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Rational approximation of famous numbers

Revision as of 10:43, 26 June 2006 by Fedja (talk | contribs) (→‎Proof of the Main Theorem)

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Introduction

The Dirichlet's theorem shows that, for each irrational number $x\in\mathbb R$, the inequality $\left|x-\frac pq\right|<\frac 1{q^2}$ has infinitely many solutions. On the other hand, sometimes it is useful to know that $x$ cannot be approximated by rationals too well, or, more precisely, that $x$ is not a Liouvillian number, i.e., that for some power $M<+\infty$, the inequality $\left|x-\frac pq\right|\ge \frac 1{q^M}$ holds for all sufficiently large denominators $q$. So, how does one show that a number is not Liouvillian? The answer is given by the following

Main theorem

Suppose that there exist $\beta>\mu>1$, $Q>1$ and a sequence of rational numbers $\frac {P_n}{Q_n}$ such that for all $n$, $Q_n\le Q^n$ and $Q^{-\beta n}< \left|x-\frac {P_n}{Q_n}\right|<Q^{-\mu n}$. Then, for every $M>\frac\beta{\mu-1}$, the inequality $\left|x-\frac pq\right|<\frac 1{q^M}$ has only finitely many solutions.

The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of $\pi$ but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that $x$ cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of $x$.

Proof of the Main Theorem

Choose the least $n$ such that $Q^{(\mu-1) n}\ge 2q$, i.e., that $Q^n\ge (2q)^{1/(\mu-1)}$. Note that for such choice of $n$, we have $Q^n< Q(2q)^{1/(\mu-1)}$. There are two possible cases:

Case 1: $\frac pq=\frac {P_n}{Q_n}$. Then $\left|x-\frac pq\right|=\left|x-\frac {P_n}{Q_n}\right|>Q^{-\beta n}>Q^{-\beta}\frac 1{(2q)^{\frac \beta{\mu-1}}}>\frac 1{q^M}$ if $q$ is large enough.

Case 2: $\frac pq\ne \frac {P_n}{Q_n}$. Then $\left|x-\frac pq\right|\ge \left|\frac pq-\frac{P_n}{Q_n}\right|-\left|x-\frac {P_n}{Q_n}\right| >\frac 1{qQ_n}-Q^{-\mu n}\ge Q^{-n}\left(\frac 1q-\frac 1{Q^{(\mu-1)n}}\right)\ge \frac 1{2qQ^n}\ge Q^{-1}\frac 1{(2q)^{\frac \mu{\mu-1}}}>\frac 1{q^M}$ if $q$ is large enough (recall that $\mu<\beta$).

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