# Division Problem of a Regular Form: The Casex2u=λxv

Authors: M. Mejri
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We present a systematic study of a regular linear functionalv to find all regular forms u which satisfy the equation
x2u=λxv, λ∈ℂ−{0}. We also give the second-order recurrence relation of
the orthogonal polynomial sequence with respect to u and study the semiclassical
character of the found families. We conclude by treating some
examples.


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## Body

## 1. Introduction

In the present paper, we intend to study the following problem: letv be a regular form (linear functional), and R and D nonzero polynomials. Find all regular forms u satisfying(1.1)Ru=Dv.
This problem has been studied in some particular cases. In fact the product of a linear form by a polynomial (R(x)=1) is studied in [1–3] and the inverse problem (D(x)=λ, λ∈ℂ-{0}) is considered in [4–7]. More generally, when R and D have nontrivial common factor the authors of [8] found necessary and sufficient conditions for u to be a regular form. The case where R=D is treated in [4, 9–11]. The aim of this contribution is to analyze the case in which R(x)=x2 and D(x)=λx, λ∈ℂ-{0}. We remark that R and D have a common factor and R≠D (see also [7]). In fact, the inverse problem is studied in [12]. On the other hand, this situation generalize the case treated in [13] (see (2.9)). In Section 1, we will give the regularity conditions and the coefficients of the second-order recurrence relation satisfied by the monic orthogonal polynomial sequence (MOPS) with respect to u. We will study the case where v is a symmetric form; thus regularity conditions become simpler. The particular case when v is a symmetric positive definite form is analyzed. The second section is devoted to the case where v is semi-classical form. We will prove that u is also semi-classical and some results concerning the class of u are given. In the last section, some examples will be treated. The regular forms u found in theses examples are semi-classical of class s∈{1,2,3} [14]. The integral representations of these regular forms and the coefficients of the second-order recurrence satisfied by the MOPS with respect to u are given.


