Timeline for Polynomial with all zeros on a circle and many real coefficients
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2013 at 13:08 | comment | added | Wolfgang | With a more or less straightforward construction, I don't think much less than half is possible. That is why the question intrigues me so much, because in the light of my intuitive argument, much "better" polynomials should exist. Assuming it's possible to find rational polynomials, one could of course try to solve a system of diophantine equations, but I haven't the courage to set up such a system and don't think even a computer could solve it. It would be nice if someone knows about a Java applet somewhere on the internet that calculates non-real coefficients from a set of graphical points. | |
| Feb 27, 2013 at 7:33 | comment | added | Aaron Meyerowitz | What is the best you have been able to do? Can you get much below half the coeffcients? | |
| Feb 26, 2013 at 1:56 | answer | added | Alexandre Eremenko | timeline score: 3 | |
| Feb 25, 2013 at 18:26 | comment | added | Wolfgang | My idea is the latter. I think that's already difficult enough. Once this is settled, I wonder if it is feasible to try to characterize which circles only allow less real coefficients. | |
| Feb 25, 2013 at 18:08 | comment | added | David E Speyer | @Wolfgang Could you clarify the order of quantifiers here? I interpreted the question as "For a given $\Omega$, what is the maximum number of real coefficients (as a function of $\Omega$ and $n$)?" I think other people are reading it as "What is the absolute maximum number, where we can choose any $\Omega$ as long as it is not circular over the real axis." | |
| Feb 25, 2013 at 18:06 | comment | added | David E Speyer | @Aaron Golden: Well, if the circle meets the real line. I think the question is interesting when it doesn't meet the real line, as well. | |
| Feb 25, 2013 at 17:31 | history | edited | Wolfgang | CC BY-SA 3.0 |
added "distinct"
|
| Feb 25, 2013 at 17:30 | comment | added | Wolfgang | yes, of course. Otherwise we could as well start with a circle that has 0 on it. OK, I'll add "distinct". | |
| Feb 25, 2013 at 17:14 | comment | added | Aaron Golden | You want all the roots to be simple right? Otherwise you can pick any suitable circle and take as many real roots as you like from the one or two points where that circle intersects the real line. | |
| Feb 25, 2013 at 15:53 | comment | added | Wolfgang | how's that? If they were all real, any non-real zeros would come in conjugate pairs and so $\Omega$ would be symmetric to the real line, which is excluded! The constraint of the points being on a circle sort of removes 1 degree of freedom. But only "sort of", that's why I have said vaguely "more or less" :) | |
| Feb 25, 2013 at 15:35 | comment | added | Anthony Quas | Why not $n$ degrees of freedom? You're choosing $n$ points (one degree of freedom each) and trying to satisfy $n$ constraints: $\Re(a_i)=0$ for $i=0,\ldots,n-1$. Shouldn't you expect to be able to get them all real? | |
| Feb 25, 2013 at 14:07 | history | asked | Wolfgang | CC BY-SA 3.0 |