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

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4Proof. By previous notation, we can write(L + ξ)(L + 2 + ξ) n − (L − ξ)(L + 2 − ξ) nn∑ (= (L + ξ) n)n∑k (L + 2) n−k ξ k − (L − ξ) (−1) k( )nk (L + 2) n−k ξ k== 2n∑k=0k=0(1 − (−1)k )( nk)L(L + 2) n−k ξ k +[(n−1)/2]∑i=0⎧⎨= 2ξ⎩∑[(n−1)/2]i=0n∑k=0( n2i+1)L(L + 2) n−2i−1 ξ 2i+1 + 2( n2i+1)L(L + 2) n−2i−1 ξ 2i +k=0( )( )1 + (−1)k nk (L + 2) n−k ξ k+1[n/2]∑i=0[n/2]∑wherefrom immediately follows the expression for h n .✸k=0i=02. The generating function( n2i)(L + 2) n−2i ξ 2i+1⎫( n) ⎬2i (L + 2) n−2i ξ 2i ⎭ ,The Jacobi polynomials are given byP n (a,b) (x) = 1 n∑( )( )n + a n + b(x − 1) n−k (x + 1) k (a, b > −1).2 n k n − kAlso, they can be written in the form( x − 1) nn∑( )( ) n + a n + b (P n (a,b)x+1(x) =2k n − kFrom the factL = x + 1x − 1we conclude thatandk=0⇔ x = L + 1L − 1T (2n, n; L) = (L − 1) n · P (0,0)nT (2n + 2, n; L) = (L − 1) n · P (2,0)nx−1) k.(x ≠ 1, L ≠ 1).The generating function G(x, t) for the Jacobi polynomials is∞∑G (a,b) (x, t) = P n (a,b) (x)t n 2 a+b=φ · (1 − t + φ) a · (1 + t + φ) , (13)bwheren=0(L+1L−1()L+1L−1).φ = φ(x, t) = √ 1 − 2xt + t 2 .
5Now,Also,∞∑T (2n, n; L) t n =n=0∞∑T (2n + 2, n; L) t n =n=0∞∑n=0∞∑n=0P (0,0)nP (2,0)n((L+1L−1L+1L−1∞∑T (2n, n − 1; L) t n = t ·n=0∞∑T (2n + 2, n + 1; L) t n = 1 t ·n=0) ((L− 1)t) n= G(0,0)(L+1L−1 , (L − 1)t ),) ((L− 1)t) n= G(2,0)(L+1L−1 , (L − 1)t ).() }{G (2,0) L+1, (L − 1)t − 1 ,L−1() }{G (0,0) L+1, (L − 1)t − 1 .L−1The generating function G(t; L) for the sequence {a n } n≥0 is given byG(t; L) =∞∑a n t nn=0(14)= t + 1 ( )( )G (0,0) L+1L−1t, (L − 1)t − (t + 1)G (2,0) L+1, (L − 1)t − 1 L−1t .The function()L+1ρ(t; L) = φ , (L − 1)t = √ 1 − 2(L + 1)t + (L − 1)L−1 2 t 2 (15)has domainorD ρ =(−∞, 1 − 2√ L + L) (√1 + 2 L + L)∪1 − 2L + L 2 1 − 2L + L , +∞ 2D ρ = (−∞, 1/4) (L = 1).(L ≠ 1),Theorem 2.1. The generating function G(t; L) for the sequence {a n } n≥0 isG(t; L) = t + 1 { }1ρ(t; L) t − 4− 1 (1 − (L − 1)t + ρ(t; L)) 2 t . (16)Example 2.1. For L = 1, we getG(t; 1) =and for L = 2, we find∞∑a n (1) t n = 1 tn=0( (1 −√ 1 − 4t)(1 + t)2t)− 1 . (17)G(t; 2) =∞∑n=0a n (2) t n = − 1 { }t + t + 1 1√t2 − 6t + 1 t − 4(1 − t + √ . (18)t 2 − 6t + 1) 2
Page 1 and 2: The Hankel Transform of the Sum ofC
Page 3: 3is given by−n)/2h n =L(n2 2 n+1
Page 7 and 8: 7Example 3.2. From (??), we have:{
Page 9 and 10: 9Example 4.1. For L = 4 we getwhere
Page 11 and 12: 11Example 4.2. For L = 4 we getwher
Page 13: We will apply the mathematical indu

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