Rational approximation of famous numbers

Revision as of 13:47, 26 January 2008 by Temperal (talk | contribs) (Proof of the Main Theorem: sub)

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

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

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

$P(x)\frac 1{n!}\left(\frac d{dx}\right)^n [x^n(1-x)^n]$

Its coefficients can be easily computed using the binomial theorem:

$P(x)=\sum_{k=0}^n (-1)^k{n+k\choose n}{n\choose k}x^k.$

The important points are that all the coefficients are integer and the sum of their absolute values does not exceed $\max_{0\le k\le n}{n+k\choose n}\sum_{0\le k\le n}{n\choose  k}\le {2n\choose n}2^n\le 8^n$.

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

$\int_0^1 F(x)P(x)\,dx=(-1)^n\frac 1{n!}\int_0^1 F^{(n)}(x) x^n(1-x)^n\,dx.$

Example

This section will prove that the number $\ln 2$ 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 $\ln 2$ is much easier than that for $\pi$, but somewhat more difficult than that for $e$.

A Useful Integral

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

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

Let now $D_n$ be the least common multiple of the numbers $1,2,\dots, n$. Then

$D_n\int_0^1\frac{P(x)}{1+x}\,dx=P_n-Q_n\ln 2$

with some integer $P_n$ and $Q_n=-D_nP(-1)$. It remains to choose a polynomial $P(x)$ that makes the integral small.

Polynomial $P(x)$

We shall just take our "magic polynomial" $P(x):=\frac{1}{n!}\left(\frac{d}{dx}\right)^n[x^n(1-x)^n]$ (see the parent article for its properties).

Estimates of the Integral

Integrating by parts $n$ times, we see that $\int_0^1\frac{P(x)}{1+x}\,dx= (-1)^n\int_0^1\left[\frac{x(1-x)}{1+x}\right]^n\,\frac{dx}{1+x}$. Since $\max_{0<x<1}x(1-x)=\frac 14$ and since it is attained only at the point $x=\frac 12$ where $1+x>1$, we can conclude that $\Gamma=\max_{0<x<1}\frac{x(1-x)}{1+x}<\frac 14$. The absolute value of our integral does not exceed $\Gamma^n$ for all $n$. To estimate it from below, just notice that the integrand is at least $\frac 12\left(\frac 2 {15}\right)^n$ for $\frac 1 3 < x< \frac 2 3$, so the absolute value of our integral is at least $\frac 16\left(\frac 2 {15}\right)^n>\left(\frac 1 8\right)^n$ for large $n$.

Estimate of $D_n$

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

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

according to Chebyshev's estimate. Also, we clearly have $D_n\ge 1$. Thus the absolute value of the product $D_n\int_0^1\frac{P(x)}{1+x}\,dx$ is between $(1/8)^n$ and $(4\Gamma)^n\cdot n^{\sqrt n}$ for large $n$. Note that $n^{\sqrt n}$ grows slower than any geometric progression, so the upper bound can be replaced by $\gamma^n$ with any $\gamma\in(4\Gamma, 1)$. In order to apply the main theorem from the parent article, it remains to show that $Q_n$ do not grow too fast.

Estimate of $Q_n$

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

See Also

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