Den torsdagen den 1:e augusti 2013 kl. 04:30:12 UTC+2 skrev William Elliot: > > > > If you create a hexagon with the radius of one, and recursively double the > > > > > number of vertices of the hexagon(have same radius as side length) > > > > > Maybe this is a valid way to calculate Pi? > > > > Given a regular n sided polygon, with distance from center to the vertices > > > > of r, the length of one side is 2r.sin pi/n and the total length of all the > > > > sides is 2rn.sin pi/n = > > > > Now lim(n->oo) 2r.pi.(sin pi/n)/(pi/n) > > > > = lim(n->oo) 2r.pi.(sin pi/n)/(pi/n) = 2pi.r > > > > For n = 6, sin pi/6 = 1/2, so 2*6r(1/2) = 6r is a crude first approximation > > > > for the circumference of a circle of radius r, giving approximately 3 for pi. > > > Well if you iterate in a loop double up the vertices you do a lot better > > > then 3, smooth. > > You go iterate off to a computer newsgroup. This is math newsgroup. > > Use lim(n->oo) 2^(n+1)r.pi.(sin pi/2^n) and don't be so dumb as to > > limp around on a google goose. Get a real bird to fly the newsgroups. > > > > pi = lim(n->oo) n.sin pi/n = lim(n->oo) 6*2^n sin pi/6*2^n
Well i do not intend to use any trigonometric functions, and honestly i always thought the ancients, did geometry *the correct way* not losing precision. By double up vertices and using fractions.