Difference between revisions of "Rational approximation of famous numbers"
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Suppose that there exist <math>\beta>\mu>1</math>, <math>Q>1</math> | Suppose that there exist <math>\beta>\mu>1</math>, <math>Q>1</math> | ||
| − | and a sequence of rational numbers <math>\frac {P_n}{Q_n}</math> such that for all <math>n</math>, <math>Q_n\le Q^n</math> and | + | and a sequence of rational numbers <math>\frac {P_n}{Q_n}</math> such that for all sufficiently large <math>n</math>, <math>Q_n\le Q^n</math> and |
<math>Q^{-\beta n}< \left|x-\frac {P_n}{Q_n}\right|<Q^{-\mu n}</math>. Then, for every <math>M>\frac\beta{\mu-1}</math>, the inequality <math>\left|x-\frac pq\right|<\frac 1{q^M}</math> has only finitely many solutions. | <math>Q^{-\beta n}< \left|x-\frac {P_n}{Q_n}\right|<Q^{-\mu n}</math>. Then, for every <math>M>\frac\beta{\mu-1}</math>, the inequality <math>\left|x-\frac pq\right|<\frac 1{q^M}</math> has only finitely many solutions. | ||
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| − | The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math> but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of <math>x</math>. | + | The exact formulation of the main theorem in this article is fitted to the Beukers proof of the non-Liouvillian character of <math>\pi</math> but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that <math>x</math> cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of <math>x</math>. |
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==Proof of the Main Theorem== | ==Proof of the Main Theorem== | ||
Choose the least <math>n</math> such that <math>Q^{(\mu-1) n}\ge 2q</math>, i.e., that <math>Q^n\ge (2q)^{1/(\mu-1)}</math>. Note that for such choice of <math>n</math>, we have <math>Q^n< Q(2q)^{1/(\mu-1)}</math>. There are two possible cases: | Choose the least <math>n</math> such that <math>Q^{(\mu-1) n}\ge 2q</math>, i.e., that <math>Q^n\ge (2q)^{1/(\mu-1)}</math>. Note that for such choice of <math>n</math>, we have <math>Q^n< Q(2q)^{1/(\mu-1)}</math>. There are two possible cases: | ||
Revision as of 10:45, 26 June 2006
This article is a stub. Help us out by expanding it.
Introduction
The Dirichlet's theorem shows that, for each irrational number 
, the inequality 
has infinitely many solutions. On the other hand, sometimes it is useful to know that 
cannot be approximated by rationals too well, or, more precisely, that is not a Liouvillian number, i.e., that for some power 
, the inequality 
holds for all sufficiently large denominators 
. So, how does one show that a number is not Liouvillian? The answer is given by the following
Main theorem
Suppose that there exist 
, 
and a sequence of rational numbers 
such that for all sufficiently large 
, 
and
. Then, for every 
, the inequality 
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 
but the general spirit of all such theorems is the same: roughly speaking, they tell you that in order to show that cannot be approximated by rationals too well, one needs to find plenty of good but not too good rational approximations of .
Proof of the Main Theorem
Choose the least such that 
, i.e., that 
. Note that for such choice of , we have 
. There are two possible cases:
Case 1: 
. Then 
if is large enough.
Case 2: 
. Then 
if is large enough (recall that 
).
