# Extended Laguerre Polynomials Associated with Hermite, Bernoulli, and Euler Numbers and Polynomials

Authors: Taekyun Kim; Dae San Kim
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

LetPn={p(x)∈ℝ[x]∣degp(x)≤n} be an inner product space with the inner product 〈p(x),q(x)〉=∫0∞xαe-xp(x)q(x)dx, where p(x),q(x)∈Pn and α∈ℝ with α>-1. In this paper we study the properties of the extended Laguerre polynomials which are an orthogonal basis for Pn. From those properties, we derive some interesting relations and identities of the extended Laguerre polynomials associated with Hermite, Bernoulli, and Euler numbers and polynomials.


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## Body

## 1. Introduction/Preliminaries

Forα∈ℝ with α>-1, the extended Laguerre polynomials are defined by the generating function as follows:
(1.1)exp⁡(-xt/(1-t))(1-t)α+1=∑n=0∞‍Lnα(x)tn,
see [1–6].From (1.1), we can derive the following:
(1.2)Lnα(x)=∑r=0n‍(-1)r(n+αn-r)r!xr,
see [1–9].As is well known, Rodrigues' formula forLnα(x) is given by
(1.3)Lnα(x)=1n!x-αex(dndxne-xxn+α),
see [1–6, 8, 9].From (1.3), we note that
(1.4)∫0∞‍xαe-xLmα(x)Lnα(x)dx=1n!Γ(α+n+1)δm,n,(α>-1),
where δm,n is the Kronecker symbol.From (1.1), (1.2), and (1.3), we can derive the following identities:
(1.5)(n+1)Ln+1α(x)+(x-α-2n-1)Lnα(x)+(n+α)Ln-1α(x)=0,(n∈N),(1.6)ddxLnα(x)-ddxLn-1α(x)+Ln-1α(x)=0,forn≥1,(1.7)xddxLnα(x)=nLnα(x)-(n+α)Ln-1α(x)=0,(n≥1),
and Lnα(x) is a solution of xy′′+(α+1-x)y′+xy=0.The derivatives of general Laguerre polynomials are given by(1.8)ddxLnα(x)=-Ln-1α+1(x),ddx(xαLnα(x))=(n+α)xα-1Lnα-1(x),ddx(e-xLnα(x))=-e-xLnα+1(x),ddx(xαe-xLnα(x))=(n+1)xα-1e-xLn+1α-1(x).
The nth Bernoulli polynomials, Bn(x), are defined by the generating function to be
(1.9)tet-1ext=eB(x)t=∑n=0∞‍Bn(x)tnn!,
see [10–17], with the usual convention about replacing Bn(x) by Bn(x). In the special case, x=0, Bn(0)=Bn are called the nth Bernoulli numbers.It is well known that thenth Euler polynomials are also defined by the generating function to be
(1.10)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!,
see [18–22], with the usual convention about replacing En(x) by En(x).The Hermite polynomials are given by(1.11)Hn(x)=(H+2x)n=∑l=0n‍(nl)2lxlHn-l,
see [23, 24], with the usual convention about replacing Hn by Hn. In the special case, x=0, Hn(0)=Hn are called the nth Hermite numbers.From (1.11), we note that
(1.12)ddxHn(x)=2n(H+2x)n-1=2nHn-1(x),
see [23, 24], and Hn(x) is a solution of Hermite differential equation which is given by
(1.13)y′′-2xy′+ny=0,
(see [1–6, 23–32]).Throughout this paper we assume thatα∈ℝ with α>-1. Let Pn={p(x)∈ℝ[x]|degp(x)≤n}. Then Pn is an inner product space with the inner product 〈p(x),q(x)〉=∫0∞‍xαe-xp(x)q(x)dx, where p(x),q(x)∈Pn. By (1.4) the set of the extended Laguerre polynomials {L0α(x),L1α(x),…,Lnα(x)} is an orthogonal basis for Pn. In this paper we study the properties of the extended Laguerre polynomials which are an orthogonal basis for Pn. From those properties, we derive some new and interesting relations and identities of the extended Laguerre polynomials associated with Hermite, Bernoulli and Euler numbers and polynomials.


