Precise asymptotic of diophantine approximation

Question

I know that every irrational number $\xi$ can be approximated by rational numbers in such a way that $$ \left| \xi - \frac{p}{q} \right| \leq \frac{c}{q^2} $$ for infinitely many choices of $p$ and $q$. Here $c>0$ is some constant that depends on $\xi$.

My question is: for what (irrational) numbers can we have a precise asymptotic estimate, namely $$ \frac{c_1}{q^2} \leq \left| \xi - \frac{p}{q} \right| \leq \frac{c_2}{q^2} $$ as $q \to +\infty$?

Edit

My question comes from the following problem:

Does there exist a cluster point for the sequence $n \mapsto 2n \{n\xi \}-n$? Here $\xi$ is an irrational number and $\{.\}$ denotes the fractional part of a real number.

If I assume that $x \in \mathbb{R}$ is a cluster point, in particular for every $a<x$ and every $b>x$ there are infinitely many integers $p$ and $q$ such that $$ \frac{2a}{q^2} < \xi - \frac{p}{q} < \frac{2b}{q^2}. $$ Furthermore, $p = [q\xi]$, the integer part of $q\xi$.

At this point I am kind of stuck. I can't understand if the answer is positive or negative, although I suspect that it is more probably negative.

Answer 1

One useful characterization is: if and only if the coefficients in the continued fraction expansion of $\xi$ are bounded ($c_1$ and $c_2$ can be bounded in terms of the minimal and maximal coefficient). In particular, this is true when the expansion is periodic, i.e. when $\xi$ is a quadratic irrational.

One more useful pieceful information: if $c_1>1/3$, then there are countably many such $\xi$'s (each equivalent to a quadratic irrational occurring in the Markoff spectrum), while if $c_1=1/3$, then there are already continuum many (pairwise inequivalent) such $\xi$'s. Here equivalence is meant under the action of $\mathrm{SL}_2(\mathbb{Z})$ by fractional linear transformations.

Clarification. My response focused on a classical version of the problem, in which we try to achieve the lower bound for all $q$'s with $c_1$ as large as possible, while we try to achieve the upper bound for infinitely many $q$'s with $c_2$ as small as possible.

Answer 2

Given $(p,q)$ and $\alpha$, let the normalized error of the approximation be $q^2|\xi - \frac{p}{q}|$. My interpretation of the question is whether there are $c_1,c_2 \gt 0$ so that every irrational number has infinitely many approximations whose normalized errors are in $[c_1,c_2]$.

As Wojowu comments, given one of the infinitely many convergents $p/q$ to the simple continued fraction for $\xi$, if $p/q$ is too close, you can make it worse by multiplying $p$ and $q$ by the same factor $d$. This doesn't change the quotient but it increases the normalized error of the approximation by a factor of $d^2$. If you choose $c_2$ so that it is achievable infinitely often (e.g., $1/\sqrt{5}$) then any convergent that is too good can be spoiled to have a normalized error in $[c_2/4,c_2]$.

If you require that $p/q$ is reduced, you can still construct infinitely many rational approximations whose normalized errors are in a uniform fixed interval, even when the coefficients of the simple continued fraction increase such as

$$\frac{e+1}{e-1} = 2+\cfrac{1}{6 + \cfrac{1}{10+\cfrac{1}{14+...}}}$$

There are infinitely many convergents to the simple continued fraction. If the coefficient $a_n$ is $1$ so $p_{n+1}/q_{n+1} = (p_n+p_{n-1})/(q_n+q_{n-1})$, then the convergent $p_n/q_n$ has a normalized error between $1/3$ and $1$: Since $\alpha$ is between $(p_n+p_{n-1})/(q_n+q_{n-1})$ and $(2p_n+p_{n-1})/(2q_n+q_{n-1})$, $|\alpha - p_n/q_n|$ is at least $1/(2q_n^2+q_nq_{n-1}) \ge 1/(3q_n^2)$, and of course it is at most $1/q_n^2$. If the coefficient $a_n$ is not $1$, then $(p_n+p_{n-1})/(q_n+q_{n-1})$ is not a convergent, so its normalized error is at least $1/2$, but its normalized error is not too large. The first $n$ convergents to $\xi$ are also convergents to $(p_n+p_{n-1})/(q_n+q_{n-1})$, so it is not far away and the denominator is less than twice $q_n$.

$$\begin{eqnarray}\left|\xi -\frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right| &\le& \left|\xi - \frac{p_n}{q_n}\right|+\left|\frac{p_n}{q_n} - \frac{p_n+p_{n-1}}{q_n+q_{n-1}}\right|\newline &\le& \frac{1}{q_n(2q_n+q_{n-1})} + \frac{1}{q_n(q_n+q_{n-1})}\newline &\le& \frac{2}{q_n(q_n+q_{n-1})} \newline &\le& \frac{4}{(q_n+q_{n-1})^2}.\end{eqnarray}$$

So, if $a_n \gt 1$, the relative error of $(p_n+p_{n-1})/(q_n+q_{n-1})$ is between $1/2$ and $4$. For any irrational $\xi$, there are infinitely many reduced approximations with normalized errors between $1/3$ and $4$.