## 2. The Problemx2u=λxv

Let𝒫 be the vector space of polynomials with coefficients in 𝒞 and 𝒫′ its algebraic dual. We denote by 〈u,f〉 the action of u∈𝒫′ on f∈𝒫. In particular, we designate by (u)n∶=〈u,xn〉, n≥0, the moments of u. For any form u, any polynomial g, any c∈ℂ, a∈ℂ-{0}, let u′, hau, gu, and (x-c)-1u be the forms defined by duality:(2.1)〈u′,p〉∶=-〈u,p′〉;〈hau,p〉∶=〈u,hap〉;〈gu,p〉∶=〈u,gp〉;〈(x-c)-1u,p〉∶=〈u,θcp〉,p∈P,
where (θcp)(x)=(p(x)-p(c))/(x-c); (hap)(x)=p(ax).We define a left multiplication of a formu by a polynomial p as(2.2)(up)(x)∶=〈u,xp(x)-ξp(ξ)x-ξ〉,u∈P′,p∈P.
Let us recall that a form u is called regular if there exists a monic polynomial sequence {Pn}n≥0, degPn=n, such that(2.3)〈u,PnPm〉=rnδn,m,n,m≥0,rn≠0,n≥0.
We have the following result.Lemma 2.1 (see [15]).
Letu∈𝒫′, f∈𝒫, and c∈𝒞. The following formulas hold:
(2.4)(vf)′(x)=(v′f)(x)+(vf′)(x)+(vθ0f)(x),f∈P.(2.5)(δf)(x)=f(x),f∈P.(2.6)(x-c)-1((x-c)u)=u-(u)0δc,
where 〈δc,p〉=p(c), p∈𝒫.We consider the following problem: given a regular formv, find all regular forms u satisfying(2.7)x2u=λxv,λ∈C-{0},
with constraints (u)0=1, (v)0=1. From (2.6) we can deduce that(2.8)xu=((u)1-λ)δ+λv,(2.9)u=δ+(λ-(u)1)δ′+λx-1v.
Then the form u depends on two arbitrary parameters (u)1 and λ.We notice that when(u)1=λ, we encounter the problem studied in [13] again.We suppose that the formv has the following integral representation:(2.10)〈v,f〉=∫-∞+∞V(x)f(x)dx,foreachpolynomialf,
where V is a locally integrable function with rapid decay, continuous at the origin; then the form u is represented by(2.11)〈u,f〉=(1-λP∫-∞+∞V(x)xdx)f(0)+((u)1-λ)f′(0)+λP∫-∞+∞V(x)f(x)xdx,
where [16, 17](2.12)P∫-∞+∞V(x)xdx=lim⁡ϵ→0+(∫-∞-ϵV(x)xdx+∫ϵ+∞V(x)xdx).
Let {Sn}n≥0 denote the sequence of monic orthogonal polynomials with respect to v; we have(2.13)S0(x)=1,S1(x)=x-ξ0,Sn+2(x)=(x-ξn+1)Sn+1(x)-σn+1Sn(x),n≥0,
with(2.14)ξn=〈v,xSn2(x)〉〈v,Sn2〉,σn+1=〈v,Sn+12〉〈v,Sn2〉,n≥0.
When u is regular, let {Zn}n≥0 be the corresponding MOPS:(2.15)Z0(x)=1,Z1(x)=x-β0.Zn+2(x)=(x-βn+1)Zn+1(x)-γn+1Zn(x),n≥0.
From (2.7), we know that the existence of the sequence {Zn}n≥0 is among all the strictly quasi-orthogonal sequences of order two with respect to λxv=w (w is not necessarily a regular form) [15, 18–20]. That is,(2.16)xZ0(x)=S1(x)+c0,xZ1(x)=S2(x)+c1S1(x)+b0.xZn+2(x)=Sn+3(x)+cn+2Sn+2(x)+bn+1Sn+1(x)+anSn(x),n≥0,
with an≠0, n≥0.From (2.16), we have(2.17)Z1(x)=(θ0S2)(x)+c1,(2.18)Zn+2(x)=(θ0Sn+3)(x)+cn+2(θ0Sn+2)(x)+bn+1(θ0Sn+1)(x)+an(θ0Sn)(x),n≥0.Lemma 2.2.
Let{Zn}n≥0 be a sequence of polynomials satisfying (2.16) where an, bn, and cn are complex numbers such that an≠0 for all n≥0. The sequence {Zn}n≥0 is orthogonal with respect to u if and only if
(2.19)〈u,Zn〉=0,n≥1,〈u,xZn(x)〉=0,n≥2,〈u,xZ1(x)〉≠0.Proof.
The conditions (2.19) are necessary from the definition of the orthogonality of {Zn}n≥0 with respect to u.
Fork≥2, we have (by (2.7))
(2.20)〈u,xkZn+2(x)〉=〈x2u,xk-2Zn+2(x)〉=λ〈v,xk-1Zn+2(x)〉,n≥0,
and from (2.16), we get
(2.21)〈u,xkZn+2(x)〉=λ〈v,xk-2Sn+3(x)〉+λcn+2〈v,xk-2Sn+2(x)〉+λbn+1〈v,xk-2Sn+1(x)〉+λan〈v,xk-2Sn(x)〉,n≥0.
Taking into account the orthogonality of {Sn}n≥0, we obtain
(2.22)〈u,xkZn+2(x)〉=0,2≤k≤n+1,n≥1,〈u,xn+2Zn+2(x)〉=λan〈v,Sn2〉≠0,n≥0.
By (2.19), it follows that
(2.23)〈u,Z1〉=0,〈u,xZ1(x)〉≠0,〈u,Zn+2〉=〈u,xZn+2(x)〉=0,n≥0.
Consequently, the previous relations and (2.22) prove that {Zn}n≥0 is orthogonal with respect to u, which proves the Lemma.Remark 2.3.
Whenu is regular, from Theorem 5.1 in [21], there exist complex numbers rn+2≠0, tn+2 and vn+2≠0 such that
(2.24)Zn+2(x)+rn+2Zn+1(x)=Sn+2(x)+tn+2Sn+1(x)+vn+2Sn(x),n≥0.From (2.16), (2.24), and (2.15) we obtain the following relations:(2.25)rn+2-tn+2+cn+2-ξn+2=0,n≥0,rn+2cn+1-tn+2ξn+1-vn+2+bn+1-σn+2=0,n≥0,rn+2bn-tn+2σn+1-vn+2ξn+an=0,n≥0,rn+2an-1-vn+2σn=0,n≥1.
Taking into account (2.16), (2.18) and (2.19), we get(2.26)0=〈u,xZn+2(x)〉=〈u,Sn+3〉+cn+2〈u,Sn+2〉+bn+1〈u,Sn+1〉+an〈u,Sn〉=0,n≥0,0=〈u,Zn+2〉=〈u,θ0Sn+3〉+cn+2〈u,θ0Sn+2〉+bn+1〈u,θ0Sn+1〉+an〈u,θ0Sn〉,n≥0,0=Sn+3(0)+cn+2Sn+2(0)+bn+1Sn+1(0)+anSn(0),n≥0,
with the initial conditions:(2.27)0=S1(0)+c0,0=S2(0)+c1S1(0)+b0,0=〈u,Z1〉=〈u,(θ0S2)〉+c1,0≠〈u,xZ1(x)〉=〈u,S2〉+c1〈u,S1〉+b0.
If we denote(2.28)Δn∶=|Sn+2(0)Sn+1(0)Sn(0)〈u,Sn+2〉〈u,Sn+1〉〈u,Sn〉〈u,θ0Sn+2〉〈u,θ0Sn+1〉〈u,θ0Sn〉|,n≥0,
from the Cramer rule we have (2.29)Δnan=-Δn+1,n≥0,(2.30)Δnbn+1=|Sn+2(0)-Sn+3(0)Sn(0)〈u,Sn+2〉-〈u,Sn+3〉〈u,Sn〉〈u,θ0Sn+2〉-〈u,θ0Sn+3〉〈u,θ0Sn〉|,n≥0,(2.31)Δncn+2=|-Sn+3(0)Sn+1(0)Sn(0)-〈u,Sn+3〉〈u,Sn+1〉〈u,Sn〉-〈u,θ0Sn+3〉〈u,θ0Sn+1〉〈u,θ0Sn〉|,n≥0.Lemma 2.4.
