Quoting Robert Baillie <rjbaillie@frii.com>:
Can one work backwards, from the coefficients to the function?
That is: given an expression in n, say, c(n) = (-1)^(n+1)/n, can one find the function that has c(n) as its coefficients?
This seems difficult, especially because in the second example, the function consists of two distinct pieces over [-Pi, Pi], and as part of th calculation, you'd have to find the point (x = 0) where the function jumps.
Engineers, physicists, numerical analysts, and Oliver Heaviside know or knew things about Fourier Series that would greatly distress a proper mathematician: 1) If the Fourier coefficients decrease as 1/(n^k) the function and its derivatives have continuity just to order k-2. When k=1 the function would have a discontinuity somewhere. For k=2, the function could be continuous yet have a discontinuity in the derivative. Even for k=0, neither the function nor its integral are continuous at all points. 2) The variance (or better, the standard deviation) of a function is inverse to the variance of its Fourier Transform. Thus, many large coefficients are needed to confine a function to a small region and conversely. 3) Gaps in the coefficients affect analytic continuability. And so on. This information doesn't sketch out the function, but together with a pencil and paper sketch of some sines and cosines, can give some insight. Try sin x +1/3 sin 3x, or sin x -1/3 sin 3x to get an idea of what can happen. - hvm ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos