# Thread: Formal power series: inverse element

1. ## 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!

2. 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

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

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