The following formulas hold:(2.32)〈u,Sn〉=Sn(0)+((u)1-λ)Sn′(0)+λSn-1(1)(0),n≥0,(2.33)〈u,xSn(x)〉=((u)1-λ)Sn(0),n≥1,(2.34)〈u,(θ0Sn)〉=Sn′(0)+12((u)1-λ)Sn′′(0)+λ(Sn-1(1))′(0),n≥0,(2.35)Sn(1)(0)Sn(0)-Sn-1(1)(0)Sn+1(0)=〈v,Sn2〉,n≥0,
where Sn(1)(x)∶=(vθ0Sn+1)(x), n≥0, and S-1(1)(x)=0.Proof.
Equations (2.32) and (2.33) are deduced, respectively, from (2.9) and (2.8).
We have(2.36)〈v,θ02Sn〉=〈v,θ0Sn′-(θ0Sn)′〉=〈v,θ0Sn′〉+〈v′,θ0Sn〉=(xθ0Sn′)(0)+(v′θ0Sn)(0),n≥0.
Using (2.4), we get
(2.37)〈v,θ02Sn〉=(vθ0Sn)′(0)=(Sn-1(1))′(0),n≥0.
From (2.9), we obtain
(2.38)〈u,θ0Sn〉=〈δ,θ0Sn〉+((u)1-λ)〈δ,(θ0Sn)′〉+λ〈v,θ02Sn〉,n≥0.
According to (2.5) and (2.37), we can deduce (2.34).
We have(2.39)S0(1)(x)=1,S1(1)(x)=x-ξ2,Sn+2(1)(x)=(x-ξn+2)Sn+1(1)(x)-σn+2Sn(1)(x),n≥0.
Then (by (2.39))
(2.40)Sn(1)(0)Sn(0)-Sn-1(1)(0)Sn+1(0)=σnSn-1(1)(0)Sn-1(0)+Sn(0)(Sn(1)(0)+ξnSn-1(1)(0))=σn(Sn-1(1)(0)Sn-1(0)-Sn-2(1)(0)Sn(0)).
It follows that
(2.41)Sn(1)(0)Sn(0)-Sn-1(1)(0)Sn+1(0)=∏μ=0nσμ=〈v,Sn2〉,n≥0,
hence (2.35).Proposition 2.5.
One has(2.42)Δn=Enλ2+Fnλ+Gn,n≥0,
where
(2.43)En=Sn+1(0){μn(0)+12χn′(0)}+{Sn(1)(0)-Sn+1′(0)}{χn(0)-〈v,Sn2〉},n≥0,Fn=-Sn+1(0){(u)1(μn(0)+χn′(0))+〈v,Sn2〉}-(u)1{Sn(1)(0)χn(0)-2Sn+1′(0)χn(0)+Sn+1′(0)〈v,Sn2〉},n≥0,Gn=(u)12{12Sn+1(0)χn′(0)-Sn+1′(0)χn(0)},n≥0,
with
(2.44)χn(x)=Sn(x)Sn+1′(x)-Sn+1(x)Sn′(x),n≥0,μn(x)=Sn+1(x)(Sn-1(1))′(x)-Sn(x)(Sn(1))′(x),n≥0.Proof.
Using (2.13), we, respectively, obtain
(2.45)Sn+2(0)=-ξn+1Sn+1(0)-σn+1Sn(0),n≥0,〈u,Sn+2〉=〈u,xSn+1(x)〉-ξn+1〈u,Sn+1〉-σn+1〈u,Sn〉,n≥0,〈u,θ0Sn+2〉=〈u,Sn+1〉-ξn+1〈u,θ0Sn+1〉-σn+1〈u,θ0Sn〉,n≥0.
Taking into account previous relations, we obtain for (2.28) the following:
(2.46)Δn=|0Sn+1(0)Sn(0)〈u,xSn+1(x)〉〈u,Sn+1〉〈u,Sn〉〈u,Sn+1〉〈u,θ0Sn+1〉〈u,θ0Sn〉|,n≥0,
that is,
(2.47)Δn=-〈u,xSn+1(x)〉|Sn+1(0)Sn(0)〈u,θ0Sn+1〉〈u,θ0Sn〉|+〈u,Sn+1〉|Sn+1(0)Sn(0)〈u,Sn+1〉〈u,Sn〉|,n≥0.
Let n≥0; based on the relations (2.32)–(2.34), it follows that
(2.48)|Sn+1(0)Sn(0)〈u,θ0Sn+1〉〈u,θ0Sn〉|={μn(0)+12χn′(0)}λ-χn(0)-12(u)1χn′(0),|Sn+1(0)Sn(0)〈u,Sn+1〉〈u,Sn〉|={χn(0)-〈v,Sn2〉}λ-(u)1χn(0).
From (2.48) and (2.47), we obtain the desired results.Proposition 2.6.
The formu is regular if and only if Δn≠0, n≥0. Then, the coefficients of the three-term recurrence relation (2.15) are given by
(2.49)γ1=Δ0,γ2=-λΔ1Δ0-2,(2.50)γn+3=ΔnΔn+2Δn+12σn+1,n≥0,(2.51)β0=(u)1,β1=c1-ξ0-ξ1+λb0Δ0-1,(2.52)βn+2=cn+2-ξn+1-ξn+2-bn+1ΔnΔn+1-1σn+1,n≥0.Proof
Necessity
From (2.27) and Lemma 2.4, we get
(2.53)〈u,xZ1(x)〉=〈u,S2〉+〈u,θ0S2〉(S1(0)-〈u,S1〉)-S2(0)=λS1(0)-(u)12,
and again with (2.27) and (2.42), we can deduce that
(2.54)Δ0=〈u,S2〉+〈u,θ0S2〉(S1(0)-〈u,S1〉)-S2(0)=〈u,xZ1(x)〉≠0.
Moreover, {Zn}n≥0 is orthogonal with respect to u, therefore it is strictly quasiorthogonal of order two with respect to xv, and then it satisfies (2.16) with an≠0, n≥0. This implies Δn≠0, n≥0. Otherwise, if there exists an n0≥1 such that Δn0=0, from (2.29), Δ0=0, which is a contradiction.
Sufficiency
Let(2.55)c0=-S1(0)=ξ0,(2.56)c1=-〈u,(θ0S2)〉,(2.57)b0=Δ0-〈u,S2〉-c1〈u,S1〉.
We get
(2.58)〈u,xZ1(x)〉=〈u,S2〉+c1〈u,S1〉+b0=Δ0≠0.
We have 〈u,Z1〉=c1+〈u,θ0S2〉=0.
From (2.56) and (2.57) we get
(2.59)S2(0)+c1S1(0)+b0=S2(0)-〈u,S2〉-〈u,θ0S2〉(S1(0)-〈u,S1〉)+Δ0.
On account of (2.54), we can deduce that S2(0)+c1S1(0)+b0=0.
Then we had just proved that the initial conditions (2.27) are satisfied.
Furthermore, the system (2.26) is a Cramer system whose solution is given by (2.29), (2.30), and (2.31); with all these numbers an, bn, and cn (n≥0), define a sequence polynomials {Zn}n≥0 by (2.16). Then it follows from (2.26) and Lemma 2.2 that u is regular and {Zn}n≥0 is the corresponding MOPS.
Moreover, by (2.22) we get
(2.60)〈u,Zn+22〉=λan〈v,Sn2〉,n≥0.
Making n=0 in (2.60), it follows that
(2.61)〈u,Z22〉=λa0.
Based on relations (2.58), (2.60), (2.61), and (2.29), we, respectively, obtain
(2.62)γ1=〈u,xZ1(x)〉=Δ0;γ2=〈u,Z22〉〈u,xZ1(x)〉=-λΔ1Δ0-2,γn+3=〈u,Zn+32〉〈u,Zn+22〉=ΔnΔn+2Δn+12σn+1,n≥0.
We have proved (2.49) and (2.50).
When{Zn}n≥0 is orthogonal, we have
(2.63)β0=(u)1.
By (2.16) and the orthogonality of {Zn}n≥0, we get
(2.64)〈u,xZ12(x)〉=c1〈u,Z12〉+〈u,S2Z1〉.
By virtue of (2.13) and the regularity of u we obtain
(2.65)〈u,S2Z1〉=〈x2u,Z1〉-(ξ0+ξ1)〈u,Z12〉=λ〈v,xZ1(x)〉-(ξ0+ξ1)〈u,Z12〉=λb0-(ξ0+ξ1)〈u,Z12〉,
and consequently, we get the second result in (2.51) from (2.58), and (2.64).
From (2.16), and the orthogonality of {Zn}n≥0, we have
(2.66)βn+2〈u,Zn+22〉=cn+2〈u,Zn+22〉+〈u,Sn+3Zn+2〉,n≥0.
Using (2.13), (2.16), and the the orthogonality of {Sn}n≥0, we have
(2.67)〈u,Sn+3Zn+2〉=λbn+1〈v,Sn+12〉-(ξn+1+ξn+2)〈u,Zn+22〉,n≥0.
Taking into account the previous relation, (2.66) becomes
(2.68)βn+2=cn+2-ξn+1-ξn+2+λbn+1〈v,Sn+12〉〈u,Zn+22〉,n≥0.
