Division Problem of a Regular Form: The Case 𝑥^𝟐𝑢=𝜆𝑥𝑣

Abstract

We present a systematic study of a regular linear functional 𝑣 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.

1. Introduction

In the present paper, we intend to study the following problem: let 𝑣 be a regular form (linear functional), and 𝑅 and 𝐷 nonzero polynomials. Find all regular forms 𝑢 satisfying𝑅𝑢=𝐷𝑣.(1.1) 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 Problem 𝑥²𝑢=𝜆𝑥𝑣

Let 𝒫 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,𝑥^𝑛⟩, 𝑛≥0, the moments of 𝑢. For any form 𝑢, any polynomial 𝑔, any 𝑐∈ℂ, 𝑎∈ℂ−{0}, let 𝑢^, ℎ_𝑎𝑢, 𝑔𝑢, and (𝑥−𝑐)⁻¹𝑢 be the forms defined by duality:𝑢^,𝑝∶=−𝑢,𝑝(𝑥^−;𝑐)₋₁⟨ℎ_𝑎𝑢,𝑝⟩∶=⟨𝑢,ℎ_𝑎𝑝⟩;⟨𝑔𝑢,𝑝⟩∶=⟨𝑢,𝑔𝑝⟩;𝑢,𝑝∶=⟨𝑢,𝜃_𝑐𝑝⟩,𝑝∈𝒫,(2.1) where (𝜃_𝑐𝑝)(𝑥)=(𝑝(𝑥)−𝑝(𝑐))/(𝑥−𝑐); (ℎ_𝑎𝑝)(𝑥)=𝑝(𝑎𝑥).

We define a left multiplication of a form 𝑢 by a polynomial 𝑝 as(𝑢𝑝)(𝑥)∶=𝑢,𝑥𝑝(𝑥)−𝜉𝑝(𝜉)𝑥−𝜉,𝑢∈𝒫^,𝑝∈𝒫.(2.2) Let us recall that a form 𝑢 is called regular if there exists a monic polynomial sequence {𝑃_𝑛}_𝑛≥0, deg𝑃_𝑛=𝑛, such that⟨𝑢,𝑃_𝑛𝑃_𝑚⟩=𝑟_𝑛𝛿_𝑛,𝑚,𝑛,𝑚≥0,𝑟_𝑛≠0,𝑛≥0.(2.3) We have the following result.

Lemma 2.1 (see [15]). Let 𝑢∈𝒫^, 𝑓∈𝒫, and 𝑐∈𝒞. The following formulas hold: (𝑣𝑓)^(𝑥)=𝑣^𝑓(𝑥)+𝑣𝑓^(𝑥)+𝑣𝜃₀𝑓(𝑥),𝑓∈𝒫.(2.4)(𝛿𝑓)(𝑥)=𝑓(𝑥),𝑓∈𝒫.(2.5)(𝑥−𝑐)⁻¹((𝑥−𝑐)𝑢)=𝑢−(𝑢)₀𝛿_𝑐,(2.6) where ⟨𝛿_𝑐,𝑝⟩=𝑝(𝑐), 𝑝∈𝒫.

We consider the following problem: given a regular form 𝑣, find all regular forms 𝑢 satisfying𝑥²𝑢=𝜆𝑥𝑣,𝜆∈ℂ−{0},(2.7) with constraints (𝑢)₀=1, (𝑣)₀=1. From (2.6) we can deduce that𝑥𝑢=(𝑢)₁−𝜆𝛿+𝜆𝑣,(2.8)𝑢=𝛿+𝜆−(𝑢)₁𝛿^+𝜆𝑥⁻¹𝑣.(2.9) 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 form 𝑣 has the following integral representation:⟨𝑣,𝑓⟩=^+∞−∞𝑉(𝑥)𝑓(𝑥)𝑑𝑥,foreachpolynomial𝑓,(2.10) where 𝑉 is a locally integrable function with rapid decay, continuous at the origin; then the form 𝑢 is represented by⟨𝑢,𝑓⟩=1−𝜆𝑃^+∞−∞𝑉(𝑥)𝑥𝑑𝑥𝑓(0)+(𝑢)¹−𝜆𝑓^(0)+𝜆𝑃^+∞−∞𝑉(𝑥)𝑓(𝑥)𝑥𝑑𝑥,(2.11) where [16, 17]𝑃^+∞−∞𝑉(𝑥)𝑥𝑑𝑥=lim_𝜖→0⁺^{−𝜖−∞}𝑉(𝑥)𝑥𝑑𝑥+^𝜖+∞𝑉(𝑥)𝑥𝑑𝑥.(2.12) Let {𝑆_𝑛}_𝑛≥0 denote the sequence of monic orthogonal polynomials with respect to 𝑣; we have𝑆_𝑛+2𝑆₀(𝑥)=1,𝑆₁(𝑥)=𝑥−𝜉₀,(𝑥)=𝑥−𝜉_𝑛+1𝑆_𝑛+1(𝑥)−𝜎_𝑛+1𝑆_𝑛(𝑥),𝑛≥0,(2.13) with𝜉_𝑛=𝑣,𝑥𝑆^2𝑛(𝑥)𝑣,𝑆^2𝑛,𝜎^𝑛+1=𝑣,𝑆^2𝑛+1𝑣,𝑆^2𝑛,𝑛≥0.(2.14) When 𝑢 is regular, let {𝑍_𝑛}_𝑛≥0 be the corresponding MOPS:𝑍_𝑛+2𝑍₀(𝑥)=1,𝑍₁(𝑥)=𝑥−𝛽₀.(𝑥)=𝑥−𝛽_𝑛+1𝑍_𝑛+1(𝑥)−𝛾_𝑛+1𝑍_𝑛(𝑥),𝑛≥0.(2.15) From (2.7), we know that the existence of the sequence {𝑍_𝑛}_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,𝑥𝑍_𝑛+2𝑥𝑍₀(𝑥)=𝑆₁(𝑥)+𝑐₀,𝑥𝑍₁(𝑥)=𝑆₂(𝑥)+𝑐₁𝑆₁(𝑥)+𝑏₀.(𝑥)=𝑆_𝑛+3(𝑥)+𝑐_𝑛+2𝑆_𝑛+2(𝑥)+𝑏_𝑛+1𝑆_𝑛+1(𝑥)+𝑎_𝑛𝑆_𝑛(𝑥),𝑛≥0,(2.16) with 𝑎_𝑛≠0, 𝑛≥0.

