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Let $P$ and $Q$ be polynomials over $\mathbb R$ such that the rational function $P/Q$ is not constant. Let $\alpha$ be an irrational number. Is it possible that $(P/Q)^\alpha$ is a rational function? I expect not, but on the other hand, it is well known that a rational number raised to an irrational power can be rational. So perhaps something similar is possible in this case.

EDIT:

Definition. A function $f$ is said to be a rational function if it can be written as $f = R/S$ for two polynomials $R$ and $S$.

Ultimately, I would like to answer this question for multivariate polynomials, but I would already be quite happy to understand the univariate case.

Inzinity
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2 Answers2

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A univariate rational function is asymptotic to $cx^z$ where $z$ is an integer number. When raised to $\alpha$, it is asymptotic to $cx^{\alpha z}$. This holds for polynomials as well.

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By Yves Daoust's answer, the degrees of $P$ and $Q$ are equal. Assume $P$ and $Q$ have no common factors, i.e., their roots are distinct.

Then, your question is whether there are polynomials $R$ and $S$ (again wlog without common factors) such that $$\left(\frac{P}{Q}\right)^\alpha=\frac{R}{S}\\\Longleftrightarrow \quad P^\alpha S=Q^\alpha R\,.$$ So $R$ has exactly the same roots as $P^\alpha$. Around one of these roots, $R\sim x^p$ for some multiplicity $p$, while $P^\alpha\sim x^{q\alpha}$. is $\alpha$ is irrational, this doesn't work.

Hence, the answer is no.

Toffomat
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  • The function $R:\mathbb{R}\to\mathbb{R}$ $R(x)=\tfrac{x^2-x+1}{x^2+x+1}$ doesn't have any zeros, so that one cannot arrive at a contradiction by comparing asymptotic behavior around the zeros – K B Dave Nov 09 '24 at 20:51
  • @KBDave Be $R$, I mean a polynomial, not the rational function. And the argument works for complex roots as well. – Toffomat Nov 09 '24 at 21:06