|
|
C6L1V@xxxxxxx wrote:
> 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?
Indeed, it does. The result is in accordance with my experiments (for
n=4 which is the case I am actually interested in). Thank you.
> Happy New Year.
Happy New Year.
Ralf
|
|