----- Original Message ---- From: Greg Fee <gfee@cecm.sfu.ca> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, December 3, 2007 11:11:14 AM Subject: Re: [math-fun] extend exp(x) for quickly convergent Fourier series Greg Fee wrote:

Define a function f(x) on the interval -1 <= x < 0, to create a function g(x), such that:

1. g(x)=f(x) for -1 <= x < 0 2. g(x)=exp(x) for 0 <= x <= 1 3. The period 2 Fourier series of g(x) on the interval -1 <= x <= 1, of the form: a[0] + sum(a[i]*cos(i*Pi*x),i=1..infinity) + sum(b[i]*sin(i*Pi*x),i=1..infinity)

converges to exp(x) on the interval 0 <= x <= 1, faster than choosing f(x)=exp(x)

Can you find an f(x) such that g(x) has the quickest converging Fourier series to exp(x) on [0,1] ?

If you are contented with a function that is continuous, but has discontinuous slope (so that a[n] ~ n^-2), then g(x) = exp|x| has the advantage of being symmetric. Thus you have only cosine terms. Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