6
$\begingroup$

It is well known that for almost every $c \in \mathbb{R} / \mathbb{Z}$ there exists $(q_n)_{n \geq 1}$ and $(a_n)_{n \geq 1}$ such that $$|c - a_n / q_n| \leq 1/ q_n^2,$$ where $q_n < q_{n+1} \leq q_n^{1+ \epsilon}$ for every $\epsilon > 0$.

The problem with the above is that we have no idea when the sequence $(q_n)_{n \geq 1}$ starts. That is, how large is $q_1$ (assume that we take $q_1$ to be minimal so that the above holds). Do we know anything about the set of $c$ for which $q_1 \leq T$ for some fixed large $T$? Has this set been studied before?

$\endgroup$
5
  • 1
    $\begingroup$ I think you need to impose strict inequality in $q_n<q_{n+1}$ since $q_1=q_2=\ldots =1$ is presumably not the sequence you want. $\endgroup$ Jul 29, 2017 at 5:17
  • 2
    $\begingroup$ $q_1=1$ works for every real number $c$: there is always an integer $a_1$ within distance $1/2$ of $c$, and therefore $$|c-a_1/1| \le 1/2 < 1/1^2$$ So the sequence starts at $1$. $\endgroup$
    – Lee Mosher
    Jul 29, 2017 at 14:14
  • $\begingroup$ Can you compare what you are asking with what you get from the continued fraction? $\endgroup$ Jul 29, 2017 at 14:40
  • 1
    $\begingroup$ @LeeMosher that does not work. Choosing $q_2$ is impossible when $q_1 = 1$. To address the other comment, I'm asking about the growth rate of denominators of the convergents in the continued fraction expansion of $c$. $\endgroup$ Jul 29, 2017 at 20:28
  • $\begingroup$ Ah, you are right, I did not pay close enough attention to your inequality "$q_1 < q_2 \le q_1^{1+\epsilon}$ for every $\epsilon>0$". $\endgroup$
    – Lee Mosher
    Jul 29, 2017 at 21:46

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.