Numer Algor (2008) 49:251–282DOI 10.1007/s11075-008-9170-2

ORIGINAL PAPER

The symmetric Dω-semi-classical orthogonalpolynomials of class one

P. Maroni · M. Mejri

Received: 18 October 2007 / Accepted: 17 January 2008 /Published online: 20 February 2008© Springer Science + Business Media, LLC 2008

Abstract We give the system of Laguerre–Freud equations associated with theDω-semi-classical functionals of class one, where Dω is the divided differenceoperator. This system is solved in the symmetric case. There are essentially twocanonical cases. The corresponding integral representations are given.

Keywords Discrete semi-classical orthogonal polynomials ·Difference operator

Mathematics Subject Classifications (2000) 42C05 · 33C45

1 Introduction

There are many papers whose interest is semi-classical orthogonal polynomialsrelated either to the differential operator D or to the divided difference opera-tor Dω [1, 2, 6, 8, 10, 12, 14]. In [6], M. Bachène established the system satisfiedby the coefficients of the recurrence relation of D-semi-classical orthogonal

In memory of Professor Luigi Gatteschi.

P. MaroniLaboratoire Jacques-Louis Lions, Université Pierre et Marie Curie,Centre National de la Recherche Scientifique (CNRS),Boite courrier 187, 75252 Paris cedex05, Francee-mail: maroni@ann.jussieu.fr

M. Mejri (B)Institut Supérieur des Sciences Appliquées et de Technologie,Rue Omar Ibn El Khattab, Gabes 6072, Tunisiae-mail: manoubi.mejri@issatgb.rnu.tn

252 Numer Algor (2008) 49:251–282

sequences of class one; in [7], S. Belmehdi found again the same system,but those authors did not solve it. By carrying out the complete descriptionof the symmetric D-Laguerre–Hahn forms of class one, in [2], the authors,in particular, obtained the canonical cases of all symmetric D-semi-classicalforms of class one. Analogously in [8], the authors established the systemsatisfied by the coefficients of the recurrence relation of Dω-semi-classicalorthogonal sequences of class one, but their method is short of luminousness.

So, the aim of this paper is threefold. First, to establish the Laguerre–Freudequations corresponding to Dω-semi-classical orthogonal sequences of classone in a simpler manner and for the sake of completeness. Secondly to solvethe system in the symmetric case. Thus we exhaustively describe the solutionswhich arise. Thirdly to identify the integral representation of each symmetricDω-classical-form of class one by taking into account its difference equation.Finally we investigate the limiting case: ω → 0. We recover again the threecanonical cases put forth previously [2, 5].

The first section contains material of preliminary and some results regardingthe class of Dω-semi-classical forms. In the second section, the system ofLaguerre–Freud equations is built. In the third section, we give some prop-erties of symmetric Dω-semi-classical forms of class s. The fourth section isdevoted to the resolution of the system in the symmetric case. The canonicalcases are described in the fifth section. The sixth section deals with integralrepresentations of the so-called canonical forms found in the previous section.The last section is devoted to the limiting case ω = 0.

2 Preliminaries

Let P be the vector space of polynomials with coefficients in C and let P ′be its dual. We denote by 〈u, f 〉 the effect of u ∈ P ′ on f ∈ P . In particular,we denote by (u)n := 〈u, xn〉, n ≥ 0 the moments of u. For any form u, anypolynomial g, any b ∈ C, a ∈ C − {0}, let gu, τb u, hau be the forms defined byduality

〈gu, f 〉 := 〈u, gf 〉, 〈τb u, f 〉 := 〈u, τ−b f 〉, 〈hau, f 〉 := 〈u, haf 〉,

where (τ−b f )(x) = f (x + b), (haf )(x) = f (ax).We define (x − c)−1u by 〈(x − c)−1u, f 〉 := 〈u, θc f 〉 where (θc f )(x) =

f (x)− f (c)x−c , c ∈ C.We call polynomial sequence (PS), the sequence of polynomials {Pn}n≥0

when deg Pn = n, n ≥ 0. Then any polynomial Pn can be supposed monic andthe sequence becomes a monic polynomial sequence (MPS). Let {un}n≥0 be itsdual sequence defined by 〈un, Pm〉 = δn,m, n, m ≥ 0.

The MPS {Pn}n≥0 is orthogonal with respect to u ∈ P ′ when the followingconditions hold

〈u, Pm Pn〉 = rnδn,m, n, m ≥ 0, rn �= 0, n ≥ 0.

Numer Algor (2008) 49:251–282 253

Then the monic orthogonal sequence (MOPS) {Pn}n≥0 fulfils the standardrecurrence relation

P0(x) = 1 , P1(x) = x − β0,Pn+2(x) = (x − βn+1)Pn+1(x) − γn+1 Pn(x), n ≥ 0. (2.1)

In this case, the form u is said regular and it is necessarly proportional to u0.Let us introduce the Hahn’s operator

(Dω f )(x) := f (x + ω) − f (x)ω

, f ∈ P , ω �= 0.

We have Dω = 1ω

(τ−ω − IP) where IP is the identity operator in P . Thetransposed t Dω of Dω is t Dω = 1

ω(τω − IP ′) = −D−ω. Thus, we have

〈D−ωu, f 〉 = −〈u, Dω f 〉 , u ∈ P ′ , ω ∈ C − {0}.When ω → 0, we meet again the derivative D.

Lemma 2.1 [1, 11] Let a ∈ C − {0}, c ∈ C, f ∈ P and u ∈ P ′ we haveha ◦ D−ωu = aD−aω ◦ hau. (2.2)ha(

f u) = (ha−1 f

)(hau

). (2.3)

D−ω( f u) = (τω f )(D−ωu) + (D−ω f )u. (2.4)(x − c)−1((x − c)u) = u − (u)0δc. (2.5)

Definition A MOPS {Pn}n≥0 is called D−ω-semi-classical sequence if u0 fulfilsan equation

Dω(φu0) + ψu0 = 0, (2.6)where φ, ψ are polynomials, φ monic and deg ψ ≥ 1. In this case, the form u0is called Dω-semi-classical.

Let us introduce the integer s(φ, ψ) = max(deg φ − 2, deg ψ − 1). Then s =min s(φ, ψ) where the minimum is taken over all the pairs (φ, ψ) occurring in(2.6) is called the class of u0. By extension, the integer s is also the class of{Pn}n≥0 [11].

Remark The case s = 0 has been handled in [1].

Lemma 2.2 [1] Let {Pn}n≥0 be a D−ω-semi-classical sequence. Then the se-quence {P̃n}n≥0 where P̃n = a−n(ha ◦ τ−b Pn)(x) is orthogonal with respect toũ0 = (ha−1 ◦ τ−b )u0. Moreover ũ0 fulfils the equation

Dωa−1(φ̃ũ0

)+ ψ̃ ũ0 = 0,

254 Numer Algor (2008) 49:251–282

where

φ̃(x) = a− deg φφ(ax + b) , ψ̃(x) = a1−deg φψ(ax + b).

Lemma 2.3 The functional equation (2.6) is equivalent to

D−ω((φ − ωψ)u0

)+ ψu0 = 0. (2.7)

Proof Indeed, from the definition of the operator Dω the (2.6) is equivalent to

1

ω

(τω(φu0) − φu

)+ ψu0 = 0,Applying the operator τ−ω, the previous equation becomes

1

−ω(τ−ω

((φ − ωψ)u0

)− (φ − ωψ)u0)

+ ψu0 = 0,by the definition of the operator D−ω we get the desired result. �

The question is whether the integer s(φ, ψ) is the class of u0.

Proposition 2.4 (compare with [8, 10] ) The form u0 fulfiling (2.6) is of classs = s(φ, ψ) if and only if

∏

c∈ Z (φ)

(| ψ(c − ω) + (θcφ)(c − ω) | + | 〈u0, θc−ω

(ψ + θcφ

)〉 |)

> 0,

where Z (φ) := {z ∈ C, φ(z) = 0}. When there exists c ∈ Z (φ) such thatψ(c − ω) + (θcφ)(c − ω) = 0, 〈u0, θc−ω

(ψ + θcφ

)〉 = 0,equation (2.6) becomes

Dω((θcφ)u0) +{θc−ω(ψ + θcφ)

}u0 = 0. (2.8)

Proof The condition is necessary. Let c be a root of φ, we can write

φ(x) = (x − c)(θcφ)(x).Then on account of (2.4) Eq. (2.6) becomes

(x − c + ω)Dω((θcφ)u0

)+ (ψ + θcφ)u0 = 0. (2.9)Carrying out the Euclidean division of ψ + θcφ by x − c + ω, we obtain

ψ(x) + (θcφ)(x) = (x − c + ω)ψc−ω(x) + rc−ω,where

ψc−ω(x) =(θc−ω

(ψ + θcφ

))(x) , rc−ω = ψ(c − ω) + (θcφ)(c − ω).

