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Thread: Value of nth derivative using Malcurian Series

  1. #1
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    Value of nth derivative using Malcurian Series

    I have a question:
    Given: $\displaystyle f(x)=x^7 \cos(x^9)$

    Find $\displaystyle f^{79}(0)$, the 79-th derivative of f(x) at x=0

    I know the answer is $\displaystyle \frac{79!}{8!}$ but I am not sure of all the steps.

    I found the Taylor Series representation to be:
    $\displaystyle \sum\frac{(-1)^nx^{18n+7}}{2n!}$

    Any enlightenment would be much appreciated.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Planeman View Post
    I have a question:
    Given: $\displaystyle f(x)=x^7 \cos(x^9)$

    Find $\displaystyle f^{79}(0)$, the 79-th derivative of f(x) at x=0

    I know the answer is $\displaystyle \frac{79!}{8!}$ but I am not sure of all the steps.

    I found the Taylor Series representation to be:
    $\displaystyle \sum\frac{(-1)^nx^{18n+7}}{2n!}$

    Any enlightenment would be much appreciated.
    We know that $\displaystyle \frac{f^{79}(0)}{n!}$ is the coefficient of the term containing $\displaystyle x^{79}$ term in the power series. Noting though that $\displaystyle 18n+7=79\implies n=4$. Thus $\displaystyle \frac{f^{79}(0)}{79!}=\frac{(-1)^4}{(2\cdot4)!}\implies f^{79}(0)=\frac{79!}{8!}$.
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    Thank you so much, now it all makes sense.
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