From (2.16), we have𝑍₁(𝑥)=𝜃₀𝑆₂(𝑥)+𝑐₁,(2.17)𝑍_𝑛+2(𝑥)=𝜃₀𝑆_𝑛+3(𝑥)+𝑐_𝑛+2𝜃₀𝑆_𝑛+2(𝑥)+𝑏_𝑛+1𝜃₀𝑆_𝑛+1(𝑥)+𝑎_𝑛𝜃₀𝑆_𝑛(𝑥),𝑛≥0.(2.18)

Lemma 2.2. Let {𝑍_𝑛}_𝑛≥0 be a sequence of polynomials satisfying (2.16) where 𝑎_𝑛, 𝑏_𝑛, and 𝑐_𝑛 are complex numbers such that 𝑎_𝑛≠0 for all 𝑛≥0. The sequence {𝑍_𝑛}_𝑛≥0 is orthogonal with respect to 𝑢 if and only if ⟨𝑢,𝑥𝑍_𝑛⟨𝑢,𝑍_𝑛⟩=0,𝑛≥1,(𝑥)⟩=0,𝑛≥2,⟨𝑢,𝑥𝑍₁(𝑥)⟩≠0.(2.19)

Proof. The conditions (2.19) are necessary from the definition of the orthogonality of {𝑍_𝑛}_𝑛≥0 with respect to 𝑢.
For 𝑘≥2, we have (by (2.7)) 𝑢,𝑥^𝑘𝑍_𝑛+2(𝑥)=𝑥²𝑢,𝑥^𝑘−2𝑍_𝑛+2(𝑥)=𝜆𝑣,𝑥^𝑘−1𝑍_𝑛+2(𝑥),𝑛≥0,(2.20) and from (2.16), we get 𝑢,𝑥^𝑘𝑍_𝑛+2(𝑥)=+𝜆𝜆𝑏𝑣,𝑥^𝑘−2_𝑛+1𝑆_𝑛+3(𝑥)+𝜆𝑐_𝑛+2𝑣,𝑥^𝑘−2𝑆_𝑛+2(𝑥)𝑣,𝑥^𝑘−2𝑆_𝑛+1(𝑥)+𝜆𝑎_𝑛𝑣,𝑥^𝑘−2𝑆_𝑛(𝑥),𝑛≥0.(2.21) Taking into account the orthogonality of {𝑆_𝑛}_𝑛≥0, we obtain 𝑢𝑢,,𝑥𝑥^𝑘𝑍_𝑛+2𝑛+2(𝑥)=0,2≤𝑘≤𝑛+1,𝑛≥1,𝑍_𝑛+2(𝑥)=𝜆𝑎_𝑛𝑣,𝑆^2𝑛≠0,𝑛≥0.(2.22) By (2.19), it follows that 𝑢,𝑍⟨𝑢,_𝑛+2𝑍₁⟩=0,⟨𝑢,𝑥𝑍₁(𝑥)⟩≠0,=𝑢,𝑥𝑍_𝑛+2(𝑥)=0,𝑛≥0.(2.23) Consequently, the previous relations and (2.22) prove that {𝑍_𝑛}_𝑛≥0 is orthogonal with respect to 𝑢, which proves the Lemma.

Remark 2.3. When 𝑢 is regular, from Theorem 5.1 in [21], there exist complex numbers 𝑟_𝑛+2≠0, 𝑡_𝑛+2 and 𝑣_𝑛+2≠0 such that 𝑍_𝑛+2(𝑥)+𝑟_𝑛+2𝑍_𝑛+1(𝑥)=𝑆_𝑛+2(𝑥)+𝑡_𝑛+2𝑆_𝑛+1(𝑥)+𝑣_𝑛+2𝑆_𝑛(𝑥),𝑛≥0.(2.24)

From (2.16), (2.24), and (2.15) we obtain the following relations:𝑟𝑟_𝑛+2−𝑡_𝑛+2+𝑐_𝑛+2−𝜉_𝑛+2=0,𝑛≥0,_𝑛+2𝑐𝑟_𝑛+1−𝑡_𝑛+2𝜉_𝑛+1−𝑣_𝑛+2+𝑏_𝑛+1−𝜎_𝑛+2=0,𝑛≥0,_𝑛+2𝑏_𝑛−𝑟𝑡_𝑛+2𝜎_𝑛+1−𝑣_𝑛+2𝜉_𝑛+𝑎_𝑛=0,𝑛≥0,_𝑛+2𝑎_𝑛−1−𝑣_𝑛+2𝜎_𝑛=0,𝑛≥1.(2.25) Taking into account (2.16), (2.18) and (2.19), we get0==𝑢𝑢,,𝑥𝑆𝑍_𝑛+2𝑛+3(𝑥)0=𝑢,𝑍_𝑛+2+𝑐_𝑛+2𝑢,𝑆_𝑛+2+𝑏_𝑛+1𝑢,𝑆_𝑛+1+𝑎_𝑛⟨𝑢,𝑆_𝑛⟩=0,𝑛≥0,=𝑢,𝜃₀0=𝑆_𝑛+3𝑆_𝑛+3+𝑐_𝑛+2𝑢,𝜃₀𝑆_𝑛+2+𝑏_𝑛+1𝑢,𝜃₀𝑆_𝑛+1+𝑎_𝑛⟨𝑢,𝜃₀𝑆_𝑛⟩,𝑛≥0,(0)+𝑐_𝑛+2𝑆_𝑛+2(0)+𝑏_𝑛+1𝑆_𝑛+1(0)+𝑎_𝑛𝑆_𝑛(0),𝑛≥0,(2.26) with the initial conditions:0=𝑆0₂=𝑆₁(0)+𝑐₀,0=⟨𝑢,𝑍₁(0)+𝑐₁𝑆₁(0)+𝑏₀,0≠⟨𝑢,𝑥𝑍₁⟩=𝑢,𝜃₀𝑆₂+𝑐₁,(𝑥)⟩=⟨𝑢,𝑆₂⟩+𝑐₁⟨𝑢,𝑆₁⟩+𝑏₀.(2.27) If we denoteΔ_𝑛∶=||||||𝑆^𝑛+2𝑢,𝑆(0_𝑛+2)𝑆_𝑛+1(0)𝑆_𝑛(0)𝑢,𝜃₀𝑢,𝑆_𝑛+1⟨𝑢,𝑆_𝑛⟩𝑆_𝑛+2𝑢,𝜃₀𝑆_𝑛+1⟨𝑢,𝜃₀𝑆_𝑛⟩||||||,𝑛≥0,(2.28) from the Cramer rule we have Δ_𝑛𝑎_𝑛=−Δ_𝑛+1,𝑛≥0,(2.29)Δ_𝑛𝑏_𝑛+1=||||||𝑆^𝑛+2⟨𝑢,𝑆(0_𝑛+2)−𝑆_𝑛+3(0)𝑆_𝑛(0)⟨𝑢,𝜃₀⟩−⟨𝑢,𝑆_𝑛+3⟩⟨𝑢,𝑆_𝑛⟩𝑆_𝑛+2⟩−⟨𝑢,𝜃₀𝑆_𝑛+3⟩⟨𝑢,𝜃₀𝑆_𝑛⟩||||||,𝑛≥0,(2.30)Δ_𝑛𝑐_𝑛+2=||||||−𝑆^𝑛+3−⟨𝑢,𝑆(0_𝑛+3)𝑆_𝑛+1(0)𝑆_𝑛(0)−⟨𝑢,𝜃₀⟩⟨𝑢,𝑆_𝑛+1⟩⟨𝑢,𝑆_𝑛⟩𝑆_𝑛+3⟩⟨𝑢,𝜃₀𝑆_𝑛+1⟩⟨𝑢,𝜃₀𝑆_𝑛⟩||||||,𝑛≥0.(2.31)

