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Determinant Inequalities for Sieved Ultraspherical Polynomials

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Determinant Inequalities for Sieved Ultraspherical Polynomials IJMMS 25:11 (2001) 745–751 PII. S0161171201004896 http://ijmms.hindawi.com © Hindawi Publishing Corp. DETERMINANT INEQUALITIES FOR SIEVED ULTRASPHERICAL POLYNOMIALS J. BUSTOZ and I. S. PYUNG (Received 28 March 2000) Abstract. Paul Turan first observed that the Legendre polynomials satisfy the inequality 2 Pn(x) − Pn−1(x)Pn(x) > 0, −1 <x<1. Inequalities of this type have since been proved for both classical and nonclassical orthogonal polynomials. In this paper, we prove such an inequality for sieved orthogonal polynomials of the second kind. 2000 Mathematics Subject Classification. Primary 33C47; Secondary 33E30. 1. Introduction. It was observed by Paul Turán [9] that the Legendre polynomials satisfy the determinant inequality 2 Pn(x)−Pn+1(x)Pn−1(x) > 0, −1 <x<1,n= 0,1,.... (1.1) G. Szegö [8] gave two very beautiful proofs of Turán’s inequality. In the years since Szegö’s paper appeared, it has been proved by various authors [5, 6, 7] that inequality (1.1) is satisfied by the classical orthogonal polynomials. In general, let {Pn(x)} be a sequence of polynomials orthogonal in an interval [a,b]. Then the polynomials must satisfy a recursion Pn+1(x) = Anx +Bn Pn(x)−CnPn−1(x), n = 0,1,..., (1.2) where we define P−1(x) = 0. We begin with a very simple result shows that, inequalities of Turán type are satis- fied by any sequence of orthogonal polynomials. Theorem 1.1. If the polynomials {Pn(x)} are orthogonal on a ≤ x ≤ b, then for each n there exists cn,a≤ cn ≤ b, such that P 2(x) P (x) P (x) n n+1 n−1 2 − ≥ 0,a≤ x ≤ b. (1.3) Pn cn Pn+1 cn Pn−1 cn Proof. Consider the quotient Pn+1(x)Pn−1(x) fn(x) = 2 . (1.4) Pn(x) Obviously, the roots of Pn(x) are singularities and apart from these points fn(x) is continuous. Also, for each root xk,n of Pn(x), k = 1,2,...,n, there is an open interval In centered at xkn in which Pn+1(x)Pn−1(x) < 0. This follows from the recursion (1.2). 746 J. BUSTOZ AND I. S. PYUNG Consequently, fn(x) is bounded above and must take on a positive maximum value at a point cn ∈ [a,b]. Thus P c P c P (x)P (x) n+1 n n−1 n n+1 n−1 2 ≥ 2 ,a≤ x ≤ b. (1.5) Pn cn Pn(x) This proves the theorem. Although Theorem 1.1 asserts that Turán’s inequality is a simple consequence of orthogonality, it is generally quite difficult to determine the point cn referred to in Theorem 1.1. In the case of the classical orthogonal polynomials of Jacobi and their special case, the Gegenbauer polynomials, the point cn = 1 is an endpoint of the in- terval of orthogonality [−1,1]. Turán’s inequality for these classical polynomials is established by using differential identities that are characteristic of classical polyno- mials. This seems to be the only case that lends itself to that technique. Obviously, if Pn+1(x)Pn−1(x) δn > max 2 , (1.6) a≤x≤b Pn(x) then 2 δnPn(x)−Pn+1(x)Pn−1(x) > 0,a≤ x ≤ b. (1.7) Inequalities of the form (1.7) will be called weak Turán inequalities to distinguish them from (1.3) which will simply be referred to as Turán inequalities. 