Timeline for Polynomial $g:\mathbb R^n \rightarrow\mathbb R^n$ with no critical point may have no root
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2015 at 15:19 | answer | added | Jean Duchon | timeline score: 3 | |
| Nov 12, 2015 at 8:22 | history | edited | Denis Serre | CC BY-SA 3.0 |
deleted 11 characters in body; edited title
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| Nov 12, 2015 at 6:32 | history | edited | Guo Qi | CC BY-SA 3.0 |
added 8 characters in body; edited title
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| S Nov 10, 2015 at 15:06 | history | suggested | BigM | CC BY-SA 3.0 |
Latex corrections
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| Nov 10, 2015 at 14:46 | review | Suggested edits | |||
| S Nov 10, 2015 at 15:06 | |||||
| Nov 9, 2015 at 20:28 | answer | added | David E Speyer | timeline score: 27 | |
| Nov 9, 2015 at 16:31 | comment | added | David E Speyer | This reminds me of mathoverflow.net/questions/38019/… | |
| Nov 9, 2015 at 16:30 | history | edited | David E Speyer | CC BY-SA 3.0 |
edited body
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| Nov 9, 2015 at 16:29 | comment | added | David E Speyer | I don't understand the votes to close: Both (1) and (2) are reasonable questions; a "yes" answer to (1) implies a "yes" answer to (2). My guess is that they are both false, though... | |
| Nov 9, 2015 at 16:28 | comment | added | Guo Qi | I did some changes. Thank you, Jean. | |
| Nov 9, 2015 at 16:25 | history | edited | Guo Qi | CC BY-SA 3.0 |
edited body; edited title
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| Nov 9, 2015 at 16:19 | history | edited | Guo Qi | CC BY-SA 3.0 |
edited body; edited title
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| Nov 9, 2015 at 16:17 | comment | added | Jean Duchon | Then version 2 should read: Version 2: If $f$ : $\mathbb R^n \rightarrow \mathbb R$ is a polynomial, $D^2f(x)$ is non-degenerate for every $x$, then $f$ has at least one critical point. (It is a consequence of version 1 if version 1 is true, not equivalent to version 1). | |
| Nov 9, 2015 at 15:54 | comment | added | Guo Qi | In version 1, we can prove g is an open mapping. However, g maybe not a closed mapping. | |
| Nov 9, 2015 at 15:48 | comment | added | Guo Qi | Sorry about my mistake, f in version 2 should be a polynomial functional on R^n, which means f:R^n \rightarrow R. | |
| S Nov 9, 2015 at 14:51 | history | suggested | Jean Duchon | CC BY-SA 3.0 |
formatting (latex)
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| Nov 9, 2015 at 14:42 | review | Suggested edits | |||
| S Nov 9, 2015 at 14:51 | |||||
| Nov 9, 2015 at 14:38 | review | Close votes | |||
| Nov 9, 2015 at 23:59 | |||||
| Nov 9, 2015 at 14:22 | comment | added | Liviu Nicolaescu | Is $f$ a map $\mathbb{R}^n\to\mathbb{R}$ or a map $\mathbb{R}^n\to\mathbb{R}^n$. In the second case, what is the meaning of the nondegeneracy of $D^2f$? | |
| Nov 9, 2015 at 14:13 | review | First posts | |||
| Nov 9, 2015 at 15:34 | |||||
| Nov 9, 2015 at 14:13 | history | asked | Guo Qi | CC BY-SA 3.0 |