glutynowy44

2022-07-12

It is easy to show $\mathrm{log}$ isn't a polynomial (no continuous extension to $\mathbb{R}$). More challenging is showing it isn't rational.
Suppose it were a rational function. Then write, the fraction in lowest terms
$\mathrm{log}x=\frac{G\left(x\right)}{Q\left(x\right)}⟺\frac{G\left(x\right)}{\mathrm{log}x}=Q\left(x\right)$
Clearly, as $x\to 0$, $Q\left(x\right)\to 0$. However, $Q\left(x\right)$ then has $x$ as factor so that
$\frac{G\left(x\right)}{x\mathrm{log}x}={Q}_{2}\left(x\right)$
It is well known that $x\mathrm{log}x\to 0$ as $x\to 0$, so for ${Q}_{2}\left(x\right)$ to have a finite limit as $x\to 0$, which it must since it is a polynomial, $G\left(x\right)\to 0$ as $x\to 0$ so that $x$ is a factor of $G\left(x\right)$. This contradicts the assumption that $\frac{G}{Q}$ was in lowest terms.
If there is something wrong with this, please comment, but my main question is
What are some other ways to prove that logx isn't a rational function?

pampatsha

Expert

One reason is that a rational function is defined over all $\mathbb{R}$ except for a finite number of points, but $\mathrm{log}$ is not.
Let's prove that $\mathrm{log}$ is not even a rational function restricted to $\left(0,+\mathrm{\infty }\right)$.
If $\mathrm{log}=\frac{G}{Q}$, then $\frac{1}{x}={\mathrm{log}}^{\prime }=\frac{{G}^{\prime }Q-G{Q}^{\prime }}{{Q}^{2}}$.
This implies that $G$ and $Q$ have the same degree and therefore $\underset{x\to \mathrm{\infty }}{lim}\frac{G\left(x\right)}{Q\left(x\right)}$ is finite.
But $\underset{x\to \mathrm{\infty }}{lim}\mathrm{log}\left(x\right)=\mathrm{\infty }$.

civilnogwu

Expert

A rational function $r$ vanishes at $\mathrm{\infty }$ or there is an integer power $q$ so that $r\left(x\right)\sim c\cdot {x}^{q}$ as $x\to \mathrm{\infty }$. The log function exhibits none of these behaviors.