Classical orthogonal polynomials with respect to a

lowering operator generalizing the Laguerre Operator

Baghdadi Alouib, Francisco Marcellana,∗, Ridha Sfaxic

aDepartamento de Matematicas, Universidad Carlos III de Madrid, Avenida de laUniversidad 30, 28911 Leganes, Spain

bFaculte des Sciences de Gabes, Departement de Mathematiques, Cite Erriadh,6072Gabes, Tunisie

c Department of Mathematics, University of Dammam, Faculty of Education of Girls ofHafr Al-Batin, Saudi Arabia.

Abstract

We introduce the lowering operator Ω1,c = Ω1− cD2, where c is an arbitrary

complex number and Ω1 is the Generalized Laguerre Operator introducedby G. Dattoli and P. E. Ricci. Then, we establish an intertwining relationbetween the operators Ω1,c and the standard derivative D. On the otherhand, an analogue of the Hahn characterization of D- classical orthogonalpolynomials is given for the operator Ω1,c . As a consequence, some integralrelations between the corresponding polynomials are deduced. Finally, someexpansions in series of Laguerre polynomials are studied.

Keywords: Laguerre polynomials, Classical polynomials, Appell’s property,Lowering, Transfer and Raising operators, Integral formulas.

1. Introduction

Let L(1)n n≥0 be the monic (normalized) Laguerre polynomial sequence

with parameter α = 1. It is a sequence of polynomials orthogonal with respectto the weight function xe−x on (0,+∞), i.e.,

∫ +∞0

xe−xL(1)n (x)L

(1)m (x) dx =

(m+1)δn,m, n,m ≥ 0, where δn,m is the Kronecker delta. These polynomials

∗Corresponding authorEmail addresses: Baghdadi.Aloui@fsg.rnu.tn (Baghdadi Aloui),

pacomarc@ing.uc3m.es (Francisco Marcellan), Ridha.Sfaxi@isggb.rnu.tn (RidhaSfaxi)

Preprint submitted to Elsevier October 23, 2012

can be represented as an hypergeometric function [11]

L(1)n (x) =

n∑ν=0

(−1)n−ν(n

ν

)(n+ 1)!

(ν + 1)!xν . (1)

It is well-known that L(1)n n≥0 can be characterized taking into account its

orthogonality as well as the following differential properties (see [14])-The Second-order linear differential equation

xL(1)n+1

′′(x)− (x− 2)L

(1)n+1

′(x) + (n+ 1)L

(1)n+1(x) = 0, n ≥ 0. (2)

-The first structure relation

xL(1)n+1

′(x) = (n+ 1)L

(1)n+1(x) + (n+ 1)(n+ 2)L(1)

n (x), n ≥ 0. (3)

Substituting (3) in (2), we get

L(1)n (x) =

Ω1L(1)n+1(x)

(n+ 1)(n+ 2), n ≥ 0, (4)

where Ωm = DxD + mD. This means that the above family of standardorthogonal polynomials is an Appell sequence with respect to the operatorΩ1. Polynomial Appell sequences with respect to the standard derivativeoperator D were introduced in [3]. For instance, in [7] the authors haveconsidered this family of operators with parameter m 6= −n, n ≥ 1. In [11]and [12], polynomial Appell sequence with respect to the second order lineardifferential operator Fε = 2xD2 + (ε + 2)D have studied. They deduce thatthe orthogonal polynomial Appell sequences with respect to such an operatorare the Laguerre sequences with parameter α = ε

2. Notice that Ω1 = 2F2.

Moreover, L(1)n n≥0 is an Ω1−classical polynomial sequence, since it satisfies

the Hahn property for the lowering operator Ω1, i.e., it is an orthogonalpolynomial sequence whose sequence of Ω1−derivatives is also orthogonal,[8]. This Hahn’s property can de considered for other differential operators.In Section 4.2 of the Doctoral Dissertation by A. F. Loureiro, for the secondorder (Laguerre) differential Fε the characterization and the description ofthe Fε-classical orthogonal polynomials is given (see [11], Theorem 4.2.4,and subsection 4.2.3)). More precisely, Laguerre sequences with parameterα = ε

2and some particular Jacobi sequences with parameters ( ε

2,−1

2− ε

4+ b)

2

are the only Fε-classical orthogonal polynomial sequences. This result pointsout that these Fε-classical orthogonal polynomial constitute a more restrictedfamily than those D- classical families in the Hahn’s sense.

For a given c ∈ C, let consider Ω1,c : P→ P, the linear operator definedby in the linear space P of polynomials with complex coefficients

Ω1,c := Ω1 − cD2, (Ω1,0 = Ω1).

Notice that our new operator is different of Fε but they share the fact thatthey are lowering operators, i.e., the image of a polynomial of degree n, n ≥ 1is a polynomial of degree n− 1. For the canonical basis xnn≥0 we have

Ω1,c(xn) = n(n+ 1)xn−1 − cn(n− 1)xn−2 ; Fε(xn) = n(2n+ ε)xn−1.