From (2.60) and (2.29), we have
(2.69)〈v,Sn+12〉〈u,Zn+22〉=-λ-1ΔnΔn+1-1σn+1,n≥0.
Last equation and (2.68) give (2.52).Moreover, if the formu is regular, for (2.29), (2.30), and (2.31), we get(2.70)an=-Δn+1Δn,n≥0,(2.71)bn+1=(Dnλ2+Hnλ+In)Δn-1+σn+2,n≥0,(2.72)cn+2=-(Jnλ2+Lnλ+Kn)Δn-1+ξn+2,n≥0,
where(2.73)Dn=Sn(0)(〈v,Sn+12〉-χn+1(0))-ξn+1Sn+2(0)(μn(0)+12χn′(0))-ξn+1(Sn+1(1)(0)-Sn+2′(0))(χn(0)-〈v,Sn2〉),n≥0,Hn=(u)1Sn(0)(2χn+1(0)-〈v,Sn+12〉)+ξn+1Sn+2(0)(χn(0))+(u)1(χn′(0)+μn(0))+(u)1ξn+1χn(0)(Sn+1(1)(0)-Sn+2′(0))+ξn+1(〈v,Sn2〉-χn(0))(Sn+2(0)+(u)1Sn+2′(0)),n≥0,In=-(u)12{Sn(0)χn+1(0)+12ξn+1(Sn+2(0)χn′(0)-Sn+2′(0)χn(0))},n≥0,Jn=Sn+2(0)(μn(0)+12χn′(0))+(Sn+1(1)(0-Sn+2′(0))(χn(0)-〈v,Sn2〉),n≥0,Ln=(u)1χn(0)(2Sn+2′(0)-Sn+1(1)(0))-(u)1Sn+2(0)(μn(0)+χn′(0))-〈v,Sn2〉(Sn+2(0)+(u)1Sn+2′(0)),n≥0,Kn=(u)12{12Sn+2(0)χn′(0)-χn(0)Sn+2′(0)},n≥0.
In the sequel, we will assume that v is a symmetric linear form.We need the following lemmas.Lemma 2.7.
If{yn}n≥0 and {bn}n≥0 are sequences of complex numbers fulfilling
(2.74)yn+1+anyn=bn+1,n≥0,an≠0,n≥0,y0=b0,
then
(2.75)yn=(-1)nan-1(∏μ=0naμ)∑ν=0n(-1)νaν(∏μ=0νaμ-1)bν,n≥0.Lemma 2.8.
When{Sn}n≥0 given by (2.13) is symmetric, one has
(2.76)S2n(0)=(-1)nσ2n+1∏μ=0nσ2μ+1,n≥0,S2n+1(0)=0,n≥0,S2n(1)(0)=(-1)n∏μ=0nσ2μ,n≥0,S2n+1(1)(0)=0,n≥0,S2n+1′(0)=(-1)n(∏μ=0nσ2μ)Λn,n≥0,S2n′(0)=0,n≥0,(S2n(1))′(0)=0,n≥0,S2n+1′′(0)=0,n≥0.Proof.
Asv is symmetric, then ξn=0, n≥0, and therefore from (2.13) we have
(2.77)S0(0)=1,S1(0)=0,S0(1)(0)=1,S1(1)(0)=0,Sn+2(0)=-σn+1Sn(0),n≥0,Sn+2(1)(0)=-σn+2Sn(1)(0),n≥0,S0′(0)=0,S1′(0)=1,Sn+2′(0)=-σn+1Sn′(0)+Sn+1(0),n≥0,(S0(1))′(0)=0,(Sn+2(1))′(0)=-σn+2(Sn(1))′(0)+Sn+1(1)(0),n≥0,S0′′(0)=0,S1′′(0)=0,Sn+2′′(0)=-σn+1Sn′′(0)+2Sn+1′(0),n≥0.
Now, it is sufficient to use Lemma 2.7 in order to obtain the desired results.Let(2.78)ω=λ-1(u)1.Corollary 2.9.
Ifv is a symmetric form, one has
(2.79)Δ2n=λ2(-1)n+1σ2n+1(∏μ=0nσ2μ+1)(∏μ=0nσ2μ)2{(ω-1)Λn+1}2,n≥0,Δ2n+1=λ(-1)n(∏μ=0nσ2μ+1)2(∏μ=0nσ2μ),n≥0,
where
(2.80)Λn=∑ν=0n1σ2ν+1∏μ=0νσ2μ+1σ2μ,n≥0,σ0=1.Proof.
Following Lemma2.8, for (2.43) we have
(2.81)E2n=(-1)n+1σ2n+1(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)(1-Λn),n≥0;E2n+1=0,n≥0,F2n=2ωλ(-1)n+1σ2n+1(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)(1-Λn)Λn+1,n≥0,F2n+1=(-1)n(∏μ=0nσ2μ)(∏μ=0nσ2μ+1)2,n≥0,G2n=ω2λ2(-1)n+1σ2n+1(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)Λn2,n≥0;G2n+1=0,n≥0.
As a consequence, relations (2.81) and (2.42) yield (2.79).Theorem 2.10.
The formu is regular if and only if (ω-1)Λn+1≠0, n≥0, where Λn is defined in (2.80).
In this case one has(2.82)a2n=σ2n+1λΘn((ω-1)Λn+1)2,a2n+1=-λσ2n+22Θn((ω-1)Λn+1)2,n≥0,(2.83)b2n=σ2n+1,n≥0,b2n+1=σ2n+2(ω-1)Λn+1+1(ω-1)Λn+1,n≥0,(2.84)c0=0,c1=-ωλ,c2n+2=1λΘn((ω-1)Λn+1)2,n≥0,c2n+3=-λσ2n+2Θn((ω-1)Λn+1+1)((ω-1)Λn+1),n≥0,γ1=-λ2ω2,γ2=-σ12λ2ω4,γ2n+4=1λ2Θn+12((ω-1)Λn+1+1)2,n≥0,γ2n+3=λ2σ2n+22Θn2((ω-1)Λn+1)2((ω-1)Λn+1+1)2,n≥0,β0=λω,β1=-λω-σ1λω2,(2.85)β2n+2=1λΘn((ω-1)Λn+1)2+λσ2n+2Θn((ω-1)Λn+1)((ω-1)Λn+1+1),β2n+3=1λΘn+1((ω-1)Λn+1+1)2-λσ2n+2Θn((ω-1)Λn+1)((ω-1)Λn+1+1),n≥0,
where Θn=∏μ=0nσ2μ/σ2μ+1, n≥0.Proof.
From Proposition2.6 and Corollary 2.9, we can deduce that u is regular if and only if (ω-1)Λn+1≠0, n≥0.
Moreover, from (2.70) we can deduce (2.82).
By (2.49), (2.51), (2.78), and (2.79), for (2.55), (2.56), and (2.57) we get
(2.86)c0=0,c1=-(u)1=-ωλ,b0=σ1.
When n≥0 by Lemma 2.8, for (2.73) we get
(2.87)D2n=(-1)nσ2n+1(∏μ=0nσ2μ)(∏μ=0nσ2μ+1)2(1-Λn);D2n+1=0,H2n=ωλ(-1)nσ2n+1(∏μ=0nσ2μ)(∏μ=0nσ2μ+1)2(2Λn-1);H2n+1=0,I2n=ω2λ2(-1)n+1σ2n+1(∏μ=0nσ2μ)(∏μ=0nσ2μ+1)2Λn;I2n+1=0,J2n=0;J2n+1=(-1)nσ2n+2(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)(1-Λn)(1-Λn+1),L2n=(-1)n+1σ2n+1(∏μ=0nσ2μ)(∏μ=0nσ2μ+1)2,L2n+1=ωλ(-1)nσ2n+2(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)(Λn+1+(1-2Λn+1)Λn),K2n=0,n≥0;K2n+1=ω2λ2(-1)nσ2n+2(∏μ=0nσ2μ)2(∏μ=0nσ2μ+1)ΛnΛn+1.
Taking into account (2.79), (2.80), and (2.86)-(2.87), relations (2.70), (2.71) and (2.72) give (2.82)–(2.84).
As a result of relations (2.82)–(2.84) and Proposition 2.6 we get (2.85).Corollary 2.11.
(1) If v is a symmetric positive definite form, then the form u is regular when ω∈ℂ-]-∞,1[.
(2) When u is regular, it is positive definite form if and only if
(2.88)λω2<0,σ12ω2>0,1λ2Θn+12((ω-1)Λn+1+1)2,n>0,λ2σ2n+22Θn2((ω-1)Λn+1)2((ω-1)Λn+1+1)2,n>0.Proof.
(1) Ifv is positive definite, then σn+1>0, n≥0, therefore Λn>0, n≥0 and so (ω-1)Λn+1≠0, n≥0 under the hypothesis of the corollary.
(2) Ifu is regular, it is positive definite if and only if γn+1>0, n≥0. By Theorem 2.10, we conclude the desired results.


## 3. Some Results on the Semiclassical Case

Let us recall that a formv is called semiclassical when it is regular and its formal Stieltjes function S(·;v) satisfies [15](3.1)ϕ(z)S′(z;v)=C(z)S(z;v)+D(z),
where ϕ monic, C, and D are polynomials with(3.2)D(z)=-(vθ0ϕ)′(z)+(vθ0C)(z),S(z;v)=-∑n≥0(v)nzn+1.
The class of the semi-classical form v is s=max(degϕ-2,degC-1) if and only if the following condition is satisfied [22]:(3.3)∏c(|C(c)|+|D(c)|)>0,
where c∈{x:ϕ(x)=0}, that is, ϕ, C, and D are coprime.In the sequel, we will suppose that the formv is semi-classical of class s satisfying (3.1).Proposition 3.1.
Whenu is regular, it is also semi-classical and satisfies
(3.4)ϕ̃(z)S′(z;u)=C̃(z)S(z;u)+D̃(z),
where
(3.5)ϕ̃(z)=z3ϕ(z),C̃(z)=z3C(z)-z2ϕ(z),D̃(z)=z(z+(u)1-λ)C(z)+λz2D(z)+((u)1-λ)ϕ(z).
Moreover, the class of u depends on the zero x=0 of ϕ.Proof.
We need the following formula:(3.6)S(z;fw)=fS(z;w)+(wθ0f)(z),w∈P′,f∈P.
From (2.7), we have S(z;x2u)=λS(z;xv). Using (3.6), we get
(3.7)z2S(z;u)+z+(u)1=λzS(z;v)+λ.
Differentiating the previous equation, we obtain
(3.8)z2S′(z;u)+2zS(z;u)+1=λzS′(z;v)+λS(z;v).
By (3.1) we can deduce (3.4) and (3.5).
Sincev is a semi-classical, S(z;v) satisfies (3.1) where ϕ, C and D are coprime.
Letc be a zero of ϕ̃ different from 0, which implies that ϕ(c)=0. We know that |C(c)|+|D(c)|≠0.
IfC(c)≠0, then C̃(c)≠0. if C(c)=0, then D̃(c)=λc2D(c)≠0. Hence |C̃(c)|+|D̃(c)|≠0.Corollary 3.2.
Introducing(3.9)ϑ1∶=((u)1-λ)ϕ(0),ϑ2∶=((u)1-λ)(C(0)+ϕ′(0)),ϑ3∶=C(0)+((u)1-λ)(C′(0)+ϕ′′(0))+λD(0),(1)
ifϑ1≠0,  then s̃=s+3;(2)
ifϑ1=0  and ϑ2≠0, then s̃=s+2;(3)
ifϑ1=ϑ2=0  and ϕ(0)≠0 or ϑ3≠0,  then s̃=s+1.Proof.
(1) From (3.9) and (3.5), we obtain C̃(0)=0, D̃(0)=ϑ1≠0. Therefore, it is not possible to simplify, which means that the class of u is s+3.
(2) Ifϑ1=0, then from (3.5) we have C̃(0)=D̃(0)=0. Consequently, (3.4)–(3.6) is divisible by z. Thus, u fulfils (3.4) with
(3.10)ϕ̃(z)=z2ϕ(z),C̃(z)=z2C(z)-zϕ(z),D̃(z)=(z+(u)1-λ)C(z)+λzD(z)+((u)1-λ)θ0ϕ(z).
If D̃(0)=ϑ2≠0, it is not possible to simplify, which means that the class of u is s+2.
(3) Whenϑ1=ϑ2=0, then it is possible to simplify (3.4)–(3.10) by z. Thus, u fulfils (3.4) with
(3.11)ϕ̃(z)=zϕ(z),C̃(z)=zC(z)-ϕ(z),D̃(z)=((u)1-λ)(θ0C(z)+θ02ϕ(z))+λD(z)+C(z).
Since we have C̃(0)=-ϕ(0), D̃(0)=ϑ3, then we can deduce that if ϕ(0)≠0 or ϑ3≠0, it is not possible to simplify, which means that the class of u is s+1.


## 4. Some Examples

In the sequel the examples treated generalize some of the cases studied in [13].


### 4.1.v the Generalized Hermite Form

Let us describe the casev∶=ℋ(τ), where ℋ(τ) is the generalized Hermite form. Here is [1] (4.1)ξn=0,n≥0,σn+1=12(n+1+τ(1+(-1)n)),n≥0.
From (4.1), we get(4.2)∏μ=0nσ2μ+1=Γ(n+τ+3/2)Γ(τ+1/2),n≥0,∏μ=0nσ2μ=Γ(n+1),n≥0.
We want Λn=∑ν=0n1/σ2ν+1∏μ=0νσ2μ+1/σ2μ, n≥0.But from (4.1) and (4.2), we have 1/σ2ν+1∏μ=0νσ2μ+1/σ2μ=(1/Γ(τ+1/2))hν, with(4.3)hn=Γ(n+τ+1/2)Γ(n+1),n≥0,
fulfilling(4.4)(n+1)hn+1-nhn=(τ+12)hn,n≥0,
and so(4.5)Λn=1(τ+1/2)Γ(τ+1/2)∑ν=0n(ν+1)hν+1-νhν=1Γ(τ+3/2)Γ(n+τ+3/2)Γ(n+1),n≥0.
Then we get Table 1.Table 1ΔnΔ2n=(-1)n+1λ2Γ(τ+1/2)Γ(n+τ+1/2)Γ2(n+1)((ω-1)Λn+1)2,n≥0,Δ2n+1=(-1)nλΓ2(τ+1/2)Γ2(n+τ+3/2)Γ(n+1),n≥0.ana2n=(n+τ+1/2)2λΓ(τ+1/2)hn((ω-1)Λn+1)2,n≥0,a2n+1=-λΓ(τ+1/2)n+1hn+1((ω-1)Λn+1)2,n≥0.bnb2n=n+τ+1/2,n≥0,b2n+1=(n+1)(ω-1)Λn+1+1(ω-1)Λn+1,n≥0.cnc0=0,c1=-ωλ,c2n+2=1λn+τ+1/2Γ(τ+1/2)hn((ω-1)Λn+1)2,n≥0,c2n+3=-λ(n+1)Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1+1)((ω-1)Λn+1),n≥0.γ1=-λ2ω2,γ2=-(τ+1/2)2λ2ω4,γn+1γ2n+3=-λ2Γ2(τ+1/2)hn+12((ω-1)Λn+1+1)2((ω-1)Λn+1)2,n≥0,γ2n+4=-1λ2Γ2(τ+1/2)(n+τ+3/2)2hn+12((ω-1)Λn+1+1)4,n≥0.β0=ωλ,β1=-ωλ-τ+1/2λω2,βnβ2n+3=-λ(n+1)Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1+1)((ω-1)Λn+1)-n+τ+3/2λΓ(τ+1/2)hn+1((ω-1)Λn+1+1)2,n≥0,β2n+2=1λ1Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1)2+λΓ(τ+1/2)hn+1((ω-1)Λn+1+1)((ω-1)Λn+1),n≥0.Proposition 4.1.
Ifv=ℋ(τ) is the generalized Hermite form, then the form u(τ,ω,λ) given by (2.9) has the following integral representation:
(4.6)〈u(τ,ω,λ),f〉=f(0)+λ(ω-1)f′(0)+λΓ(τ+1/2)P∫-∞+∞|x|2τxe-x2f(x)dx,∀f∈P.
It is a quasi-antisymmetric ((u(τ,ω,λ))2n+2=0, n≥0) and semi-classical form of class s satisfying the following functional equation:
(4.7)τ=0,ω≠1,z3S′(z;u(0,ω,λ))=-z2(2z2+1)S(z;u(0,ω,λ))-2z3-2λωz2+λ(ω-1),s=3,τ=0,ω=1,zS′(z;u(0,1,λ))=-(2z2+1)S(z;u(0,1,λ))-2z-2λ,s=1,(4.8)τ≠0,ω≠1,z3S′(z;u(τ,ω,λ))=-z2(2z2-2τ+1)S(z;u(τ,ω,λ))-2z3-2λωz2+2τz+2τλ(ω-1)+λ(ω-1),s=3,τ≠0,ω=1,z2S′(z;u(τ,1,λ))=z(-2z2+2τ-1)S(z;u(τ,1,λ))-2z2-2λz+2τ,s=2.Proof.
It is well known that the generalized Hermite form possesses the following integral representation [1]:
(4.9)〈v,f〉=∫-∞+∞1Γ(τ+1/2)|x|2τe-x2f(x)dx,R(τ)>-12,∀f∈P.
Following (2.11), we obtain (4.6). Also the form u is quasi-antisymmetric because it satisfies
(4.10)〈u,x2n+2〉=λ〈v,x2n+1〉=0,n≥0,
since v is symmetric by hypothesis.
Whenτ=0, v is classical and satisfies (3.4) with [22]
(4.11)ϕ(x)=1,C(z)=-2z,D(z)=-2.
Then, ϑ1=λ(ω-1), ϑ2=0.
Now, it is sufficient to use Corollary3.2 and Proposition 3.1 in order to obtain (4.7).
Ifτ≠0, the form v is semi-classical of class one and satisfies (3.4) with [23]
(4.12)ϕ(x)=x,C(z)=-2z2+2τ,D(z)=-2z.
Therefore ϑ1=0, ϑ2=λ(ω-1)(2τ+1), ϑ3=2τ.
By Proposition3.1 and Corollary 3.2 we can deduce (4.8).


