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Math Help - Computing a sum using Parseval's theorem

  1. #1
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    Computing a sum using Parseval's theorem

    Hi,

    I have the following 2\pi-periodic function:

    \displaystyle g(t)=e^{-t}   ,\,\,0 < t < 2\pi

    I've computed the Exponential Fourier Series for the given function:

    \displaystyle{g(t)=\sum_{k=-\infty}^{\infty}c_{k}e^{ik\Omega t}}

    \displaystyle c_{k}={\frac{1-e^{-2\pi}}{{2\pi}(1+ik)}

    \displaystyle\implies{g(t)=\sum_{k=-\infty}^{\infty}({\frac{1-e^{-2\pi}}{{2\pi}(1+ik)})e^{ikt}} (1)
    ----------------------------------------------------------------------

    I need to compute the following sum:

    \displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}}

    -----------------------------------------------------------------------

    The Parseval's identity is:

    \displaystyle \frac{1}{2\pi} \int_{0}^{2\pi}|f(t)|^{2}dt=\sum_{k=-\infty}^{\infty}|c_{k}(f)|^{2}

    \displaystyle \frac{1}{2\pi} \int_{0}^{2\pi}|f(t)|^{2}dt=\frac{1}{2\pi} \int_{0}^{2\pi}e^{-2t}dt={\frac{1}{4\pi}(1-e^{-4\pi}) (2)

    \displaystyle  |c_{k}(f)|={\frac{1-e^{-2\pi}}{{2\pi}(1+k)}(3)

    \displaystyle |c_{k}(f)|^{2}=({\frac{1-e^{-2\pi}}{{2\pi}(1+k)})^{2}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+k)^{2}}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+k)(1+k)}={\frac{(1-e^{-2\pi})^{2}}{{4\pi^{2}}(1+2k+k^{2})}(4)
    ---------------------------------------------------------------------------------------------------------------------------


    What is:

    \displaystyle |{{2\pi}(1+ik)}|

    Isn't it:

    \displaystyle {{2\pi}(1+k)}

    Is (3) and (4) correct?


    I would greatly appriciate guidance to compute the sum.

    Thank you


    Last edited by 4Math; January 9th 2011 at 02:51 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If \sigma= \alpha + i\ \beta is any complex number, then |\sigma|^{2} = \sigma\ \sigma^{*} = \alpha^{2} + \beta^{2}. If...

    \displaystyle \sigma=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} (1)

    ... what is \sigma^{*} and |\sigma|^{2} ?...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post

    ... what is \sigma^{*} and |\sigma|^{2} ?...
    \sigma^{*}=(1-ik)

    and

    \displaystyle|(1-ik)|^{2}=\sqrt{1^{2}+k^{2}}=(1+k)^{2}

    therefore:

    |\displaystyle\sigma|^{2}={\frac{(1-e^{-2\pi})^{2}(1+k)^{2}}{{4\pi^{2}}(1+k^{2})^{2}}

    Kind regards
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    If \sigma= \alpha + i\ \beta is any complex number, then |\sigma|^{2} = \sigma\ \sigma^{*} = \alpha^{2} + \beta^{2}. If...

    \displaystyle \sigma=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} (1)

    ... what is \sigma^{*} and |\sigma|^{2} ?...

    Kind regards

    \chi \sigma
    \displaystyle \sigma=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} \implies \sigma^{*}=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1- i\ k)} \implies |\sigma|^{2} = \sigma\ \sigma^{*} =  \frac{(1-e^{-2\ \pi})^{2}}{4\ \pi^{2}\ (1+k^{2})}

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    \displaystyle \sigma=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1+ i\ k)} \implies \sigma^{*}=  \frac{1-e^{-2 \pi}}{2\ \pi\ (1- i\ k)} \implies |\sigma|^{2} = \sigma\ \sigma^{*} =  \frac{(1-e^{-2\ \pi})^{2}}{4\ \pi^{2}\ (1+k^{2})}
    If the denominator is multiplied by  \ (1- i\ k) don't I need to also multiply the numerator with  \ (1- i\ k) ??...

    Kind regards
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  6. #6
    MHF Contributor chisigma's Avatar
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    Once You have \sigma and \sigma^{*} the only You have to do is multiplying them, 'numerator x numerator' and 'denominator x denominator' independently...

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by chisigma View Post
    Once You have \sigma and \sigma^{*} the only You have to do is multiplying them, 'numerator x numerator' and 'denominator x denominator' independently...
    Ok, thank you kindly for clarifying that.

    I get the following resulat:

    \displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{1+k^{2}}={\frac{{\pi}(1+e  ^{-2\pi})}{1-e^{-2\pi}} (5)


    In (5) the sum goes from {k=-\infty} to {k=\infty} according to Parseval's identity.

    ...but the given sum goes from {k=0 to {k=\infty}.

    How do I compute:

    \displaystyle \sum_{k=0}^{\infty}\frac{1}{1+k^{2}}

    Kind regards
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by 4Math View Post
    Ok, thank you kindly for clarifying that.

