# The first return time of an irrational rotation

For the probability space $([0,1),\mathcal{B},\lambda)$ and for an irrational $\theta \in (0,1)$ we have the map $Tx = x + \theta \bmod 1$. I'm trying to find an expression for $n(x):=\inf\{n \in \mathbb{N}_{>0}: T^nx \in [0,\theta)\}$ for $x \in [0,\theta)$.

What I've been looking for is the smallest $n$ such that $x + n\theta \geq1$. My problem with finding this is that irrational numbers are slippery guys. I've tried using continued fraction representations and Diophantine approximations, but the problem I run into is that the required accuracy of the approximation seems to depend on $x$.

Any tips on how I should view this problem, or tips what techniques could be useful. Just to be clear, I really don't want an outright answer. Just some help in the right direction.

EDIT: Fixed inequality.

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I think you want the smallest $n$ such that $x+n\theta\geq 1$. This can be written as $\lceil \frac{1-x}{\theta}\rceil$ - see http://en.wikipedia.org/wiki/Floor_and_ceiling_functions.

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Note that this approach works only because you're guaranteed to get a return within one revolution - it's impossible to 'miss' the interview you've defined because no $\theta$-sized step can jump over it. If you were looking for e.g. the first $n$ s.t. $T^nx\in [0,\theta/2)$ then that would already be much trickier. – Steven Stadnicki Oct 31 '12 at 21:09
Huh, it does seem to be that simple. I'm trying to find the induced map, which would then be $T'x = x + \lceil\frac{1-x}{\theta}\rceil\theta \bmod 1$ – BallzofFury Oct 31 '12 at 22:52

Take the case $\boxed{0 < \theta < \frac12}$.

Let $a = \max\{k\geq 0 \mid k\theta < 1-\theta\} = \lfloor \frac{1}{\theta} \rfloor -1 \geq 1$.

Take $x \in (0,\theta)$. Then after $a-1$ iterations, $y:=T^{(a-1)} x \in ((a-1)\theta, a\theta)$. At the next iteration, $Ty$ falls in $(a\theta, (a+1) \theta) = (a\theta,1-\theta)\dot{\cup}(1-\theta, (a+1)\theta)$. It falls in the first interval $(a\theta,1-\theta)$ when $x \in (0,\theta')$ where $\boxed{\theta'=(1-\theta)-a\theta}<\theta$, otherwise it falls in the second interval $(1-\theta, (a+1)\theta)$.

If $Ty \in (a\theta,1-\theta)$, then $T^2y \in ((a+1)\theta,1)$ and $T^3y\in(0,\theta)$. If $Ty \in (1-\theta, (a+1)\theta)$, then $T^2y\in(0,\theta)$.

Finally, the return time $n(x)$ takes only two possible values: $$n(x)=\begin{cases} a+2 & \text{if } x \in (0,\theta') \\ a+1 & \text{if } x \in (\theta', \theta) \end{cases}.$$ We can check the result by checking that Kac's lemma holds: $$\int n(x) d\lambda(x) = (a+2)\theta' + (a+1)(\theta-\theta') = (a+1)\theta+\theta'= 1.$$

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