From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Prove sequence dense in R Date: 24 Aug 1995 07:24:08 GMT In article <41ealq$73m@news.onramp.net>, Dennis Thompson wrote: >regarding comments on proving that the sequence given below > >{m ! / 2 to the power of n | m, n belongs to N} >is dense in R plus.. Yes, the set of these numbers is dense in the positive real half-line. I find it easier to establish that their logs (base 2) are dense in R; since f(x)=2^x is a continuous function, this density implies the other. So we must show that the set { log(m!) - n : m, n in N } is dense in R, i.e., that the numbers log(m!) are dense in the circle R/Z. I will be cavalier about the distinction between R and R/Z below when there is no harm; I hope this meets with general approval. Given a point x in the circle and a distance e > 0, we will find an m so that log(m!) is within e of x. The trick is to choose m to be a little more than a power of 2. Begin by choosing r so large that (1/2)^r < e. Let m1 = 2^(2r-1); log( m1 !) is somewhere else in the circle. Now, for k = 0, 1, ..., 2^r -1, we will compute a_k = log( m1 + k )= log(m1) + log(1 + k/m1); as m1 is a power of 2, log(m1) is an integer, so that a_k = log(1 + k/m1) mod 1 . Since 0 < log(1+z) < z for z>0, we have 0 < a_k < k/m1. Thus the sum of all these a_k is positive but less than 1, so that the partial sums march counterclockwise around the circle and don't lap the circle. On the other hand, the next such partial sum (through k=2^r) is about (2^r)(2^r + 1)/2 * (1/2^(2r-1)), which is just larger than 1. In fact, it's not hard to show from log(1+z) > z - z^2/2 that this estimate is closer to the true sum than to 1; I'll spare the reader the details. The point is that the next sum _does_ lap the circle. Now, the displacement between any two consecutive sums is just a_k, which is never more than a_(2^r), which in turn is at most (1/2)^r < e. So every interval of length e around the circle has to meet at least one of these partial sums. Adding log( m1 ! ) to the partial sums simply rotates them around the circle, so it's still true that all short intervals contain some sum. Now, however, these sums are of the form log( m1 ! ) + log( m1 + 1 ) + ... + log( m1 + k) = log( (m1+k)! ) so we have shown that the interval contains log( m! ) for some m=m1+k. dave ============================================================================== From: Rene Grothmann Newsgroups: sci.math Subject: Re: Prove sequence dense in R Date: Thu, 24 Aug 1995 10:40:40 +0200 On Wed, 23 Aug 1995, Dennis Thompson wrote: > A friend of mine without direct Internet access asked that I inquire > regarding comments on proving that the sequence given below > > {m ! / 2 to the power of n | m, n belongs to N} > is dense in R plus.. The problem is equivalent to log_2(m!) mod 1 being dense in [0,1). For then any real x would be approximated as close as one wants by expressions of the type log_2(m!) - n with an integer n. Taking 2^x yields the denseness of the above set in [0,infty). Looking at the sums log_2(1)+...+log_2(m) for growing m modulo 1: We know (Theory of Equidistribution by Weil) that the numbers log_2(3^l) mod 1, l=1...N_eps cover [0,1) with denseness eps/2 for any eps>0 as small as we want, where we have to choose N_eps big enough. Taking a very huge M_eps, we see that |log_2(3 2^M_eps + l) - log_2(3)| < eps/2 for all l=1,...,N_eps. This implies that log_2(m!) covers [0,1) with denseness eps using m=1,...,3 2^M + N_eps. I hope, I have not done a mistake here. Rene.