A Class of Non-Symmetric Semi-Classical Linear Forms. Zaatra Mohamed [a],*

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1 Progress in Applied Mathematics Vol. 7, No. 1, 014, pp. [48 64] DOI: /4630 ISSN X [Print] ISSN [Online] A Class of Non-Symmetric Semi-Classical Linear Forms Zaatra Mohamed [a],* [a] Algèbre de Lie et Fonction speciale University/Institution, Sfax. * Corresponding author. Address: Institut Supèrieur de Gestion de Gabès Av. Habib Jilani - Gabès - 600, Tunisia; medzaatra@yahoo.fr Received: September 14, 013 / Accepted: December 15, 013 / Published online: January 4, 014 Abstract: We show that if v is a regular semi-classical form linear functional, then the form u defined by x τ σu = λv and σx τu = 0 where σu is the even part of u, is also regular and semi-classical form for every complex λ except for a discrete set of numbers depending on v. We give explicitly the recurrence coefficients and the structure relation coefficients of the orthogonal polynomials sequence associated with u and the class of the form u knowing that of v. We conclude with illustrative examples. Key words: Orthogonal polynomials; Semi-classical linear forms; Integral representation; Structure relation Mohamed, Z A Class of Non-Symmetric Semi-Classical Linear Forms. Progress in Applied Mathematics, 7 1, Available from: URL: index.php/pam/article/. DOI: 1. INTRODUCTION Semi-classical orthogonal polynomials O.P were introduced in [14]. They are naturel generalization of the classical polynomials Hermite, Laguerre, Jacobi and Bessel. Maroni [8, 10] has worked on the linear form of moments and has given a unified theory of this kind of polynomials. The form u is called semi-classical form if its formal Stieltjes function Suz satisfies the Riccati differential equation ΦzS uz = C 0 zsuz + D 0 z, where Φ 0, C 0 and D 0 are polynomials. In [5, 7], the authors determine all the semi-classical monic orthogonal polynomials sequence MOPS of class one satisfying a three terms recurrence relation with β n = 1 n τ, n 0, τ C {0}. See also [1] for a special case. 48

2 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 The whole idea of the following work is to build a new construction process of semi-classical form, which has not yet been treated in the literature on semi-classical polynomials. The problem we tackle is as follows. We study the form u, fulfilling x τ σu = λv, λ 0, u n+1 = τu n, where σu is the even part of u, τ C and v is a given semi-classical form. This paper is arranged in sections: The first provides a focus on the preliminary results and notations used in the sequel. We will also give the regularity condition and the coefficients of the three-term recurrence relation satisfied by the new family of O.P. In the second, we compute the exact class of the semi-classical form obtained by the above modification and the structure relation of the O.P. sequence relatively to the form u will follow. In the final section, we apply our results to some examples. The regular linear functional found in the examples are semi-classical linear functional of class s {1, } and we present their integral representations. Let P be the vector space of polynomials with coefficients in C and let P be its dual. We denote by v, f the action of v P on f P. In particular, we denote by v n := v, x n, n 0, the moments of v. For any form v and any polynomial h let Dv = v, hv, δ c, and x c 1 v be the forms defined by: v, f := v, f, hv, f := v, hf, δ c, f := fc, and x c 1 v, f := v, θ c f, where θ c f x = fx fc, c C, f P. x c Then, it is straightforward to prove that for f P and v P, we have x c 1 x cv = v v 0 δ c, 1 x cx c 1 u = v. Let us define the operator σ : P P by σfx := fx. Then, we define the even part σv of v by σv, f := v, σf. Therefore, we have [6, 9] fxσv = σfx v, 3 σv n = v n, n 0. 4 The form v will be called regular if we can associate with it a polynomial sequence {S n } n 0 degsn n such that v, S n S m = r n δ n,m, n, m 0, r n 0, n 0. Then degs n = n, n 0, and we can always suppose each S n monic i.e. S n x = x n +. The sequence {S n } n 0 is said to be orthogonal with respect to v. It is a very well known fact that the sequence {S n } n 0 satisfies the recurrence relation see, for instance, the monograph by Chihara [6] S n+ x = x ξ n+1 S n+1 x ρ n+1 S n x, n 0, S 1 x = x ξ 0, S 0 x = 1, 49 5

