Liouville's Theorem in Diophantine Approximation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:15:10Z http://mathoverflow.net/feeds/question/100799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100799/liouvilles-theorem-in-diophantine-approximation Liouville's Theorem in Diophantine Approximation Captain Darling 2012-06-27T18:55:26Z 2012-06-27T21:43:14Z <p>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 $$|\alpha-\frac{p}{q}|>\frac{c}{q^n}$$ for any $p \in \mathbb{Z}$ and $q \in \mathbb{N}.$</p> <p>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 </p> <p><a href="http://www.math.uni-bremen.de/~bos/dioph.pdf" rel="nofollow">http://www.math.uni-bremen.de/~bos/dioph.pdf</a></p> <p>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.</p> http://mathoverflow.net/questions/100799/liouvilles-theorem-in-diophantine-approximation/100811#100811 Answer by Robert Israel for Liouville's Theorem in Diophantine Approximation Robert Israel 2012-06-27T21:23:57Z 2012-06-27T21:23:57Z <p>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 $\left|\alpha - \dfrac{p}{q}\right| &lt; \dfrac{1}{2q^2}$ is a convergent of $\alpha$, and for the $n$'th convergent $$\dfrac{1}{q_n^2 (a_{n+1}+2)} &lt; \left| \alpha - \dfrac{p_n}{q_n} \right| \le \dfrac{1}{q_n^2 a_{n+1}}$$ 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 $\alpha$. </p> http://mathoverflow.net/questions/100799/liouvilles-theorem-in-diophantine-approximation/100813#100813 Answer by Nikita Sidorov for Liouville's Theorem in Diophantine Approximation Nikita Sidorov 2012-06-27T21:43:14Z 2012-06-27T21:43:14Z <p>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. </p> <p>A nice exposition can be found in [Cassels, An Introduction to Diophantine Approximation]. You may also start with a <a href="http://en.wikipedia.org/wiki/Markov_spectrum" rel="nofollow">Wikipedia article</a>. </p>