In this contribution we will analyze the Ω1,c− classical polynomials and thenwe will provide a full description of them. We will emphasize two basic facts.The first is that the Ω1,c-classical character is preserved by shifting. The

second is that the Ω1,c- classical polynomial sequences constitute a subfamilyof the well-known D−classical ones (Hermite, Laguerre, Bessel, and Jacobi).

More precisely, the Ω1,c-classical polynomial sequences are, after suitable

shifting, the Laguerre polynomial sequence L(1)n n≥0 , when c = 0, or the

Jacobi polynomial sequence P (α−2,1)n n≥0 with parameter α 6= −n+2, n ≥ 1,

when c = 1. Notice that they are related to particular cases of the operatorFε for different values of ε. Our proof approach is completely different of theproof provided in [11] and, even more, shorter than it.

The paper is organized as follows. In Section 2, we introduce the basicbackground to be used throughout the paper. Section 3 gives the descriptionof the Ω1,c−classical sequences. As a consequence, we establish a new integralrelation between the monic Bessel and Jacobi polynomials. Finally, in Section4 some expansions in series of Laguerre polynomials are presented.

2. Preliminaries and notations

2.1. Basic definitions

Let P be the linear space of polynomials with complex coefficients andlet P′ be its dual. We denote by 〈u, f〉 the action of u ∈ P′ on f ∈ P andby (u)n = 〈u, xn〉, n ≥ 0, the moments of u. Let us define the following

3

operations on P′ [13]. For any a ∈ C\0, b, c ∈ C, f, p ∈ P, and u ∈ P′,

〈fu, p〉 = 〈u, fp〉, 〈Du, f〉 = −〈u, f ′〉,〈hau, p〉 = 〈u, hap〉 = 〈u, p(ax)〉, 〈τbu, p〉 = 〈u, τ−bp〉 = 〈u, p(x+ b)〉,

〈(x− c)−1u, p〉 = 〈u, θc(p)〉 = 〈u, p(x)− p(c)x− c

〉, 〈δc, p〉 = p(c), p ∈ P.

Here δc denotes the Dirac linear functional at c ∈ C.Let Pnn≥0 be a sequence of monic polynomials (MPS) with degPn = n

and let unn≥0 be its dual sequence, un ∈ P′, defined by 〈un, Pm〉 = δn,m,n, m ≥ 0. Recall that any u ∈ P′ can be represented as u =

∑+∞n=0〈u, Pn〉un.

So, if u[1]n n≥0 denotes the dual sequence of the MPS P [1]n n≥0 where P

[1]n :=

(n+ 1)−1P ′n+1, n ≥ 0, then Du[1]n = −(n+ 1)un+1, n ≥ 0. Likewise, the dual

sequence unn≥0 of the shifted MPS Pnn≥0, where Pn(x) := a−nPn(ax+b)with (a, b) ∈ C\0 × C, is given by un = an(ha−1 τ−b)un, n ≥ 0.

An MPS Pnn≥0 is said to be orthogonal (MOPS) with respect to u ∈ P′if 〈u, PnPm〉 = 0, n 6= m, and 〈u, P 2

n〉 6= 0, n ≥ 0. In this case, u is said tobe quasi-definite (regular) [6]. Notice that u = (u)0u0, with (u)0 6= 0.

Proposition 2.1. [13]. An MPS Pnn≥0 with dual sequence unn≥0, isorthogonal if and only if one of the following statements hold.

(i) un = 〈u0, P 2n〉−1Pnu0, n ≥ 0.

(ii) Pnn≥0 satisfies a Three-Term Recurrence Relation

(TTRR)

P0(x) = 1, P1(x) = x− β0,Pn+2(x) = (x− βn+1)Pn+1(x)− γn+1Pn(x), n ≥ 0,

(5)

where βn = 〈u0, xP 2n〉/〈u0, P 2

n〉 ∈ C and γn+1 = 〈u0, P 2n+1〉/〈u0, P 2

n〉 ∈ C\0.

When Pnn≥0 is an MOPS with respect to u0, then Pnn≥0 is alsoorthogonal with respect to u0 = (ha−1 τ−b)u0 and satisfies [6]

(TTRR)

P0(x) = 1, P1(x) = x− β0,Pn+2(x) = (x− βn+1)Pn+1(x)− γn+1Pn(x), n ≥ 0,

where βn = a−1(βn − b) and γn+1 = a−2γn+1.A linear functional u is said to be D−classical when it is quasi-definite

and there exist two polynomials Φ and Ψ, Φ monic, deg Φ = t ≤ 2, and

4

deg Ψ = 1, such that u satisfies a Pearson equation, (PE): (Φu)′ + Ψu = 0,(see [5, 13, 14]). In such a case, the corresponding MOPS Pnn≥0 is said tobe D−classical. Any shift leaves invariant the D−classical character. Indeed,the shifted linear functional u = (ha−1 τ−b)u fulfils (Φu)′ + Ψu = 0, where

Φ(x) = a−tΦ(ax+b) and Ψ(x) = a1−tΨ(ax+b). Any D−classical polynomialsequence Pnn≥0 can be characterized taking into account its orthogonalityas well as a First Structure Relation (FSR), or a Second Structure Relation(SSR), or Second-Order Differential Equation (SODE) as follows.

