Timeline for Banach-Mazur distance between the cube and the octahedron
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2016 at 23:42 | history | edited | Wlodek Kuperberg | CC BY-SA 3.0 |
misspelling corrected
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| May 3, 2016 at 10:59 | vote | accept | tkobos | ||
| Apr 29, 2016 at 10:54 | comment | added | tkobos | That is a very good insight! And that would be actually a really nice answer. As I said, for $n=2^k$ it can be proved that the distance is at most $\sqrt{n}$ using Walsh matrices (although I don't recall the details at this moment). Maybe it is actually $\sqrt{n}$? For $n=4$ it seems so. I will wait for a few days before accepting this as an answer, as someone could possibly find a specific reference or post the proof. | |
| Apr 28, 2016 at 23:44 | comment | added | Robert Israel | $n=5$ also seems difficult for the Global Optimization Toolbox. | |
| Apr 28, 2016 at 19:15 | comment | added | j.c. | For what it's worth, this Mathematica code gist.github.com/anonymous/8f2bbb5ddb9a463a592ae6e6bfe363d6 returns the same values of s (though my T is the inverse of yours, confusingly). When I try using your conventions for T, Mathematica returns s=2.15 for n=4, and it's not clear to me why numerical error is so much larger in that case. I couldn't get a numerically reliable answer for n=5, there the best I've found is s=2.32871 with a floating point matrix with some entries suspiciously close to +-1 but the other entries messy. | |
| Apr 28, 2016 at 17:50 | history | edited | Robert Israel | CC BY-SA 3.0 |
added 22 characters in body
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| Apr 28, 2016 at 17:43 | history | edited | Robert Israel | CC BY-SA 3.0 |
added 22 characters in body
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| Apr 28, 2016 at 17:22 | history | answered | Robert Israel | CC BY-SA 3.0 |