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Math Help - A sum

  1. #1
    Math Engineering Student
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    A sum

    Evaluate \sum_{k=0}^\infty\frac1{(2k+1)^2}.

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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Evaluate \sum_{k=0}^\infty\frac1{(2k+1)^2}.

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    Solution 1: Let f(z) = \frac{1}{(2z+1)^2} it has a singularity at z=-1/2. That means \sum_{n\in \mathbb{Z}} \frac{1}{(2n+1)^2} = - \mbox{res}\left( \pi f(z) \cot\pi z, -1/2 \right).

    The residue is \lim_{z\to -1/2} \left( \frac{\pi \cot \pi z(z+1/2)^2}{(2z+1)^2}\right)'  = \frac{\pi}{4} \lim_{z\to -1/2} \left( \cot \pi z \right)' = - \frac{\pi^2}{4} \csc^2 \pi z.

    So the sum is (if I did not make any computational mistakes) \frac{\pi^2}{8}.
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  3. #3
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    Solution 2: We know \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}. This means \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{\pi^2}{6}.
    But \sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{24}. This means \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}.
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  4. #4
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    Okay.

    Here's a solution that I really love:

    \sum\limits_{k = 0}^\infty  {\frac{1}<br />
{{(2k + 1)^2 }}}  = \int_0^1 {\int_0^1 {\left\{ {\sum\limits_{k = 0}^\infty  {\left( {x^2 y^2 } \right)^k } } \right\}\,dx} \,dy}  = \int_0^1 {\int_0^1 {\frac{{dx\,dy}}<br />
{{1 - x^2 y^2 }}} } .

    Substitute (x,y) = \left( {\frac{{\sin u}}<br />
{{\cos v}},\frac{{\sin v}}<br />
{{\cos u}}} \right). The Jacobian matrix is 1-x^2y^2, and the integral becomes

    \iint_A {du\,dv}, where A is the triangle with coordinates (0,0),\,\bigg( {\frac{\pi }<br />
{2},0} \bigg),\,\bigg( {0,\frac{\pi }<br />
{2}} \bigg). Finally

    \begin{aligned}<br />
  \sum\limits_{k = 0}^\infty  {\frac{1}<br />
{{(2k + 1)^2 }}}  &= \int_0^{\pi /2} {\int_0^{\pi /2 - v} {\,du} \,dv}\\<br />
   &= \int_0^{\pi /2} {\bigg( {\frac{\pi }<br />
{2} - v} \bigg)\,dv}\\<br />
   &= \frac{{\pi ^2 }}<br />
{8},<br />
\end{aligned}

    as required.
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  5. #5
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    Haha, I know where you got that from. My professor used that too when we were doing the Jacobian.
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  6. #6
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    Can this be solved using the Riemann Zeta function?
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  7. #7
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    Quote Originally Posted by colby2152 View Post
    Can this be solved using the Riemann Zeta function?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Sorry I'm bored.

    Consider some function f. Let f(x)=|x|\quad\forall{x}\in[-\pi,\pi]. Let the function also posses the charcteristic f\left(x+2\pi\right)=f(x)\quad\forall{x}\in\mathbb  {R}. So now we seek a Fourier Series to describe this function. This series will be of the form

    f(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 \delta is a number such that f\left(x+2\delta\right)=f(x)

    A_0=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)dx

    A_n=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)\cos\left(\frac{n\pi{x}}{\delt  a}\right)dx

    B_n=\frac{1}{\delta}\int_{-\delta}^{\delta}f(x)\sin\left(\frac{n\pi{x}}{\delt  a}\right)dx

    So for f(x)=|x| and \delta=\pi we have

    f(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx+\sum_{n=1}^{\infty}\bigg[\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos(nx)dx\cos(n{x})+\frac{1}{\pi}\in  t_{-\pi}^{\pi}|x|\sin(nx)dx\sin(nx)\bigg]

    Now \frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx=\frac{\pi}{2}

    \frac{1}{\pi}\int_{-\pi}^{\pi}|x|\cos(nx)dx

    =\frac{2}{\pi}\cdot\frac{\cos(n\pi)-1}{n^2}

    {{=}}{{0\quad{n\text{ is even}}}\brace{\frac{-4}{\pi{n^2}}\quad{n}\text{ is odd}}}

    And |x|\cdot{\sin(nx)} is odd which implies that

    B_n=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|\sin(nx)dx=0

    \therefore|x|=\frac{\pi}{2}+\frac{2}{\pi}\sum_{n=0  }^{\infty}\frac{(-1)^n-1}{n^2}\cos(nx)

    =\frac{\pi}{2}+\frac{2}{\pi}\sum_{n=1,3,5\cdots}^{  \infty}\frac{(-1)^n-1}{n^2}\cos(nx)+\frac{2}{\pi}\sum_{n=2,4,6\cdots}^  {\infty}\frac{(-1)^n-1}{n^2}\cos(nx)

    =\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1,3,5,\cdots}^{\infty}\frac{\  cos(nx)}{n^2}+\sum_{n=2,4,6,\cdots}^{\infty}0

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

    Therefore we can see that

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

    Solving gives

    \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}  {8}\quad\blacksquare

    I'm sure this is solution is common...I'm not sure though...I found this today while tinkering.
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  9. #9
    Moo
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    I'm sure this is solution is common...I'm not sure though...I found this today while tinkering.
    Two or three teachers used Fourier series, last year, to show us that \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}, so nothing that uncommon :P
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Two or three teachers used Fourier series, last year, to show us that \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}, so nothing that uncommon :P
    Yeah, I found that while tinkering too. I just found the fourier series for some functions and these presented themselves as particulars of these. I ended up calculating the even values \zeta(x) up to 10. Anyway, I have always been more fan of comparing the coefficents of

    \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x\prod_{n=1}^{\infty}\bigg(  1-\frac{x^2}{n^2\pi^2}\bigg)
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  11. #11
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    Quoting MOO:

    Two or three teachers used Fourier series, last year, to show us that , so nothing that uncommon :P
    Last edited by ThePerfectHacker; November 12th 2008 at 05:05 AM.
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