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International Journal of Mathematics and Mathematical Sciences
VolumeΒ 2011, Article IDΒ 794593, 23 pages
http://dx.doi.org/10.1155/2011/794593
Research Article

Division Problem of a Regular Form: The Case π‘₯πŸπ‘’=πœ†π‘₯𝑣

Department of Mathematics, Institut Supérieur des Sciences Appliquées et de Technologie, Rue Omar Ibn El Khattab, Gabès 6072, Tunisia

Received 13 December 2010; Revised 25 February 2011; Accepted 17 March 2011

Academic Editor: HeinrichΒ Begehr

Copyright Β© 2011 M. Mejri. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We present a systematic study of a regular linear functional 𝑣 to find all regular forms 𝑒 which satisfy the equation π‘₯2𝑒=πœ†π‘₯𝑣, πœ†βˆˆβ„‚βˆ’{0}. We also give the second-order recurrence relation of the orthogonal polynomial sequence with respect to 𝑒 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 𝑒 satisfying𝑅𝑒=𝐷𝑣.(1.1) This problem has been studied in some particular cases. In fact the product of a linear form by a polynomial (𝑅(π‘₯)=1) is studied in [1–3] and the inverse problem (𝐷(π‘₯)=πœ†, πœ†βˆˆβ„‚βˆ’{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 𝑒 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(π‘₯)=π‘₯2 and 𝐷(π‘₯)=πœ†π‘₯, πœ†βˆˆβ„‚βˆ’{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 𝑒. 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 𝑒 is also semi-classical and some results concerning the class of 𝑒 are given. In the last section, some examples will be treated. The regular forms 𝑒 found in theses examples are semi-classical of class π‘ βˆˆ{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 𝑒 are given.

2. The Problem π‘₯2𝑒=πœ†π‘₯𝑣

Let 𝒫 be the vector space of polynomials with coefficients in π’ž and π’«ξ…ž its algebraic dual. We denote by βŸ¨π‘’,π‘“βŸ© the action of π‘’βˆˆπ’«ξ…ž on π‘“βˆˆπ’«. In particular, we designate by (𝑒)π‘›βˆΆ=βŸ¨π‘’,π‘₯π‘›βŸ©, 𝑛β‰₯0, the moments of 𝑒. For any form 𝑒, any polynomial 𝑔, any π‘βˆˆβ„‚, π‘Žβˆˆβ„‚βˆ’{0}, let π‘’ξ…ž, β„Žπ‘Žπ‘’, 𝑔𝑒, and (π‘₯βˆ’π‘)βˆ’1𝑒 be the forms defined by duality:ξ«π‘’ξ…žξ¬ξ«,π‘βˆΆ=βˆ’π‘’,π‘ξ…žξ¬;βŸ¨β„Žπ‘Žπ‘’,π‘βŸ©βˆΆ=βŸ¨π‘’,β„Žπ‘Žξ«π‘βŸ©;βŸ¨π‘”π‘’,π‘βŸ©βˆΆ=βŸ¨π‘’,π‘”π‘βŸ©;(π‘₯βˆ’π‘)βˆ’1𝑒,π‘βˆΆ=βŸ¨π‘’,πœƒπ‘π‘βŸ©,π‘βˆˆπ’«,(2.1) where (πœƒπ‘π‘)(π‘₯)=(𝑝(π‘₯)βˆ’π‘(𝑐))/(π‘₯βˆ’π‘); (β„Žπ‘Žπ‘)(π‘₯)=𝑝(π‘Žπ‘₯).

We define a left multiplication of a form 𝑒 by a polynomial 𝑝 as(𝑒𝑝)(π‘₯)∢=𝑒,π‘₯𝑝(π‘₯)βˆ’πœ‰π‘(πœ‰)ξƒ’π‘₯βˆ’πœ‰,π‘’βˆˆπ’«ξ…ž,π‘βˆˆπ’«.(2.2) Let us recall that a form 𝑒 is called regular if there exists a monic polynomial sequence {𝑃𝑛}𝑛β‰₯0, deg𝑃𝑛=𝑛, such thatβŸ¨π‘’,π‘ƒπ‘›π‘ƒπ‘šβŸ©=π‘Ÿπ‘›π›Ώπ‘›,π‘š,𝑛,π‘šβ‰₯0,π‘Ÿπ‘›β‰ 0,𝑛β‰₯0.(2.3) We have the following result.

Lemma 2.1 (see [15]). Let π‘’βˆˆπ’«ξ…ž, π‘“βˆˆπ’«, and π‘βˆˆπ’ž. The following formulas hold: (𝑣𝑓)ξ…žξ€·π‘£(π‘₯)=ξ…žπ‘“ξ€Έξ€·(π‘₯)+π‘£π‘“ξ…žξ€Έξ€·(π‘₯)+π‘£πœƒ0𝑓(π‘₯),π‘“βˆˆπ’«.(2.4)(𝛿𝑓)(π‘₯)=𝑓(π‘₯),π‘“βˆˆπ’«.(2.5)(π‘₯βˆ’π‘)βˆ’1((π‘₯βˆ’π‘)𝑒)=π‘’βˆ’(𝑒)0𝛿𝑐,(2.6) where βŸ¨π›Ώπ‘,π‘βŸ©=𝑝(𝑐), π‘βˆˆπ’«.

We consider the following problem: given a regular form 𝑣, find all regular forms 𝑒 satisfyingπ‘₯2𝑒=πœ†π‘₯𝑣,πœ†βˆˆβ„‚βˆ’{0},(2.7) with constraints (𝑒)0=1, (𝑣)0=1. From (2.6) we can deduce thatξ€·π‘₯𝑒=(𝑒)1ξ€Έβˆ’πœ†π›Ώ+πœ†π‘£,(2.8)𝑒=𝛿+πœ†βˆ’(𝑒)1ξ€Έπ›Ώξ…ž+πœ†π‘₯βˆ’1𝑣.(2.9) Then the form 𝑒 depends on two arbitrary parameters (𝑒)1 and πœ†.

We notice that when (𝑒)1=πœ†, we encounter the problem studied in [13] again.

We suppose that the form 𝑣 has the following integral representation:ξ€œβŸ¨π‘£,π‘“βŸ©=+βˆžβˆ’βˆžπ‘‰(π‘₯)𝑓(π‘₯)𝑑π‘₯,foreachpolynomial𝑓,(2.10) where 𝑉 is a locally integrable function with rapid decay, continuous at the origin; then the form 𝑒 is represented byξ‚΅ξ€œβŸ¨π‘’,π‘“βŸ©=1βˆ’πœ†π‘ƒ+βˆžβˆ’βˆžπ‘‰(π‘₯)π‘₯𝑓𝑑π‘₯(0)+(𝑒)1ξ€Έπ‘“βˆ’πœ†ξ…žξ€œ(0)+πœ†π‘ƒ+βˆžβˆ’βˆžπ‘‰(π‘₯)𝑓(π‘₯)π‘₯𝑑π‘₯,(2.11) where [16, 17]π‘ƒξ€œ+βˆžβˆ’βˆžπ‘‰(π‘₯)π‘₯𝑑π‘₯=limπœ–β†’0+ξ‚΅ξ€œβˆ’πœ–βˆ’βˆžπ‘‰(π‘₯)π‘₯ξ€œπ‘‘π‘₯+πœ–+βˆžπ‘‰(π‘₯)π‘₯ξ‚Ά.𝑑π‘₯(2.12) Let {𝑆𝑛}𝑛β‰₯0 denote the sequence of monic orthogonal polynomials with respect to 𝑣; we have𝑆0(π‘₯)=1,𝑆1(π‘₯)=π‘₯βˆ’πœ‰0,𝑆𝑛+2ξ€·(π‘₯)=π‘₯βˆ’πœ‰π‘›+1𝑆𝑛+1(π‘₯)βˆ’πœŽπ‘›+1𝑆𝑛(π‘₯),𝑛β‰₯0,(2.13) withπœ‰π‘›=𝑣,π‘₯𝑆2𝑛(π‘₯)𝑣,𝑆2𝑛,πœŽπ‘›+1=𝑣,𝑆2𝑛+1𝑣,𝑆2𝑛,𝑛β‰₯0.(2.14) When 𝑒 is regular, let {𝑍𝑛}𝑛β‰₯0 be the corresponding MOPS:𝑍0(π‘₯)=1,𝑍1(π‘₯)=π‘₯βˆ’π›½0.𝑍𝑛+2ξ€·(π‘₯)=π‘₯βˆ’π›½π‘›+1𝑍𝑛+1(π‘₯)βˆ’π›Ύπ‘›+1𝑍𝑛(π‘₯),𝑛β‰₯0.(2.15) From (2.7), we know that the existence of the sequence {𝑍𝑛}𝑛β‰₯0 is among all the strictly quasi-orthogonal sequences of order two with respect to πœ†π‘₯𝑣=𝑀 (𝑀 is not necessarily a regular form) [15, 18–20]. That is,π‘₯𝑍0(π‘₯)=𝑆1(π‘₯)+𝑐0,π‘₯𝑍1(π‘₯)=𝑆2(π‘₯)+𝑐1𝑆1(π‘₯)+𝑏0.π‘₯𝑍𝑛+2(π‘₯)=𝑆𝑛+3(π‘₯)+𝑐𝑛+2𝑆𝑛+2(π‘₯)+𝑏𝑛+1𝑆𝑛+1(π‘₯)+π‘Žπ‘›π‘†π‘›(π‘₯),𝑛β‰₯0,(2.16) with π‘Žπ‘›β‰ 0, 𝑛β‰₯0.

