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Math Help - Functions as power series

  1. #1
    Junior Member symstar's Avatar
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    Functions as power series

    My real question does not have to do with the problem itself so much as it has to do with a concept within problems like these... or, perhaps, series in general.

    f(x)=\frac{x^3}{(x-2)^2}

    Breaking it down into pieces, starting first with the following...
    \frac{1}{x-2}= \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}x^n

    Now to the next step (where my problems begin)...
    \frac{1}{(x-2)^2}
    =\frac{d}{dx}\Big[\frac{1}{x-2}\Big]
    =\frac{d}{dx}\Big[
    \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}x^n\Big]

    And now my problem...
    =\sum_{n=1}^{\infty}
    \frac{1}{2^{n+1}}nx^{n-1}

    My problem is in not knowing how we went from starting the sum at n=0 to suddenly starting it at n=1.

    Why does this happen and when should it happen?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by symstar View Post
    My real question does not have to do with the problem itself so much as it has to do with a concept within problems like these... or, perhaps, series in general.

    f(x)=\frac{x^3}{(x-2)^2}

    Breaking it down into pieces, starting first with the following...
    \frac{1}{x-2}= \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}x^n

    Now to the next step (where my problems begin)...
    \frac{1}{(x-2)^2}
    =\frac{d}{dx}\Big[\frac{1}{x-2}\Big]
    =\frac{d}{dx}\Big[
    \sum_{n=0}^{\infty} \frac{1}{2^{n+1}}x^n\Big]

    And now my problem...
    =\sum_{n=1}^{\infty}
    \frac{1}{2^{n+1}}nx^{n-1}

    My problem is in not knowing how we went from starting the sum at n=0 to suddenly starting it at n=1.

    Why does this happen and when should it happen?
    Because
    =\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}nx^{n-1}
    and
    =\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}nx^{n-1} are the same sum! There is an "n" multiplied in each term. When n= 0, that term is 0.
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  3. #3
    Junior Member symstar's Avatar
    Joined
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    Berkeley, CA
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    Quote Originally Posted by HallsofIvy View Post
    Because
    =\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}nx^{n-1}
    and
    =\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}nx^{n-1} are the same sum! There is an "n" multiplied in each term. When n= 0, that term is 0.
    Ok, so essentially whenever n=0 makes the term 0, I should make n=1? ...such that:

    =\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}nx^{n-1}
    =\sum_{n=1}^{\infty}\frac{1}{2^{n+1}}nx^{n-1}
    =\sum_{n=0}^{\infty}\frac{n+1}{2^{n+2}}nx^n

    Whenever I don't do this, I'm getting the problems wrong.
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