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Math Help - Mc-Laurin series of sin^3 sqrt(x).

  1. #1
    Newbie dttah's Avatar
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    Mc-Laurin series of sin^3 sqrt(x).

    Hello everyone, I would like to ask a question.

    Knowing that:

    sinx = x - \frac{x^3}{6} +o(x^3)

    How do I find..:
    sin^3 \sqrt(x)

    sin \sqrt(x) = \sqrt(x) - \frac{\sqrt(x^3)}{6}

    Woops... I just realized how it should be done, I just had to do that series cubed.
    i can't delete the post, sorry for bothering guys. >.<'
    Last edited by dttah; December 11th 2011 at 10:14 AM. Reason: I understood how >.< sorry.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Mc-Laurin series of sin^3 sqrt(x).

    The function  f(x)= \sin \sqrt{x} is not analytic in x=0 because its derivative doesn't exist here... the same is for f^{3}(x)...

    Kind regards

    \chi \sigma
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