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Re: Closed expression for coefficients Chebychev polynomial
Posted:
Jun 14, 1996 8:10 PM


In article <4prj8q$2pq@rzsun02.rrz.unihamburg.de>, Hauke Reddmann (fc3a501@GEOMAT.math.unihamburg.de) writes: >I could err (must look up my math dictionary), >but wasn't simply Tn(x)=sin(n*arccos(x)),Un(x)=cos(thesame)?
i think you switched the Tn and Un.
T[n](x) = cos( n*arccos(x) ) for x <= 1 = cosh( n*arccosh(x) ) for x >= 1
it is what you say but it is also a polynomial (because T[0](x) = 1, T[1](x) = x, and there is this cool recursion formula, T[n+1](x) = 2*x*T[n](x)  T[n1]) and the original poster (and myself) want to know if there is a closed form expression for the kth coefficient of T[n](x) that is a simple function of n and k. i suppose there is (that can be inductively proven using the recursion formula and boundary conditions above) but i have never seen it.
r bj wave mechanics, inc. robert@audioheads.com or robert@wavemechanics.com



