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Timeline for "Local" Gauss-Lucas theorem?

Current License: CC BY-SA 3.0

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Aug 3, 2017 at 2:02 answer added Trevor J Richards timeline score: 6
Jul 30, 2017 at 14:22 comment added Andreas Rüdinger Sendov's conjecture could be worthwhile looking at: mathworld.wolfram.com/SendovConjecture.html,https://arxiv.org/…
Jul 30, 2017 at 14:20 comment added Andreas Rüdinger Some comments: a) Taking $z^3-1$ and adding one further zero (say at $z=a$), i.e taking $p = (z^3-1)(z-a) = z^4 + \ldots -z -a$ it is easy to see that there must be at least one zero of $p'$ with absolute value $\le 4^{-1/3}$. b) For a general triangle of zeros and one further zero, my conjecture is that there is at least one zero of $p'$ within the (closed) triangle. c) Conjecture: For each triple of zeros the union of the circumcircle of this triangle with the three circles that have the three sides of the triangle as diameter contains at least one zero of $p'$? Any counterexamples?
Jul 30, 2017 at 5:47 comment added Andreas Rüdinger My last comment is not correct, I had something different in mind. Sorry. I wasn't able to delete it.
Jul 29, 2017 at 20:09 comment added Andreas Rüdinger A small observation: for $p(z) = z^n -1$ a root of $f'$ is located not inside the circle through three zeroes, but on its boundary.
Jul 29, 2017 at 19:58 comment added Andreas Rüdinger I would assume the following paper on approximate Gauss-Lucas theorems (see already page 1) is relevant: arxiv.org/pdf/1706.05410.pdf
Jul 29, 2017 at 17:43 comment added Trevor J Richards @CarlSchildkraut Yes, if $p(z)$ has zeros at the third roots of unity, and the other zeros of $p$ are sufficiently far away, then $p$ will have two critical points arbitrarily close to $0$. On the other hand, if the other roots of $p$ are close to (or in) the unit disk, then $p$ will have at least two, or all, of its critical points in the unit disk. The question is what happens in the "intermediate case", when at least some of the other zeros of $p$ are (relatively) close to, but not in, the unit disk.
Jul 29, 2017 at 16:07 comment added Carl Schildkraut @GerryMyerson as $a$ grows, the roots of $z^3-1$ act pretty much like a triple root, so there should be two roots of the polynomials derivative near them.
Jul 29, 2017 at 12:27 comment added Gerry Myerson Here's a suggestion, using just polynomials, and avoiding collinearity. Consider $f(z)=(z^2+1)(z-1)(z-a)^n$. This has zeros at $z=\pm i$ and at $z=1$. By playing with $n$ and $a$, you may be able to convince $f'$ to have all of its zeros far away from these three zeros of $f$. Maybe even $(z^3-1)(z-a)^n$ can be made to work.
Jul 29, 2017 at 11:59 history edited Harry Richman CC BY-SA 3.0
footnote on collinear case modified, still unresolved
Jul 29, 2017 at 5:31 comment added Gerry Myerson @Trevor, my information wasn't really relevant – OP wanted a polynomial, and my example was $(z^3-z)e^{az}$ for suitable $a$.
Jul 28, 2017 at 16:20 comment added Trevor J Richards Could you give an indication of Gerry Myerson's explanation that information about colinear roots gives no information about the zeros of the derivative?
Jul 28, 2017 at 4:43 history edited Harry Richman CC BY-SA 3.0
noted the exception when roots are collinear
Jul 28, 2017 at 4:39 comment added Harry Richman [edited to make note of the degenerate case when roots are collinear]
Jul 28, 2017 at 3:37 history edited Harry Richman CC BY-SA 3.0
added 4 characters in body
Jul 27, 2017 at 22:13 history edited Harry Richman CC BY-SA 3.0
added 53 characters in body
Mar 22, 2017 at 21:40 comment added Trevor J Richards I will think more about your question. There is a paper with some partial results which I will try to find. Meanwhile, I will point you to a slightly related question which you may find interesting which I asked on this site some time ago: mathoverflow.net/questions/189245/…
Feb 23, 2017 at 8:17 history asked Harry Richman CC BY-SA 3.0