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Math Help - Fourier Series Coefficients and Symetry

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    Fourier Series Coefficients and Symetry

    Given the series



    and that f(x) = f(pi - x), what can one say about the coefficients an and bn? I know that due to orthogonality an=0 for odd functions and bn=0 for even functions, but I'm not sure what symmetry the function f(x) = f(pi-x) has.
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    Quote Originally Posted by StaryNight View Post
    Given the series



    and that f(x) = f(pi - x), what can one say about the coefficients an and bn? I know that due to orthogonality an=0 for odd functions and bn=0 for even functions, but I'm not sure what symmetry the function f(x) = f(pi-x) has.
    Here are a few hints:
    First the function is symmetric across the line x=\frac{\pi}{2}

    Also since the inside of your trig function is of the form

    nx The generic form is \displaystyle \frac{n \pi x}{L} This gives the function must be 2L peroidic.

    Using this try to figure out if the function is odd or even!
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    Quote Originally Posted by TheEmptySet View Post
    Here are a few hints:
    First the function is symmetric across the line x=\frac{\pi}{2}

    Also since the inside of your trig function is of the form

    nx The generic form is \displaystyle \frac{n \pi x}{L} This gives the function must be 2L peroidic.

    Using this try to figure out if the function is odd or even!
    So the function f is even about the point x=pi/2. This means that a(n) will be 0 when cos(nx) is odd about x=pi/2 and b(n)=0 when sin(nx) is odd about x=pi/2.



    A similar argument for cos gives a(2n+1) = b(2n) = 0. Is this the best way to tackle the problem?
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