From (2.16), we have𝑍1ξ€·πœƒ(π‘₯)=0𝑆2ξ€Έ(π‘₯)+𝑐1,(2.17)𝑍𝑛+2(ξ€·πœƒπ‘₯)=0𝑆𝑛+3ξ€Έ(π‘₯)+𝑐𝑛+2ξ€·πœƒ0𝑆𝑛+2ξ€Έ(π‘₯)+𝑏𝑛+1ξ€·πœƒ0𝑆𝑛+1ξ€Έ(π‘₯)+π‘Žπ‘›ξ€·πœƒ0𝑆𝑛(π‘₯),𝑛β‰₯0.(2.18)

Lemma 2.2. Let {𝑍𝑛}𝑛β‰₯0 be a sequence of polynomials satisfying (2.16) where π‘Žπ‘›, 𝑏𝑛, and 𝑐𝑛 are complex numbers such that π‘Žπ‘›β‰ 0 for all 𝑛β‰₯0. The sequence {𝑍𝑛}𝑛β‰₯0 is orthogonal with respect to 𝑒 if and only if βŸ¨π‘’,π‘π‘›βŸ©=0,𝑛β‰₯1,βŸ¨π‘’,π‘₯𝑍𝑛(π‘₯)⟩=0,𝑛β‰₯2,βŸ¨π‘’,π‘₯𝑍1(π‘₯)βŸ©β‰ 0.(2.19)

Proof. The conditions (2.19) are necessary from the definition of the orthogonality of {𝑍𝑛}𝑛β‰₯0 with respect to 𝑒.
For π‘˜β‰₯2, we have (by (2.7)) 𝑒,π‘₯π‘˜π‘π‘›+2=π‘₯(π‘₯)2𝑒,π‘₯π‘˜βˆ’2𝑍𝑛+2(π‘₯)=πœ†π‘£,π‘₯π‘˜βˆ’1𝑍𝑛+2(π‘₯),𝑛β‰₯0,(2.20) and from (2.16), we get 𝑒,π‘₯π‘˜π‘π‘›+2(π‘₯)=πœ†π‘£,π‘₯π‘˜βˆ’2𝑆𝑛+3(π‘₯)+πœ†π‘π‘›+2𝑣,π‘₯π‘˜βˆ’2𝑆𝑛+2(π‘₯)+πœ†π‘π‘›+1𝑣,π‘₯π‘˜βˆ’2𝑆𝑛+1(π‘₯)+πœ†π‘Žπ‘›ξ«π‘£,π‘₯π‘˜βˆ’2𝑆𝑛(π‘₯),𝑛β‰₯0.(2.21) Taking into account the orthogonality of {𝑆𝑛}𝑛β‰₯0, we obtain 𝑒,π‘₯π‘˜π‘π‘›+2(π‘₯)=0,2β‰€π‘˜β‰€π‘›+1,𝑛β‰₯1,𝑒,π‘₯𝑛+2𝑍𝑛+2(π‘₯)=πœ†π‘Žπ‘›ξ«π‘£,𝑆2𝑛≠0,𝑛β‰₯0.(2.22) By (2.19), it follows that βŸ¨π‘’,𝑍1⟩=0,βŸ¨π‘’,π‘₯𝑍1(π‘₯)βŸ©β‰ 0,𝑒,𝑍𝑛+2=𝑒,π‘₯𝑍𝑛+2(π‘₯)=0,𝑛β‰₯0.(2.23) Consequently, the previous relations and (2.22) prove that {𝑍𝑛}𝑛β‰₯0 is orthogonal with respect to 𝑒, which proves the Lemma.

Remark 2.3. When 𝑒 is regular, from Theorem 5.1 in [21], there exist complex numbers π‘Ÿπ‘›+2β‰ 0, 𝑑𝑛+2 and 𝑣𝑛+2β‰ 0 such that 𝑍𝑛+2(π‘₯)+π‘Ÿπ‘›+2𝑍𝑛+1(π‘₯)=𝑆𝑛+2(π‘₯)+𝑑𝑛+2𝑆𝑛+1(π‘₯)+𝑣𝑛+2𝑆𝑛(π‘₯),𝑛β‰₯0.(2.24)

From (2.16), (2.24), and (2.15) we obtain the following relations:π‘Ÿπ‘›+2βˆ’π‘‘π‘›+2+𝑐𝑛+2βˆ’πœ‰π‘›+2π‘Ÿ=0,𝑛β‰₯0,𝑛+2𝑐𝑛+1βˆ’π‘‘π‘›+2πœ‰π‘›+1βˆ’π‘£π‘›+2+𝑏𝑛+1βˆ’πœŽπ‘›+2π‘Ÿ=0,𝑛β‰₯0,𝑛+2π‘π‘›βˆ’π‘‘π‘›+2πœŽπ‘›+1βˆ’π‘£π‘›+2πœ‰π‘›+π‘Žπ‘›π‘Ÿ=0,𝑛β‰₯0,𝑛+2π‘Žπ‘›βˆ’1βˆ’π‘£π‘›+2πœŽπ‘›=0,𝑛β‰₯1.(2.25) Taking into account (2.16), (2.18) and (2.19), we get0=𝑒,π‘₯𝑍𝑛+2=(π‘₯)𝑒,𝑆𝑛+3+𝑐𝑛+2𝑒,𝑆𝑛+2+𝑏𝑛+1𝑒,𝑆𝑛+1+π‘Žπ‘›βŸ¨π‘’,π‘†π‘›ξ«βŸ©=0,𝑛β‰₯0,0=𝑒,𝑍𝑛+2=𝑒,πœƒ0𝑆𝑛+3+𝑐𝑛+2𝑒,πœƒ0𝑆𝑛+2+𝑏𝑛+1𝑒,πœƒ0𝑆𝑛+1+π‘Žπ‘›βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©,𝑛β‰₯0,0=𝑆𝑛+3(0)+𝑐𝑛+2𝑆𝑛+2(0)+𝑏𝑛+1𝑆𝑛+1(0)+π‘Žπ‘›π‘†π‘›(0),𝑛β‰₯0,(2.26) with the initial conditions:0=𝑆1(0)+𝑐0,0=𝑆2(0)+𝑐1𝑆1(0)+𝑏0,0=βŸ¨π‘’,𝑍1ξ«ξ€·πœƒβŸ©=𝑒,0𝑆2+𝑐1,0β‰ βŸ¨π‘’,π‘₯𝑍1(π‘₯)⟩=βŸ¨π‘’,𝑆2⟩+𝑐1βŸ¨π‘’,𝑆1⟩+𝑏0.(2.27) If we denoteΔ𝑛||||||π‘†βˆΆ=𝑛+2(0)𝑆𝑛+1(0)𝑆𝑛(0)𝑒,𝑆𝑛+2𝑒,𝑆𝑛+1ξ¬βŸ¨π‘’,π‘†π‘›βŸ©ξ«π‘’,πœƒ0𝑆𝑛+2𝑒,πœƒ0𝑆𝑛+1ξ¬βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||||,𝑛β‰₯0,(2.28) from the Cramer rule we have Ξ”π‘›π‘Žπ‘›=βˆ’Ξ”π‘›+1,𝑛β‰₯0,(2.29)Δ𝑛𝑏𝑛+1=||||||𝑆𝑛+2(0)βˆ’π‘†π‘›+3(0)𝑆𝑛(0)βŸ¨π‘’,𝑆𝑛+2βŸ©βˆ’βŸ¨π‘’,𝑆𝑛+3βŸ©βŸ¨π‘’,π‘†π‘›βŸ©βŸ¨π‘’,πœƒ0𝑆𝑛+2βŸ©βˆ’βŸ¨π‘’,πœƒ0𝑆𝑛+3βŸ©βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||||,𝑛β‰₯0,(2.30)Δ𝑛𝑐𝑛+2=||||||βˆ’π‘†π‘›+3(0)𝑆𝑛+1(0)𝑆𝑛(0)βˆ’βŸ¨π‘’,𝑆𝑛+3βŸ©βŸ¨π‘’,𝑆𝑛+1βŸ©βŸ¨π‘’,π‘†π‘›βŸ©βˆ’βŸ¨π‘’,πœƒ0𝑆𝑛+3βŸ©βŸ¨π‘’,πœƒ0𝑆𝑛+1βŸ©βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||||,𝑛β‰₯0.(2.31)

