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Thread: Calculus Power Series

  1. #1
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    Calculus Power Series

    My question looks like this:

    f(x)= 5/ (1-5x)^2

    Find
    C0=
    c1=
    C2=
    C3=
    C4=

    Any help is much appreciated.
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  2. #2
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    Quote Originally Posted by BKennedy View Post
    My question looks like this:

    f(x)= 5/ (1-5x)^2
    .
    If $\displaystyle - 1 < \xi < 1$ then $\displaystyle \frac{1}{1-\xi } = \sum_{n=0}^{\infty} \xi^n$.

    Term-by-term differenciation implies,
    $\displaystyle \frac{1}{(1-\xi)^2} = \sum_{n=0}^{\infty}(n+1)\xi ^n$.

    Thus if $\displaystyle -1 < 5x < 1 \implies -\frac15 < x <\frac15$ then,
    $\displaystyle \frac{1}{(1-5x)^2} = \sum_{n=0}^{\infty}5^n(n+1)x^n$

    Thus,
    $\displaystyle \frac{5}{(1-5x)^2} = \sum_{n=0}^{\infty} 5^{n+1} (n+1)x^n$.

    Its Taylor series is $\displaystyle \sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}$.

    This means, $\displaystyle f^{(n)}(0) = (n+1)! 5^{n+1} $.
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