(FSR) Φ(x)P ′n+1(x) = r(x;n)Pn+1(x) + snPn(x), n ≥ 0, (6)

(SSR) Pn(x) = P [1]n (x) + anP

[1]n−1(x) + bnP

[1]n−2(x), n ≥ 0, (7)

(SODE) Φ(x)P ′′n+1(x)−Ψ(x)P ′n+1(x) = ωnPn+1(x), n ≥ 0, (8)

where for the fourth canonical situations, we have (see [1, 5, 13, 14]).Table 1

(C1) Hermite: Pn(x) = Hn(x), n ≥ 0.

βn = 0, n ≥ 0, γn+1 = n+12 , n ≥ 0, Φ(x) = 1, Ψ(x) = 2x.

r(x;n) = 0, sn = n+ 1, n ≥ 0, an = bn = 0, n ≥ 0, ωn = −2(n+ 1), n ≥ 0.

(C2) Laguerre: Pn(x) = L(α)n (x), n ≥ 0, (α 6= −n, n ≥ 1).

βn = 2n+ α+ 1, n ≥ 0, γn+1 = (n+ 1)(n+ α+ 1), n ≥ 0, Φ(x) = x, Ψ(x) = x− α− 1,r(x;n) = n+ 1, sn = γn+1, n ≥ 0, an = n, bn = 0, n ≥ 0, ωn = −(n+ 1), n ≥ 0.

(C3) Bessel Pn(x) = B(α)n (x), n ≥ 0, (α 6= −n2 , n ≥ 0).

β0 = − 1α , βn = 1−α

(n+α−1)(n+α) , n ≥ 0, γn = − n(n+2α−2)(2n+2α−3)(n+α−1)2(2n+2α−1) , n ≥ 1.

Φ(x) = x2, Ψ(x) = −2(αx+ 1).r(x;n) = (n+ 1)(x− 1

n+α ), sn = −(2n+ 2α+ 1)γn+1, n ≥ 0, an = n(n+α−1)(n+α) , n ≥ 0,

bn = (n−1)n(2n+2α−3)(n+α−1)2(2n+2α−1) , n ≥ 2, b0 = b1 = 0, ωn = (n+ 1)(n+ 2α), n ≥ 0.

(C4) Jacobi Pn(x) = P(α,β)n (x), n ≥ 0, (α, β 6= −n, α+ β 6= −n− 1, n ≥ 1).

β0 = α−βα+β+2 , βn = α2−β2

(2n+α+β)(2n+α+β+2) , γn = 4n(n+α+β)(n+α)(n+β)(2n+α+β−1)(2n+α+β)2(2n+α+β+1) , n ≥ 1.

Φ(x) = x2 − 1, Ψ(x) = −(α+ β + 2)x+ α− β.r(x;n) = (n+ 1)(x− α−β

2n+α+β+2 ), sn = −(2n+ α+ β + 3)γn+1, n ≥ 0.

an = − 2n(α−β)(2n+α+β)(2n+α+β+2) , n ≥ 1, a0 = 0, bn = − 4(n−1)n(n+α)(n+β)

(2n+α+β−1)(2n+α+β)2(2n+α+β+1) , n ≥ 2,

b0 = b1 = 0, ωn = (n+ 1)(n+ α+ β + 2), n ≥ 0.

Further, we need the following properties [6, 9].

B(α)n (x) =

n∑ν=0

(n

ν

)2n−νΓ(n+ 2α + ν − 1)

Γ(2n+ 2α− 1)xν , n ≥ 0. (9)

5

P (α,β)n (x) =

n∑ν=0

(n

ν

)2n−νΓ(n+ α + β + ν + 1)Γ(n+ β + 1)

Γ(2n+ α + β + 1)Γ(ν + β + 1)(x− 1)ν , (10)

P (α,β)n (−x) = (−1)nP (β,α)

n (x), n ≥ 0, (symmetry). (11)

2.2. Some properties of the operator Ω1,c

In the sequel, we will denote by “” the composition law between linearoperators on the linear space of polynomials. The following formulas are astraightforward consequence of the definition of the operators.

Lemma 2.1. For any c in C, we have

(i) Ω1,c(x− c)n+1 = (n+ 1)(n+ 2)(x− c)n, n ≥ 0, and Ω1,c(1) = 0.

(ii) Ω1,c Πa,b = aΠa,b Ω1,ac+b and Πa,b Ω1,c = a−1Ω1, c−ba Πa,b, where

Πa,bf(x) = f(ax+ b), for every f ∈ P and where (a, b) ∈ C\0 × C.(iii) Ω1,c(pq) = qΩ1,c(p) + pΩ1,c(q) + 2(x− c)p′q′, for every p and q in P.

