(x)∈Pn. By (2.3) and (2.4), we get
(2.20)p(x)=∑k=0n‍Ek(x)En-k(x)=∑k=0n‍CkLk(x),
where
(2.21)Ck=1k!∫0∞‍(dkdxke-xxk)p(x)dx.
It is known (see [15]) that
(2.22)∑k=0n‍Ek(x)En-k(x)=∑k=0n-1‍(n+1)(nk)n-k+1(∑l=kn‍El-kEn-l-2En-k)Ek(x)+2(n+1)En(x).
From (2.20), (2.21), and (2.22), we can derive the following equations ((2.23)-(2.24)):
(2.23)Cn=(-1)nn!2(n+1)(n!)2=2(-1)n(n+1)!.
For 0≤k≤n-1, we have
(2.24)Ck=∑l=kn-1‍(n+1)(nl)(n-l+1)!(∑m=ln‍Em-lEn-m-2En-l)∑p=0l-k‍(-1)kl!(k+pk)El-k-p(l-k-p)!+2(n+1)!(-1)k∑p=0n-k‍(k+pk)En-k-p(n-k-p)!.
Therefore, by (2.20) and (2.24), we obtain the following theorem.Theorem 2.5.
Forn∈ℤ+, one has
(2.25)∑k=0n‍Ek(x)En-k(x)=∑k=0n-1‍{∑l=kn-1‍(n+1)(nl)(n-l+1)!(∑m=ln‍Em-lEn-m-2En-l)∑p=0l-k‍(-1)kl!(k+pk)∑k=0n-1‍22×El-k-p(l-k-p)!+2(n+1)!(-1)k∑p=0n-k‍(k+pk)En-k-p(n-k-p)!}Lk(x)+2(-1)n(n+1)!Ln(x).It is known that(2.26)∑k=0n‍Ek(x)En-k(x)=p(x)=-4n+2∑k=0n‍(n+2k)En-k+1Bk(x)
(see [15]). From (2.20), (2.21), and (2.23), we have
(2.27)Ck=1k!∫0∞‍(dkdxke-xxk)p(x)dx=-4n+2∑l=kn‍(n+2l)En-l+11k!∫0∞‍(dke-xxkdxk)Bl(x)dx=-4n+2∑l=kn‍∑m=0l-k‍(n+2l)(-1)kEn-l+1Bl-k-m(l-k-m)!l!(m+kk).
Therefore, by (2.20) and (2.27), we obtain the following theorem.Theorem 2.6.
Forn∈ℤ+, one has
(2.28)∑k=0n‍Ek(x)En-k(x)=-4n+2∑k=0n‍∑l=kn‍∑m=0l-k‍(n+2l)(-1)kEn-l+1Bl-k-m(l-k-m)!l!(m+kk)Lk(x).Remark 2.7.
Laguerre's differential equation(2.29)ty′′+(1-t)y′+ny=0
is known to possess polynomial solutions when n is a nonnegative integer. These solutions are naturally called Laguerre polynomials and are denoted by Ln(t). That is, y=Ln(t) are solutions of (2.29) which are given by
(2.30)y=Ln(t)=∑r=0n‍(nr)(-1)rr!tr,L0(1)=1.
From (2.30), we note that Laplace transform of y=Ln(t) is given by
(2.31)L(y)=L(Ln(t))=1s∑r=0n‍(nr)(-1)r(1s)r=(s-1)nsn+1.
It is not difficult to show that
(2.32)L(etn!(dndtne-ttn))=L(y)=(s-1)nsn+1.
Thus, we conclude that
(2.33)Ln(t)=etn!(dndtne-ttn),forn∈Z+.



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