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Math Help - Fourier Series

  1. #1
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    Fourier Series

    Sorry that I don't know how to form piecewise functions using Latex, so I'll just describe it using f_{1} and f_{2}

    The piecewise function is as follows:

    f_{1}=1+2x ..... -1\leq x \leq 0
    f_{2}=1-2x ..... 0\leq x \leq 1

    In the solution that my professor posted, he states that we can ignore the sinx part of the series because it's an even function. That part I understand. However, when he goes on to solve for a_{0} and a_{n} he ignores the f_{1} part of the original function. Is this because only x's greater than zero are significant, or is there another reason for this?

    If anyone could answer this as soon as possible, I have a final a 8am EST tomorrow morning.

    Much appreciated.

    Edit...Could it possibly have something to do with the symmetry of the two functions, since they sort of form a \sqrt{x} when they are both graphed together?
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  2. #2
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    Quote Originally Posted by Kalter Tod View Post
    Sorry that I don't know how to form piecewise functions using Latex, so I'll just describe it using f_{1} and f_{2}

    The piecewise function is as follows:

    f_{1}=1+2x ..... -1\leq x \leq 0
    f_{2}=1-2x ..... 0\leq x \leq 1

    In the solution that my professor posted, he states that we can ignore the sinx part of the series because it's an even function. That part I understand. However, when he goes on to solve for a_{0} and a_{n} he ignores the f_{1} part of the original function. Is this because only x's greater than zero are significant, or is there another reason for this?

    If anyone could answer this as soon as possible, I have a final a 8am EST tomorrow morning.

    Much appreciated.

    Edit...Could it possibly have something to do with the symmetry of the two functions, since they sort of form a \sqrt{x} when they are both graphed together?
    \int_{-1}^{0} 1 + 2x \, dx = \int_{0}^{1} 1 - 2x \, dx.

    So \int_{-1}^{1} f(x) \, \cos (nx) \, dx = 2 \int_{0}^{1} f_2 \, \cos(nx) dx .....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \int_{-1}^{0} 1 + 2x \, dx = \int_{0}^{1} 1 - 2x \, dx.

    So \int_{-1}^{1} f(x) \, \cos (nx) \, dx = 2 \int_{0}^{1} f_2 \, \cos(nx) dx .....
    Okay, yeah I wasn't sure if was that simple or not.

    Thanks a lot.
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