Timeline for Largest number of vectors with pairwise negative dot product
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 6 at 16:48 | history | edited | LSpice | CC BY-SA 4.0 |
`\label`+`\eqref`, while this is on the front page
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| Aug 3, 2021 at 19:37 | comment | added | Michael | @WillJagy, you can get $\cos^{-1}(-1/n)$ from noticing that "medians" of any $(n+1)$-simplex intersect in proportion $n:1$, and for a regular simplex you get a right triangle with one side $1/(n+1)$ of the median and the hypotenuse $n/(n+1)$. | |
| Aug 3, 2021 at 19:32 | comment | added | Michael | Could you elaborate about why the 2D space is giving enough room to manoeuvre and 1D is not? | |
| Apr 20, 2020 at 17:13 | comment | added | Wlod AA | This is a very simple theorem (see @YuichiroFujiwara). For this reason, the shortcut "This gives enough room for maneuver" is a hole in the proof under the given circumstances. | |
| Apr 20, 2020 at 16:25 | comment | added | Steven Stadnicki | @WillJagy I think that comes basically for free by just computing dot products on the canonical coordinates in $\mathbb{R}^{n+1}$. | |
| Jul 27, 2010 at 0:25 | vote | accept | CommunityBot | ||
| Jul 12, 2010 at 1:25 | comment | added | Will Jagy | Side note, I get that the (central) angle between two of the $n+1$ vectors at the regular simplex vertices is $$ \arccos \frac{-1}{n}.$$ It is probably somewhere in Coxeter's "Regular Polytopes" but I couldn't find it. | |
| Jul 11, 2010 at 18:12 | comment | added | Benoît Kloeckner | Nice argument! | |
| Jul 11, 2010 at 18:08 | history | answered | Robin Chapman | CC BY-SA 2.5 |