## 2. On the Extended Laguerre Polynomials Associated with Hermite, Bernoulli, and Euler Polynomials

Forp(x)∈Pn, p(x) is given by
(2.1)p(x)=∑k=0n‍CkLkα(x),foruniquelydeterminedrealnumbersCk.
From (1.3), (1.4), and (2.1), we note that
(2.2)〈p(x),Lkα(x)〉=Ck〈Lkα(x),Lkα(x)〉=Ck∫0∞‍xαe-xLkα(x)Lkα(x)dx=CkΓ(α+k+1)k!.
Thus, by (2.2), we get
(2.3)Ck=k!Γ(α+k+1)〈p(x),Lkα(x)〉=k!Γ(α+k+1)1k!∫0∞‍(dkdxkxk+αe-x)p(x)dx=1Γ(α+k+1)∫0∞‍(dkdxkxk+αe-x)p(x)dx.
Therefore, by (2.1) and (2.3), we obtain the following proposition.Proposition 2.1.
Forp(x)∈Pn, let
(2.4)p(x)=∑k=0n‍CkLkα(x),(α>-1).
Then one has the following:
(2.5)Ck=1Γ(α+k+1)∫0∞‍(dkdxkxk+αe-x)p(x)dx.To derive inverse formula of (1.2), let take one p(x)=xn∈Pn. Then, by Proposition 2.1, one gets
(2.6)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)xndx=(-1)kn(n-1)⋯(n-k+1)Γ(α+k+1)∫0∞‍e-xxα+ndx=(-1)kn(n-1)⋯(n-k+1)Γ(α+k+1)Γ(α+n+1)=(-1)kn!(α+n)⋯αΓ(α)(α+k)⋯αΓ(α)(n-k)!=(-1)kn!(α+n)⋯(α+k+1)(n-k)!=(-1)kn!(α+nn-k).
Therefore, by (2.6), we obtain the following corollary.Corollary 2.2 (Inverse formula ofLnα(x)).
Forn∈Z+, one has
(2.7)xn=n!∑k=0n‍(α+nn-k)(-1)kLkα(x).Let one takes Bernoulli polynomials of degreen with p(x)=Bn(x)∈Pn. Then Bn(x) can be written as
(2.8)Bn(x)=∑k=0n‍CkLkα(x),(α∈Rwithα>-1).
From Proposition 2.1, one has
(2.9)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)Bn(x)dx=(-1)kn(n-1)⋯(n-k+1)Γ(α+k+1)∫0∞‍e-xxk+αBn-k(x)dx=(-1)kn(n-1)⋯(n-k+1)Γ(α+k+1)∑l=0n-k‍(n-kl)Bn-k-l∫0∞‍e-xxk+α+ldx=(-1)kn(n-1)⋯(n-k+1)Γ(α+k+1)∑l=0n-k‍(n-kl)Bn-k-lΓ(α+k+l+1).
By the fundamental property of gamma function, one gets
(2.10)Γ(α+k+l+1)Γ(α+k+1)(n-k)!(n-kl)=(α+l+k)⋯(α+k+1)Γ(α+k+1)(n-k)!Γ(α+k+1)(n-k)!(n-k-l)!l!=(α+k+ll)(n-k-l)!.
Therefore, by (2.8), (2.9), and (2.10), we obtain the following theorem.Theorem 2.3.
Forn∈ℤ+, α∈ℝ with α>-1, one has
(2.11)Bn(x)=n!∑k=0n‍∑l=0n-k‍(-1)k(α+k+ll)Bn-k-l(n-k-l)!Lkα(x).As is known, relationships between Hermite and Laguerre polynomials are given by(2.12)H2m(x)=(-1)m22mm!Lm-1/2(x2),(2.13)H2m+1(x)=(-1)m22m+1m!Lm-1/2(x2),
see [1–6]. In the special case α=-1/2, by (2.12) and (2.13), we obtain the following corollary.Corollary 2.4.
