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Math Help - Formal power series: identity problem

  1. #1
    Junior Member Greg98's Avatar
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    Formal power series: identity problem

    Hello,
    the problem is to show by calculating that:
    \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k.

    I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Greg98 View Post
    Hello,
    the problem is to show by calculating that:
    \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k.

    I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!
    Very formally we have

    \displaystyle \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k

    Now consider

    What is

    \displaystyle x\frac{d}{dx}\left( x\frac{d}{dx}x^k\right)=?

    Why does this help?
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Greg98 View Post
    Hello,
    the problem is to show by calculating that:
    \displaystyle\frac{x(1+x)}{(1-x)^3}=\sum_{i=0}^{\infty}k^2x^k.

    I know the solution has to do something formal power series. I found following pieces: link 1 and link 2 , but even those couldn't get me any further. Also WolframAlpha didn't gave me any clue, though there's some information. I'm completely stuck, which means any help is welcome. Thank you!
    Proceeding [tediously...] 'step by step' we have...

    \displaystyle \sum_{k=0}^{\infty} k^{2}\ x^{k} = x\ \sum_{k=0}^{\infty} k^{2}\ x^{k-1} = x\ \frac{d}{dx} \sum_{k=0}^{\infty} k\ x^{k}=  x\ \frac{d}{dx} (x\ \sum_{k=0}^{\infty} k\ x^{k-1})=

    \displaystyle = x\ \frac{d}{dx} (x\ \frac{d}{dx} \frac{1}{1-x}) = x\ \frac{d}{dx} \frac{x}{(1-x)^{2}}= x\ \frac{(1-x)^{2} +2 x (1-x)}{(1-x)^{4}} =

    \displaystyle = x\ \frac{1-x+2 x}{(1-x)^{3}} = x\ \frac{1+x}{(1-x)^{3}}

    Kind regards

    \chi \sigma
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