3 A Class of Non-Symmetric Semi-Classical Linear Forms with ξ n, ρ n+1 C C {0}, n 0, by convention we set ρ0 = v 0 = 1. In this case, let {S n 1 } n 0 be the associated sequence of first kind for the sequence {S n } n 0 satisfying the three-term recurrence relation [6] S 1 n+ x = x ξ n+s 1 n+1 x ρ n+s 1 n x, n 0, S 1 1 x = x ξ 1, S 1 0 x = 1, S 1 1 x = 0. 6 Another important representation of S 1 n is, see [6] S n 1 x := v, S n+1x S n+1 ζ x ζ, n 0. 7 Also, let {S n., µ} n 0 be co-recursive polynomials for the sequence {S n } n 0 satisfying [6] S n x, µ = S n x µs 1 n 1, n 0. 8 We recall that a form v is called symmetric if v n+1 = 0, n 0. The conditions v n+1 = 0, n 0 are equivalent to the fact the corresponding MOPS {S n } n 0 satisfies the recurrence relation 5 with ξ n = 0, n 0 [6]. Now let v be a regular, normalized form i.e. v 0 = 1 and {S n } n 0 be its corresponding sequence of polynomials. For a τ C and λ C, we can define a new form u as following: Equivalently, u n+ τ u n = λv n, u n+1 = τu n, u 0 = 1, n 0. 9 From 1 and 10, we have Remarks. i 10 is equivalent to where the form w defined by x τ σu = λv, σ x τu = σu = λx τ 1 v + δ τ. 11 x τ u = λw, 1 σw = v, σx τw = 0. Notice that w is not necessarily a regular form in the problem understudy. In [], the authors have solved where w is regular and τ = 0 and in [3], the problem 1 is solved when τ 0 and w is regular. ii The case τ = 0 is treated in [13], so henceforth we assume τ 0. Proposition 1. The form u is regular if and only if λ λ n, n 0 where λ 0 = 0, λ n+1 = S n+1τ, n S n 1 τ 50

4 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 To prove the above proposition, we need the following lemma: Lemma. [9] The form u defined by 10 is regular if and only if σu and x τ σu are regular. Proof of Proposition 1. We have u is defined by 10. Then, according to Lemma, u is regular if and only if x τ σu and σu are regular. But x τ σu = λv is regular since λ 0 and v is regular. So u is regular if and only if σu = λx τ 1 σv + δ τ is regular. Or, {S n } n 0 is the corresponding orthogonal sequence to v, and it was shown in [11] that σu = λx τ 1 v + δ τ is regular if and only if λ 0, and S n τ, λ 0, n 0. Then we deduce the desired result. When u is regular let {Z n } n 0 be its corresponding sequence of polynomials satisfying the recurrence relation Z n+ x = x 1 n+1 τz n+1 x γ n+1 Z n x, n 0, Z 1 x = x τ, Z 0 x = 1. Let us consider its quadratic decomposition [6, 9]: 14 Z n x = P n x, Z n+1 x = x τr n x. 15 The sequences {P n } n 0 and {R n } n 0 are respectively orthogonal with respect to σu and x τ σu. From 10, we have R n x = S n x, n Proposition 3. We may write where γ 1 = λ, γ n+ = a n, γ n+3 = ρ n+1 a n, n 0, 17 a n = S n+1τ, λ S n τ,, λ n For the proof, we need the following lemma: Lemma 4. [4] We have Z 1 n x = R nx, λ, Z 1 1 n+1 x = x + τp n x, n Proof of Proposition 3. Using 10 and the condition u, Z = 0, we obtain γ 1 = λ. From 6 and 14 where n n and taking 16 and 19 into account, we get S n+1 x, γ 1 = x τz 1 n+1 x γ n+s n x, γ 1. Substituting x by τ in the above equation, we obtain γ n+ = a n. From 14, we have u, Z n+ u, Z n+3 u, Z n+3 γ n+ γ n+3 = =. 0 u, Z n+1 u, Z n+ u, Z n+1 Using 5, 10 and 15 16, Equation 0 becomes then we deduce γ n+3 = ρ n+1 a n. γ n+ γ n+3 = ρ n+1, 1 51