Forn∈ℤ+, one has
(2.14)Bn(x2)=n!∑k=0n‍∑l=0n-k‍H2k(x)22kk!(-12+k+ll)Bn-k-l(n-k-l)!.By the same method as Theorem2.3, one gets
(2.15)En(x)=n!∑k=0n‍∑l=0n-k‍(-1)k(α+k+ll)En-k-l(n-k-l)!Lkα(x),
where En(x) are the nth Euler polynomials. In the special case, x=0, En(0)=En are called the nth Euler numbers.Let one considers thenth Hermite polynomials with p(x)=Hn(x)∈Pn. Then Hn(x) can be written as
(2.16)Hn(x)=∑k=0n‍CkLkα(x),(α∈Rwithα>-1).
From Proposition 2.1, one notes that
(2.17)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)Hn(x)dx=(-2n)Γ(α+k+1)∫0∞‍(dk-1dxk-1e-xxk+α)Hn-1(x)dx=⋯=(-2n)(-2(n-1))⋯(-2(n-k+1))Γ(α+k+1)∫0∞‍e-xxk+αHn-k(x)dx=(-1)k2kn!Γ(α+k+1)(n-k)!∑l=0n-k‍(n-kl)Hn-k-l2l∫0∞‍e-xxk+α+ldx=(-1)k2kn!Γ(α+k+1)(n-k)!∑l=0n-k‍(n-kl)Hn-k-l2lΓ(α+k+l+1).
It is not difficult to show that
(2.18)(n-kl)Γ(α+k+l+1)Γ(α+k+1)(n-k)!=(n-k)!(α+k+l)⋯(α+k+1)Γ(α+k+1)(n-k-l)!l!Γ(α+k+1)(n-k)!=(α+k+ll)(n-k-l)!.
Therefore, by (2.16), (2.17), and (2.18), we obtain the following theorem.Theorem 2.5.
Forn∈ℤ+, α∈ℝ with α>-1, one has
(2.19)Hn(x)=n!∑k=0n‍∑l=0n-k‍(-1)k2k+l(α+k+ll)Hn-k-l(n-k-l)!Lkα(x).In the special case,α=-1/2, we obtain the following corollary.Corollary 2.6.
Forn∈ℤ+, one has
(2.20)Hn(x2)=n!∑k=0n‍∑l=0n-k‍2l-kH2k(x)k!(-12+k+ll)Hn-k-l(n-k-l)!.Forβ∈ℝ with β>-1, let one takes
(2.21)p(x)=Lnβ(x)∈Pn.
Then Lnβ(x) is also written as
(2.22)Lnβ(x)=∑k=0n‍CkLkα(x).
From Proposition 2.1, one can determine the coefficients of (2.22) as follows:
(2.23)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)Lnβ(x)dx=1Γ(α+k+1)∫0∞‍(dk-1dxk-1e-xxk+α)Ln-1β+1(x)dx=⋯=1Γ(α+k+1)∫0∞‍e-xxk+αLn-kβ+k(x)dx=1Γ(α+k+1)∑r=0n-k‍(-1)r(n+βn-k-r)r!∫0∞‍e-xxk+α+rdx=1Γ(α+k+1)∑r=0n-k‍(-1)r(n+βn-k-r)r!Γ(k+α+r+1).
By the fundamental property of gamma function, one gets
(2.24)Γ(k+α+r+1)r!Γ(α+k+1)=(k+α+r)⋯(α+k+1)Γ(α+k+1)r!Γ(α+k+1)=(k+α+rr).
Therefore, by (2.22), (2.23), and (2.24), we obtain the following theorem.Theorem 2.7.
Forβ∈ℝ with β>-1, and n∈ℤ+, one has
(2.25)Lnβ(x)=∑k=0n‍∑r=0n-k‍(-1)r(n+βn-k-r)(α+k+rr)Lkα(x).In the special case,α=β, one has
(2.26)∑k=0n-1‍(∑r=0n-k‍(-1)r(n+αn-k-r)(α+k+rr))Lkα(x)=0.
