Timeline for Measure of the Attractor of Critical Points of a Manifold
Current License: CC BY-SA 3.0
26 events
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| Apr 29, 2016 at 21:32 | comment | added | Ben McKay | I am afraid that this web site has the fundamental problem that if people don't suddenly see a clear insight into your problem that is at the same time a fun insight for them to pursue, then they lose interest immediately. I suspect that Ryan Budney (who is a great source of help to many people on this site) just didn't have anything exciting come to his mind. That doesn't mean it isn't a good problem, or that he isn't clever enough to make progress with it, but only that it didn't immediately set his brain on fire. This site depends on the rare instances when we set brains on fire. | |
| Apr 29, 2016 at 20:47 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 27, 2016 at 20:40 | comment | added | Blake | @RyanBudney Why do you even bother answering questions if you're not going to explain what you mean in full? In the future please stop "answering" questions that I ask. Your answers are never helpful and all it does is stop other people from answering. | |
| Apr 27, 2016 at 18:40 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 27, 2016 at 16:55 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 27, 2016 at 16:39 | comment | added | Blake | @RyanBudney Could you at least provide a source for that result? | |
| Apr 26, 2016 at 18:04 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 26, 2016 at 0:42 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 23, 2016 at 16:23 | comment | added | Jaap Eldering | @Ryan Budney: wouldn't you need that bound to be relative to the eigenvalues of the Hessian of $f$ at the critical points? Otherwise, you could simply take any bounded $f$ and then rescale it to make its fourth derivative as small as you like without changing the basins of attraction. | |
| Apr 23, 2016 at 16:00 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 22, 2016 at 2:36 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 22, 2016 at 0:32 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 21, 2016 at 20:11 | comment | added | Blake | Yes, I could assume that I have access to 4th derivatives. Could you elaborate further on how you turn that into a bound on the derivative of the gradient and how that gives a lower bound on the diameter of a ball contained in the stable manifold of a critical point? | |
| Apr 21, 2016 at 18:00 | comment | added | Ryan Budney | You are free to make assumptions through a fairly wide universe of possible assumptions that would lead to consequences. For example, if you had a bound on the 4-th derivative of $f$ you could convert that into a lipschitz bound on the derivative of the gradient. . . which would give you a lower bound on the diameter of a ball contained in the stable manifold of a critical point. In which ever context you are in, does the fourth derivative of $f$ come up? | |
| Apr 21, 2016 at 17:46 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 20, 2016 at 14:32 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 20, 2016 at 0:24 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 19, 2016 at 16:08 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 19, 2016 at 2:42 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 18, 2016 at 20:48 | comment | added | Blake | What type of assumption would I have to make to get somewhere on this problem? Lovely reference, btw. | |
| Apr 18, 2016 at 20:46 | comment | added | Ben McKay | You can still make small valleys and large valleys in a smooth mountain range, and there is no bound on how much or how little of the rain water falls into some chosen valleys, if all you know is how many valleys you get to choose: xkcd.com/681 | |
| Apr 18, 2016 at 20:32 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 18, 2016 at 20:31 | comment | added | Blake | So $f > 0$ everywhere and assume that it has at least 1 local minima. In general it's nonconvex. I'd like the lower bound to depend on properties like $d$ and the number of critical points with the property $P$ | |
| Apr 18, 2016 at 20:28 | comment | added | Ben McKay | Since $f$ might have no critical points, the only lower bound (in absence of other information) that will hold for all such problems is zero. What data do you want your lower bound to depend on? Keep in mind that if $f(x)$ gets large negative as $|x|$ gets large, gradient flow will tend to move you away from critical points. | |
| Apr 18, 2016 at 17:28 | history | edited | Blake | CC BY-SA 3.0 |
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| Apr 18, 2016 at 17:22 | history | asked | Blake | CC BY-SA 3.0 |