Does a Fourier Series Have an Infinite Dimensional Parameter Space?

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SUMMARY

The Fourier series, represented as s(x) = a_0/2 + ∑(a_n cos(nx) + b_n sin(nx)), possesses an infinite dimensional parameter space due to the infinite number of coefficients a_n and b_n. This contrasts with simpler linear equations like y = mx + b, which have a two-dimensional parameter space defined by the parameters m and b. The discussion clarifies that the term "high dimensional PDE" refers to the number of independent parameters in a function, confirming that the Fourier series indeed has an infinite dimensional parameter space.

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Want to understand a concept here about dimensions of a function.

Using example 1: a simple Fourier series from http://en.wikipedia.org/wiki/Fourier_series

s(x) = \frac{a_0}{2} + \sum ^{\infty}_{0}[a_n cos(nx) + b_n sin(nx)]

So do we now say that s(x) has an infinite dimensional parameter space?

When I think of x being one dimensional, I think of y = mx + b which to me is a 2 dimensional parameter problem for y...

Trying to figure out what exactly is meant by a "high dimensional pde" which lives in 2-d (as an example)...

I assume this would be the same answer for a problem that considers the Karhunen-Loeve transform.
 
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The "dimension of the parameter space" is the number of independent, real valued parameters. Yes, "y= mx+ b" depends upon the two "parameters", m and b, and has parameter space of dimension two. In particular, any point (a, b) in R^2, gives a function y= ax+ b.

And, yes,
s(x)= \frac{a_0}{2}+ \sum_0^\infty \left[a_ncos(nx)+ b_nsin(nx)\right]
has infinite dimensional parameter space because there are an infinite number of parameters.

(In fact some people would say it has a "doubly infinite parameter space" since both the a_n and b_n sequences have an infinite number of parameters. But "infinite" should be enough for anybody!)
 

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