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Thread: Fourier Series Expansion

  1. #1
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    Fourier Series Expansion

    Hello everyone

    I am considering the following equation:

    f(x) = - \sin( x)^2 - \cos( x)

    Which looks like this:

    Fourier Series Expansion-curve.jpg

    And need to expand it in terms of a Fourier series. I know the above function is continuous, differentiable and 2 \pi-periodic, hence there should exist a convergent Fourier series of the form:

    \frac{1}{2} a_0 + \sum \limits_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)]

    With coefficients given as:

    a_n =  \frac{1}{\pi} \int \limits_{- \pi}^{\pi} f(x) \cos(n x) \mathrm{d}x

    b_n =  \frac{1}{\pi} \int \limits_{- \pi}^{\pi} f(x) \sin(n x) \mathrm{d}x

    However I get, in both cases, a_n = 0 and b_n=0. The only thing I find to be non-zero is a_0 = -1.
    Can anybody help me realizing where I am going wrong?

    Kind Regards
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  2. #2
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    Re: Fourier Series Expansion

    The b_n will indeed be zero because f(x) is an even function.

    \sin^2 x + \cos x = \tfrac12(1-\cos 2x) + \cos x

    So the Fourier series is -\tfrac12 - \cos x + \tfrac12\cos 2x

    Your a_0 = -1 is correct though (because the series uses \tfrac12a_0).
    Last edited by Archie; Jan 10th 2017 at 12:53 PM.
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  3. #3
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    Re: Fourier Series Expansion

    Thank you very much! Is \sin^2(x) = \frac{1}{2}(1-\cos(2x)) a trigonometrical identity?
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  4. #4
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    Re: Fourier Series Expansion

    Yes, from \cos 2A = \cos^2 A - \sin^2 A.

    Note that you can find the Fourier Series for all purely trigonometric functions simply by using trigonometric identities.
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  5. #5
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    Re: Fourier Series Expansion

    That is nice to know thank you
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