Lemma 2.4. The following formulas hold: ⟨𝑢,𝑆_𝑛⟩=𝑆_𝑛(0)+(𝑢)₁−𝜆𝑆^𝑛(0)+𝜆𝑆^(1)𝑛−1(0),𝑛≥0,(2.32)⟨𝑢,𝑥𝑆_𝑛(𝑥)⟩=(𝑢)₁−𝜆𝑆_𝑛(0),𝑛≥1,(2.33)𝑢,𝜃₀𝑆_𝑛=𝑆^𝑛(0)+12(𝑢)¹−𝜆𝑆^𝑛(0)+𝜆𝑆^(1)𝑛−1^(0),𝑛≥0,(2.34)𝑆^𝑛(1)(0)𝑆_𝑛(0)−𝑆^(1)𝑛−1(0)𝑆_𝑛+1(0)=𝑣,𝑆^2𝑛,𝑛≥0,(2.35) where 𝑆^𝑛(1)(𝑥)∶=(𝑣𝜃₀𝑆_𝑛+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 𝑣,𝜃²⁰𝑆_𝑛==⟨𝑣𝑥,𝜃𝜃₀₀𝑆^𝑛−𝜃₀𝑆_𝑛^⟩=⟨𝑣,𝜃₀𝑆^𝑛⟩+𝑣^,𝜃₀𝑆_𝑛𝑆^𝑛(0)+𝑣^𝜃₀𝑆_𝑛(0),𝑛≥0.(2.36) Using (2.4), we get 𝑣,𝜃²⁰𝑆_𝑛=𝑣𝜃₀𝑆_𝑛^(0)=𝑆^(1)𝑛−1^(0),𝑛≥0.(2.37) From (2.9), we obtain ⟨𝑢,𝜃₀𝑆_𝑛⟩=⟨𝛿,𝜃₀𝑆_𝑛⟩+(𝑢)₁−𝜆𝛿,𝜃₀𝑆_𝑛^+𝜆𝑣,𝜃²⁰𝑆_𝑛,𝑛≥0.(2.38) According to (2.5) and (2.37), we can deduce (2.34).
We have 𝑆_(1)𝑛+2𝑆⁰⁽¹⁾(𝑥)=1,𝑆¹⁽¹⁾(𝑥)=𝑥−𝜉₂,(𝑥)=𝑥−𝜉_𝑛+2𝑆^(1)𝑛+1(𝑥)−𝜎_𝑛+2𝑆^𝑛(1)(𝑥),𝑛≥0.(2.39) Then (by (2.39)) 𝑆^𝑛(1)(0)𝑆_𝑛(0)−𝑆^(1)𝑛−1(0)𝑆_𝑛+1(0)==𝜎𝜎_𝑛𝑛𝑆^(1)𝑛−1(0)𝑆_𝑛−1(0)+𝑆_𝑛(0)𝑆^𝑛(1)(0)+𝜉_𝑛𝑆^(1)𝑛−1(0)𝑆^(1)𝑛−1(0)𝑆_𝑛−1(0)−𝑆^(1)𝑛−2(0)𝑆_𝑛(0).(2.40) It follows that 𝑆^𝑛(1)(0)𝑆_𝑛(0)−𝑆^(1)𝑛−1(0)𝑆_𝑛+1(0)=^𝑛_𝜇=0𝜎_𝜇=𝑣,𝑆^2𝑛,𝑛≥0,(2.41) hence (2.35).

Proposition 2.5. One has Δ_𝑛=𝐸_𝑛𝜆²+𝐹_𝑛𝜆+𝐺_𝑛,𝑛≥0,(2.42) where 𝐸_𝑛=𝑆_𝑛+1(0)𝜇_𝑛(0)+12𝜒^𝑛𝐹(0)+𝑆^𝑛(1)(0)−𝑆^𝑛+1(0)𝜒_𝑛(0)−𝑣,𝑆^2𝑛,𝑛≥0,_𝑛=−−𝑆(𝑢)_𝑛+11(0)(𝑢)₁𝜇_𝑛(0)+𝜒^𝑛(0)+𝑣,𝑆^2𝑛𝐺𝑆^𝑛(1)(0)𝜒_𝑛(0)−2𝑆^𝑛+1(0)𝜒_𝑛(0)+𝑆^𝑛+1(0)𝑣,𝑆^2𝑛,𝑛≥0,_𝑛=(𝑢)²¹12𝑆^𝑛+1(0)𝜒^𝑛(0)−𝑆^𝑛+1(0)𝜒_𝑛(0),𝑛≥0,(2.43) with 𝜇𝜒_𝑛(𝑥)=𝑆_𝑛(𝑥)𝑆^𝑛+1(𝑥)−𝑆_𝑛+1(𝑥)𝑆^𝑛(𝑥),𝑛≥0,_𝑛(𝑥)=𝑆_𝑛+1(𝑥)𝑆^(1)𝑛−1^(𝑥)−𝑆_𝑛(𝑥)𝑆^𝑛(1)^(𝑥),𝑛≥0.(2.44)

Proof. Using (2.13), we, respectively, obtain 𝑢,𝑆_𝑛+2𝑆_𝑛+2(0)=−𝜉_𝑛+1𝑆_𝑛+1(0)−𝜎_𝑛+1𝑆_𝑛(0),𝑛≥0,𝑢,𝜃₀=𝑢,𝑥𝑆_𝑛+1(𝑥)−𝜉_𝑛+1𝑢,𝑆_𝑛+1−𝜎_𝑛+1⟨𝑢,𝑆_𝑛⟩,𝑛≥0,𝑆_𝑛+2=𝑢,𝑆_𝑛+1−𝜉_𝑛+1𝑢,𝜃₀𝑆_𝑛+1−𝜎_𝑛+1⟨𝑢,𝜃₀𝑆_𝑛⟩,𝑛≥0.(2.45) Taking into account previous relations, we obtain for (2.28) the following: Δ_𝑛=||||||0𝑆^𝑛+1𝑢,𝑥𝑆_𝑛+1(0)𝑆_𝑛(0)𝑢,𝑆_𝑛+1(𝑥)𝑢,𝑆_𝑛+1⟨𝑢,𝑆_𝑛⟩𝑢,𝜃₀𝑆_𝑛+1⟨𝑢,𝜃₀𝑆_𝑛⟩||||||,𝑛≥0,(2.46) that is, Δ_𝑛=−𝑢,𝑥𝑆_𝑛+1(𝑥)||||𝑆^𝑛+1𝑢,𝜃₀(0)𝑆_𝑛(0)𝑆_𝑛+1⟨𝑢,𝜃₀𝑆_𝑛⟩||||+𝑢,𝑆^𝑛+1||||𝑆^𝑛+1𝑢,𝑆(0_𝑛+1)𝑆_𝑛(0)⟨𝑢,𝑆_𝑛⟩||||,𝑛≥0.(2.47) Let 𝑛≥0; based on the relations (2.32)–(2.34), it follows that ||||𝑢𝑆,𝜃_𝑛+10(0)𝑆_𝑛(0)𝑆_𝑛+1⟨𝑢,𝜃₀𝑆_𝑛⟩||||=𝜇^𝑛(0)+12𝜒^𝑛(0)𝜆−𝜒_𝑛(0)−12(𝑢)¹||||𝑆𝜒^𝑛(0),_𝑛+1𝑢,𝑆(0_𝑛+1)𝑆_𝑛(0)⟨𝑢,𝑆_𝑛⟩||||=𝜒^𝑛(0)−𝑣,𝑆^2𝑛𝜆−(𝑢)₁𝜒_𝑛(0).(2.48) From (2.48) and (2.47), we obtain the desired results.

