Timeline for Gauss--Lucas type theorem for tracts and higher derivatives of a polynomial
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 5, 2015 at 23:19 | comment | added | Bobby Ocean | Well, I knew it had to create a non-convex level set (otherwise Gauss-Lucas would apply). Hence, a rotationally symmetric distribution of zeros would fail. Likewise, I was pretty sure the conjecture held for polynomials with only real zeros. Thus, I needed a kind of C-shaped level set. Hence, the above. | |
| May 5, 2015 at 11:59 | comment | added | Lasse Rempe | Nice example - how did you go about finding it? | |
| May 4, 2015 at 17:41 | comment | added | Bobby Ocean | I believe, I can prove your conjecture is true for all real valued cubic polynomials; i.e., the second derivative's zeros will be contained in the tract as well as the first derivative's zeros. Oh well, time for work. I will definitely give this some more thought. | |
| May 4, 2015 at 17:01 | comment | added | Bobby Ocean | I believe your conjecture is true for polynomials with only real zeros. Since any level set that contains all of the zeros must also contain the critical points, it follows that the interval $[x_0,x_n]$ must be contained in the level set as well (where $x_0$ is the smallest zero and $x_n$ is the largest zero). Since it is know that derivatives interlace, then your conjecture follows. | |
| May 4, 2015 at 16:08 | comment | added | Bobby Ocean | I did spend some time trying to put the second derivative's zeros outside the tract, but could not successfully. Need to spend some more time with it. | |
| May 4, 2015 at 15:31 | vote | accept | Trevor J Richards | ||
| May 4, 2015 at 15:31 | comment | added | Trevor J Richards | Thanks very much for the counter-example. The conjecture definitely holds for the first deriv. If, $x_0$ is a crit. point of $p$, and $\lambda$ is the level curve of $p$ containing $x_0$, then the max. mod. thm. implies that each face of $\lambda$ contains a zero of $p$. Thus any level curve of $p$ containing the zeros of $p$ in its bounded face must also contain $\lambda$ in its bounded face, and thus contain $x_0$ in its bounded face. For the second derivative, it looks like if you tighten up the parameters on the counter-example, the second deriv. zeros might be outside the level curve. | |
| May 4, 2015 at 7:37 | history | answered | Bobby Ocean | CC BY-SA 3.0 |