The Hankel Transform of the Sum of Consecutive Generalized ...

9Example 4.1. For L = 4 we getwherefromQ 0 (x) = 1, ‖Q 0 ‖ 2 = 5,Q 1 (x) = x − 24 5 , ‖Q 1‖ 2 = 1045 ,Q 2 (x) = x 2 − 12713 x + 25613 , ‖Q 2‖ 2 = 108813 ,Q 3 (x) = x 3 − 54117 x2 + 109617 x − 134417 , ‖Q 3‖ 2 = 569617 ,α 0 = 24 5 , β 0 = 5, α 1 = 32365 , β 1 = 10425 , α 2 = 1104221 , β 2 = 680169 .Hence( 104) 2h 1 = a 0 = 5, h 2 = a 2 0β 1 = 104, h 3 = a 3 0β1β 2 2 = 5 3 68025 169 = 8704.At the beginning, we will notice that in the definition of the weight appears thesquare root member.That’s why, let us consider the monic orthogonal polynomials {S n (x)} with respectto the p (1/2,1/2) (x) = √ 1 − x 2 on the interval (−1, 1). These polynomials aremonic Chebyshev polynomials of the second kind:S n (x) = sin( (n + 1) arccos x )2 n · √1− x 2They satisfy the three-term recurrence relation (Chihara [?]):with initial valueswhereS n+1 (x) = (x − α ∗ n) S n (x) − β ∗ nS n−1 (x) (n = 0, 1, . . .), (29)S −1 (x) = 0, S 0 (x) = 1,αn ∗ = 0 (n ≥ 0) and β0 ∗ = π 2 , β∗ n = 1 (n ≥ 1).4If we use the weight function ŵ(x) = (x − c) p (1/2,1/2) (x), then the correspondingcoefficients ˆα n and ˆβ n can be evaluated as follows (see, for example, Gautschi [?])λ n = S n (c),λ nˆα n = c − λ n+1− βn+1∗ ,λ n λ n+1ˆβ n = βn ∗ λ n−1 λ n+1(n ∈ Nλ 2 0 ).nFrom the relation (21), we conclude that the sequence {λ n } n∈N satisfies the followingrecurrence relation:(30)4λ n+1 − 4cλ n + λ n−1 = 0 (λ −1 = 0; λ 0 = 1). (31)