Through an appropriate linear isomorphism called intertwining operator,we can establish a relationship between the operators Ω1,c and the standardderivative D. We first recall the following formulas [9]:

n! =

∫ +∞

0

tne−tdt, n = 0, 1, 2, ..., (12)

1

n!=−1

2πi

∫C

(−z)−n−1e−zdz, n = 0, 1, 2, ..., (13)

where C is the following contour in the complex plane:

C

Proposition 2.2. For any c ∈ C, we have Sc Ω1,c = D Sc, where theoperator Sc : P→ P and its reciprocal operator S−1c are linear isomorphisms,

Sc(p)(x) =

∫ +∞

0

te−tp(t(x− c) + c

)dt, p ∈ P,

S−1c (p)(x) =

∫C

z−2e−zp(− z−1(x− c) + c

)dz, p ∈ P,

and C is the same contour as above.

6

Proof. The proof is a consequence of Lemma 2.1 (i), (12) and (13).Notice that the operator Sc can be characterized taking into account

its linearity as well as the fact

Sc

((x− c)n

)= (n+ 1)!(x− c)n, n ≥ 0. (14)

By transposition of the operator Ω1,c, we get tΩ1,c = Ω−1−cD2. So, Ω−1,c :P′ → P′ is the operator defined by

Ω−1,c = Ω−1 − cD2 = (x− c)D2. (15)

For any MPS Pnn≥0 we define

Qn(x; c) :=Ω1,c Pn+1(x)

(n+ 1)(n+ 2), n ≥ 0. (16)

Clearly, Qn( ; c)n≥0 is an MPS, degQn( ; c) = n. If vn(c)n≥0 denotes thedual sequence of Qn( ; c)n≥0, then we have

Ω−1,c(vn(c)) = (n+ 1)(n+ 2)un+1, n ≥ 0. (17)

3. The Ω1,c−classical orthogonal polynomials

Definition 3.1. An MOPS Pnn≥0 (orthogonal with respect to u0) is said

to be Ω1,c−classical, when it satisfies the Hahn’s property with respect to the

operator Ω1,c, i.e., the MPS Qn( ; c)n≥0 given by (16) is also orthogonal.

In this case u0 is also said to be an Ω1,c−classical linear functional.

Any shift leaves invariant the Ω1,c−classical character.

Lemma 3.1. When Pnn≥0 is Ω1,c−classical, then for any (a, b) ∈ C∗ × Cthe shifted polynomial sequence Pnn≥0 given by Pn(x) = a−nPn(ax + b),

n ≥ 0, is Ω1,c−classical, where c = a−1(c− b).

Proof. Assume that Pnn≥0 is Ω1,c−classical. By Definition 3.1, the MPSQn( ; c)n≥0 given by (16) is orthogonal. Notice that, we can write

(n+ 1)(n+ 2)Qn(x; c) = (x− c)P ′′n+1(x) + 2P ′n+1(x), n ≥ 0, (18)

For any fixed (a, b) ∈ C\0 × C, let Pnn≥0 and Qnn≥0 be the shifted

MOPS given by Pn(x) = a−nPn(ax + b) and Qn(x) = a−nQn(ax + b; c).

7

Replacing x by ax+ b in (18), we get (n+ 1)(n+ 2)Qn(x) = (x− c)P ′′n+1(x) +

2P ′n+1(x), where c = a−1(c − b), i.e., (n + 1)(n + 2)Qn(x) = Ω1,c Pn+1(x),

n ≥ 0. Hence, Pnn≥0 is Ω1,c −classical.In the sequel, we write Qn(x) := Qn(x; c), n ≥ 0, if there is no ambiguity.

Our next goal is to describe all the Ω1,c−classical polynomial sequences.Assume that Pnn≥0 and Qnn≥0 are MOPS satisfying

P0(x) = 1, P1(x) = x− β0,Pn+2(x) = (x− βn+1)Pn+1(x)− γn+1Pn(x), γn+1 6= 0, n ≥ 0,

(19)

Q0(x) = 1, Q1(x) = x− ξ0,Qn+2(x) = (x− ξn+1)Qn+1(x)− λn+1Qn(x), λn+1 6= 0, n ≥ 0.

(20)

The dual sequences of Pnn≥0 and Qnn≥0 will be denoted by unn≥0 andvnn≥0, respectively. By Proposition 2.1 (i), we get

un =Pn

〈u0, P 2n〉u0, n ≥ 0 ; vn =

Qn

〈v0, Q2n〉v0, n ≥ 0. (21)

Let us start with some auxiliary results.

Lemma 3.2. The MOPS Qnn≥0 is classical and we have

(i) (SSR) Qn(x) = Q[1]n (x) + cnQ

[1]n−1(x) + dnQ

[1]n−2(x), n ≥ 0, where

cn =n

2

(βn+1−ξn

), n ≥ 0, dn =

n− 1

2

( n

n+ 2γn+1−λn

), n ≥ 1, d0 = 0.

(ii) (PE) (Φv0)′ + Ψv0 = 0, where κΦ = d2(λ1λ2)

−1Q2 + c1λ−11 Q1 + 1,

Ψ = (κλ1)−1Q1 and κ is a normalization factor.

