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] ?

Third try f(x)= 1+(-exp(1)+1)*x The Fourier series is: g(x) = -1/4+3/4*exp(1) + sum( (exp(1)*(-1)^n-(-1)^n+1-exp(1))/n^2/Pi^2+(-1+exp(1)*(-1)^n)/(1+n^2*Pi^2) * cos(n*Pi*x) + (exp(1)*n*Pi*(-1)^n-n*Pi)/n^2/Pi^2-(exp(1)*n*Pi*(-1)^n-n*Pi)/(1+n^2*Pi^2) * sin(n*Pi*x) . n=1..infinity); This seems to look like exp(x) on my computer screen for x in [0,1] for N terms, when N=600 (1 constant term plus N cos terms plus N sin terms).