I know that I'll get this right :-). For 0 <= x <= pi/2 we have x >= sin(x) >= (2/pi)x (use the fact that (d/dx)^2 sin(x) < 0 for x in that range) If ||x|| denotes the distance from x to the nearest integer then for all x, |sin(x)| = sin(pi || x/pi || ) since (x/pi +/- ||x/pi||) is an integer for suitable sign, and |sin(x)| = | sin(x - pi(x/pi +/- ||x/pi||)) | = |sin(-/+ pi || x/pi ||) | = sin(pi || x/pi ||) So for all positive integers n we have pi || n/pi || >= |sin(n)| >= 2 || n/pi ||. So this shows that lim |sin(n)|^(1/n) exists if and only if lim || n/pi ||^(1/n) exists, and if either limit exists they have the same value. But by the results that I quoted before, there is an e (currently the best is about 7.7, but finite is all that's necessary) so that || n/ pi || > C / n^e, for some positive constant C. Taking n-th roots finishes it off. Victor On Sat, Dec 28, 2013 at 4:13 PM, Bill Gosper <billgosper@gmail.com> wrote:
VM>
W[]oops, for (1) I meant that for -pi/2 <= x <= pi/2, that |sin(x)| >= |x| or that |sin(x)| >= pi || x/pi || for *all* real x. The rest of what I said follows without change. <VM
Isn't this *still* backwards? I.e., For -pi/2 <= x <= pi/2, |sin(x)| <= |x|.
WDS> Returning to pi, http://mathworld.wolfram.com/IrrationalityMeasure.html indicates pi has a most a finite set of rational approxs p/q with |pi-p/q| < q^(-7.7). Therefore, in view of limit(N-->oo) N^(7.7/N)=1, the limit exists and is 1. <WDS
Can someone confirm or correct this rephrasing?:
"For every r>1 and q> 7.7, the nth term of π's continued fraction exceeds
r^q^n at most finitely often."
In any event, this is pathetically lame, since we probably only need q>1.
--rwg
REPHRASING
SPRINGHARE
RANGERSHIP _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun