Quadratic decomposition of a Laguerre-Hahn polynomial sequence II

Transcription

1 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II Belgacem Bouras and Francisco Marcellán Abstract. In [4] the authors proved that a quasi-symmetric orthogonal polynomial sequence {R n } n 0 is a Laguerre-Hahn sequence if and only if the component {P n} n 0 in its quadratic decomposition is also a Laguerre-Hahn sequence. In this paper and under these conditions, we deduce the class s of the Laguerre-Hahn sequence {R n } n 0. More precisely, if s is the class of {P n } n 0 then s s s + 3. On the other hand the polynomial coefficients of the Riccati equation satisfied by the Stieltjes function corresponding to {R n } n 0 are given in terms of those of {P n} n 0. As an application, we determine all non-symmetric quasi-symmetric Laguerre-Hahn sequences of class one. Mathematics Subject Classification (010). Primary 33C45; Secondary 4C05. Keywords. Orthogonal polynomials, recurrence coefficients, Laguerre- Hahn polynomials, structure relation coefficients. 1. Introduction Let {R n } n 0 be a monic orthogonal polynomial sequence (MOPS in short) with respect to a linear functional u in the linear space of polynomials with complex coefficients such that the following three-term recurrence relation holds (1.1) R 0 (x) = 1, R 1 (x) = x β 0, R n+ (x) = (x ( 1) n+1 β 0 )R n+1 (x) γ n+1 R n (x), n 0, where β 0 IC, and γ n+1 IC, n 0. Such a MOPS is characterized by the following quadratic decomposition (see The work of the second author has been supported by Dirección General de Investigación, Ministerio de Ciencia e Innovación of Spain, grant MTM C03-01.

2 Bouras and Marcellán [8] and [11]) (1.) R n (x) = P n (x ), n 0, R n+1 (x) = (x β 0 )P n(x ), n 0, where {P n } n 0 is a MOPS and {P n} n 0 is the sequence of monic kernel polynomials of K-parameter β 0 associated with {P n } n 0 defined by (see [5] and [8]) (1.3) Pn(x) 1 = x β0 [P n+1 (x) P n+1(β 0) P n (β 0 ) P n(x)], n 0. The family of sequences {R n } n 0 satisfying (1.1) is wide enough to accommodate all symmetric sequences of orthogonal polynomials (β 0 = 0) as well as non-symmetric sequences (β 0 0). From a structural and constructive point of view, the Laguerre-Hahn polynomials constitute a very remarkable family of orthogonal polynomials taking into account most of the MOPS considered in the literature belong to this family. In particular, semiclassical orthogonal polynomials are Laguerre- Hahn MOPS (see [1]). The Laguerre-Hahn set of linear functional is invariant under the standard perturbations of linear functionals (see [1], [6], [7], [9], and [10] among others). In a previous paper (see [4]) we dealt with a perturbation of linear functionals such that the corresponding MOPS satisfies (1.1). We have shown that {R n } n 0 is a Laguerre-Hahn sequence of polynomials if and only if {P n } n 0 is also a Laguerre-Hahn sequence. In this contribution, our main goal is to complete the results of [4], focussing our attention in the analysis of the class of the MOPS {R n } n 0 in terms of the class of the MOPS {P n } n 0. Thus the description of all the possible situations is given. As a consequence, a constructive approach to new families of Laguerre-Hahn linear functionals in the non-classical sense is presented. The structure of the manuscript is as follows. In Section, we introduce the basic background about the algebra of linear functionals, orthogonal polynomials, and Stieltjes functions to be used in the sequel. In particular, we will remind the basic results of [4]. In Section 3, a complete analysis of the class s of the Laguerre-Hahn polynomial sequence {R n } n 0 is done. More precisely, we show that s s s + 3, where s is the class of {P n } n 0. In Section 4, we express the polynomial coefficients A, B, C, and D of the Riccati equation satisfied by the Stieltjes formal series associated with {R n } n 0, in terms of the polynomial coefficients of the Riccati equation satisfied by the Stieltjes formal series corresponding to {P n } n 0. Finally, in Section 5 we give explicitly the coefficients of the three-term recurrence relation coefficients as well as the polynomial coefficients of the structure relation for the Laguerre-Hahn polynomial sequences {R n } n 0 of class one satisfying (1.1). This is done by analyzing the case when {P n } n 0, the component of {R n } n 0 in the quadratic decomposition, is a Laguerre-Hahn sequence of class zero.

3 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 3. Notations and preliminary results Let P be the linear space of polynomials with complex coefficients and let P be its dual space. We denote by u, f the action of u P on f P and S(u)(z) = (u) n n 0 z the formal Stieltjes function of u where n+1 (u) n = u, x n, n 0, are the moments of u. Let introduce the following operations on P. For more details, see [1]. (1) The left multiplication of a linear functional by a polynomial is defined by (.1) gu, f = u, gf, f, g P, u P. () The right multiplication of a linear functional by a polynomial is given by xf(x) ξf(ξ) (.) (uf)(x) = u,, f P, u P, x ξ where u is acting over ξ. (3) The product of two linear functionals reads as (.3) vu, f = u, vf, f P, u,v P. (4) The dilation of a linear functional is defined as (.4) h a u, f = u, h a f, a IC {0}, f P, u P, where (.5) (h a f) (x) = f(ax). (5) The shift of a linear functional is (.6) τ b u, f = u, τ b f, b IC, f P, u P, where (.7) (τ b f)(x) = f(x b). (6) The even part of a linear functional is given by (.8) σu, f = u, σf, where (.9) σf(x) = f(x ). (7) The division of a linear functional by a polynomial of first degree is the linear functional such that (.10) (x c) 1 u, f = u, θ c f, c IC, f P, u P, where (.11) (θ c f)(x) = f(x) f(c). x c The following result can be easily proved

4 4 Bouras and Marcellán Lemma.1. (.1) (θ c σf)(x) = (x + c)(σθ c f)(x), c IC, f P, and (.13) u, θ a f θ 0 f = a u, θ 0 θ a f, a IC, f P, u P. Definition.. (See [5]) A sequence of polynomials {R n } n 0 is said to be a monic orthogonal polynomial sequence (MOPS) with respect to a linear functional w if i) deg R n = n and the leading coefficient of B n (x) is equal to 1, n 0. ii) w, R n R m = r n δ n,m, n, m 0, r n 0, n 0. It is well known that a sequence of monic orthogonal polynomials satisfies a three-term recurrence relation (see [5]) (.14) with R 0 (x) = 1, R 1 (x) = x β 0, R n+ (x) = (x β n+1 )R n+1 (x) γ n+1 R n (x), n 0, (β n, γ n+1 ) IC IC, n 0. In such conditions w is said to be regular or quasi-definite (see [5] and [1]). In the sequel we will consider regular linear functionals w with the normalization (w) 0 = 1. The sequence {R n (1) } n 0 of associated polynomials of the first kind for the MOPS {R n } n 0 is defined by (.15) R n (1) (x) = w, R n+1(x) R n+1 (ξ), n 0. x ξ It satisfies the shifted recurrence relation (see [5] and [1]) (.16) R (1) 0 (x) = 1, R(1) 1 (x) = x β 1, R (1) n+ (x) = (x β n+)r (1) n+1 (x) γ n+r n (1) (x), n 0. The orthogonality is preserved by a shifting in the variable. Indeed, for the shifted sequence { R n } n 0 defined by R n (x) = a n R n (ax + b), n 0, the following recurrence relation holds (see [1]) (.17) R 0 (x) = 1, R 1 (x) = x β 0 b a, R n+ (x) = (x β n b a ) R n+1 (x) γ n+1 a Rn (x), n 0. Such a sequence of monic polynomials is the MOPS with respect to the linear functional ŵ = (h a 1 τ b )w. A linear functional w is said symmetric if and only if (w) n+1 = 0, n 0, or, equivalently, in (.16) β n = 0, n 0. If the sequence {R n } n 0 satisfies (1.1) then there exists a MOPS {P n } n 0

5 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 5 such that (1.) holds (see [11]). Furthermore the sequence {P n } n 0 the following recurrence relation (.18) where (.19) P 0 (x) = 1, P 1 (x) = x β P 0, P n+ (x) = (x β P n+1) P n+1 (x) γ P n+1p n (x), n 0, β P 0 = γ 1 + β 0, β P n+1 = γ n+ + γ n+3 + β 0, n 0, γ P n+1 = γ n+1 γ n+, n 0. satisfies Denoting by w, u, and v the linear functionals associated with {R n } n 0, {P n } n 0, and {P n} n 0 respectively, we get (.0), u = σw (.1), β 0 u = σ(xw) (.) v = γ 1 1 (x β 0)u. The regularity of v and (.) means that (see [5], [6], [7], [1]) (.3) P n (β 0) 0, n 0. Conversely, if β 0 IC and u is a linear functional such that the corresponding MOPS {P n } n 0 satisfies (.18) and (.3) then the linear functional v defined in (.) where γ 1 = β P 0 β 0, is regular (see [5], [6], [7], [1]). In such a case, the MOPS corresponding to v is the sequence {P n} n 0 given in (1.3). Let us define the sequence of monic polynomials {R n } n 0 by (1.). Then it satisfies (1.1) with (.19) and its corresponding linear functional w satisfies (.0) and (.1). Definition.3. ([9], [10]). Let {R n } n 0 be a MOPS with respect to the linear functional w. {R n } n 0 is said to be a Laguerre-Hahn orthogonal polynomial sequence (respectively, w is called Laguerre-Hahn linear functional) of class s if the following conditions hold. There exist three polynomials Φ of degree t, Ψ of degree p, and B of degree r such that (.4) (Φw) + Ψw + B(x 1 w ) = 0, as well as (.5) { Ψ(c) + Φ (c) + B(c) + w, θc Ψ + θcφ + wθ 0 θ c B } = 0, c Z Φ where Z Φ denotes the set of zeros of Φ. The class s of {R n } n 0 is given by s = max{t, p 1, r }.

6 6 Bouras and Marcellán Remark.4. 1.If c 1 is a zero of Φ such that Ψ(c 1 ) + Φ (c 1 ) = 0, B(c 1 ) = 0, and w, θc1 Ψ + θ c 1 Φ + wθ 0 θ c1 B = 0, then the functional equation (.4) can be simplified dividing by x c 1. Thus, it becomes (.6) (Φ 1 w) + Ψ 1 w + B 1 (x 1 w ) = 0, where (.7) Φ 1 = θ c1 Φ, Ψ 1 = θ c 1 Φ + θ c1 Ψ, B 1 = θ c1 B. To see if (.6) can also be simplified by x c, where c is a zero of Φ 1, we must find B 1 (c), Ψ 1 (c)+φ 1(c), and w, θ c Ψ 1 + θ cφ 1 + wθ 0 θ c B 1. This yields a tedious computational work in particular when we must simplify many times the functional equation (.4). In the following lemma, whose proof is straightforward, we show that the two last expressions can be obtained by dividing Ψ(c) + Φ (c) and w, θ c Ψ + θ cφ + wθ 0 θ c B, respectively, by c c 1. Lemma.5. Let c 1 IC be such that (.8) Φ(c 1 ) = 0, (.9) Ψ(c 1 ) + Φ (c 1 ) = 0, (.30) B(c 1 ) = 0, and (.31) w, θc1 Ψ + θ c 1 Φ + wθ 0 θ c1 B = 0. For all c IC we have (.3) w, θc Ψ 1 + θcφ w, θc Ψ + θ 1 + wθ 0 θ c B 1 = cφ + wθ 0 θ c B, c c 1 (.33) (Φ 1 ) (c) + Ψ 1 (c) = Φ (c) + Ψ(c) c c 1. If w is a Laguerre-Hahn linear functional of class s fulfilling (.4) then, ŵ = (h a 1 τ b )w is also a Laguerre-Hahn linear functional of the same class and satisfies (see [] and [3]) (.34) ( Φŵ) + Ψŵ + B(x 1 ŵ ) = 0, where (.35) Φ(z) = a t Φ(az + b), B(z) = a t B(az + b), Ψ(z) = a 1 t Ψ(az + b). A Laguerre-Hahn orthogonal polynomial sequence {R n } n 0 satisfies the following structure relation (see [1], [], [3], [6], and [7]) where Φ(x)R n+1(x) B(x)R n (1) (x) = C n+1(x) C 0 (x) R n+1 (x) γ n+1 D n+1 (x)r n (x), n 0, (.36) C 0 (x) = Φ (x) Ψ(x),

7 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 7 (.37) χ 0 (x) = B(x), (.38) D 0 (x) = (wθ 0 Φ) (z) (wθ 0 Ψ)(z) (w θ 0B)(z), (.39) χ n+1 (x) = γ n+1 D n (x), n 0, (.40) C n+1 (x) = C n (x) + (x β n )D n (x), n 0, (.41) γ n+1 D n+1 (x) = Φ(x) + χ n (x) + (x β n ) D n (x) (x β n )C n (x), n 0. Remark.6. If c is a common zero of Φ, B, C 0, and D 0 then it is also a common zero of χ n, C n, and D n, n 0. Therefore, in both sides of (.4) we can divide by x c. where The sequence { R n } n 0 given in (.17) satisfies Φ(x) R n+1(x) (1) B(x) R n+1 (x) = Ĉn+1(x) Ĉ0(x) R n+1 (x) γ n+1 a Dn+1 (x) R n (x), n 0, (.4) Ĉ n (x) = a 1 t C n (ax + b), n 0, (.43) χ n (x) = a t χ n (ax + b), n 0, (.44) Dn (x) = a t D n (ax + b), n 0. In the sequel we suppose that {R n } n 0, {P n } n 0, {P n} n 0, u, w, and v satisfy (1.1)-(1.3), (.16), (.18) (.3). The structure relation plays an important role in order to find the fourth order linear differential equation with polynomial coefficients that every Laguerre-Hahn MOPS satisfies (see [6]). To conclude this section, we recall the following well known results. Theorem.7. ([4]) The sequence {P n } n 0 is a Laguerre-Hahn MOPS if and only if {R n } n 0 is a Laguerre-Hahn MOPS. Furthermore, if the class of u is s and satisfies (.45) (Φ P u) + Ψ P u + B P (x 1 u ) = 0, then w is of class s s + 3 and satisfies (.4) with (.46) Φ(x) = (x + β 0 )Φ P (x ), (.47) B(x) = xb P (x ), and (.48) Ψ(x) = x(x + β 0 )Ψ P (x ) Φ P (x ).

8 8 Bouras and Marcellán Theorem.8. ([4]) If the sequence {P n } n 0 satisfies the structure relation Φ P (x)p n+1(x) B P (x)p n (1) (x) = CP n+1(x) C0 P (x) P n+1 (x) γn+1d P n+1(x)p P n (x), n 0, then, the sequence {R n } n 0, for n s +, satisfies (.36) with (.49) C n (x) = Φ P (x ) + x(x + β 0 )(C P n (x ) + γp n D P n (x ) γ n 1 ), (.50) C n+1 (x) = Φ P (x ) x(x+β 0 )C P n (x )+4x(x+β 0 )(x β 0 γp n γ n 1 )D P n (x ), (.51) D n (x) = x(x + β 0 ) D P n (x ), (.5) D n+1 (x) = x(c P n+1(x ) C P n (x ) + γp n+1 γ n+1 D P n+1(x ) +(x β 0 γp n γ n 1 )Dn P (x )). 3. The class of {R n } n 0 As we have shown, Theorem.5 does not precise the class s of the sequence {R n } n 0. The aim of this section is to study in detail this class. We will assume that the equation (.47) satisfied by the linear functional w can not be simplified. In other words, s = max{deg Φ P, deg Ψ P 1, deg B P }. To find s we use condition (.5). Lemma 3.1. For all f in P we have (3.1) w, xf(x ) = β 0 u, f(x), (3.) w, f(x ) = u, f(x), (3.3) (wσf)(x) = (uf)(x ) + β 0 x(uθ 0 f)(x ), (3.4) (w(xσf))(x) = (x + β 0 )(uf)(x ). Proof. (3.1) follows from (.8), (.9), and (.1). From (.0), we get (3.). (3.3) and (3.4) are deduced in a straightforward way using (.). Lemma 3.. For all c IC we have (3.5) Ψ(c) + Φ (c) = Φ P (c ) + c(c + β 0 )(Ψ P (c ) + (Φ P ) (c )), (3.6) w, θc Ψ + θ cφ + wθ 0 θ c B = c(c + β 0 ) u, θ c Ψ P + θ c ΦP + uθ 0 θ c B P.

9 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 9 Proof. Formula (3.5) follows in a straightforward way from (.48) and (.50). On the other hand, from (.48) (.56) we get (3.7) w, θc Ψ + θ cφ + wθ 0 θ c B = w, θ c (x(x + β 0 )Ψ P (x ) + w, θ c((x + β 0 )Φ P (x ) + θ c ( Φ P (x )) + w, wθ 0 θ c (xb P (x )). Using the definition of the operator θ c, it is easy to see that, for two polynomials f and g, we have (3.8) θ c (fg) = g(x)θ c (f)(x) + f(c)(θ c g)(x). Taking g(x) = Ψ P (x ) and f(x) = x(x + β 0 ) in (3.8), and using (.1), from (3.1) and (3.) we get (3.9) w, θc (x(x + β 0 )Ψ P (x ) = (β 0 + c) u, Ψ P (x) +c(c + β 0 ) u, (θ c Ψ P )(x). Replacing g(x) = Φ P (x ) and f(x) = x + β 0 in (3.8), and using (.1) we deduce (3.10) θ c((x + c)φ P (x )) = (x + c)(θ c Φ P )(x ) + (c + β 0 )θ c ((x + c)(θ c Φ P )(x )). Next, let consider (3.8) with g(x) = (θ c Φ P )(x ) and f(x) = x+c. From (.1) we get θ c ((x + c)(θ c Φ P )(x )) = (θ c Φ P )(x ) + c(x + c)(θ c Φ P )(x )). From (3.1), (3.), and (3.10) we get (3.11) w, θ c ((x + β 0 )Φ P (x )) + θ c ( Φ P (x )) = c(c + β 0 ) u, (θ c ΦP )(x). If we consider in (3.8) g(x) = B P (x ) and f(x) = x, then, θ c (xb P (x )) = B P (x ) + c(x + c)(θ c B P )(x ). But, from (.1, ) θ 0 θ c (xb P (x )) = x(θ 0 B P )(x ) + cθ 0 ((x + c)(θ c B P )(x )). Hence, applying (3.8) to g(x) = (θ c B P )(x ) and f(x) = x + c, we get θ 0 θ c (xb P (x )) = x(θ 0 B P )(x ) + c(θ c B P )(x ) + c x(θ 0 θ c B P )(x ). Therefore, by (3.3) and (3.4), and taking into account (3.1) and (3.) we get

10 10 Bouras and Marcellán (3.1) w, wθ0 θ c (xb P (x )) = 4β 0 u, (θ 0 B P )(x) + c u, (θ c B P )(x) +β 0c u, (θ 0 θ c B P )(x) Using (.13) we deduce +4β 0 c u, (θ 0 θ c B P )(x). u, (θ c B P )(x) = c u, (θ 0 θ c B P )(x) + u, (θ 0 B P )(x). As a consequence, (3.1) becomes (3.13) w, θ 0 θ c (xb P (x )) = (β 0 + c) u, (θ 0 B P )(x) +c(c + β 0 ) u, (θ 0 θ c B P )(x). Finally, replacing (3.9), (3.11), and (3.13) in (3.7) w, θc Ψ + θcφ + wθ 0 θ c B = c(c + β 0 ) u, θc Ψ P + θc Φ P + uθ 0 θ c B P +(β o + c)( u, (θ 0 B P )(x) + u, Ψ P (x) ). This yields (3.6) since u, (θ 0 B P )(x) + u, Ψ P (x) = 0. Proposition 3.3. The class of w depends only on the zeros x = 0 and x = β 0. Proof. Let c be a zero of Φ such that c 0 and c β 0. If Φ (c) + Ψ(c) = 0 and B(c) = 0, then from (.54), (.55) and (3.5), c is a common zero of Φ P, B P, and (Φ P ) + Ψ P. But, since (.48) can not be simplified, then u0, θ c Ψ P + θ c Φ P + u 0 θ 0 θ c B P 0. Therefore, from (3.6) we have w, θc Ψ + θ cφ + wθ 0 θ c B 0. So we cannot divide in (.4) by x c. Notice that if c = 0 or c = β 0, then we can divide by x c in the functional equation. Thus, we will analyze the cases β 0 = 0 and β 0 0, separately. CASE 1.- β 0 = 0. According to Theorem.5, the linear functional w satisfies (.4) with Φ(x) = xφ P (x ), Ψ(x) = x Ψ P (x ) Φ P (x ), B(x) = xb P (x ). The class of w is at most s + 3. From (3.5) we have Φ(0) = 0, Φ (0) + Ψ(0) = Φ P (0), B(0) = 0. But from (3.6) (3.14) w, θc Ψ + θ cφ + wθ 0 θ c B = c 3 u, θ c Ψ P + θ c ΦP + uθ 0 θ c B P.

11 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 11 Thus, w, θ0 Ψ + θ 0Φ + wθ 0B = 0. 1.A. If Φ P (0) 0 then (.4) can not be simplified and we have s = s B. If Φ P (0) = 0 then (.4) can be simplified and, according to (.7) and (.1), w satisfies (.4) with Φ(x) = Φ P (x ), Ψ(x) = xψ P (x ) x(θ 0 Φ P )(x ), B(x) = B P (x ). Application of Lemma.4 yields w, θc Ψ + θ cφ + wθ 0 θ c B = c u, θ c Ψ P + θ c ΦP + uθ 0 θ c B P. Because c = 0, we get Φ(0) = 0, Φ (0) + Ψ(0) = 0, B(0) = B P (0), and w, θ0 Ψ + θ 0Φ + wθ 0B = 0. 1.B.-1. If B P (0) 0, then (.4) can not be simplified and we have s = s +. 1.B.-. If B P (0) = 0, then the functional equation can be simplified and w satisfies (.4) with Φ(x) = x(θ 0 Φ P )(x ), Ψ(x) = Ψ P (x ), B(x) = x(θ 0 B P )(x ). Using Lemma.4 we get w, θc Ψ + θ cφ + wθ 0 θ c B = c u, θ c Ψ P + θ c ΦP + uθ 0 θ c B P. Hence Φ(0) = 0, Φ (0) + Ψ(0) = Ψ P (0) + (Φ P ) (0), B(0) = 0, and w, θ 0 Ψ + θ 0Φ + wθ 0B = 0 1.B If Ψ P (0) + (Φ P ) (0) 0, then the functional equation can not be simplified and thus s = s B.-.-. If Ψ P (0) + (Φ P ) (0) = 0, then w satisfies (.4) with Φ(x) = (θ 0 Φ P )(x ), Ψ(x) = x(θ 0 Ψ P )(x ) + x(θ 0Φ P )(x ), B(x) = (θ 0 B P )(x ), and w, θc Ψ + θ cφ + wθ 0 θ c B = u, θ c Ψ P + θ c Φ P + uθ 0 θ c B P. In this case, the functional equation can not be simplified. Indeed, in the contrary, we have u, θ 0 Ψ P + θ 0Φ P + uθ 0B P = 0 and (θ 0 Φ P )(0) = 0.

12 1 Bouras and Marcellán The last equality is equivalent to (Φ P ) (0) = 0. Taking into account that Ψ P (0)+(Φ P ) (0) = 0, we obtain Ψ P (0) = 0. Therefore Ψ P (0)+(Φ P ) (0) = 0. But we also have Φ P (0) = 0 and B P (0) = 0. Then we can divide in (.47) by x and this yields a contradiction. Hence s = s. CASE.- β 0 0..A. If Φ P (0) 0, then in (.4) we can not divide by x. Thus we can analyze the possible division by x + β 0..A.-1. If Φ P (β 0) 0 or B P (β 0) 0, then s = s + 3..A.-. If Φ P (β 0) = B P (β 0) = 0, then w satisfies (.4), with Φ(x) = Φ P (x ), Ψ(x) = xψ P (x ) (x β 0 )(θ β 0 Φ P )(x ), B(x) = x(x β 0 )(θ β 0 B) P (x ). From Lemma.4 we get w, θc Ψ + θ cφ + wθ 0 θ c B = c(c + β 0 ) u, θ c Ψ P + θ c ΦP + uθ 0 θ c B P..A If (B P ) (β 0) 0 or Ψ P (β 0) 0, then s = s +..A.-.-. If (B P ) (β 0) = Ψ P (β 0) = 0, then w satisfies (.4) with Φ(x) = (x β 0 )(θ β 0 Φ P )(x ), Ψ(x) = x(x β 0 )(θ β 0 Ψ P )(x ), B(x) = (x β 0 )(θ β 0 B) P (x ) + 4β 0(x β 0 )(θ β 0 B) P (x ). Thus, the functional equation can not be simplified and s = s + 1..B. If Φ P (0) = 0, then w satisfies (.4) with Φ(x) = x(x + β 0 )(θ 0 Φ P )(x ), B(x) = B P (x ), Ψ(x) = (x + β 0 )Ψ P (x ) (x β 0 )(θ 0 Φ P )(x )..B.-1. If Φ P (β 0) 0 or B P (β 0) 0, then s = s +..B.-. If Φ P (β 0) = B P (β 0) = 0, then w satisfies (.4), with Φ(x) = x(θ 0 Φ P )(x ), B(x) = (x β 0 )(θ β 0 B) P (x ),

13 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 13 Ψ(x) = Ψ P (x ) + β 0 (x β 0 )(θ β 0 θ 0 Φ P )(x )..B If (B P ) (β 0) 0 or Ψ P (β 0) 0, then s = s + 1..B.-.-. If (B P ) (β 0) = Ψ P (β 0) =, 0 then w satisfies (.6), with Φ(x) = x(x β 0 )(θ β 0 θ 0 Φ P )(x ), B(x) = (x β 0 ) (θ β 0 B) P (x ), Ψ(x) = (x β 0 )(θ β 0 Ψ P )(x ) + (x β 0 )(θ β 0 θ 0 Φ P )(x ). Thus, the functional equation can not be simplified and we have s = s. The previous discussion can be summarized as follows. Proposition 3.4. We distinguish the following cases. (A) β 0 = 0. (A 1 ) Φ P (0) 0. Then s = s + 3. (A ) Φ P (0) = 0 and B P (0) 0. Then s = s +. (A 3 ) Φ P (0) = 0, B P (0) = 0 and Ψ P (0) + (Φ P ) (0) 0. Then s = s + 1. (A 4 ) Φ P (0) = 0, B P (0) = 0 and Ψ P (0) + (Φ P ) (0) = 0. Then s = s. (B) β 0 0. (B 1 ) Φ P (0) 0 and Φ P (β 0) + B P (β 0) = 0. Then s = s + 3. (B ) Either Φ P (0) 0, Φ P (β 0) = 0, B P (β 0) = 0, (B P ) (β 0) + Ψ P (β 0) = 0, or Φ P (0) = 0, Φ P (β 0) + B P (β 0) = 0. Then s = s +. (B 3 ) Either Φ P (0) 0, Φ P (β 0) = 0, B P (β 0) = 0, (B P ) (β 0) = 0, Ψ P (β 0) = 0, or Φ P (0) = 0, Φ P (β 0) = 0, B P (β 0) = 0, (B P ) (β 0) + Ψ P (β 0) = 0. Then s = s + 1. (B 4 ) Φ P (0) = 0, Φ P (β 0) = 0, B P (β 0) = 0, (B P ) (β 0) = 0, Ψ P (β 0) = 0. Then s = s. 4. Riccati equation It is well known that the linear functional w satisfies (.4) if and only if its formal Stieltjes function S(w)(z) = n 0 satisfies the following (w) n z n+1

14 14 Bouras and Marcellán Riccati equation (see [], [3], [6], [9],[10]) (4.1) A(z)S (w)(z) = B(z)S (w)(z) + C(z)S(w)(z) + D(z) with (4.) A(z) = Φ(z), C(z) = Φ (z) Ψ(z), D(z) = (wθ 0 Φ) (z) (wθ 0 Ψ)(z) (w θ 0B)(z). Condition (.5) is equivalent to the fact that polynomials A, B, C, and D are coprime. Let assume that {P n } n 0 is a Laguerre-Hahn sequence. According to Theorem.5, {R n } n 0 is also a Laguerre-Hahn sequence and, as a consequence, its corresponding Stieltjes function satisfies a Riccati equation. In this section our aim is to express the polynomial coefficients A, B, C, and D of the Riccati equation (4.1) corresponding to w in terms of those of u which we will denote A P, B P, C P, and D P. We need the following Lemma whose proof is a straightforward exercise. Lemma 4.1. (4.3) S(w)(z) = (z + β 0 )S(u)(z ). As a consequence, we get Proposition 4.. If S(u)(z) satisfies (4.4) A P (z)s (u)(z) = B P (z)s (u)(z) + C P (z)s(u)(z) + D P (z) then, S(w)(z) satisfies (4.1) with A(z) = (z + β 0 )A P (z ), B(z) = zb P (z ), C(z) = A P (z ) + z(z + β 0 )C P (z ), D(z) = z(z + β 0 ) D P (z ). Proof. Taking formal derivatives in (4.3) we get Therefore zs (u)(z ) = (z + β 0)S (w)(z) S(w)(z) (z + β 0 ). (4.5) S (u)(z ) = (z + β 0)S (w)(z) S(w)(z) z(z + β 0 ). In (4.4) the change of variable z z yields (4.6) A P (z )S (u)(z ) = B P (z )S (u)(z ) + C P (z )S(u)(z ) + D P (z ). Replacing (4.3) and (4.5) in (4.6) we get A P (z )( (z + β 0)S (w)(z) S(w)(z) z(z + β 0 ) ) = B P (z )( S(w)(z) (z + β 0 ) ) +C P (z ) S(w)(z) (z + β 0 ) + DP (z ).

15 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 15 Hence, (z + β 0 ) A P (z ) S (w)(z) = zb P (z )S (w)(z) +(A P (z ) + z (z + β 0 ) C P (z ))S(w)(z) + z (z + β 0 ) D P (z ). 5. Example In this section we will give the coefficients of the three term recurrence relation and those of the structure relation coefficients for all non-symmetric Laguerre- Hahn orthogonal polynomial sequences {R n } n 0 of class s = 1 satisfying (1.1). This method can be used to determine all symmetric Laguerre-Hahn sequences of class s = 1. But we don t do it here because the coefficients of such sequences are obtained in an alternative way in [1] by solving a non linear system. Let assume β 0 0. Using a suitable dilation in the variable that, for a sake of simplicity, yields β 0 = 1, from Theorem.5 and Proposition 3.4, {R n } n 0 is a Laguerre-Hahn sequence of class s = 1 if and only if {P n } n 0 is a Laguerre-Hahn sequence of class s = 0 and one of the following conditions hold (1) Φ P (0) = 0, Φ P (1) = 0, B P (1) = 0, (B P ) (1) + Ψ P (1) = 0, () Φ P (0) 0, Φ P (1) = 0, B P (1) = 0, (B P ) (1) = 0, Ψ P (1) = 0. First we will analyze the conditions in (1). Since s = 0 implies deg Φ P, then from the conditions Φ P (0) = 0, Φ P (1) = 0 we have (5.1) Φ P (x) = x x. Therefore, from (.34) and (.35), u can be obtained by shifting the linear functional satisfying (.47) with Φ P (x) = x 1. Indeed, we use the shift h 1 τ 1. The sequence {P n } n 0 corresponding to the above linear functional appears in [], [3], and [13]. These polynomials are related to some perturbations of Jacobi polynomials and their associated polynomials of a real order τ (a shift by τ in the parameters of the recurrence relation). Thus, using the previous results, we can give the characteristic elements of the sequence {R n } n 0. The main difficulty is that given γn P and βn P, n 0, the recurrence coefficients of the sequence {P n } n 0, it is not easy, by solving directly the non linear system (.19), to give explicitly γ n, n 0, the coefficients of the recurrence relation of {R n } n 0. To solve this problem, we state the following Proposition 5.1. Let denote by C P n (x) and D P n (x), n 0, the polynomial coefficients of the structure relation of {P n } n 0. Then (5.) γ n+1 = γp n+1dn+1(1) P Cn+1 P (1) n 0, CP 0 (1),