It follows for (2.9)

(x − c + ω){Dω((θcφ)u0

)+ ψc−ωu0}+ rc−ωu0 = 0,

Numer Algor (2008) 49:251–282 255

therefore by virtue of (2.5)

Dω((θcφ)u0

)+ ψc−ωu0 = 〈u0, θc−ω(ψ + θcφ

)〉δc−ω−{ψ(c − ω) + (θcφ)(c − ω)

}(x − c + ω)−1u0. (2.10)

Now suppose 〈u0, θc−ω(ψ + θcφ

)〉 = 0 and ψ(c − ω) + (θcφ)(c − ω) = 0.Then from (2.10) u0 would be of class s̃ ≤ max

(deg(θcφ) − 2, deg(ψc−ω) −

1) = s − 1, which is contradictory.

The condition is sufficient. Let us suppose u0 to be of class s̃ ≤ s withDω(φ̃u0) + ψ̃u0 = 0.

Then there exists a polynomial such that

φ = φ̃ and ψ = (τ−ω

)ψ̃ − (Dω

)φ̃

Indeed for obtaining ψ , we can write(τ−ω

)Dω(φ̃u0

)+ (τ−ω

)ψ̃u0 = 0,

and in accordance with (2.4), we have

Dω(

φ̃u0

)+ {(τ−ω)ψ̃ −(Dω

)φ̃}u0 = 0,

hence ψ .Now suppose s̃ < s, consequently deg() ≥ 1 and there exists c such that

(c) = 0, thus (x) = (x − c)(θc)(x). This impliesψ(x) + (θcφ)(x) = (x − c + ω)

{(τ−ω(θc)

)(x)ψ̃(x) − (Dω(θc)

)(x)φ̃(x)

},

therefore rc−ω = 0 and〈u0, ψc−ω〉=〈D−ω(φ̃u0)+τω(ψ̃u0), θc〉=〈D−ω

((φ̃−ωψ̃)u0

)+ψ̃u0, θc〉=0,by virtue of Lemma 2.3. It is contradictory with the assumption. So s̃ = s. �

Remark Notice that there are misprints in the Proposition 2 given in [10],σ(x) = φ(x) − ωψ(x) must be read as σ(x) = ψ(x) − (θcφ)(x).

3 The Laguerre–Freud equations

In the sequel we assume that {Pn} is D−ω-semi-classical sequence of class one,we have

P0(x) = 1 , P1(x) = x − β0,Pn+2(x) = (x − βn+1)Pn+1(x) − γn+1 Pn(x) , n ≥ 0. (3.1)

Dω(φu0) + ψu0 = 0, (3.2)

256 Numer Algor (2008) 49:251–282

with

φ(x) = b 3x3 + b 2x2 + b 1x + b 0 , ψ(x) = a2x2 + a1x + a0. (3.3)

Let

In,k(ω) = 〈u0, xk Pn(x)Pn(x − ω)〉, 0 ≤ k ≤ 3, (3.4)Jn,k(ω) = 〈u0, xk Pn+1(x)Pn(x − ω)〉, 0 ≤ k ≤ 3, (3.5)

Kn,k(ω) = 〈u0, xk Pn(x)Pn+1(x − ω)〉, 0 ≤ k ≤ 3. (3.6)

Lemma 3.1 For n ≥ 0, we have the following results:

b 3(In,3(ω) − In,3(−ω)

)+ b 2(In,2(ω) − In,2(−ω)

)

+ b 1(In,1(ω) − In,1(−ω)

)+ b 0(In,0(ω)− In,0(−ω)

)

+ ωa2 In,2(−ω) + ωa1 In,1(−ω)+ωa0 In,0(−ω)=0. (3.7)b 3(Kn,3(ω) − Kn,3(−ω) + Jn,3(ω) − Jn,3(−ω)

)+ b 2(Kn,2(ω) − Kn,2(−ω)

)

+ b 2(Jn,2(ω) − Jn,2(−ω)

)+ b 1(Kn,1(ω) − Kn,1(−ω)

+ Jn,1(ω) − Jn,1(−ω))+ b 0

(Kn,0(ω) − Kn,0(−ω)

+ Jn,0(ω) − Jn,0(−ω))+ ωa2

(Jn,2(−ω) + Kn,2(−ω)

)

+ ωa1(Jn,1(−ω) + Kn,1(−ω)

)+ ωa0(Jn,0(−ω) + Kn,0(−ω)

) = 0. (3.8)

Proof By (3.2), we get 〈Dω(φu0) + ψu0, Pn(x)Pn(x + ω)〉 = 0, n ≥ 0 it isequivalent to

〈φu0, Pn(x)Pn(x − ω)〉 + 〈(ωψ − φ)u0, Pn(x)Pn(x + ω)〉 = 0, n ≥ 0,

then we can deduce (3.7).We have 〈Dω(φu0)+ψu0, Pn(x)Pn+1(x+ω)+Pn(x + ω)Pn+1(x)〉 = 0, n ≥ 0,

then

〈φu0, Pn(x − ω)Pn+1(x)〉 + 〈(ωψ − φ)u0, Pn(x)Pn+1(x + ω)〉+〈φu0, Pn(x)Pn+1(x − ω)〉 + 〈(ωψ − φ)u0, Pn+1(x)Pn(x + ω)〉 = 0, n ≥ 0,

it follows (3.8). �

Numer Algor (2008) 49:251–282 257

In order to determine {In,k(ω)}n≥0, {Jn,k(ω)}n≥0, {Kn,k(ω)}n≥0 , 0 ≤ k ≤ 3, weneed the following results:

Lemma 3.2 We have the following formulas:

(u0)1 = β0.(u0)2 = γ1 + β20 .(u0)3 = β30 + (2β0 + β1)γ1.

Lemma 3.3 Let {an}n≥0 with an �= 0 , n ≥ 0, {b n}n≥0 two sequences and {xn}n≥0the sequence satisfying the recurrence relation:

xn+1 = anxn + b n , n ≥ 0 , x0 = a ∈ C − {0}.

We have

xn+1 =n∏

k=0ak

⎧⎪⎨

⎪⎩a +

n∑

k=0

⎛

⎝k∏

μ=0aμ

⎞

⎠

−1

b k

⎫⎪⎬

⎪⎭, n ≥ 0.

Lemma 3.4 Let {an}n≥0 be a sequence we have the following formulas:n∑

ν=0

ν∑

μ=0aμ = (n + 1)

n∑

ν=0aν −

n∑

ν=0νaν, n ≥ 0. (3.9)

n∑

ν=0ν

ν∑

μ=0aμ = 1

2n(n + 1)

n∑

ν=0aν − 1

2

n∑

ν=0ν(ν − 1)aν, n ≥ 0. (3.10)

Lemma 3.5 We have

In,0(ω) = 〈u0, P2n〉, n ≥ 0. (3.11)In,1(ω) = (βn − nω)〈u0, P2n〉, n ≥ 0. (3.12)I0,2(ω) = γ1 + β20 . (3.13)

In,2(ω) ={

γn + γn+1 + β2n − nωβn − ωn−1∑

ν=0βν + 1

2n(n − 1)ω2

}〈u0, P2n

〉,

n ≥ 1. (3.14)I0,3(ω) =

(2β0 + β1

)γ1 + β30 . (3.15)

258 Numer Algor (2008) 49:251–282

In,3(ω) ={

−2ωn−1∑

ν=0γν+1 +

(βn−1 + 2βn − (n − 2)ω

)γn

+ (2βn + βn+1 − nω)γn+1 − ω

n−1∑

ν=0β2ν + (n − 1)ω2

n−1∑

ν=0βν

− ωβnn∑

ν=0βν + β3n − (n − 1)ωβ2n +

1

2n(n − 1)ω2βn

− 16

n(n − 1)(n − 2)ω3}〈u0, P2n

〉, n ≥ 1. (3.16)

Jn,0(ω) = 0, n ≥ 0. (3.17)Jn,1(ω) = γn+1

〈u0, P2n

〉, n ≥ 0. (3.18)

Jn,2(ω) = (βn + βn+1 − nω) γn+1〈u0, P2n

〉, n ≥ 0. (3.19)

J0,3(ω) = γ1{γ1 + γ2 + β20 + β0β1 + β21

}. (3.20)

Jn,3(ω) ={

γn + γn+1 + γn+2 + βn(βn − nω) + βn+1(βn + βn+1 − nω)

−ωn−1∑

ν=0βν + 1

2n(n − 1)ω2

}

γn+1〈u0, P2n

〉, n ≥ 1. (3.21)

Kn,0(ω) = −(n + 1)ω〈u0, P2n

〉, n ≥ 0. (3.22)

Kn,1(ω) ={

γn+1 − ωn∑

ν=0βν + 1

2n(n + 1)ω2

}〈u0, P2n

〉, n ≥ 0. (3.23)

K0,2(ω) = (β0 + β1 − ω)γ1 − ωβ20 , (3.24)

Kn,2(ω) ={

−2ωn−1∑

ν=0γν+1 +

(βn + βn+1 − (n + 1)ω)γn+1 − ω

n∑

ν=0β2ν

+nω2n∑

ν=0βν − 1

6n(n2 − 1)ω3

}〈u0, P2n

〉, n ≥ 1. (3.25)

K0,3(ω) − K0,3(−ω) = −2ω(β30 + (2β0 + β1)γ1

), (3.26)

Kn,3(ω) − Kn,3(−ω) = −2ω{

γn+1

(

(n + 1)(βn + βn+1) +n∑

ν=0βν

)

+n∑

ν=0β3ν + 3

n−1∑

ν=0γν+1(βν +βν+1)

+12

n(n−1)ω2n∑

ν=0βν

}〈u0, P2n

〉, n≥1. (3.27)

Numer Algor (2008) 49:251–282 259

Proof From the orthogonality of {Pn}n≥0, we can deduce (3.11), (3.17) and(3.18). We have I0,1(ω) = (u0)1 by the Lemma 3.2, we get

I0,1(ω) = β0. (3.28)For n ≥ 0, by (3.1) we have

In+1,1(ω) = 〈u0, {Pn+2(x) + βn+1 Pn+1(x) + γn+1 Pn(x)}Pn+1(x − ω)〉 ,taking the orthogonality of {Pn}n≥0 into account, we can deduce that

In+1,1(ω) = βn+1〈u0, P2n+1

〉+ γn+1 Kn,0(ω), n ≥ 0. (3.29)By (3.1) and the orthogonality of {Pn}n≥0, we have

Kn,0(ω) = In,1(ω) − (ω + βn)〈u0, P2n

〉, n ≥ 0. (3.30)

By virtue of the last relation, (3.29) can be written

In+1,1(ω) = γn+1 In,1(ω) +(βn+1 − βn − ω

) 〈u0, P2n+1

〉, n ≥ 0,

consequently from the Lemma 3.3 and (3.28), we can deduce (3.12).We remark that by (3.12), (3.30) give (3.22).We have I0,2(ω) =

〈u0, x2

〉 = (u0)2, from the Lemma 3.3, we get (3.13).For n ≥ 0, by (3.1), we can write

In+1,2(ω) = 〈u0, x{Pn+2(x) + βn+1 Pn+1(x) + γn+1 Pn(x)}Pn+1(x − ω)〉 ,by the orthogonality of {Pn}n≥0, we obtain

In+1,2(ω) =〈u0, P2n+2

〉+ βn+1 In+1,1(ω) + γn+1 Kn,1(ω), n ≥ 0. (3.31)We can write K0,1(ω) = (u0)2 − (ω + β0)(u0)1, by the Lemma 3.3, we obtain

K0,1(ω) = γ1 − ωβ0. (3.32)Making n = 0 in (3.31), by (3.12) and (3.32), it follows that

I1,2(ω) = γ1{γ1 + γ2 + β21 − ω(β0 + β1)

}. (3.33)

When n ≥ 1, by (3.1) and the orthogonality of {Pn}n≥0, we getKn,1(ω) = In,2(ω) − (ω + βn)In,1(ω) − γn

〈u0, P2n

〉,

by (3.12), we can deduce that

Kn,1(ω) = In,2(ω) +{(nω − βn)(ω + βn) − γn

} 〈u0, P2n

〉, n ≥ 1. (3.34)

By virtue of (3.34), (3.31) becomes

In+1,2(ω) = γn+1 In,2(ω) +{β2n+1 − (n + 1)ωβn+1 − β2n + (n − 1)ωβn+ nω2 + γn+2 − γn

} 〈u0, P2n+1

〉, n ≥ 1,

by (3.33) and the Lemma 3.3, we can deduce (3.14).We can remark that by the relation (3.14), (3.34) and (3.32) give (3.23).We have I0,3(ω) =

〈u0, x3

〉 = (u0)3, from the Lemma 3.3, we get (3.15).

260 Numer Algor (2008) 49:251–282

When n ≥ 0, from (3.1) we have

In+1,3(ω) =〈u0, x2{Pn+2(x) + βn+1 Pn+1(x) + γn+1 Pn(x)}Pn+1(x − ω)

〉,

taking the orthogonality of {Pn}n≥0 into account, we can deduce that

In+1,3(ω) = Jn+1,2(ω) + βn+1 In+1,2(ω) + γn+1 Kn,2(ω), n ≥ 0. (3.35)

By (3.1), Jn,2(ω) = βn+1〈u0, P2n+1

〉+ γn+1 In,1(ω), n ≥ 0, by (3.12) we get (3.19).On the other hand K0,2(ω) = (u0)3 − (ω + β0)(u0)2, by the Lemma 3.3 we get(3.24).

Making n = 0 in (3.35) and taking (3.14), (3.19), (3.24) into account, weobtain

I1,3(ω) = γ1{(β0 + 2β1 − ω)γ1 + (2β1 + β2 − ω)γ2 − ωβ0β1 − ωβ20 − ωβ21 + β31

}.

(3.36)

From (3.1) and the orthogonality of {Pn}n≥0, it follows that

Kn,2(ω) = In,3(ω) − (ω + βn)In,2(ω) − γn Jn−1,2(ω), n ≥ 1. (3.37)

On account of (3.14), (3.19), (3.35), (3.37) can be written

In+1,3(ω) = γn+1 In,3(ω)

+ 〈u0, P2n+1〉{γn+2

(2βn+1 + βn+2 − (n + 1)ω

)

− γn+1(2βn + βn+1 − nω

)

+ γn+1(βn + 2βn+1 − (n + 1)ω

)

− γn(βn−1 + 2βn − nω

)− 2ωγn + β3n+1 − β3n− (n + 1)ωβ2n+1 + nωβ2n − ωβ2n + nω2βn+ 1

2n(n + 1)ω2βn+1 − 1

2n(n − 1)ω2βn

− ωβn+1n∑

ν=0βν + ωβn

n−1∑

ν=0βν + ω2

n−1∑

ν=0βν

− 12

n(n − 1)ω3}, n ≥ 1,

Numer Algor (2008) 49:251–282 261

taking the Lemma 3.3 into account, we can deduce that

In+1,3(ω) =〈u0, P2n+1

〉

×{

γn+1(βn + 2βn+1 − (n + 1)ω

)+ γn+2(2βn+1 + βn+2 − (n + 1)ω

)

− 2ωn∑

ν=0γν + β3n+1 − (n + 1)ωβ2n+1 − ω

n∑

ν=0β2ν

+12

n(n + 1)ω2βn+1 + ω2n∑

ν=0νβν − ωβn+1

n∑

ν=0βν

+ω2n−1∑

ν=0

ν∑

μ=0βμ − 1

6n(n2 − 1)ω3

⎫⎬

⎭, n ≥ 1,

by virtue of (3.9) and (3.36), we get (3.16).

By (3.14), (3.18) and (3.16) the relation (3.37) give (3.25).

Let n ≥ 0, by (3.1) and the orthogonality of {Pn}n≥0, we get

Jn,3(ω) =〈u0, P2n+2

〉+ βn+1 Jn,2(ω) + γn+1 In,2(ω)

by virtue of (3.19), (3.13) and (3.14), we obtain (3.20) and (3.21).We have K0,3(ω) − K0,3(−ω) = −2ω(u0)3, from the Lemma 3.3 , we obtain

(3.26).When n ≥ 0 by (3.1) and the orthogonality of {Pn}n≥0 we can write

Kn+1,3(ω)= In+2,2(ω)+βn+1 Kn+1,2(ω)+γn+1〈u0, x2 Pn(x)Pn+2(x−ω)

〉, n≥0,

(3.38)

but, by (3.1) and the orthogonality of {Pn}n≥0, we get〈u0, x2 Pn(x)Pn+2(x − ω)

〉 = Kn,3(ω) − (ω + βn+1)Kn,2(ω) − γn+1 In,2(ω), n ≥ 0,

then (3.38) becomes

Kn+1,3(ω) = γn+1 Kn,3(ω) + In+2,2(ω) − γ 2n+1 In,2(ω) + βn+1 Kn+1,2(ω)−(ω + βn+1)γn+1 Kn,2(ω), n ≥ 0, (3.39)

on account of (3.13), (3.14), (3.24–3.26) and (3.39) it follows that

K1,3(ω) − K1,3(−ω) = −2ωγ1{β30 + β31 + 3γ1(β0 + β1) + γ2(β0 + 3β1 + 2β2)},

262 Numer Algor (2008) 49:251–282

and

Kn+1,3(ω) − Kn+1,3(−ω) = γn+1(Kn,3(ω) − Kn,3(−ω))

