Division in the ring of formal power series I was wondering if the following simplification is allowed in the ring of formal power series:
If I have $ \displaystyle\frac{1+x}{1+2x+x^2}$ , can I simplify it so that it is equal to $\displaystyle\frac{1}{1+x}$.
I think that it can't be simplified, but i would appreciate if someone clarifies my doubt.
Thanks
 A: The formal power series $\frac{1+x}{1+2x+x^2}$ can be simplified. We consider the ring $\mathbb{C}[[x]]$ of formal power series, where we take as ring the field $\mathbb{C}$. The formal power series
\begin{align*}
A(x)=1+x\in\mathbb{C}[[x]]
\end{align*}
has a constant term $1\ne 0$ which guarantees the existence of a multiplicative inverse $A^{-1}(x)$ with
\begin{align*}
A(x)A^{-1}(x)=1
\end{align*}
The multiplicative inverse $A^{-1}(x)=1-x+x^2-x^3+\cdots=\sum_{n=0}^\infty (-1)^nx^n$ and we can easily show
\begin{align*}
\color{blue}{A(x)A^{-1}(x)}&=(1+x)\sum_{n=0}^\infty (-1)^nx^n\\
&=\sum_{n=0}^\infty (-1)^nx^n+\sum_{n=0}^\infty (-1)^nx^{n+1}\\
&=\sum_{n=0}^\infty (-1)^nx^n-\sum_{n=1}^\infty (-1)^nx^{n}\\
&\,\,\color{blue}{=1}
\end{align*}
Two more aspects:

*

*We recall the multiplicative inverse of formal power series is unique provided it exists.


*We commonly write the inverse multiplicative of a formal power series $C(x)$ as
\begin{align*}
\frac{1}{C(x)}:=C^{-1}(x)
\end{align*}
which enables us to write
\begin{align*}
A^{-1}(x)=\frac{1}{1-x}=1-x+x^2-x^3+\cdots
\end{align*}

We can now show purely algebraically that OPs two formal power series are equal. We thereby apply well-known algebraic rules which are also valid in the ring of formal power series. We obtain
\begin{align*}
\color{blue}{\frac{1+x}{1+2x+x^2}}&=(1+x)\left(1+2x+x^2\right)^{-1}\\
&=(1+x)\left((1+x)(1+x)\right)^{-1}\\
&=(1+x)\left((1+x)^{-1}(1+x)^{-1}\right)\\
&=\left((1+x)(1+x)^{-1}\right)(1+x)^{-1}\\
&=1\cdot (1+x)^{-1}\\
&=(1+x)^{-1}\\
&\,\,\color{blue}{=\frac{1}{1+x}}
\end{align*}
and the claim follows.

Note we do not use any Taylor series expansion which considers a power series as analytic object where coefficients are defined via (analytic) derivatives. We do also not consider any radius of convergence since $x\in\mathbb{C}[[x]]$ is just a formal placeholder for a sequence $(0,\color{blue}{1},0,0,\ldots)$ and is not subject of evaluation at certain points of $\mathbb{C}$.
