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

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8In the case when x /∈(( √ L − 1) 2 , ( √ L + 1) 2 ), value R(x; L) is real. Therefore wecan calculate imaginary part of F(x; L) = lim y→0 + F(x + iy; L):IF(x; L) = I[l 2 (x) − l 1 (x)] = 0.Otherwise, if x ∈(( √ L − 1) 2 , ( √ )L + 1) 2 we have that:l 1 (x)⎧⎨Il 1 (x) =⎩= 2(3L + 1) log[x − (L + 1) ± i √ ]4L − (x − L − 1) 2√4L−(x−L−1) 22(3L + 1) arctan, x ≥ L + 1;x−(L+1)( √ )2(3L + 1) π + arctan, x < L + 14L−(x−L−1) 2x−(L+1)[ −(L−1) 2 +2x(L+1)−i(L−1)√4L−(x−L−1) 2l 2 (x) = 2(L − 1) logIl 1 (x)⎧⎪⎨ 2(L − 1)(2π + arctan √x(L+1)−(L−1)2=)⎪⎩ 2(L − 1)(π + arctan √x(L+1)−(L−1)24L−(x−L−1) 2x 2 (L−1) 3 ]4L−(x−L−1) 2 ), x ≥ (L−1)2L+1 ;, x < (L−1)2After substituting all considered cases in (7) we finally obtain valueIF(x; L) = limy→0 + IF(x + iy; L) = Il 2(x) − Il 1 (x) − (x − L + 1) √ 4L − (x − L − 1) 2From the relation (16) we conclude thatand finally we obtain ω(x; L):ω(x; L) = ψ ′ (x; L) = − 1 πω(x; L) = 12π (1 + 1 x )√ 4L − (x − L − 1) 2 =√LπL+1dIF(x; L) (27)dx(1 + 1 x) √ 1 −Previous formula holds for x ∈ (a, b), and otherwise we have ω(x; L) = 0.( x − L − 1) 22 √ (28)L4. Determining the three-term recurrence relationThe crucial moment in our proof of the conjecture is to determine the sequence ofpolynomials {Q n (x)} orthogonal with respect to the weight w(x; L) given by (??)on the interval (a, b) and to find the sequences {α n (x)} {β n (x)} in the three-termrecurrence relation.
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)
Page 1 and 2: The Hankel Transform of the Sum ofC
Page 3 and 4: 3is given by−n)/2h n =L(n2 2 n+1
Page 5 and 6: 5Now,Also,∞∑T (2n, n; L) t n =n
Page 7: 7Example 3.2. From (??), we have:{
Page 11 and 12: 11Example 4.2. For L = 4 we getwher
Page 13: We will apply the mathematical indu

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