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Simple question about dimension

  1. Dec 4, 2013 #1
    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

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

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

    When I think of x being one dimensional, I think of [itex] y = mx + b [/itex] 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.
  2. jcsd
  3. Dec 9, 2013 #2


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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 [itex]R^2[/itex], gives a function y= ax+ b.

    And, yes,
    [tex]s(x)= \frac{a_0}{2}+ \sum_0^\infty \left[a_ncos(nx)+ b_nsin(nx)\right][/tex]
    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 [itex]a_n[/itex] and [itex]b_n[/itex] sequences have an infinite number of parameters. But "infinite" should be enough for anybody!)
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