+ 2ω 〈u0, P2n+1〉{

−γn+2{

(n + 3)βn+1 + (n + 2)βn+2 +n∑

ν=0βν

}

+ γn+1{

(n + 2)βn + (n + 1)βn+1 +n−1∑

ν=0βν

}

− 3γn+1(βn + βn+1) − β3n+1 −1

2ω2n(n + 1)βn+1

− nω2n∑

ν=0βν

}

, n ≥ 1,

on account of the Lemma 3.3 and (3.26), we obtain

Kn+1,3(ω) − Kn+1,3(−ω)

= 2ω 〈u0, P2n+1〉×

{

−γn+2{

(n + 3)βn+1 + (n + 2)βn+2 +n+1∑

ν=0βν

}

−3n+1∑

ν=1γν(βν−1 + βν) − 1

2ω2

n+1∑

ν=0ν(ν − 1)βν

−ω2n∑

ν=0ν

ν∑

μ=0βμ −

n+1∑

ν=0β3ν

⎫⎬

⎭, n ≥ 0,

by (3.10), we obtain (3.27). �

Proposition 3.6 We have the following system

a2γ1 = −ψ(β0). (3.40)

(a2 − 2nb 3

)(γn + γn+1) − 4b 3

n−2∑

ν=0γν+1 = −ψ(βn) +

n−1∑

ν=0

(θβn(2φ − ωψ)

)(βν)

+ 13

n(n − 1)(n − 2)ω2b 3 − 12

n(n − 1)ω2a2, n ≥ 1, (3.41)

Numer Algor (2008) 49:251–282 263

with−1∑

ν=0= 0.

{2(a2 − b 3)β1 − 2(2b 3 − a2)β0 − 2b 2 + ωa2 + 2a1

}γ1 = (2φ − ωψ)(β0).

(3.42)

nγn+1 − 2(2b 2 − ωa2)n−1∑

ν=0γν+1 − 6b 3

n−1∑

ν=0γν+1(βν + βν+1)

=n∑

ν=0(2φ − ωψ)(βν) + ω2n

((n − 1)b 3 − a2

) n∑

ν=0βν

− 12

n(n + 1)ω2a1 + 16

n(n2 − 1)ω2(2b 2 − ωa2), n ≥ 1, (3.43)

where

n = 2(a2 − (2n + 1)b 3

)βn+1 + 2

(a2 − 2nb 3

)βn − 4b 3

n∑

ν=0βν

−(2n + 1)(2b 2 − ωa2) + 2a1, n ≥ 1. (3.44)

Proof By virtue of the previous lemma, we can deduce that

In,0(ω) − In,0(−ω) = 0, n ≥ 0. (3.45)In,1(ω) − In,1(−ω) = −2nω

〈u0, P2n

〉, n ≥ 0. (3.46)

I0,2(ω) − I0,2(−ω) = 0. (3.47)

In,2(ω) − In,2(−ω) = −2ω(

nβn +n−1∑

ν=0βν

)〈u0, P2n

〉, n ≥ 1. (3.48)

I0,3(ω) − I0,3(−ω) = 0, (3.49)

In,3(ω)− In,3(−ω) = −2ω{

2n−1∑

ν=0γν+1+(n−2)γn+nγn+1+

n−1∑

ν=0β2ν +βn

n∑

ν=0βν

+(n − 1)β2n +1

6n(n − 1)(n − 2)ω2

}

× 〈u0, P2n〉, n ≥ 1. (3.50)

Jn,0(ω) − Jn,0(−ω) = 0, n ≥ 0. (3.51)Jn,1(ω) − Jn,1(−ω) = 0, n ≥ 0. (3.52)Jn,2(ω) − Jn,2(−ω) = −2nωγn+1

〈u0, P2n

〉, n ≥ 0, (3.53)

264 Numer Algor (2008) 49:251–282

J0,3(ω) − J0,3(−ω) = 0, (3.54)

Jn,3(ω)− Jn,3(−ω) =−2ω{

nβn+nβn+1+n−1∑

ν=0βν

}

γn+1〈u0, P2n

〉, n≥1. (3.55)

Kn,0(ω) − Kn,0(−ω) = −2(n + 1)ω〈u0, P2n

〉, n ≥ 0. (3.56)

Kn,1(ω) − Kn,1(−ω) = −2ω〈u0, P2n

〉 n∑

ν=0βν, n ≥ 0. (3.57)

K0,2(ω) − K0,2(−ω) = −2ω(β20 + γ1). (3.58)

Kn,2(ω) − Kn,2(−ω) = −2ω〈u0, P2n

〉{

2n−1∑

ν=0γν+1 + (n + 1)γn+1 +

n∑

ν=0β2ν

+ 16

n(n2 − 1)ω2

}

, n ≥ 1. (3.59)

Making n = 0 in (3.7) and taking the relations (3.11)–(3.13), (3.45)–(3.47),(3.49) into account, we can deduce (3.40).

Let n ≥ 1, by virtue of the relations (3.11), (3.12), (3.14), (3.45), (3.46), (3.48)and (3.50), the equation (3.7) becomes

−2b 3{

2n−1∑

ν=0γν+1 + (n − 2)γn + nγn+1 +

n−1∑

ν=0β2ν + βn

n∑

ν=0βν + (n − 1)β2n

+16

n(n − 1)(n − 2)ω2}

−2b 2(

nβn +n−1∑

ν=0βν

)

− 2nb 1 + a2{

γn + γn+1 + β2n + nωβn + ωn−1∑

ν=0βν

+12

n(n − 1)ω2}

+ a1(βn + nω) + a0 = 0,

it is equivalent to

(a2 − 2nb 3

)(γn + γn+1) − 4b 3

n−2∑

ν=0γν+1 = −a2β2n − a1βn − a0

+n−1∑

ν=0

{2b 3

(β2ν + βνβn + β2n

)+ (2b 2 − ωa2)(βν + βn) + 2b 1 − ωa1}

× 13

n(n − 1)(n − 2)ω2b 3 − 12

n(n − 1)a2ω2,

Numer Algor (2008) 49:251–282 265

but(θβn(2φ − ωψ)

)(βν) = 2b 3

(β2ν + βνβn + β2n

)+ (2b 2 − ωa2)(βν + βn)+2b 1 − ωa1,

then we can deduce (3.41).Let n = 0 in (3.8), by virtue of (3.17)–(3.19), (3.22)–(3.24), (3.26), (3.51)–

(3.58), we get (3.42).When n ≥ 1, on account of (3.17)–(3.19), (3.22), (3.23), (3.25), (3.27), (3.51)–

(3.53), (3.55)–(3.57) and (3.59), (3.8) becomes

−2b 3{

(2n + 1)(βn + βn+1)γn+1 + 2γn+1n∑

ν=0βν − γn+1βn +

n∑

ν=0β3ν

+3n−1∑

ν=0γν+1(βν + βν+1 + 1

2n(n − 1)ω2

n∑

ν=0βν

}

−2b 2{

2n−1∑

ν=0γν+1 + (2n + 1)γn+1 +

n∑

ν=0β2ν +

1

6n(n2 − 1)ω2

}

− 2b 1n∑

ν=0βν

−2b 0(n + 1) + a2{(2βn + 2βn+1 + (2n + 1)ω

)γn+1 + 2ω

n−1∑

ν=0γν+1 + nω2

n∑

ν=0β2ν

+16

n(n2 − 1)ω3}

a1

{

2γn+1 + ωn∑

ν=0βν + 1

2n(n + 1)ω2

}

+ ωa0(n + 1) = 0,

this leads to{

2(a2 − (2n + 1)b 3

)βn+1 + 2

(a2 − 2nb 3

)βn − 4b 3

n∑

ν=0βν

− (2n + 1)(2b 2 − ωa2) + 2a1}

γn+1 − 6b 3n−1∑

ν=0γν+1(βν + βν+1)

− 2(2b 2 − ωa2)n−1∑

ν=0γν+1 =

n∑

ν=0

{2b 3β3ν + (2b 2 − ωa2)β2ν + (2b 1 − ωa1)βν

+ 2b 0−ωa0}+ n((n−1)b 3−a2

)ω2

n∑

ν=0βν

− 12

n(n+1)a1ω2+ 16

n(n2−1)(2b 2−ωa2),

then we can deduce (3.43). �

266 Numer Algor (2008) 49:251–282

4 Some properties of the symmetric Dω-semi-classical forms of class s

In the sequel we assume that u0 is a symmetric Dω-semi-classical formsatisfying

Dω(φu0) + ψu0 = 0. (4.1)

Lemma 4.1 The following functional equations hold:

Dω((

φ(−x) − ωψ(−x))u0)

− ψ(−x)u0 = 0, (4.2)

Dω((

φ(x) + φ(−x) − ωψ(−x))u0)

+ (ψ(x) − ψ(−x))u0 = 0, (4.3)

Dω((

φ(x) − φ(−x) + ωψ(−x))u0)

+ (ψ(x) + ψ(−x))u0 = 0. (4.4)

Proof On account of the Lemma 2.3, the equation (4.1) is equivalent to

D−ω(φ − ωψ)u0

)+ ψu0 = 0.Applying the operator h−1 to the previous functional equation we get

h−1 ◦ D−ω((

φ(x) − ωψ(x))u0)

+ h−1(ψ(x)u0) = 0,

taking (2.2) and (2.3) into account we obtain

Dω((

φ(−x) − ωψ(−x))h−1u0)

− ψ(−x)h−1u0 = 0,but u is a symmetric form then h−1u0 = u0, the previous equation give (4.2).

