# Thread: Formal power series: inverse element

1. ## Formal power series: inverse element

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

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 $1-x$ in the ring $\mathbb{R}[[x]]$.

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

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

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

So, any help is appreciated. Thanks very much!

2. The formal series $S(x)=\sum_{n\geq 0}a_nx^n\in\mathbb{R}[[x]]$ has inverse element iff $a_0\neq 0$ .

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

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

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

Fernando Revilla

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

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