Thus, by (2.26), we obtain the following corollary.Corollary 2.8.
For0≤k≤n-1, α∈ℝ with α>-1, one has
(2.27)∑r=0n-k‍(-1)r(n+αn-k-r)(α+k+rr)=0.Let one assumes that(2.28)p(x)=∑l=0n‍Bl(x)Bn-l(x)∈Pn.
Then p(x) can be rewritten as a linear combination of L0α(x),L1α(x),…,Lnα(x) as follows:
(2.29)p(x)=∑l=0n‍Bl(x)Bn-l(x)=∑k=0n‍CkLkα(x).
By Proposition 2.1, one can determine the coefficients of (2.29) as follows:
(2.30)Ck=1Γ(α+k+1)∑l=0n‍∫0∞‍(dkdxke-xxk+α)Bl(x)Bn-l(x)dx.
It is known that
(2.31)∑l=0n‍Bl(x)Bn-l(x)=2n+2∑l=0n-2‍(n+2l)Bn-lBl(x)+(n+1)Bn(x),
see [25].From (2.30) and (2.31), one notes that
(2.32)Cn=n+1Γ(α+n+1)∫0∞‍(dndxne-xxn+α)Bn(x)dx=n+1Γ(α+n+1)(-1)nn!∫0∞‍e-xxn+αdx=(n+1)!(-1)nΓ(α+n+1)Γ(n+α+1)=(n+1)!(-1)n,Cn-1=n+1Γ(α+n)∫0∞‍(dn-1dxn-1e-xxn-1+α)Bn(x)dx=n+1Γ(α+n)(-1)n-1n!∫0∞‍e-xxn-1+αB1(x)dx=n+1Γ(α+n)(-1)n-1n!{Γ(α+n+1)-12Γ(α+n)}=(n+1)!(-1)n-1(n+α-12).
For 0≤k≤n-2, one has
(2.33)Ck=1Γ(α+k+1){2n+2∑l=0n-2‍(n+2l)Bn-l∫0∞‍(dkdxke-xxk+α)Bl(x)dx+(n+1)∫0∞‍(dkdxke-xxk+α)Bn(x)dx∑l=0n-2‍(n+2l)}=1Γ(α+k+1){2n+2∑l=kn-2‍(n+2l)Bn-l(-1)kl(l-1)⋯(l-k+1)∫0∞‍e-xxk+αBl-k(x)dx+(n+1)(-1)kn(n-1)⋯(n-k+1)∫0∞‍e-xxk+αBn-k(x)dx∑l=kn-2‍(n+2l)}=1Γ(α+k+1){2n+2∑l=kn-2‍(n+2l)Bn-l(-1)kl!(l-k)!∑j=0l-k‍(l-kj)Bl-k-j∫0∞‍e-xxk+α+jdx+(n+1)(-1)kn!(n-k)!∑j=0n-k‍(n-kj)Bn-k-j∫0∞‍e-xxk+α+jdx}=1Γ(α+k+1){2n+2∑l=kn-2‍∑j=0l-k‍(n+2l)Bn-l(-1)kl!(l-k)!(l-kj)Bl-k-jΓ(α+k+j+1)+(n+1)(-1)kn!(n-k)!∑j=0n-k‍(n-kj)Bn-k-jΓ(α+k+j+1)}=2n+2∑l=kn-2‍∑j=0l-k‍(n+2l)Bn-l(-1)kl!(α+k+j)⋯(α+k+1)Bl-k-jj!(l-k-j)!+(n+1)(-1)kn!∑j=0n-k‍(α+k+j)⋯(α+k+1)j!(n-k-j)!Bn-k-j=2n+2∑l=kn-2‍∑j=0l-k‍(n+2l)Bn-l(-1)kl!(α+k+jj)Bl-k-j(l-k-j)!+(n+1)(-1)kn!∑j=0n-k‍(α+k+jj)Bn-k-j(n-k-j)!.
Therefore, by (2.29) and (2.32), we obtain the following theorem.Theorem 2.9.