16 16 Bouras and Marcellán (5.3) γ n+ = ( CP n+1(1) C P 0 (1) D P n+1 (1) ), n 0. Proof. From (1.1) we have Setting x = 1, we obtain R n+ (x) = (x + 1)R n+1 (x) γ n+1 R n (x). Taking into account (1.), we have Hence R n+ ( 1) = γ n+1 R n ( 1). P n+1 (1) = γ n+1 P n (1). (5.4) γ n+1 = P n+1(1) P n (1), n 0, and from (.19) we have (5.5) γ n+ = γn+1 P P n (1) P n+1 (1), n 0. Substituting x = 1 in the structure relation (.51) and taking into account (5.1) as well as B P (1) = 0, we obtain Therefore (5.6) Cn+1(1) P C0 P (1) P n+1 (1) = γ P n+1dn+1(1)p P n (1), n 0. P n+1 (1) P n (1) Using (5.4) (5.6), we get (5.) and (5.3). = γp n+1dn+1(1) P Cn+1 P (1) n 0. CP 0 (1), According to [] (see also [3]) as well as (.34) and (.35), we distinguish two cases (1.A) ( B P (x) = 1 4 (1 α)( x + 1) + (αλ µ(α 1) α )( x + 1) µ ) α +λµ 1 + (1 + α). In this case, (5.7) ψ P (x) = 1 ((α )( x + 1) + µ), (5.8) C0 P (x) = 1 (α( x + 1) + µ), ( Cn+1(x) P C0 P (x) = 1 (n + α)( x + 1) (5.9) +α (λ + µ µ(α 1) α ) (αλ α ) αµ( 1 α 1 n+α ), ) n 0, (5.10) D P 0 (x) = 0, D P n+1(x) = n + α + 1, n 0,

17 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 17 (5.11) γ P 1 = 4, (5.1) γ P n+1 = n(n + α)(n + α µ)(n + α + µ) 4(n + α + 1)(n + α 1)(n + α), n 1, where α n, n 0 and α ±µ n, n 1. Condition B P (1) = 0 is equivalent to = (µ α)(µ α+ αλ+λ) (α+1)(α ). Hence, from (.48), (.49), and (.50) we obtain, after division by (x + 1)x, (5.13) Φ(x) = x 3 x, (5.14) Ψ(x) = x 3 + ( 1 + α) x + x α µ, B(x) = ( α ( α + ) x 3 + ( α ) x λ + α λ + µ α µ ) x + α α λ + α λ + µ α µ. α Using (5.), (5.3), and (5.9) (5.1) we get (5.15) γ 1 = (µ α + αλ + λ), (α ) (5.16) γ n+1 = 1 n(n + α + µ) (n + α 1)(n + α), n 1, (5.17) γ n+ = 1 (n + α)(n + α µ) (n + α)(n + α + 1), n 0. From (.37), (.39), (5.13), and (5.14), dividing by x(x + 1) we get (5.18) C 0 (x) = (1 α)x x + α + µ, (5.19) D 0 (x) = 0. From (.5) (.55), (5.8) (5.1), and (5.1) we obtain after a division by x(x + 1) (5.0) C n+1 (x) = (4n 1 + α)x + x 4n α µ, n 0, (5.1) C n+ (x) = (1 + 4n + α)x x 4n 3α + µ, n 0, (5.) D n+1 (x) = (n + α)(x 1), n 0, (5.3) D n+ (x) = (n + α + 1)(x + 1), n 0. We can summarize our results about the basic information for the linear functional in the following Table 1.

18 18 Bouras and Marcellán Φ(x) = x 3 x, Table 1 Ψ(x) = x 3 + ( 1 + α) x + x α µ, B(x) = ( α + ) x 3 + ( α ) x + ( α λ+ α λ+ µ α µ)x α α λ+ α λ+ µ α µ α, γ 1 = (µ α+ αλ+λ) ( (α ) γ n+1 = 1 4 n+(α 1 )(1 ( 1)n ) )( n+α+µ+(1+µ)( ( 1)n 1 ) (n+α)(n+α 1), n 1, ) C 0 (x) = (1 α)x x + α + µ, C n+1 (x) = ( 1 + α + n)x + ( 1) n x n α µ +(1 ( 1) n )(µ α + 1), n 0, D 0 (x) = 0, D n+1 (x) = (n + α)(x ( 1) n ), n 0. (1.B) with B P (x) = 1 4 [b ( x + 1) + b 1 ( x + 1) + b 0 ], b = ( 1 1)(τ + α + β + 1), α β b 1 = (1 1 )(1 + 1 τ + α + β )( τ + α + β + ) +λ[(1 )(τ + α + β) ( 1 1)], b 0 = {1 b }((β0 P ) ) b 1 β0 P 1 (1 + τ)(1 + τ + α)(1 + τ + β)(1 + τ + α + β) +4 (τ + α + β + 1)(τ + α + β + ), β P 0 = λ + α β (τ + α + β + )(τ + α + β).

19 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 19 In this case, (5.4) ψ P (x) = [(1 )(τ + α + β) + β + τ + 1 ]x + α β 0 + µ, (5.5) C0 P (x) = 1 ({( 1)(τ + α + β) + }( x + 1) + β α τ + α + β λ(τ + α + β + 1) + (1 1 )(1 + 1 τ+α+β )( α β τ+α+β+ ), ) (5.6) C P n+1(x) = C P 0 (x) 1 ( (n + 1 b )( x + 1) + (1 b )β 0 b 1 +(α β )( 1 τ+α+β+ 1 n+τ+α+β+ ) ), (5.7) D P 0 (x) = 1 (τ + α + β + 1), DP n+1(x) = (n + τ + α + β + 3), n 0, (5.8) γ P 1 = (5.9) γ P n+1 = (1 + τ)(1 + τ + α)(1 + τ + β)(1 + τ + α + β) (τ + α + β + 1)(τ + α + β + 3)(τ + α + β + ), (n τ)(n τ + α)(n τ + β)(n τ + α + β) (n + τ + α + β + 1)(n + τ + α + β + ) (n + τ + α + β + 3), n 1, where τ n 1, τ + α n 1, τ + β n 1, n 0 and τ + α + β n, n. Taking into account that B P (1) = 0 becomes a quadratic equation in λ, we can consider two values for λ which we denote λ 1 and λ. For λ = λ 1, we get (5.30) Φ(x) = x 3 x, (5.31) Ψ(x) = x 3 ( τ + α + β 8τ 4α 4β) x + x + τ 4τ + β β + α, (5.3) B(x) = ( 1 + ) (α β + τ) x 3 ( 1 + ) (α β + τ) + x ( α 1 β + + β + τ τ) (α β + τ τ). + x (α β + τ) ( α 1 β + + β + τ τ) (α β + τ τ). (α β + τ)

20 0 Bouras and Marcellán Using (5.) and (5.3) we get (1 + τ)(1 + τ + β) (5.33) γ 1 = (τ + α + β + 1)(τ + α + β + ), (n τ)(n τ + β) (5.34) γ n+1 = (n + τ + α + β + 1)(n + τ + α + β + ), n 1, (n τ + α)(n τ + α + β) (5.35) γ n+ = (n + τ + α + β + )(n + τ + α + β + 3), n 0. From (.37), (.39), (5.30) (5.3) if we divide by x(x + 1) we get (5.36) τ + α + β 8τ 4α 4β C 0 (x) = ( )x x τ 4τ + β + β α, (5.37) D 0 (x) = (τ + α + β + 1) (x + 1). Using (.58) (.61) the division by x(x + 1) yields (5.38) C n+1 (x) = (3 + 4n + α + 4τ + β)x + x 4n 4 4τ β, n 0 (5.39) C n+ (x) = (5+4n+4τ +α+β)x x 4n β 4α 4τ 4, n 0, (5.40) D n+1 (x) = (n + τ + α + β + )(x 1), n 0, (5.41) D n+ (x) = (n + τ + α + β + 3)(x + 1), n 0. In the following table we summarize our results concerning this case.

