Suppose $\{\alpha n\}$ is the fractional part of $\alpha n$. Put $$A_{\alpha}(n) = \#\{\{\alpha k\} < 1/\sqrt{k} : k \leq n\}.$$ If $\alpha$ is irrational, can I find some constant $K$ such that $A_{\alpha}(n) < K \sqrt{n}$ for all $n$? Does the order of $A_{\alpha}(n)$ depend on $\alpha$? Suppose $1/\sqrt{k}$ is replaced by some function $f(k)$. What can I say about the number of $\{\alpha n\}$ less than $f(n)$ as $n$ tends to infinity?
Estimating $\#\{\{\alpha k\} < 1/\sqrt{k} : k \leq n\}$ for irrational $\alpha$
1 Answers
The Weyl equidistribution theorem says that for irrational $\alpha$ and sufficiently many $k$, the fractional parts $\{\alpha k\}$ will be equidistributed in the interval $[0,1]$.
To apply this to your particular problem, consider a large interval $k\in[N,2N]$. Because of equidistribution, the numbers $k$ will "hit" the condition $\{\alpha k\} < 1/\sqrt{N}$ roughly $1/\sqrt{N}$ of the time, i.e. there will be roughly $N·1/\sqrt{N} = \sqrt{N}$ numbers $k$ from the interval that fulfill the condition.
Of course, we are actually interested in the condition $\{\alpha k\} < 1/\sqrt{k}$, so we have overestimated things a bit, but it will still work out.
Now, piecing intervals together by choosing $N=2^M$ as a sequence of powers of two, we obtain an estimate along the lines of
$$A_\alpha(n=2^M) \lesssim \sqrt{1} + \sqrt{2^1} + \sqrt{2^2} + .. + \sqrt{2^{M-1}} \leq K \sqrt{2}^M = K\sqrt{n} $$
as desired. You might have to fill in some epsilons and stuff to make the proof precise, but this is the core argument.
In the general case, a similar argument will yield a good bound as long as the function $f(k)$ doesn't vanish too fast. If it does vanish very fast, then it can only get better, though a better bound might be harder to prove.
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0Thanks for the answer. What could I read to learn more? – 2011-02-13
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1Unfortunately, I don't know a good reference myself. Results of this kind are usually attributed to the topic of [diophantine approximation](http://en.wikipedia.org/wiki/Diophantine_approximation) and are often proven with methods from [ergodic theory](http://en.wikipedia.org/wiki/Ergodic_theory), but the latter topic is so vast that there may well be books on ergodic theory that don't mention Weyl's theorem at all. Maybe the recent [Ergodic Theory with a view towards Number Theory](http://www.uea.ac.uk/menu/acad_depts/mth/ergodic/) is suitable, though. – 2011-02-14
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0Thanks for the references. – 2011-02-15