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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:

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

    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 $\displaystyle 2 \pi$-periodic, hence there should exist a convergent Fourier series of the form:

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

    With coefficients given as:

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

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

    However I get, in both cases, $\displaystyle a_n = 0$ and $\displaystyle b_n=0$. The only thing I find to be non-zero is $\displaystyle a_0 = -1$.
    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 $\displaystyle b_n$ will indeed be zero because $\displaystyle f(x)$ is an even function.

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

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

    Your $\displaystyle a_0 = -1$ is correct though (because the series uses $\displaystyle \tfrac12a_0$).
    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 $\displaystyle \sin^2(x) = \frac{1}{2}(1-\cos(2x))$ a trigonometrical identity?
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  4. #4
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    Re: Fourier Series Expansion

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

    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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