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Is it possible that $\ln(x)=\frac{p(x)}{q(x)}$ for all $x>0,$ where $p$ and $q$ are polynomials with real coefficients?

I think the answer is no. Suppose two such polynomials did exist. Take the limit as $x$ goes to infinity. This gives that $\deg(p)>\deg(q),$ as $\lim_{x\to\infty}\ln(x)=\infty.$ Let $m=\deg(p)$ and $n=\deg(q).$ Differentiate both sides to get $$\frac{1}{x}=\frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}.$$

Rearrange (this step is valid as $x>0$ and $q(x)^2>0$ by hypothesis) to get $q(x)^2=xp'(x)q(x)-xp(x)q'(x).$

Let the leading coefficient of $p$ be $a$ and the leading coefficient of $q$ be b. The leading coefficient of $xp'(x)q(x)$ then, is $amb.$ Similarly, the leading coefficient of $-xp(x)q'(x)$ is $-bna.$ Suppose their sum were $0.$ Then, $ab(m-n)=0.$ But, $ab≠0$ (as $a$ and $b$ are leading coefficients). So, $m-n=0.$ This contradicts $m>n.$ Hence, the coefficient of $x^{m+n}$ in the RHS is non-zero. Now, compare degrees to get $2n=m+n.$ This contradicts $m>n.$

Is my approach right? What other methods can we use to show this?

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    $\begingroup$ This seems right! Alternatively, consider the limit of both sides as you did, then divide both sides by $x$ and consider the new limits. $\endgroup$ Dec 6, 2023 at 7:24
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    $\begingroup$ You can also use growth rate arguments (which would be similar to Greg's idea) $\endgroup$
    – Calvin Lin
    Dec 6, 2023 at 7:29
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    $\begingroup$ Why is the degree of the numerator exactly $m+n-1$? Adding two polynomials of the same degree may result in a polynomial of a lower degree. $\endgroup$
    – lhf
    Dec 6, 2023 at 11:27
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    $\begingroup$ @lhf, yes I missed that. See my new edit; it should have hopefully fixed the error. $\endgroup$
    – aqualubix
    Dec 6, 2023 at 11:39
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    $\begingroup$ Your proof is correct now, as far as I can tell. – As indicated in the other comments, you can also use that the logarithm does not have “polynomial growth,” i.e. there is no integer exponent $k$ such that $\ln(x) \sim x^k$ for $x \to \infty$. $\endgroup$
    – Martin R
    Dec 7, 2023 at 8:21

1 Answer 1

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In this answer no derivatives nor growth rate of $\ln x$ are used.

If $\ln x={p(x)\over q(x)}$ then $\deg p>\deg q$ as the limit at $\infty $ is equal $\infty.$ We get $${p(x)\over q(x)}=\ln x={1\over 2}\ln (x^2)={p(x^2)\over 2 q(x^2)}$$ Thus $2p(x)q(x^2)=p(x^2)q(x).$ Hence $$\deg p+2\deg q=2\deg p+\deg q $$ This implies $\deg p=\deg q,$ a contradiction.

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    $\begingroup$ +1 An elegant solution. $\endgroup$ Dec 8, 2023 at 1:57

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