What follows is a less hand-wavy proof that lim |sin(n)|^(1/n) = 1. The Weyl criterion was a bit of a red herring, though it lead me to realize that if the limit existed, then the answer had to be 1. The only way that the limit couldn't be 1 would be if pi had an unusually close sequence of rational approximations. More specifically 1) For all real x we have |sin(x)| <= |x| (pretty easy exercise) If || x || denotes the distance to the nearest integer from x, then we have 2) |sin(x)| <= pi* ||x/pi|| (applying (1) to x mod pi) Thus, if n is an integer we have 0 <= - log |sin(n)| <= -log pi - log || n/pi || However, by what I said before there is a positive real C, and a positive e (known to be around 8, is sufficient) such that ||n/pi || >= C/n^e Taking logs, and plugging into the above we get 0 <= -log|sin(n)| <= \log pi - log C + e*log(n) We then divide by n and take limits, and see that -(1/n) log |sin(n)| is squeezed to become 0. On Fri, Dec 27, 2013 at 1:36 PM, Victor Miller <victorsmiller@gmail.com>wrote:
I said
So you should have Prob(|sin(n)| in [a,b]) = Prob(n in [arcsin(a),arcsin(b)]) which isn't unform,
I meant
So you should have Prob(|sin(n)| in [a,b]) = Prob(n mod pi in [arcsin(a),arcsin(b)]) which isn't unform,
On Fri, Dec 27, 2013 at 1:09 PM, Victor Miller <victorsmiller@gmail.com>wrote:
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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