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Math Help - Power series

  1. #1
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    Power series

    Find a power series representation for the function

    f(x) = x^3 / (x - 2)^2
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    Find a power series representation for the function

    f(x) = x^3 / (x - 2)^2
    Ok first we note that


    \frac{d}{dx}\bigg[\frac{1}{x-2}\bigg]=\frac{-1}{(x-2)^2}

    But we also know that


    \frac{1}{x-2}=\frac{-1}{2}\frac{1}{1-\frac{x}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\ri  ght)^n=-\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}

    So we have

    -x^3\cdot\bigg[-\sum_{n=0}^{\infty}\frac{x^n}{2^{n+1}}\bigg]=x^3\cdot\sum_{n=1}^{\infty}\frac{nx^{n-1}}{2^{n+1}}=\sum_{n=0}^{\infty}\frac{nx^{n+2}}{2^  {n+1}}

    \therefore\frac{x^3}{(x-2)^2}=\sum_{n=0}^{\infty}\frac{nx^{n+2}}{2^{n+1}}
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  3. #3
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    Ok, so when the n becomes 1 in the Sigma, that means that you took the derivative? And can you only multiply the x^3 after you do that?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by LeoBloom. View Post
    Ok, so when the n becomes 1 in the Sigma, that means that you took the derivative? And can you only multiply the x^3 after you do that?
    What do you mean? I multiplied after I took the derivative because if I would have done so before hand I would be saying the following is true

    f(x)\cdot{g'(x)}=\bigg(f(x)\cdot{g(x)}\bigg)'

    Which I hope is not true, or I got an A+ in calculus when I shouldnt have
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  5. #5
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    Good man, you've answered my question, you deserve your A+
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