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Given the set of all primes. From this one builds subsets of $N$-almost primes according to a certain partition, e.g. $\lambda=(3,1)$, so the set is $$ M=\{2^33,2^35,2^37,...,3^32,3^35,...\}. $$ Is the set of natural logarithms of elements of $M$ equidistributed$\mod 1$?

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What does it means for a set of integers to be equidistributed mod 1? and what is the logarithm of a set? and what does it mean for a finite set to be equidistributed mod 1? and why do you call $M$ a set, but write it as a sequence? –  Gerry Myerson Nov 21 '11 at 22:34
    
for definition of equidistributed$\mod 1$ see en.wikipedia.org/wiki/Equidistributed_mod_1 –  draks ... Nov 22 '11 at 6:56
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Dear Andreas, *Equidistributed mod $1$* is a property of the fractional parts of the numbers in question (as your link shows). Your sets consist of natural numbers, whose fractional parts are thus always $0$. Hence @Gerry Myerson's quesion. Regards, –  Matt E Nov 22 '11 at 7:11
    
Damn, I was actually thinking more over the logs and so i totally forgot about that. –  draks ... Nov 22 '11 at 7:19
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Andreas, I don't need wikipedia to tell me what equidistributed mod 1 means, I can get that by rereading the papers I've published about it. What I need is for you to tell me what you mean when you talk about a finite set being equidistributed mod 1. Hint: you won't find it on that wikipedia page. –  Gerry Myerson Nov 22 '11 at 11:19

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Equidistribution is a property of sequences, not of sets. von Neumann proved that any set that is dense in $[0,1)$ can be ordered in such a way as to be a uniformly distributed sequence. The sets you are talking about are surely dense in $[0,1)$ (although I wouldn't be able to give you a proof of that off the top of my head), so it comes down to a question of how you order them.

The order implicit in the way you've written the question - all the ones with $2^2$ first, then, after that infinity of numbers, all the ones with $3^2$, etc. - well, that's not a sequence, so it gives us nothing to go on.

I guess the simplest order is just increasing order of the numbers, so, $$\log24,\log40,\log54,\log56,\log88,\log104,\log135,\dots$$

Now it's an exercise to show that the sequence $\log1,\log2,\log3,\log4,\dots$ is not uniformly distributed mod $1$, and that's basically because it grows too slowly. Once it gets into a subinterval $[a,b)$ it just stays there too long. Now I'm confident that the same is true for the sequence of logs of primes, and the Prime Number Theorem should be strong enough to show that (although again that's not a proof I'm prepared to write down on the spur of the moment). If that's true, then it's also true for the sequence of squares of primes, of cubes of primes, of any fixed power of primes, because multiplying all the terms in a sequence that isn't u.d. by a fixed number can't make the sequence u.d.

So the next thing is to show that the numbers $p^3q$ grow more slowly than, say, the numbers $p^4$. That ought to be enough to conclude (if everything else I've written here holds up to careful scrutiny) that the sequence in question is not u.d., and once you've done $\lambda=(3,1)$ you should be able to generalize to any $\lambda$ you like.

Sorry for being so sketchy and leaving all the details as exercises.

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Thanks for your roadmap, how to get the proof. Let's see how far a non-mathematician (me) can get... –  draks ... Nov 22 '11 at 13:34
    
To show that the sequence $(x_j)=\log 1,\log 2,...$ is not u.d., I would like to show that the Weyl's Criterion is not fullfilled for $l=1$: Re-writing it as $$\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^N n^{2\pi i}=\lim_{N\to \infty} \frac{N^{1+2\pi i}}{N} \sum_{n=1}^N (\frac{n}{N})^{2\pi i} \frac{1}{N}=$$ $$\lim_{N\to \infty} \frac{N^{1+2\pi i}}{N} \int_0^1 x^{2\pi i}dx =\lim_{N\to \infty} \frac{N^{1+2\pi i}}{N} (\frac{x^{1+2\pi i}}{1+2\pi i})|_0^1=\lim_{N\to \infty} \frac{N^{2\pi i}}{1+2\pi i} \neq 0$$ proves it! Right? –  draks ... Nov 23 '11 at 7:52
    
Could one also use the fact that $\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^N n^{2\pi i}=\lim_{N\to \infty} \frac{\zeta(-2\pi i)}{N} $ –  draks ... Nov 23 '11 at 7:55
    
I wouldn't use Weyl, I'd go right to the definition. Say $m\lt e^{100}\lt m+1$. Then $0\le\log n\lt0.1$ (modulo 1) for $e^{100}\le n\lt e^{100.1}=e^{0.1}e^{100}$ which is more than one-tenth of the numbers up to $e^{100}$, so the proportion of terms $\log n$ in that interval of length one-tenth can't converge to one-tenth. –  Gerry Myerson Nov 23 '11 at 12:11
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No contradiction. Density is a property of sets; u.d. is a property of sequences. A set comes without an ordering; a (countably infinite) set can be viewed as a sequence by imposing an ordering on it, but there are (uncountably) many orderings available. If a set is dense, you can impose an ordering in such a way that the resulting sequence is u.d., but you can also impose a different ordering in such a way that the resulting sequence is not u.d. So, you can't ask whether a set is u.d.; all you can do is choose an ordering and ask whether the resulting sequence is u.d. –  Gerry Myerson Mar 23 '12 at 0:42

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