From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Newsgroups: sci.math Subject: Re: Legendre polynomal order 6 Date: 23 Oct 1997 13:20:54 -0400 In article <877574579.26957@dejanews.com>, wrote: :In article <62ljq5$mbg$1@strauss.udel.edu>, : rawlins@strauss.udel.edu (Michael A Rawlins) wrote: :> :> :> Greetings, :> :> Could someone please post/reply the algebraic formula for the 6th :> order Legendre polynomial. For example: :> :> P1 = x :> 2 :> P2 = 1/2 (3x - 1) :> :> . :> :> . :> :> P6 = ????? :> :> Thanks for any help. :> :> Michael Rawlins email: rawlins@udel.edu : :You may find it in several ways: : :1) Differentiate (x^2 - 1)^6 six times, then divide the result by : 2^6*6! = 46080. : :2) Beginning with P(x,0) = 1 and P(x,1) = x apply five times the : recurrence: : : 2n+1 n : P(x,n+1) = ----- xP(x,n) - --- P(x,n-1) : n+1 n+1 : :3) Use a symbolic computation system. : Example with MAPLE: : :> P:= (x,n) -> diff((x^2-1)^n,x$n)/(2^n*n!); : : 2 n : diff((x - 1) , x$n) : P := (x,n) -> -------------------- : n : 2 n! : : :> P(x,6); : : 6 2 4 2 2 2 2 3 : x + 15/2 (x - 1) x + 45/8 (x - 1) x + 5/16 (x - 1) : :> simplify("); : : 231 6 315 4 105 2 : --- x - --- x + --- x - 5/16 : 16 16 16 : : :José H. Nieto In addition, you can expand the expression ("generating function") G(x,z) = (1 - 2*x*z + z^2)^(-1/2) in powers of z, and pick the coefficient at z^6 (same as): Find the 6-th partial derivative of G(x,z), six times by z, at z=0, and divide by 6!. References: many textbooks and handbooks on special functions; I had this one at hand: Nico M. Temme, Special Functions, An Introduction... John Wiley & Sons, New York etc. 1996 ISBN 0-471-11313-1 Cheers, ZVK (Slavek)