Smallest denominator of a rational in the Machin intervals for $\pi$
问题内容
Let
$$ t_n = {16 \over (2n+1)5^{2n+1}} - {4 \over (2n+1)239^{2n+1}}. $$
Define
$$ S_N = \sum_{n=0}^N (-1)^n t_n. $$
For each integer $m \geq 0$, define
$$ I_m = [S_{2m+1}, S_{2m}]. $$
By Machin's formula,
$$ \pi = 16\arctan(1/5) - 4\arctan(1/239), $$
so
$$ S_{2m+1} \leq \pi \leq S_{2m}. $$
Also,
$$ |I_m| = S_{2m} - S_{2m+1} = t_{2m+1}. $$
So
$$ |I_m| \sim {16 \over (4m+3)5^{4m+3}}. $$
Now define $Q_m$ to be the smallest denominator of a rational number in $I_m$. In other words,
$$ Q_m = \min\{q \geq 1 : \text{ there is an integer } p \text{ such that } (p,q)=1 \text{ and } p/q \in I_m\}. $$
A random interval of length $\delta$ should first contain a rational with denominator about $\delta^{-1/2}$. Since
$$ |I_m| \sim {16 \over (4m+3)5^{4m+3}}, $$
Numerical evidence suggests
$$ Q_m \approx m^{1/2}5^{2m}. $$
So I would like to know whether one can prove the weaker lower bound
$$ \liminf_{m\to\infty} {\log_5 Q_m \over m} \geq 2. $$
Equivalently, is it true that for every $\epsilon > 0$, there is an integer $M$ such that for all $m \geq M$,
$$ Q_m > 5^{(2-\epsilon)m}? $$
Even a weaker exponential lower bound would be interesting. For example, can one prove that there is some constant $c>0$ such that, for all sufficiently large $m$,
$$ Q_m > 5^{cm}? $$
Equivalently, can one prove that the interval $I_m$ eventually contains no reduced rational $p/q$ with
$$ q \leq 5^{(2-\epsilon)m}? $$
Are there known continued-fraction or Farey-sequence methods that apply to this kind of deterministic sequence of shrinking intervals?
回答 (1)
Look at it from the other end. For a rational $\frac{p}{q}$ close to $\pi$, what is the largest $m$ such that $\frac{p}{q} \in I_{m}$? This question is not equivalent, but related, and answering it can help answering your question.
Since we're not equivalent, we can also fudge things a bit to simplify. Ignoring small $m$, we can neglect the $239$-term, because that decreases so much faster. Thus, for not too small $n$, $S_{n}$ is essentially $\pi + (-1)^{n} \frac{25}{26}\cdot\frac{16}{(2n+3)5^{2n+3}}$. Hence for sufficiently large $m$, \begin{equation} J_{m} = \biggl(\pi, \pi + \frac{1}{2(4m+3)5^{4m+1}}\biggr) \subset I_{m}\,. \end{equation} Now if $r$ is an odd positive integer and $\frac{p_{r}}{q_{r}}$ the $r^{\text{th}}$ convergent to $\pi$, we know $0 < \frac{p_{r}}{q_{r}} - \pi < \frac{1}{q_{r}^{2}}$. Thus $\frac{p_{r}}{q_{r}} \in J_{m}$ for \begin{equation} (8m+6)\cdot 5^{4m+1} \leqslant q_{r}^{2} \tag{1} \end{equation} and quite possibly some larger $m$. The largest $m$ satisfying $(1)$ is $(1/2 + o(1))\cdot \log_{5} q_{r}$, and evidently $Q_{m} \leqslant q_{r}$.
This shows \begin{equation} \liminf_{m \to \infty} \frac{\log_{5} Q_{m}}{m} \leqslant 2\,, \end{equation} the direction of the inequality being opposed to that you asked for. Whether that holds depends on the irrationality measure of $\pi$. Using the same type of argument, one can show that \begin{equation} \liminf_{m \to \infty} \frac{\log_{5} Q_{m}}{m} = \frac{4}{\mu}\,, \tag{$\ast$} \end{equation} where $\mu$ is the irrationality measure of $\pi$. There are positive $a,b,c,d$ such that \begin{equation} \biggl(\pi - \frac{a}{m5^{4m}}, \pi + \frac{b}{m5^{4m}}\biggr) \subset I_{m} \subset \biggl(\pi - \frac{c}{m5^{4m}}, \pi + \frac{d}{m5^{4m}}\biggr) \end{equation} for all $m$. Now, given $\epsilon > 0$, there are infinitely many rational solutions to \begin{equation} \biggl\lvert \frac{p}{q} - \pi\biggr\rvert < \frac{1}{q^{\mu - \epsilon}} \tag{2} \end{equation} and only finitely many rational solutions to \begin{equation} \biggl\lvert \frac{p}{q} - \pi\biggr\rvert \leqslant \frac{1}{q^{\mu + \epsilon}}\,. \tag{3} \end{equation} As above, from $(2)$ we obtain \begin{equation} \liminf_{m \to \infty} \frac{\log_{5} Q_{m}}{m} \leqslant \frac{4}{\mu - \epsilon} \end{equation} while \begin{equation} \liminf_{m \to \infty} \frac{\log_{5} Q_{m}}{m} < \frac{4}{\mu + \epsilon} \end{equation} would imply infinitely many solutions to $(3)$. Hence $(\ast)$ follows.
The current best upper bound for $\mu$ is still larger than $7$, so we are quite far from even proving a lower bound of $1$ for your $\liminf$.