    I get the following resulat:

    \displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{1+k^{2}}={\frac{{\pi}(1+e  ^{-2\pi})}{1-e^{-2\pi}} (5)


    In (5) the sum goes from {k=-\infty} to {k=\infty} according to Parseval's identity.

    ...but the given sum goes from {k=0 to {k=\infty}.

    How do I compute:

    \displaystyle \sum_{k=0}^{\infty}\frac{1}{1+k^{2}}

    Kind regards
    If the result You have found...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha (1)

    ... is correct, the procedure to obtain the final result is...

    \displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2} (2)

    Kind regards

    \chi \sigma
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  9. #9
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    Quote Originally Posted by chisigma View Post
    If the result You have found...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha (1)

    ... is correct, the procedure to obtain the final result is...

    \displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2} (2)
    Is it:

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}=\sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}}+ \sum_{k= 0}^{+ \infty} \frac{1}{1+k^{2}}=\sum_{n= 1}^{+ \infty} \frac{1}{1+n^{2}}+\sum_{k=0}^{\infty} \frac{1}{1+k^{2}} =\sum_{n= 0}^{+ \infty} \frac{1}{1+n^{2}}+\frac{1}{2}+\sum_{k=0}^{\infty} \frac{1}{1+k^{2}}

    I am just trying to understand how to get to:

    \displaystyle\frac{\alpha+1}{2}

    Kind regards
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  10. #10
    MHF Contributor chisigma's Avatar
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    One and half year ago I opened this thread...

    http://www.mathhelpforum.com/math-he...ies-95343.html

    ... with the task of verify the result I obtained in the Fourier series expansion of the function \cosh t in -\pi < t < \pi. Opalg and DeMath confirmed the following result...

    \displaystyle \cosh t = \frac{2\ \sinh \pi}{\pi}\ \{\frac{1}{2} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{1+n^{2}}\ \cos n t \} (1)

    Now if You set in (1) t= \pi with some easy steps You obtain...

    \displaystyle \sum_{n=0}^{\infty} \frac{1}{1+n^{2}} = \frac{1 + \pi\ \coth \pi}{2} = 2.07667404746858... (2)

    A simple computer program can verify the convergence of (2) to that value. It seems to me however that with the approach used by 4Math we arrive at a different result , and I'm not able to explain why. Regarding the series (2) it is remarkable the fact that in F. Sheid, Numerical Analysis, McGraw-Hill 1968 in the prob. 17.15 [pag. 162] You can read regarding series (2)...

    ... Stegun & Abramowitz [Journal of SIAM, 1956] demonstrated that 5745 terms are needed for three exact decimals...

    Well!... my computer for three exact decimals needs 'only' 1484 terms... did something change in the last 55 years? ...

    Kind regards

    \chi \sigma
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    If the result You have found...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}}= \frac{{\pi}(1+e^{-2\pi})}{1-e^{-2\pi}} = \alpha (1)

    ... is correct, the procedure to obtain the final result is...

    \displaystyle \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2} (2)

    Kind regards

    \chi \sigma
    In detail...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha \implies

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha-1}{2} \implies \sum_{k=0}^{\infty} \frac{1}{1+k^{2}} = \frac{\alpha+1}{2}

    Kind regards

    \chi \sigma
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  12. #12
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    Quote Originally Posted by chisigma View Post
    In detail...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha
    I don't follow...

    Where does 1 in the right hand side comes in and why do we double the sum: \displaystyle2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}

    I greatly appriciate for an explanation.

    Thank you
    Kind regards
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  13. #13
    MHF Contributor chisigma's Avatar
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    ... because k and - k both generate the same contribute \displaystyle \frac{1}{1+k^{2}} ...

    Kind regards

    \chi \sigma
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  14. #14
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    Quote Originally Posted by chisigma View Post

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}}= \alpha
    I would like to ask you kindly if you could please write out the sums that results in 1+2\sum_{n=1}^{\infty} \frac{1}{1+k^{2}} in detail:

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = .........=1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \alpha

    Kind regards
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  15. #15
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by 4Math View Post
    I would like to ask you kindly if you could please write out the sums that results in 1+2\sum_{n=1}^{\infty} \frac{1}{1+k^{2}} in detail:

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = .........=1 + 2\ \sum_{n=1}^{\infty} \frac{1}{1+k^{2}} = \alpha

    Kind regards
    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} =\sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}} + 1 + \sum_{1}^{+ \infty} \frac{1}{1+k^{2}}

    ... but is...

    \displaystyle \sum_{k= -\infty}^{-1} \frac{1}{1+k^{2}} = \sum_{k= 1}^{+ \infty} \frac{1}{1+k^{2}}

    ... so that...

    \displaystyle \sum_{k= -\infty}^{+ \infty} \frac{1}{1+k^{2}} = 1 + 2 \sum_{k= 1}^{\infty} \frac{1}{1+k^{2}}

    Kind regards

    \chi \sigma
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