Proposition 2.6. The form 𝑢 is regular if and only if Δ_𝑛≠0, 𝑛≥0. Then, the coefficients of the three-term recurrence relation (2.15) are given by 𝛾₁=Δ₀,𝛾₂=−𝜆Δ₁Δ⁰⁻²,(2.49)𝛾_𝑛+3=Δ^𝑛Δ_𝑛+2Δ^2𝑛+1𝜎_𝑛+1,𝑛≥0,(2.50)𝛽₀=(𝑢)₁,𝛽₁=𝑐₁−𝜉₀−𝜉₁+𝜆𝑏₀Δ⁰⁻¹,(2.51)𝛽_𝑛+2=𝑐_𝑛+2−𝜉_𝑛+1−𝜉_𝑛+2−𝑏_𝑛+1Δ_𝑛Δ^−1𝑛+1𝜎_𝑛+1,𝑛≥0.(2.52)

Proof
Necessity
From (2.27) and Lemma 2.4, we get ⟨𝑢,𝑥𝑍₁(𝑥)⟩=⟨𝑢,𝑆₂⟩+⟨𝑢,𝜃₀𝑆₂⟩𝑆₁(0)−⟨𝑢,𝑆₁⟩−𝑆₂(0)=𝜆𝑆₁(0)−(𝑢)²¹,(2.53) and again with (2.27) and (2.42), we can deduce that Δ₀=⟨𝑢,𝑆₂⟩+⟨𝑢,𝜃₀𝑆₂⟩𝑆₁(0)−⟨𝑢,𝑆₁⟩−𝑆₂(0)=⟨𝑢,𝑥𝑍₁(𝑥)⟩≠0.(2.54) Moreover, {𝑍_𝑛}_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 𝑎_𝑛≠0, 𝑛≥0. This implies Δ_𝑛≠0, 𝑛≥0. Otherwise, if there exists an 𝑛₀≥1 such that Δ_𝑛0=0, from (2.29), Δ₀=0, which is a contradiction.
Sufficiency
Let 𝑐₀=−𝑆₁(0)=𝜉₀,(2.55)𝑐₁=−𝑢,𝜃₀𝑆₂,(2.56)𝑏₀=Δ₀−⟨𝑢,𝑆₂⟩−𝑐₁⟨𝑢,𝑆₁⟩.(2.57) We get ⟨𝑢,𝑥𝑍₁(𝑥)⟩=⟨𝑢,𝑆₂⟩+𝑐₁⟨𝑢,𝑆₁⟩+𝑏₀=Δ₀≠0.(2.58) We have ⟨𝑢,𝑍₁⟩=𝑐₁+⟨𝑢,𝜃₀𝑆₂⟩=0.
From (2.56) and (2.57) we get 𝑆₂(0)+𝑐₁𝑆₁(0)+𝑏₀=𝑆₂(0)−⟨𝑢,𝑆₂⟩−⟨𝑢,𝜃₀𝑆₂⟩𝑆₁(0)−⟨𝑢,𝑆₁⟩+Δ₀.(2.59) 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 𝑎_𝑛, 𝑏_𝑛, and 𝑐_𝑛 (𝑛≥0), define a sequence polynomials {𝑍_𝑛}_𝑛≥0 by (2.16). Then it follows from (2.26) and Lemma 2.2 that 𝑢 is regular and {𝑍_𝑛}_𝑛≥0 is the corresponding MOPS.
Moreover, by (2.22) we get 𝑢,𝑍^2𝑛+2=𝜆𝑎_𝑛𝑣,𝑆^2𝑛,𝑛≥0.(2.60) Making 𝑛=0 in (2.60), it follows that 𝑢,𝑍²²=𝜆𝑎₀.(2.61) Based on relations (2.58), (2.60), (2.61), and (2.29), we, respectively, obtain 𝛾₁=⟨𝑢,𝑥𝑍₁(𝑥)⟩=Δ₀;𝛾₂=𝑢,𝑍²²⟨𝑢,𝑥𝑍₁(𝑥)⟩=−𝜆Δ¹𝛾_𝑛+3Δ⁰⁻²,=𝑢,𝑍^2𝑛+3𝑢,𝑍^2𝑛+2=Δ^𝑛Δ_𝑛+2Δ^2𝑛+1𝜎_𝑛+1,𝑛≥0.(2.62) We have proved (2.49) and (2.50).
When {𝑍_𝑛}_𝑛≥0 is orthogonal, we have 𝛽₀=(𝑢)₁.(2.63) By (2.16) and the orthogonality of {𝑍𝑛}_𝑛≥0, we get 𝑢,𝑥𝑍²¹(𝑥)=𝑐₁𝑢,𝑍²¹+⟨𝑢,𝑆₂𝑍₁⟩.(2.64) By virtue of (2.13) and the regularity of 𝑢 we obtain ⟨𝑢,𝑆₂𝑍₁⟩==𝜆𝑏𝑥²₀𝑢,𝑍₁−𝜉₀+𝜉₁𝑢,𝑍²¹=𝜆⟨𝑣,𝑥𝑍₁(𝑥)⟩−𝜉₀+𝜉₁𝑢,𝑍²¹−𝜉₀+𝜉₁𝑢,𝑍²¹,(2.65) and consequently, we get the second result in (2.51) from (2.58), and (2.64).
From (2.16), and the orthogonality of {𝑍_𝑛}_𝑛≥0, we have 𝛽_𝑛+2𝑢,𝑍^2𝑛+2=𝑐_𝑛+2𝑢,𝑍^2𝑛+2+𝑢,𝑆_𝑛+3𝑍_𝑛+2,𝑛≥0.(2.66) Using (2.13), (2.16), and the the orthogonality of {𝑆_𝑛}_𝑛≥0, we have 𝑢,𝑆_𝑛+3𝑍_𝑛+2=𝜆𝑏_𝑛+1𝑣,𝑆^2𝑛+1−𝜉_𝑛+1+𝜉_𝑛+2𝑢,𝑍^2𝑛+2,𝑛≥0.(2.67) Taking into account the previous relation, (2.66) becomes 𝛽_𝑛+2=𝑐_𝑛+2−𝜉_𝑛+1−𝜉_𝑛+2+𝜆𝑏_𝑛+1𝑣,𝑆^2𝑛+1𝑢,𝑍^2𝑛+2,𝑛≥0.(2.68) From (2.60) and (2.29), we have 𝑣,𝑆^2𝑛+1𝑢,𝑍^2𝑛+2=−𝜆⁻¹Δ_𝑛Δ^−1𝑛+1𝜎_𝑛+1,𝑛≥0.(2.69) Last equation and (2.68) give (2.52).

