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May 6, 2019 at 14:22 comment added Joshua P. Swanson "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.
Dec 20, 2018 at 9:41 history edited Wolfgang
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Aug 22, 2015 at 0:43 history edited Louis Deaett CC BY-SA 3.0
I fixed a typo in one equation, and made the title a bit more succinct.
Aug 21, 2015 at 20:02 comment added Louis Deaett @Josep: The reference you gave is extremely interesting. I was totally unaware of Suffridge's result.
Aug 21, 2015 at 19:49 history edited Louis Deaett CC BY-SA 3.0
I changed the format of the citation and also added some additional info regarding the cyclotomic case.
Aug 19, 2015 at 14:44 comment added user75485 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.
Aug 18, 2015 at 20:42 comment added user75485 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.
Aug 15, 2015 at 1:51 comment added Suvrit 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 $\{z : |\text{arg}z| \le \pi - \pi/n\}$....
Aug 14, 2015 at 14:23 history asked Louis Deaett CC BY-SA 3.0