# 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 | + | ==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 | 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 | ||

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<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> | <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> | ||

− | + | ==Example== | |

+ | This section will prove that the number <math>\ln 2</math> is not [[Liouvillian number|Liouvillian]]. It can be read right after its parent article [[rational approximation of famous numbers]]. The proof of the non-Liouvillian character of <math>\ln 2</math> is much easier than that for <math>\pi</math>, but somewhat more difficult than that for <math>e</math>. | ||

+ | |||

+ | ===A Useful Integral=== | ||

+ | Everyone knows an integral representation for <math>\ln 2</math>. It comes right from the definition of the natural [[logarithm]]: <math>\ln 2=\int_0^1\frac {1}{1+x}\,dx</math>. Let us look at what will happen if we replace <math>1</math> in the numerator by some polynomial <math>P(x)</math> of degree <math>n</math> with integer coefficients. Since <math>P(x)-P(-1)=(1+x)R(x)</math> where <math>R(x)=\sum_{k=0}^{n-1} c_k x^k</math> is some polynomial of degree <math>n-1</math> with integer coefficients, we see that | ||

+ | |||

+ | <math>\int_0^1\frac{P(x)}{1+x}\,dx=\sum_{k=0}^{n-1}\frac {c_k}{k+1}+P(-1)\ln 2.</math> | ||

+ | |||

+ | Let now <math>D_n</math> be the [[least common multiple]] of the numbers <math>1,2,\dots, n</math>. Then | ||

+ | |||

+ | <math>D_n\int_0^1\frac{P(x)}{1+x}\,dx=P_n-Q_n\ln 2</math> | ||

+ | |||

+ | with some integer <math>P_n</math> and <math>Q_n=-D_nP(-1)</math>. | ||

+ | It remains to choose a polynomial <math>P(x)</math> that makes the integral small. | ||

+ | |||

+ | ===Polynomial <math>P(x)</math>=== | ||

+ | We shall just take our "magic polynomial" <math>P(x):=\frac{1}{n!}\left(\frac{d}{dx}\right)^n[x^n(1-x)^n]</math> | ||

+ | (see the [[rational approximation of famous numbers|parent article]] for its properties). | ||

+ | |||

+ | ===Estimates of the Integral=== | ||

+ | Integrating by parts <math>n</math> times, we see that | ||

+ | <math>\int_0^1\frac{P(x)}{1+x}\,dx= | ||

+ | (-1)^n\int_0^1\left[\frac{x(1-x)}{1+x}\right]^n\,\frac{dx}{1+x}</math>. | ||

+ | Since <math>\max_{0<x<1}x(1-x)=\frac 14</math> and since it is attained only at the point <math>x=\frac 12</math> where <math>1+x>1</math>, we can conclude that <math>\Gamma=\max_{0<x<1}\frac{x(1-x)}{1+x}<\frac 14</math>. The absolute value of our integral does not exceed <math>\Gamma^n</math> for all <math>n</math>. To estimate it from below, just notice that the integrand is at least <math>\frac 12\left(\frac 2 {15}\right)^n</math> for <math>\frac 1 3 < x< \frac 2 3</math>, so the absolute value of our integral is at least <math>\frac 16\left(\frac 2 {15}\right)^n>\left(\frac 1 8\right)^n</math> for large <math>n</math>. | ||

+ | |||

+ | ===Estimate of <math>D_n</math>=== | ||

+ | Since the largest possible power of a given prime <math>p\le n</math> that can divide one of the numbers <math>1,2,\dots,n</math> is <math>\left\lfloor\frac{\log n}{\log p}\right\rfloor</math> where <math>\lfloor\cdot\rfloor</math> is the floor function, we have | ||

+ | |||

+ | <math>D_n=\prod_{p\mathrm{\ prime,}\,p\le n}p^{\left\lfloor\frac{\log n}{\log p}\right\rfloor}\le \left(\prod_{p\mathrm{\ prime,}\,p\le \sqrt n}n\right)\cdot\left( \prod_{p\mathrm{\ prime,}\,\sqrt n<p\le n}p\right)\le n^{\sqrt n}\cdot\prod_{p\mathrm{\ prime,}\,p\le n}p\le n^{\sqrt n}\cdot 4^n</math> | ||

+ | |||

+ | according to [[Prime Number Theorem|Chebyshev's estimate]]. Also, we clearly have <math>D_n\ge 1</math>. Thus the absolute value of the product <math>D_n\int_0^1\frac{P(x)}{1+x}\,dx</math> | ||

+ | is between <math>(1/8)^n</math> and <math>(4\Gamma)^n\cdot n^{\sqrt n}</math> for large <math>n</math>. Note that <math>n^{\sqrt n}</math> grows slower than any geometric progression, so the upper bound can be replaced by <math>\gamma^n</math> with any <math>\gamma\in(4\Gamma, 1)</math>. In order to apply the main theorem from the [[rational approximation of famous numbers|parent article]], it remains to show that <math>Q_n</math> do not grow too fast. | ||

+ | |||

+ | ===Estimate of <math>Q_n</math>=== | ||

+ | We already saw that <math>D_n</math> grows not much faster than <math>4^n</math>. As to <math>|P(-1)|</math>, it does not exceed the sum of the absolute values of the coefficients of <math>P(x)</math>, which is not greater than <math>8^n</math>. Thus <math>|Q_n|</math> grow not much faster than <math>32^n</math>, and we are done. | ||

==See Also== | ==See Also== |

## Revision as of 12:47, 26 January 2008

## 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:

## Example

This section will prove that the number is not Liouvillian. It can be read right after its parent article rational approximation of famous numbers. The proof of the non-Liouvillian character of is much easier than that for , but somewhat more difficult than that for .

### A Useful Integral

Everyone knows an integral representation for . It comes right from the definition of the natural logarithm: . Let us look at what will happen if we replace in the numerator by some polynomial of degree with integer coefficients. Since where is some polynomial of degree with integer coefficients, we see that

Let now be the least common multiple of the numbers . Then

with some integer and . It remains to choose a polynomial that makes the integral small.

### Polynomial

We shall just take our "magic polynomial" (see the parent article for its properties).

### Estimates of the Integral

Integrating by parts times, we see that . Since and since it is attained only at the point where , we can conclude that . The absolute value of our integral does not exceed for all . To estimate it from below, just notice that the integrand is at least for , so the absolute value of our integral is at least for large .

### Estimate of

Since the largest possible power of a given prime that can divide one of the numbers is where is the floor function, we have

according to Chebyshev's estimate. Also, we clearly have . Thus the absolute value of the product is between and for large . Note that grows slower than any geometric progression, so the upper bound can be replaced by with any . In order to apply the main theorem from the parent article, it remains to show that do not grow too fast.

### Estimate of

We already saw that grows not much faster than . As to , it does not exceed the sum of the absolute values of the coefficients of , which is not greater than . Thus grow not much faster than , and we are done.