Moreover, if the form 𝑢 is regular, for (2.29), (2.30), and (2.31), we get𝑎_𝑛=−Δ^𝑛+1Δ_𝑛,𝑛≥0,(2.70)𝑏_𝑛+1=𝐷_𝑛𝜆²+𝐻_𝑛𝜆+𝐼_𝑛Δ^𝑛−1+𝜎_𝑛+2,𝑛≥0,(2.71)𝑐_𝑛+2=−𝐽_𝑛𝜆²+𝐿_𝑛𝜆+𝐾_𝑛Δ^𝑛−1+𝜉_𝑛+2,𝑛≥0,(2.72) where𝐷_𝑛=𝑆_𝑛(0)𝑣,𝑆^2𝑛+1−𝜒_𝑛+1(0)−𝜉_𝑛+1𝑆_𝑛+2(0)𝜇_𝑛(0)+12𝜒^𝑛−𝜉_𝑛+1(0)𝐻𝑆^(1)𝑛+1(0)−𝑆^𝑛+2(0)𝜒_𝑛(0)−𝑣,𝑆^2𝑛,𝑛≥0,_𝑛=+(𝑢()𝑢₁)𝑆_𝑛1(0)2𝜒_𝑛+1(0)−𝑣,𝑆^2𝑛+1+𝜉_𝑛+1𝑆_𝑛+2(0)𝜒_𝑛(0)+𝜉_𝑛+1𝜒^𝑛(0)+𝜇_𝑛(0)+(𝑢)₁𝜉_𝑛+1𝜒_𝑛(0)𝑆^(1)𝑛+1(0)−𝑆^𝑛+2(0)𝐼𝑣,𝑆^2𝑛−𝜒_𝑛(0)𝑆_𝑛+2(0)+(𝑢)₁𝑆^𝑛+2(0),𝑛≥0,_𝑛=−(𝑢)²¹𝑆^𝑛(0)𝜒_𝑛+1(0)+12𝜉^𝑛+1𝑆_𝑛+2𝐽(0)𝜒^𝑛(0)−𝑆^𝑛+2(0)𝜒_𝑛(0),𝑛≥0,_𝑛=𝑆_𝑛+2(0)𝜇_𝑛(0)+12𝜒^𝑛𝐿(0)+(𝑆^(1)𝑛+10−𝑆^𝑛+2(0)𝜒_𝑛(0)−𝑣,𝑆^2𝑛,𝑛≥0,_𝑛=−(𝑢)₁𝑣𝜒,𝑆_𝑛(_2𝑛0)2𝑆^𝑛+2(0)−𝑆^(1)𝑛+1(0)−(𝑢)₁𝑆_𝑛+2(0)𝜇_𝑛(0)+𝜒^𝑛(0)𝐾𝑆_𝑛+2(0)+(𝑢)₁𝑆^𝑛+2(0),𝑛≥0,_𝑛=(𝑢)²¹12𝑆^𝑛+2(0)𝜒^𝑛(0)−𝜒_𝑛(0)𝑆^𝑛+2(0),𝑛≥0.(2.73) In the sequel, we will assume that 𝑣 is a symmetric linear form.

We need the following lemmas.

Lemma 2.7. If {𝑦_𝑛}_𝑛≥0 and {𝑏_𝑛}_𝑛≥0 are sequences of complex numbers fulfilling 𝑦_𝑛+1+𝑎_𝑛𝑦_𝑛=𝑏𝑦_𝑛+1,𝑛≥0,𝑎_𝑛≠0,𝑛≥0,₀=𝑏₀,(2.74) then 𝑦_𝑛=(−1)^𝑛𝑎^𝑛−1^𝑛_𝜇=0𝑎_𝜇^𝑛𝜈=0(−1)^𝜈𝑎_𝜈^𝜈_𝜇=0𝑎^𝜇−1𝑏^𝜈,𝑛≥0.(2.75)

Lemma 2.8. When {𝑆_𝑛}_𝑛≥0 given by (2.13) is symmetric, one has 𝑆_2𝑛(0)=(−1)^𝑛𝜎_𝑛2𝑛+1_𝜇=0𝑆𝜎_2𝜇+1,𝑛≥0,𝑆_2𝑛+1(0)=0,𝑛≥0,_(1)2𝑛(0)=(−1)^𝑛𝑛_𝜇=0𝑆𝜎_2𝜇,𝑛≥0,𝑆^(1)2𝑛+1(0)=0,𝑛≥0,_2𝑛+1(0)=(−1)^𝑛^𝑛_𝜇=0𝜎_2𝜇Λ^𝑛𝑆,𝑛≥0,𝑆^2𝑛(0)=0,𝑛≥0,_(1)2𝑛^(0)=0,𝑛≥0,𝑆^{2𝑛+1}(0)=0,𝑛≥0.(2.76)

Proof. As 𝑣 is symmetric, then 𝜉_𝑛=0, 𝑛≥0, and therefore from (2.13) we have 𝑆𝑆₀(0)=1,𝑆₁(0)=0,𝑆⁰⁽¹⁾(0)=1,𝑆¹⁽¹⁾(0)=0,_𝑛+2𝑆(0)=−𝜎_𝑛+1𝑆_𝑛(0),𝑛≥0,𝑆^(1)𝑛+2(0)=−𝜎_𝑛+2𝑆^𝑛(1)(0),𝑛≥0,_0𝑆(0)=0,𝑆^1(0)=1,𝑆^𝑛+2(0)=−𝜎_𝑛+1𝑆^𝑛(0)+𝑆_𝑛+1(0),𝑛≥0,₀₍₁₎𝑆^(0)=0,𝑆^(1)𝑛+2^(0)=−𝜎_𝑛+2𝑆^𝑛(1)^(0)+𝑆^(1)𝑛+1(0),𝑛≥0,_0(0)=0,𝑆^1(0)=0,𝑆^𝑛+2(0)=−𝜎_𝑛+1𝑆^𝑛(0)+2𝑆^𝑛+1(0),𝑛≥0.(2.77) Now, it is sufficient to use Lemma 2.7 in order to obtain the desired results.

Let 𝜔=𝜆⁻¹(𝑢)₁.(2.78)

Corollary 2.9. If 𝑣 is a symmetric form, one has Δ_2𝑛=𝜆²(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇+1^𝑛_𝜇=0𝜎_2𝜇²Δ(𝜔−1)Λ_𝑛+1²,𝑛≥0,_2𝑛+1=𝜆(−1)^𝑛^𝑛_𝜇=0𝜎_2𝜇+1²^𝑛_𝜇=0𝜎_2𝜇,𝑛≥0,(2.79) where Λ_𝑛=^𝑛𝜈=01𝜎_𝜈2𝜈+1_𝜇=0𝜎_2𝜇+1𝜎_2𝜇,𝑛≥0,𝜎₀=1.(2.80)

Proof. Following Lemma 2.8, for (2.43) we have 𝐸_2𝑛=(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+11−Λ^𝑛𝐹,𝑛≥0;𝐸_2𝑛+1=0,𝑛≥0,_2𝑛=2𝜔𝜆(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+11−Λ^𝑛𝐹Λ_𝑛+1,𝑛≥0,_2𝑛+1=(−1)^𝑛^𝑛_𝜇=0𝜎_2𝜇^𝑛_𝜇=0𝜎_2𝜇+1²𝐺,𝑛≥0,_2𝑛=𝜔²𝜆²(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+1Λ^2𝑛,𝑛≥0;𝐺_2𝑛+1=0,𝑛≥0.(2.81) As a consequence, relations (2.81) and (2.42) yield (2.79).

