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Math Help - Maclaurin series question

  1. #1
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    Smile Maclaurin series question

    How can I expand these expressions with Maclaurin series :

    sin(sinx)

    sin(cosx)
    Last edited by mr fantastic; December 27th 2009 at 03:58 PM. Reason: Clarified question
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  2. #2
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    Quote Originally Posted by gilyos View Post
    How can I expand these expressions with Maclaurin series :

    sin(sinx)

    sin(cosx)
    The same way you would expand anything else.

    Find  f'(x), f''(x), f'''(x),...,f^n(x) using the chain rule.

    For example,  \frac{d}{dx} \sin(\cos(x)) = \cos(\cos(x)) \times (-\sin(x)) = - \sin(x) \cos(\cos(x))

    Then from these, find  f'(0), f''(0), f'''(0),...,f^n(0)

    For example,  f'(0) = -\sin(0) \cos((\cos(0)) = 0

    And then insert them into the basic formula:

     f(x) = f(0) + f'(0) x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6} x^3... + \frac{f^n (0)}{n!} x^n

    Which part of this process are you having trouble with?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by gilyos View Post
    How can I expand these expressions with Maclaurin series :

    sin(sinx)

    sin(cosx)
    This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin(  x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    This is not easy in the way that you'd think. Finding the exact Maclaurin series for this is not easy. But, note that \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}x^{2n+1}}{(2n+1)!}\implies\sin\left(\sin(  x)\right)=\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}\sin^{2n+1}(x)}{(2n+1)!}
    But that last is not a Maclaurin series. What is your point?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    But that last is not a Maclaurin series. What is your point?
    I never said it was? I was merely pointing out another possible root of attack.
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