On Thursday 02 February 2006 09:10, dasimov@earthlink.net wrote:
Given integer n > 0, draw a line segment in C connecting each pair of nth roots of unity. Then as n -> oo, does the set of intersection points in C (assume each is given equal weight and the weights sum to 1) approach a continuous density on the unit disk? (Note: we care only about the intersection points, not the rest of the line segments.)
0. Instead of using rectangles (-oo,a) x (-oo,b) you can use various other families of sets. I'm not sure what the exact criterion is, but I bet you can work with, say, "polar sectors" of the form S(a,b1,b2) := { r exp(it) : 0 <= r < a, b1 < t < b2 }. 1. Consider any two fixed S(-) having the same values of a and b2-b1. For large n, everything is fixed under a set of rotations about 0 that comes arbitrarily close to taking one S(-) onto the other. Therefore (mumble, wave hands) the limiting distribution or non-distribution has circular symmetry, and it suffices to look at the S(a) := S(a,0,2pi). 2. So, when will an intersection occur within S(a)? Working that out in detail is a bit fiddly, but we don't need to do it. Line segments exp(it1)..exp(it2) and exp(it3)..exp(it4) intersect iff ... *some* condition on the signs and sizes of the various tj-tk holds, which means that there's some region of (t1 t2 t3 t4)-space where it holds, and then the number of such intersections for large n is just (n/2pi)^4 times the measure of that region plus an error term of smaller order; and the fraction of all intersections for which it holds is just the measure of that region plus some error term that -> 0 as n -> oo. So, modulo a whole lot of handwaving: yes, there is a limiting distribution, it's rotation-invariant, and (though you'd need to fill in more details than I have done to be sure) it seems pretty likely that it's well enough behaved to have a pdf that's at worst a continuous function plus some delta functions. Or have I waved my hands *too* much? :-) -- g