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Math Help - Sum (kind of difficult)

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Sum (kind of difficult)

    Compute \sum_{n=0}^{\infty}\frac{1}{\pi-en^2}

    And if you cannot find it see how closely you can bound it.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Compute \sum_{n=0}^{\infty}\frac{1}{\pi-en^2}

    And if you cannot find it see how closely you can bound it.
    Let \mathcal{F}(x) be function such that \forall{x}\in\left[-\delta,\delta\right]~\mathcal{F}(x)=\sin^2(ax)~a\notin\mathbb{Z}. Furthermore let \mathcal{F}(x) have the charcteristic that \mathcal{F}(x)\in{P}_{2\delta}\implies\mathcal{F}\  left(x+2\delta\right)=\mathcal{F}(x)~\forall{x}\in  \mathbb{R}. Now it is pretty clear that \mathcal{F}(x)\in\mathcal{C}\left(\mathbb{R}\right  ) except possibly at \delta\mathbb{Z}, but in fact \mathcal{F}(x) is continuous at all these points due to its eveness. Next note that \mathcal{F}(x)\in{D^1}, in other words its posses a right and left hand limit everywhere. So since noting all of these things we may conclude that S_n(x)\to\mathcal{F}(x)~\forall{x}\in\mathbb{R} where S_n(x) is the Fourier Series given by.

    S_n(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\bigg[A_n\cos\left(\frac{n\pi{x}}{\delta}\right)+B_n\sin  \left(\frac{n\pi{x}}{\delta}\right)\bigg]

    Where

    \begin{aligned}A_0&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)dx\\<br />
&=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)dx\\<br />
&=\frac{2a\delta-\sin\left(2a\delta\right)}{2a\delta}<br />
\end{aligned}


    \begin{aligned}A_n&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)\cos\left(\frac{n\pi  {x}}{\delta}\right)dx\\<br />
&=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)\cos\left(\frac{n\pi{x}}  {\delta}\right)dx\\<br />
&=-2a\delta\sin\left(2a\delta\right)\frac{(-1)^n}{(2a\delta)^2-n^2\pi^2}~\forall{n\in\mathbb{N}}<br />
\end{aligned}


    \begin{aligned}B_n&=\frac{1}{\delta}\int_{-\delta}^{\delta}\mathcal{F}(x)\sin\left(\frac{n\pi  {x}}{\delta}\right)dx\\<br />
&=\frac{1}{\delta}\int_{-\delta}^{\delta}\sin^2(ax)\sin\left(\frac{n\pi{x}}  {\delta}\right)dx\\<br />
&=0~\because\sin^2(ax)\sin\left(\frac{n\pi{x}}{\de  lta}\right)\text{ is odd}<br />
\end{aligned}

    \therefore~S_n(x)=\frac{2a\delta-\sin\left(4a\delta\right)}{2a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft  y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)

    And as was stated above S_n(x)\to\mathcal{F}(x)

    \therefore~\mathcal{F}(x)=\frac{2a\delta-\sin\left(2a\delta\right)}{4a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft  y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)

    So \forall{x}\in\left[-\delta,\delta\right]

    \sin^2(ax)=\frac{2a\delta-\sin\left(2a\delta\right)}{4a\delta}-2a\delta\sin\left(2a\delta\right)\sum_{n=1}^{\inft  y}\frac{(-1)^n}{\left(2a\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)

    Now letting 2a=z gives

    \sin^2\left(\frac{zx}{2}\right)=\frac{z\delta-\sin\left(z\delta\right)}{2z\delta}-z\delta\sin\left(z\delta\right)\sum_{n=1}^{\infty}  \frac{(-1)^n}{\left(z\delta\right)^2-n^2\pi^2}\cos\left(\frac{n\pi{x}}{\delta}\right)

    So now consider when x=\delta=\pi we are left with

    \sin^2\left(\frac{z\pi}{2}\right)=\frac{z\pi-\sin(z\pi)}{2z\pi}-z\pi\sin\left(z\pi\right)\sum_{n=1}^{\infty}\frac{  1}{z^2\pi^2-n^2\pi^2}

    Solving gives

    \forall{z}\notin2\mathbb{Z}~\frac{-\pi\sin^2\left(\frac{z\pi}{2}\right)^2}{z\sin(z\pi  )}+\frac{\pi}{2z\sin(z\pi)}-\frac{1}{2z^2}=\sum_{n=1}^{\infty}\frac{1}{z^2-n^2}

    So then we get finally that

    \forall{z}\notin2\mathbb{Z}~\frac{-\pi\sin^2\left(\frac{{z}\pi}{2}\right)^2}{z\sin(z\  pi)}+\frac{\pi}{2z\sin(z\pi)}+\frac{1}{2z^2}=\sum_  {n=0}^{\infty}\frac{1}{z^2-n^2}

    \boxed{\therefore~\sum_{n=0}^{\infty}\frac{1}{\pi-en^2}=\frac{\sin^2\left(\frac{\pi^{\frac{3}{2}}}{2  \sqrt{e}}\right)\sqrt{\pi}}{\sqrt{e}\sin\left(\fra  c{\pi^{\frac{3}{2}}}{\sqrt{e}}\right)}-\frac{\sqrt{\pi}}{2\sqrt{e}\sin\left(\frac{\pi^{\f  rac{3}{2}}}{\sqrt{e}}\right)}-\frac{1}{2\pi}}

    Note from this we can also glean that

    \forall{z}\notin2\mathbb{Z}~\sum_{n=0}^{\infty}\fr  ac{(-1)^n}{z^2-n^2}=\frac{\pi}{2z\sin(z\pi)}+\frac{1}{2z^2}
    Last edited by Mathstud28; November 22nd 2008 at 01:47 PM. Reason: typo
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Because it's not very different from the others you've posted.
    Hey! Don't be a Negative Nancy! Haha, your right though
    We have seen that: <br />
\sum\limits_{n = - \infty }^{ + \infty } {\tfrac{1}<br />
{{n^2 + x^2 }}} = \tfrac{{\pi \cdot \coth \left( {\pi \cdot x} \right)}}<br />
{x}<br />
which actually holds whenever <br />
x \in \mathbb{C}<br />
and <br />
n^2 + x^2 \ne 0{\text{ }}\forall n \in \mathbb{Z}<br />

    Change <br />
x \to i \cdot x<br />
and then set <br />
x = \sqrt {\tfrac{\pi }<br />
{e}} <br />

    See the Herglotz trick
    I like this solution because it is simplistic and powerful and in so beautiful. I dislike it because it doesn't take much thought. Picking a Fourier Series is much harder than one would imagine. Also, it can become repetitive, not that my Fourier Series are that much different each time though...and FS looks cooler Regardless, thanks again Paul for a nice solution.
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