Re: distribution of n point on the unit circle


C6L1V@xxxxxxx wrote:

Ralf Goertz wrote:
C6L1V@xxxxxxx wrote:

Ralf Goertz wrote:

In general, alpha must be smaller than or equal to 2pi*(n-1)/n.

This lookss wrong: alpha must be <= 2pi/n, because if alpha (the
smallest angle) is > 2pi/n,
the sum of all the angles is > 2pi. Anyway, as I said, Feller solves
the problem.

Thank you for your comments, I think I can now figure out the recursive
relation myself. There seems to be a misunderstanding though. alpha
denotes the smallest angle that will *cover* all points.

My apologies: I mis-read the question. However, Feller also solves your
actual problem, too, I think. It seem to me that the smallest angle
that covers all the points must equal 2*pi minus the largest gap. The
distribution function of the largest gap is given by Feller's formula
(9.9) on page 28 (as remarked on page 29). For a circle of
circumference b, the distribution P{largest gap <=y} is
sum_k=0^n (-1)^k (n choose k) 1{ky/b <=1} (1 - ky/b)^(n-1)
= 1 - n*1{y <= b} (1-y/b)^(n-1) + n(n-1)/2 * 1{y <= b/2} (1-2y/b)^(n-1)
+ ... .
Does this seem OK to you?


Checks out for n = 2, 3 and 4 and looking good! It's pleasing that it
unravels into an expression with such relatively simple if/then
clauses. I must say I expected more of a tangle.

.