The equation (4.3) is obtained by adding the equations (4.1) and (4.2).Subtracting both sides of the equations (4.1) and (4.2) we obtain (4.4). �

Proposition 4.2 Let s̃ be the class of u0 we have the following results:

a) If s̃ is odd then 2φ − ωψ is odd and ψ is even.b) If s̃ is even then 2φ − ωψ is even and ψ is odd.

Proof We can write

φ(x) = φe(x2) + xφo(x2) , ψ(x) = ψe(x2) + xψo(x2). (4.5)Let

deg(φe) = t1, deg(φo) = t2, deg(ψe) = p1, deg(ψo) = p2, (4.6)by the definition of s̃ we can deduce that

max(2t1, 2t2 + 1) ≤ s̃ + 2, max(2p1, 2p2 + 1) ≤ s̃ + 1. (4.7)Taking (4.5) into account, the equations (4.3) and (4.4) become respectively

Dω(φ0(x)u0

)+ ψ0(x)u0 = 0, (4.8)Dω(φ1(x)u0

)+ ψ1(x)u0 = 0, (4.9)

Numer Algor (2008) 49:251–282 267

where

φ0(x) = 2φe(x2)− ωψe(x2)+ ωxψo(x2), ψ0(x) = 2xψo

(x2), (4.10)

φ1(x) = 2xφo(x2)+ ωψe(x2)− ωxψo(x2), ψ1(x) = 2ψe

(x2). (4.11)

a) In this case we can write s̃ = 2s + 1, from (4.7) we gett1 ≤ s + 1, t2 ≤ s + 1, p1 ≤ s + 1, p2 ≤ s. (4.12)

On account of (4.12), it follows

deg(φ0) − 2 ≤ 2s < s̃, deg(ψ0) − 1 ≤ 2s < s̃. (4.13)By virtue of the definition of s̃ , from (4.8) we can deduce that

φ0(x) = 0 , ψ0(x) = 0. (4.14)then we get

2xψo(x2) = 0 , 2φe(x2)− ωψe(x2)− ωxψo(x2) = 0,

we can deduce that

ψo(x) = 0 , 2φe(x) − ωψe(x) = 0. (4.15)By the previous relations we can deduce that 2φ − ωψ is odd and ψ iseven. Hence the desired results.

b) We have s̃ = 2s from (4.7), it followst1 ≤ s + 1, t2 ≤ s, p1 ≤ s, p2 ≤ s. (4.16)

By virtue of the last relations, we obtain

deg(φ1) − 2 ≤ 2s < s̃, deg(ψ1) − 1 ≤ 2s < s̃. (4.17)Taking the definition of s̃ and (4.17) and (4.9) into account, we get

φ1(x) = 0 , ψ1(x) = 0,which means that

2xψe(x2) = 0 , 2xφo(x2)+ ωψe(x2)− ωxψo(x2) = 0,

then

ψe(x) = 0 , 2φo(x) − ωψo(x) = 0. (4.18)On account of the previous relations, we can deduce the desired results.

�

5 The symmetric case when s = 1

In the sequel we assume that {Wn}n≥0 is a symmetric D−ω-semi-classicalorthogonal sequence of class one and {wn}n≥0 its dual sequence.

268 Numer Algor (2008) 49:251–282

Then we have

W0(x) = 1 , W1(x) = x,Wn+2(x) = xWn+1(x) − γn+1Wn(x) , n ≥ 0. (5.1)

By virtue of the Proposition 4.2, it follows that

Dω(φw0) + ψw0 = 0, (5.2)with

φ(x) = b 3x3 + 12ωa2x2 + b 1x + 1

2ωa0 , ψ(x) = a2x2 + a0. (5.3)

In this case the system (3.40)–(3.43) becomes

a2γ1 = −a0. (5.4)(a2 − 2nb 3

)(γn + γn+1) − 4b 3

n−2∑

ν=0γν+1 = −a0 + 2nb 1 + 1

3n(n − 1)(n − 2)ω2b 3

−12

n(n − 1)ω2a2 , n ≥ 1. (5.5)

Let

Tn =n∑

ν=0γν+1, n ≥ 0. (5.6)

Then

Tn − Tn−2 = γn + γn+1, n ≥ 1 , T−1 = 0, (5.7)Tn − Tn−1 = γn+1, n ≥ 0. (5.8)

Taking the relations (5.6), (5.7) and (5.8) into account, the system (5.4)–(5.5)becomes

T0 = −a0a2 , (5.9)(a2 − 2nb 3

)(Tn − Tn−2) − 4b 3Tn−2 = − a0 + 2nb 1 − 1

2n(n − 1)ω2a2

+13

n(n−1)(n−2)ω2b 3, n≥1. (5.10)

We need the following result:

Lemma 5.1 Let

Sn(m) =n∑

ν=1νm , n ≥ 1 , m ≥ 0. (5.11)

Numer Algor (2008) 49:251–282 269

We have

Sn(1) = 12

n(n + 1) , n ≥ 1 ; Sn(2) = 16

n(n + 1)(2n + 1) , n ≥ 1;

Sn(3) = 14

n2(n + 1)2 , n ≥ 1.

Proposition 5.2 We have

T2n = (n + 1)a0−2nb 1+16 n(4n−1)ω2a2− 23 n2(n−1)ω2b 3

4nb 3−a2 , n≥0. (5.12)

T2n+1 = (n + 1)a0 − 2(n + 1)b 1 +16 n(4n + 5)ω2a2 − 13 n

(2n2 + 2n − 1)ω2b 3

2(2n + 1)b 3 − a2 ,

n ≥ 0. (5.13)

Proof Making n = 1 in (5.10), we obtain(a2 − 2b 3)T1 = 2b 1 − a0. (5.14)

The equation (5.10) can be written

(2nb 3 − a2

)Tn −

(2(n − 2)b 3 − a2

)Tn−2 = a0 − 2nb 1 + 1

2n(n − 1)ω2a2

−13

n(n − 1)(n − 2)ω2b 3 , n ≥ 1.(5.15)

From the previous equation, we get(4nb 3 − a2

)T2n − (4(n − 1)b 3 − a2) T2n−2 = a0 − 4nb 1 + n(2n − 1)ω2a2

−43

n(n − 1)(2n − 1)ω2b 3 , n ≥ 1,

so we obtain

(4nb 3 − a2

)T2n + a2T0 =

n∑

ν=1

{a0 − 4νb 1 + ν(2ν − 1)ω2a2

− 43ν(ν − 1)(2ν − 1)ω2b 3

}

= na0 − 4b 1Sn(1) +(2Sn(2) − Sn(1)

)ω2a2

−43

{2Sn(3) − 3Sn(2) + Sn(1)

}ω2b 3 , n ≥ 1.

270 Numer Algor (2008) 49:251–282

From the relation (5.9) we have a2T0 = −a0, then we get(4nb 3 − a2

)T2n = (n + 1)a0 − 4b 1Sn(1) +

(2Sn(2) − Sn(1)

)ω2a2

−43

{2Sn(3) − 3Sn(2) + Sn(1)} ω2b 3, n ≥ 1.

Taking the Lemma 5.1 into account, we obtain

T2n = (n + 1)a0 − 2nb 1 +16 n(4n − 1)ω2a2 − 23 n2(n − 1)ω2b 3

4nb 3 − a2 , n ≥ 1,

by (5.9), we can deduce that the last relation is valid for n = 0. Hence (5.12).Likewise, making n → 2n + 1 in (5.15), we get

(2(2n + 1)b 3 − a2

)T2n+1 −

(2(2n − 1)b 3 − a2

)T2n−1 = a0 − 2(2n + 1)b 1

+ n(2n + 1)ω2a2 − 23

n(4n2 − 1)ω2b 3 , n ≥ 1,

it follows that

(2(2n + 1)b 3 − a2)T2n+1 −(2b 3 − a2

)T1

=n∑

ν=1

{a0 − 2(2ν + 1)b 1 + ν(2ν + 1)ω2a2 − 2

3ν(4ν2 − 1)ω2b 3

}

= na0 −(4Sn(1) + 2n

)b 1 +

(2Sn(2) + Sn(1)

)ω2a2

−23

(4Sn(3) − Sn(1)

)ω2b 3 , n ≥ 1.

