You're right. Weyl's criterion says that m*pi mod 1 is uniformly distributed. So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform, You then get the formula that you gave. However, that doesn't invalidate my rough argument since this is just a mild distortion of uniform. Victor On Fri, Dec 27, 2013 at 1:01 PM, Dan Asimov <dasimov@earthlink.net> wrote:
The conclusion of this statement surprises me, since I'd expect that for any c such with c/pi irrational, the values of {|sin(nc)| : n = 1,2,3,...} would be distributed with density rho(y) (y in (0,1)) proportional to 1/|slope| of the unit circle at height = y, i.e., proportional to y/sqrt(1-y^2). (The proportionality constant is 1, so this should be the exact expression.)
Meaning, that in the limit, the fraction of values of |sin(nc)| lying in the interval (a,b) (0 < a < b < 1) ought to be
frac(a,b) = Integral_a^b y/sqrt(1-y^2) dy, i.e., frac(a,b) = sqrt(1-a^2) - sqrt(1-b^2).
No?
--Dan
On 2013-12-27, at 8:48 AM, Victor Miller wrote:
Since pi is irrational, Weyl's criterion tells us the values of |sin(n)| are uniformly distributed in [0,1].
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