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Showing fourier coefficients reduce mean square error best

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    So I'm supposed to show that a finite fourier approximation is the optimal approximation for a given function.
    I am to suppose we have a given set of functions [tex]\phi _k(x),k=1,2,\text{...}N[/tex] defined on [tex]a\leq x\leq b[/tex].

    They are orthogonal [tex]\int _a^b\phi _m(x)\phi _n(x)dx=0 \text{ for } m\neq n[/tex]

    and are normalized [tex]\int _a^b\left[\phi _m(x)\right]{}^2dx=1[/tex]

    a general approximation for f(x) in terms of these N functions is
    [tex]f_{\text{app}} (x)=\sum _{m=1}^N \gamma _m\phi _m(x)[/tex]

    One possible choice of coefficients is the fourier coefficients defined by:
    [tex]f_m=\int _a^bf(x)\phi _m(x)dx[/tex]

    The mean square error is defined as:

    [tex]E=\int_a^b \left[f(x)-f_{\text{app}} (x)\right]{}^2 \, dx=\int_a^b \left[f(x)-\sum _{m=1}^N \gamma _m\phi _m(x)\right]{}^2 \, dx[/tex]

    I am supposed to show that the fourier coefficients would be the optimal choice of [tex]\gamma _m[/tex] to minimize E

    3. The attempt at a solution
    Thus far, I have carried out the square in the integrand of the error term, used the idea of orthogonality, and substituted the fourier coefficients in for gamma, but from there I am stuck! Here is what I have...

    [tex]E=\int _a^b[f(x)]^2dx-2\int _a^bf(x)\left[\sum _{m=1}^N \int _a^bf(x)\phi _m(x)dx\right]\phi _m(x)dx+\left[\sum _{m=1}^N \int _a^bf(x)\phi _m(x)dx\right]{}^2[/tex]

    From here, I am stuck! is there some kind of simplification that I am missing? this is very frustrating. Any help/nudge would be appreciated.

    -Eric
     
  2. jcsd
  3. Sep 24, 2009 #2
    Double check your squaring of [tex]\left[f(x)-\sum _{m=1}^N \gamma _m\phi _m(x)\right]{}^2[/tex] because your last formula for E looks suspect. Did you confuse f_m and gamma_m?



    The key trick (eventually) is to use [tex]-2\gamma_m f_m+\gamma_m^2 = -f_m^2 + (f_m - \gamma_m)^2 [/tex]
     
  4. Sep 24, 2009 #3
    if I am understanding the problem correctly, which is not an assumption I would bet any substantial amount of money on, I thought that the point was to show that [tex]f_m[/tex] is the best substitution for [tex]\gamma_m[/tex], e.g. show that [tex]f_m=\gamma_m[/tex] reduces E better than any other substitution for [tex]\gamma_m[/tex].

    So I everywhere I see [tex]\gamma_m[/tex] i want to substitute the expression for [tex]f_m[/tex] and show somehow that this is minimizes E, correct?
     
  5. Sep 24, 2009 #4
    No, wait until the end to imagine replacing gamma_m with f_m. Do the algebra I suggested and you should end up with an expression

    E = expression involving both gamma_m and f_m

    If you do enough algebra (correctly), you will see that this E is clearly minimized when gamma_m = f_m.

    Sorry I'm being vague. It's not really too hard.
     
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