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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?
Happy New Year.
R.G. Vickson
> And since the
> biggest gap angle cannot be smaller than 2pi/n alpha must be smaller
> than 2pi-2pi/n=2pi*(n-1)/n.
>
> Ralf
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