# Is a rational function a Laurent polynomial?

I was trying to convert a rational function $$\mathbb{R}(X)$$ to a polynomial of type $$\mathbb{R}[X,X^{-1}]$$ but I failed. I searched internet and $$\mathbb{R}[X,X^{-1}]$$ has its own name!: "Laurent polynomials".

My questions:

1- Every Laurent polynomial is a combination of rational functions. Is every Laurent polynomial possible to be equal to a single rational function?

2- Are there methods to convert rational functions to Laurent polynomials? How are they?

• $R[X,X^{-1}]=R[X]_X$ – Mustafa Oct 26 '18 at 8:31
• @Mustafa, I don't know the meaning for $R[X]_X$ – user231343 Oct 26 '18 at 8:32
• $R[X]_X=\{ \frac{f(X)}{X^n}; f(X)\in R[X] \}$ – Mustafa Oct 26 '18 at 8:34

1. Yes. A Laurent polynomial is an expression of the type$$a_n{X^n}+a_{n+1}X^{n+1}+\cdots+a_{n+k}X^{n+k},\tag1$$with $$n\in\mathbb Z$$ and $$k\in\mathbb N$$. This is a polynomial (and therefore a rational function) is $$n\geqslant0$$. Otherwise$$(1)=\frac{a_n+a_{n+1}X^n+\cdots+a_{n+k}X^k}{X^{-n}},$$which is a rational function.
2. No. For instance, $$\frac1{1+X}$$ is a rational function, but you can't express it as an element of $$\mathbb{R}[X,X^{-1}]$$.
• If $\frac{P(X)}{Q(X)}$ is an irreducible fraction, then that quotient can be expressed as a Laurent polynomial if and only if $Q(X)=X^n$ for some $n\in\mathbb N$. – José Carlos Santos Oct 26 '18 at 8:41
A rational function $$f(X)$$ can be a Laurent polynomial if and only if $$X^mf(X)$$ is an ordinary polynomial for some positive integer $$m$$.