Lemma 2.4. The following formulas hold: βŸ¨π‘’,π‘†π‘›βŸ©=𝑆𝑛(0)+(𝑒)1ξ€Έπ‘†βˆ’πœ†ξ…žπ‘›(0)+πœ†π‘†(1)π‘›βˆ’1(0),𝑛β‰₯0,(2.32)βŸ¨π‘’,π‘₯𝑆𝑛(π‘₯)⟩=(𝑒)1ξ€Έπ‘†βˆ’πœ†π‘›(0),𝑛β‰₯1,(2.33)ξ«ξ€·πœƒπ‘’,0𝑆𝑛=π‘†ξ…žπ‘›1(0)+2ξ€·(𝑒)1ξ€Έπ‘†βˆ’πœ†π‘›ξ…žξ…žξ‚€π‘†(0)+πœ†(1)π‘›βˆ’1ξ‚ξ…ž(0),𝑛β‰₯0,(2.34)𝑆𝑛(1)(0)𝑆𝑛(0)βˆ’π‘†(1)π‘›βˆ’1(0)𝑆𝑛+1(0)=𝑣,𝑆2𝑛,𝑛β‰₯0,(2.35) where 𝑆𝑛(1)(π‘₯)∢=(π‘£πœƒ0𝑆𝑛+1)(π‘₯), 𝑛β‰₯0, and 𝑆(1)βˆ’1(π‘₯)=0.

Proof. Equations (2.32) and (2.33) are deduced, respectively, from (2.9) and (2.8).
We have 𝑣,πœƒ20𝑆𝑛=βŸ¨π‘£,πœƒ0π‘†ξ…žπ‘›βˆ’ξ€·πœƒ0π‘†π‘›ξ€Έξ…žβŸ©=βŸ¨π‘£,πœƒ0π‘†ξ…žπ‘›ξ«π‘£βŸ©+ξ…ž,πœƒ0𝑆𝑛=ξ€·π‘₯πœƒ0π‘†ξ…žπ‘›ξ€Έξ€·π‘£(0)+ξ…žπœƒ0𝑆𝑛(0),𝑛β‰₯0.(2.36) Using (2.4), we get 𝑣,πœƒ20𝑆𝑛=ξ€·π‘£πœƒ0π‘†π‘›ξ€Έξ…žξ‚€π‘†(0)=(1)π‘›βˆ’1ξ‚ξ…ž(0),𝑛β‰₯0.(2.37) From (2.9), we obtain βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©=βŸ¨π›Ώ,πœƒ0π‘†π‘›ξ€·βŸ©+(𝑒)1ξ€Έξ‚¬ξ€·πœƒβˆ’πœ†π›Ώ,0π‘†π‘›ξ€Έξ…žξ‚­ξ«+πœ†π‘£,πœƒ20𝑆𝑛,𝑛β‰₯0.(2.38) According to (2.5) and (2.37), we can deduce (2.34).
We have 𝑆0(1)(π‘₯)=1,𝑆1(1)(π‘₯)=π‘₯βˆ’πœ‰2,𝑆(1)𝑛+2ξ€·(π‘₯)=π‘₯βˆ’πœ‰π‘›+2𝑆(1)𝑛+1(π‘₯)βˆ’πœŽπ‘›+2𝑆𝑛(1)(π‘₯),𝑛β‰₯0.(2.39) Then (by (2.39)) 𝑆𝑛(1)(0)𝑆𝑛(0)βˆ’π‘†(1)π‘›βˆ’1(0)𝑆𝑛+1(0)=πœŽπ‘›π‘†(1)π‘›βˆ’1(0)π‘†π‘›βˆ’1(0)+𝑆𝑛𝑆(0)𝑛(1)(0)+πœ‰π‘›π‘†(1)π‘›βˆ’1(0)=πœŽπ‘›ξ‚€π‘†(1)π‘›βˆ’1(0)π‘†π‘›βˆ’1(0)βˆ’π‘†(1)π‘›βˆ’2(0)𝑆𝑛.(0)(2.40) It follows that 𝑆𝑛(1)(0)𝑆𝑛(0)βˆ’π‘†(1)π‘›βˆ’1(0)𝑆𝑛+1(0)=π‘›ξ‘πœ‡=0πœŽπœ‡=𝑣,𝑆2𝑛,𝑛β‰₯0,(2.41) hence (2.35).

Proposition 2.5. One has Δ𝑛=πΈπ‘›πœ†2+πΉπ‘›πœ†+𝐺𝑛,𝑛β‰₯0,(2.42) where 𝐸𝑛=𝑆𝑛+1ξ‚†πœ‡(0)𝑛1(0)+2πœ’ξ…žπ‘›ξ‚‡+𝑆(0)𝑛(1)(0)βˆ’π‘†ξ…žπ‘›+1ξ‚‡ξ€½πœ’(0)𝑛(0)βˆ’π‘£,𝑆2𝑛𝐹,𝑛β‰₯0,𝑛=βˆ’π‘†π‘›+1ξ€½(0)(𝑒)1ξ€·πœ‡π‘›(0)+πœ’ξ…žπ‘›ξ€Έ+(0)𝑣,𝑆2π‘›ξ¬ξ€Ύβˆ’(𝑒)1𝑆𝑛(1)(0)πœ’π‘›(0)βˆ’2π‘†ξ…žπ‘›+1(0)πœ’π‘›(0)+π‘†ξ…žπ‘›+1(0)𝑣,𝑆2𝑛𝐺,𝑛β‰₯0,𝑛=(𝑒)2112𝑆𝑛+1(0)πœ’ξ…žπ‘›(0)βˆ’π‘†ξ…žπ‘›+1(0)πœ’π‘›ξ‚‡(0),𝑛β‰₯0,(2.43) with πœ’π‘›(π‘₯)=𝑆𝑛(π‘₯)π‘†ξ…žπ‘›+1(π‘₯)βˆ’π‘†π‘›+1(π‘₯)π‘†ξ…žπ‘›πœ‡(π‘₯),𝑛β‰₯0,𝑛(π‘₯)=𝑆𝑛+1(𝑆π‘₯)(1)π‘›βˆ’1ξ‚ξ…ž(π‘₯)βˆ’π‘†π‘›(𝑆π‘₯)𝑛(1)ξ‚ξ…ž(π‘₯),𝑛β‰₯0.(2.44)

Proof. Using (2.13), we, respectively, obtain 𝑆𝑛+2(0)=βˆ’πœ‰π‘›+1𝑆𝑛+1(0)βˆ’πœŽπ‘›+1𝑆𝑛(0),𝑛β‰₯0,𝑒,𝑆𝑛+2=𝑒,π‘₯𝑆𝑛+1(π‘₯)βˆ’πœ‰π‘›+1𝑒,𝑆𝑛+1ξ¬βˆ’πœŽπ‘›+1βŸ¨π‘’,π‘†π‘›ξ«βŸ©,𝑛β‰₯0,𝑒,πœƒ0𝑆𝑛+2=𝑒,𝑆𝑛+1ξ¬βˆ’πœ‰π‘›+1𝑒,πœƒ0𝑆𝑛+1ξ¬βˆ’πœŽπ‘›+1βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©,𝑛β‰₯0.(2.45) Taking into account previous relations, we obtain for (2.28) the following: Δ𝑛=||||||0𝑆𝑛+1(0)𝑆𝑛(0)𝑒,π‘₯𝑆𝑛+1(π‘₯)𝑒,𝑆𝑛+1ξ¬βŸ¨π‘’,π‘†π‘›βŸ©ξ«π‘’,𝑆𝑛+1𝑒,πœƒ0𝑆𝑛+1ξ¬βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||||,𝑛β‰₯0,(2.46) that is, Δ𝑛=βˆ’π‘’,π‘₯𝑆𝑛+1||||𝑆(π‘₯)𝑛+1(0)𝑆𝑛(0)𝑒,πœƒ0𝑆𝑛+1ξ¬βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||+𝑒,𝑆𝑛+1||||𝑆𝑛+1(0)𝑆𝑛(0)𝑒,𝑆𝑛+1ξ¬βŸ¨π‘’,π‘†π‘›βŸ©||||,𝑛β‰₯0.(2.47) Let 𝑛β‰₯0; based on the relations (2.32)–(2.34), it follows that ||||𝑆𝑛+1(0)𝑆𝑛(0)𝑒,πœƒ0𝑆𝑛+1ξ¬βŸ¨π‘’,πœƒ0π‘†π‘›βŸ©||||=ξ‚†πœ‡π‘›1(0)+2πœ’ξ…žπ‘›ξ‚‡(0)πœ†βˆ’πœ’π‘›1(0)βˆ’2(𝑒)1πœ’ξ…žπ‘›||||𝑆(0),𝑛+1(0)𝑆𝑛(0)𝑒,𝑆𝑛+1ξ¬βŸ¨π‘’,π‘†π‘›βŸ©||||=ξ€½πœ’π‘›(0)βˆ’π‘£,𝑆2π‘›ξ¬ξ€Ύπœ†βˆ’(𝑒)1πœ’π‘›(0).(2.48) From (2.48) and (2.47), we obtain the desired results.

