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Thread: Difficulty finding power series representation of a function

  1. #1
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    Difficulty finding power series representation of a function

    As the title says, I'm having a lot of trouble finding a power series representation of a function. Here it is:

    Difficulty finding power series representation of a function-scan0013.jpg


    And here's my attempt at a solution:


    Difficulty finding power series representation of a function-scan0012.jpg


    I had a lot of trouble trying to reindex the summations such that I'd have a single summation at the end. This is what I ultimately got. Is this correct? If so, how would I go about finding the interval of convergence here? Would it just be with the ratio test as usual? It seems like it would be quite difficult to do that here, considering the number of terms in the final summation.
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  2. #2
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    Re: Difficulty finding power series representation of a function

    You are making this so much more difficult on yourself than you have to.

    Notice that $\displaystyle \begin{align*} \frac{1}{\left( 1 - 2\,x \right) ^3} = \left[ 1 + \left( -2\,x \right) \right] ^{-3} \end{align*}$, which can be expanded using the Binomial Series $\displaystyle \begin{align*} \left( 1 + X \right) ^{\alpha} = \sum_{k = 0}^{\infty}{ {\alpha\choose{k}}\,X^k }= 1 + \alpha\,X + \frac{\alpha \,\left( \alpha - 1 \right) }{2!}\,X^2 + \frac{\alpha \, \left( \alpha - 1 \right) \, \left( \alpha - 2 \right) }{3!}\,X^3 + \dots \end{align*}$

    Once you have that, multiply through by $\displaystyle \begin{align*} \left( x - x^2 \right) \end{align*}$.

    Notice that as $\displaystyle \begin{align*} \alpha < -1 \end{align*}$ this will only be convergent where $\displaystyle \begin{align*} \left| X \right| < 1 \end{align*}$, i.e. where $\displaystyle \begin{align*} \left| x \right| < \frac{1}{2} \end{align*}$.
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