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] ? --------------------------------------------------------------------------- First try f(x)=exp(x), then the Fourier series is: g(x) = 1/2*(exp(1)-exp(-1)) + sum( (-1)^n*(-exp(-1)+exp(1))/(1+n^2*Pi^2)*cos(n*Pi*x) -n*Pi*(-1)^n*(-exp(-1)+exp(1)/(1+n^2*Pi^2)*sin(n*Pi*x) ,n=1..infinity); This has Gibbs phenomenon, at x=1. ---------------------------------------------------------------------------- Second try f(x)={exp(x+2) for x=-1/2 {exp(x) for x>-1/2 The Fourier series is: g(x) = 1/2*(exp(3/2)-exp(-1/2)) + sum( ((-exp(1)*(-1)^n+exp(3/2)*cos(1/2*n*Pi)-exp(3/2)*n*Pi*sin(1/2*n*Pi))/(1+n^2*Pi ^2)+(-exp(-1/2)*cos(1/2*n*Pi)+exp(-1/2)*n*Pi*sin(1/2*n*Pi)+exp(1)*(-1)^n)/(1+n ^2*Pi^2))*cos(n*Pi*x)+((exp(1)*n*Pi*(-1)^n-exp(3/2)*n*Pi*cos(1/2*n*Pi)-exp(3/2 )*sin(1/2*n*Pi))/(1+n^2*Pi^2)-(-exp(-1/2)*n*Pi*cos(1/2*n*Pi)-exp(-1/2)*sin(1/2 *n*Pi)+exp(1)*n*Pi*(-1)^n)/(1+n^2*Pi^2))*sin(n*Pi*x) , n=1..infinity); This has shifted the Gibbs phenomenon to x=-1/2. It seems to look like exp(x) on my computer screen for x in [0,1] for N terms, when N=2^10 (1 constant term plus 2^10 cos terms plus 2^10 sin terms).