Timeline for Is this single-variable function being called with two variables (and if so, how do I handle that), or am I misreading the notation? [closed]
Current License: CC BY-SA 4.0
14 events
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| Feb 16 at 7:32 | comment | added | Willie Wong | You are definitely misreading the result. $r_0$ need not be any specific support function: the construction holds for any arbitrary known support function of a surface of constant width. Basically, the point is that if you start with a known example with very little symmetry, the summation allows you to produce a new example that is more symmetric, so that $\mathscr{G}$ is included in its symmetric group. | |
| Feb 16 at 4:12 | history | closed |
Steven Landsburg David White Andy Putman Sam Hopkins abx |
Not suitable for this site | |
| Feb 16 at 2:35 | comment | added | Carlo Beenakker | since this discussion is cumbersome for a string of comments, I have copied my answer to the answer box; if you point out what you have tried and why it does not "behave in a way that makes sense" I will try to respond to that. | |
| Feb 16 at 2:34 | answer | added | Carlo Beenakker | timeline score: 2 | |
| Feb 16 at 1:15 | comment | added | Lawton | @CarloBeenakker I've tried implementing it in GeoGebra the way you suggest, but I can't make it behave in a way that makes sense. | |
| Feb 15 at 22:10 | comment | added | Carlo Beenakker | just look at section 4.1: it refers to a rotationally symmetric function around the $x_3$ axis in terms of the radial parameter $R$, which must imply $R^2=x_1^2+x_2^2$, because that is what radial symmetry means | |
| Feb 15 at 21:33 | comment | added | Lawton | @CarloBeenakker If that is how $R$ should be interpreted, it isn't obvious at all to me from the rest of the paper. I'll try implementing it that way and see how it works. | |
| Feb 15 at 21:07 | comment | added | Carlo Beenakker | isn't the answer simply that $r_0(R)$ means a function of the two variables $x_1$ and $x_2$ that depends only on the combination $R=\sqrt{x_1^2+x_2^2}$, so you can write it equivalently as a function of a single variable; the equation you write down in terms of $r_0(g(\xi),g(\bar{\xi}))$ applies to any function $r_0(x_1,x_2)$, while the specific example 4.2 is for a radially symmetric function. | |
| Feb 15 at 20:03 | review | Close votes | |||
| Feb 16 at 4:15 | |||||
| Feb 15 at 19:59 | comment | added | Lawton | @StevenLandsburg See my comment to Carlo. | |
| Feb 15 at 19:56 | comment | added | Lawton | @CarloBeenakker The old question had been edited so much that the comments weren't at all connected to the latest content, and the comments hadn't been helpful in solving my problem anyway. Someone commented that due to all the changes and the lack of relevancy with old comments I should start over with a new question, so I did. | |
| Feb 15 at 19:48 | comment | added | Steven Landsburg | Voted to close because this is a duplicate of an easily un-deletable question. | |
| Feb 15 at 19:31 | comment | added | Carlo Beenakker | you asked this before, why delete and ask again, losing all the comments you received? mathoverflow.net/q/463816/11260 --- if you wish to improve the questions, you do that by editing, not by deleting and reposting. | |
| Feb 15 at 19:06 | history | asked | Lawton | CC BY-SA 4.0 |