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Math Help - sum of taylor series

  1. #1
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    sum of taylor series

    find the taylor sum of sigma from 0 to infinity of ((-1)^n)(2)^(n +3)/(3^(2n + 1)
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  2. #2
    MHF Contributor chisigma's Avatar
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    The general term is...

    a_{n}=(-1)^{n}\cdot \frac{2^{n+3}}{3^{2n+1}}= \frac{8}{3}\cdot (-1)^{n}\cdot (\frac{2}{9})^{n}

    ... so that is a geometric series and is...

    \sum_{n=0}^{\infty}a_{n} = \frac{8}{3}\cdot \frac{1}{1+\frac{2}{9}}=\frac{24}{11}

    Kind regards

    \chi \sigma
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by twilightstr View Post
    find the taylor sum of sigma from 0 to infinity of ((-1)^n)(2)^(n +3)/(3^(2n + 1)
    \sum_{n=0}^{\infty}\frac{(-1)^n 2^{n+3}}{3^{2n+1}} = \sum_{n=0}^{\infty}\frac{(-1)^n\cdot 2^3\cdot 2^n}{3\cdot 3^{2n}} = \frac{8}{3}\sum_{n=0}^{\infty}\frac{(-1)^n 2^n}{(3^2)^n} = \frac{8}{3}\sum_{n=0}^{\infty}\left(\frac{-2}{9}\right)^n = \frac{8}{3}\cdot\frac{1}{1-(-2/9)} = \frac{8}{3}\cdot\frac{9}{11} = \frac{72}{33} = \boxed{\frac{24}{11}}

    EDIT: I was beat to the chase!
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  4. #4
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    taylor series

    find the taylor sum of sigma from 0 to infinity of ((-1)^n)(2)^(n +3)/(3^(2n + 1)(2n + 1) sorry i forget to add what was in bold. so that is the actual problem
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by twilightstr View Post
    find the taylor sum of sigma from 0 to infinity of ((-1)^n)(2)^(n +3)/(3^(2n + 1)(2n + 1) sorry i forget to add what was in bold. so that is the actual problem
    If this is what you mean then

    \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \frac{2^{n+3}}{3^{2n+1}}

    Then we can fix it up by rewriting it as

    \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \frac{(\sqrt{2})^{2n+6}}{3^{2n+1}}

    \sum_{n=0}^{\infty}\frac{(-1)^{n}(4\sqrt{2})}{(2n+1)}\cdot \left( \frac{\sqrt{2}}{3} \right)^{2n+1}

    So we can identify this with the function

    f(x)= (4\sqrt{2})\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)}\cdot \left( x \right)^{2n+1}

    So now we can take the derivative to get

    f'(x)= (4\sqrt{2})\sum_{n=0}^{\infty}(-1)^{n}\cdot \left( x \right)^{2n}

    This is a geometric series so

    f'(x)= (4\sqrt{2})\sum_{n=0}^{\infty}(-1)^{n}\cdot \left( x \right)^{2n} =4\sqrt{2} \cdot \frac{1}{1+x^2}

    So integrating we get

    f(x)=4\sqrt{2}\tan^{-1}(x)

    So now just evaluate this at x=\frac{\sqrt{2}}{3}
    Last edited by TheEmptySet; May 3rd 2009 at 09:58 AM. Reason: Typo Thanks Redsoxfan325 :)
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  6. #6
    Super Member redsoxfan325's Avatar
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    You have a typo near the end. It's f(x) = 4\sqrt{2}\tan^{-1}(x). Clever solution though.
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