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While Descartes' rule of sign works for polynomials, does it work for 'polynomials' where the exponents are real instead of natural? That is does it work for

$$5x^{\alpha} -2x^{\beta} + 3x^{\gamma} + 3 =0$$

where, assuming I have ordered them such that $\alpha>\beta>\gamma$, $\alpha, \beta, \gamma \in \mathbb{R}$?

Thank you for any help or reference.

danny dan
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1 Answers1

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Interesting question. Probably yes.

This makes sense only for $x>0$.

Approximate the exponents by rationals with a large denominator $n$. Then replace $x$ by $x^n$. That changes no roots and Descartes' rule of signs applies.

Ethan Bolker
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    Just to nitpick on a +1 answer, the tacit assumption here is that the positive roots depend continuously on the exponents. This holds true, except in the case of a double (or higher order) positive root. But in that case a perturbation of the exponents either changes the double root to a pair of real roots, or eliminates the real roots altogether - so the number of positive roots either stays the same or goes down, meaning the rule of signs still applies. – dxiv Jan 10 '22 at 23:08
  • @dxiv Thanks for the nitpick. I suspected possible gaps in the argument, hence the "probably yes". – Ethan Bolker Jan 10 '22 at 23:14