21 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 1 Φ(x) = x 3 x, Tabla Ψ(x) = x 3 + ( 4+3+4τ+α+β 8τ 4α 4β) x +x τ 4τ+β β+ α, B(x) = ( 1+)(α+1+β+τ) x 3 + ( 1+)(α+1+β+τ) x + ( α 1 β++β+τ τ)(α+1+β + τ τ) (α+1+β+τ) x ( α 1 β++β+τ τ)(α+1+β +τ τ) (α+1+β+τ), γ 1 = γ n+1 = 1 4 (1+τ)(1+τ+β) (τ+α+β+1)(τ+α+β+), ( )( ) n+τ++(α 1 )(1 ( 1)n ) n+τ+β++(α 1 )(1 ( 1)n ) (n+τ+α+β+1)(n+τ+α+β+), C 0 (x) = ( 4+3+4τ+α+β 8τ 4α 4β )x x + τ 4τ + β+β α, C n+1 (x) = (n + β + α + 4τ + 3)x +( 1) n x n 4τ β 4 D 0 (x) = (τ+α+β+1) (x + 1), D n+1 (x) = (n + α + β + τ + )(x ( 1) n ), n 0, n 0. For λ = λ we have (5.4) Φ(x) = x 3 x, (5.43) Ψ(x) = x 3 ( α + 4τ + β 4α 8τ 4β) + x + x α + α + β + τ + β 4τ,

22 Bouras and Marcellán ( 1) (α β + τ) B(x) = x 3 ( 1) (α β + τ) + x bx + b. where ( β 1 + τ + τ + α α) (β β 1 + τ + τ + α α) b = (α β + τ) Using (5.) and (5.3) we get (1 + τ + α)(1 + τ + α + β) (5.44) γ 1 = (τ + α + β + 1)(τ + α + β + ), (n τ + α)(n τ + α + β) (5.45) γ n+1 = (n + τ + α + β + 1)(n + τ + α + β + ), n 1, (n τ)(n τ + β) (5.46) γ n+ = (n + τ + α + β + )(n + τ + α + β + 3), n 0. From (.37) and (.39), the division by x(x + 1) leads to (5.47) τ + α + β 8τ 4α 4β C 0 (x) = ( )x x τ 4τ + β + β α +, (5.48) D 0 (x) = (τ + α + β + 1) (x + 1). Using (.5) (.55) we obtain, after simplification by x(x + 1), (5.49) C n+1 (x) = (3+4n+α+4τ +β)x +x 4n 4α 4 4τ β, n 0, (5.50) C n+ (x) = (5 + 4n + 4τ + α + β)x x 4n β 4τ 4, n 0, (5.51) D n+1 (x) = (n + τ + α + β + )(x 1), n 0, (5.5) D n+ (x) = (n + τ + α + β + 3)(x + 1), n 0.

23 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 3 Φ(x) = x 3 x, Tabla 3 Ψ(x) = x 3 ( 4+3+α+4τ+ β 4α 8τ 4β) + x + x α+α+β+τ + β 4τ, ( 1)(α+1+β+ τ) x ( β 1+ τ+τ+α α)(β β 1+ τ+τ+α α) B(x) = ( 1)(α+1+β+τ) x (α+1+β+τ) ( β 1+ τ+τ+α α)(β β 1+ τ+τ+α α) (α+1+β+τ), x γ 1 = γ n+1 = 1 4 n 1, (1+τ+α)(1+τ+α+β) (τ+α+β+1)(τ+α+β+), ( )( ) n+τ+1+(α+ 1 )(1+( 1)n ) n+τ+β+1+(α+ 1 )(1+( 1)n ) (n+τ+α+β+1)(n+τ+α+β+), C 0 (x) = ( 4+3+4τ+α+β 8τ 4α 4β )x x C n+1 (x) = (n + β + α + 4τ + 3)x +( 1) n x n 4τ β D 0 (x) = (τ+α+β+1) (x + 1), D n+1 (x) = (n + α + β + τ + )(x ( 1) n ), n 0. The sequences described in Tables 1,, and 3 are the unique Laguerre- Hahn polynomial sequences satisfying (1.1). In fact, we will show that conditions in () are not compatible with the regularity of u. Indeed, let assume that there exists a regular linear functional u satisfying (.47) with conditions in (). Then, using (.34) and (.35), its shift τ 1 u satisfies (.47) with ( ) Φ P (0) = 0, B P (0) = 0, (B P ) (0) = 0, Ψ P (0) = 0, We will analyze three cases.

24 4 Bouras and Marcellán 1. First case: Φ P (x) = x. Following [] as well as [3] and [13], we distinguish two subcases B P (x) = x + ((1 α) λ)x + α(α 1 λ). Here, we have Ψ P (x) = x + α 1, and γ P 1 = 0. From ( ) we get α = 1, λ = 0, = 0 which yields a contradiction. 1.. B P (x) = (1 1 )x + {( 1 1)(τ + α + 1) + λ( 1)}x + (τ + α λ){(1 1 )(τ + α + 1) + 1 λ } (1 + τ)(1 + τ + α). Here, Ψ P (x) = ( 1)x + (1 )(τ + α + 1) λ, and γp 1 = (1 + τ)(1 + τ + α), 0. From ( ) we have (5.53) λ = ( 1)(τ + α + 1), (5.54) {( 1 1)(τ + α + 1) + λ( 1)} = 0, and (5.55) (τ +α+1+λ){(1 1 )(τ +α+1)+1 λ } (1+τ)(1+τ +α) = 0. Replacing (5.53) in (5.54) we get (τ + α + 1) = 0. If = 0, then γ1 P = 0, that contradicts the regularity of the linear functional. If τ + α + 1 = 0, then from (5.53) and (5.55) we get λ = 0 and (1 + τ)(1 + τ + α) = 0. Therefore, γ1 P = 0, a contradiction with the regularity of the linear functional.. Second case: Φ P (x) = x..1. Following [13] we distinguish two subcases. B P (x) = ( 1 1)(α 1)x + { α + [α(1 ) + ]βp 0 }x + α 1 (β0 P ) α βp 0 + α (α 1) with β P 0 = α(α ) + λ.