Proposition 2.6. The form 𝑒 is regular if and only if Δ𝑛≠0, 𝑛β‰₯0. Then, the coefficients of the three-term recurrence relation (2.15) are given by 𝛾1=Ξ”0,𝛾2=βˆ’πœ†Ξ”1Ξ”0βˆ’2,(2.49)𝛾𝑛+3=Δ𝑛Δ𝑛+2Ξ”2𝑛+1πœŽπ‘›+1,𝑛β‰₯0,(2.50)𝛽0=(𝑒)1,𝛽1=𝑐1βˆ’πœ‰0βˆ’πœ‰1+πœ†π‘0Ξ”0βˆ’1,(2.51)𝛽𝑛+2=𝑐𝑛+2βˆ’πœ‰π‘›+1βˆ’πœ‰π‘›+2βˆ’π‘π‘›+1Ξ”π‘›Ξ”βˆ’1𝑛+1πœŽπ‘›+1,𝑛β‰₯0.(2.52)

Proof
Necessity
From (2.27) and Lemma 2.4, we get βŸ¨π‘’,π‘₯𝑍1(π‘₯)⟩=βŸ¨π‘’,𝑆2⟩+βŸ¨π‘’,πœƒ0𝑆2βŸ©ξ€·π‘†1(0)βˆ’βŸ¨π‘’,𝑆1βŸ©ξ€Έβˆ’π‘†2(0)=πœ†π‘†1(0)βˆ’(𝑒)21,(2.53) and again with (2.27) and (2.42), we can deduce that Ξ”0=βŸ¨π‘’,𝑆2⟩+βŸ¨π‘’,πœƒ0𝑆2βŸ©ξ€·π‘†1(0)βˆ’βŸ¨π‘’,𝑆1βŸ©ξ€Έβˆ’π‘†2(0)=βŸ¨π‘’,π‘₯𝑍1(π‘₯)βŸ©β‰ 0.(2.54) Moreover, {𝑍𝑛}𝑛β‰₯0 is orthogonal with respect to 𝑒, therefore it is strictly quasiorthogonal of order two with respect to π‘₯𝑣, and then it satisfies (2.16) with π‘Žπ‘›β‰ 0, 𝑛β‰₯0. This implies Δ𝑛≠0, 𝑛β‰₯0. Otherwise, if there exists an 𝑛0β‰₯1 such that Δ𝑛0=0, from (2.29), Ξ”0=0, which is a contradiction.

Sufficiency
Let 𝑐0=βˆ’π‘†1(0)=πœ‰0,(2.55)𝑐1ξ«ξ€·πœƒ=βˆ’π‘’,0𝑆2,(2.56)𝑏0=Ξ”0βˆ’βŸ¨π‘’,𝑆2βŸ©βˆ’π‘1βŸ¨π‘’,𝑆1⟩.(2.57) We get βŸ¨π‘’,π‘₯𝑍1(π‘₯)⟩=βŸ¨π‘’,𝑆2⟩+𝑐1βŸ¨π‘’,𝑆1⟩+𝑏0=Ξ”0β‰ 0.(2.58) We have βŸ¨π‘’,𝑍1⟩=𝑐1+βŸ¨π‘’,πœƒ0𝑆2⟩=0.
From (2.56) and (2.57) we get 𝑆2(0)+𝑐1𝑆1(0)+𝑏0=𝑆2(0)βˆ’βŸ¨π‘’,𝑆2βŸ©βˆ’βŸ¨π‘’,πœƒ0𝑆2βŸ©ξ€·π‘†1(0)βˆ’βŸ¨π‘’,𝑆1βŸ©ξ€Έ+Ξ”0.(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 π‘Žπ‘›, 𝑏𝑛, and 𝑐𝑛 (𝑛β‰₯0), define a sequence polynomials {𝑍𝑛}𝑛β‰₯0 by (2.16). Then it follows from (2.26) and Lemma 2.2 that 𝑒 is regular and {𝑍𝑛}𝑛β‰₯0 is the corresponding MOPS.
Moreover, by (2.22) we get 𝑒,𝑍2𝑛+2=πœ†π‘Žπ‘›ξ«π‘£,𝑆2𝑛,𝑛β‰₯0.(2.60) Making 𝑛=0 in (2.60), it follows that 𝑒,𝑍22=πœ†π‘Ž0.(2.61) Based on relations (2.58), (2.60), (2.61), and (2.29), we, respectively, obtain 𝛾1=βŸ¨π‘’,π‘₯𝑍1(π‘₯)⟩=Ξ”0;𝛾2=𝑒,𝑍22ξ¬βŸ¨π‘’,π‘₯𝑍1(π‘₯)⟩=βˆ’πœ†Ξ”1Ξ”0βˆ’2,𝛾𝑛+3=𝑒,𝑍2𝑛+3𝑒,𝑍2𝑛+2=Δ𝑛Δ𝑛+2Ξ”2𝑛+1πœŽπ‘›+1,𝑛β‰₯0.(2.62) We have proved (2.49) and (2.50).
When {𝑍𝑛}𝑛β‰₯0 is orthogonal, we have 𝛽0=(𝑒)1.(2.63) By (2.16) and the orthogonality of {𝑍𝑛}𝑛β‰₯0, we get 𝑒,π‘₯𝑍21(π‘₯)=𝑐1𝑒,𝑍21+βŸ¨π‘’,𝑆2𝑍1⟩.(2.64) By virtue of (2.13) and the regularity of 𝑒 we obtain βŸ¨π‘’,𝑆2𝑍1π‘₯⟩=2𝑒,𝑍1ξ¬βˆ’ξ€·πœ‰0+πœ‰1𝑒,𝑍21=πœ†βŸ¨π‘£,π‘₯𝑍1ξ€·πœ‰(π‘₯)βŸ©βˆ’0+πœ‰1𝑒,𝑍21=πœ†π‘0βˆ’ξ€·πœ‰0+πœ‰1𝑒,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 {𝑍𝑛}𝑛β‰₯0, we have 𝛽𝑛+2𝑒,𝑍2𝑛+2=𝑐𝑛+2𝑒,𝑍2𝑛+2+𝑒,𝑆𝑛+3𝑍𝑛+2,𝑛β‰₯0.(2.66) Using (2.13), (2.16), and the the orthogonality of {𝑆𝑛}𝑛β‰₯0, we have 𝑒,𝑆𝑛+3𝑍𝑛+2=πœ†π‘π‘›+1𝑣,𝑆2𝑛+1ξ¬βˆ’ξ€·πœ‰π‘›+1+πœ‰π‘›+2𝑒,𝑍2𝑛+2,𝑛β‰₯0.(2.67) Taking into account the previous relation, (2.66) becomes 𝛽𝑛+2=𝑐𝑛+2βˆ’πœ‰π‘›+1βˆ’πœ‰π‘›+2+πœ†π‘π‘›+1𝑣,𝑆2𝑛+1𝑒,𝑍2𝑛+2,𝑛β‰₯0.(2.68) From (2.60) and (2.29), we have 𝑣,𝑆2𝑛+1𝑒,𝑍2𝑛+2=βˆ’πœ†βˆ’1Ξ”π‘›Ξ”βˆ’1𝑛+1πœŽπ‘›+1,𝑛β‰₯0.(2.69) Last equation and (2.68) give (2.52).

