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S Nov 3, 2022 at 18:37 history bounty ended C Squared
S Nov 3, 2022 at 18:37 history notice removed C Squared
Nov 3, 2022 at 8:03 vote accept jwguan
Nov 1, 2022 at 22:17 comment added mr_e_man "and $|\phi_i(\mathbf x)-\phi_i(\mathbf y)|$ is exactly the length of the arc" - No, the arc is generally shorter. E.g. near the north pole, two points at the same latitude can have very different longitudes, but the arc length is small.
Nov 1, 2022 at 0:09 history edited Apass.Jack CC BY-SA 4.0
Define y_i explicitly.
Oct 30, 2022 at 4:35 history edited Apass.Jack CC BY-SA 4.0
Fixed a typo.
Oct 30, 2022 at 2:41 answer added Apass.Jack timeline score: 6
Oct 28, 2022 at 15:00 history tweeted twitter.com/StackMath/status/1586009901687513092
Oct 28, 2022 at 9:35 comment added Ѕᴀᴀᴅ @jwguan It's not a matter about naming. The point is that even for $S^1$ the definition is not natural, e.g. $φ(\cos θ,-\sin θ)=2\mathrm π-θ$ for small $θ$ and $\|φ(\cos θ,-\sin θ)-φ(1,0)\|=2\mathrm π-θ$, but the distance should be more naturally defined to be $θ$.
Oct 28, 2022 at 9:00 comment added jwguan @Ѕᴀᴀᴅ Thanks. Can you suggest it a better name?
Oct 28, 2022 at 8:54 answer added Christophe Leuridan timeline score: 4
Oct 28, 2022 at 7:03 answer added Idontgetit timeline score: 1
Oct 28, 2022 at 6:49 comment added Ѕᴀᴀᴅ That's an usual definition of "spherical distance."
S Oct 28, 2022 at 6:35 history bounty started C Squared
S Oct 28, 2022 at 6:35 history notice added C Squared Draw attention
Sep 27, 2022 at 9:20 comment added jwguan @CalvinKhor Yes, I do mean this.
Sep 27, 2022 at 9:10 history edited jwguan CC BY-SA 4.0
deleted 1 character in body
Sep 27, 2022 at 9:08 history edited jwguan CC BY-SA 4.0
added 240 characters in body
Sep 27, 2022 at 9:06 comment added Calvin Khor Just checking, by $\|v\|_1$ do you mean $|v_1|+\dots +|v_n|$?
Sep 27, 2022 at 9:03 history edited jwguan
edited tags
S Sep 27, 2022 at 9:01 review First questions
Sep 27, 2022 at 9:01
S Sep 27, 2022 at 9:01 history asked jwguan CC BY-SA 4.0