Mathematical Methods for Physics and Engineering - Matematica.NET

16.7 HINTS AND ANSWERS(c)Show that the corresponding non-terminating series solutions S m (z) have astheir first few termsS 0 (z) =a 0(z + 1 3! z3 + 9 )5! z5 + ··· ,S 1 (z) =a 0(1 − 1 2! z2 − 3 )4! z4 − ··· ,(S 2 (z) =a 0 z − 3 3! z3 − 15 )5! z5 − ··· ,S 3 (z) =a 0(1 − 9 2! z2 + 454! z4 + ···16.16 Obtain the recurrence relations for the solution of Legendre’s equation (18.1) ininverse powers of z, i.e.sety(z) = ∑ a n z σ−n ,witha 0 ≠ 0. Deduce that, if l is aninteger, then the series with σ = l will terminate and hence converge for all z,whilst the series with σ = −(l + 1) does not terminate and hence converges onlyfor |z| > 1.).16.7 Hints and answers16.1 Note that z = 0 is an ordinary point of the equation.For σ =0,a n+2 /a n =[n(n +2)− λ]/[(n +1)(n + 2)] and, correspondingly, forσ =1,U 2 (z) =a 0 (1 − 4z 2 )andU 3 (z) =a 0 (z − 2z 3 ).16.3 σ =0and3;a 6m /a 0 =(−1) m /(2m)! and a 6m /a 0 =(−1) m /(2m + 1)!, respectively.y 1 (z) =a 0 cos z 3 and y 2 (z) =a 0 sin z 3 .TheWronskianis±3a 2 0 z2 ≠0.16.5 (b) a n+1 /a n =[l(l +1)− n(n +1)]/[2(n +1) 2 ].(c) R = 2, equal to the distance between z = 1 and the closest singularity atz = −1.(−1) n z 2n16.7 A typical term in the series for y(σ, z) is[(σ +2)(σ +4)···(σ +2n)] . 216.9 The origin is an ordinary point. Determine the constant of integration by examiningthe behaviour of the related functions for small x.y 2 (z) =(expz 2 ) ∫ z0 exp(−x2 ) dx.16.11 Repeated roots σ =2.∞∑y(z) =az 2 − 4az 3 +6bz 3 (n +1)(−2z) n+2 { a}+n! 4 + b [ln z + g(n)] ,n=2whereg(n) = 1n +1 − 1 n − 1n − 1 − ···− 1 2 − 2.16.13 The transformed equation is xy ′′ +2y ′ + y =0;a n =(−1) n (n +1) −1 (n!) −2 a 0 ;du/dz = A[ y 1 (z)] −2 .16.15 (a) (i) a n+2 =[a n (n 2 − m 2 )]/[(n +2)(n + 1)],(ii) a n+2 = {a n [(n +1) 2 − m 2 ]}/[(n +3)(n + 2)]; (b) 1, z, 2z 2 − 1, 4z 3 − 3z.553