Timeline for Isodiametric Inequality
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2010 at 13:54 | answer | added | Victor Miller | timeline score: 0 | |
| Aug 27, 2010 at 17:25 | answer | added | Skippy | timeline score: 8 | |
| Oct 22, 2009 at 20:05 | answer | added | Hung Tran | timeline score: 1 | |
| Oct 22, 2009 at 4:13 | comment | added | j.c. | Looks like the informal statement in words of that inequality is: Any subset A of R^n has volume bounded by above by the ball with diameter diam(A). | |
| Oct 22, 2009 at 4:10 | comment | added | user962 | For all sets A the subset of R_n, the n-dimensional Lebesgue measure is less or equal to alpha(n)(diamA/2)^n , where alpha(n) is (pi^n/2 )/gamma fn(n/2 +1) Probably it is messy but that the inequality. | |
| Oct 22, 2009 at 4:08 | answer | added | Anton Geraschenko | timeline score: 13 | |
| Oct 22, 2009 at 4:07 | answer | added | Alon Amit | timeline score: 2 | |
| Oct 22, 2009 at 4:06 | comment | added | j.c. | Consider the three vertices of an equilateral triangle with side length a. The diameter of the set is a, but the smallest ball which can contain all three points has radius \sqrt{3}a/3 and hence diameter 2\sqrt{3}a/3 > a | |
| Oct 22, 2009 at 4:01 | comment | added | Kim Morrison | I know it's not directly relevant to your question, but why not state the inequality too, for the sake of background? | |
| Oct 22, 2009 at 3:52 | history | asked | user962 | CC BY-SA 2.5 |