Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with spherical-geometry
Search options answers only
not deleted
user 108884
6
votes
Maximal distance between $2d+1$ points on the $(d-1)$-sphere
Turns out kodlu's idea works in all dimensions, regardless of the existence of any Hadamard matrices.
Consider all coordinate permutations of
$$(1,...,1,-d)\in\Bbb R^{d+1}\quad\text{and}\quad (-1,.. …
- 13.6k
7
votes
Accepted
Maximal distance between $2d+1$ points on the $(d-1)$-sphere
I finally came to do the computations on Yoav Kallus' comment. After quite some tedious work one finds a cubic polynomial:
$$p_d(x)\,=\,d(d-2)^2 x^3 - d^2 x^2 - dx + 1$$
which has exactly one zero in …
- 13.6k