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Thread: Formal power series: inverse element

  1. #1
    Junior Member Greg98's Avatar
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    Formal power series: inverse element

    Hello,
    the problem is to calculate inverse element (reciprocal) of $\displaystyle 1-x^3$ in ring $\displaystyle $\mathbb{R}[[x]]$.

    I skimmed some Wikipedia and found following formula:
    Inverting series

    But I don't know to use it. Also the following model answer didn't help me either:
    "Task is to calculate inverse element (reciprocal) of $\displaystyle 1-x$ in the ring $\displaystyle $\mathbb{R}[[x]]$.

    So, we have to calculate $\displaystyle A(x)=a_0+a_1x+a_2x^2+...,$ that
    $\displaystyle (1-x)(a_0+a_1x+a_2x^2+...)=1$.

    When the $\displaystyle a_0$ is solved, and then $\displaystyle a_1$..., we see that inverse element is $\displaystyle 1+x+x^2+...$"

    I don't understand how those $\displaystyle a_0$ etc. are solved. Only thing, that is clear to me is, that $\displaystyle 1+x+x^2+...$ is generated by $\displaystyle \frac{1}{1-x}$, which equals $\displaystyle (1-x)^{-1}$.

    So, any help is appreciated. Thanks very much!
    Last edited by Greg98; Feb 14th 2011 at 09:39 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The formal series $\displaystyle S(x)=\sum_{n\geq 0}a_nx^n\in\mathbb{R}[[x]]$ has inverse element iff $\displaystyle a_0\neq 0$ .

    If $\displaystyle T(x)=\sum_{n\geq 0}b_nx^n$ is the inverse of the formal series $\displaystyle 1-x^3$ then, the equality $\displaystyle (1-x^3)T(x)=1$ will provide:

    $\displaystyle \begin{Bmatrix}b_0=1\\b_1=0\\b_2=0\\b_3-b_0=1\\b_n-b_{n-3}=0\;(n\geq 4)\end{matrix} $

    So, you can determine $\displaystyle T(x)$ .


    Fernando Revilla
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  3. #3
    MHF Contributor

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    Another way to do this is to think of $\displaystyle \frac{1}{1- x^3}$ as the sum of a geometric series.

    $\displaystyle \sum_{n=0}^\infty r^n= \frac{1}{1- r}$.
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