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The question in brief:   When does a subset $S$ of the complex $n$th roots of unity have the property that $$\prod_{\alpha\, \in \,S} (z-\alpha)$$ gives a polynomial in $\mathbb R[z]$ with nonnegative coefficients?

Some trivial necessary conditions include $1\not\in S$ and also that $S$ is self-conjugate so that the coefficients will at least be real.

A sufficient condition that is not obvious (to me) is that the roots of unity in $S$ are precisely those lying outside of some wedge-shaped region of the complex plane, i.e., those with $\left|\arg(z)\right| > t$ for some fixed $t > 0$. This follows from the main theorem of

Barnard, R. W., W. Dayawansa, K. Pearce, and D. Weinberg. "Polynomials with nonnegative coefficients." Proceedings of the American Mathematical Society 113, no. 1 (1991): 77-85. Journal link

which says that, given any polynomial with nonnegative coefficients, dividing it by the linear factors corresponding to exactly those of its roots lying in such a region produces a polynomial with only positive coefficients. Applying this to $z^{n-1}+\cdots+z+1$ gives the desired result.

But that theorem is not number-theoretic; it doesn't care at all that we are dealing with roots of unity. So maybe there is a nicer proof in this special case?

More generally, what other necessary and/or sufficient conditions exist? Is there some reasonable number-theoretic or combinatorial characterization of such sets? What tools seem likely to shed some light on this question?

ADDED LATER: A couple of very interesting suggestions have been made in the comments so far, and I'm eager to see where they lead. However, I remain quite curious as to what can be said from a number-theoretic perspective. For example, if $n=p^r > 1$ for some prime $p$, then the cyclotomic polynomial $\Phi_n(x)=\Phi_p(x^{n/p})$ has only nonnegative coefficients. So a sufficient condition that I did not mention above is that $S$ is the set of primitive $n$th roots of unity for some prime power $n$. I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial, but that doesn't imply that nothing else can be gained from a number-theoretic perspective on this question – however I myself lack the expertise necessary to make the most of such a perspective.

I should also mention that (ironically, given my desire for a number-theoretic perspective on this) the cyclotomic result actually follows from a more general, not number-theoretic result of

Evans, Ronald, and John Greene. "Polynomials with nonnegative coefficients whose zeros have modulus one." SIAM Journal on Mathematical Analysis 22, no. 4 (1991): 1173-1182. Author's link

which gives that for any proper divisor $d$ of $n$, $(x^n-1)/(x^d-1)$ has only nonnegative coefficients. So that generalizes the cyclotomic case, but not via number theory!

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    $\begingroup$ Maybe, the following related result of Kellog is helpful: Let $A$ be a complex $n\times n$ matrix. If all its elementary symmetric functions are positive (so that the characteristic polynomial has alternating signs), then the spectrum of $A$ lies in the set $\{z : |\text{arg}z| \le \pi - \pi/n\}$.... $\endgroup$ – Suvrit Aug 15 '15 at 1:51
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    $\begingroup$ In the wedge case you can compute explicitly the coefficients and see they are positive. It's a particular case of Suffridge's extremal polynomials- you have the expression deduced in Sheil-Small, Complex polynomials, p. 251-252. $\endgroup$ – user75485 Aug 18 '15 at 20:42
  • $\begingroup$ In fact the Suffridge polynomials cover the more general case when $1\not\in S,$ S self-conjugate and the arguments of the roots in S are separated by the same angle, except for a pair. $\endgroup$ – user75485 Aug 19 '15 at 14:44
  • $\begingroup$ @Josep: The reference you gave is extremely interesting. I was totally unaware of Suffridge's result. $\endgroup$ – Louis Deaett Aug 21 '15 at 20:02
  • $\begingroup$ "I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial": yes, you're correct. The easiest way to see this is plugging in x=1: it's easy to show that $\Phi_m(1) = \begin{cases}0 & m=1 \\ p & m=p^k \\ 1 & \text{otherwise}\end{cases}.$ This essentially says there must be cancellation outside of the prime-power case. $\endgroup$ – Joshua P. Swanson May 6 '19 at 14:22

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