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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?

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  • $\begingroup$ $R[X,X^{-1}]=R[X]_X $ $\endgroup$ – Mustafa Oct 26 '18 at 8:31
  • $\begingroup$ @Mustafa, I don't know the meaning for $R[X]_X$ $\endgroup$ – user231343 Oct 26 '18 at 8:32
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    $\begingroup$ $R[X]_X=\{ \frac{f(X)}{X^n}; f(X)\in R[X] \}$ $\endgroup$ – Mustafa Oct 26 '18 at 8:34
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  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 $\mathbb{R}[X,X^{-1}]$.
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  • $\begingroup$ If some rational function convertible to a laurent pol then are there methods available? I mean any link or text would be useful. $\endgroup$ – user231343 Oct 26 '18 at 8:39
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    $\begingroup$ 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 $n\in\mathbb N$. $\endgroup$ – José Carlos Santos Oct 26 '18 at 8:41
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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 $m$.

Clearly one can find rational functions failing this condition.

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