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

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

+ | ==Magic polynomial== | ||

+ | Before proceeding to the applications of the main theorem, let us introduce one very useful polynomial that often appears in proofs of irrationality. It is the polynomial | ||

+ | |||

+ | <math>P(x)\frac 1{n!}\left(\frac d{dx}\right)^n [x^n(1-x)^n]</math> | ||

+ | |||

+ | Its coefficients can be easily computed using the [[Binomial Theorem|binomial theorem]]: | ||

+ | |||

+ | <math>P(x)=\sum_{k=0}^n (-1)^k{n+k\choose n}{n\choose k}x^k.</math> | ||

+ | |||

+ | The important points are that all the coefficients are integer and | ||

+ | the sum of their absolute values does not exceed | ||

+ | <math>\max_{0\le k\le n}{n+k\choose n}\sum_{0\le k\le n}{n\choose k}\le {2n\choose n}2^n\le 8^n</math>. | ||

+ | |||

+ | Another useful remark is that the first <math>n-1</math> derivatives of <math>x^n(1-x)^n</math> vanish at <math>0</math> and <math>1</math>, which makes the integration by parts extremely convenient: | ||

+ | |||

+ | <math>\int_0^1 F(x)P(x)\,dx=(-1)^n\frac 1{n!}\int_0^1 F^{(n)}(x) x^n(1-x)^n\,dx.</math> | ||

+ | |||

+ | And now everything is ready for three | ||

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

− | |||

* 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]] | ||

+ | * Beukers' proof that [[pi is not Liouvillian]] |

## Revision as of 22:51, 26 June 2006

## Contents

## 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 ).

## Magic polynomial

Before proceeding to the applications of the main theorem, let us introduce one very useful polynomial that often appears in proofs of irrationality. It is the polynomial

Its coefficients can be easily computed using the binomial theorem:

The important points are that all the coefficients are integer and the sum of their absolute values does not exceed .

Another useful remark is that the first derivatives of vanish at and , which makes the integration by parts extremely convenient:

And now everything is ready for three

## Applications

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