Difference between revisions of "Rational approximation of famous numbers"

m (proofreading)
(changed the main theorem to make a better fit with the Beukers proof)
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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>\mu>1</math>, <math>Q>1</math>
+
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 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>\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>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.  
 
''
 
 
----
 
----
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 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>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>\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=\frac {P_n}{Q_n}</math>. Then <math>\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}</math> if <math>q</math> is large enough.
+
'''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 \frac {P_n}{Q_n}</math>. Then <math>\left|x-\frac pq\right|\ge
+
if <math>q</math> is large enough.
\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
+
'''Case 2:''' <math>P_n-Q_n\frac pq\ne 0</math>. Then  
Q^{-1}\frac 1{(2q)^{\frac \mu{\mu-1}}}>\frac 1{q^M}</math> if <math>q</math> is large enough (recall that <math>\mu<\beta</math>).
+
 
 +
<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 $x\in\mathbb R$, 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 $0<\beta<\gamma<1$, $1<Q<+\infty$ and a sequence of pairs of integers $(P_n,Q_n)$ such that for all sufficiently large $n$, we have $|Q_n|\le Q^n$ and $\beta^n< \left|P_n-Q_n x\right|<\gamma^n$. Then, for every $M>\frac{\log(Q/\beta)}{\log(1/\gamma)}$, 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 small, but not too small, linear combinations of $x$ and $1$ with not too large integer coefficients.

Proof of the Main Theorem

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

Case 1: $P_n-Q_n\frac pq=0$. Then $\left|x-\frac pq\right|=\left|x-\frac {P_n}{Q_n}\right|>\frac{\beta^n}{|Q_n|}>(\beta/Q)^n =(\gamma^n)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}> \left(\frac\gamma{2q}\right)^{\frac{\log(Q/\beta)}{\log(1/\gamma)}}>\frac 1{q^M}$

if $q$ is large enough.

Case 2: $P_n-Q_n\frac pq\ne 0$. Then

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

Hence, in this case,

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

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

Applications

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