Journal of Computational and Applied Mathematics 133 (2001) 489–493 www.elsevier.com/locate/cam
A new bound for the Laguerre polynomials M. Michalska, J. Szynal ∗ Maria Curie-Sklodowska University, Sq. M. Curie-Sklodowska, 20-031 Lublin, Poland Received 4 November 1999; received in revised form 28 June 2000
Abstract c 2001 Published by Elsevier A new uniform bound for the Laguerre polynomials L() n (x), ∈ R is determined. Science B.V. Keywords: Laguerre polynomials; Gegenbauer polynomials
1. Introduction The Laguerre polynomials Ln() (x), ∈ R, x¿0, and n = 0; 1; 2; : : : can be de5ned in many ways, for instance, by the generating function [11] (1 − z)
−−1
xz exp − 1−z
=
∞ n=0
Ln() (x)z n ;
|z| ¡ 1:
(1)
In this note we present a new uniform bound for the Laguerre polynomials which improve the bounds already known. We use the connection between Laguerre and Gegenbauer polynomials as well as some ideas and estimates from the paper by Loh;ofer [8]. There are well-known (see, e.g., [1]) two classical global uniform (w.r.t. n, x and ) estimates given by Szeg;o: ( + 1)n x=2 e ; ¿0; x¿0; n = 0; 1; 2; : : : ; n! ( + 1)n x=2 e ; −1 ¡ 60; x¿0; n = 0; 1; 2; : : : : |Ln() (x)|6 2 − n! |Ln() (x)|6
∗
Corresponding author. E-mail address:
[email protected] (J. Szynal). c 2001 Published by Elsevier Science B.V. 0377-0427/01/$ - see front matter PII: S 0 3 7 7 - 0 4 2 7 ( 0 0 ) 0 0 6 7 0 - 1
(S1) (S2)
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The estimate (S2) was improved by Rooney [10], who proved using the Askey formula that |Ln() (x)|62− qn ex=2 ; 6 − 12 ; and
x¿0;
√ qn ( + 1)n x=2 e ; ¿ − 12 ; |Ln() (x)|6 2 ( 12 )n
where
n = 0; 1; 2; : : :
(R1)
x¿0;
(R2)
n = 0; 1; 2; : : : ;
√
(2n)! 1 ; qn ∼ √ ; n → ∞: 4 2n+1=2 n! 4n However, by his method Rooney could not improve the estimate (S1). Using the Koornwinder formula (K) (Lemma 1 below), in [7] we extended and improved the estimate (S1) of Szeg;o as follows: qn =
Theorem A. For ¿ − 12 ; x¿0 and n = 0; 1; 2; : : : we have ( + 1)n () |Ln() (x)|6 (LS) n (exp x); n! where n() ( ∞ aro mean of the formal series ∞ k=0 ak ) denotes the Ces3 k=0 ak and is given by the formula
n()
∞ k=0
ak
n ( + 1)n−k n! = ak ; ( + 1)n k=0 (n − k)!
¿ − 1;
(2)
and ()n = ( + 1) : : : ( + n − 1) is the Pochhammer’s symbol. Using the result from [8], we improve our estimate (LS) below [7]. Our motivation came out from the Krzy˙z Conjecture (see, e.g., [6]). Moreover, there exists the substantial interest in upper bounds for other classes of orthogonal polynomials (see, e.g., [2– 4,8]). 2. Lemmas Lemma 1 (Koornwinder formula [5]). For any ¿ − 12 ; x¿0; and n = 0; 1; 2; : : : ; we have ∞ √ 2 2(−1)n (x − r 2 + i2 xr cos )n e−r r 2+1 sin2 d dr: Ln() (x) = √ 1 ( + 2 )n! 0 0
(K)
Application of the Laplace integral for the Gegenbauer polynomials Cn() (x) [9] gives Lemma 2 (Lewandowski [7]). For ¿ − 12 ; x¿0 and n = 0; 1; 2; : : : ; we have ∞ x − t −t (−1)n () (+1=2) Ln (x) = Cn e t (t + x)n dt; (2 + 1)n ( + 1) 0 x+t
(3)
where Cn() (y); ¿ 0; y ∈ [ − 1; 1] denotes the Gegenbauer polynomials of degree n and order .
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Lemma 3 (Loh;ofer inequality [8]). Let n¿2; ¿ 0; and x ∈ [ − 1; 1]. Then the Gegenbauer polynomials Cn() (x) satisfy the inequality 2= =2 2 2 2 |Cn() (x)|6[x2 d2= 2n; 2 + (1 − x )dn; ] 6x d2n; 2 + (1 − x )dn; ;
(4)
where dn; =
(n=2 + ) : ()(n=2 + 1)
Lemma 4 (Lewandowski [7]). For any ¿ − 12 ; x¿0 and n = 0; 1; 2; : : : ; we have (n; ; x) :=
∞
0
e−t t (x + t)n dt
= ( + 1 + n)
n ( + 1)n−k xk n! = ( + 1 + n)n() (exp x): ( + 1)n k=0 (n − k)! k!
