Timeline for Is there a uniform bound on the number of solutions to ${\partial p \over \partial x_i} (x_1,...,x_n) = c_i$ outside a set of measure zero?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2022 at 20:09 | vote | accept | Zarrax | ||
| Jun 15, 2022 at 20:05 | comment | added | Zarrax | Gotcha. While the $(c_1,..,c_n)$ coming from the critical set are excluded using Bezout, I think I have a way around that in the situation I'm looking at. Thanks. | |
| Jun 15, 2022 at 19:54 | comment | added | Kevin Casto | Yeah, I'd worry about $e^{-1/x}$ or something similar | |
| Jun 15, 2022 at 19:52 | comment | added | Kevin Casto | You're only applying Bezout to the preimages of $c$'s outside of $Z$, where there won't be any common components for the reason I mentioned | |
| Jun 15, 2022 at 19:52 | comment | added | Zarrax | Basically I start with a function real analytic near the origin and I'd need a neighborhood of the origin where this would work. The neighborhood could be as small as you want though. | |
| Jun 15, 2022 at 19:51 | comment | added | Kevin Casto | @Zarrax re:locally, depends what you want. Locally on the domain, where you can pick the nbhd after you pick the function -- sure, just pick somewhere where everything is nice. If you have to pick the point before F, I would worry about bump functions | |
| Jun 15, 2022 at 19:48 | comment | added | Zarrax | So Bezout's theorem still gives the desired bound at points that are not in the zero set of the Jacobian? The versions of Bezout theorem I found online just didn't allow common components at all. | |
| Jun 15, 2022 at 19:47 | comment | added | pinaki | Re: "the assumption" in Bezout's theorem, see the text of this question: mathoverflow.net/q/417812/1508 | |
| Jun 15, 2022 at 19:38 | comment | added | Kevin Casto | The Jacobian determinant will vanish at points where they have a common component. Otherwise the intersection would be codimension less than $n$ and so it couldn't be the preimage of a regular value. | |
| Jun 15, 2022 at 19:30 | history | edited | Kevin Casto | CC BY-SA 4.0 |
added 12 characters in body
|
| Jun 15, 2022 at 18:04 | comment | added | Zarrax | By the way, the reason I was hesitant to use Bezout's theorem is the assumption in that theorem that the zero sets have no common component, which can happen in some cases here. Are you saying there's a way around this? | |
| Jun 15, 2022 at 17:51 | comment | added | Zarrax | Ok thanks. For the real analytic case it would have to be a local theorem. Is this possible? | |
| Jun 15, 2022 at 17:18 | comment | added | Kevin Casto | Sorry, actually the simplest thing I can think of bounds it by the product of the degrees, which in your case would be $(n-1)^n$. I can try to think if there's a better bound in your case. It definitely does not extend to real analytic, both bc you need to switch Z to the codomain bc of bump functions, and bc there could be infinitely many preimages (consider sine) | |
| Jun 15, 2022 at 17:15 | history | edited | Kevin Casto | CC BY-SA 4.0 |
added 140 characters in body
|
| Jun 15, 2022 at 16:35 | history | edited | Kevin Casto | CC BY-SA 4.0 |
added 17 characters in body
|
| Jun 15, 2022 at 16:32 | comment | added | Zarrax | Can you provide a reference for saying the size of the finite set is uniformly bounded by the degree of $F$? Also, would this extend to real analytic functions? | |
| Jun 15, 2022 at 16:24 | history | undeleted | Kevin Casto | ||
| Jun 15, 2022 at 16:24 | history | edited | Kevin Casto | CC BY-SA 4.0 |
added 164 characters in body
|
| Jun 15, 2022 at 15:55 | history | deleted | Kevin Casto | via Vote | |
| Jun 15, 2022 at 15:55 | history | answered | Kevin Casto | CC BY-SA 4.0 |