Proof. Let us introduce the MPS Znn≥0 given by

(n+ 1)Zn(x) := (x− c)P ′n(x) + Pn(x), n ≥ 0. (22)

Taking derivatives in both hand sides of (22), where n is replaced by n + 1,and using (16), we get

Z [1]n (x) = Qn(x), n ≥ 0. (23)

Notice that Zn(x) is a monic primitive of Qn(x).From (19) and (22), we get (n + 3)Zn+2 = (n + 2)(x − βn+1)Zn+1 − (n +

8

1)γn+1Zn + (x − c)Pn+1, n ≥ 0. Differentiating in both hand sides of theprevious identity and inserting (23), (n+ 3)(n+ 2)Qn+1 = (n+ 1)(n+ 2)(x−βn+1)Qn − n(n+ 1)γn+1Qn−1 + 2(n+ 2)Zn+1. Then, by (20), it follows that

Zn(x) = Qn(x) + anQn−1(x) + bnQn−2(x), n ≥ 0, (24)

where an = n2(βn − ξn−1), n ≥ 1, a0 = 0, bn = n

2(n−1n+1

γn − λn−1), n ≥ 2,b0 = b1 = 0.By differentiating both hand sides of (24) and using (23), (i) holds.

Let v[1]n n≥0 be the dual sequence of Q[1]n n≥0. By (i), 〈v[1]0 , Qn〉 = 0, n ≥ 3,

〈v[1]0 , Q2〉 = d2, 〈v[1]0 , Q1〉 = c1, and 〈v[1]0 , Q0〉 = 1. So, v[1]0 = d2v2 + c1v1 + v0,

and by (21), we get v[1]0 = κΦv0, where κΦ = d2λ

−11 λ−12 Q2 + c1λ

−11 Q1 + 1

and κ is a normalization factor. Since (v[1]0 )′ = −v1 = −λ−11 Q1v0, then

(Φv0)′ + Ψv0 = 0, where Ψ = (κλ1)

−1Q1. Hence, (ii) holds.

Lemma 3.3. There exist two non-zero polynomials F and G, with degF ≤ 3and degG ≤ 2 such that

(i) (x− c)v0 = Fu0.

(ii) FQ′′n +GQ′n + ρ0P1Qn = ρnPn+1, n ≥ 0, where

F = (1/2)(ρ2P3 −GQ′2 − ρ0P1Q2

), G = ρ1P2 − ρ0P1Q1, and

ρn = (n+ 1)(n+ 2)〈v0, Q2

n〉〈u0, P 2

n+1〉=F (3)(0)

6n(n− 1) +

G(2)(0)

2n+ ρ0.

(iii) The following relations hold

(a) AF = ρ0P1Φ2,

(b) −2(Φ′ + Ψ

)F = ΦG,

(c) γ1Φ2 =

[((ρ−10 − ρ−11 )x− β1ρ−10 + ξ0ρ

−11

)A− ρ−11 B

]F, where

A = (2Φ′ + Ψ)(Φ′ + Ψ)− (Φ′′ + Ψ′)Φ, and B = −2Φ(Φ′ + Ψ).

Proof. From (15), (17) and (21), we obtain

(x− c)Qnv′′0 + 2(x− c)Q′nv′0 + (x− c)Q′′nv0 = ρnPn+1u0, n ≥ 0, (25)

where ρn = (n+ 1)(n+ 2)〈v0, Q2n〉〈u0, P 2

n+1〉−1, n ≥ 0.From (25) with n = 0, we obtain

(x− c)v′′0 = ρ0P1u0. (26)

9

Using (25) and (26), it follows that

(x− c)Q′′nv0 + 2(x− c)Q′nv′0 = (ρnPn+1 − ρ0P1Qn)u0. (27)

For n = 1 , (27) becomes

2(x− c)v′0 = Gu0, (28)

where G = ρ1P2 − ρ0P1Q1.By inserting (28) in (27), we obtain

(x− c)Q′′nv0 = (ρnPn+1 − ρ0P1Qn −GQ′n)u0. (29)

Hence, taking n = 2 in (29), (i) holds.So, by substituting (x − c)v0 = Fu0 in (29) and taking into account the

quasi-definiteness of u0, we deduce (ii).By using Lemma 3.2 (ii), we can write

Φv′0 = −(Φ′ + Ψ)v0, Φ2v′′0 = Av0, (30)

where A = (2Φ′ + Ψ)(Φ′ + Ψ)− (Φ′′ + Ψ′)Φ.So, if we multiply (26) by Φ2 and we take into account (30), (i) as well as thequasi-definiteness of u0, we get (a). If we multiply (28) by Φ and we use (30),(i) and the quasi-definiteness of u0 then (b) follows. Finally, multiplying (25)by Φ2 and using (30), (i) as well as the quasi-definiteness of u0, we get

(Φ2Q′′n +BQ′n + AQn)F = ρnΦ2Pn+1, n ≥ 0, (31)

where B = −2Φ(Φ′ + Ψ).For n = 1 in (31), n = 0 in (19), n = 0 in (31), and (20), we obtain (c).

