glutynowy44

Answered

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

$\mathrm{log}x=\frac{G(x)}{Q(x)}\u27fa\frac{G(x)}{\mathrm{log}x}=Q(x)$

Clearly, as $x\to 0$, $Q(x)\to 0$. However, $Q(x)$ then has $x$ as factor so that

$\frac{G(x)}{x\mathrm{log}x}={Q}_{2}(x)$

It is well known that $x\mathrm{log}x\to 0$ as $x\to 0$, so for ${Q}_{2}(x)$ to have a finite limit as $x\to 0$, which it must since it is a polynomial, $G(x)\to 0$ as $x\to 0$ so that $x$ is a factor of $G(x)$. 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?

Suppose it were a rational function. Then write, the fraction in lowest terms

$\mathrm{log}x=\frac{G(x)}{Q(x)}\u27fa\frac{G(x)}{\mathrm{log}x}=Q(x)$

Clearly, as $x\to 0$, $Q(x)\to 0$. However, $Q(x)$ then has $x$ as factor so that

$\frac{G(x)}{x\mathrm{log}x}={Q}_{2}(x)$

It is well known that $x\mathrm{log}x\to 0$ as $x\to 0$, so for ${Q}_{2}(x)$ to have a finite limit as $x\to 0$, which it must since it is a polynomial, $G(x)\to 0$ as $x\to 0$ so that $x$ is a factor of $G(x)$. 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?

Answer & Explanation

pampatsha

Expert

2022-07-13Added 15 answers

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 $(0,+\mathrm{\infty})$.

If $\mathrm{log}={\displaystyle \frac{G}{Q}}$, then $\frac{1}{x}}={\mathrm{log}}^{\prime}={\displaystyle \frac{{G}^{\prime}Q-G{Q}^{\prime}}{{Q}^{2}}$.

This implies that $G$ and $Q$ have the same degree and therefore $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{G(x)}{Q(x)}}$ is finite.

But $\underset{x\to \mathrm{\infty}}{lim}\mathrm{log}(x)=\mathrm{\infty}$.

Let's prove that $\mathrm{log}$ is not even a rational function restricted to $(0,+\mathrm{\infty})$.

If $\mathrm{log}={\displaystyle \frac{G}{Q}}$, then $\frac{1}{x}}={\mathrm{log}}^{\prime}={\displaystyle \frac{{G}^{\prime}Q-G{Q}^{\prime}}{{Q}^{2}}$.

This implies that $G$ and $Q$ have the same degree and therefore $\underset{x\to \mathrm{\infty}}{lim}{\displaystyle \frac{G(x)}{Q(x)}}$ is finite.

But $\underset{x\to \mathrm{\infty}}{lim}\mathrm{log}(x)=\mathrm{\infty}$.

civilnogwu

Expert

2022-07-14Added 6 answers

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

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