Timeline for Gauss--Lucas type theorem for tracts and higher derivatives of a polynomial
Current License: CC BY-SA 3.0
10 events
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| Dec 15, 2014 at 11:12 | comment | added | Lasse Rempe | @TrevorRichards a) The convex hull of the set of zeros of an entire function will in general be much larger than the "tracts" in question, so $f$ would normally be expected to be unbounded on this convex hull (e.g. consider the zeros of the function e^{z^3}-1); b) There may be some limiting zeros of polynomials that "disappear", so that the convex hull of the zeros of the polynomials may not converge to the convex hull of the zeros of the limiting functions. | |
| Dec 12, 2014 at 17:55 | comment | added | Trevor J Richards | But surely one could find an entire function with finitely many singular values for which that result fails (ie there is a higher order crit point $z$ at which $|f|$ takes a value bigger than $M_f=\max(|f(x)|:x\in ConvexHull(Zeros\ of\ f))$), and then approximate with polys. But the failure for $f$ does not pass to the approximating polys by Hurwitz's theorem in this case...? | |
| Dec 12, 2014 at 17:54 | comment | added | Trevor J Richards | thank you again for your responses. I do not know why I did not think of Hurwitz's theorem. But I am still confused, because it seems that the same sort of argument might be made against the regular Gauss--Lucas theorem. The regular G-L theorem implies that the $|p|$ values at each higher order crit point of a poly $p$ are less/equal than $M$, where now $M$ is the max of $|p|$ on the convex hull of the zeros of $p$. (continued) | |
| Dec 12, 2014 at 9:55 | comment | added | Lasse Rempe | @TrevorRichards Re: Question 2) - en.wikipedia.org/wiki/… | |
| Dec 12, 2014 at 9:52 | comment | added | Lasse Rempe | @TrevorRichards Alternatively, if you are looking for explicit examples of polynomials, I suggest looking at some Shabat polynomials (= polynomial Belyi functions), i.e. polynomials with two critical values (or -1 and 1 may be the best normalisation for your question). Given any tree, with an embedding in the plane, there is a Shabat polynomial realising this tree. There are some programs for computing these (eg Don Marshall's "zipper", and Laurent Bartholdi also has a program). Not sure they're publicly available, but they exist. Just draw some "complicated" trees and experiment .. | |
| Dec 12, 2014 at 9:47 | comment | added | Lasse Rempe | @TrevorRichards: The trouble with the "easy" examples is that they all essentially look like e^z, which does not have higher-order critical points ... One case of entire functions with a finite set of singular values that it's relatively easy to get your hands on, and that have quite different tracts from exponential maps, are Poincaré (linearising) functions of post-critically finite polynomials around repelling periodic points. | |
| Dec 12, 2014 at 2:51 | comment | added | Trevor J Richards | Question 2) I am not so sure now that I understand the application of your statement that if a higher deriv. of $f$ had a zero outside the disk in question, then the higher deriv of the approximating polys would as well. After all, while the function $f$ has a fixed tract containing all of its crit. pts., the tracts which contain the crit. points of the approximating polys may be growing very large, since they only have the same crit values as $f$, not necessarily the same crit. points as $f$, is this right? | |
| Dec 12, 2014 at 2:41 | comment | added | Trevor J Richards | Question 1) Is there an easy example you can think of where the higher derivs of $f$ have roots in $V$? (Obviously if the answer were immediate, you would have given it earlier, but perhaps my examples above my suggest something.) | |
| Dec 12, 2014 at 2:39 | comment | added | Trevor J Richards | Thank you very much for the response, and the references. I follow your answer, and have been working through some examples. Several (naive) observations: An easy collection of examples are the functions $f(z)=p(z)e^{q(z)}$, where $p$ and $q$ are polynomials. For this case it appears that $V$ has a single component, and the simple examples I have already done (ie $ze^z$, $e^{z^2+z}$) work out fine. Alternatively, a simple example where $V$ has several (here $2$) components is $f(z)=\cos(z)$. Here also the higher derivatives of $f$ have no roots in $V$. (Question in the next commment.) | |
| Dec 11, 2014 at 16:37 | history | answered | Lasse Rempe | CC BY-SA 3.0 |