Using (5.14), the last equation can be written

(2(2n + 1)b 3 − a2

)T2n+1 = (n + 1)a0 −

(4Sn(1) + 2n + 2

)b 1

+(2Sn(2) + Sn(1))ω2a2

−23

(4Sn(3) − Sn(1)

)ω2b 3 , n ≥ 1.

On account of the Lemma 5.1, we obtain

T2n+1 = (n + 1)a0 − 2(n + 1)b 1 +16 n(4n + 5)ω2a2 − 13 n

(2n2 + 2n − 1)ω2b 3

2(2n + 1)b 3 − a2 ,n ≥ 1,

by (5.14), we see that the last equation is valid for n = 0. Hence (5.13). �

Numer Algor (2008) 49:251–282 271

Corollary 5.3 The sequence {γn+1}n≥0 is defined by

γ2n+1 =(

2(n − 1)b 3 − a2)(

a0 − 2nb 1 − n2(2nb 3 − a2)ω2)

(4nb 3 − a2

)((4n − 2)b 3 − a2

) , n ≥ 0, (5.16)

γ2n+2 =(n+1)

(−2a0b 3+2(a2−2nb 3)b 1−n(2nb 3−a2)2ω2

)

(4nb 3−a2

)((4n+2)b 3−a2

) , n ≥ 0. (5.17)

Proof From (5.6) we get γ2n+1 = T2n − T2n−1, n ≥ 0. By the Proposition 5.2,we get

(4nb 3 − a2

)((4n − 2)b 2 − a2

)γ2n+1

= (n + 1)((4n − 2)b 2 − a2){

a0 − 2nb 1 + 23

n2(a2 − nb 3)ω2

+ 16

n(4nb 3 − a2

)ω2}

− n(4nb 3 − a2){

a0 − 2nb 1 + 23

n2(a2 − nb 3)ω2

− 16

((−8n2 + 2n + 2)b 3 + (3n + 1)a2)ω2}

= (2(n − 1)b 3 − a2){

a0 − 2nb 1 + 23

n2(a2 − nb 3)ω2}

−13

n2(4nb 3 − a2

)(2(n − 1)b 3 − a2

), n ≥ 0,

then we can deduce (5.16).On account of (5.6), we have γ2n+2 = T2n+1 − T2n, n ≥ 0, by (5.12) and

(5.13) we obtain

(4nb 3 − a2

)(((4n + 2)b 3 − a2

)γ2n+2

= (n + 1)(4nb 3 − a2){

a0 − 2(n + 1)b 1 + 16

n(4n + 5)ω2a2

− 13

n(2n2 + 2n − 1)ω2b 3

}

−(n + 1)((4n + 2)b 3 − a2){

a0 − 2nb 1 + 16

n(4n − 1)ω2a2

− 23

n2(n − 1)ω2b 3}

272 Numer Algor (2008) 49:251–282

= (n + 1){

− 2a0b 3 + 2(a2 − 2nb 3

)b 1 + 1

3n((8n + 1)b 3 − 3a2

)ω2a2

− 13

n(12n2b 3 + (−4n + 1)a2

)ω2b 3

}, n ≥ 0,

this is(4nb 3 − a2

)(((4n + 2)b 3 − a2

)γ2n+2

= (n + 1) {−2a0b 3 + 2(a2 − 2nb 3

)b 1 − n

(a22 − 4na2b 3 + 4n2b 23

)}, n ≥ 0.

Hence (5.17). �

6 The canonical cases

Before quoting the different canonical situations, let us proceed to the generaltransformation see the Lemma 2.2

W̃n = A−nWn(Ax) , n ≥ 0. (6.1)γ̃n+1 = A−2γn+1, n ≥ 0. (6.2)

The form w̃0 = hA−1w0 fulfilsD ω

A

(A− deg φφ(Ax)w̃0

)+ A1−deg φψ(Ax)w̃0 = 0. (6.3)Any so-called canonical case will be denoted by γ̂n+1, ŵ0.On account of (5.3) and the Corollary 5.3, we get the general situation⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ2n+1 =(2(n − 1)b 3 − a2

)(a0 − 2nb 1 − n2(2nb 3 − a2)ω2

)

(4nb 3 − a2

)((4n − 2)b 3 − a2

) , n ≥ 0,

γ2n+2 =(n + 1)

(−2a0b 3 + 2

(a2 − 2nb 3

)b 1 − n(2nb 3 − a2)2ω2

)

(4nb 3 − a2

)((4n + 2)b 3 − a2

) , n ≥ 0,

Dω((

b 3x3 + 12ωa2x2 + b 1x + 12ωa0)w0

)+ (a2x2 + a0)w0 = 0, (w0)1 = 0.

(6.4)

Theorem 6.1 The following canonical cases arise:

a) When deg φ̂ = 2, we have⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

γ̂2n+1 = (n + α)(n + β), n ≥ 0,γ̂2n+2 = (n + 1)(n + α + β), n ≥ 0,D−i

((x + iα)(x + iβ)ŵ0(α, β)

)+ 2i(x2 − αβ)ŵ0(α, β) = 0,(

ŵ0(α, β))

1 = 0.

(6.5)

Numer Algor (2008) 49:251–282 273

b) The case when deg φ̂ = 3, we obtain the canonical case below⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ̂2n+1 = (n + τ)(n + μ)(n + δ)(n + τ + μ + δ − 1)(2n + τ + μ + δ − 1)(2n + τ + μ + δ) , n ≥ 0.

γ̂2n+2 = (n + 1)(n + τ + μ)(n + τ + δ)(n + μ + δ)(2n + τ + μ + δ)(2n + τ + μ + δ + 1) , n ≥ 0.

Di((x − iτ)(x − iμ)(x − iδ))ŵ0(τ, μ, δ)

)

− 2((τ + μ + δ)x2 − τμδ)ŵ0(τ, μ, δ) = 0,(ŵ0(τ, μ, δ)

)1 = 0.

(6.6)

Proof

a) In this case (6.4) becomes⎧⎪⎪⎨

⎪⎪⎩

γ2n+1 = −ω2(n2 − 2nb 1a−12 ω−2 + a0a−12 ω−2

), n ≥ 0,

γ2n+2 = −ω2(n + 1)(n − 2b 1a−12 ω−2

), n ≥ 0,

Dω(( 12ωa2x

2 + b 1x + 12ωa0)w0)

+(

a2x2 + a0)w0 = 0, (w0)1 = 0.

That is⎧⎪⎪⎪⎨

⎪⎪⎪⎩

γ2n+1 = −ω2(n2 − 2nb 1a−12 ω−2 + a0a−12 ω−2

), n ≥ 0,

γ2n+2 = −ω2(n + 1)(n − 2b 1a−12 ω−2), n ≥ 0,Dω((x2 + 2b 1a−12 ω−1x + a0a−12 )w0

)+ 2(ω−1x2 + a0a−12 ω−1)w0 = 0,

(w0)1 = 0.With the choice A = iω and putting α + β = −2b 1a−12 ω−2, αβ = a0a−12 ω−2,we get (6.5).

b) In this case (6.4) can be written⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ2n+1 =−ω2(n − 1 − 12 a2b−13

)(n3 − 12 a2b−13 n2 + nb 1b−13 ω−2 − 12 a0b−13 ω−2

)

(2n − 1 − 12 a2b−13

)(2n − 12 a2b−13

) ,

n ≥ 0.γ2n+2 =

−ω2(n+1)

(n3 − a2b−13 n2+

( 14 a

22b

−23 +b 1b−13 ω−2

)n+ 12 a0b−13 ω−2− 12 a2b1b−23 ω−2

)

(2n − 12 a2b−13

)(2n + 1 − 12 a2b−13

) ,

n ≥ 0.Dω((

x3 + 12 ωa2b−13 x2 + b1b−13 x + 12 ωa0b−13)w0

)+ (a2b−13 x2 + a0b−13

)w0 = 0,

(w0)1 = 0.The choice A = −iω and putting τ + μ + δ = − 12 a2b−13 , τμ + τδ + δμ =b 1b−13 ω

−2 and τμδ = − 12 a0b−13 ω−2, we get (6.6). �

Remark

1. The form ŵ0(α, β) given by (6.5) is regular if and only if α �= −n, n ≥ 0,β �= −n, n ≥ 0 and α + β �= −n, n ≥ 0. When α > 0 and β > 0, the formŵ0(α, β) is positive definite.

