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Thread: Fourier Series and Mean Error

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    Super Member Aryth's Avatar
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    Fourier Series and Mean Error

    Suppose the function f(x) is defined on the interval (-L,L) and $\displaystyle \int_{-L}^L f^2(x)~dx$ is finite.

    We want to approximate f(x) by a finite series of (co-)sines:

    $\displaystyle f(x) \approx g_N(x) = A_0 + \sum_{n=1}^N A_n\cos{\left(\frac{n\pi}{L}x\right)} + B_n\sin{\left(\frac{n\pi}{L}x\right)}$

    where parameters $\displaystyle A_0$, $\displaystyle A_n$, and $\displaystyle B_n$ are freely variable.

    The mean square error of the approximation is defined as:

    $\displaystyle E_N = \int_{-L}^L (f(x) - g_N(x))^2~dx$

    a) Show that the choice: $\displaystyle A_0 = a_0$, $\displaystyle A_n = a_n$ and $\displaystyle B_n = b_n$, where $\displaystyle a_0$, $\displaystyle a_n$, and $\displaystyle b_n$ are the Fourier coefficients of f, minimizes $\displaystyle E_N$

    I think I can figure the rest of it out when I get this, so I won't bother typing it... I have never worked with the mean error before so I'm a bit confused as to how to start. Any help would be appreciated.
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    Quote Originally Posted by Aryth View Post
    Suppose the function f(x) is defined on the interval (-L,L) and $\displaystyle \int_{-L}^L f^2(x)~dx$ is finite.

    We want to approximate f(x) by a finite series of (co-)sines:

    $\displaystyle f(x) \approx g_N(x) = A_0 + \sum_{n=1}^N A_n\cos{\left(\frac{n\pi}{L}x\right)} + B_n\sin{\left(\frac{n\pi}{L}x\right)}$

    where parameters $\displaystyle A_0$, $\displaystyle A_n$, and $\displaystyle B_n$ are freely variable.

    The mean square error of the approximation is defined as:

    $\displaystyle E_N = \int_{-L}^L (f(x) - g_N(x))^2~dx$

    a) Show that the choice: $\displaystyle A_0 = a_0$, $\displaystyle A_n = a_n$ and $\displaystyle B_n = b_n$, where $\displaystyle a_0$, $\displaystyle a_n$, and $\displaystyle b_n$ are the Fourier coefficients of f, minimizes $\displaystyle E_N$

    I think I can figure the rest of it out when I get this, so I won't bother typing it... I have never worked with the mean error before so I'm a bit confused as to how to start. Any help would be appreciated.
    To save writing, put $\displaystyle c_n(x) = \cos\left(\frac{n\pi}{L}x\right)$ and $\displaystyle s_n(x) = \sin\left(\frac{n\pi}{L}x\right)$.

    Here's how to start the calculation:

    $\displaystyle E_N = \int_{-L}^L \Bigl(f(x) -\sum(A_mc_m(x) + B_ms_m(x)\Bigr) \Bigl(f(x) -\sum(A_nc_n(x) + B_ns_n(x)\Bigr)\,dx$. (You'll notice that I'm being a bit vague about the sums. They go from 0 to m or n for the cosines, and from 1 to m or n for the sines. It gets messy if you write this all out really carefully.) Now multiply out those parentheses and notice that a lot of the terms disappear or simplify when you integrate them from L to L. For example, $\displaystyle \int_{-L}^Lc_m(x)c_n(x)\,dx$ is zero unless m=n, in which case it is 1. Also, $\displaystyle \int_{-L}^Lf(x)c_n(x)\,dx = a_n$, with a similar result for $\displaystyle s_n(x)$ and $\displaystyle b_n$.

    When you have finished that calculation, you should be able to write it in the form

    $\displaystyle E_N = \int_{-L}^Lf(x)^2dx - \sum(a_n^2+b_n^2) + \sum\bigl((a_n-A_n)^2 + (b_n-B_n)^2)\bigr).$

    That is obviously minimised by taking $\displaystyle A_n=a_n$ and $\displaystyle B_n=b_n$ for each n.
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    Super Member Aryth's Avatar
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    I really appreciate this, it got me on the right track... But I'm having trouble getting the sums to simplify to the given final expression... It's a lot of tedious typing that I'll do if you really want me to... But I ended up with:

    $\displaystyle \int_{-L}^L f^2(x) - 2\sum (a_n^2 + b_n^2) + \sum (A_n^2 + B_n^2) ~dx$

    Would you show a couple of steps in the integration?? You don't have to show them all, I know it is a bit tedious.
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    Quote Originally Posted by Aryth View Post
    I really appreciate this, it got me on the right track... But I'm having trouble getting the sums to simplify to the given final expression... It's a lot of tedious typing that I'll do if you really want me to... But I ended up with:

    $\displaystyle \int_{-L}^L f^2(x) - 2\sum (a_n^2 + b_n^2) + \sum (A_n^2 + B_n^2) ~dx$

    Would you show a couple of steps in the integration?? You don't have to show them all, I know it is a bit tedious.
    That's very much along the right lines, except that you should have $\displaystyle - 2\sum (a_nA_n + b_nB_n)$ instead of $\displaystyle - 2\sum (a_n^2 + b_n^2)$. Also, all the terms except for $\displaystyle f^2(x)$ are constants and should not be integrated any further.

    Now use the fact that $\displaystyle (a_n-A_n)^2 = a_n^2 - 2a_nA_n + A_n^2$, which you can write in the form $\displaystyle - 2a_nA_n + A_n^2 = -a_n^2 + (a_n-A_n)^2$. Substitute that (and the similar identity for $\displaystyle b_n$ and $\displaystyle B_n$) into the expression $\displaystyle \int_{-L}^L f^2(x)\,dx - 2\sum (a_nA_n + b_nB_n) + \sum (A_n^2 + B_n^2)$ and you'll get the answer.
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    Super Member Aryth's Avatar
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    Awesome, I got it. Just one last question, what are the indices of the sums in the final expression? Zero to N?
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    Quote Originally Posted by Aryth View Post
    Awesome, I got it. Just one last question, what are the indices of the sums in the final expression? Zero to N?
    As always with Fourier series, the cosine terms ($\displaystyle a_n$ and $\displaystyle A_n$) go from 0 to N, the sine terms ($\displaystyle b_n$ and $\displaystyle B_n$) go from 1 to N.
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    Super Member Aryth's Avatar
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    Alright, I'm stumped on another part of this... It says:

    State the value of the minimum of $\displaystyle E_N$
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    Quote Originally Posted by Opalg View Post
    $\displaystyle E_N = \int_{-L}^Lf(x)^2dx - \sum(a_n^2+b_n^2) + \sum\bigl((a_n-A_n)^2 + (b_n-B_n)^2)\bigr).$

    That is obviously minimised by taking $\displaystyle A_n=a_n$ and $\displaystyle B_n=b_n$ for each n.
    Quote Originally Posted by Aryth View Post
    Alright, I'm stumped on another part of this... It says:

    State the value of the minimum of $\displaystyle E_N$
    So the minimum occurs when $\displaystyle A_n=a_n$ and $\displaystyle B_n=b_n$ for each n, and then $\displaystyle \boxed{E_N = \int_{-L}^Lf(x)^2dx - \sum(a_n^2+b_n^2).}$
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    Super Member Aryth's Avatar
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    That's what I thought... But is it the same as $\displaystyle \min{(E_N)}$?
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