5 A Class of Non-Symmetric Semi-Classical Linear Forms We suppose that the form v has the following integral representation: v, f = 0 V xfxdx, f P, with v 0 = v, f = 0 V xdx, where V is a locally integrable function with rapid decay and continuous at the point x = τ. It is obvious that Therefore, fx = f e x + xf o x, f P. u, fx = u, f e x + τf o x = σu, f e x + τf o x, since u satisfies 10. Using 11 and taking into account that f e τ + τf o τ = fτ, we obtain { V x } u, f = fτ 1+λP x τ χ [0,+ [xdx λp V x x τ χ [0,+ [xf e + τf o xdx, where P V x { τ ε fx = lim x τ ɛ 0 V x + x τ fxdx + τ +ɛ V x } x τ fxdx and χ [a,b] denotes the characteristic function of the interval [a, b], i.e. χ [a,b] x = 1 when x [a, b] and zero otherwise. Using the fact that f e x = f x+f x and f o x = f x f x for x > 0 x and making the change of variables t = x, we get P V x + x τ χ [0,+ [xf e + τf o xdx =P +P V t t τ χ [0,+ [tftdt V t t + τ χ [0,+ [tf tdt. Inserting the last equation into, we get after a change variables in the obtained equation { V t } u, f = fτ 1+λP t τ χ [0,+ [tdt +λp λp 5 V t t τ χ ],0]tftdt V t t τ χ [0,+ [tftdt. 3

6 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014. THE SEMI-CLASSICAL CASE Let us recall that a form v is called semi-classical when it is regular and satisfies a linear non-homogeneous differential equation [10] ΦzS v z = C 0 zs v z + D 0 z, 4 where Φ monic, C 0 and D 0 are polynomials with Svz = n 0 v n, 5 zn+1 and D 0 x = vθ 0 Φ x + vθ0 C 0 x. 6 It was shown in [10] that equation 4 is equivalent to Φxv + Ψxv = 0 7 with Ψx = Φ x C 0 x. 8 The triple Φ, C 0, D 0 of the differential equation is not unique, then 4 can simplified if and only if there exists a root c of Φ such that C 0 c = 0 and D 0 c = 0. Then v fulfils the differential equation θ c ΦzS v z = θ c C 0 zs v z + θ c D 0 z. We call the class of the linear form v, the minimum value of the integer max degφ, degc 0 1 for all triples satisfying 4. The class of the semi-classical form v is s = max degφ, degc 0 1 if and only if the following condition is satisfied [8] C 0 c + D 0 c 0, 9 c Z where Z denotes the set of zeros of Φ. The corresponding orthogonal sequence {S n } n 0 is also called semi-classical of class s. The semi-classical character is invariant by shifting. Indeed, the shifted form ˆv = h a 1ot b v, a C {0}, b C satisfies ˆΦzS ˆv z = Ĉ0zS ˆv z + ˆD 0 z, 30 with ˆΦz = a k Φaz + b, ˆD 0 z = a k D 0 az + b, Ĉ 0 z = a 1 k C 0 az + b, k = degφ. The forms t b v translation of v and h a v dilation of v are defined by t b v, f := v, fx + b, h a v, f := v, fax, f P. 53

7 A Class of Non-Symmetric Semi-Classical Linear Forms The sequence {Ŝnx = a n S n ax + b} n 0 is orthogonal with respect to ˆv and fulfils 5 with ˆξ n = ξ n b a, ˆρ n+1 = ρ n+1, n a In the sequel the form v will be supposed semi-classical linear form of class s satisfying 4 and 7 and using a dilation in the variable τ, we can take him equal to one. Proposition 5. For every λ C {0} such that S n 1, λ 0, n 0, the form u defined by 10 is regular and semi-classical. It satisfies ΦzS u z = C 0 zs u z + D 0 z, 3 where Φz = z 1Φz, C 0 z = zz 1C 0 z Φz, D 0 z = z λd 0 z C 0 z, 33 and u is of class s such that s s + 3. Proof. From 10 and 5, we have S v z = λ 1 z 1S u z λ Make a change of variable z z in 4, multiply by λz and substitute 34 in the obtained equation, we get Then, deg Φ s + 5 and deg C 0 s + 4. Thus, s = maxdeg Φ, deg C 0 1 s + 3. As an immediate consequence of 3 33, the form u satisfies the functional equation Φu + Ψu = 0, 35 where Φ is the polynomial defined by 33 and Ψx = Φ x C 0 x = xx 1Ψx. 36 Proposition 6. The class of u depends only on the zeros x = 0 and x = 1 of Φ. Proof. Since v is a semi-classical form of class s, Svz satisfies 4, where the polynomials Φ, C 0 and D 0 are coprime. Let Φ, C0 and D 0 be as in Proposition 5. Let d be a zero of Φ different from 0 and 1, this implies that Φd = 0. We know that C 0 d + D 0 d 0 i if C 0 d 0, then C 0 d 0, ii if C 0 d = 0, then D 0 d 0, whence C 0 d + D 0 d 0. Concerning the class of u, we have the following result see Proposition 8. But first, let us this technical lemma. 54

