2
$\begingroup$

O(1) or o(logn) discrepancy for multiples of an irrational for at least one sub interval.

Using $\{x\}$ to denote the fraction part of $x$ we can define for any $I\subset [0,1]$,

$$E(n,\theta, I) ={ \left|\{\,\{\theta\},\{2\theta\},\dots,\{n\theta\} \,\} \cap I \right|}-n|I|$$ $ $ $$\Delta_{sup}(n,\theta)=\sup_I |E(n,\theta,I)|$$.

The Equidistribution Theorem says that the sequence $a_i=(i\theta)$, $i\in\mathbb{Z}_{\geq1}$ is equidistributed modulo 1 when $\theta$ is irrational hence

$$\Delta_{sup}(n,\theta)=o(n)$$ for all irrational $\theta$.

Furthermore if $\theta$ has bounded partial denominators in its continued fraction expansion then we have

$$\Delta_{sup}(n,\theta)\ll \log n$$

(See Theorem 1.B of W Schmidt "Lectures on irregularities of distribution" $^*$

My question asks how small can $E$ be as a function of $n$.

1) Does there exist an irrational number $\theta$ and a fixed interval $I\subsetneq [0,1]$ with $E(n,\theta, I)=O(1)$?

If this false or too difficult:

2) Does there exist an irrational number $\theta$ and a fixed interval $I\subsetneq [0,1]$ with $E(n,\theta, I)=o(\log n)$?

If this false or too difficult:

3) Does there exist an irrational number $\theta$ and an interval $I\subsetneq [0,1]$ of fixed size whose position is allowed to vary with $n$, s.t. $E(n,\theta, I)=o(\log n)$?

$*$ "Schmidt, Wolfang M., Lectures on irregularities of distribution. (Notes by T. N. Shorey), Tata Institute of Fundamental Research, Lectures on Mathematics and Physics: Mathematics, 56. Bombay: Tata Institute of Fundamental Research. vi, 128 p. (1977). ZBL0434.10031.")

$\endgroup$
5
  • $\begingroup$ Isn't it known that the golden ratio minimizes the irregularities of distribution across all irrationals? (per the discussion about the coefficients of the continued fraction expansion). I don't imagine that $I$ really has any bearing on this, as I would expect all fixed intervals to perform asymptotically the same. (This doesn't cover your (3), admittedly) $\endgroup$ – Steven Stadnicki Mar 26 '20 at 20:21
  • $\begingroup$ @StevenStadnicki Thanks for your comment Steven. Yes I think it's possible that the interval doesn't matter at least for certain ranges of the error. This isn't true for general equidistributed sequences - for the Van der Corput sequence you have the analogous discrepancy to be $\leq1$ for any elementary interval $[w/2^k, (w+1)/2^k]$ so it would be interesting to have a proof. $\endgroup$ – Ivan Meir Mar 26 '20 at 21:34
  • $\begingroup$ Maybe van Aardenne-Ehrenfest, T. "Proof of the Impossibility of a Just Distribution of an Infinite Sequence Over an Interval." Proc. Kon. Ned. Akad. Wetensch. 48, 3-8, 1945 is relevant here, or mathworld.wolfram.com/DiscrepancyTheorem.html $\endgroup$ – Gerry Myerson Mar 26 '20 at 23:12
  • $\begingroup$ @GerryMyerson Thank you for the reference. I think this is more related to showing that there is at least one interval with discrepancy $\geq C \log n$ whereas I am interested in seeing if there is at least one interval with $\leq C \log n/\log \log n$ say. However if we could show that for the irrational number sequence that all intervals of the same size have approximately the same discrepancy (as Steven proposed) then this result would show that 1) , 2) and 3) are all false. $\endgroup$ – Ivan Meir Mar 27 '20 at 0:27
  • $\begingroup$ I'm not sure how to do this though! $\endgroup$ – Ivan Meir Mar 27 '20 at 0:29
1
$\begingroup$

I think you are asking for so-called "bounded remainder sets". Given $\theta$, there exist intervals $I$ having discrepancy $O(1)$, namely those whose length is in $\mathbb{Z} + \theta \mathbb{Z}$. The classical reference is a paper of Kesten:

H. Kesten,On a conjecture of Erdös and Szüsz related to uniform distribution mod 1, Acta Arith. 12(1966), 193–212.

$\endgroup$
1
  • $\begingroup$ Thank you that is an extremely elegant and satisfactory result! To summarise this shows that for any irrational, 1) and hence 2) and 3) are true for intervals of sizes $\{n\theta\}$ only. Fantastic! $\endgroup$ – Ivan Meir Apr 2 '20 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.