Timeline for Extremum placement for two-variable function
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 3 at 3:02 | comment | added | IscoBerlin | @Iosif Pinelis: Thanks a lot for your answer. I agree with you. | |
| May 3 at 3:00 | vote | accept | IscoBerlin | ||
| Jan 10 at 17:05 | comment | added | Gerry Myerson | "While teaching Calculus 2, one of my students asked me...." So, one of your students was teaching Calculus 2? | |
| Jan 9 at 13:36 | comment | added | Iosif Pinelis | @PietroMajer : On the other thought, $\mathbb R^2$ is not diffeomorphic (or even homeomorphic) to (say) $\mathbb R^2\setminus\{(0,0)\}$, because $\mathbb R^2$ is simply connected but $\mathbb R^2\setminus\{(0,0)\}$ is not. | |
| Jan 9 at 12:20 | answer | added | fedja | timeline score: 2 | |
| Jan 9 at 5:39 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Fully Math Jaxed and formatted
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| Jan 8 at 21:05 | comment | added | Pietro Majer | Or also if G is the finite set of all the others critical points, we can connect them by a path (or a tree) $\Gamma$ that avoids F, and take $\Omega$ as a nbd of $\Gamma$ | |
| Jan 8 at 20:31 | comment | added | Iosif Pinelis | @PietroMajer : I see now: for a finite $F$, you can push points in $F$ to inifinity one-by-one. | |
| Jan 8 at 20:10 | comment | added | Iosif Pinelis | @PietroMajer : I understand this. But how do you construct a diffeo from $\mathbb R^n$ onto a given open subset $U$ of $\mathbb R^n$, say $U:=\mathbb R^n\setminus F$, where $F$ is a finite set? | |
| Jan 8 at 19:50 | comment | added | Pietro Majer | Maybe the expression was too vague/colored; I mean just this: Suppose $f:\mathbb R^n\to\mathbb R$ has many critical points with various indices. If $\phi:\mathbb R^n\to\Omega\subset \mathbb R^n$ is a diffeo, then $g:=f\circ \phi$ has critical points in bijection with the critical points of $f$ in $\Omega$, with same index. | |
| Jan 8 at 19:19 | comment | added | Iosif Pinelis | @PietroMajer : Can you explain how to push them to infinity by a diffeo? | |
| Jan 8 at 18:44 | comment | added | Pietro Majer | As to the existence, it is no problem realizing any configuration. You can start with a function with all critical points and morse indices you want, and maybe more; then you push to infinity all extra critical points by a diffeo | |
| Jan 8 at 17:03 | review | Close votes | |||
| Jan 16 at 12:31 | |||||
| Jan 8 at 16:48 | history | edited | YCor |
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| Jan 8 at 16:42 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Jan 8 at 4:27 | history | asked | IscoBerlin | CC BY-SA 4.0 |