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Math Help - Show the Fourier series of a function

  1. #1
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    Show the Fourier series of a function

    Hello,

    I can't solve the following problem involving the Fourier series of a function and need som guidance. I appriciate any suggestion. Thank you kindly.
    Attached Thumbnails Attached Thumbnails Show the Fourier series of a function-fourier-series.jpg  
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  2. #2
    MHF Contributor chisigma's Avatar
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    The even function in [-\pi,\pi] is...

    \displaystyle f(x)=\left\{\begin{array}{ll}2-x ,\,\,0 < x < 2\\{}\\0 ,\,\, 2 < x< \pi\end{array}\right. (1)

    ... and its cosine Fourier coefficients are...

    \displaystyle a_{n} = \frac{2}{\pi} \int_{0}^{2} (2-x)\ \cos nx \ dx =

    \displaystyle = \frac{4}{\pi} |\frac{\sin n x }{n}|_{0}^{2} - \frac{2}{\pi} |\frac{n x \sin nx + \cos nx}{n^{2}}|_{0}^{2} =

    \displaystyle = \frac{2}{\pi} \frac{1- \cos 2n}{n^{2}} = \frac{4}{\pi} \frac{\sin^{2} n}{n^{2}} (2)

    Kind regards

    \chi \sigma
    Last edited by chisigma; August 18th 2010 at 06:47 AM.
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  3. #3
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    So when can the rule: odd function times an even function becomes zero, applies when solving for Fourier coefficients??
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by 4Math View Post
    So when can the rule: odd function times an even function becomes zero, applies when solving for Fourier coefficients??
    Please, can You explain with other words the question?...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Please, can You explain with other words the question?...

    Kind regards

    \chi \sigma
    When solving for a_{n} :

    \displaystyle a_{n} = \frac{2}{\pi} (({2}\int_{0}^{2} cos nx  \ dx)-(\int_{0}^{2} x\ \cos nx \ dx))

    the second integral:

    -(\int_{0}^{2} x\ \cos nx \ dx)

    x is an odd function and cos nx  is an even function. And the result of that is an odd function times an even function becomes zero.

    Why can this not be applied here?

    By the way, thank you kindly for your response.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by 4Math View Post
    When solving for a_{n} :

    \displaystyle a_{n} = \frac{2}{\pi} (({2}\int_{0}^{2} cos nx  \ dx)-(\int_{0}^{2} x\ \cos nx \ dx))

    the second integral:

    -(\int_{0}^{2} x\ \cos nx \ dx)

    x is an odd function and cos nx  is an even function. And the result of that is an odd function times an even function becomes zero. Why can this not be applied here?...
    Because the integral is from 0 to 2 and not from -a to +a...

    Kind regards

    \chi \sigma
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