8 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 Lemma 7. Let Xz = C 0 z λd 0 z and Y z = C 0 z Φ z, where the polynomials Φ, C 0 and D 0 are defined in 4. We have the following properties. R 1. The equation 3 33 is irreducible in 0 if and only if Φ0 0. R. The equation 3 33 is divisible by z but not by z if and only if Φ0 = 0. R 3. The equation 3 33 is irreducible in 1 if and only if Φ1, X1 0, 0. R 4. The equation 3 33 is divisible by z 1 and not by z 1 if and only if Φ1, X1 = 0, 0 and X 1, Y 1 0, 0. R 5. The equation 3 33 is divisible by z 1 and not by z 1 3 if and only if Φ1, X1 = X 1, Y 1 = 0, 0. Proof. From 33, we have Φ0 = Φ0. So by virtue of 9, we get R 1. Now, if Φ0 = 0, the equation 3 33 is divisible by z according to 9. Thus Suz satisfies 3 with Φz = zz 1θ 0 Φz, C 0 z = z 1C 0 z zθ 0 Φz, 37 D 0 z = C 0 z λd 0 z. Then, C0 0 = C 0 0. If C 0 0 = 0, thus the equation 3 37 is irreducible in 0. If C 0 0 = 0, so from 37, we obtain D 0 0 = λd since v is semi-classical form of class s and so satisfies 9. Therefore, we deduce R. From 33, we get C 0 1 = Φ1 and D 0 1 = X1. We can deduce that C D if and only if Φ1, X1 0, 0. Thus R 3 is proved. If Φ1, X1 = 0, 0, then the equation 3 33 can be divided by z 1 according to 9. In this case, Suz satisfies 3 with Φz = Φz, C 0 z = zc 0 z z + 1θ 1 Φz, D 0 z = C 0 z λd 0 z + z + 1 θ 1 C 0 λd 0 38 z. Substituting z by 1 in 38, we obtain C 0 1 = Y 1 and D 0 1 = X 1. Then 3 38 is irreducible in 1 if and only if X 1, Y 1 0, 0. Hence R 4. If X 1, Y 1 0, 0, then the equation 3 38 can be divided by z 1 according to 9. Therefore Suz satisfies 3 with Φz = z + 1θ 1 Φz, C 0 z = C 0 z + z + 1 θ 1 C0 θ 1 Φ z θ 1 Φz, D 0 z = z + θ 1 λd 0 C 0 z 4 θ1λd 0 C 0 39 z. From the above equation, we have Φ1 = Φ 1 = 0. If Φ 1 = 0, then from the condition Y 1 = 0 we obtain C 0 1 = 0. Thus from the last result and the condition X1 = 0, we get D 0 1 = 0. Impossible, since v is semi-classical form of class s and so satisfies 9. Thus R 5 is proved. 55

9 A Class of Non-Symmetric Semi-Classical Linear Forms Proposition 8. Under the conditions of Proposition 5, for the class of u, we have the two different cases: 1 Φ0 0. i s = s + 3 if Φ1, X1 0, 0. ii s = s + if Φ1, X1 = 0, 0 and Y 1, X 1 0, 0. iii s = s + 1 if Φ1, X1 = Y 1, X 1 = 0, 0. Φ0 = 0. i s = s + if Φ1, X1 0, 0. ii s = s + 1 if Φ1, X1 = 0, 0 and Y 1, X 1 0, 0. iii s = s if Φ1, X1 = Y 1, X 1 = 0, 0. Proof. From Proposition 6, the class of u depends only on the zeros 0 and 1. For the zero 0 we consider the following situations: A Φ0 0. In this case the equation 3 33 is irreducible in 0 according to R 1. But what about the zero 1? We will analyze the following cases: i Φ1, X1 0, 0, the equation 3 33 is irreducible in 1 according to R 3. Then 3 33 is irreducible and s = s + 3. Thus we proved 1 i. ii Φ1, X1 = 0, 0 and Y 1, X 1 0, 0. From R 4., 3 33 is divisible by z 1 but not by z 1 and thus the order of the class of u decreases in one unit. In fact, Suz satisfies the irreducible equation 3 38 and then s = s + and 1 ii is also proved. iii Φ1, X1 = Y 1, X 1 = 0, 0. From R 5., 3 33 is divisible by z 1 but not by z 1 3 and thus the order of the class of u decreases in two units. In fact, Suz satisfies the irreducible equation 3 39 and then s = s + 1. Thus 1 iii is proved. B Φ0 0. In this condition, 3 33 is divisible by z but not by z according to R. But what about the zero 1? We have the three following cases: i Φ1, X1 0, 0, the equation 3 33 is irreducible in 1 according to R 3. Then Suz satisfies the irreducible equation 3 37 and then s = s +. Thus we proved i. ii Φ1, X1 = 0, 0 and Y 1, X 1 0, 0. From R 4., 3 33 is divisible by z 1 but not by z 1 and thus the order of the class of u decreases in one unit. In fact, Suz satisfies the irreducible Equation 3 with Φz = zθ 0 Φz, C 0 z = C 0 z θ 0 Φz z + 1θ 0 θ 1 Φz, D 0 z = z + 1 θ 1 λd 0 C 0 40 z. 56

10 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 Thus s = s + 1 and ii is proved. iii Φ1, X1 = Y 1, X 1 = 0, 0. From R 5., 3 33 is divisible by z 1 but not by z 1 3. So, Suz satisfies the irreducible Equation 3 with Φz = z + 1θ 0 θ 1 Φz + θ 0 Φz, C 0 z = z + 1 θ 1 C0 θ 0 θ 1 Φ z z + θ 0 θ 1 Φz, D 0 z = θ 1 λd 0 C 0 z 4z + 1 θ1λd 0 C 0 z. Therefore s = s and iii is also proved. Note that the sequence of orthogonal polynomials OPS relatively to a semiclassical form has a structure relation written in a compact form[10]. Then, if we consider that the form v is semi-classical, its OPS {S n } n 0 fulfils the following structure relation: 41 ΦxS n+1x = 1 Cn+1 x C 0 x S n+1 x ρ n+1 D n+1 xs n x, n 0, 4 with C n+1 x = C n x + x ξ n D n x, n 0, ρ n+1 D n+1 x = Φx + ρ n D n 1 x x ξ n C n x + x ξ n D n x, n 0, where Φ, C 0 x and D 0 x are the same polynomials as in 4; ξ n, ρ n are the coefficients of the three term recurrence relation 5. Notice that D 1 x = 0, deg C n s + 1 and deg D n s, n 0 [10]. According to Proposition 5, the form u is also semi-classical and its OPS {Z n } n 0 satisfies a structure relation. In general, {Z n } n 0 fulfils 43 ΦxZ n+1x = 1 Cn+1 x C 0 x Z n+1 x γ n+1 Dn+1 xz n x, n 0, 44 with C n+1 x = C n x + x 1 n D n x, n 0, γ n+1 Dn+1 x = Φx + γ n Dn 1 x x 1 n C n x + x 1 n Dn x, n 0, 45 where Φ, C0 x and D 0 x are the same polynomials as in Equation 3. We are going to establish the expression of C n and D n, n 0 in terms of those of the sequence {S n } n 0. Proposition 9. The sequence {Z n } n 0 fulfils 44 with for n 0 { Cn+1 x = Φx + xx 1C n x + 4γ n+1 xx 1D n x, D n+1 x = xx 1 D n x. 46 { Cn+ x = Φx + xx 1C n+1 x + 4xx 1γ n+ D n x, D n+ x = xγ n+ D n x + xγ n+3 D n+1 x + xc n+1 x

