I am curious about making a laurent series for a rational function, if possible by long division of polynomials. For example, $$\frac{3}{x+2}.$$ If I do long division here, I'd first multiply the divisor $(x+2)$ by $3x^{-1}$ and subtract this from $3$ to get $-6x^{-1}$ and then multiply the divisor $(x+2)$ by $-6x^{-2}$ and subtract this from $-6x^{-1}$ to get $12x^{-2}$ and so on. Is this a way to get a kind of Laurent series representation for the rational function? So I would get my "series representation" to be something like $$\frac{3}{x+2} = 3x^{-1}-6x^{-2}+12x^{-3}-24x^{-4}+\cdots.$$ Or is this nonsense?
I am looking at a problem in this paper where we are looking at the fraction field of a group ring and taking two non-zero element $p,q$ of the group ring and are looking at $\frac{p}{q}$ and by way of coming to an equivalent form of $\frac{p}{q}$ the author says we do long division of polynomials to write $\frac{p}{q}$ as a Laurent power series. I can provide more details if needed about this, at least lets see if what I am thinking above makes any sense =p thanks
This is not entirely correct. Formal Laurent series are only allowed finitely many negative terms, to avoid problems with defining multiplication (the definition works because their domains are wellordered; you can check that the definition won't work if you allow infinitely small and large exponents, as you'd need to define infinite sums which are not allowed. We could, in principle, instead allow only finitely many positive terms, but that would clash with the idea that Laurent series should contain regular formal power series).
So you want to do it in the other direction, to get $$\frac{3}{x+2}=\frac{3}{2}-\frac{3}{4}x+\frac{3}{8}x^2-\frac{3}{16}x^3+\ldots =\sum_{n\geq 0} \frac{3}{2}\left(-\frac{x}{2}\right)^n$$ Notice that the result is actually a formal power series. In general, a formal power series with the constant term invertible (such as $(x+2)$) is invertible within the ring of formal power series.