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Math Help - Help finding high order derivatives using Taylor series?

  1. #1
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    Help finding high order derivatives using Taylor series?

    I need to find the 7th derivative f^{(7)}(0):

    f(x)=(1/14)x cos(2x)

    I know it has something to do with Taylor/Maclaurin series, but I don't know the process. Can someone fill me in?
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  2. #2
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    Quote Originally Posted by Mattpd View Post
    I need to find the 7th derivative f^{(7)}(0):

    f(x)=(1/14)x cos(2x)

    I know it has something to do with Taylor/Maclaurin series, but I don't know the process. Can someone fill me in?

    If you know the MacClaurin series for \cos 2x then you're done:

    \cos 2x =\sum^\infty_{k=0}(-1)^k\,\frac{(2x)^{2k}}{(2k)!}\Longrightarrow \frac{1}{14}x\cos 2x= \frac{1}{14}\sum^\infty_{k=0}(-1)^k\,\frac{2^{2k}x^{2k+1}}{(2k)!} ...

    Tonio
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  3. #3
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     \cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - ...

     \cos 2x = 1 - \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} - \frac{(2x)^{6}}{6!} +  \frac{(2x)^{8}}{8!} - ...

     \frac{1}{14} x \cos 2x = \frac{1}{14} \Big(x - \frac{2^{2}x^{3}}{2!} + \frac{2^{4}x^{5}}{4!} - \frac{2^{6}x^{7}}{6!} -   \frac{2^{8}x^{9}}{8!} - ...\Big)

    the coefficient of the  x^{7} term is  \frac{f^{(7)}(0)}{7!}

    so \frac{f^{(7)}(0)}{7!} = - \frac{1}{14} \frac{2^{6}}{6!}

    which implies that  f^{(7)}(0) = -32
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    I understand up until you state the coefficient of the x^{7} term is \frac{f^{(7)}(0)}{7!}. Can you more simply explain what this means and how you ended up with the term you did?
    Last edited by Mattpd; April 29th 2010 at 08:49 PM.
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  5. #5
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    Quote Originally Posted by Mattpd View Post
    I understand up until you state the coefficient of the x^{7} term is \frac{f^{(7)}(0)}{7!}. Can you more simply explain what this means and how you ended up with the term you did?
    The Taylor series of a function f(x) about x=0 is  {f(0)} + {f'(0)}x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + ...

    so for the  x^{7} term the coefficient is  \frac{f^{(7)}(0)}{7!}

    the coefficient of the  x^{7} term in our series is  -\frac{1}{14} \frac{2^{6}}{6!} , but this has to be equal to  \frac{f^{(7)}(0)}{7!}
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