Theorem 2.10. The form 𝑢 is regular if and only if (𝜔−1)Λ_𝑛+1≠0, 𝑛≥0, where Λ_𝑛 is defined in (2.80).
In this case one has 𝑎_2𝑛=𝜎^2𝑛+1𝜆Θ_𝑛(𝜔−1)Λ_𝑛+1²,𝑎_2𝑛+1=−𝜆𝜎^22𝑛+2Θ_𝑛(𝜔−1)Λ_𝑛+1²,𝑛≥0,(2.82)𝑏_2𝑛=𝜎_2𝑛+1,𝑛≥0,𝑏_2𝑛+1=𝜎_2𝑛+2(𝜔−1)Λ_𝑛+1+1(𝜔−1)Λ_𝑛+1,𝑛≥0,(2.83)𝑐₀=0,𝑐₁=−𝜔𝜆,𝑐_2𝑛+2=1𝜆Θ_𝑛(𝜔−1)Λ_𝑛+1²𝑐,𝑛≥0,_2𝑛+3𝛾=−𝜆𝜎_2𝑛+2Θ_𝑛(𝜔−1)Λ_𝑛+1+1(𝜔−1)Λ_𝑛+1,𝑛≥0,₁=−𝜆²𝜔²,𝛾₂=−𝜎²¹𝜆²𝜔⁴,𝛾_2𝑛+4=1𝜆²Θ^2𝑛+1𝛾(𝜔−1)Λ_𝑛+1+1²,𝑛≥0,_2𝑛+3𝛽=𝜆²𝜎^22𝑛+2Θ^2𝑛(𝜔−1)Λ_𝑛+1²(𝜔−1)Λ_𝑛+1+1²,𝑛≥0,₀=𝜆𝜔,𝛽₁=−𝜆𝜔−𝜎¹𝜆𝜔²,(2.84)𝛽_2𝑛+2=1𝜆Θ_𝑛(𝜔−1)Λ_𝑛+1²𝛽+𝜆𝜎_2𝑛+2Θ_𝑛(𝜔−1)Λ_𝑛+1(𝜔−1)Λ_𝑛+1+1_2𝑛+3,=1𝜆Θ_𝑛+1(𝜔−1)Λ_𝑛+1+1²−𝜆𝜎_2𝑛+2Θ_𝑛(𝜔−1)Λ_𝑛+1(𝜔−1)Λ_𝑛+1+1,𝑛≥0,(2.85) where Θ_𝑛=∏^𝑛𝜇=0𝜎_2𝜇/𝜎_2𝜇+1, 𝑛≥0.

Proof. From Proposition 2.6 and Corollary 2.9, we can deduce that 𝑢 is regular if and only if (𝜔−1)Λ_𝑛+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 𝑐₀=0,𝑏𝑐₁=−(𝑢)₁=−𝜔𝜆,₀=𝜎₁.(2.86) When 𝑛≥0 by Lemma 2.8, for (2.73) we get 𝐷_2𝑛=(−1)^𝑛𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇^𝑛_𝜇=0𝜎_2𝜇+1²𝐻1−Λ_𝑛;𝐷_2𝑛+1=0,_2𝑛=𝜔𝜆(−1)^𝑛𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇^𝑛_𝜇=0𝜎_2𝜇+1²𝐼2Λ_𝑛−1;𝐻_2𝑛+1=0,_2𝑛=𝜔²𝜆²(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇^𝑛_𝜇=0𝜎_2𝜇+1²𝐽Λ_𝑛;𝐼_2𝑛+1=0,_2𝑛=0;𝐽_2𝑛+1=(−1)^𝑛𝜎_2𝑛+2^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+11−Λ^𝑛𝐿_2𝑛1−Λ_𝑛+1,=(−1)^𝑛+1𝜎_2𝑛+1^𝑛_𝜇=0𝜎_2𝜇^𝑛_𝜇=0𝜎_2𝜇+1²𝐿_2𝑛+1,=𝜔𝜆(−1)^𝑛𝜎_2𝑛+2^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+1Λ^𝑛+1𝐾_2𝑛+1−2Λ_𝑛+1Λ_𝑛,=0,𝑛≥0;𝐾_2𝑛+1=𝜔²𝜆²(−1)^𝑛𝜎_2𝑛+2^𝑛_𝜇=0𝜎_2𝜇²^𝑛_𝜇=0𝜎_2𝜇+1Λ^𝑛Λ_𝑛+1.(2.87) 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 𝜆𝜔²<0,𝜎²¹𝜔²>0,1𝜆²Θ^2𝑛+1𝜆(𝜔−1)Λ_𝑛+1+1²,𝑛>0,₂𝜎^22𝑛+2Θ^2𝑛(𝜔−1)Λ_𝑛+1²(𝜔−1)Λ_𝑛+1+1²,𝑛>0.(2.88)

Proof. (1) If 𝑣 is positive definite, then 𝜎_𝑛+1>0, 𝑛≥0, therefore Λ_𝑛>0, 𝑛≥0 and so (𝜔−1)Λ_𝑛+1≠0, 𝑛≥0 under the hypothesis of the corollary.
(2) If 𝑢 is regular, it is positive definite if and only if 𝛾_𝑛+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 form 𝑣 is called semiclassical when it is regular and its formal Stieltjes function 𝑆(⋅;𝑣) satisfies [15]𝜙(𝑧)𝑆^(𝑧;𝑣)=𝐶(𝑧)𝑆(𝑧;𝑣)+𝐷(𝑧),(3.1) where 𝜙 monic, 𝐶, and 𝐷 are polynomials with𝐷(𝑧)=𝑆−(𝑧𝑣;𝜃𝑣₀)𝜙=^−(𝑧)+_𝑛≥0𝑣𝜃₀𝐶(𝑧),(𝑣)_𝑛𝑧^𝑛+1.(3.2) 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]:_𝑐||𝐶(𝑐)||+||𝐷(𝑐)||>0,(3.3) where 𝑐∈{𝑥∶𝜙(𝑥)=0}, that is, 𝜙, 𝐶, and 𝐷 are coprime.

In the sequel, we will suppose that the form 𝑣 is semi-classical of class 𝑠 satisfying (3.1).

Proposition 3.1. When 𝑢 is regular, it is also semi-classical and satisfies 𝜙(𝑧)𝑆^(𝑧;𝑢)=𝐶(𝑧)𝑆(𝑧;𝑢)+𝐷(𝑧),(3.4) where 𝐷(𝑧)=𝜙𝑧(𝑧)𝑧=+𝑧(𝑢³𝜙)₁(𝑧),𝐶(𝑧)=𝑧³𝐶(𝑧)−𝑧²𝜙(𝑧),−𝜆𝐶(𝑧)+𝜆𝑧²𝐷(𝑧)+(𝑢)₁−𝜆𝜙(𝑧).(3.5) Moreover, the class of 𝑢 depends on the zero 𝑥=0 of 𝜙.

Proof. We need the following formula: 𝑆(𝑧;𝑓𝑤)=𝑓𝑆(𝑧;𝑤)+𝑤𝜃₀𝑓(𝑧),𝑤∈𝒫^,𝑓∈𝒫.(3.6) From (2.7), we have 𝑆(𝑧;𝑥²𝑢)=𝜆𝑆(𝑧;𝑥𝑣). Using (3.6), we get 𝑧²𝑆(𝑧;𝑢)+𝑧+(𝑢)₁=𝜆𝑧𝑆(𝑧;𝑣)+𝜆.(3.7) Differentiating the previous equation, we obtain 𝑧²𝑆^(𝑧;𝑢)+2𝑧𝑆(𝑧;𝑢)+1=𝜆𝑧𝑆^(𝑧;𝑣)+𝜆𝑆(𝑧;𝑣).(3.8) By (3.1) we can deduce (3.4) and (3.5).
Since 𝑣 is a semi-classical, 𝑆(𝑧;𝑣) satisfies (3.1) where 𝜙, 𝐶 and 𝐷 are coprime.
Let 𝑐 be a zero of 𝜙 different from 0, which implies that 𝜙(𝑐)=0. We know that |𝐶(𝑐)|+|𝐷(𝑐)|≠0.
If 𝐶(𝑐)≠0, then 𝐶(𝑐)≠0. if 𝐶(𝑐)=0, then 𝐷(𝑐)=𝜆𝑐²𝐷(𝑐)≠0. Hence |𝐶(𝑐)|+|𝐷(𝑐)|≠0.