2. The form ŵ0(τ, μ, δ) is regular if and only if τ �= −n, n ≥ 0, μ �= −n,n ≥ 0, δ �= −n, n ≥ 0, τ + μ �= −n, n ≥ 0, τ + δ �= −n, n ≥ 0, μ + δ �= −n,

274 Numer Algor (2008) 49:251–282

n ≥ 0, τ + μ + δ − 1 �= −n, n ≥ 0. When τ > 0, μ > 0, δ > 0, τ + μ + δ −1 > 0, it is positive definite.

Corollary 6.2

a) The form ŵ0(α, β) given by (6.5) is D−i-semi-classical of class one if andonly if α �= 1/2 and β �= 1/2.

b) The form ŵ0(τ, μ, δ) is Di-semi-classical of class one if and only if τ �=1/2, μ �= 1/2 and δ �= 1/2.

Proof

a) We have φ̂(x) = (x + iα)(x + iβ), ψ̂(x) = 2i(x2 − αβ) and ω = −i.Let c = −iα, we can deduce that

θcφ̂(x) = x + iβ.ψ(c − ω) + (θcφ̂

)(c − ω) = i{−2(α − 1)2 − 2αβ + 1 + β − α}. (6.7)

θc−ω(ψ̂ + θcφ̂

)(x) = 2ix + 2α − 1. (6.8)

From (6.8), it follows that〈ŵ0(α, β), θc−ω

(ψ̂ + θcφ̂

)〉 = 2i(ŵ0(α, β))

1 + 2α − 1.

But, (ŵ0(α, β))1 = 0 since ŵ0(α, β) is a symmetric form, then〈ŵ0(α, β), θc−ω

(ψ̂ + θcφ̂

)〉 = 2α − 1. (6.9)

On account of (6.7), (6.9) and the Proposition 2.4 we get the desired result.b) We have φ̂(x) = (x − iτ)(x − iμ)(x − iδ), ψ̂(x) = −2((τ + μ + δ)x2 − τμδ)

and ω = i. Let c = iτ , thenθcφ̂(x) = (x − iμ)(x − iδ).

ψ(c − ω) + (θcφ̂)(c − ω) = (2τ + 2μ + 2δ − 1)(τ + 1)2

− (μ + δ)(τ + 1) + μδ(2τ − 1). (6.10)θc−ω

(ψ̂ + θcφ̂

)(x) = −(2τ + 2μ + 2δ − 1)x − i(μ + δ)

+ i(2τ + 2μ + 2δ − 1)(τ + 1). (6.11)By (6.11), we obtain〈ŵ0(τ, μ, δ), θc−ω

(ψ̂ + θcφ̂

)〉 = i(2τ + 2μ + 2δ − 1)(τ + 1) − i(μ + δ).(6.12)

Therefore, by (6.10), (6.12) and the Proposition 2.4, we can deduce thedesired result. �

Numer Algor (2008) 49:251–282 275

Corollary 6.3

a) When α = 1/2 in (6.5), we obtain⎧⎨

⎩

γ̂n+1 = 14(n + 1)(n + 2β), n ≥ 0.

D−i((x + iβ)ŵ0(1/2, β)

)+ 2ixŵ0(1/2, β) = 0.

(6.13)

The form ŵ0(1/2, β) is D−i-classical.b) Let τ = 1/2 in (6.6), we have

⎧⎪⎨

⎪⎩

γ̂n+1 = 14

(n + 1)(n + 2μ)(n + 2δ)(n + 2μ + 2δ − 1)(2n + 2μ + 2δ − 1)(2n + 2μ + 2δ + 1) , n ≥ 0.

Di((x − iμ)(x − iδ)ŵ0(1/2, μ, δ)

)− 2(μ + δ)xŵ0(1/2, μ, δ) = 0.(6.14)

The form ŵ0(1/2, μ, δ) is Di- classical form.

Proof

a) When α = 1/2, by virtue of Corollary 6.2, the functional equation in (6.5)can be simplified by the factor x + i2 . In accordance of Proposition 5.1 wecan deduce (6.13).

b) If τ = 1/2, according to Corollary 6.2, the functional equation in (6.5) canbe simplified by the factor x − i2 . Therefore the Proposition 5.1 give thedesired result. �

Remark

1. The form ŵ0(1/2, β) is the symmetric Meixner–Pollaczek polynomials [5],see also [1, pp. 15]

2. The form ŵ0(1/2, α/2 + 1/2, β/2 + 1/2) is given in [1, pp. 17, (3.22)].3. The form ŵ0(1/2, δ/2 + 1/2, α + 1/2 − δ/2) is studied in [1, pp. 17, (3.21)].4. The form ŵ0(1/2, α/2, α/2) is given in [1, pp. 16, (4.10) for μ = 0].

7 Integral representations

The scope of this section is to determine an integral representation for anycanonical case.

When ω → iω , ω ∈ IR, we are looking for a weight function U such that

〈ŵ0, f

〉 =+∞∫

−∞U(x) f (x)dx, f ∈ P, (7.1)

where we suppose that U is regular as far as it is necessary.

276 Numer Algor (2008) 49:251–282

In the case where ŵ0 is a symmetric form we have ŵ0 = h−1ŵ0, consequently

〈ŵ0, f

〉 = 〈h−1ŵ0, f〉 =

+∞∫

−∞U(−x) f (x)dx, f ∈ P,

on account of (7.1) we can deduce that

〈ŵ0, f

〉 =+∞∫

−∞

1

2

(U(x) + U(−x)

)f (x)dx, f ∈ P. (7.2)

On account of (7.1), we get [1]

+∞∫

−∞

{(Diω(φU)

)(x) + ψ(x)U(x)

}f (x)dx = 0, f ∈ P,

with the additional condition

+∞+iω∫

−∞+iωU(x + iω)φ(x + iω) f (x)dx =

+∞∫

−∞U(x + iω)φ(x + iω) f (x)dx, f ∈ P.

(7.3)Therefore

(Diω(φU)

)(x) + ψ(x)U(x) = λg(x), (7.4)

where λ ∈ C and g is a locally integrable function with rapid decay representingthe null form. For instance

g(x) ={

0 , x ≤ 0 ,exp(−x1/4) sin(x1/4) , x > 0 ,

was given by Stieltjes [13]. When λ = 0, the equation (7.4), becomes

φ(x + iω)U(x + iω) =(φ(x) − iωψ(x)

)U(x),

so, that, if ω = 1, we have

U(x + i) = φ(x) − iψ(x)φ(x + i) U(x) , x ∈ IR, (7.5)

and if ω = −1, with x → x + i, we have

U(x + i) = φ(x)φ(x + i) + iψ(x + i)U(x), x ∈ IR. (7.6)

Now we are able to give the integral representations for any canonical form.

Numer Algor (2008) 49:251–282 277

Theorem 7.1

a) The form ŵ0(α, β) given in (6.5) possesses the following integral represen-tation:

〈ŵ0(α, β), f

〉 = K1+∞∫

−∞

| �(ix) |2 �(α + ix) |2| �(β + ix) |2| �(2ix) |2

× f (x)dx, f ∈ P, α, β > 0, (7.7)with

K−11 = π�(α)�(β)�(α + β). (7.8)b) When τ > 0, μ > 0, δ > 0 and τ + μ + δ − 1 > 0, the form ŵ0(τ, μ, δ)

defined in (6.6) have the following integral representation:

〈ŵ0(τ, μ, δ), f

〉 = K2+∞∫

−∞

| �(ix) |2| �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2| �(2ix) |2

× f (x)dx, f ∈ P, (7.9)with

K−12 = π�(τ)�(μ)�(δ)�(τ + μ)�(τ + δ)�(μ + δ)

�(τ + μ + δ) . (7.10)

Proof We need the following formulas [3, 4, 9]

+∞∫

0

| �(a + ix) |2| �(b + ix) |2| �(c + ix) |2| �(2ix) |2 dx

= 2π�(a + b)�(a + c)�(b + c), a, b , c ≥ 0. (7.11)+∞∫

0

| �(a + ix) |2| �(b + ix) |2| �(c + ix) |2| �(d + ix) |2| �(2ix) |2 dx

= 2π�(a+b)�(a+c)�(a+d)�(b +c)�(b +d)�(c+d)�(a+b +c+ d) , a, b , c, d≥0.