### 4.2.v the Corecursive of the Second Kind Chebychev Form

Let us describe the casev∶=𝒥(-1/2,1/2); it is the corecursive of the second kind Chebychev functional. Here is [1](4.13)ξ0=-12,ξn+1=0,n≥0,σn+1=14,n≥0.
In this case we have the following result.Lemma 4.2.
Forn≥0, one has
(4.14)S2n(0)=(-1)n22n,S2n+1(0)=(-1)n22n+1,S2n(1)(0)=(-1)n22n,S2n+1(1)(0)=0,S2n′(0)=n(-1)n+122n-1,S2n+1′(0)=(n+1)(-1)n22n,(S2n(1))′(0)=0,(S2n+1(1))′(0)=(n+1)(-1)n22n,S2n′′(0)=n(n+1)(-1)n+122n-2,S2n+1′′(0)=n(n+1)(-1)n+122n-1.Proof.
The proof is analogous for the demonstration of Lemma2.8.Following Lemma4.2, for (2.44) we have (4.15)χ2n(0)=2n+124n,n≥0;χ2n+1(0)=n+124n+1,n≥0;χ2n′(0)=0,n≥0;χ2n+1′(0)=n+124n,n≥0;μ2n(0)=-n24n-1,n≥0;μ2n+1(0)=-n+124n+1,n≥0.
Therefore, we get for (2.42)(4.16)Δ2n=n(2n+1)(-1)n+126nλ2+(8n(n+1)(u)1-1)(-1)n26n+1λ+(n+1)(2n+1)(u)12(-1)n+126n,n≥0,Δ2n+1=(n+1)(2n+1)(-1)n+126n+3λ2+(8(n+1)2(u)1+1)(-1)n26n+4λ(n+1)(2n+3)(u)12(-1)n+126n+3,n≥0.
Then we obtain(4.17)Δ2n=4(-1)n+126n+1(tn-x1)(tn-x2),n≥0,Δ2n+1=4(-1)n+126n+4(tn-x3)(tn-x4),n≥0,
where(4.18)x1=14{-3t-2λ+(t2-4λt-4λ2-4λ)1/2},x2=14{-3t-2λ-(t2-4λt-4λ2-4λ)1/2},x3=14{-5t-2λ+((t+2λ)2+4λ)1/2},x4=14{-5t-2λ-((t+2λ)2+4λ)1/2},(u)1=t+λ.
On account of Proposition 2.6, we can deduce that the form u given by (2.9) is regular if and only if tn-xi≠0, n≥0, 1≤i≤4.In the sequel, we suppose that the last condition is satisfied.By virtue of (4.17) and Lemma 4.2, relations (2.49)–(2.52), and (2.55)–(2.57), (2.70)–(2.72) give Table 2.Table 2ana2n=-18(tn-x3)(tn-x4)(tn-x1)(tn-x2),n≥0,a2n+1=18(tn-x1)(tn-x2)(tn-x3)(tn-x4), n≥0.bnb0=-2x1x2+14-t2+(t+λ+12)2,b2n+1=14+t82(n+1)t-λ(tn-x1)(tn-x2),n≥0,b2n+2=14+t8(2n+3)t+λ(tn-x3)(tn-x4),n≥0.cnc0=-12,c1=-12-t-λ,c2n+3=182t(2n+1)((n+1)t-λ)-λ(tn-x3)(tn-x4),n≥0,c2n+2=-182t(2n+1)((n+1)t+λ)-λ(tn-x1)(tn-x2),n≥0.γ1=-2x1x2,γ2=λ16x3x4x12x22,γn+1γ2n+3=-14(tn-x1)(t(n+1)-x1)(tn-x2)(t(n+1)-x2)(tn-x3)2(tn-x4)2,n≥0,γ2n+4=-14(tn-x3)(t(n+1)-x3)(tn-x4)(t(n+1)-x4)(t(n+1)-x1)2(t(n+1)-x2)2,n≥0.β0=t+λ,β1=-t-λ-λ2x1x2{-2x1x2+14-t2+(t+λ+12)2},β2n+3=182t(2n+1)((n+1)t-λ)-λ(tn-x3)(tn-x4)+12(tn-x3)(tn-x4)(t(n+1)-x1)(t(n+1)-x2)βn+t4(2n+3)t+λ(t(n+1)-x1)(t(n+1)-x2),n≥0,β2n+2=-182t(2n+1)((n+1)t+λ)-λ(tn-x1)(tn-x2)-12(tn-x1)(tn-x2)(tn-x3)(tn-x4)-t42(n+1)t-λ(tn-x3)(tn-x4),n≥0.Proposition 4.3.
Ifv=𝒥(-1/2,1/2) is the corecursive of the second kind Chebychev form, then the form u(t,λ) given by (2.9) has the following integral representation:
(4.19)〈u(t,λ),f〉=(1-λ)f(0)+tf′(0)+λπP∫-111x1-x1+xf(x)dx,∀f∈P.
It is a semi-classical form of class s satisfying the following functional equation:
(4.20)t≠0,z3(z2-1)S′(z;u(t,λ))=-z2(z2-z-1)S(z;u(t,λ))+(t-2λ+1)z2+tz-t,s=3t=0,z(z2-1)S′(z;u(0,λ))=(-z2+z+1)S(z;u(0,λ))-2λ+1,s=1.Proof.
It is well known thatv=𝒥(-/2,1/2) possesses the following integral representation [1]:
(4.21)〈v,f〉=∫-111π1-x1+xf(x)dx,f∈P.
From (2.11) we easily obtain (4.19).
The formv satisfies (3.4) with [15]
(4.22)ϕ(x)=x2-1,C(z)=1,D(z)=-2.
Therefore, ϑ1=-t, ϑ2=t, ϕ(0)≠0.
Now, we can simply use Proposition3.1 and Corollary 3.2 in order to obtain (4.20).Corollary 4.4.
Whent=0 and λ=-1, one has
(4.23)βn=(-1)n+1,n≥0,γ1=-12,γn+2=-14,n≥0,z(z2-1)S′(z;u(0,-1))=(-z2+z+1)S(z;u(0,-1))+3,s=1.Proof.
From Table2, we reach the desired results.Remarks 4.2.
(1) One has the formh-1u(0,-1)=ℒ(-3/2,1/2), where ℒ(α,β) is studied in [24].
(2) Let{Zn(1)}n≥0 [15, 19] be the first associated sequence of {Zn}n≥0 orthogonal with respect to u(0,-1) and βn(1), γn+1(1) the coefficients of the three-term recurrence relations; we have
(4.24)βn(1)=βn+1=(-1)n,n≥0;γn+1(1)=γn+2=-14,n≥0.The sequence{Zn(1)}n≥0 is a second-order self-associated sequence; that is, {Zn(1)}n≥0 is identical to its associated orthogonal sequence of second kind (see [25]).


