Difference between revisions of "Rational approximation of famous numbers"

Revision as of 16:55, 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

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

@@ 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.
 ==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==
-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==
 * Beuker's proof that [[pi is not Liouvillian]]
 * Proof that [[e is not Liouvillian]]
 * Proof that [[ln 2 is not Liouvillian]]

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