Right. Forgive me for the interruption, but I have a slight question.
Say you consider the stereographic projection of the unit sphere onto the extended complex plane.
If you have 2 points on the plane, neither corresponding to the north pole, the spherical line through them is a circle that intersects the unit circle at negatives.
My only question is, is there any easy way to know whether the line connecting the 2 points is the minor or major arc of the circle?
For instance, I know that if both points are inside the unit circle, you take the minor arc.
Well, @AlpArslan, I've never thought about this before. So, for sure, if the two points are in the southern hemisphere, the shorter arc will project to something inside the unit circle in the plane.
If the two points are in the northern hemisphere, the shorter arc will project to something outside the unit circle.
And since, the great circles project to circles that meet the unit circle at antipodes (which happen to be negatives of each other here), the former case is the minor arc, and the latter, the major arc.
I want to write an algorithm that finds an optimal vertex cover of a tree in linear time O(n), where n is the number of the vertices of the tree.
A vertex cover of a graph G=(V,E) is a subset W of V such that for every edge (a,b) in E, a is in W or b is in W.
In a vertex cover we need to have ...
Because, for instance, if we draw our triangle such that geodesic on the sphere passes close to the north pole, that will distort to something enormous on the plane, for instance, no?
Wait a minute. If we take two points in the upper hemisphere, we get a shorter path by staying in the upper hemisphere, not going around into the lower hemisphere.
That's the rule. If the two points are outside, always stay outside. If they are inside, always stay inside. And if one is outside and one is inside, take the shorter arc in the plane.
It would be interesting to prove this by integrating with respect to the chordal metric in the plane ... :)
Hey only on for 5 min, but just so I can think about it, what did you mean by " If the first $k$ diagonals of $C$ and $D$ agree, then so do the first $k$ diagonals of $C^{-1}$ and $D^{-1}$." What do you mean agree @Ted?
Yes, by symmetry, @AlpArslan. Because reflection across the perpendicular bisector of your diameter gives an isometry in both places. So you'll have the same arc plus extra going around the other way.
Ah, I thought about stuff like this writing some stuff on projective geometry based on thinking of the sphere modded out by the antipodal map. But never this particular question.
Right, minor arcs aren't necessarily right if both points are in the northern hemisphere.
@iwriteonbananas The presentation, at least as given, of that group does not "naturally" form a $C'(1/6)$ presentation. So I am not sure if it is a $C'(1/6)$ group, probably not.
and what are thoughts on starting something like this (codeforamerica.org) but for the outreach of mathematics and for the love of doing it? any takes anyone? like starting linked brigades of communities. (just a thought :) )
Well, since my algebra backround was (finite) group-heavy, I picked it up for Galois and modules, etc, as well as to see the categorical language unfold in familiar territory
But I keep getting distracted in the group theory section
what are your thoughts on starting something like this (codeforamerica.org) but for the outreach of mathematics and for the love of doing it? like starting linked brigades of communities. (just a thought :) )
I thought of this after looking over some things on Facebook.
user61230
Hmm... isn't Code For America a political movement?
user61230
I don't know too much about it; sorry if that's a naïve question.
But you meant that its not important as what you do
Mathematics is a linked subject @Mew what you do might have evolved from abstract mathematics itself.....it must have evolved and statistics is not the place where the money is if you have the correct way of thinking money is just an atrocious need