# Quantitative bound on irrational rotation recurrence time

Given an irrational $$a$$, the sequence $$b_n := na$$ is dense and equidistributed in $$\mathbb S^1$$ where we view $$\mathbb S^1$$ as $$[0, 1]$$ with its endpoints identified.

Given a point $$p$$ in $$\mathbb S^1$$, can we obtain a quantitative upper bound (that can depend on $$a, p, e$$) on the smallest $$n$$ such that $$na$$ is in $$B_e (p)$$?

This is probably suboptimal, but one can use the convergents of the continued fraction expansion of $$a$$ to efficiently find $$n \in \mathbb{N}$$ such that $$||na||_\mathbb{Z} \leq e$$, where $$||\cdot||_\mathbb{Z}$$ denotes the distance to the nearest integer. Then one can find $$m \in \mathbb{N}$$ with $$m \leq ||na||_\mathbb{Z}^{-1}$$ such that $$|mna-p| \leq e$$.

• This is exactly what I need! Thanks! – James Baxter Jan 12 '19 at 2:53
• Btw, could you elaborate just a little more on the construction itself? – James Baxter Jan 12 '19 at 2:59
• Let $\frac{p_n}{q_n}$ be the convergents of the continued fraction of $a$, defined as in the Wikipedia article on continued fractions. One has $|a - \frac{p_n}{q_n}| \leq \frac{1}{q_n^2}$ so that $|q_na-p_n| \leq \frac{1}{q_n}$ and therefore $||q_na||_{\mathbb{Z}} \leq \frac{1}{q_n}$. – burtonpeterj Jan 12 '19 at 4:56
• One also has $\frac{1}{q_{n+1}+q_n} \leq |q_na - p_n|$ so this gives an upper bound on $m$. – burtonpeterj Jan 12 '19 at 5:38

I think that the answer from @burtonpeterj is in fact pretty much best possible. Note that it does not utilize $$p.$$ I don't think there is a way to work general values of $$p$$ into the estimate.

If you take the fractional parts of $$0,a,2a,3a,\cdots,na$$ they divide $$[0,1]$$ into sub-intervals. The three gap theorem states that there will be at most three distinct lengths. The particular lengths can be determined from the convergents mentioned above.

This allows us to find the smallest $$n$$ so that all the sub-intervals have length $$2e$$ or less. Then this would work for all values of $$p$$ in the unit interval and be the right one for $$p$$ being the middle of the last interval of longer length which get split (and any points close enough to it.)

Here are a few special cases which illustrate what can happen.

• If $$a=\alpha$$ is positive but much smaller than $$e$$ ($$\alpha \ll e)$$ then at about $$n=\frac{1-2e}{\alpha}$$ we have lots of intervals of length $$a$$ and one just barely under $$2e.$$

• The case that $$a=\frac{p}{q}$$ is rational doesn't really fit the question but, ignoring that, we see that at $$n=q-1$$ there is an interval of length $$\frac2{q}$$ which gets split exactly in half. Then we just get the points we had before.

• If $$a=\frac{p}{q}+\alpha$$ is irrational with $$|\alpha| \ll e \ll \frac1q$$ then up to about $$n=q$$ it is just about like for $$a=\frac{p}{q}$$ and at that point we have $$q$$ subintervals all of length about $$\frac1q.$$ Then the following fractional parts start chipping off new subintervals of length $$|\alpha|.$$