2. A weak Turán inequality for sieved ultraspherical polynomials of the second kind. The sieved ultraspherical polynomials were discovered by Al-Salam et al. [1]. Ismail [2, 4] investigated them at great length. If k ≥ 2 is an integer, then the sieved λ ultraspherical polynomials of the second kind, Bn(x;k), satisfy λ λ λ Bn+1(x;k) = 2xBn(x;k)−Bn−1(x;k), n+1 = mk, (2.1) λ λ λ mBmk(x;k) = 2x(m+λ)Bmk−1(x;k)−(m+2λ)Bmk−2(x;k), m > 0, λ λ where B0 (x;k) = 1, B1 (x;k) = 2x, k ≥ 2. Charris and Ismail [4] proved the following remarkable formula that is critical in deriving a weak Turán inequality for these polynomials, λ λ+1 λ+1 Bmk+j (x;k) = Uj (x)Cm Tk(x) +Uk−j−2(x)Cm−1 Tk(x) , (2.2) j = 0,1,...,k−1; m = 0,1,2,.... In (2.2), Uj (x) and Tk(x) are the Chebychev polynomials of the second kind and the λ first kind, respectively, and Cm(x) are the ultraspherical polynomials. We recall that if x = cosθ, then sin(j +1)θ T (x) = coskθ, U (x) = . (2.3) k j sinθ We note that U−1(x) = 0 and U−j (x) =−(sin(j − 1)θ/sinθ) =−Uj−2(x), j = 1. First, we remark that the sieved ultraspherical polynomials do not in general satisfy ∆n(x) = 2 (Bn(x;k)) − Bn+1(x;k)Bn−1(x;k) > 0. The inequality fails when n = mk + k − 1 DETERMINANT INEQUALITIES FOR SIEVED ULTRASPHERICAL POLYNOMIALS 747 because a lengthy calculation, which we omit, shows that (2λ+2) (2λ+1) (2λ−2k+1) ∆ (1) = m−1 m+1 . (2.4) mk+k−1 (m+1)!m! This last quantity may become negative depending on k and λ. Note that (a)n is defined by (a)n = a(a+1)···(a+n−1). λ 2 λ λ Theorem 2.1. Let ∆n =δn[Bn(x;k)] −Bn+1(x;k)Bn−1(x;k), where k≥2 and δn =1 if n = mk+j, j = 0,1,...,k−2, m = 0,1,..., and δmk+k−1 = (m+λ+1)/(m+1). Then ∆n > 0 for −1 ≤ x ≤ 1, n = 0,1,.... Proof. First, we prove that ∆n > 0 for n = mk+j, j = 0,1,...,k−2. Using elemen- tary trigonometricidentities, we canshow that 2 Uj −Uj−1Uj+1 = 1, 2Uj Uk−j−2 −Uj+1Uk−j−1 −Uj−1Uk−j−3 =−2Tk. (2.5) Using (2.2), we have λ+1 λ+1 2 λ+1 λ+1 ∆mk+j (x) = Uj Cm Tk +Uk−j−2Cm−1 Tk − Uj+1Cm Tk +Uk−j−3Cm−1 Tk λ+1 λ+1 · Uj−1Cm Tk +Uk−j−1Cm−1 Tk 2 λ+1 2 2 λ+1 2 = Uj −Uj−1Uj+1 Cm Tk + Uk−j−2−Uk−j−3Uk−j−1 Cm−1 Tk λ+1 λ+1 + 2Uj Uk−j−2 −Uj+1Uk−j−1 −Uj−1Uk−j−3 Cm Tk Cm−1 Tk . (2.6) From (2.5), the above equation yields λ+1 2 λ+1 λ+1 λ+1 2 ∆mk+j (x) = Cm Tk −2TkCm Tk Cm−1 Tk + Cm−1 Tk λ+1 λ+1 2 2 λ+1 2 = Cm Tk −TkCm−1 Tk + 1−Tk Cm−1 Tk (2.7) λ+1 λ+1 2 2 λ+1 2 = Cm Tk −TkCm−1 Tk +sin kθ Cm−1 Tk > 0. For the case n = mk+k−1, we need the recursion for ultraspherical polynomials λ λ λ (n+1)Cn+1(x) = 2(n+λ)xCn(x)−(n+2λ−1)Cn−1(x), n ≥ 1, (2.8) λ λ where C0 = 1, C1 = 2λx. Using elementary trigonometricidentities, we obtain Un = Un−2 +2Tn, (2.9) 2 2 Un−1 − Un−2 −2TnUn−2 = 1. (2.10) Using (2.2), we get the following formula when j = k−1, writing θ = (m+λ+1)/(m+1) for simplicity, λ 2 λ λ ∆mk+k−1(x) = θ Bmk+k−1 −Bmk+k−2Bmk+k λ+1 2 λ+1 λ+1 λ+1 λ+1 =θ Uk−1Cm Tk − Uk−2Cm Tk +Cm−1 Tk Cm+1 Tk +Uk−2Cm Tk 2 λ+1 2 λ+1 λ+1 2 λ+1 = θ Uk−1 Cm Tk −Uk−2Cm Tk Cm+1 Tk − Uk−2 Cm Tk λ+1 λ+1 λ+1 λ+1 −Cm−1 Tk Cm+1 Tk −Uk−2Cm−1 Tk Cm Tk . (2.11) 748 J. BUSTOZ AND I. S. PYUNG From (2.8), the above equation yields 2 2 λ+1 2 λ+1 ∆mk+k−1(x) = θ Uk−1 − Uk−2 Cm Tk −Uk−2Cm Tk 2(m+λ+1) m+2λ+1 · T Cλ+1 T − Cλ+1 T m+1 k m k m+1 m−1 k 2(m+λ+1) m+2λ+1 −Cλ+1 T · T Cλ+1 T − Cλ+1 T m−1 k m+1 k m k m+1 m−1 k λ+1 λ+1 −Uk−2Cm−1 Tk Cm Tk (2.12) 2(m+λ+1) = θ U 2 − U 2 − T U Cλ+1 T 2 k−1 k−2 m+1 k k−2 m k 2λ 2(m+λ+1) + U − T Cλ+1 T Cλ+1 T m+1 k−2 m+1 k m k m−1 k m+2λ+1 + Cλ+1 T 2. m+1 m−1 k Multiplying by (m+1) on both sides and using (2.10), we now have (m+1)∆mk+k−1(x) 2 λ+1 2 = (m+λ+1)+λUk−2 Cm Tk +2 λUk−2 −(m+λ+1)Tk λ+1 λ+1 λ+1 2 ·Cm Tk Cm−1 Tk +(m+2λ+1) Cm−1 Tk 2 2 λ+1 λUk−2−(m+λ+1)Tk λ+1 = (m+λ+1)+λUk−2 Cm Tk + 2 Cm−1 Tk (m+λ+1)+λUk−2 2 2 λ+1 2 + (m+λ+1)+λUk−2 (m+2λ+1)− λUk−2 −(m+λ+1)Tk Cm−1 Tk .
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