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Thread: 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 $\displaystyle f^{(7)}(0)$:

    $\displaystyle f(x)=(1/14)x cos(2x)$

    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 $\displaystyle f^{(7)}(0)$:

    $\displaystyle f(x)=(1/14)x cos(2x)$

    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 $\displaystyle \cos 2x$ then you're done:

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

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

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

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

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

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

    which implies that $\displaystyle f^{(7)}(0) = -32 $
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  4. #4
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    I understand up until you state the coefficient of the $\displaystyle x^{7}$ term is $\displaystyle \frac{f^{(7)}(0)}{7!}$. Can you more simply explain what this means and how you ended up with the term you did?
    Last edited by Mattpd; Apr 29th 2010 at 07:49 PM.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Mattpd View Post
    I understand up until you state the coefficient of the $\displaystyle x^{7}$ term is $\displaystyle \frac{f^{(7)}(0)}{7!}$. Can you more simply explain what this means and how you ended up with the term you did?
    The Taylor series of a function $\displaystyle f(x)$ about $\displaystyle x=0$ is $\displaystyle {f(0)} + {f'(0)}x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + ... $

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

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