Timeline for Surface area of superellipsoid (dice)
Current License: CC BY-SA 3.0
13 events
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| Nov 7, 2011 at 22:15 | comment | added | Will Sawin | Also, a function $f$ whose level sets are all scaled copies of a convex body around the origin, should satisfy the property if and only if that convex body is the intersection of hyperplanes tangent to a single sphere. Thus, all regular polytopes, spheres, cylinders, and some stranger things work, but spheres are the only smooth bounded solutions. | |
| Nov 7, 2011 at 21:40 | comment | added | Will Jagy | @Will Sawin, it is also true that $1,2,\infty$ are the only nice matrix norms, where "nice" means worth calculating. I suppose this comes from the vector norms being nice. Anyway, sorry that I did not get your full intent, a comment box is a little small to get across the nature of a query...if I put an at sign and the full MO name, it is supposed to alert the person named. Maybe that will work. | |
| Nov 7, 2011 at 14:46 | comment | added | Vitali Kapovitch |
Given a function $f$ we have that $\frac{d}{dt}vol\{f\le t\}=\int_{f=t}\frac{1}{|\nabla f|}$. For $f(x)=||x||_p=(|x_1|^p+\ldots+|x_n|^p)^{1/p}$ on $\mathbb R^n$ with $n\ge 2,p\ge 1$ it's only true that $|\nabla f|=const$ for $p=1,2$ and $\infty$. That's what makes the formula jbsmoove was suggesting valid for those $p$ and no others.
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| Nov 7, 2011 at 8:00 | comment | added | Will Sawin | I know that. What's not obvious is why a certain type of symmetry should be connected to three different, other types of symmetries. It's odd. I can do calculations, but they're ugly. Slightly better is thinking about it in terms of allowable shapes (it's an interesting differential equation-ish thing), but it still leaves open the question of why those are the only times you get allowable shapes. | |
| Nov 7, 2011 at 6:16 | comment | added | Will Jagy | $m = 1,2 \infty$ are, in this problem, the regular octahedron, sphere, cube. | |
| Nov 7, 2011 at 5:35 | comment | added | Will Sawin | What sort of statement is true for $m=1$, $m=2$, and $m=\infty$ but not any other value of $m$? I mean $(m-1)(m-2)/m^3=0$, of course, but how does that show up in this problem? | |
| Nov 7, 2011 at 2:27 | comment | added | Will Jagy | Yes, removing unfortunate posts is a good idea. You may be able to see the word "delete," if so click on that, and then when it asks you whether you wish to vote to delete say yes. If you cannot see any such option, you can still edit your post down to the minimum 15 characters, such as "Well, never mind." | |
| Nov 7, 2011 at 1:57 | comment | added | Noam D. Elkies | Removing one's post must be legitimate since there are two "badges" that can only be earned by doing that (though it would take two more down-votes or four ups for you to become eligible for one by removing this post). | |
| Nov 7, 2011 at 1:00 | comment | added | jbsmoove | I am an idiot, messed up the computation. Is it considered polite to vote to remove your own post? | |
| Nov 7, 2011 at 0:47 | comment | added | Vitali Kapovitch | true but the Op restricted the range to $m\ge 2$ so I didn't mention the case $m=1$. | |
| Nov 7, 2011 at 0:22 | comment | added | Noam D. Elkies | The level sets are also equidistant at least for $m=1$ (octahedron). | |
| Nov 6, 2011 at 23:32 | comment | added | Vitali Kapovitch | this is obviously wrong as the level sets are not equidistant unless $m=\infty$(cube) or $m=2$ (sphere). | |
| Nov 6, 2011 at 22:21 | history | answered | jbsmoove | CC BY-SA 3.0 |