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Thread: find a power series representation of f(x)

  1. #1
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    find a power series representation of f(x)

    i am to use $\displaystyle \frac{1}{1-x}=\sum{x^n}=1+x+x^2+...$ provided $\displaystyle |x|<1$ to find a power series representation for $\displaystyle f(x)=\frac{x^3}{(x-2)^2}$

    so i want to foil the binomial, move the cube, and convert the statement into the form of $\displaystyle \frac{1}{1-x}$

    considering only $\displaystyle \frac{1}{(x-2)^2}$

    i get $\displaystyle \frac{1}{x^2-4x+4}=\frac{1}{4-(4x-x^2)}=\frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}$

    so $\displaystyle \frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}=\frac{1}{4}*\sum{(x-\frac{x^2}{4})^n}$

    now i'm at a loss has to how to get it into the form $\displaystyle \sum{c_n*x^n}$

    integrating/differentiating term-by-term doesn't seem to be helpful as each makes the situation far more complicated.
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  2. #2
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    Re: find a power series representation of f(x)

    actually i should be able to take the derivative of $\displaystyle \frac{1}{(x-2)^2}$

    so $\displaystyle d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$

    then $\displaystyle d/dx\sum{x^n}=\frac{-1}{2}\frac{1}{(1-x)}$

    and $\displaystyle d/dx\frac{-1}{2}\sum{x^n}=\frac{-1}{2}\sum{nx^{n-1}}$

    and then just multiply in the cube, no?
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  3. #3
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    Re: find a power series representation of f(x)

    Quote Originally Posted by bkbowser View Post
    actually i should be able to take the derivative of $\displaystyle \frac{1}{(x-2)^2}$

    so $\displaystyle d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$

    then $\displaystyle d/dx\sum{x^n}=\frac{-1}{2}\frac{1}{(1-x)}$

    and $\displaystyle d/dx\frac{-1}{2}\sum{x^n}=\frac{-1}{2}\sum{nx^{n-1}}$

    and then just multiply in the cube, no?
    sorry there's a few simple things wrong here.

    $\displaystyle d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$ should be;

    $\displaystyle \frac{1}{(x-2)^2}=d/dx\frac{1}{(x-2)}$

    $\displaystyle =d/dx\frac{1}{(x-2)}=d/dx\frac{-1}{2}*\frac{1}{(1-\frac{x}{2})}=d/dx\frac{-1}{2}*\sum\frac{x^n}{2^n}$

    then after reintroducing the cube;

    $\displaystyle x^3*\frac{-1}{2}*\sum\frac{x^n}{2^n}=-\sum{\frac{x^{n+3}}{2^{n+1}}}$
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