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Thread: [SOLVED] Fourier Series

  1. #1
    Junior Member
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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; Aug 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 $\displaystyle t=0$ to the series $\displaystyle F(t)=\frac{\pi}{4}-\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{\cos(2n+1)t} {(2n+1)^2}+\sum_{n=0}^{\infty}(-1)^{n}\frac{\sin(n+1)t}{n+1}$, we get

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

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

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

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

    $\displaystyle \mathbb{Q.E.D.}$

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

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

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

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

    $\displaystyle \mathbb{Q.E.D.}$

    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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