Timeline for A generalization of Cauchy's mean value theorem.
Current License: CC BY-SA 2.5
15 events
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| Feb 26, 2010 at 22:25 | comment | added | Petya | @Leonid Thank you! Nice statement. Here is another my generalization of Cauchy's thm: For a smooth curve $x(t)\colon [a,b] \to {\mathbf R^{n+1}}$ there exists a value $t\in]a,b$ such that vectors $x(b)-x(a), x(t)-x(a), x'(t), x''(t),...,x^{(n-2)}, x^{(n)}(t)$ are linearly dependent. | |
| Feb 26, 2010 at 17:24 | vote | accept | Petya | ||
| Feb 26, 2010 at 16:01 | answer | added | Petya | timeline score: 4 | |
| Feb 25, 2010 at 21:31 | comment | added | Anton Petrunin | Here is an other sufficient condition for disc: for any linear function $\ell:\mathbb R^3\to\mathbb R$, the function $\ell\circ f: S^1\to\mathbb R$ has at most two local mimima and maxima. | |
| Feb 25, 2010 at 13:34 | comment | added | Douglas Zare | For the 2-disk, if you have the extra condition that two immersions agree on the boundary which describes a simple closed curve of tangent plane directions, then the Gauss map gives you two disks with the same boundary in the projective plane. The boundary curve can't be contracted in the punctured projective plane, so there is no disjoint pair of immersions agreeing on the boundary. This suggests looking at solid tori immersed in R^4 with a torus boundary, since you can decompose RP^3 into two solid tori. | |
| Feb 25, 2010 at 5:37 | comment | added | Petya | Even if there is a counterexample, there are plenty of possibilities to vary the question. It is interesting to get a positive statement! Thank you. | |
| Feb 25, 2010 at 5:33 | comment | added | Anton Petrunin | After deleting my answer: I still think there should be a counterexample --- it is kind of h-principle. But I will better check it in the morning. | |
| Feb 25, 2010 at 5:04 | vote | accept | Petya | ||
| Feb 25, 2010 at 5:10 | |||||
| Feb 25, 2010 at 4:34 | answer | added | Anton Petrunin | timeline score: 5 | |
| Feb 25, 2010 at 4:17 | comment | added | Anton Petrunin | @Leonid. It does not really work this way. Look at the ruled surfaces... | |
| Feb 25, 2010 at 4:13 | comment | added | Anton Petrunin | Is it stated as an "open-problem" somewhere? | |
| Feb 25, 2010 at 2:26 | history | edited | Petya | CC BY-SA 2.5 |
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| Feb 25, 2010 at 2:25 | comment | added | Petya | You are right, thank you. M is supposed to be compact, I forget to mention it. I've tried to solve it for a disk, yes, but unsucsessfully. | |
| Feb 25, 2010 at 2:19 | comment | added | Harald Hanche-Olsen |
I suspect you need $M$ to be compact. Otherwise, $M=[0,1)$ looks like a counterexample to me. What reasons do you have for the conjecture to be true? The next simplest case after the intervals would be the disk, I think. Have you tried proving your conjecture for that case? (I am not a differential geometer, so I can't contribute to a solution. But I am curious all the same.)
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| Feb 25, 2010 at 0:32 | history | asked | Petya | CC BY-SA 2.5 |