Corollary 3.2. Introducing 𝜗₁∶=𝜗(𝑢)₁−𝜆𝜙(0),𝜗₂∶=(𝑢)₁−𝜆𝐶(0)+𝜙^(0)₃,∶=𝐶(0)+(𝑢)₁−𝜆𝐶^(0)+𝜙^(0)+𝜆𝐷(0),(3.9)(1)if 𝜗₁≠0, then 𝑠̃=𝑠+3;(2)if 𝜗₁=0 and 𝜗₂≠0, then 𝑠̃=𝑠+2;(3)if 𝜗₁=𝜗₂=0 and 𝜙(0)≠0 or 𝜗₃≠0, then 𝑠̃=𝑠+1.

Proof. (1) From (3.9) and (3.5), we obtain 𝐶(0)=0, 𝐷(0)=𝜗₁≠0. Therefore, it is not possible to simplify, which means that the class of 𝑢 is 𝑠+3.
(2) If 𝜗₁=0, then from (3.5) we have 𝐶(0)=𝐷(0)=0. Consequently, (3.4)–(3.6) is divisible by 𝑧. Thus, 𝑢 fulfils (3.4) with 𝐷(𝑧)=𝜙(𝑧𝑧+)=(𝑢𝑧)²₁𝜙(𝑧),𝐶(𝑧)=𝑧²𝐶(𝑧)−𝑧𝜙(𝑧),−𝜆𝐶(𝑧)+𝜆𝑧𝐷(𝑧)+(𝑢)₁−𝜆𝜃₀𝜙(𝑧).(3.10) If 𝐷(0)=𝜗₂≠0, it is not possible to simplify, which means that the class of 𝑢 is 𝑠+2.
(3) When 𝜗₁=𝜗₂=0, then it is possible to simplify (3.4)–(3.10) by 𝑧. Thus, 𝑢 fulfils (3.4) with 𝐷(𝑧)=𝜙((𝑧𝑢))₁=𝑧𝜙(𝑧),𝐶(𝑧)=𝑧𝐶(𝑧)−𝜙(𝑧),−𝜆𝜃₀𝐶(𝑧)+𝜃²⁰𝜙(𝑧)+𝜆𝐷(𝑧)+𝐶(𝑧).(3.11) Since we have 𝐶(0)=−𝜙(0), 𝐷(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 case 𝑣∶=ℋ(𝜏), where ℋ(𝜏) is the generalized Hermite form. Here is [1] 𝜉_𝑛=0,𝑛≥0,𝜎_𝑛+1=12(𝑛+1+𝜏(1+(−1)^𝑛)),𝑛≥0.(4.1) From (4.1), we get^𝑛_𝜇=0𝜎_2𝜇+1=Γ(𝑛+𝜏+3/2)Γ(𝜏+1/2),𝑛≥0,^𝑛_𝜇=0𝜎_2𝜇=Γ(𝑛+1),𝑛≥0.(4.2) We want Λ_𝑛=∑^𝑛𝜈=01/𝜎_2𝜈+1∏^𝜈𝜇=0𝜎_2𝜇+1/𝜎_2𝜇, 𝑛≥0.

But from (4.1) and (4.2), we have 1/𝜎_2𝜈+1∏^𝜈𝜇=0𝜎_2𝜇+1/𝜎_2𝜇=(1/Γ(𝜏+1/2))ℎ𝜈, withℎ_𝑛=Γ(𝑛+𝜏+1/2)Γ(𝑛+1),𝑛≥0,(4.3) fulfilling(𝑛+1)ℎ_𝑛+1−𝑛ℎ_𝑛=𝜏+12ℎ^𝑛,𝑛≥0,(4.4) and soΛ_𝑛=1(𝜏+1/2)Γ(𝜏+1/2)^𝑛𝜈=0(𝜈+1)ℎ_𝜈+1−𝜈ℎ_𝜈=1Γ(𝜏+3/2)Γ(𝑛Γ+(𝑛𝜏++13)/2),𝑛≥0.(4.5) Then we get Table 1.

Table 1 

Proposition 4.1. If 𝑣=ℋ(𝜏) is the generalized Hermite form, then the form 𝑢(𝜏,𝜔,𝜆) given by (2.9) has the following integral representation: ⟨𝑢(𝜏,𝜔,𝜆),𝑓⟩=𝑓(0)+𝜆(𝜔−1)𝑓^(0)+𝜆Γ(𝜏+1/2)𝑃^+∞−∞|𝑥|^2𝜏𝑥𝑒^−𝑥2𝑓(𝑥)𝑑𝑥,∀𝑓∈𝒫.(4.6) 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: 𝜏=0,𝜔≠1,𝑧³𝑆^(𝑧;𝑢(0,𝜔,𝜆))=−−𝑧2²𝑧₃2𝑧²+1𝑆(𝑧;𝑢(0,𝜔,𝜆))𝜏=0,𝜔=1,𝑧𝑆_−2𝜆𝜔𝑧²+𝜆(𝜔−1),𝑠=3,(𝑧;𝑢(0,1,𝜆))=−2𝑧²+1𝑆(𝑧;𝑢(0,1,𝜆))−2𝑧−2𝜆,𝑠=1,(4.7)𝜏≠0,𝜔≠1,𝑧³𝑆^(𝑧;𝑢(𝜏,𝜔,𝜆))=−𝑧²2𝑧²−2𝜏+1𝑆(𝑧;𝑢(𝜏,𝜔,𝜆))−2𝑧³𝜏≠0,𝜔=1,𝑧₂−2𝜆𝜔𝑧²+2𝜏𝑧+2𝜏𝜆(𝜔−1)+𝜆(𝜔−1),𝑠=3,𝑆^(𝑧;𝑢(𝜏,1,𝜆))=𝑧−2𝑧²+2𝜏−1𝑆(𝑧;𝑢(𝜏,1,𝜆))−2𝑧²−2𝜆𝑧+2𝜏,𝑠=2.(4.8)

Proof. It is well known that the generalized Hermite form possesses the following integral representation [1]: ⟨𝑣,𝑓⟩=^+∞−∞1Γ(𝜏+1/2)|𝑥|^2𝜏𝑒^−𝑥2𝑓(𝑥)𝑑𝑥,ℜ(𝜏)>−12,∀𝑓∈𝒫.(4.9) Following (2.11), we obtain (4.6). Also the form 𝑢 is quasi-antisymmetric because it satisfies 𝑢,𝑥^2𝑛+2=𝜆𝑣,𝑥^2𝑛+1=0,𝑛≥0,(4.10) since 𝑣 is symmetric by hypothesis.
When 𝜏=0, 𝑣 is classical and satisfies (3.4) with [22] 𝜙(𝑥)=1,𝐶(𝑧)=−2𝑧,𝐷(𝑧)=−2.(4.11) Then, 𝜗₁=𝜆(𝜔−1), 𝜗₂=0.
Now, it is sufficient to use Corollary 3.2 and Proposition 3.1 in order to obtain (4.7).
If 𝜏≠0, the form 𝑣 is semi-classical of class one and satisfies (3.4) with [23] 𝜙(𝑥)=𝑥,𝐶(𝑧)=−2𝑧²+2𝜏,𝐷(𝑧)=−2𝑧.(4.12) Therefore 𝜗₁=0, 𝜗₂=𝜆(𝜔−1)(2𝜏+1), 𝜗₃=2𝜏.
By Proposition 3.1 and Corollary 3.2 we can deduce (4.8).

4.2. 𝑣 the Corecursive of the Second Kind Chebychev Form

Let us describe the case 𝑣∶=𝒥_(−1/2,1/2); it is the corecursive of the second kind Chebychev functional. Here is [1]𝜉₀=−12,𝜉^𝑛+1=0,𝑛≥0,𝜎_𝑛+1=14,𝑛≥0.(4.13) In this case we have the following result.

Lemma 4.2. For 𝑛≥0, one has 𝑆_2𝑛(0)=(−1)^𝑛2^2𝑛,𝑆_2𝑛+1(0)=(−1)^𝑛2^2𝑛+1,𝑆^(1)2𝑛(0)=(−1)^𝑛2^2𝑛𝑆,𝑆^(1)2𝑛+1(0)=0,_2𝑛(0)=𝑛(−1)^𝑛+12^2𝑛−1,𝑆^2𝑛+1(0)=(𝑛+1)(−1)^𝑛2^2𝑛𝑆,𝑆^(1)2𝑛^(0)=0,_(1)2𝑛+1^(0)=(𝑛+1)(−21_2𝑛)^𝑛,𝑆^2𝑛(0)=𝑛(𝑛+1)(−21)^𝑛+1_2𝑛−2,𝑆^{2𝑛+1}(0)=𝑛(𝑛+1)(−21)^𝑛+1_2𝑛−1.(4.14)

Proof. The proof is analogous for the demonstration of Lemma 2.8.

Following Lemma 4.2, for (2.44) we have 𝜒_2𝑛(0)=2𝑛+12^4𝑛,𝑛≥0;𝜒_2𝑛+1(0)=𝑛+12^4𝑛+1𝜒,𝑛≥0;𝜒^2𝑛(0)=0,𝑛≥0;_2𝑛+1(0)=𝑛+12^4𝑛,𝑛≥0;𝜇_2𝑛(0)=−𝑛2^4𝑛−1,𝑛≥0;𝜇_2𝑛+1(0)=−𝑛+12^4𝑛+1,𝑛≥0.(4.15) Therefore, we get for (2.42)Δ_2𝑛=𝑛(2𝑛+1)(−1)^𝑛+12^6𝑛𝜆²+8𝑛(𝑛+1)(𝑢)₁−1(−1)^𝑛2^6𝑛+1+(𝑛+1)(2𝑛+1)(𝑢)₂₁𝜆(−1)^𝑛+12^6𝑛Δ,𝑛≥0,_2𝑛+1=(𝑛+1)(2𝑛+1)(−1)^𝑛+12^6𝑛+3𝜆²+8(𝑛+1)²(𝑢)₁+1(−1)^𝑛2^6𝑛+4𝜆(𝑛+1)(2𝑛+3)(𝑢)²¹(−1)^𝑛+12^6𝑛+3,𝑛≥0.(4.16) Then we obtainΔ_2𝑛=4(−1)^𝑛+12^6𝑛+1Δ𝑡𝑛−𝑥₁𝑡𝑛−𝑥₂,𝑛≥0,_2𝑛+1=4(−1)^𝑛+12^6𝑛+4𝑡𝑛−𝑥₃𝑡𝑛−𝑥₄,𝑛≥0,(4.17) where𝑥₁=14−3𝑡−2𝜆+𝑡²−4𝜆𝑡−4𝜆²−4𝜆^1/2,𝑥²=14−3𝑡−2𝜆−𝑡²𝑥₃−4𝜆𝑡−4𝜆²−4𝜆^1/2,=14−5𝑡−2𝜆+(𝑡+2𝜆)²+4𝜆^1/2,𝑥⁴=14−5𝑡−2𝜆−(𝑡+2𝜆)²(𝑢)₁+4𝜆^1/2,=𝑡+𝜆.(4.18) On account of Proposition 2.6, we can deduce that the form 𝑢 given by (2.9) is regular if and only if 𝑡𝑛−𝑥_𝑖≠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 2 

Proposition 4.3. If 𝑣=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: ⟨𝑢(𝑡,𝜆),𝑓⟩=(1−𝜆)𝑓(0)+𝑡𝑓^(0)+𝜆𝜋𝑃¹⁻¹1𝑥1−𝑥1+𝑥𝑓(𝑥)𝑑𝑥,∀𝑓∈𝒫.(4.19) It is a semi-classical form of class 𝑠 satisfying the following functional equation: 𝑡≠0,𝑧³𝑡𝑧=²0−,1𝑧𝑆𝑧^₂(𝑧;𝑢(𝑡,𝜆))=−𝑧²𝑧²−𝑧−1𝑆(𝑧;𝑢(𝑡,𝜆))+(𝑡−2𝜆+1)𝑧²+𝑡𝑧−𝑡,𝑠=3−1𝑆^(𝑧;𝑢(0,𝜆))=−𝑧²+𝑧+1𝑆(𝑧;𝑢(0,𝜆))−2𝜆+1,𝑠=1.(4.20)

Proof. It is well known that 𝑣=𝒥_(−/2,1/2) possesses the following integral representation [1]: ⟨𝑣,𝑓⟩=¹⁻¹1𝜋1−𝑥1+𝑥𝑓(𝑥)𝑑𝑥,𝑓∈𝒫.(4.21) From (2.11) we easily obtain (4.19).
The form 𝑣 satisfies (3.4) with [15] 𝜙(𝑥)=𝑥²−1,𝐶(𝑧)=1,𝐷(𝑧)=−2.(4.22) Therefore, 𝜗₁=−𝑡, 𝜗₂=𝑡, 𝜙(0)≠0.
Now, we can simply use Proposition 3.1 and Corollary 3.2 in order to obtain (4.20).

Corollary 4.4. When 𝑡=0 and 𝜆=−1, one has 𝛽_𝑛=(−1)^𝑛+1,𝑛≥0,𝛾₁=−12,𝛾^𝑛+2=−1𝑧𝑧4,𝑛≥0,₂−1𝑆^(𝑧;𝑢(0,−1))=−𝑧²+𝑧+1𝑆(𝑧;𝑢(0,−1))+3,𝑠=1.(4.23)

Proof. From Table 2, we reach the desired results.

Remarks 4.2. (1) One has the form ℎ₋₁𝑢(0,−1)=ℒ(−3/2,1/2), where ℒ(𝛼,𝛽) is studied in [24].
(2) Let {𝑍^𝑛(1)}_𝑛≥0 [15, 19] be the first associated sequence of {𝑍_𝑛}_𝑛≥0 orthogonal with respect to 𝑢(0,−1) and 𝛽^𝑛(1), 𝛾^(1)𝑛+1 the coefficients of the three-term recurrence relations; we have 𝛽^𝑛(1)=𝛽_𝑛+1=(−1)^𝑛,𝑛≥0;𝛾^(1)𝑛+1=𝛾_𝑛+2=−14,𝑛≥0.(4.24)

The sequence {𝑍^𝑛(1)}_𝑛≥0 is a second-order self-associated sequence; that is, {𝑍^𝑛(1)}_𝑛≥0 is identical to its associated orthogonal sequence of second kind (see [25]).

Acknowledgments

Sincere thanks are due to the referee for his valuable comments and useful suggestions and his careful reading of the manuscript. The author is indebted to the proofreader the English teacher Hajer Rebai who checked the language of this work.