Timeline for Approximating a convex disk by an ellipse
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 24, 2016 at 22:31 | comment | added | Jairo Bochi | @Gerhard, your suggestion ("How about this for a contradiction ...") actually applies whenever the two minimal ellipses have the same center. We can start by considering a restricted version of the problem where all sets involved are symmetric across the origin. But still it seems difficult to prove the desired inequality. :P | |
| S Dec 24, 2016 at 22:22 | history | suggested | Jairo Bochi | CC BY-SA 3.0 |
Better figure.
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| Dec 24, 2016 at 22:22 | comment | added | Jairo Bochi | I added labels to the figure. In order that the set with black boundary (which has the required area) be a counterexample (i.e., to be closer to the ellipses than to the circle), we would need the following inequality between areas: a+b+c>d+e. It should be clear now that there is no advantage in having P in the boundary of the set; actually the set in the figure seems to be the optimal choice. Still, the inequality fails to hold for all choices of epsilon, if my calculations are correct. I can't imagine a better attempt, so I suspect that no counterexample exists. Happy holidays! | |
| Dec 24, 2016 at 22:21 | comment | added | Gerhard Paseman | How about this for a contradiction (at least in the case that two minimal ellipses are axis aligned)? Do an affine transform that brings the two minimal ellipses to the same eccentricity and area. Now we have a curve of the type I posit, perhaps we can prove that the circle beats both ellipses. Gerhard "Anyone Ready To Make Lemonade?" Paseman, 2016.12.24. | |
| Dec 24, 2016 at 22:18 | review | Suggested edits | |||
| S Dec 24, 2016 at 22:22 | |||||
| Dec 24, 2016 at 22:08 | comment | added | Gerhard Paseman | Thank you @Jairo ! Even if I am wrong, and every disk does have a minimal covering ellipse, I hope the figure you provided helps the intuition and leads to a proof. Gerhard "And Happy New Year Too" Paseman, 2016.12.24. | |
| Dec 24, 2016 at 20:55 | comment | added | Jairo Bochi | I also initially thought that the are of the circle that "sticks out" might be the key for a counterexample. But that area is too small, unless epsilon is big (in which case the circle seems even more unbeatable). | |
| Dec 24, 2016 at 20:00 | comment | added | Joseph O'Rourke | "it looks like $P$ is very close to the circle": Indeed my drawing was slightly inaccurate; fixed now. Jairo's shows the situation more clearly. | |
| Dec 24, 2016 at 20:00 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Fixed location of P.
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| S Dec 24, 2016 at 17:46 | history | suggested | Jairo Bochi | CC BY-SA 3.0 |
Added a figure with the candidate example.
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| Dec 24, 2016 at 17:20 | comment | added | Jairo Bochi | I included a figure with a candidate set. It's not exactly your suggestion since P is not a boundary point, but it seems to be more efficient. Still, it's not sufficient to beat the circle (even taking smaller epsilon's). | |
| Dec 24, 2016 at 17:16 | review | Suggested edits | |||
| S Dec 24, 2016 at 17:46 | |||||
| Dec 24, 2016 at 16:42 | comment | added | Gerhard Paseman | Thank you Joseph! Unfortunately, it looks like P is very close to the circle, so more care will be needed to form the octaoscilliptical disk. Gerhard "And Merry Christmas To All" Paseman, 2016.12.24. | |
| Dec 24, 2016 at 14:15 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Attempted to draw what is described verbally.
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| Dec 24, 2016 at 5:42 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |