[math-fun] Compute a Function From its Fourier Series?
One of the first examples of Fourier series that a student encounters is something like this: f(x) = x/2, for -Pi < x < Pi. The Fourier series is Sum[ (-1)^(n+1)/n * Sin[n x] ]. A second, more complicated example is: f(x) = -Pi/2, for -Pi < x < 0, f(x) = Pi/2, for 0 < x < Pi. This Fourier series is Sum[ (1 - (-1)^n)/n * Sin[n x] ]. 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 the calculation, you'd have to find the point (x = 0) where the function jumps. One could plot 1000 terms of Sum[ c(n) * Sin[n x] ] for -Pi < x < Pi, then try to guess what the expressions and points of discontinuity of f(x) are. Or, one could compute 1000 terms for many x's and do curve fitting, but guessing would be involved here, too. But is there a systematic method to do this? Bob Baillie
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
On 9/23/07, mcintosh@servidor.unam.mx <mcintosh@servidor.unam.mx> wrote:
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.
Which is a general form of Heisenberg's uncertainty principle. -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
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