Moreover, if the form 𝑒 is regular, for (2.29), (2.30), and (2.31), we getπ‘Žπ‘›Ξ”=βˆ’π‘›+1Δ𝑛,𝑛β‰₯0,(2.70)𝑏𝑛+1=ξ€·π·π‘›πœ†2+π»π‘›πœ†+πΌπ‘›ξ€ΈΞ”π‘›βˆ’1+πœŽπ‘›+2,𝑛β‰₯0,(2.71)𝑐𝑛+2𝐽=βˆ’π‘›πœ†2+πΏπ‘›πœ†+πΎπ‘›ξ€ΈΞ”π‘›βˆ’1+πœ‰π‘›+2,𝑛β‰₯0,(2.72) where𝐷𝑛=𝑆𝑛(0)𝑣,𝑆2𝑛+1ξ¬βˆ’πœ’π‘›+1ξ€Έ(0)βˆ’πœ‰π‘›+1𝑆𝑛+2ξ‚€πœ‡(0)𝑛1(0)+2πœ’ξ…žπ‘›ξ‚(0)βˆ’πœ‰π‘›+1𝑆(1)𝑛+1(0)βˆ’π‘†ξ…žπ‘›+2ξ‚ξ€·πœ’(0)𝑛(0)βˆ’π‘£,𝑆2𝑛𝐻,𝑛β‰₯0,𝑛=(𝑒)1𝑆𝑛(0)2πœ’π‘›+1(0)βˆ’π‘£,𝑆2𝑛+1+πœ‰π‘›+1𝑆𝑛+2ξ€·πœ’(0)𝑛(0)+(𝑒)1ξ€·πœ’ξ…žπ‘›(0)+πœ‡π‘›ξ€Έ(0)+(𝑒)1πœ‰π‘›+1πœ’π‘›ξ‚€π‘†(0)(1)𝑛+1(0)βˆ’π‘†ξ…žπ‘›+2(0)+πœ‰π‘›+1𝑣,𝑆2π‘›ξ¬βˆ’πœ’π‘›π‘†(0)𝑛+2(0)+(𝑒)1π‘†ξ…žπ‘›+2𝐼(0),𝑛β‰₯0,𝑛=βˆ’(𝑒)21𝑆𝑛(0)πœ’π‘›+11(0)+2πœ‰π‘›+1𝑆𝑛+2(0)πœ’ξ…žπ‘›(0)βˆ’π‘†ξ…žπ‘›+2(0)πœ’π‘›ξ€Έξ‚‡π½(0),𝑛β‰₯0,𝑛=𝑆𝑛+2ξ‚€πœ‡(0)𝑛1(0)+2πœ’ξ…žπ‘›ξ‚+(0)(𝑆(1)𝑛+1ξ€·0βˆ’π‘†ξ…žπ‘›+2πœ’(0)𝑛(0)βˆ’π‘£,𝑆2𝑛𝐿,𝑛β‰₯0,𝑛=(𝑒)1πœ’π‘›ξ‚€(0)2π‘†ξ…žπ‘›+2(0)βˆ’π‘†(1)𝑛+1(0)βˆ’(𝑒)1𝑆𝑛+2ξ€·πœ‡(0)𝑛(0)+πœ’ξ…žπ‘›ξ€Έβˆ’ξ«(0)𝑣,𝑆2𝑛𝑆𝑛+2(0)+(𝑒)1π‘†ξ…žπ‘›+2(𝐾0),𝑛β‰₯0,𝑛=(𝑒)2112𝑆𝑛+2(0)πœ’ξ…žπ‘›(0)βˆ’πœ’π‘›(0)π‘†ξ…žπ‘›+2(0),𝑛β‰₯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 {𝑦𝑛}𝑛β‰₯0 and {𝑏𝑛}𝑛β‰₯0 are sequences of complex numbers fulfilling 𝑦𝑛+1+π‘Žπ‘›π‘¦π‘›=𝑏𝑛+1,𝑛β‰₯0,π‘Žπ‘›π‘¦β‰ 0,𝑛β‰₯0,0=𝑏0,(2.74) then 𝑦𝑛=(βˆ’1)π‘›π‘Žπ‘›βˆ’1ξƒ©π‘›ξ‘πœ‡=0π‘Žπœ‡ξƒͺξ“π‘›πœˆ=0(βˆ’1)πœˆπ‘Žπœˆξƒ©πœˆξ‘πœ‡=0π‘Žπœ‡βˆ’1ξƒͺπ‘πœˆ,𝑛β‰₯0.(2.75)

Lemma 2.8. When {𝑆𝑛}𝑛β‰₯0 given by (2.13) is symmetric, one has 𝑆2𝑛(0)=(βˆ’1)π‘›πœŽπ‘›2𝑛+1ξ‘πœ‡=0𝜎2πœ‡+1,𝑛β‰₯0,𝑆2𝑛+1𝑆(0)=0,𝑛β‰₯0,(1)2𝑛(0)=(βˆ’1)π‘›π‘›ξ‘πœ‡=0𝜎2πœ‡,𝑛β‰₯0,𝑆(1)2𝑛+1𝑆(0)=0,𝑛β‰₯0,ξ…ž2𝑛+1(0)=(βˆ’1)π‘›ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺΛ𝑛,𝑛β‰₯0,π‘†ξ…ž2𝑛𝑆(0)=0,𝑛β‰₯0,(1)2π‘›ξ‚ξ…ž(0)=0,𝑛β‰₯0,π‘†ξ…žξ…ž2𝑛+1(0)=0,𝑛β‰₯0.(2.76)

Proof. As 𝑣 is symmetric, then πœ‰π‘›=0, 𝑛β‰₯0, and therefore from (2.13) we have 𝑆0(0)=1,𝑆1(0)=0,𝑆0(1)(0)=1,𝑆1(1)𝑆(0)=0,𝑛+2(0)=βˆ’πœŽπ‘›+1𝑆𝑛(0),𝑛β‰₯0,𝑆(1)𝑛+2(0)=βˆ’πœŽπ‘›+2𝑆𝑛(1)𝑆(0),𝑛β‰₯0,ξ…ž0(0)=0,π‘†ξ…ž1(0)=1,π‘†ξ…žπ‘›+2(0)=βˆ’πœŽπ‘›+1π‘†ξ…žπ‘›(0)+𝑆𝑛+1𝑆(0),𝑛β‰₯0,0(1)ξ‚ξ…žξ‚€π‘†(0)=0,(1)𝑛+2ξ‚ξ…ž(0)=βˆ’πœŽπ‘›+2𝑆𝑛(1)ξ‚ξ…ž(0)+𝑆(1)𝑛+1𝑆(0),𝑛β‰₯0,0ξ…žξ…ž(0)=0,𝑆1ξ…žξ…ž(0)=0,π‘†ξ…žξ…žπ‘›+2(0)=βˆ’πœŽπ‘›+1π‘†π‘›ξ…žξ…ž(0)+2π‘†ξ…žπ‘›+1(0),𝑛β‰₯0.(2.77) Now, it is sufficient to use Lemma 2.7 in order to obtain the desired results.

Let πœ”=πœ†βˆ’1(𝑒)1.(2.78)

Corollary 2.9. If 𝑣 is a symmetric form, one has Ξ”2𝑛=πœ†2(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξ€½(πœ”βˆ’1)Λ𝑛+12Ξ”,𝑛β‰₯0,2𝑛+1=πœ†(βˆ’1)π‘›ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ,𝑛β‰₯0,(2.79) where Λ𝑛=ξ“π‘›πœˆ=01𝜎𝜈2𝜈+1ξ‘πœ‡=0𝜎2πœ‡+1𝜎2πœ‡,𝑛β‰₯0,𝜎0=1.(2.80)

Proof. Following Lemma 2.8, for (2.43) we have 𝐸2𝑛=(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺξ€·1βˆ’Ξ›π‘›ξ€Έ,𝑛β‰₯0;𝐸2𝑛+1𝐹=0,𝑛β‰₯0,2𝑛=2πœ”πœ†(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺξ€·1βˆ’Ξ›π‘›ξ€ΈΞ›π‘›+1𝐹,𝑛β‰₯0,2𝑛+1=(βˆ’1)π‘›ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2𝐺,𝑛β‰₯0,2𝑛=πœ”2πœ†2(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺΞ›2𝑛,𝑛β‰₯0;𝐺2𝑛+1=0,𝑛β‰₯0.(2.81) As a consequence, relations (2.81) and (2.42) yield (2.79).

