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Is there a sort of distributional estimate in Diophantine approximation which allows to estimate the number of solutions which provide a certain quality of approximation? For example, how large is the measure of the numbers $x \in [0,1]$ for which $$ \# \Big\{ 1 \leq n \leq N: \|n x\| \leq 1/N \Big\} \geq \log N? $$ (Here, of course, the $\log N$ on the right-hand side could be any other term as well, say $(\log N)^{1/2}$ or whatever. Or the $1/N$ inside the brackets could be $1/(N \log N)$. Note that the usual methods from Diophantine approximation allow to estimate the desired measure in the case when the right-hand side is 1. By $\| \cdot \|$ we denote the distance to the nearest integer.)

If one believes the events to be somewhat independent, then the distribution we look for should follow roughly a Poisson distribution. Is something known in that direction?

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  • $\begingroup$ Are you interested in the measure of this set as a function of $N$ or only in the limit as $N \rightarrow \infty$? $\endgroup$ – Rodrigo Oct 7 '17 at 22:17
  • $\begingroup$ Well, I suppose it would be fine to have something in the limit as $N \to \infty$. However, note that $N$ appears two times, once as the size of the deviation, and once as the cardinality of the set. That makes everything a bit tricky. In the meantime, I realized that one will not get a Poisson distribution here - the events are correlated, and it is more likely to see $\log N$ solutions, say, than it would be in a totally random setting. The reference Schmidt, Wolfgang: A metrical theorem in Diophantine approximation. Canad. J. Math. 12: 1960, 619-631 is closely related, but insufficient. $\endgroup$ – Kurisuto Asutora Oct 9 '17 at 8:48
  • $\begingroup$ Maybe I am missing something - picking $x$ uniformly the expected size of $ \{ 1 \leq n \leq N: \|n x\| \leq 1/N \}$ is $1$, so from Markov's inequality the measure of the set of $x$ satisfying your inequality is at most $\frac{1}{\log n}$. Maybe you are interested in the set $ \{ 1 \leq n \leq N: \|n x\| \leq 1/n \}$, which has expected size $\log N$ and this method doesn't work? $\endgroup$ – Rodrigo Oct 9 '17 at 22:13
  • $\begingroup$ Hi Rodrigo, you are right of course that Markov's inequality gives something here, but the actual probability (resp. measure) should be much much smaller. If the events $\|n x\| \leq 1/N$ were stochastically independent, then the cardinality we want to count could be approximated by a Poisson distribution with parameter 1, which would give something of order 1/((log N)!) for the measure we want to estimate. So in a sense the question is, to which extent do the sets $\|n x\| \leq 1/N$ behave like independent sets. $\endgroup$ – Kurisuto Asutora Oct 11 '17 at 17:13

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