Forn∈ℤ+, α∈ℝ with α>-1, one has
(2.34)∑k=0n‍Bk(x)Bn-k(x)=∑k=0n-2‍{2n+2∑l=kn-2‍∑0≤j≤n-k‍(-1)k(n+2l)l!Bn-l(α+k+jj)Bl-k-j(l-k-j)!+(n+1)!∑0≤j≤n-k‍(-1)k(α+k+jj)Bn-k-j(n-k-j)!}Lkα(x)+(-1)n(n+1)!{Lnα(x)-(n+α-12)Ln-1α(x)}.Let one takes the polynomialp(x) in Pn as follows:
(2.35)p(x)=∑i1+⋯+ir=n‍Bi1(x)Bi2(x)⋯Bir(x)∈Pn.
From the orthogonality of {L0α(x),…,Lnα(x)}, one notes that
(2.36)p(x)=∑i1+⋯+ir=n‍Bi1(x)Bi2(x)⋯Bir(x)=∑k=0n‍CkLkα(x),
where
(2.37)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)p(x)dx.
It is known in [25] that
(2.38)∑i1+⋯+ir=n‍Bi1(x)⋯Bir(x)=12∑k=0n-2‍(n+r-1k){∑max⁡{0,k+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-k-r‍Bi1Bi2⋯Bia+∑i1+⋯+ir=n-k‍Bi1⋯Bir∑max⁡{0,k+r-n}≤a≤r‍(ra)}Ek(x)+(n+r-1n)En(x).
From (2.35), (2.37), and (2.38), one notes that
(2.39)Cn=(n+r-1n)Γ(α+n+1)∫0∞‍(dndxne-xxn+α)En(x)dx=(n+r-1n)Γ(α+n+1)(-1)nn!∫0∞‍xn+αe-xdx=(n+r-1n)Γ(α+n+1)(-1)nn!Γ(α+n+1)=(n+r-1n)(-1)nn!,Cn-1=(n+r-1n)Γ(α+n)∫0∞‍(dn-1dxn-1e-xxn+α-1)En(x)dx=(n+r-1n)Γ(α+n)(-1)n-1n!∫0∞‍e-xxn+α-1E1(x)dx=(n+r-1n)Γ(α+n)(-1)n-1n!{Γ(α+n+1)-12Γ(α+n)}=(n+r-1n)(-1)n-1n!(n+α-12).
For 0≤k≤n-2, by (2.37) and (2.38), one gets
(2.40)Ck=1Γ(α+k+1){12∑l=0n-2‍(n+r-1l)(∑max⁡{0,l+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-l-r‍Bi1Bi2⋯Bia+∑i1+⋯+ir=n-l‍Bi1⋯Bir)∫0∞‍(dkdxke-xxn+α)El(x)dx+(n+r-1n)∫0∞‍(dkdxke-xxn+α)En(x)dx∑max⁡⁡{0,l+r-n}≤a≤r‍}=1Γ(α+k+1){12∑l=kn-2‍(n+r-1l)(∑max⁡{0,l+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-l-r‍Bi1Bi2⋯Bia+∑i1+⋯+ir=n-l‍Bi1⋯Bir∑max⁡⁡{0,l+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-l-r‍)(-1)kl(l-1)⋯(l-k+1)×∫0∞‍e-xxk+αEl-k(x)dx+(n+r-1n)(-1)kn(n-1)⋯(n-k+1)∫0∞‍e-xxk+αEn-k(x)dx∑max⁡⁡{0,l+r-n}≤a≤r‍}=1Γ(α+k+1){12∑l=kn-2‍(n+r-1l)(∑max⁡{0,l+r-n}≤a≤r‍∑i1+⋯+ia=n+a-l-r‍Bi1⋯Bia+∑i1+⋯+ir=n-l‍Bi1⋯Bir∑max⁡⁡{0,l+r-n}≤a≤r‍∑i1+⋯+ia=n+a-l-r‍)(-1)kl!(l-k)!∑j=0l-k‍(l-kj)El-k-j×∫0∞‍e-xxk+α+jdx+(n+r-1n)(-1)kn!(n-k)!∑j=0n-k‍(n-kj)En-k-j∫0∞‍e-xxk+α+jdx}=12∑l=kn-2‍(n+r-1l)(∑max⁡{0,l+r-n}≤a≤r‍∑i1+⋯+ia=n+a-l-r‍Bi1⋯Bia+∑i1+⋯+ir=n-l‍Bi1⋯Bir)×(-1)kl!∑j=0l-k‍(α+k+jj)El-k-j(l-k-j)!+(n+r-1n)(-1)kn!∑j=0n-k‍(α+k+jj)En-k-j(n-k-j)!.
Therefore, by (2.36), (2.39), and (2.40), we obtain the following theorem.Theorem 2.10.
Forn∈ℤ+, r∈ℕ, and α∈ℝ with α>-1, one has
(2.41)∑i1+⋯+ir=n‍Bi1(x)Bi2(x)⋯Bir(x)=∑k=0n-2‍(-1)k2{∑l=kn-2‍l!(n+r-1l)(∑max⁡{0,l+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-l-r‍Bi1⋯Bia+∑i1+⋯+ir=n-l‍Bi1⋯Bir∑max⁡{0,l+r-n}≤a≤r‍(ra)∑i1+⋯+ia=n+a-l-r‍)×∑j=0l-k‍(α+k+jj)El-k-j(l-k-j)!+(n+r-1n)(-1)kn!×∑j=0n-k‍(α+k+jj)En-k-j(n-k-j)!}Lkα(x)+(n+r-1n)(-1)n-1n!×(n+α-12)Ln-1α(x)+(n+r-1n)(-1)nn!Lnα(x).Form,s∈ℤ+ with m+s=n, let one assumes that p(x)=Lsα(x)Lmα(x)∈Pn.By Proposition2.1, one sees that p(x) can be written as
(2.42)p(x)=Lsα(x)Lmα(x)=∑k=0n‍CkLkα(x),α∈Rwithα>-1.
From the orthogonality of {L0α(x),L1α(x),…,Lnα(x)}, one has
(2.43)Ck=1Γ(α+k+1)∫0∞‍(dkdxke-xxk+α)p(x)dx.
By (1.2), (1.3), and (1.8), one gets
(2.44)Lsα(x)Lmα(x)=(∑r1=0s‍(-1)r1r1!(s+αs-r1)xr1)(∑r2=0m‍(-1)r2r2!(m+αm-r2)xr2)=∑r=0n‍(∑r1=0r‍(-1)r(rr1)(s+αs-r1)(n-s+αα+r-r1))xrr!.
Thus, from (2.44), one has
(2.45)Lsα(x)Lmα(x)=∑r=0n‍(∑l=0r‍(-1)r(rl)(s+αα+l)(n-s+αα+r-l))xrr!.
By (2.44) and (2.45), one gets
(2.46)Ck=1Γ(α+k+1)∑r=0n‍(-1)r{∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)}1r!×∫0∞‍(dkdxke-xxk+α)xrdx=1Γ(α+k+1)∑r=kn‍(-1)r{∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)1r!}×(-1)kr(r-1)⋯(r-k+1)∫0∞‍e-xxr+αdx=1Γ(α+k+1)∑r=kn‍(-1)r{∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)1r!}×(-1)kr!(r-k)!Γ(α+r+1)=∑r=kn‍(-1)r+k(∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)(α+r)(α+r-1)⋯(α+k+1)(r-k)!)=∑r=kn‍(-1)r+k(∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)(α+rr-k)).
Therefore, by (2.42) and (2.46), we obtain the following theorem.Theorem 2.11.
Fors,m∈ℤ+ with s+m=n, α∈ℝ with α>-1, one has
(2.47)Lsα(x)Lmα(x)=∑k=0n‍∑r=kn‍(-1)r+k(∑l=0r‍(rl)(s+αα+l)(n-s+αα+r-l)(α+rr-k))Lkα(x).



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*Source: 957350-2012-09-09.xml*