Theorem 3.1. The Ω1,c−classical polynomial sequences are, up to a suitableshifting, one of the following D−classical polynomial sequences:

(i) Pn = Qn = L(1)n , n ≥ 0, with c = 0.

(ii) Pn = P(α−2,1)n , n ≥ 0, with α 6= −n + 2, n ≥ 1, Qn = P

(α,1)n , n ≥ 0,

and c = 1.

Proof. From Lemma 3.2, Qnn≥0 is D−classical. By Lemma 3.1, we willanalyze the four canonical situations given in Table 1.

10

(C1). Qnn≥0 is the Hermite MOPS. From Table 1 (C1), A = 2(2x2−1)and B = −4x. Since Lemma 3.3 (iii) (a), 2(2x2 − 1)F = ρ0P1. This yields acontradiction, since degP1 = 1 and ρ0 6= 0.

(C2). Qnn≥0 is the Laguerre MOPS. From Table 1 (C2), A = x2 −2αx+ α(α− 1) and B = −2x(x− α). By Lemma 3.3, (iii), we get(

x2 − 2αx+ α(α− 1))F = ρ0x

2P1, (32)

−2(x− α)F = xG, (33)

γ1x2 =

(((ρ−10 − ρ−11 )x− β1ρ−10 + ξ0ρ

−11

)A− ρ−11 B

)F, (34)

From (32), degF = 1. So, F = ρ0x, by (32) and (34). Hence, (32) becomes(x2 − 2αx + α(α − 1)) = xP1. Thus, α(α − 1) = 0. Necessarily, α = 1.Otherwise, if α = 0 then G = −2ρ0x and P1 = x. But, by evaluating atx = 0 the equation given by Lemma 3.3 (ii), we get Pn+1(0) = 0, n ≥ 0,which contradicts the orthogonality of Pnn≥0.By (32) with α = 1 and F = ρ0x, we get P1 = x− 2 and then β0 = 2. So,

A = x(x− 2), B = −2x(x− 1), G = −2ρ0(x− 1). (35)

Since ξ0 = 2 (see Table 1 (C2) with α = 1) and by (35), (34) yields ρ1 = ρ0,β1 = 4, and γ1 = 2. Then, ρ0 = 2〈v0, Q2

0〉〈u0, P 21 〉−1 = 2γ−11 = 1. By Table 1

(C2), with α = 1, and by Lemma 3.2 (i), βn+1 = 2n+ 4, n ≥ 1, and γn+1 =(n+ 1)(n+ 2), n ≥ 2. By Lemma 3.3 (ii), xQ′′n − 2(x− 1)Q′n + (x− 2)Qn =ρnPn+1, n ≥ 0. Thus, ρn = 1, n ≥ 0. So, xQ′′n−2(x−1)Q′n+(x−2)Qn = Pn+1,n ≥ 0. Since 1 = ρ1 = 6〈v0, Q2

1〉〈u0, P 22 〉−1 = 6λ1(γ1γ2)

−1, then γ2 = 6,

βn = 2(n + 1) and γn+1 = (n + 1)(n + 2), n ≥ 0. Hence, Pn = Qn = L(1)n ,

n ≥ 0, and u0 = v0. Also, by Lemma 3.3 (i), we get (x − c)u0 = xu0. Thisrequires that c = 0, because of the quasi-definiteness of u0. Thus, we have

L(1)n = (n+ 1)−1(n+ 2)−1Ω1 L

(1)n+1, n ≥ 0, (36)

xL(1)n

′′(x)− 2(x− 1)L(1)

n

′(x) + (x− 2)L(1)

n (x) = L(1)n+1(x), n ≥ 0. (37)

(C3). Qnn≥0 is the Bessel MOPS with parameter α 6= −n/2, n ≥ 0.By Table 1 (C3), A = 2(1− α)(3− 2α)x2 + 4(2α − 3)x + 4. By Lemma 3.3(iii) (a), AF = ρ0x

4P1. This requires that degA = 2 and degF = 3, sincedegA ≤ 2, degF ≤ 3, and degA+degF = 5. But, by the previous equation,we must have A(0) = 0, that contradicts the fact that A(0) = 4.

11

(C4). Qnn≥0 is the Jacobi MOPS with parameters α and β satisfyingα, β 6= −n, α + β 6= −n− 1, n ≥ 1. From Table 1 (C4), we get

A = (α + β − 1)((α + β)x2 + 2(β − α)x

)+ (α− β)2 − (α + β), (38)

B = 2(x2 − 1)((α + β)x+ β − α

). (39)

By Lemma 3.3 (iii) (a), AF = ρ0(x2 − 1)2P1. Since degA + degF = 5,

degF ≤ 3 and degA ≤ 2, then degA = 2 and degF = 3. By Lemma 3.3 (iii)(c), F divides Φ2 = (x2 − 1)2 and hence there are two situations to be treat.Either F = µ(x− 1)(x+ 1)2 or F = µ(x+ 1)(x− 1)2, where µ 6= 0.

