Jamiya Weber

Answered

2022-06-25

I was just thinking that as rational functions form an ordered field you could describe analogous version of the absolute value function, but we don't quite have a 'metric' - for example |1/x| < e for all e > 0 in R but 1/x =/= 0.

I was wondering if anybody got anywhere with this 'metric' and if there are any links to papers exploring actual metrics on the rational functions?

I was wondering if anybody got anywhere with this 'metric' and if there are any links to papers exploring actual metrics on the rational functions?

Answer & Explanation

Samantha Reid

Expert

2022-06-26Added 22 answers

The field $\mathbb{R}(x)$ of real rational functions is ordered by the condition that $\frac{{r}_{o}{x}^{n}+...+{r}_{n}}{{s}_{o}{x}^{m}+...+{s}_{m}}>0$ if ${r}_{0},{s}_{o}>0$.

This gives rise to a topology, which is metrizable:The reason is that there is a denumerable set, consisting of the fractions $\frac{1}{{x}^{N}}$, which is cofinal in the sense that for every positive rational real function $\frac{p(x)}{q(x)}>0$, there exists $N$ with $0<\frac{1}{{x}^{N}}<\frac{P(x)}{q(x)}$.

This implies that the ordered field $\mathbb{R}(x)$ is metrizable by a theorem of Dobbs.

The whole paper is interesting and might serve as the reference you are looking for.

This gives rise to a topology, which is metrizable:The reason is that there is a denumerable set, consisting of the fractions $\frac{1}{{x}^{N}}$, which is cofinal in the sense that for every positive rational real function $\frac{p(x)}{q(x)}>0$, there exists $N$ with $0<\frac{1}{{x}^{N}}<\frac{P(x)}{q(x)}$.

This implies that the ordered field $\mathbb{R}(x)$ is metrizable by a theorem of Dobbs.

The whole paper is interesting and might serve as the reference you are looking for.

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