11 A Class of Non-Symmetric Semi-Classical Linear Forms C 0 x and D 0 x are given by 33 and γ n+1 by 17. Proof. Change x x, n n 1 in 4 and multiply by xx 1 we obtain by taking into account, { } x 1ΦxZ n+3x = xx 1 C n+1 x C 0 x + Φx Z n+3 x ρ n+1 xx 1D n+1 x Z n+1 x, n 0. Using 14 and 1 where n n, the last equation becomes { ΦxZ n+3x = xx 1 C n+1 x C 0 x + xx 1γ n+3 D n+1 x } + Φx Z n+3 x γ n+3 xx 1 D n+1 x Z n+ x, n 0. From 44 and the above equation, we have for n 0 { Cn+3 x C } 0 x { } X n+1 x Z n+3 x = γ n+3 Dn+3 Y n+1 x Z n+ x, with for n 0 { Xn x = C n x C 0 x + γ n+1 D n x xx 1 + Φx, Y n x = xx 1 D n x. Z n+3 and Z n+ have no common zeros, then Z n+3 divides Y n+1 x D n+3 x, which is a polynomial of degree at most equal to s + 3. Then we have necessarily Y n+1 x D n+3 x = 0 for n > s, and also Therefore, X n x = C n+1 x C 0 x, n > s. C n+3 x = X n+1 x + C 0 x and Dn+3 = Y n+1 x, n > s. Then, by 33, we get 46 for n > s. By virtue of the recurrence relation 43 and 33, we can easily prove by induction that the system 46 is valid for 0 n s. Hence 46 is valid for n 0. After a derivation of 14 where n n + 1 multiplying by x 1Φx and using 44, we obtain x 1 Φx Z n+x = C n+3x C 0x Z n+3x γ n+3 Dn+3xZ n+x x 1Φx Z n+x { Cn+1x + γ C } 0x n+ Z n+1x γ n+1 Dn+1xZ nx. 58

12 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 Applying the recurrence relation 14, we get { x 1 Φx Z n+x = x 1 C n+3x C 0x + γ n+ Dn+1x } γ n+3 Dn+3x x 1Φx Z n+x γ n+ { Cn+3x C n+1x + x + 1 D } n+1x Z n+1x. Now, using 44 and taking into account the fact that Z n+ x and Z n+1 x are coprime, we get from the last equation after simplification by x 1 47 for n > s. Finally, by virtu of the recurrence relation 43 and 46 with n = 0, we can easily prove by induction that the system 47 is valid for 0 n s. 3. ILLUSTRATIVE EXAMPLES 1 We study the problem 10, with v := Lα where Lα is the Laguerre form. This form has the following integral representation [10] v, f = 1 Γα x α e x fxdx, Rα > 1, f P. 48 Thus, using 3, we obtain the following integral representation of u { u, f = f1 1 + λp 0 + λ x α e x fxdx λp x 1 x α e x } x 1 χ [0,+ [xdx x α e x x 1 χ [0,+ [xfxdx. The form v is classical semi-classical of class s = 0, it satisfies 4 and 7 with [10] { Φx = x, Ψx = x α 1, 49 C n x = x + n + α, D n x = 1, n 0. The sequence {S n } n 0 fulfils 5 with [6] 50 ξ n = n + α + 1, ρ n+1 = n + 1n + α + 1, n The regularity condition is α n, n 1. First, we study the regularity of the form u. From 7 and.11 in [6], we have for n 0 and where S n 1 = 1 n n k=0 1 k Γn + 1Γn + α + 1 Γk + 1Γn k + 1Γα + k + 1, 5 n+1 S n 1 1 = 1 n+1 1 k Γn + Γn + α + Γk + 1Γn k + 1Γα + k + 1 b k 1α, 53 b n α = k=0 n k=0 Γα + k + 1, b 1 α = 0. Γα

