Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let the ring $R[X]$ of all power formal series $\sum_{i=-\infty}^{\infty}{a_iX^i}$, (where $a_i$ are coefficients in finite field $K$) in which $a_i=0$, for every $i>0$ with at most a finite number. Observe that $K[X]$, the ring of polynomials in $X$, is embedded in $R[X]$ in natural manner.

My question is: Let $g(X)$ and $f(X) \in R[X]$, with degree $n$ and $n-1$ respect. How i get the g(X) inverse with indeterminants $X^{-i}$?. Because in my text say

$f(X) = ((a_0X^{-1}+a_1X^{-2}+...+a_{n-1}X^{-n}) + b_nX^{-(n+1)}+...)g(X)$.

pdta: I think that $f(X)=f(X)g(X)^{-1}g(X)$ then $f(X)g(X)^{-1} = ((a_0X^{-1}+a_1X^{-2}+...+a_{n-1}X^{-n}) + b_nX^{-(n+1)}+...)$, but my problem is the inverse of $g(X)$

share|cite|improve this question
The hint is: long division. In these days of calculators, they still teach that, don't they? – GEdgar Feb 27 '12 at 21:28
no understand GEdgar – juaninf Feb 27 '12 at 23:48
If you want to get someone's attention, you have to write @GEdgar – Gerry Myerson Feb 28 '12 at 11:41


Divide $x+1$ by $x^2-x-1$, express the answer as a Laurent series at $x=\infty$.

long division

continue to get as many terms as you like! After a while, maybe you can see a pattern, then prove it.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.