4
$\begingroup$

Liouville's Theorem states that for any algebraic $\alpha \in \mathbb{R}$ of degree $n$, there exists a positive constant $c:=c(\alpha)$ such that $$\left\lvert\alpha-\frac{p}{q}\right\rvert>\frac{c}{q^n}$$ for any $p \in \mathbb{Z}$ and $q \in \mathbb{N}.$

One can find an effective lower bound for $c(\alpha).$ In the special case that $\alpha$ is a quadratic irrational, Exercise 27 in the following set of notes of Jorn Steuding

http://www.math.uni-bremen.de/~bos/dioph.pdf

yields $$c(\alpha) \gg \frac{1}{(1+|\alpha|)H(\alpha)} .$$ Here if $m_{\alpha}(x):=x^2+bx+c \in \mathbf{Q}[x]$ is the minimal polynomial of $\alpha$, the height $H(\alpha)$ is defined as the maximum of $|b|$ and $|c|.$ My question is whether one can find a better lower bound for $c(\alpha)$ when $\alpha$ is a quadratic irrational or if this is best possible.

$\endgroup$
1
  • $\begingroup$ I would start with the ultra-classic Hardy - Wright. I vaguely remember some related discussion in the final notes of the relevant chapters, with references $\endgroup$
    – user24527
    Jun 27, 2012 at 19:48

2 Answers 2

4
$\begingroup$

Let's assume $\alpha > 0$. If $\alpha$ is a quadratic irrational, its simple continued fraction $a_0 + \dfrac{1}{a_1 + \frac{1}{a_2+\ldots}}$ is eventually periodic. Every $p/q$ (in lowest terms) with $\left|\alpha - \dfrac{p}{q}\right| < \dfrac{1}{2q^2}$ is a convergent of $\alpha$, and for the $n$'th convergent $$ \dfrac{1}{q_n^2 (a_{n+1}+2)} < \left| \alpha - \dfrac{p_n}{q_n} \right| \le \dfrac{1}{q_n^2 a_{n+1}}$$ Thus $\dfrac{1}{a_M+2} \le c(\alpha) \le \dfrac{1}{a_M}$ where $a_M$ is the largest element in the continued fraction of $\alpha$.

$\endgroup$
3
  • $\begingroup$ Thank you Robert. Your answer proves that an equivalent formulation (up to absolute constants) of my question is: Is there a good upper bound about the maximum of the elements in the continued fraction of a quadratic irrational ? $\endgroup$
    – Dr. Pi
    Jun 27, 2012 at 23:23
  • $\begingroup$ In the case where the quadratic irrational is simply $\sqrt d$, the largest element in the continued fraction is twice the integer part of $\sqrt d$. $\endgroup$ Jun 27, 2012 at 23:50
  • $\begingroup$ Indeed Lagrange showed that the largest element for $(P+\sqrt D)/Q$ is less that $2\sqrt D$. $\endgroup$ Jun 28, 2012 at 0:03
-1
$\begingroup$

The constant $c=1/\sqrt5$ (with $n=2$) works for any $\alpha$. If $\alpha$ is not, roughly speaking, the golden ratio, then $c$ can be improved to $1/2\sqrt2$, etc. If one removes a certain infinite sequence of quadratic irrationals, one can take $c=1/3$, but this is the best you can do in a general setting.

A nice exposition can be found in [Cassels, An Introduction to Diophantine Approximation]. You may also start with a Wikipedia article.

$\endgroup$
3
  • $\begingroup$ Dear Nikita, I am afraid that Liouville's is about how bad the approximation of an irrational by a rational can be. Unfortunately you refer to Hurwitz's result, as well as Markoff's spectrum, both of which mention how good this approximation can be. That is the opposite of what I am asking. I am really sorry for this. $\endgroup$
    – Dr. Pi
    Jun 27, 2012 at 23:17
  • $\begingroup$ OK, noted. Still, if you look at all the quadratic irrationals in Markoff's spectrum, these constants are exact. $\endgroup$ Jun 28, 2012 at 0:19
  • $\begingroup$ Indeed you are right. However, the reason for my question was that I had a specific application in mind, where no information regarding $\alpha$ exists, apart of course from the fact that $\alpha$ is a quadratic irrational. $\endgroup$
    – Dr. Pi
    Jun 28, 2012 at 0:47

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.