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Math Help - Computing a sum involving sin

  1. #1
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    Computing a sum involving sin

    I have trouble solving the following sum:

    <br />
\displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4\frac{  sin^{2}{k}}{k^{2}}

    I've tried the following:

    \displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4(\frac  {sin{k}}{k})^{2}

    The first term is:

    <br />
\displaystyle4(\frac{sin{1}}{1})^{2}=4(\frac{\pi}{  2})^{2}=2{\pi}^{2}

    The sum is than:

    \displaystyle\frac{1}{2}\sum_{k=1}^{\infty}4\frac{  sin^{2}{k}}{k^{2}}=\frac{1}{2}2{\pi}^{2}={\pi}^{2} but this is incorrect.

    Any help to compute this sum would be appriciate it. Thank you
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  2. #2
    MHF Contributor chisigma's Avatar
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    In...

    http://www.mathhelpforum.com/math-he...on-153971.html

    ... it has been demonstrated that the 'mirror imaging' of the function...

    \displaystyle f(x)=\left\{\begin{array}{ll}2-x ,\,\,0 < x < 2\\{}\\0 ,\,\, 2 < x< \pi\end{array}\right. (1)

    ... has the following Fourier series expansion...

    \displaystyle f(x)= \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx (2)

    Setting x=0 in (2) You obtain...

    \displaystyle \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} = 2 (3)

    ... so that is...

    \displaystyle \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} =\frac{\pi -1}{2} = 1.07079632679... (4)

    Kind regards

    \chi \sigma
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  3. #3
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    ... has the following Fourier series expansion...

    \displaystyle f(x)= \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx (2)

    Setting x=0 in (2) You obtain...

    \displaystyle\frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}} = 2 (3)
    So to compute the following sum:

    \displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}


    for this fourier expansion (using Parseval):

    \displaystyle{\frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^  {2}dx=\frac{4}{\pi^2} + \frac{16}{\pi^2}\ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = 2

    ..and I get the following:

    \displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}=\frac{\pi^2 -2}{8}=\frac{\pi^2}{8}-\frac{1}{4}

    But this is incorrect answer. What am I doing wrong here?

    By the way, thank you for your response.

    Kind regards
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  4. #4
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    I think I found my mistake for computing the following sum:

    \displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}

    using the following Fourier series

    \displaystyle \frac{2}{\pi} + \frac{4}{\pi}\ \sum_{n=1}^{\infty} \frac{\sin^{2} n}{n^{2}}\ \cos nx

    \displaystyle a_{0} = \frac{2}{\pi}

    by using Parseval's theorem:

    \displaystyle{\frac{2}{\pi}= {\frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^{2}dx=\frac{4  }{\pi^2} + {\frac{1}{2}\ \sum_{n=1}^{\infty} \frac{16}{\pi^2}\frac{\sin^{4} n}{n^{4}}

    \displaystyle\implies \displaystyle{\frac{2}{\pi}= \frac{4}{\pi^2} +  \frac{8}{\pi^2}\ \sum_{n=1}^{\infty}\frac{\sin^{4} n}{n^{4}} \displaystyle\implies

    \displaystyle\implies \displaystyle\sum_{n=1}^{\infty}\frac{\sin^{4} n}{n^{4}}=\frac{2\pi -4}{8}=\frac{\pi}{4}-\frac{1}{2}

    Please let me know if this is not correct. Thank you
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  5. #5
    MHF Contributor chisigma's Avatar
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    The Parseval's identity is...

    \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n}) (1)

    ... and in Your case is...

    a_{0}=\frac{4}{\pi} , a_{n}=\frac{4}{\pi}\ \frac{\sin^{2} n}{n^{2}} , b_{n}=0 (2)

    ... and ...

    \displaystyle \int_{-\pi}^{\pi} f^{2} (x)\ dx = 2\ \int_{0}^{2} (4 - 4\ x + x^{2})\ dx = \frac{16}{3} (3)

    ... so that is...

    \displaystyle \frac{2}{3\ \pi} =\frac{1}{\pi^{2}} + \frac{2}{\pi^{2}} \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} \implies \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = \frac{\pi}{3} - \frac{1}{2} = .547197551196... (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 17th 2010 at 07:14 AM. Reason: transcription errors...
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  6. #6
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    Quote Originally Posted by chisigma View Post
    The Parseval's identity is...

    \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n}) (1)
    Isn't it: \displaystyle \frac{a_{0}^{2}}{2} in the right hand side?

    \displaystyle \int_{-\pi}^{\pi} f^{2} (x)\ dx = 2\ \int_{0}^{2} (4 - 4\ x + x^{2})\ dx = \frac{16}{3} (3)
    Why do you double the integral? Is it because of that the given function is even?

    Kind regards
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by 4Math View Post
    Isn't it: \displaystyle \frac{a_{0}^{2}}{2} in the right hand side? Yes!... thank You very much!... typical transcription error!... [/COLOR]

    Why do you double the integral? Is it because of that the given function is even? Yes again!...

    Kind regards
    Kind regards

    \chi \sigma
    Last edited by chisigma; September 18th 2010 at 12:03 AM. Reason: added signature...
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  8. #8
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    Quote Originally Posted by chisigma View Post
    a_{0}=\frac{4}{\pi}
    Since the given function is even, continuous and its periodic with period {2\pi}, I get the following when I compute a_{0} ...

    \displaystyle a_{0}= {\frac{2}{T}\int_{0}^{{\frac{T}{2}}} f(x) dx \displaystyle\implies<br />

    in this case T={2\pi} ...

    \displaystyle\implies a_{0}= {\frac{2}{2\pi}(2\int_{0}^{2} (2-x) dx)=\frac{4}{\pi}

    Am I using a correct formula to compute a_{0}?
    ------------------------------------------------------------------------------------------------------
    The Parseval's identity is...

    \displaystyle \frac{1}{\pi} \int_{-\pi}^{\pi} f^{2} (x)\ dx= \frac{a_{0}^{2}}{2} + \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n}) (1)
    In our course book says that the Parseval's identity is:
    \displaystyle \frac{1}{T} \int_{P} f^{2} (x)\ dx= {a_{0}^{2}} + \frac{1}{2} \sum_{n=1}^{\infty} (a^{2}_{n} + b^{2}_{n}) (4)

    But when I apply (4) to the the same problem above I get a different result. Why is that? But your result \displaystyle \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}} = \frac{\pi}{3}- \frac{1}{2} is the correct result according to the answer in the problem book.

    I used T=\pi and the limits for the integral [0,2]. I also tried with T=2\pi, but still get different result.
    ---------------------------------------------------------------------------------------------------------------
    \displaystyle \frac{2}{3\ \pi} =\frac{1}{\pi^{2}} + \frac{2}{\pi^{2}} \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}
    I would like to ask you kindly to write out all the terms without simplifications in other words:


    \displaystyle \frac{16}{3\ \pi} =... + ... \ \sum_{n=1}^{\infty} \frac{\sin^{4} n}{n^{4}}<br />


    Thank you
    Kind regards
    Last edited by 4Math; September 20th 2010 at 03:57 AM.
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