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Math Help - [SOLVED] Fourier Series

  1. #1
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    [SOLVED] Fourier Series

    Hi

    I am having some problems trying to answer part of this question

    I have attached file as I don't quite know how to upload the image... yet

    I have done the first part and calculated the F(t)

    I understand what happens to the series when you input t=0 as it that means cos 0 = 1 and sin 0 = 0.

    Same again with the third part - the best value i think to use is t=pi as cos pi = -1 and sin pi = 0.

    What i'm stuck on is how I apply this to evaluate the series.

    Please help!!

    Thanks
    Attached Files Attached Files
    Last edited by Ian1779; August 19th 2008 at 02:14 PM. Reason: Wrong file attachment
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Ian1779 View Post
    Hi

    I am having some problems trying to answer part of this question

    I have attached file as I don't quite know how to upload the image... yet

    I have done the first part and calculated the F(t)

    I understand what happens to the series when you input t=0 as it that means cos 0 = 1 and sin 0 = 0.

    Same again with the third part - the best value i think to use is t=pi as cos pi = -1 and sin pi = 0.

    What i'm stuck on is how I apply this to evaluate the series.

    Please help!!

    Thanks
    Applying t=0 to the series F(t)=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{\cos(2n+1)t}  {(2n+1)^2}+\sum_{n=0}^{\infty}(-1)^{n}\frac{\sin(n+1)t}{n+1}, we get

    F(0)=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}

    Since F(0)=0, we see that 0=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}.

    This implies that \frac{\pi}{4}=\frac{2}{\pi}\sum_{n=0}^{\infty}\fra  c{1}{(2n+1)^2}\implies\color{red}\boxed{\sum_{n=0}  ^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}

    So we have shown that \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}  {8} evaluating the series about t=0.

    \mathbb{Q.E.D.}

    Now we need to find the value of \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}.

    The appropriate t value here would be t=\frac{\pi}{2}.

    The nice thing about the value t=\frac{\pi}{2} is that the cosine series disappears and we are left with F\left(\frac{\pi}{2}\right)=\frac{\pi}{4}+\sum_{n=  0}^{\infty}\frac{(-1)^n}{2n+1}

    I'm not 100% sure why this is the case [maybe someone else can explain why], but if we let F\left(\frac{\pi}{2}\right)=\frac{\pi}{2}, then \frac{\pi}{2}=\frac{\pi}{4}+\sum_{n=0}^{\infty}\fr  ac{(-1)^n}{2n+1}\implies\color{red}\boxed{\sum_{n=0}^{\  infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}}

    \mathbb{Q.E.D.}

    I hope this answers your question.

    --Chris
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  3. #3
    Junior Member
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    Thanks very much - I can see clearly now the 'missing bit' I couldn't work out.
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