(C4,1). F = µ(x− 1)(x+ 1)2, µ 6= 0. By Lemma 3.3 (iii),

µA = ρ0(x− 1)P1, (40)

G = 2µ((α + β)x+ β − α

)(x+ 1), (41)

µ−1γ1(x− 1) =((ρ−10 − ρ−11 )x− β1ρ−10 + ξ0ρ

−11

)A− ρ−11 B, (42)

where ξ0 = (α− β)(α + β + 2)−1.From (38) and (40), ρ0 = µ(α+ β − 1)(α+ β), β0 = (α− 3β)(α+ β)−1, andβ(β−1) = 0. We must have β = 1. Otherwise, if β = 0, then G(1) = P1(1) =F (1) = 0. But, by evaluating at x = 1 the equation given by Lemma 3.3 (ii)we get Pn+1(1) = 0, n ≥ 0, which contradicts the orthogonality of Pnn≥0.For β = 1, we get ρ0 = µα(α+1), β0 = (α−3)(α+1)−1 and G = 2µ

((α+1)x+

1−α)(x+1). This expression of G together with that given by Lemma 3.3 (ii),

yields by identification of the coefficients β1 = (α−1)(α−3)(α+1)(α+3)

and γ1 = 8(α−1)(α+1)2(α+2)

.

Moreover, since ρ0 = 2γ−11 , we get µ = (α+1)(α+2)4α(α−1) .

From Table 1 (C4) with β = 1 and by Lemma 3.2 (i), βn = (α−2)2−1(2n+α−1)(2n+α+1)

,

γn+1 = 4(n+1)(n+2)(n+α)(n+α−1)(2n+α)(2n+α+1)2(2n+α+2)

, n ≥ 2. By Lemma 3.3 (ii), (x−1)(x+1)2Q′′n+

2((α+1)x+1−α

)(x+1)Q′n+α(α+1)(x−β0)Qn = µ−1ρnPn+1, n ≥ 0. This

allows us to deduce that ρn = µ(n2 + (2α + 1)n+ α(α + 1)

). Consequently,

(x− 1)(x+ 1)2Q′′n + 2((α + 1)x+ 1− α

)(x+ 1)Q′n+

α((α + 1)x+ 3− α

)Qn = (n+ α)(n+ α + 1)Pn+1, n ≥ 0.

Since (α+1)2(α+2)2

4α(α−1) = ρ1 = 6λ1(γ1γ2)−1, then γ2 = 24α(α+1)

(α+2)(α+3)2(α+4). So, βn =

(α−2)2−1(2n+α−1)(2n+α+1)

, γn+1 = 4(n+1)(n+2)(n+α)(n+α−1)(2n+α)(2n+α+1)2(2n+α+2)

, n ≥ 0. Thus, Pn = P(α−2,1)n ,

Qn = P(α,1)n , n ≥ 0 with α 6= −n + 2, n ≥ 1. Besides, by (18) with n = 1,

12

(19) and (20), c = −β0 − β1 + 3ξ0 = 1. By Lemma 3.3 (i), (x − 1)v0 =

µ(x− 1)(x+ 1)2u0, where µ = (α+1)(α+2)4α(α−1) . Thus, v0 = µ(x+ 1)2u0. Then,

P (α,1)n (x) = (n+ 1)−1(n+ 2)−1Ω1,1 P

(α−2,1)n+1 (x), (43)

(x− 1)(x+ 1)2P (α,1)′′

n (x) + 2((α + 1)x+ 1− α

)(x+ 1)P (α,1)′

n (x) +

α((α + 1)x+ 3− α

)P (α,1)n (x) = %nP

(α−2,1)n+1 (x), n ≥ 0, (44)

where %n = (n+ α)(n+ α + 1), n ≥ 0.(C4,2). F = µ(x + 1)(x − 1)2, µ 6= 0. By a similar computation as in

(C4,1), Pn = P(1,β−2)n , (β 6= −n + 2, n ≥ 1), as well as Qn = P

(1,β)n and

c = −1. By shifting, we get (C4,1). Indeed, by (11) and Lemma 3.1, the

MOPS Pnn≥0, Pn(x) = (−1)nPn(−x) = P(β−2,1)n (x), is Ω1,1−classical.

As we have pointed in the introduction, you can compare the proof ofthis theorem with [11]. The techniques are different as well as the length ofthis one is shorter.

In the linear space P we can introduce two operators L and Pα, whereα 6= −n+ 2, n ≥ 1, defined as follows: For any p ∈ P,

L (p) = xp′′ − 2(x− 1)p′ + (x− 2)p,

Pα(p) = (x− 1)(x+ 1)2p′′ + 2((α + 1)x+ 1− α

)(x+ 1)p′ +

α((α + 1)x+ 3− α

)p.