(7.12)

a) In the case (6.5) we have φ(x) = (x + iα)(x + iβ) , ψ(x) = 2i(x2 − αβ) andω = −1. Supposing α > 0, β > 0. Then the equation (7.6) becomes

U(x + i) = − (x + iα)(x + iβ)(x − i(α − 1))(x − i(β − 1))U(x), x ∈ IR,

278 Numer Algor (2008) 49:251–282

hence

U(x) = eπx �(α − ix)�(β − ix)�(1 − α − ix)�(1 − β − ix) A(x), x ∈ IR,

then A(x + i) = A(x) , x ∈ IR.Taking account of

�(z)�(1 − z) = πsin(πz)

, (7.13)

we have

U(x)=π−2eπx | �(α + ix) |2| �(β + ix) |2 sin(π(α + ix)) sin(π(β + ix))A(x).Choosing

A(x) = K1 π2

sin(π(α + ix)) sin(π(β + ix)) ,

we obtain

U(x) = K1eπx | �(α + ix) |2| �(β + ix) |2, x ∈ IR. (7.14)It follows that

1

2

(U(x) + U(−x)) = K1 cosh(πx) | �(α + ix) |2| �(β + ix) |2, x ∈ IR.

But

cosh(πx) = | �(ix) |2

4 | �(2ix) |2 , x ∈ IR. (7.15)

Therefore

1

2

(U(x) + U(−x)) = K1

4

| �(ix) |2| �(α + ix) |2| �(β + ix) |2| �(2ix) |2 , x ∈ IR.

(7.16)where

K−11 =1

2

+∞∫

0

| �(ix) |2 �(α + ix) |2| �(β + ix) |2| �(2ix) |2 dx. (7.17)

Finally, on account of (7.2), (7.16), (7.17) and (7.11), we get (7.7) and (7.8).b) For the case (6.6), we have φ(x) = (x − iτ)(x − iμ)(x − iδ), ψ(x) = −2(τ +

μ + δ)x2 + 2τμδ and ω = 1. Then (7.5) can be written

U(x + i) = (x + iτ)(x + iμ)(x + iδ)(x + i(1 − τ))(x + i(1 − μ))(x + i(1 − δ))U(x), x ∈ IR.

This leads to

U(x) = �(τ − ix)�(μ − ix)�(δ − ix)�(1 − τ − ix)�(1 − μ − ix)�(1 − δ − ix) A(x), x ∈ IR,

with A(x + i) = A(x), x ∈ IR

Numer Algor (2008) 49:251–282 279

Taking (7.13) into account, we obtain

U(x) = π−3 | �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2

× sin(π(τ + ix)) sin(π(μ + ix)) sin(π(δ + ix))A(x), x ∈ IR,choosing

A(x) = K2π3eπx

sin(π(τ + ix)) sin(π(μ + ix)) sin(π(δ + ix)) , x ∈ IR,we get

U(x) = K2eπx | �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2, x ∈ IR. (7.18)Consequently

1

2

(U(x) + U(−x)) = K2 cosh(πx) | �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2,

x ∈ IR,by (7.15), we obtain

1

2

(U(x) + U(−x)) = K2

4

| �(ix) |2| �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2| �(2ix) |2 ,

x ∈ IR. (7.19)with

K−12 =1

2

+∞∫

0

| �(ix) |2| �(τ + ix) |2| �(μ + ix) |2| �(δ + ix) |2| �(2ix) |2 dx. (7.20)

Then by (7.2), (7.19), (7.20) and (7.12) we can deduce (7.9) and (7.10). �

Remark In any case (7.14) and (7.18), the condition (7.3) is fulfilled by virtueof the standard asymptotic formula

| �(a + ix) |= √2πe−π |x|/2 | x |a−1/2 (1 + r(a, x)),where r(a, x) → 0, as | x |→ +∞, uniformly for bounded | a |.

8 The limiting cases: ω → 0

When ω → 0 in (6.4), we obtain the following general situation:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ2n+1 =(2(n − 1)b 3 − a2

)(a0 − 2nb 1

)

(4nb 3 − a2

)((4n − 2)b 3 − a2

) , n ≥ 0,

γ2n+2 =(n + 1)

(−2a0b 3 + 2

(a2 − 2nb 3

)b 1)

(4nb 3 − a2

)((4n + 2)b 3 − a2

) , n ≥ 0,((

b 3x3 + b 1x)w0

)′ + (a2x2 + a0)w0 = 0, (w0)1 = 0.

(8.1)

280 Numer Algor (2008) 49:251–282

A1. deg φ = 1. Then (8.1) becomes⎧⎪⎨

⎪⎩

γ2n+1 = a−12 b 1(2n − a0b−11

), n ≥ 0.

γ2n+2 = 2a−12 b 1(n + 1), n ≥ 0.(xw0)′ +

(a2b−11 x

2 + a0b−11)w0 = 0, (w0)1 = 0.

(8.2)

Choosing a2b−11 = 2 and a0b−11 = −2μ − 1 we recover the generalizedHermite form [2, 5].

⎧⎪⎨

⎪⎩

γ̂2n+1 = 12 (2n + 2μ + 1) , n ≥ 0.γ̂2n+2 = (n + 1) , n ≥ 0.(xŵ0)′ +

(2x2 − 2μ − 1)ŵ0 = 0, (ŵ0)1 = 0.

(8.3)

The form ŵ0 is regular if and only if 2μ + 1 �= −2n , n ≥ 0. The formŵ0 is semi-classical of class s = 1 if μ �= 0 and s = 0 if μ = 0.

A2. deg φ = 3. For (8.1) we get⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ2n+1 = b−13(2n − 2 − a2b−13 )(a0 − 2nb 1)(4n − a2b−13 )

(4n − 2 − a2b−13

) , n ≥ 0.

γ2n+2 = b−13(n + 1)(−2a0 + 2

(a2b−13 − 2n

)b 1)

(4n − a2b−13 )(4n + 2 − a2b−13

) , n ≥ 0.((

x3 + b 1b−13 x)w0

)′ + (a2b−13 x2 + a0b−13)w0 = 0, (w0)1 = 0.

(8.4)

Two cases ariseA21. b 1 = 0. Then we obtain for (8.4):

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

γ2n+1 = a0b−132n − 2 − a2b−13(

4n − a2b−13)(

4n − 2 − a2b−13) , n ≥ 0.

γ2n+2 = −2a0b−13n + 1

(4n − a2b−13

)(4n + 2 − a2b−13

) , n ≥ 0.((x3w0

)′ + (a2b−13 x2 + a0b−13)w0 = 0, (w0)1 = 0.

Choosing a2b−13 = −2(ν + 1) and a0b−13 = − 12 , we rediscover the fol-lowing situation [2]

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

γ̂2n+1 = −14

n + ν(2n + ν)(2n + ν + 1) , n ≥ 0.

γ̂2n+2 = 14

n + 1(2n + ν + 1)(2n + ν + 2) , n ≥ 0.(

(x3ŵ0)′ − (2(ν + 1)x2 + 12

)ŵ0 = 0, (ŵ0)1 = 0.

(8.5)

The form ŵ0 is regular if and only if ν �= −n , n ≥ 0 and semi-classicalof class one.

Numer Algor (2008) 49:251–282 281

A22. b 1 �= 0. Then, (8.4) can be written as follow⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

γ2n+1 = b 1b−13(2n − 2 − a2b−13

)(a0b

−11 − 2n

)

(4n − a2b−13 )(4n − 2 − a2b−13

) , n ≥ 0.

γ2n+2 = b 1b−13(n + 1)(+4n + 2a2b−13 − 2a0b−11

)

(4n − a2b−13

)(4n + 2 − a2b−13

) , n ≥ 0.((

x3 + b 1b−13 x)w0

)′+(a2b−13 x2 + a0b−13)w0 = 0, (w0)1 = 0.

With the choice b 1b−13 = − 1, a0b−13 = 2β + 2 and a2b−13 = − 2α −2β − 4,we obtain⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

γ̂2n+1 = (n + β + 1)(n + α + β + 1)(2n + α + β + 1)(2n + α + β + 2) , n ≥ 0.

γ̂2n+2 = (n + 1)(n + α + 1)(2n + α + β + 2)(2n + α + β + 3) , n ≥ 0.

((x(x2 − 1)ŵ0

)′ − 2((α + β + 2)x2 − β − 1)ŵ0 = 0, (ŵ0)1 = 0.

(8.6)

It is the definition of the form studied in [2, 5]. The form ŵ0 isregular if and only if α �= −n − 1, n ≥ 0, β �= −n − 1, n ≥ 0, α + β �=−n − 1, n ≥ 0 and it is semi-classical of class one if β �= −1/2.When β = −1/2, α → α + 1, we obtain the classical Gegenbauer form.

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The symmetric bold0mu mumu DDunitsDDDDbold0mu mumu units-semi-classical orthogonal polynomials of class oneAbstractIntroductionPreliminariesThe Laguerre--Freud equationsSome properties of the symmetric D-semi-classical forms of class sThe symmetric case when s=1The canonical casesIntegral representationsThe limiting cases: 0References

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