The Hankel Transform of the Sum of Consecutive Generalized ...
The Hankel Transform of the Sum of Consecutive Generalized ... The Hankel Transform of the Sum of Consecutive Generalized ...
10The characteristic equationhas the solutions4z 2 − 4 c z + 1 = 0z 1,2 = 1 2and the integral solution of (23) isλ n = E 1 z n 1 + E 2 z n 2(c ± √ )c 2 − 1 .(n ∈ N).We evaluate values E 1 and E 2 from the initial conditions (λ −1 = 0; λ 0 = 1).i.e,In other to solve our problem, we will choose c = − L+22 √ L . HenceFinally, we obtain:z k = −t k4 √ L (k = 1, 2), where t 1,2 = L + 2 ± √ L 2 + 4.λ n =(−1) n2 · 4 n L n 2√L2 + 4λ n =After replacing in (22), we obtain:( )t n+11 − t n+12(λ = −1, 0, 1, . . .),(−1)n2 · 4 n L n 2 ξ ψ n+1 (λ = −1, 0, 1, . . .).ˆα n = − L + 22 √ L + 14 √ L · ψn+2ψ n+1+ √ L · ψn+1ψ n+2, (32)ˆβ n = ψ nψ n+2. (33)4ψn+12If a new weight function ˜w(x) is introduced bythen we have˜w(x) = ŵ(ax + b)˜α n = ˆα n − b, ˜βn = ˆβ n(n ≥ 0).aa 2Now, by using x ↦→ x−L−12 √ , i.e., a = 1L 2 √ L+1and b = −L 2 √ , we have the weight functionLThusand˜w(x) = ŵ( x − L − 12 √ L ) = 1 ( x − L − 12 2 √ L+ L + 2 ) √2 √ 1 −L( x − L − 1) 2.2 √ L˜α n = −1 + 1 2 · ψn+2ψ n+1+ 2L · ψn+1ψ n+2(n ∈ N 0 ), (34)
11Example 4.2. For L = 4 we getwherefrom˜β 0 = (L + 2) π 2 , ˜βn = L ψ nψ n+2ψ 2 n+1P 0 (x) = 1, ‖P 0 ‖ 2 = 3π,P 1 (x) = x − 173 , ‖P 1‖ 2 = 32π3 ,P 2 (x) = x 2 − 434 x + 1014 , ‖P 2‖ 2 = 42π,(n ∈ N). (35)P 3 (x) = x 3 − 33121 x2 + 157921 x − 218921 , ‖P 3‖ 2 = 3520π21 ,˜α 0 = 17 3 , ˜β0 = 3π, ˜α 1 = 6112 , ˜β1 = 32 9 , ˜α 2 = 42184 , ˜β2 = 6316 .Introducing the weight˘w(x) = 2L π ˜w(x)will not change the monic polynomials and their recurrence relations, only it willmultiply the norms by the factor 2L/π, i.e.˘P k (x) ≡ P k (x), ‖ ˘P k ‖ 2˘w =∫ ba˘P k (x) ˘w(x) dx = 2L π ‖P k‖ 2 (k ∈ N 0 ),˘β 0 = L(L + 2), ˘βk = ˜β k (k ∈ N), ᾰ k = ˜α k (k ∈ N 0 ).Here is˘β 0 ˘β1 · · · ˘β n−1 = Ln2 · ψn+1ψ n. (36)In the book [?], W. Gautschi has treated the next problem: If we know all aboutthe MOPS orthogonal with respect to ˘w(x) what can we say about the sequence{Q n (x)} orthogonal with respect to a weightw d (x) = ˘w(x) (d /∈ support( ˜w)) ?x − dW. Gautshi has proved that, by the auxiliary sequence∫r −1 = − w d (x) dx, r n = d − ᾰ n − ˘β n(n = 0, 1, . . .),r n−1Rit can be determinedα d,0 = ᾰ 0 + r 0 , α d,k = ᾰ k + r k − r k−1 ,β d,0 = −r −1 , β d,k = ˘β k−1r k−1r k−2(k ∈ N).
- 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 and 8: 7Example 3.2. From (??), we have:{
- Page 9: 9Example 4.1. For L = 4 we getwhere
- Page 13: We will apply the mathematical indu
11Example 4.2. For L = 4 we getwherefrom˜β 0 = (L + 2) π 2 , ˜βn = L ψ nψ n+2ψ 2 n+1P 0 (x) = 1, ‖P 0 ‖ 2 = 3π,P 1 (x) = x − 173 , ‖P 1‖ 2 = 32π3 ,P 2 (x) = x 2 − 434 x + 1014 , ‖P 2‖ 2 = 42π,(n ∈ N). (35)P 3 (x) = x 3 − 33121 x2 + 157921 x − 218921 , ‖P 3‖ 2 = 3520π21 ,˜α 0 = 17 3 , ˜β0 = 3π, ˜α 1 = 6112 , ˜β1 = 32 9 , ˜α 2 = 42184 , ˜β2 = 6316 .Introducing <strong>the</strong> weight˘w(x) = 2L π ˜w(x)will not change <strong>the</strong> monic polynomials and <strong>the</strong>ir recurrence relations, only it willmultiply <strong>the</strong> norms by <strong>the</strong> factor 2L/π, i.e.˘P k (x) ≡ P k (x), ‖ ˘P k ‖ 2˘w =∫ ba˘P k (x) ˘w(x) dx = 2L π ‖P k‖ 2 (k ∈ N 0 ),˘β 0 = L(L + 2), ˘βk = ˜β k (k ∈ N), ᾰ k = ˜α k (k ∈ N 0 ).Here is˘β 0 ˘β1 · · · ˘β n−1 = Ln2 · ψn+1ψ n. (36)In <strong>the</strong> book [?], W. Gautschi has treated <strong>the</strong> next problem: If we know all about<strong>the</strong> MOPS orthogonal with respect to ˘w(x) what can we say about <strong>the</strong> sequence{Q n (x)} orthogonal with respect to a weightw d (x) = ˘w(x) (d /∈ support( ˜w)) ?x − dW. Gautshi has proved that, by <strong>the</strong> auxiliary sequence∫r −1 = − w d (x) dx, r n = d − ᾰ n − ˘β n(n = 0, 1, . . .),r n−1Rit can be determinedα d,0 = ᾰ 0 + r 0 , α d,k = ᾰ k + r k − r k−1 ,β d,0 = −r −1 , β d,k = ˘β k−1r k−1r k−2(k ∈ N).
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