Theorem 2.10. The form 𝑒 is regular if and only if (πœ”βˆ’1)Λ𝑛+1β‰ 0, 𝑛β‰₯0, where Λ𝑛 is defined in (2.80).
In this case one has π‘Ž2𝑛=𝜎2𝑛+1πœ†Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+12,π‘Ž2𝑛+1=βˆ’πœ†πœŽ22𝑛+2Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+12,𝑛β‰₯0,(2.82)𝑏2𝑛=𝜎2𝑛+1,𝑛β‰₯0,𝑏2𝑛+1=𝜎2𝑛+2(πœ”βˆ’1)Λ𝑛+1+1(πœ”βˆ’1)Λ𝑛+1,𝑛β‰₯0,(2.83)𝑐0=0,𝑐1=βˆ’πœ”πœ†,𝑐2𝑛+2=1πœ†Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+12𝑐,𝑛β‰₯0,2𝑛+3=βˆ’πœ†πœŽ2𝑛+2Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+1+1ξ€Έξ€·(πœ”βˆ’1)Λ𝑛𝛾+1,𝑛β‰₯0,1=βˆ’πœ†2πœ”2,𝛾2𝜎=βˆ’21πœ†2πœ”4,𝛾2𝑛+4=1πœ†2Θ2𝑛+1ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+12𝛾,𝑛β‰₯0,2𝑛+3=πœ†2𝜎22𝑛+2Θ2𝑛(πœ”βˆ’1)Λ𝑛+12ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+12𝛽,𝑛β‰₯0,0=πœ†πœ”,𝛽1𝜎=βˆ’πœ†πœ”βˆ’1πœ†πœ”2,(2.84)𝛽2𝑛+2=1πœ†Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+12+πœ†πœŽ2𝑛+2Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ,𝛽+12𝑛+3=1πœ†Ξ˜π‘›+1ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+12βˆ’πœ†πœŽ2𝑛+2Ξ˜π‘›ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+1,𝑛β‰₯0,(2.85) where Ξ˜π‘›=βˆπ‘›πœ‡=0𝜎2πœ‡/𝜎2πœ‡+1, 𝑛β‰₯0.

Proof. From Proposition 2.6 and Corollary 2.9, we can deduce that 𝑒 is regular if and only if (πœ”βˆ’1)Λ𝑛+1β‰ 0, 𝑛β‰₯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,𝑐1=βˆ’(𝑒)1𝑏=βˆ’πœ”πœ†,0=𝜎1.(2.86) When 𝑛β‰₯0 by Lemma 2.8, for (2.73) we get 𝐷2𝑛=(βˆ’1)π‘›πœŽ2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2ξ€·1βˆ’Ξ›π‘›ξ€Έ;𝐷2𝑛+1𝐻=0,2𝑛=πœ”πœ†(βˆ’1)π‘›πœŽ2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2ξ€·2Ξ›π‘›ξ€Έβˆ’1;𝐻2𝑛+1𝐼=0,2𝑛=πœ”2πœ†2(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2Λ𝑛;𝐼2𝑛+1𝐽=0,2𝑛=0;𝐽2𝑛+1=(βˆ’1)π‘›πœŽ2𝑛+2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺξ€·1βˆ’Ξ›π‘›ξ€Έξ€·1βˆ’Ξ›π‘›+1ξ€Έ,𝐿2𝑛=(βˆ’1)𝑛+1𝜎2𝑛+1ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺ2,𝐿2𝑛+1=πœ”πœ†(βˆ’1)π‘›πœŽ2𝑛+2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺΛ𝑛+1+ξ€·1βˆ’2Λ𝑛+1Λ𝑛,𝐾2𝑛=0,𝑛β‰₯0;𝐾2𝑛+1=πœ”2πœ†2(βˆ’1)π‘›πœŽ2𝑛+2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡ξƒͺ2ξƒ©π‘›ξ‘πœ‡=0𝜎2πœ‡+1ξƒͺΛ𝑛Λ𝑛+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 𝑒 is regular when πœ”βˆˆβ„‚βˆ’]βˆ’βˆž,1[.
(2) When 𝑒 is regular, it is positive definite form if and only if πœ†πœ”2𝜎<0,21πœ”21>0,πœ†2Θ2𝑛+1ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+12πœ†,𝑛>0,2𝜎22𝑛+2Θ2𝑛(πœ”βˆ’1)Λ𝑛+12ξ€·(πœ”βˆ’1)Λ𝑛+1ξ€Έ+12,𝑛>0.(2.88)

Proof. (1) If 𝑣 is positive definite, then πœŽπ‘›+1>0, 𝑛β‰₯0, therefore Λ𝑛>0, 𝑛β‰₯0 and so (πœ”βˆ’1)Λ𝑛+1β‰ 0, 𝑛β‰₯0 under the hypothesis of the corollary.
(2) If 𝑒 is regular, it is positive definite if and only if 𝛾𝑛+1>0, 𝑛β‰₯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(β‹…;𝑣) satisfies [15]πœ™(𝑧)π‘†ξ…ž(𝑧;𝑣)=𝐢(𝑧)𝑆(𝑧;𝑣)+𝐷(𝑧),(3.1) where πœ™ monic, 𝐢, and 𝐷 are polynomials with𝐷(𝑧)=βˆ’π‘£πœƒ0πœ™ξ€Έξ…žξ€·(𝑧)+π‘£πœƒ0𝐢𝑆(𝑧),(𝑧;𝑣)=βˆ’π‘›β‰₯0(𝑣)𝑛𝑧𝑛+1.(3.2) The class of the semi-classical form 𝑣 is 𝑠=max(degπœ™βˆ’2,degπΆβˆ’1) if and only if the following condition is satisfied [22]:𝑐||||+||||𝐢(𝑐)𝐷(𝑐)>0,(3.3) where π‘βˆˆ{π‘₯βˆΆπœ™(π‘₯)=0}, that is, πœ™, 𝐢, 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 𝑒 is regular, it is also semi-classical and satisfies ξ‚πœ™(𝑧)π‘†ξ…žξ‚ξ‚(𝑧;𝑒)=𝐢(𝑧)𝑆(𝑧;𝑒)+𝐷(𝑧),(3.4) where ξ‚πœ™(𝑧)=𝑧3ξ‚πœ™(𝑧),𝐢(𝑧)=𝑧3𝐢(𝑧)βˆ’π‘§2ξ‚π·ξ€·πœ™(𝑧),(𝑧)=𝑧𝑧+(𝑒)1ξ€ΈπΆβˆ’πœ†(𝑧)+πœ†π‘§2𝐷(𝑧)+(𝑒)1ξ€Έπœ™βˆ’πœ†(𝑧).(3.5) Moreover, the class of 𝑒 depends on the zero π‘₯=0 of πœ™.

Proof. We need the following formula: 𝑆(𝑧;𝑓𝑀)=𝑓𝑆(𝑧;𝑀)+π‘€πœƒ0𝑓(𝑧),π‘€βˆˆπ’«ξ…ž,π‘“βˆˆπ’«.(3.6) From (2.7), we have 𝑆(𝑧;π‘₯2𝑒)=πœ†π‘†(𝑧;π‘₯𝑣). Using (3.6), we get 𝑧2𝑆(𝑧;𝑒)+𝑧+(𝑒)1=πœ†π‘§π‘†(𝑧;𝑣)+πœ†.(3.7) Differentiating the previous equation, we obtain 𝑧2π‘†ξ…ž(𝑧;𝑒)+2𝑧𝑆(𝑧;𝑒)+1=πœ†π‘§π‘†ξ…ž(𝑧;𝑣)+πœ†π‘†(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 πœ™, 𝐢 and 𝐷 are coprime.
Let 𝑐 be a zero of ξ‚πœ™ different from 0, which implies that πœ™(𝑐)=0. We know that |𝐢(𝑐)|+|𝐷(𝑐)|β‰ 0.
If 𝐢(𝑐)β‰ 0, then 𝐢(𝑐)β‰ 0. if 𝐢(𝑐)=0, then 𝐷(𝑐)=πœ†π‘2𝐷(𝑐)β‰ 0. Hence |𝐢(𝑐)|+|𝐷(𝑐)|β‰ 0.