## 4.1.v the Generalized Hermite Form

Let us describe the casev∶=ℋ(τ), where ℋ(τ) is the generalized Hermite form. Here is [1] (4.1)ξn=0,n≥0,σn+1=12(n+1+τ(1+(-1)n)),n≥0.
From (4.1), we get(4.2)∏μ=0nσ2μ+1=Γ(n+τ+3/2)Γ(τ+1/2),n≥0,∏μ=0nσ2μ=Γ(n+1),n≥0.
We want Λn=∑ν=0n1/σ2ν+1∏μ=0νσ2μ+1/σ2μ, n≥0.But from (4.1) and (4.2), we have 1/σ2ν+1∏μ=0νσ2μ+1/σ2μ=(1/Γ(τ+1/2))hν, with(4.3)hn=Γ(n+τ+1/2)Γ(n+1),n≥0,
fulfilling(4.4)(n+1)hn+1-nhn=(τ+12)hn,n≥0,
and so(4.5)Λn=1(τ+1/2)Γ(τ+1/2)∑ν=0n(ν+1)hν+1-νhν=1Γ(τ+3/2)Γ(n+τ+3/2)Γ(n+1),n≥0.
Then we get Table 1.Table 1ΔnΔ2n=(-1)n+1λ2Γ(τ+1/2)Γ(n+τ+1/2)Γ2(n+1)((ω-1)Λn+1)2,n≥0,Δ2n+1=(-1)nλΓ2(τ+1/2)Γ2(n+τ+3/2)Γ(n+1),n≥0.ana2n=(n+τ+1/2)2λΓ(τ+1/2)hn((ω-1)Λn+1)2,n≥0,a2n+1=-λΓ(τ+1/2)n+1hn+1((ω-1)Λn+1)2,n≥0.bnb2n=n+τ+1/2,n≥0,b2n+1=(n+1)(ω-1)Λn+1+1(ω-1)Λn+1,n≥0.cnc0=0,c1=-ωλ,c2n+2=1λn+τ+1/2Γ(τ+1/2)hn((ω-1)Λn+1)2,n≥0,c2n+3=-λ(n+1)Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1+1)((ω-1)Λn+1),n≥0.γ1=-λ2ω2,γ2=-(τ+1/2)2λ2ω4,γn+1γ2n+3=-λ2Γ2(τ+1/2)hn+12((ω-1)Λn+1+1)2((ω-1)Λn+1)2,n≥0,γ2n+4=-1λ2Γ2(τ+1/2)(n+τ+3/2)2hn+12((ω-1)Λn+1+1)4,n≥0.β0=ωλ,β1=-ωλ-τ+1/2λω2,βnβ2n+3=-λ(n+1)Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1+1)((ω-1)Λn+1)-n+τ+3/2λΓ(τ+1/2)hn+1((ω-1)Λn+1+1)2,n≥0,β2n+2=1λ1Γ(τ+1/2)(n+τ+1/2)hn((ω-1)Λn+1)2+λΓ(τ+1/2)hn+1((ω-1)Λn+1+1)((ω-1)Λn+1),n≥0.Proposition 4.1.
Ifv=ℋ(τ) is the generalized Hermite form, then the form u(τ,ω,λ) given by (2.9) has the following integral representation:
(4.6)〈u(τ,ω,λ),f〉=f(0)+λ(ω-1)f′(0)+λΓ(τ+1/2)P∫-∞+∞|x|2τxe-x2f(x)dx,∀f∈P.
It is a quasi-antisymmetric ((u(τ,ω,λ))2n+2=0, n≥0) and semi-classical form of class s satisfying the following functional equation:
(4.7)τ=0,ω≠1,z3S′(z;u(0,ω,λ))=-z2(2z2+1)S(z;u(0,ω,λ))-2z3-2λωz2+λ(ω-1),s=3,τ=0,ω=1,zS′(z;u(0,1,λ))=-(2z2+1)S(z;u(0,1,λ))-2z-2λ,s=1,(4.8)τ≠0,ω≠1,z3S′(z;u(τ,ω,λ))=-z2(2z2-2τ+1)S(z;u(τ,ω,λ))-2z3-2λωz2+2τz+2τλ(ω-1)+λ(ω-1),s=3,τ≠0,ω=1,z2S′(z;u(τ,1,λ))=z(-2z2+2τ-1)S(z;u(τ,1,λ))-2z2-2λz+2τ,s=2.Proof.
It is well known that the generalized Hermite form possesses the following integral representation [1]:
(4.9)〈v,f〉=∫-∞+∞1Γ(τ+1/2)|x|2τe-x2f(x)dx,R(τ)>-12,∀f∈P.
Following (2.11), we obtain (4.6). Also the form u is quasi-antisymmetric because it satisfies
(4.10)〈u,x2n+2〉=λ〈v,x2n+1〉=0,n≥0,
since v is symmetric by hypothesis.
Whenτ=0, v is classical and satisfies (3.4) with [22]
(4.11)ϕ(x)=1,C(z)=-2z,D(z)=-2.
Then, ϑ1=λ(ω-1), ϑ2=0.
Now, it is sufficient to use Corollary3.2 and Proposition 3.1 in order to obtain (4.7).
Ifτ≠0, the form v is semi-classical of class one and satisfies (3.4) with [23]
(4.12)ϕ(x)=x,C(z)=-2z2+2τ,D(z)=-2z.
Therefore ϑ1=0, ϑ2=λ(ω-1)(2τ+1), ϑ3=2τ.
By Proposition3.1 and Corollary 3.2 we can deduce (4.8).


