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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.

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  • $\begingroup$ I suppose you mean for the double inequality to hold for infinitely many fractions $p/q$? $\endgroup$
    – Wojowu
    Apr 23, 2016 at 16:21
  • $\begingroup$ Yes, along some subsequence of integers. $\endgroup$
    – Siminore
    Apr 23, 2016 at 16:22
  • $\begingroup$ Unless you assume $p,q$ relatively prime, you can choose $c_2=c$ from the first inequality and $c_1=c/4$. To sketch why it works, take any fraction satisfying $|\xi-p/q|<c/q^2$. If $|\xi-p/q|<c/4q^2$, then replace $p/q$ by $2p/2q$, and repeat until the double inequality works. $\endgroup$
    – Wojowu
    Apr 23, 2016 at 16:25
  • $\begingroup$ @Wojowu It seems to me that $c_1$ and $c_2$ cannot be both close to a fixed number different than zero, isn't it? $\endgroup$
    – Siminore
    Apr 23, 2016 at 18:11

2 Answers 2

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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.

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    $\begingroup$ Per what OP said in the comments, they meant the double inequality to hold for infinitely many $p/q$, not that the left inequality holds for all $p/q$. However, your answer still contains neat results, so I advise you to keep it, but add how you understood OP's question. $\endgroup$
    – Wojowu
    Apr 23, 2016 at 16:40
  • $\begingroup$ @Wojowu: Thanks, I will clarify my response. $\endgroup$
    – GH from MO
    Apr 23, 2016 at 16:50
  • $\begingroup$ @GHfromMO It is very interesting. Can you please read my edit, where I try to be more precise about my question? $\endgroup$
    – Siminore
    Apr 23, 2016 at 18:10
  • $\begingroup$ Can't you have uncountably many $\xi$ corresponding to the same element of the Markov spectrum? $\endgroup$ Apr 24, 2016 at 4:14
  • $\begingroup$ @DouglasZare: I think no, since $\mathrm{SL}_2(\mathbb{Z})$ is countable. Two $\xi$'s are equivalent if and only if their continued fraction expansions are the same apart from finitely many coefficients and a shift of the coefficients. $\endgroup$
    – GH from MO
    Apr 24, 2016 at 14:16
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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 $4$.

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  • $\begingroup$ According to my edit, my question seems to be equivalent to the following: given $a$ and $b$ arbitrarily close to some (positive) number $x$, do there exist infinitely many integers $m$ and $n$ such that $$a \leq n^2 \left( \xi - \frac{m}{n} \right)$$ and $$n \left( \xi - \frac{m}{n} \right) < \frac{1}{2}?$$ Hence the question is related to your normalized error, and - roughly - I am asking if this error can converge to some positive number. $\endgroup$
    – Siminore
    Apr 24, 2016 at 8:29
  • $\begingroup$ @Siminore: Since there are infinitely many normalized errors in an interval, there must be limit points. $\endgroup$ Apr 24, 2016 at 8:52
  • $\begingroup$ Okay, but I don't think this is enough. To reverse your argument, it seems to me that $p$ must be something special: the integer part of $q \xi$. $\endgroup$
    – Siminore
    Apr 24, 2016 at 9:12
  • $\begingroup$ @Siminore: That's not a particularly special condition on a good approximation to $\xi$. For convergents, it means that the index $n$ is even, since those are the approximations slightly below $\xi$ while the odd convergents are slightly above $\xi$. There are infinitely many even indices. $\endgroup$ Apr 24, 2016 at 9:55

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