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Consider a sequence of complex polynomials $f \in \mathbb{C}[z]$, $f(0) \neq 0$, that are composed of a negligible fraction $o(\deg{f})$ of monomials. Are the zeros of such polynomials necessarily equidistributed in angle, for the uniform measure $d\theta/2\pi$ on $S^1 = \mathbb{C}^{\times} / \mathbb{R}^{> 0}$?

Certainly the zeros are equidistributed in angle when the number of monomials is bounded while the degree goes to infinity. This follows from much more general results of A. G. Khovanskii exposed in his book Fewnomials (Transl. Math. Monographs, vol. 88). But growing like $o(\deg{f})$?

Not knowing the answer, it might make more sense to look at this question contrapositively. Clearly, for any $\varepsilon > 0$ we can construct an $\varepsilon$-sparse sequence with $\deg{f} \to \infty$ and not equidistributed in angle; just consider $h(z^n)$ as $n$ is fixed subject to $1/n < \varepsilon$ while $h$ ranges over the $\mathbb{R}$-split real polynomials. Can we do better than this?

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  • $\begingroup$ Do you have a more specific reference in Khovansky's book? $\endgroup$
    – Igor Rivin
    Commented Oct 3, 2015 at 17:24
  • $\begingroup$ @IgorRivin: My understanding is that Theorem 1 of Chapter III, 3.13 (of the AMS translated edition) contains, in the special case $n = 1$, an explicit lower bound on the number $k$ of monomials that comprise a polynomial whose zeros set is not equidistributed in angle -- viz., the computable function $\varphi(k,1)$ must not be too small in such a situation, giving some lower bound on $k$, though obviously not nearly as strong as $c \deg{f}$. On the other hand I am not seeing any obvious way to improve over the trivial construction; hence the question. $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 3, 2015 at 17:58
  • $\begingroup$ There was also a Bourbaki Seminar exposition from by Khovanskii from the 1980s, on the subject of simple topology of $\mathbb{R}$-solutions and angular equidistribution of $\mathbb{C}$-solutions of a system of fewnomial equations, but I do not seem to find it on numdam. $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 3, 2015 at 18:00
  • $\begingroup$ Also, if we take the number of monomials as an algebraic measure of complexity (being the number of algebraic operations needed to write down the polynomial), it could be compared to Mahler's arithmetic complexity measure $m(f) := \int_{S^1} \log{|f|} \, d\theta/2\pi$ of an irreducible integer polynomial $f \in \mathbb{Z}[x]$ (which refines / is majorized by $\log{\sum |a_i|}$). For the latter, we know by Bilu that the condition guaranteeing the equidistribution of the zeros is precisely $m(f) = o(\deg{f})$. $\endgroup$
    – Vesselin Dimitrov
    Commented Oct 3, 2015 at 18:38
  • $\begingroup$ Which result of Bilu? I could not find it by cursory mathscinet search... $\endgroup$
    – Igor Rivin
    Commented Oct 3, 2015 at 20:06

1 Answer 1

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A lot depends on how exactly to understand this question. The crudest form (for an $m$-nomial of degree $n$ with non-zero free term, the number of all zeroes in a sector of aperture $2\pi\theta$ is $\theta n$ with an error at most $m$) is an exercise in elementary complex analysis. Indeed, consider a sector of aperture $2\pi\theta$ with the vertex at $0$ and of huge radius $R$. The increment of the argument of $p$ over the arc is then $2\pi\theta n$ for all practical purposes. On the other hand, each loop around the origin on the interval $\{zt:0\le t\le R]$ creates $2$ real zeroes of $\Re p(zt)$, so the "side" increment of the argument is at most $\pi$ times the total number of real roots of $2$ real $m$-nomials (one for each side). However, by the Descartes rule of signs, an $m$-nomial can have at most $m$ real roots, and the story is over.

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