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Thread: Fourier series problem

  1. #1
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    Fourier series problem

    Here I got a question of Fourier series attached with this. Also I atemped some part of the question (that is attached here with) but I need to know is this answer is correct.. or not
    and how I prove the part b) from the above result.
    Attached Thumbnails Attached Thumbnails Fourier series problem-l4-math-q01-qes.jpg   Fourier series problem-l4-math-q01.jpg  
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  2. #2
    Senior Member Shanks's Avatar
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    No, you should extend the domain of f to $\displaystyle [-\pi, \pi]$ such that the extension of f is even function .
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  3. #3
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    How it is in even function.. please show me..
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Try calculating $\displaystyle a_n $ again.
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  5. #5
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    When now i calcalate this again as considering even function
    then $\displaystyle b_n=0$

    $\displaystyle a_0=\frac{2}{\pi}\int\limits_{0}^{\pi} f(x) dx $

    $\displaystyle =\frac{2}{\pi}\int\limits_{0}^{\pi/2}(1)dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi/2}0dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi}(-1)dx$

    $\displaystyle =(1-\frac{\pi}{2})$

    $\displaystyle a_n=\frac{2}{\pi}\int\limits_{0}^{\pi}f(x)Cos nx dx$

    $\displaystyle =\frac{2}{\pi}\int\limits_{0}^{\pi/2}1.Cos.nx.dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi/2}.0.Cos.nx.dx+\frac{2}{\pi}\int\limits_{\pi/2}^{\pi}(-1).Cos.nx.dx$

    $\displaystyle =\frac{2}{\pi}(\frac{Sinn(\pi/2)}{n})+\frac{2}{\pi}(\frac{-Sinn\pi}{n}+\frac{Sinn\pi}{2n})$

    =0

    Is this calculation is true? otherwise plz show me the incorrect place..
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