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Math Help - maclaurian's expansion of 4 sinh x

  1. #1
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    maclaurian's expansion of 4 sinh x

    hie ...
    I would like to know how do you solve maclaurian's expansion of 4 sinh x. Is it the same as maclaurian's expansion of 4 sin x or not ?
    thankx..
    please reply as soon as possible
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by aero-me View Post
    hie ...
    I would like to know how do you solve maclaurian's expansion of 4 sinh x. Is it the same as maclaurian's expansion of 4 sin x or not ?
    thankx..
    please reply as soon as possible
    The answer is 4\sinh(x)=4\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+  1)!}

    To derive it either use the fact that \sinh(x)=\frac{e^x-e^{-x}}{2} and use both those Maclaurin series and add them together...or even more slickly note that \frac{\sin(ix)}{i}=\sinh(x)
    Last edited by Mathstud28; December 20th 2008 at 10:38 AM. Reason: Typo
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    The answer is 4\sinh(x)=4\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+  1)!}

    To derive it either use the fact that \sinh(x)=\frac{e^x+e^{-x}}{2} and use both those Maclaurin series and add them together...or even more slickly note that \frac{\sin(ix)}{i}=\sinh(x)
    No,

    \sinh(x)=\frac{e^x-e^{-x}}{2}

    But the series still looks OK

    CB
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  4. #4
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    hey thankx fer yur post......its relli worked out now..
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