25 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 5 Here From ( ) we have Ψ P (x) = [(1 )(α 1) 1]x α + α 1 β0 P, and γ1 P = α, α, 0. (α 1)(α + 1) (5.56) β P 0 = and (5.57) α(α 1) α + (α(1 ) + )βp 0 = 0. Replacing (5.56) in (5.57) we get α = 0, a contradiction... B P (x) = ( 1 1)[(τ + θ) 1]x + {[(τ + θ)(1 ) + 1 ]βp θ τ+θ }x + ( τ+θ 1 )(β0 P ) + θ 1 τ+θ βp 0 + (τ+1)(τ+θ 1) [(τ+θ) 1](τ+θ), with β P 1 θ 0 = (τ+θ)(τ+θ 1) + λ. and γ P 1 Thus, Ψ P (x) = [(τ + θ)(1 ) + 1 1]x + (τ + θ 1)βP 0 (τ+1)(τ+θ 1) = [(τ+θ)+1][(τ+θ) 1](τ+θ), τ + θ, 0. + θ 1 τ+θ, From ( ) (5.58) β0 P 1 θ = (τ + θ)(τ + θ 1), (5.59) ( and τ + θ 1 )(β0 P ) + θ 1 (τ + 1)(τ + θ 1) τ + θ βp 0 + ((τ + θ) 1)(τ + θ) = 0, (5.60) ((τ + θ)(1 ) + 1 )βp θ τ + θ = 0 Replacing (5.58) in (5.60) we get (τ + θ)(1 θ) = 0. Since (τ + θ) 0 then θ = 1. Therefore, using (5.58) and (5.59), β0 P = 0 and (τ+1)(τ+θ 1) ((τ+θ) 1)(τ+θ) = 0. So, γ1 P = 0, a contradiction with the regularity of the linear functional. 3. Third case: Φ P (x) = x x. We distinguish two subcases B P (x) = 1 4 {(1 α)( x + 1) (µ α 1 µ α αλ)( x + 1) + α + λµ + (α + 1) 1}.

26 6 Bouras and Marcellán In these conditions we get Ψ P (x) = 1 {(α )( x + 1) + µ} together with γ1 P = 4 and γp n+1 = n(n+α)(n+α µ)(n+α+µ) 4(n+α+1)(n+α 1)(n+α), n 1. From ( ) we have (5.61) α + µ =, (5.6) (1 α) (µ α 1 µ αλ) + + λµ + (α + 1) 1 = 0, α α and (5.63) (1 α) (µ α 1 αλ) = 0. α Replacing (5.61) in (5.63) we get αλ = 0. If α = 0, then µ =. Therefore γ P = 0, in contradiction with the regularity of the linear functional. If λ = 0, then, from (5.6), we obtain (α + 1) = 0. Thus γ1 P γ P = 0, in contradiction with the regularity of the linear functional. 3.. B P (x) = 1 4 {( 1 1)(τ + α + β + 1)( x + 1) + {[(α + β + τ)(1 ) 1)]K µ}( x + 1) ( 1 + α+β+τ+1 K +µk +(τ +α+β+3)4 (τ+1)(τ+1+α)(τ+1+β)(τ+1+α+β) α and K = β (α+β+(τ+1))(α+β+τ) + λ. Thus, Ψ P (x) = 1 {[(1 )(τ +α+β) γ P 1 = (τ+1)(τ+1+α)(τ+1+β)(τ+1+α+β) (τ+α+β+1)(τ+α+β+3)(τ+α+β+). (τ+α+β+1)(τ+α+β+3)(τ+α+β+) 1} α+β+τ+1 ]( x+1)+ K+µ}, Ψ P (0) = 0 yields a linear equation in λ. For such a value of λ the condition B P (0) = 0 is equivalent to 1 (β 1)(β+1) 4 (τ+α+β+1) = 0. If β = 1 then (B P ) (0) = (τ+1)(τ+1+α) (τ+α+). If β = 1 then (B P ) (0) = τ(τ+α) (τ+α). In both cases the condition (B P ) (0) = 0 is not compatible with the regularity of u. Acknowledgment The authors thank the deep revision by the referees. Their suggestions and remarks have contributed to improve substantially the presentation of the manuscript. References [1] J. Alaya and P. Maroni, Symmetric Laguerre-Hahn forms of class s = 1. Integral Transforms Spec. Funct. 4 (1996), [] H. Bouakkaz, Les polynômes orthogonaux de Laguerre-Hahn de classe zéro. Thèse de Troisième Cycle. Université Pierre et Marie Curie, Paris, 1990.

27 Quadratic decomposition of a Laguerre-Hahn polynomial sequence II 7 [3] H. Bouakkaz and P. Maroni, Description des polynômes orthogonaux de Laguerre-Hahn de classe zéro, in Orthogonal Polynomials and Their Applications, C. Brezinski et al. Editors. Annals Comput. Appl. Math. 9. Baltzer, Basel (1991), [4] B. Bouras and F. Marcellán, Quadratic decomposition of a Laguerre-Hahn polynomial sequence I. Bull. Belg. Math. Soc. Simon Stevin (010). In press. [5] T. S. Chihara, An Introduction to Orthogonal Polynomials. Gordon and Breach, New York, [6] J. Dini, Sur les formes linéaires et les polynômes orthogonaux de Laguerre-Hahn. Thèse de Troisième Cycle. Université Pierre et Marie Curie, Paris, [7] J. Dini and P. Maroni, Sur la multiplication d une forme linéaire par une fraction rationnelle. Applications aux formes de Laguerre-Hahn. Ann. Polon. Math. 5 (1990), [8] F. Marcellán and J. Petronilho, Eigenproblem for tridiagonal -Toeplitz matrices and quadratic polynomial mappings. Linear Algebra Appl. 60 (1997), [9] F. Marcellán and E. Prianes, Orthogonal polynomials and Stieltjes functions: The Laguerre-Hahn case. Rend. Mat. Appl. (7) 16 (1996), no. 1, [10] F. Marcellán and E. Prianes, Perturbations of Laguerre-Hahn linear functionals. J. Comput. Appl. Math. 105 (1999), [11] P. Maroni, Sur la décomposition quadratique d une suite de polynômes orthogonaux. I. Riv. di Mat. Pura ed Appl. 6 (1990), [1] P. Maroni, Une théorie algébrique des polynômes orthogonaux. Application aux polynômes orthogonaux semi-classiques, in Orthogonal Polynomials and Their Applications, C. Brezinski et al. Editors. Annals Comput. Appl. Math. 9. Baltzer, Basel (1991), [13] M. Zaatra, Polynômes orthogonaux de Laguerre-Hahn, Mémoire de Master de Mathématiques, Faculté des Sciences de Gabès, Tunisie, 007. In French. Belgacem Bouras Department of Mathematics, College of Sciences, Quassim University, Kingdom of Saudi Arabia. belgacem.bouras@issatgb.rnu.tn Francisco Marcellán Departamento de Matemáticas, Universidad Carlos III de Madrid, Spain. pacomarc@ing.uc3m.es