Corollary 3.2. Introducing πœ—1ξ€·βˆΆ=(𝑒)1ξ€Έπœ™βˆ’πœ†(0),πœ—2ξ€·βˆΆ=(𝑒)1πΆβˆ’πœ†ξ€Έξ€·(0)+πœ™ξ…žξ€Έ,πœ—(0)3ξ€·(∢=𝐢(0)+𝑒)1πΆβˆ’πœ†ξ€Έξ€·ξ…ž(0)+πœ™ξ…žξ…ž(ξ€Έ0)+πœ†π·(0),(3.9)(1)if πœ—1β‰ 0, then ̃𝑠=𝑠+3;(2)if πœ—1=0 and πœ—2β‰ 0, then ̃𝑠=𝑠+2;(3)if πœ—1=πœ—2=0 and πœ™(0)β‰ 0 or πœ—3β‰ 0, then ̃𝑠=𝑠+1.

Proof. (1) From (3.9) and (3.5), we obtain 𝐢(0)=0, 𝐷(0)=πœ—1β‰ 0. Therefore, it is not possible to simplify, which means that the class of 𝑒 is 𝑠+3.
(2) If πœ—1=0, then from (3.5) we have 𝐢(0)=𝐷(0)=0. Consequently, (3.4)–(3.6) is divisible by 𝑧. Thus, 𝑒 fulfils (3.4) with ξ‚πœ™(𝑧)=𝑧2ξ‚πœ™(𝑧),𝐢(𝑧)=𝑧2𝐷𝐢(𝑧)βˆ’π‘§πœ™(𝑧),(𝑧)=𝑧+(𝑒)1ξ€ΈπΆξ€·βˆ’πœ†(𝑧)+πœ†π‘§π·(𝑧)+(𝑒)1ξ€Έπœƒβˆ’πœ†0πœ™(𝑧).(3.10) If 𝐷(0)=πœ—2β‰ 0, it is not possible to simplify, which means that the class of 𝑒 is 𝑠+2.
(3) When πœ—1=πœ—2=0, then it is possible to simplify (3.4)–(3.10) by 𝑧. Thus, 𝑒 fulfils (3.4) with ξ‚ξ‚ξ‚π·ξ€·πœ™(𝑧)=π‘§πœ™(𝑧),𝐢(𝑧)=𝑧𝐢(𝑧)βˆ’πœ™(𝑧),(𝑧)=(𝑒)1πœƒβˆ’πœ†ξ€Έξ€·0𝐢(𝑧)+πœƒ20πœ™ξ€Έ(𝑧)+πœ†π·(𝑧)+𝐢(𝑧).(3.11) Since we have 𝐢(0)=βˆ’πœ™(0), 𝐷(0)=πœ—3, then we can deduce that if πœ™(0)β‰ 0 or πœ—3β‰ 0, it is not possible to simplify, which means that the class of 𝑒 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 π‘£βˆΆ=β„‹(𝜏), where β„‹(𝜏) is the generalized Hermite form. Here is [1] πœ‰π‘›=0,𝑛β‰₯0,πœŽπ‘›+1=12(𝑛+1+𝜏(1+(βˆ’1)𝑛)),𝑛β‰₯0.(4.1) From (4.1), we getπ‘›ξ‘πœ‡=0𝜎2πœ‡+1=Ξ“(𝑛+𝜏+3/2)Ξ“(𝜏+1/2),𝑛β‰₯0,π‘›ξ‘πœ‡=0𝜎2πœ‡=Ξ“(𝑛+1),𝑛β‰₯0.(4.2) We want Λ𝑛=βˆ‘π‘›πœˆ=01/𝜎2𝜈+1βˆπœˆπœ‡=0𝜎2πœ‡+1/𝜎2πœ‡, 𝑛β‰₯0.

But from (4.1) and (4.2), we have 1/𝜎2𝜈+1βˆπœˆπœ‡=0𝜎2πœ‡+1/𝜎2πœ‡=(1/Ξ“(𝜏+1/2))β„Žπœˆ, withβ„Žπ‘›=Ξ“(𝑛+𝜏+1/2)Ξ“(𝑛+1),𝑛β‰₯0,(4.3) fulfilling(𝑛+1)β„Žπ‘›+1βˆ’π‘›β„Žπ‘›=ξ‚€1𝜏+2ξ‚β„Žπ‘›,𝑛β‰₯0,(4.4) and soΛ𝑛=1(ξ“πœ+1/2)Ξ“(𝜏+1/2)π‘›πœˆ=0(𝜈+1)β„Žπœˆ+1βˆ’πœˆβ„Žπœˆ=1Ξ“(𝜏+3/2)Ξ“(𝑛+𝜏+3/2)Ξ“(𝑛+1),𝑛β‰₯0.(4.5) Then we get Table 1.

tab1
Table 1

Proposition 4.1. If 𝑣=β„‹(𝜏) is the generalized Hermite form, then the form 𝑒(𝜏,πœ”,πœ†) given by (2.9) has the following integral representation: βŸ¨π‘’(𝜏,πœ”,πœ†),π‘“βŸ©=𝑓(0)+πœ†(πœ”βˆ’1)π‘“ξ…žπœ†(0)+Ξ“π‘ƒξ€œ(𝜏+1/2)+βˆžβˆ’βˆž|π‘₯|2𝜏π‘₯π‘’βˆ’π‘₯2𝑓(π‘₯)𝑑π‘₯,βˆ€π‘“βˆˆπ’«.(4.6) It is a quasi-antisymmetric ((𝑒(𝜏,πœ”,πœ†))2𝑛+2=0, 𝑛β‰₯0) and semi-classical form of class 𝑠 satisfying the following functional equation: 𝜏=0,πœ”β‰ 1,𝑧3π‘†ξ…ž(𝑧;𝑒(0,πœ”,πœ†))=βˆ’π‘§2ξ€·2𝑧2ξ€Έ+1𝑆(𝑧;𝑒(0,πœ”,πœ†))βˆ’2𝑧3βˆ’2πœ†πœ”π‘§2+πœ†(πœ”βˆ’1),𝑠=3,𝜏=0,πœ”=1,π‘§π‘†ξ…ž(𝑧;𝑒(0,1,πœ†))=βˆ’2𝑧2ξ€Έ+1𝑆(𝑧;𝑒(0,1,πœ†))βˆ’2π‘§βˆ’2πœ†,𝑠=1,(4.7)πœβ‰ 0,πœ”β‰ 1,𝑧3π‘†ξ…ž(𝑧;𝑒(𝜏,πœ”,πœ†))=βˆ’π‘§2ξ€·2𝑧2ξ€Έβˆ’2𝜏+1𝑆(𝑧;𝑒(𝜏,πœ”,πœ†))βˆ’2𝑧3βˆ’2πœ†πœ”π‘§2+2πœπ‘§+2πœπœ†(πœ”βˆ’1)+πœ†(πœ”βˆ’1),𝑠=3,πœβ‰ 0,πœ”=1,𝑧2π‘†ξ…ž(𝑧;𝑒(𝜏,1,πœ†))=π‘§βˆ’2𝑧2ξ€Έ+2πœβˆ’1𝑆(𝑧;𝑒(𝜏,1,πœ†))βˆ’2𝑧2βˆ’2πœ†π‘§+2𝜏,𝑠=2.(4.8)

Proof. It is well known that the generalized Hermite form possesses the following integral representation [1]: ξ€œβŸ¨π‘£,π‘“βŸ©=+βˆžβˆ’βˆž1Ξ“(𝜏+1/2)|π‘₯|2πœπ‘’βˆ’π‘₯2𝑓1(π‘₯)𝑑π‘₯,β„œ(𝜏)>βˆ’2,βˆ€π‘“βˆˆπ’«.(4.9) Following (2.11), we obtain (4.6). Also the form 𝑒 is quasi-antisymmetric because it satisfies 𝑒,π‘₯2𝑛+2=πœ†π‘£,π‘₯2𝑛+1=0,𝑛β‰₯0,(4.10) since 𝑣 is symmetric by hypothesis.
When 𝜏=0, 𝑣 is classical and satisfies (3.4) with [22] πœ™(π‘₯)=1,𝐢(𝑧)=βˆ’2𝑧,𝐷(𝑧)=βˆ’2.(4.11) Then, πœ—1=πœ†(πœ”βˆ’1), πœ—2=0.
Now, it is sufficient to use Corollary 3.2 and Proposition 3.1 in order to obtain (4.7).
If πœβ‰ 0, the form 𝑣 is semi-classical of class one and satisfies (3.4) with [23] πœ™(π‘₯)=π‘₯,𝐢(𝑧)=βˆ’2𝑧2+2𝜏,𝐷(𝑧)=βˆ’2𝑧.(4.12) Therefore πœ—1=0, πœ—2=πœ†(πœ”βˆ’1)(2𝜏+1), πœ—3=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 π‘£βˆΆ=π’₯(βˆ’1/2,1/2); it is the corecursive of the second kind Chebychev functional. Here is [1]πœ‰01=βˆ’2,πœ‰π‘›+1=0,𝑛β‰₯0,πœŽπ‘›+1=14,𝑛β‰₯0.(4.13) In this case we have the following result.

