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

12In our case it is enough to take d = 0 to get the final weightHencer −1 = −(L + 1),w(x) = ˘w(x)x .(r n = − ᾰ n +˘β )nr n−1Lemma 4.1. The parameters r n have the explicit form(n = 0, 1, . . .). (37)r n = − ψ n+1ψ n+2· Lψ n+2 + ξϕ n+2Lψ n+1 + ξϕ n+1(n ∈ N 0 ). (38)Proof. We will use the mathematical induction. For n = 0, we really get theexpected valuer 0 = − L2 + 2L + 2(L + 1)(L + 2) .Suppose that it is true for k = n. Now, by the properties for ϕ n and ψ n , we have˜α n+1 · r n + ˜β n+1 = − ψ n+1ψ n+3· Lψ n+3 + ξϕ n+3Lψ n+1 + ξϕ n+1.Dividing with r n , we conclude that the formula is valid for r n+1 .Example 4.3. For L = 4 we getwherefromα 0 = 24r −1 = −5, r 0 = − 1315 , r 1 = − 5152 , r 2 = − 356357 ,5 , β 0 = 5, α 1 = 32365 , β 1 = 10425 , α 2 = 1104221 , β 2 = 680169 ,just the same as in the Example 4.1.Proof of the main result. Let us start from Krattenthaler formulah 1 = a 0 , h n = a n 0β1 n−1 β2 n−2 · · · βn−2β 2 n−1 (n = 2, 3, . . .). (39)Here is a 0 = β 0 = L + 1. This formula can be also written in the formh 1 = a 0 , h n = β 0 β 1 β 2 · · · β n−2 β n−1 · h n−1 . (40)From the theory of orthogonal polynomials, it is known thatwherefromHere,‖Q n−1 ‖ 2 = β 0 β 1 β 2 · · · β n−2 β n−1 (n = 2, 3, . . .), (41)h 1 = a 0 , h n = ‖Q n−1 ‖ 2 · h n−1 (n = 2, 3, . . .). (42)‖Q n−1 ‖ 2 = β 0r n−2r −1n−2∏k=0˘β k = Ln−12·Lψ n + ξϕ nLψ n−1 + ξϕ n−1. (43)