(5)
3. The main result Theorem 1. For the Laguerre polynomials Ln() (x); ¿− 12 ; x¿0; n=2; 3; : : : ; we have the following bound: ( |Ln() (x)|6
+ 1)n n!
4x ( + 1)((n + 1)=2) − 1− √ (+1) (exp x) : (n + ) n−2 (n=2 + + 1)
n() (exp x)
(6)
Proof. Using the formula (3), the Loh;ofer inequality (4) for = ( + 12 ), and the formula (5) we can write the following chain of inequalities: ∞ 1 4xt () |Ln (x)| 6 d2n; 2+1 − [d2n; 2+1 − dn; +1=2 ] e−t t (t + x)n dt (2 + 1)n ( + 1) 0 (x + t)2 d2n; 2+1 (n; ; x) − 4x[d2n; 2+1 − dn; +1=2 ](n − 2; + 1; x) ; (2 + 1)n ( + 1) (n + 2 + 1) ( + 1 + n) () (exp x) |Ln() (x)| 6 (2 + 1)n ( + 1) (2 + 1)(n + 1) n =
(+1) (n=2 + + 12 ) (exp x) 4x(n + ) n−2 (n + 2 + 1) : − − (2 + 1)n ( + 1) (2 + 1)(n + 1) ( + 12 )(n=2 + 1)
The basic formulas for Gamma Euler function give |Ln() (x)| 6
( + 1)n () n (exp x) n! 4x ( + 1)n − (n + ) n!
(n=2 + + 12 )(2 + 1)(n + 1) 1− ( + 12 )(n=2 + 1)(n + 2 + 1)
(+1) n−2 (exp x)
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( + 1)n = n!
n() (exp x)
( + 1)((n + 1)=2) 4x (+1) (exp x) ; − 1− √ (n + ) n−2 (n=2 + + 1)
which ends the proof of (6). Corollary. The bound (6) in Theorem 1 is better than the corresponding bound (LS) in Theorem A for all n = 2; 3; : : : ; and ¿ − 12 ; and x ¿ 0. Proof. In order to show that our estimate (6) is better than (LS) in Theorem A it is enough to prove the inequality ( + 1)((n + 1)=2) ¿ 0; ¿ − 12 ; n = 2; 3; : : : ; 1− √ (n=2 + + 1) which is equivalent to ( + 1)((n + 1)=2) √ ¡ 1; ¿ − 12 ; (n=2 + + 1)
n = 2; 3; : : : :
Let us consider two cases: √ 1. If n = 2k, k = 1; 2; 3; : : :, then using the duplicate formula (2x) = 22x−1 (x)(x + 12 ) with x = k we obtain ( + 1)((2k + 1)=2) √ (k + + 1) =
(2k − 1)! ( + 1)(2k) = (k + + 1)(k)22k−1 (k − 1)!( + 1)k 22k−1
=
(2k − 1)!!(2k − 2)!! (2k − 1)!!2k−1 (k − 1)! (2k − 1)!! = = (k − 1)!( + 1)k 22k−1 (k − 1)!( + 1)k 22k−1 ( + 1)k 2k
=
1 · 3 · : : : · (2k − 1) 1 · 3 · : : : · (2k − 1) = ¡1 2k ( + 1) · ( + 2) · : : : · ( + k) (2 + 2) · (4 + 2) · : : : · (2k + 2) for ¿ − 12 ; k = 1; 2; 3; : : : :
2. If n = 2k + 1, k = 1; 2; 3; : : : then we have to consider two cases ¿0 and ∈ (− 12 ; 0). If ¿0 then k + + 32 ¿ 32 . Because for x¿ 32 the Gamma Euler function (x) is increasing then (k + + 32 ) ¿ (k + + 1); which implies ( + 1)(k + 1) k! ( + 1)((2k + 2)=2) √ ¡ √ =√ ¡ 1: 3 (k + + 1) ( + 1)k (k + + 2 ) (0; 1) the Gamma Euler function (x) Assume now that − 12 ¡ ¡ 0 ( 12 ¡ +1 ¡ 1). Because for x ∈√ 1 is decreasing, therefore in our case we have (+1) ¡ ( 2 )= . As above (k +1) ¡ (k ++ 32 ) which gives ( + 1)k! ( + 1)((2k + 2)=2) √ =√ ¡ 1; 3 (k + + 2 ) (k + + 32 ) and ends the proof of the corollary.
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