Timeline for Gently falling functions
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2017 at 9:42 | history | edited | CommunityBot |
replaced http://people.csail.mit.edu/ with https://people.csail.mit.edu/
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| Dec 10, 2011 at 2:58 | comment | added | Mariano Suárez-Álvarez | A Joycean title! | |
| Dec 10, 2011 at 2:23 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Images moved to a different server.
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| Jun 20, 2011 at 16:53 | vote | accept | Joseph O'Rourke | ||
| Jun 20, 2011 at 16:53 | comment | added | Joseph O'Rourke | Although the class of all gently-falling functions remains unclear, much has been clarified by the combination of Gerhard's remarks and Scott's analysis and roller-coaster pointers. So I've accepted Scott's answer. Thanks! | |
| Jun 20, 2011 at 10:49 | comment | added | Joseph O'Rourke | @Mark: An interesting thought! It is tempting to consider it equivalent by time reversal, but I think in fact the notions are different. | |
| Jun 20, 2011 at 3:20 | answer | added | S. Carnahan♦ | timeline score: 9 | |
| Jun 19, 2011 at 20:15 | comment | added | Mark Bennet | An interesting related problem is what happens if you try to launch a particle up the curve so it comes to rest on the peak. | |
| Jun 19, 2011 at 19:29 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added a 3rd example.
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| Jun 18, 2011 at 13:12 | comment | added | Joseph O'Rourke | @Alon, Scott: Thanks, yes, I meant "arbitrarily close to the max." Corrected now. | |
| Jun 18, 2011 at 13:04 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 2 characters in body
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| Jun 18, 2011 at 10:49 | comment | added | S. Carnahan♦ | @Will: The equation is highly singular with your initial conditions. The particle is happy to drop like a rock when it has no initial speed. | |
| Jun 18, 2011 at 10:43 | comment | added | S. Carnahan♦ | @Alon: It seems that you can start a stationary particle anywhere, so the limit of separation points as your starting point approaches the top equilibrium point is what is indicated. | |
| Jun 18, 2011 at 6:59 | comment | added | Alon Amit | Sorry if that's a silly question but in example 1), a particle starting at (0,1) won't go anywhere unless you give it some initial horizontal velocity. Are you suggesting that the separation point tends to the indicated point as that velocity tends to 0? | |
| Jun 18, 2011 at 6:53 | answer | added | Gerhard Paseman | timeline score: 7 | |
| Jun 18, 2011 at 2:22 | comment | added | Will Jagy | Worth getting, at least, a numerical solution with $g(0) = 1$ and $ \dot{g}(0) = 0$ to the differential equation... | |
| Jun 18, 2011 at 0:49 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |