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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:
[email protected]
M. Mejri (B)Institut Supérieur des Sciences Appliquées et de
Technologie,Rue Omar Ibn El Khattab, Gabes 6072, Tunisiae-mail:
[email protected]
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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.
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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,
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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,
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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)
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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). �
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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)
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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)
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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).
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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,
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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)},
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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)
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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)
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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,
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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). �
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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)
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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.
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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)
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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.
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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). �
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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}
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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)
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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,
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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. �
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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.
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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.
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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,
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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
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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)
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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.
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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.
References
1. Abdelkarim, F., Maroni, P.: The Dω-classical orthogonal
polynomials. Result. Math. 32, 1–28(1997)
2. Alaya, J., Maroni, P.: Symmetric Laguerre–Hahn forms of class
s = 1. Integral TransformsSpec. Funct. 4, 301–320 (1996)
3. Askey, R., Wilson, J.: A set of hypergeometric orthogonal
polynomials. SIAM J. Math. Anal.13, 651–655 (1982)
4. Askey, R., Wilson, J.: Some Basic Hypergeomertic Orthogonal
Polynomials that GeneralizeJacobi Polynomials. Memoirs of the
American Mathematical Society 319. Providence, RholeIsland
(1985)
5. Chihara, T.S.: An Introduction to Orthogonal Polynomials.
Gordon and Breach, New York(1978)
6. Bachène, M.: Les polynômes semi-classiques de classe zéro et
de classe un. Thesis of thirdcycle, Univ. P. and M. Curie, Paris
(1985)
7. Belmehdi, S.: On semi-classical linear functional of class s
= 1. Classification and integralrepresentations. Indag. Math. 3,
253–275 (1992)
8. Foupouagnigni, M., Hounkonnou, M.N., Ronveaux, A.:
Laguerre–Freud equations for therecurrence coefficients of
Dω-semi-classical orthogonal polynomials of class one. J.
Comp.Appl. Math. 99, 143–154 (1998)
9. Koekoek, R., Swarttow, R.F.: The ASkey-scheme of
hypergeometric orthogonal polynomialsand its q-analogue. Report
98–17, TU Delft (1998)
10. Marcellan, F., Salto, L.: Discrete semi-classical orthogonal
polynomials. J. Differ. Equ. Appl.4, 463–490 (1998)
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282 Numer Algor (2008) 49:251–282
11. Maroni, P.: Variations around classical orthogonal
polynomials. Connected problems. J.Comp. Appl. Math. 48, 133–155
(1993)
12. Ronveaux, A.: Discrete semi-classical orthogonal
polynomials: generalized Meixner polyno-mials. J. Approx. Theory
46(4), 403–407 (1986)
13. Stieltjes, T.J.: Recherche sur les fractions continues. Ann.
Fac. Sci. Toulouse 8, J1–122 (1894);9, A 1–47 (1895)
14. Van Assche, W., Foupouagnigni, M.: Analysis of non-linear
recurrence relations for therecurrence coefficients of generalized
Charlier polynomials. J. Nonlin. Math. Phys. 10(Suppl. 2), 231–237
(2003)
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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