What are some methods to show $\log$ is not a rational function?

Question

It is easy to show $\log$ 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

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

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

$$\frac{G(x)}{x\log x} = Q_2(x)$$

It is well known that $x \log x \to 0$ 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 $\log x$ isn't a rational function?

Answer 1

One reason is that a rational function is defined over all $\mathbb R$ except for a finite number of points, but $\log$ is not.

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

If $\log = \dfrac GQ$, then $\dfrac 1x = \log' = \dfrac{G'Q-GQ'}{Q^2}$.

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

But $\displaystyle\lim_{x\to\infty} \log(x) = \infty$.

Answer 2

More generally we can prove that $\log x$ is not an algebraic function using the following property of $\log x$: $$\lim_{x \to 0^{+}}x^{a}\log x = 0\tag{1}$$ for all positive numbers $a$.

Let's assume on the contrary that $y = \log x$ is an algebraic function. Then we have polynomial functions $a_{1}(x), a_{2}(x), \ldots, a_{n}(x)$ such that $$a_{0}(x)y^{n} + a_{1}(x)y^{n - 1} + \cdots + a_{n - 1}(x)y + a_{n}(x) = 0\tag{2}$$ for all values of $x > 0$. Also note that in the above equation we have both $a_{0}(x), a_{n}(x)$ as non-zero polynomials. Now taking limits for both sides of equation $(2)$ as $x \to 0^{+}$ and using $(1)$ we see that if $b_{i}$ are constant terms of $a_{i}(x)$ then $$\lim_{x \to 0^{+}}(b_{0}y^{n} + b_{1}y^{n - 1} + b_{n - 1}y + \cdots + b_{n}) = 0\tag{3}$$ Dividing by $y$ and noting that $y = \log x \to -\infty$ as $x \to 0^{+}$ we get $$\lim_{x \to 0^{+}}(b_{0}y^{n - 1} + b_{1}y^{n - 2} + \cdots + b_{n - 1}) = 0$$ Repeating the same argument we finally get that $b_{0}, b_{1}, \ldots, b_{n} = 0$. Thus the equation $(2)$ can be divided by $x$ to obtain a similar equation and we can apply the same argument on this new equation. Carrying on this procedure we ultimately get that all the coefficients of the polynomials $a_{i}(x)$ are $0$ and hence all these polynomials are zero polynomials. This contradicts the fact that both $a_{0}(x), a_{n}(x)$ are non-zero polynomials.

For the current question (i.e. prove that $\log x$ is not a rational function) it suffices to take $n = 1$ in the preceding argument.

Answer 3

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

Answer 4

If $r(x) =\dfrac{a(b)}{b(x)} $, then, as $x \to \infty $, $|r(x)| \sim c|x|^{\deg(a)-\deg(b)} $ for some $c$.

But $\log$ does not go to $\infty$ like this.

Answer 5

Here's a proof that no antiderivative $F(x)$ of $1/x$ with domain $\Bbb R \setminus \{0\}$ is a rational function.

Such a function is -- as is often forgotten even in calculus textbooks -- of the form $$ F_{C,D}(x) = \begin{cases} \log(-x) + C \quad \, \text{ for } x<0\\ \log(x) +D \qquad\text{ for } x>0\end{cases}$$

with $C,D \in \Bbb R$. It has exactly one vertical asymptote, at $x=0$. So if it were of the form $p(x)/q(x)$ with polynomials $p,q \in \Bbb R[x]$, which wlog we can assume to be without common factor, then $q(x) = x^n$ for some $n \in \Bbb N$. Hence in particular $$x^n \log(x)$$ is a polynomial on $(0,\infty)$. Hence its derivative $$nx^{n-1}\log(x) + x^n \cdot x^{-1} = nx^{n-1}\log(x) + x^{n-1}$$ is a polynomial on $(0,\infty)$ and hence so is $x^{n-1}\log(x)$. Inductively, we get that $\log(x)$ is a polynomial on $(0,\infty)$ which is absurd because of the vertical asymptote at $0$.

Answer 6

Here is a completely algebraic proof, using the definition $\log 1 = 0$, $ \log' x = 1/x $. It may look rather familiar if you think about the proof that $\sqrt{2} \notin \mathbb{Q}$.

Suppose that $\log x = P/Q$ is rational, $P,Q$ coprime polynomials, $Q \neq 0$. Then differentiating, $$ \frac{1}{x} = \frac{P'Q-Q'P}{Q^2} , $$ or $$ Q^2 = x(P'Q-Q'P) . \tag{*} $$ Hence $x \mid Q^2$, and since $x$ is irreducible, $x \mid Q$. So we can write $Q = x^k R$ where $R \neq 0$, $x \nmid R$ and $k \geq 1$. Inserting this into (*), $$ x^{2k} R^2 = x^{k+1} P'R - x^{k+1}PR \Big( \frac{1}{x} + \frac{R'}{R} \Big) = x^{k+1} P' R - kx^k PR - x^{k+1} PR' . $$ Cancelling, $$ x^k R^2 = xP'R - kPR - xPR' , $$ so $$ kPR = x(P'R - PR' - x^{k-1}R^2) $$ Since the polynomials are an integral domain, the left-hand side cannot be $0$, so neither bracket on the right-hand side is $0$, and therefore since $ x \nmid R$ by definition, $x \nmid P$. Hence $P,Q$ are not coprime, contradicting the original assumption. Therefore $\log x$ cannot be a rational function.

---

Nothing particularly special about $x$ was used here: suppose that $S,T$ are coprime polynomials, with $T$ a squarefree polynomial in $F[X]$. We can show that any antiderivative of $S/T$ is not rational: suppose it was, and equal to $P/Q$, then $$ SQ^2 = T(P'Q-Q'P) $$ as before. So $T \mid SQ^2$, and since $T$ is squarefree and $S,T$ are coprime, $T \mid Q$. Write $Q = T^k R$ where $T \nmid R$ and $k\geq 1$. Then $$ ST^{2k} R^2 = T^{k+1}P'R - kT^{k}T'PR -T^{k+1}PR' , $$ and cancelling, $$ ST^k R^2 = T(P'R-PR') - kT'PR . $$ Hence $$ kT'PR = T( P'R-PR' - ST^{k-1}R ) . $$ None of the factors can be $0$ as before, so $ T \mid T'PR $. But $T$ is squarefree, so it is coprime to its derivative $T'$, so $T \mid PR$. But $T \nmid R$, so the GCD of $T$ and $R$, say $G$, is a proper divisor of $T$, $T = GH$ with $H$ coprime to $R/G$. But $GH \mid PR \implies H \mid P(R/G)$, and $H$ is coprime to $R/G$, so $H \mid P$. But $H \mid T$ and $T \mid Q$, so $H \mid Q$, contradicting that $P,Q$ are coprime.

I suspect the same analysis can be carried out for $S/(UV)$, where $S$ is coprime to $UV$ and $V$ is squarefree, but I haven't worked through it. There's not really much point, given the existence of the partial fraction decomposition.