According to (37) and (44)

L (L(1)n ) = L

(1)n+1, n ≥ 0, (45)

Pα(P (α,1)n ) = (n+ α)(n+ α + 1)P

(α,1)n+1 , n ≥ 0. (46)

Notice that L and Pα are creation/raising operators ([4]). Applying Ω1,

(resp. Ω1,1) to both hand sides of (45) (resp. (46)) and using (36) (resp.(43)), we get

Ω1 L (L(1)n ) = (n+ 1)(n+ 2)L

(1)n , n ≥ 0, (47)

Ω1,1 Pα(P(α,1)n ) = (n+ 1)(n+ 2)(n+ α)(n+ α + 1)P

(α,1)n , n ≥ 0. (48)

The operator Ω1 L , (resp. Ω1,1 Pα) preserves the degree of polynomials

and also the orthogonality of the MPS L(1)n n≥0, (resp. P (α,1)

n n≥0). Such

13

operators are called transfer operators (see [4, 10]).Secondly, by (14) and (1), the intertwining operator S0 satisfies

S0(L(1)n ) = (n+ 1)!(x− 1)n, n ≥ 0. (49)

From (9), (10), and (14), the intertwining operator S1 satisfies

S1(P(α−2,1)n )(x) = (n+ 1)! B

(α+12

)n (x− 1), n ≥ 0, where (50)

B(α+12

)n n≥0 is the Bessel MOPS with parameter α+1

2, (see Table 1 (C3)).

Then, the following new integral relation holds

B(α+1

2)

n (x) =1

(n+ 1)!

∫ +∞

0

te−tP (α−2,1)n

(tx+1

)dt, n ≥ 0, (α 6= −n+2, n ≥ 1).

4. Some expansions in series of Laguerre polynomials

Our purpose here is to give some new expansion in series of Laguerrepolynomials. First, let us establish the following result:

Theorem 4.1. Let f(x) =∑

n≥0 anxn be a real function defined in the real

line. The following statements are equivalent.

(i) f is 1−periodic.

(ii)∑

n≥0an

(n+1)!xn =

∑n≥0

an(n+1)!

L(1)n (x).

(iii) (an)n≥0 satisfies: aν =∑

n≥0(−1)n(n+νν

)an+ν , ν ≥ 0.

Proof. (i)⇒(ii). Let g(x) =∑

n≥0an

(n+1)!xn −

∑n≥0

an(n+1)!

L(1)n (x). By

applying S0 to g and using (14) and (49), we get S0(g)(x) = f(x)−f(x−1).Since f is 1−periodic, then S0(g) = 0. So, g = 0, since S0 is one-to-one.

(ii)⇒(iii). From (1) and (ii), we get∑

n≥0an

(n+1)!xn =∑

n≥0 an∑n

ν=0(−1)n−ν(ν+1)!

(nν

)xν =

∑ν≥0

1(ν+1)!

∑n≥0(−1)nan+ν

(n+νν

)xν . By iden-

tification, we find (iii).(iii)⇒(i). From (iii), we get f(x) =

∑n≥0∑

m≥0(−1)m(m+nn

)am+n x

n =∑n≥0∑

m≥0(−1)m−n(mn

)am xn =

∑m≥0 am(x− 1)m = f(x− 1).

Examples. Let us consider the following 1−periodic functions.

• For all x ∈ R, cos(2πx) =∑

n≥0 cnxn, with

c2n+1 = 0, n ≥ 0,

c2n = (−1)n(2π)2n(2n)!

, n ≥ 0.

14

• For all x ∈ R, sin(2πx) =∑

n≥0 snxn, with

s2n = 0, n ≥ 0,

s2n+1 = (−1)n(2π)2n+1

(2n+1)!, n ≥ 0.

• For all x ∈ R such that x 6= π2

+ lπ, l ∈ Z,

tan(πx) =∑n≥0

tnxn, with

t2n = 0, n ≥ 0,

t2n+1 = (−1)nπ2n+122n+2(22n+2−1)(2n+2)!

B2n+2, n ≥ 0,

and Bk, k ≥ 0, the Bernoulli numbers satisfying∑n−1

k=0

(nk

)Bk = δn,1, n ≥ 1.

According to Theorem 4.1 (ii),

<e(

0F1(2πix, 2))

=∑n≥0

c2n(2n+ 1)!

x2n =∑n≥0

c2n(2n+ 1)!

L(1)2n (x),

=m(

0F1(2πix, 2))

=∑n≥0

s2n+1

(2n+ 2)!x2n+1 =

∑n≥0

s2n+1

(2n+ 2)!L(1)2n+1(x),

where 0F1(z, 2) =∑

n≥0zn

(n+1)!n!is the confluent hypergeometric function.

∑n≥0

t2n+1

(2n+ 2)!B2n+2 x

2n+1 =∑n≥0

t2n+1

(2n+ 2)!B2n+2 L

(1)2n+1(x).

Acknowledgements

The authors thank the referee for the careful report that contributed toimprove the presentation of the manuscript. In particular, it was very usefulto point out us the contributions [11] and [12]. The work of the second author(FM) was supported by Direccion General de Investigacion, Ministerio deCiencia e Innovacion of Spain, under grant MTM2009-12740- C03-01. Thework of the third author (RS) was supported by Faculty of Education of Girlsof Hafr Al-Batin, Saudi Arabia.

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