13 A Class of Non-Symmetric Semi-Classical Linear Forms By virtue of 8 and 5 53, we deduce where S n 1, λ = 1 n Γn + 1Γn + α + 1c n α, λ, n 0, 54 c n α, λ = n k=0 1 k 1 λb k 1 α Γk + 1Γn k + 1Γα + k + 1, n 0. Then, u is regular for every λ 0 such that c n α, λ 0, n and 54 give a n = n + 1n + α + 1 c n+1α, λ, n c n α, λ Therefore, with 17, we obtain for n 0 γ 1 = λ, γ n+ = n + 1n + α + 1 c n+1α, λ c n α, λ, γ n+3 = c nα, λ c n+1 α, λ. 56 Taking into account that the form v is semi-classical and by virtue of Proposition 5, Proposition 8 and 50, the form u is semi-classical of class s = and fulfils 3 and 35 with { Φx = xx 1, Ψx = x 1x α 1, C 0 x = x 3 + x + α 1x α, D0 x = x + α + λ. 57 Now, we are going the elements of the structure relation of the sequence {Z n } n 0. C 0 x = x 3 + x + α 1x α, C 1 x = x 1 x + α + λ + x, C n+ x = x 1 x + n + α + n + 1n + α + 1 c n+1α, λ x, c n α, λ C n+3 x = x 1 x + n + α + c nα, λ x, c n+1 α, λ D 0 x = x + α + λ, D n+1 x = x 1, D n+ = x + n + α + 1 n + 1n + α + 1 c n+1α, λ c n α, λ c nα, λ. c n+1 α, λ We study the problem 10, with v := h 1 oτ 1J α, β where J α, β is the Jacobi form. This form has the following integral representation [10] v, f = Γα + β + Γα + 1Γβ x α 1 x β fxdx, Rα, Rβ > 1, f P

14 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 Thus, using 3, we obtain the following integral representation of u Γα + β + u, f = λ Γα + 1Γβ sgn x x α 1 x β 1 x + 1fxdx + 1 λ α + β + 1 f1, Rα > 1, Rβ > 0, f P. β The form v is classical, it satisfies 4 and 7 with [10] Φx = xx 1, Ψx = α + β + x + α + 1, n + αα + β C n x = n + α + βx n n + α + β, D n x = n + α + β + 1, n The sequence {S n } n 0 fulfils 5 with [6] ξ 0 = α+1 α+β+, ξ n+1 = 1 β α n+α+β+n+α+β+4 + 1, n 0, ρ n+1 = n+1n+α+1n+β+1n+α+β+1 n+α+β+1n+α+β+ n+α+β+3, n The regularity conditions are α, β n, α + β n, n 1. Using 5 and 61, we get S n 1 = From 6 and 61, we obtain by induction Γβ + n + 1Γα + β + n + 1, n 0. 6 Γβ + 1Γα + β + n + 1 S n 1 α + β = Γα + β + n + 3 d nα, β, n 0, 63 where for n 0 1 Γα + n + Γα + β + n + Γα + β + 1Γn + 1Γβ + n +, α 0, α Γα + 1 Γβ + 1 d n α, β= n 1 Γn + 1Γn + β + k , α = 0. β + k + 1 k=0 By virtue of 8 and 6 63, we deduce where for n 0 S n 1, λ = Γβ + n + 1Γα + β + n + 1 e n λ, α, β, n Γβ + 1Γα + β + n + 1 α + β + 1Γβ + 1 e n λ, α, β = 1 λ Γβ + n + 1Γn + α + β + 1 d n 1α, β, d 1 α, β = 0. Then, u is regular for every λ 0 such that α + β + 1Γβ + 1 1, λ Γβ + n + 1Γn + α + β + 1 d n 1α, β n