Lemma 4.2. For 𝑛β‰₯0, one has 𝑆2𝑛(0)=(βˆ’1)𝑛22𝑛,𝑆2𝑛+1(0)=(βˆ’1)𝑛22𝑛+1,𝑆(1)2𝑛(0)=(βˆ’1)𝑛22𝑛,𝑆(1)2𝑛+1𝑆(0)=0,ξ…ž2𝑛(0)=𝑛(βˆ’1)𝑛+122π‘›βˆ’1,π‘†ξ…ž2𝑛+1(0)=(𝑛+1)(βˆ’1)𝑛22𝑛,𝑆(1)2π‘›ξ‚ξ…žξ‚€π‘†(0)=0,(1)2𝑛+1ξ‚ξ…ž(0)=(𝑛+1)(βˆ’1)𝑛22𝑛,π‘†ξ…žξ…ž2𝑛(0)=𝑛(𝑛+1)(βˆ’1)𝑛+122π‘›βˆ’2,π‘†ξ…žξ…ž2𝑛+1(0)=𝑛(𝑛+1)(βˆ’1)𝑛+122π‘›βˆ’1.(4.14)

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

Following Lemma 4.2, for (2.44) we have πœ’2𝑛(0)=2𝑛+124𝑛,𝑛β‰₯0;πœ’2𝑛+1(0)=𝑛+124𝑛+1,𝑛β‰₯0;πœ’ξ…ž2π‘›πœ’(0)=0,𝑛β‰₯0;ξ…ž2𝑛+1(0)=𝑛+124𝑛,𝑛β‰₯0;πœ‡2𝑛𝑛(0)=βˆ’24π‘›βˆ’1,𝑛β‰₯0;πœ‡2𝑛+1(0)=βˆ’π‘›+124𝑛+1,𝑛β‰₯0.(4.15) Therefore, we get for (2.42)Ξ”2𝑛=𝑛(2𝑛+1)(βˆ’1)𝑛+126π‘›πœ†2+ξ€·8𝑛(𝑛+1)(𝑒)1ξ€Έβˆ’1(βˆ’1)𝑛26𝑛+1πœ†+(𝑛+1)(2𝑛+1)(𝑒)21(βˆ’1)𝑛+126𝑛Δ,𝑛β‰₯0,2𝑛+1=(𝑛+1)(2𝑛+1)(βˆ’1)𝑛+126𝑛+3πœ†2+ξ€·8(𝑛+1)2(𝑒)1ξ€Έ+1(βˆ’1)𝑛26𝑛+4πœ†(𝑛+1)(2𝑛+3)(𝑒)21(βˆ’1)𝑛+126𝑛+3,𝑛β‰₯0.(4.16) Then we obtainΞ”2𝑛=4(βˆ’1)𝑛+126𝑛+1ξ€·π‘‘π‘›βˆ’π‘₯1ξ€Έξ€·π‘‘π‘›βˆ’π‘₯2ξ€ΈΞ”,𝑛β‰₯0,2𝑛+1(=4βˆ’1)𝑛+126𝑛+4ξ€·π‘‘π‘›βˆ’π‘₯3ξ€Έξ€·π‘‘π‘›βˆ’π‘₯4ξ€Έ,𝑛β‰₯0,(4.17) whereπ‘₯1=14ξ‚†ξ€·π‘‘βˆ’3π‘‘βˆ’2πœ†+2βˆ’4πœ†π‘‘βˆ’4πœ†2ξ€Έβˆ’4πœ†1/2,π‘₯2=14ξ‚†ξ€·π‘‘βˆ’3π‘‘βˆ’2πœ†βˆ’2βˆ’4πœ†π‘‘βˆ’4πœ†2ξ€Έβˆ’4πœ†1/2,π‘₯3=14ξ‚†ξ€·βˆ’5π‘‘βˆ’2πœ†+(𝑑+2πœ†)2ξ€Έ+4πœ†1/2,π‘₯4=14ξ‚†ξ€·βˆ’5π‘‘βˆ’2πœ†βˆ’(𝑑+2πœ†)2ξ€Έ+4πœ†1/2,(𝑒)1=𝑑+πœ†.(4.18) On account of Proposition 2.6, we can deduce that the form 𝑒 given by (2.9) is regular if and only if π‘‘π‘›βˆ’π‘₯𝑖≠0, 𝑛β‰₯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.

tab2
Table 2

Proposition 4.3. If 𝑣=π’₯(βˆ’1/2,1/2) is the corecursive of the second kind Chebychev form, then the form 𝑒(𝑑,πœ†) given by (2.9) has the following integral representation: βŸ¨π‘’(𝑑,πœ†),π‘“βŸ©=(1βˆ’πœ†)𝑓(0)+π‘‘π‘“ξ…žπœ†(0)+πœ‹π‘ƒξ€œ1βˆ’11π‘₯ξ‚™1βˆ’π‘₯1+π‘₯𝑓(π‘₯)𝑑π‘₯,βˆ€π‘“βˆˆπ’«.(4.19) It is a semi-classical form of class 𝑠 satisfying the following functional equation: 𝑑≠0,𝑧3𝑧2ξ€Έπ‘†βˆ’1ξ…ž(𝑧;𝑒(𝑑,πœ†))=βˆ’π‘§2𝑧2ξ€Έβˆ’π‘§βˆ’1𝑆(𝑧;𝑒(𝑑,πœ†))+(π‘‘βˆ’2πœ†+1)𝑧2𝑧+π‘‘π‘§βˆ’π‘‘,𝑠=3𝑑=0,𝑧2ξ€Έπ‘†βˆ’1ξ…žξ€·(𝑧;𝑒(0,πœ†))=βˆ’π‘§2𝑆+𝑧+1(𝑧;𝑒(0,πœ†))βˆ’2πœ†+1,𝑠=1.(4.20)

Proof. It is well known that 𝑣=π’₯(βˆ’/2,1/2) possesses the following integral representation [1]: ξ€œβŸ¨π‘£,π‘“βŸ©=1βˆ’11πœ‹ξ‚™1βˆ’π‘₯1+π‘₯𝑓(π‘₯)𝑑π‘₯,π‘“βˆˆπ’«.(4.21) From (2.11) we easily obtain (4.19).
The form 𝑣 satisfies (3.4) with [15] πœ™(π‘₯)=π‘₯2βˆ’1,𝐢(𝑧)=1,𝐷(𝑧)=βˆ’2.(4.22) Therefore, πœ—1=βˆ’π‘‘, πœ—2=𝑑, πœ™(0)β‰ 0.
Now, we can simply use Proposition 3.1 and Corollary 3.2 in order to obtain (4.20).

Corollary 4.4. When 𝑑=0 and πœ†=βˆ’1, one has 𝛽𝑛=(βˆ’1)𝑛+1,𝑛β‰₯0,𝛾11=βˆ’2,𝛾𝑛+21=βˆ’4𝑧𝑧,𝑛β‰₯0,2ξ€Έπ‘†βˆ’1ξ…ž(𝑧;𝑒(0,βˆ’1))=βˆ’π‘§2ξ€Έ+𝑧+1𝑆(𝑧;𝑒(0,βˆ’1))+3,𝑠=1.(4.23)

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

Remarks 4.2. (1) One has the form β„Žβˆ’1𝑒(0,βˆ’1)=β„’(βˆ’3/2,1/2), where β„’(𝛼,𝛽) is studied in [24].
(2) Let {𝑍𝑛(1)}𝑛β‰₯0 [15, 19] be the first associated sequence of {𝑍𝑛}𝑛β‰₯0 orthogonal with respect to 𝑒(0,βˆ’1) and 𝛽𝑛(1), 𝛾(1)𝑛+1 the coefficients of the three-term recurrence relations; we have 𝛽𝑛(1)=𝛽𝑛+1=(βˆ’1)𝑛,𝑛β‰₯0;𝛾(1)𝑛+1=𝛾𝑛+21=βˆ’4,𝑛β‰₯0.(4.24)

The sequence {𝑍𝑛(1)}𝑛β‰₯0 is a second-order self-associated sequence; that is, {𝑍𝑛(1)}𝑛β‰₯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.

References

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  2. E. B. Christoffel, β€œΓœber die Gaussiche quadratur und eine Verallgemeinerung derselben,” Journal fΓΌr die reine und angewandte Mathematik, vol. 55, pp. 61–82, 1858.