Timeline for Iterated antiderivatives of polynomials having many real roots
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Apr 8, 2021 at 2:54 | history | bounty ended | Stefan Steinerberger | ||
| S Apr 8, 2021 at 2:54 | history | notice removed | Stefan Steinerberger | ||
| Apr 7, 2021 at 5:34 | vote | accept | Stefan Steinerberger | ||
| Apr 7, 2021 at 1:35 | comment | added | Richard Stanley | @StefanSteinerberger, for the thrill of receiving my first bounty, I will convert my comment to an answer. | |
| Apr 7, 2021 at 1:34 | answer | added | Richard Stanley | timeline score: 5 | |
| Apr 6, 2021 at 16:28 | comment | added | Stefan Steinerberger | Ah, this is fantastic, this leads to a complete characterization! This is tremendously helpful -- thank you! I would award the bounty if it was an answer, I am not sure I can award it to a comment. | |
| Apr 6, 2021 at 2:20 | comment | added | Richard Stanley | The question seems to be answered at the "Untitled" link from canvas.wisc.edu when you do a google search on "appell sequence real zeros". I don't see how $F(x)=\sum \frac{x^n}{n!^2}$ fits into the picture. If this doesn't actually work, what is the least $n$ for which $P_n(t)$ has a nonreal zero? | |
| Apr 6, 2021 at 1:50 | comment | added | Richard Stanley | Even more generally, power series like $F(x)=\sum_{n\geq 0}\frac{x^n}{n!(2n+1)!(5n+3)!}$ also seem to work (if Maple can be trusted). | |
| Apr 6, 2021 at 0:38 | comment | added | Richard Stanley | The question is equivalent to asking for which power series $F(x)\in\mathbb{C}[[x]]$ is it true that if $e^{tx}F(x)=\sum_{n\geq 0}P_n(t)\frac{x^n}{n!}$, then all the roots (or zeros) of $P_n(t)$ are real. This seems to be quite a rare property. Possible examples other than $e^{-x^2}$ and some related ones like $(1+x)e^{-x^2}$ are $\cosh\sqrt{x}$ and $\sqrt{x}\sinh\sqrt{x}$. | |
| S Apr 4, 2021 at 14:48 | history | bounty started | Stefan Steinerberger | ||
| S Apr 4, 2021 at 14:48 | history | notice added | Stefan Steinerberger | Draw attention | |
| Apr 4, 2021 at 6:26 | comment | added | Kevin Casto | Oops, missed the 'real', you're right! | |
| Apr 4, 2021 at 4:05 | comment | added | Stefan Steinerberger | I don't think it's always possible. If we take (-1 + x) (-0.8 + x) (0.8 + x) (1 + x), then no matter what constant we choose for the antiderivative, the antiderivative cannot have more than 3 real roots. The problem is not only about the roots being distinct, they should also all be real. | |
| Apr 4, 2021 at 3:24 | comment | added | Kevin Casto | The discriminant of $p$ is polynomial in the coefficients of $p$ and vanishes just when $p$ has multiple roots. So given $p$, the discriminant of $p + C$ is a polynomial in $C$, and so only finitely many choices of $C$ will have $p + C$ have multiple roots. So you should always be able to do this. | |
| Apr 4, 2021 at 1:10 | history | edited | Stefan Steinerberger | CC BY-SA 4.0 |
simplified exposition
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| Apr 2, 2021 at 4:18 | history | asked | Stefan Steinerberger | CC BY-SA 4.0 |