15 A Class of Non-Symmetric Semi-Classical Linear Forms 18 and 64 give β + n + 1α + β + n + 1 a n = α + β + n + 1α + β + n + e n+1 λ, α, β, n e n λ, α, β Then,with 17, we obtain for n 0 γ 1 = λ, β + n + 1α + β + n + 1 e n+1 λ, α, β γ n+ = α + β + n + 1α + β + n + e n λ, α, β, n + 1α + n + 1 e n λ, α, β γ n+3 = α + β + n + α + β + n + 3 e n+1 λ, α, β. 67 Taking into account that the form v is classical and by virtue of Proposition 5, the form u is also semi-classical. It satisfies 3 and 35 with From 60, we have Φx = xx 1x 1, Ψx = x 1 α + β 3x + α + 1, C 0 x = x 1 α + β + 1x + x + α, D 0 x = α + βx α λα + β + 1. Φ0 = 0, Φ1 = 0, X1 = β λα + β + 1, X 1 = α + β, Y 1 = β 1. Now it is enough to use Proposition 8 in order to obtain the following results: i If λ satisfies 65 and λ ii If λ = Remarks. β α+β+1, then the class of u is s =. β α+β+1, then the class of u is s = 1 since X 1, Y 1 0, i The semi-classical orthogonal polynomials of class one satisfies 14 have been described in [5, 7]. ii If λ = β α+β+1, then from 59, we get for Rα > 1, Rβ > 0 u, f = Γα + β sgn x x α 1 x β 1 x + 1fxdx. 69 Γα + 1Γβ 1 This result exist in [7, 1]. 6

16 Mohamed, Z./Progress in Applied Mathematics, 7 1, 014 According to Proposition 9, 60 and 67, we have, for n 0 C 0 x = x 1 α + β + 1x + x + α, C 1 x = x 1 α + β + 1x x α 4λα + β + 1, C n+ x = x 1 4n + α + β + 3x + x n n + α + 1α + β 4 n + α + β + C n+3 x = x 1 4n + α + β + 5x x n n + α + 1α + β 4 n + α + β + D 0 x = α + βx α λα + β + 1, D n+1 x = x 1 α + β + n + 1, D n+ = n + α + β + x n 1 REFERENCES β + n + 1α + β + n + 1 α + β + n + n + 1α + n + 1 α + β + n + β + n + 1α + β + n + 1 e n+1 λ, α, β α + β + n + e n λ, α, β n + 1α + n + 1 e n λ, α, β α + β + n + e n+1 λ, α, β. e n λ, α, β e n+1 λ, α, β n + α + 1α + β n + α + β + e n+1 λ, α, β, e n λ, α, β, [1] Alaya, A., Bouras, B., & Marcellan, F. 01. A non-symmetric second degree semi-classical form of class one. Integral Transforms Spec. Funct., 3, [] Alaya, J., & Maroni, P Semi-classical and Laguerre-Hahn forms defined by pseudo-functions. Methods and Applications of Analysis, 3 1, [3] Branquinho, A., & Marcellan, F Generating new class of orthogonal polynomials. Int. J. Math. Sci., 19, [4] Bouras, B., & Marcellan, F Quadratic decomposition of a Laguerre- Hahn polynomial sequence I. Bull. Belg. Math. Soc. Simon Stevin, 17 4, [5] Bouras, B., & Alaya, A A large family of semi-classical polynomials of class one. Integral Transforms Spec. Funct., 18 1, [6] Chihara, T. S An introduction to orthogonal polynomials. New York: Gordon and Breach. [7] Maroni, P., & Mejri, M Some semiclassical orthogonal polynomials of classe one. Eurasian Mathematical Journal,, [8] Maroni, P Variations around classical orthogonal polynomials. Connected problems. In 7 th Symp. Sobre Polinomios orthogonales y Appliciones, Proc., Granada 1991, J. Comput. Appl. Math., 48,

17 A Class of Non-Symmetric Semi-Classical Linear Forms [9] Maroni, P Sur la décomposition quadratique d une suite de polynômes orthogonaux. I, Rivista di Mat. Pura ed Appl., 6, [10] Maroni, P Une théorie algébrique des polynômes orthogonaux. Application aux polynômes orthogonaux semi-classiques. In C. Brezinski et al. Eds., Orthogonal polynomials and their applications pp IMACS Ann. Comput. Appl. Math. 9. Baltzer, Basel. [11] Maroni, P Sur la suite de polynômes orthogonaux associée à la forme u = δ c + λ x c 1 L. Period. Math. Hung., 1 3, [1] Sghaier, M Some new results about a set of semi-classical polynomials of class s = 1. Integral Transforms Spec. Funct., 1 7, [13] Sghaier, M Some symmetric semi-classical polynomials sets. Filomat, 5 3, [14] Shohat, J A differential equation for orthogonal polynomials. Duke Math. J., 5,