# Difference between revisions of "Rational approximation of famous numbers"

m (proofreading) |
(changed the main theorem to make a better fit with the Beukers proof) |
||

Line 2: | Line 2: | ||

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

==Main theorem== | ==Main theorem== | ||

− | + | Suppose that there exist <math>0<\beta<\gamma<1</math>, <math>1<Q<+\infty</math> | |

− | Suppose that there exist <math>\beta | + | and a sequence of pairs of integers <math>(P_n,Q_n)</math> such that for all sufficiently large <math>n</math>, we have <math>|Q_n|\le Q^n</math> and |

− | and a sequence of | + | <math>\beta^n< \left|P_n-Q_n x\right|<\gamma^n</math>. Then, for every <math>M>\frac{\log(Q/\beta)}{\log(1/\gamma)}</math>, the inequality <math>\left|x-\frac pq\right|<\frac 1{q^M}</math> has only finitely many solutions. |

− | <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 | + | 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 small, but not too small, linear combinations of <math>x</math> and <math>1</math> with not too large integer coefficients. |

==Proof of the Main Theorem== | ==Proof of the Main Theorem== | ||

− | Choose the least <math>n</math> such that <math> | + | Choose the least <math>n</math> such that <math>\gamma^n\le 2q</math>. Note that for such choice of <math>n</math>, we have <math>\gamma^n> \frac {\gamma}{2q}</math>. Also note that <math>Q_n\ne 0</math> (otherwise <math>|P_n|</math> would be an integer strictly between <math>0</math> and <math>1</math>. Now, there are two possible cases: |

− | '''Case 1:''' <math>\frac pq= | + | '''Case 1:''' <math>P_n-Q_n\frac pq=0</math>. |

+ | Then <math>\left|x-\frac pq\right|=\left|x-\frac {P_n}{Q_n}\right|>\frac{\beta^n}{|Q_n|}>(\beta/Q)^n | ||

+ | =(\gamma^n)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}> | ||

+ | \left(\frac\gamma{2q}\right)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}>\frac 1{q^M}</math> | ||

− | '''Case 2:''' <math>\frac pq\ne | + | if <math>q</math> is large enough. |

− | \left|\frac pq | + | |

− | + | '''Case 2:''' <math>P_n-Q_n\frac pq\ne 0</math>. Then | |

− | + | ||

+ | <math>\frac 1q\le\left|P_n-Q_n\frac pq\right|\le \left|P_n-Q_n x\right|+|Q_n|\cdot\left|x-\frac pq\right|\le \frac 1{2q}+Q^n\left|x-\frac pq\right| | ||

+ | </math> | ||

+ | |||

+ | Hence, in this case, | ||

+ | |||

+ | <math>\left|x-\frac pq\right|\ge \frac 1{2q}Q^{-n}\ge \frac \gamma{2q}Q^{-n}=\frac \gamma{2q}(\gamma^n)^{\frac{\log Q}{\log(1/\gamma)}}\ge \left(\frac\gamma{2q}\right)^{1+\frac{\log(Q}{\log(1/\gamma)}}>\frac 1{q^M}</math> | ||

+ | |||

+ | if <math>q</math> is large enough. (recall that <math>\beta<\gamma</math>, so <math>1+\frac{\log(Q}{\log(1/\gamma)} | ||

+ | =\frac{\log(Q/\gamma)}{\log(1/\gamma)}<\frac{\log(Q/\beta)}{\log(1/\gamma)} </math>). | ||

==Applications== | ==Applications== | ||

* Beuker's proof that [[pi is not Liouvillian]] | * Beuker's proof that [[pi is not Liouvillian]] | ||

* Proof that [[e is not Liouvillian]] | * Proof that [[e is not Liouvillian]] | ||

* Proof that [[ln 2 is not Liouvillian]] | * Proof that [[ln 2 is not Liouvillian]] |

## Revision as of 17:55, 26 June 2006

## 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 pairs of integers such that for all sufficiently large , we have 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 small, but not too small, linear combinations of and with not too large integer coefficients.

## Proof of the Main Theorem

Choose the least such that . Note that for such choice of , we have . Also note that (otherwise would be an integer strictly between and . Now, there are two possible cases:

**Case 1:** .
Then

if is large enough.

**Case 2:** . Then

Hence, in this case,

if is large enough. (recall that , so ).

## Applications

- Beuker's proof that pi is not Liouvillian
- Proof that e is not Liouvillian
- Proof that ln 2 is not Liouvillian