How do you find the inverse of a polynomial in Laurent series division ring? In the Laurent series division ring, how can we find the inverse of a given polynomial? For example how can you find the inverse of $3x^{-2} +x^{-1}+5x+7x^4$. Is there a certain formula to find the inverse?
I've tried to find the inverse for the above polynomial with a formula used in proving the existence of the units, but it gives me a lot of unknowns.
 A: $${1\over7x^4+5x+x^{-1}+3x^{-2}}=a_{-4}x^{-4}+a_{-3}x^{-3}+\cdots$$ where the $a_i$ satisfy the recurrence $$3a_n+a_{n-1}+5a_{n-3}+7a_{n-6}=0$$ for $n\ge3$, together with certain initial conditions (conditions on $a_{-4},\dots,a_2$). 
If you can solve that recurrence relation, you can find a formula for the $a_i$; if not, not. 
A: You know how to find the inverse of $x$ and $x^{-1}$. It suffices to find the inverse of a polynomial of the form $1 - x f(x)$ where $f$ is a polynomial. But this is just $1 + x f(x) + x^2 f(x)^2 + ...$ (which converges in the $x$-adic topology). 
A: $$ \frac{1}{7 x^4 + 5 x + x^{-1} + 3 x^{-2}} = \frac{x^2}{7 x^6 + 5 x^3 + x + 3}$$
The polynomial $P(x) = 7 x^6 + 5 x^3 + x + 3$ is irreducible over the rationals, with six
distinct complex roots $r_1$ to $r_6$ (none of them real, by the way).  We have
a partial fraction decomposition
$$ \frac{x^2}{7 x^6 + 5 x^3 + x + 3} = \sum_{j=1}^6 \frac{r_j^2}{P'(r_j) (x - r_j)} $$
since $r_j^2/P'(r_j)$ is the residue of $x^2/P(x)$ at $x=r_j$.
Since $\dfrac{1}{x -r_j} = \dfrac{-r_j^{-1}}{1 - r_j^{-1} x} = - \sum_{k=0}^\infty r_j^{-k-1} x^k$, we obtain
$$  \frac{1}{7 x^4 + 5 x + x^{-1} + 3 x^{-2}} = -\sum_{k=0}^\infty \sum_{j=1}^6
\frac{r_j^{1-k}}{P'(r_j)} x^k$$
(the $k=0$ and $k=1$ terms will be $0$).