## 4.2.v the Corecursive of the Second Kind Chebychev Form

Let us describe the casev∶=𝒥(-1/2,1/2); it is the corecursive of the second kind Chebychev functional. Here is [1](4.13)ξ0=-12,ξn+1=0,n≥0,σn+1=14,n≥0.
In this case we have the following result.Lemma 4.2.
Forn≥0, one has
(4.14)S2n(0)=(-1)n22n,S2n+1(0)=(-1)n22n+1,S2n(1)(0)=(-1)n22n,S2n+1(1)(0)=0,S2n′(0)=n(-1)n+122n-1,S2n+1′(0)=(n+1)(-1)n22n,(S2n(1))′(0)=0,(S2n+1(1))′(0)=(n+1)(-1)n22n,S2n′′(0)=n(n+1)(-1)n+122n-2,S2n+1′′(0)=n(n+1)(-1)n+122n-1.Proof.
The proof is analogous for the demonstration of Lemma2.8.Following Lemma4.2, for (2.44) we have (4.15)χ2n(0)=2n+124n,n≥0;χ2n+1(0)=n+124n+1,n≥0;χ2n′(0)=0,n≥0;χ2n+1′(0)=n+124n,n≥0;μ2n(0)=-n24n-1,n≥0;μ2n+1(0)=-n+124n+1,n≥0.
Therefore, we get for (2.42)(4.16)Δ2n=n(2n+1)(-1)n+126nλ2+(8n(n+1)(u)1-1)(-1)n26n+1λ+(n+1)(2n+1)(u)12(-1)n+126n,n≥0,Δ2n+1=(n+1)(2n+1)(-1)n+126n+3λ2+(8(n+1)2(u)1+1)(-1)n26n+4λ(n+1)(2n+3)(u)12(-1)n+126n+3,n≥0.
Then we obtain(4.17)Δ2n=4(-1)n+126n+1(tn-x1)(tn-x2),n≥0,Δ2n+1=4(-1)n+126n+4(tn-x3)(tn-x4),n≥0,
where(4.18)x1=14{-3t-2λ+(t2-4λt-4λ2-4λ)1/2},x2=14{-3t-2λ-(t2-4λt-4λ2-4λ)1/2},x3=14{-5t-2λ+((t+2λ)2+4λ)1/2},x4=14{-5t-2λ-((t+2λ)2+4λ)1/2},(u)1=t+λ.
On account of Proposition 2.6, we can deduce that the form u given by (2.9) is regular if and only if tn-xi≠0, n≥0, 1≤i≤4.In the sequel, we suppose that the last condition is satisfied.By virtue of (4.17) and Lemma 4.2, relations (2.49)–(2.52), and (2.55)–(2.57), (2.70)–(2.72) give Table 2.Table 2ana2n=-18(tn-x3)(tn-x4)(tn-x1)(tn-x2),n≥0,a2n+1=18(tn-x1)(tn-x2)(tn-x3)(tn-x4), n≥0.bnb0=-2x1x2+14-t2+(t+λ+12)2,b2n+1=14+t82(n+1)t-λ(tn-x1)(tn-x2),n≥0,b2n+2=14+t8(2n+3)t+λ(tn-x3)(tn-x4),n≥0.cnc0=-12,c1=-12-t-λ,c2n+3=182t(2n+1)((n+1)t-λ)-λ(tn-x3)(tn-x4),n≥0,c2n+2=-182t(2n+1)((n+1)t+λ)-λ(tn-x1)(tn-x2),n≥0.γ1=-2x1x2,γ2=λ16x3x4x12x22,γn+1γ2n+3=-14(tn-x1)(t(n+1)-x1)(tn-x2)(t(n+1)-x2)(tn-x3)2(tn-x4)2,n≥0,γ2n+4=-14(tn-x3)(t(n+1)-x3)(tn-x4)(t(n+1)-x4)(t(n+1)-x1)2(t(n+1)-x2)2,n≥0.β0=t+λ,β1=-t-λ-λ2x1x2{-2x1x2+14-t2+(t+λ+12)2},β2n+3=182t(2n+1)((n+1)t-λ)-λ(tn-x3)(tn-x4)+12(tn-x3)(tn-x4)(t(n+1)-x1)(t(n+1)-x2)βn+t4(2n+3)t+λ(t(n+1)-x1)(t(n+1)-x2),n≥0,β2n+2=-182t(2n+1)((n+1)t+λ)-λ(tn-x1)(tn-x2)-12(tn-x1)(tn-x2)(tn-x3)(tn-x4)-t42(n+1)t-λ(tn-x3)(tn-x4),n≥0.Proposition 4.3.
Ifv=𝒥(-1/2,1/2) is the corecursive of the second kind Chebychev form, then the form u(t,λ) given by (2.9) has the following integral representation:
(4.19)〈u(t,λ),f〉=(1-λ)f(0)+tf′(0)+λπP∫-111x1-x1+xf(x)dx,∀f∈P.
It is a semi-classical form of class s satisfying the following functional equation:
(4.20)t≠0,z3(z2-1)S′(z;u(t,λ))=-z2(z2-z-1)S(z;u(t,λ))+(t-2λ+1)z2+tz-t,s=3t=0,z(z2-1)S′(z;u(0,λ))=(-z2+z+1)S(z;u(0,λ))-2λ+1,s=1.Proof.
It is well known thatv=𝒥(-/2,1/2) possesses the following integral representation [1]:
(4.21)〈v,f〉=∫-111π1-x1+xf(x)dx,f∈P.
From (2.11) we easily obtain (4.19).
The formv satisfies (3.4) with [15]
(4.22)ϕ(x)=x2-1,C(z)=1,D(z)=-2.
Therefore, ϑ1=-t, ϑ2=t, ϕ(0)≠0.
Now, we can simply use Proposition3.1 and Corollary 3.2 in order to obtain (4.20).Corollary 4.4.
Whent=0 and λ=-1, one has
(4.23)βn=(-1)n+1,n≥0,γ1=-12,γn+2=-14,n≥0,z(z2-1)S′(z;u(0,-1))=(-z2+z+1)S(z;u(0,-1))+3,s=1.Proof.
From Table2, we reach the desired results.Remarks 4.2.
(1) One has the formh-1u(0,-1)=ℒ(-3/2,1/2), where ℒ(α,β) is studied in [24].
(2) Let{Zn(1)}n≥0 [15, 19] be the first associated sequence of {Zn}n≥0 orthogonal with respect to u(0,-1) and βn(1), γn+1(1) the coefficients of the three-term recurrence relations; we have
(4.24)βn(1)=βn+1=(-1)n,n≥0;γn+1(1)=γn+2=-14,n≥0.The sequence{Zn(1)}n≥0 is a second-order self-associated sequence; that is, {Zn(1)}n≥0 is identical to its associated orthogonal sequence of second kind (see [25]).



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