6
$\begingroup$

A Liouville number is an irrational number $x$ with the property that, for every positive integer $n$, there exist integers $p$ and $q$ with $q > 1$ and such that $0 < \mid x - \frac{p}{q} \mid < \frac{1}{q^n} $.

I'm looking for either hints or a complete proof for the fact that $e$ is not a Liouville number. I can prove that $e$ is irrational and even that it is transcendental, but I'm a bit stuck here.

Here's my research:

The wikipedia article about Liouville numbers states:

[...] not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of $e$, one can show that e is an example of a transcendental number that is not Liouville.

However, theres clearly more to the argument then the boundedness of the continued fraction of $e$, because the terms of $e$'s continued fraction expansion are unbounded and yet it is not a Liouville number. Also, if possible, i would like to avoid using continued fractions at all.

This book has the following as an exercise:

Prove that $e$ is not a Liouville number. (Hint: Follow the irrationality proof of $e^n$ given in the supplements to Chapter 1.)

Unfortunately, the supplements to Chapter 1 are not publically available in the sample and I do not want to buy that book.

This book states:

Given any $\varepsilon > 0$, there exists a constant $c(e,\varepsilon) > 0$ such that for all $p/q$ there holds $\frac{c(e,\varepsilon)}{q^{2+\varepsilon}} < \mid e - \frac{p}{q} \mid$. [...] Using [this] inequality, show that $e$ is not a Liouville number.

Which, given the inequality, I managed to do. But I do not have any idea of how one would go about proving that inequality.

I greatly appreciate any help!

$\endgroup$
  • 1
    $\begingroup$ 1) Should "Using [this] inequality, show that $e$ is a Liouville number" be "is not a Liouville number?" 2) The desired inequality by definition means that $e$ has irrationality measure $\mu(e)=2$. Wikipedia notes this in passing, and Mathworld's page on irrationality measure cites Borwein & Borwein to that effect; alas, there's no preview of that book on Google books. $\endgroup$ – Semiclassical Sep 29 '16 at 14:24
  • $\begingroup$ Thanks, I corrected that. Thanks for pointing out, that the equality in question essentially means that $e$ has irrationality measure of 2. However this seem to be hard to prove and from a quick search I could not come up with a freely available proof. $\endgroup$ – cdwe Sep 29 '16 at 14:47
5
$\begingroup$

Using Gauss continued fraction for $\tanh$, it is not difficult to show that the continued fraction of $e$ has the following structure: $$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,\ldots]\tag{1}$$ then by studying the sequence of convergents $\left\{\frac{p_n}{q_n}\right\}_{n\geq 1}$ through $$\left|\frac{p_n}{q_n}-\frac{p_{n+1}}{q_{n+1}}\right|=\frac{1}{q_n q_{n+1}}=\frac{1}{q_n(\alpha_{n+1}q_n+q_{n-1})}\tag{2}$$ and $$ \left|e-\frac{p_n}{q_n}\right| = \left|\sum_{k\geq n}\frac{(-1)^k}{q_k q_{k+1}}\right| \tag{3} $$ we may easily get that there is no rational approximation such that $$ \left|e-\frac{p_n}{q_n}\right|\leq \frac{1}{q_n^4}\tag{4} $$ hence $e$ is not a Liouville number. It is not difficult to use $(1)$ to prove the stronger statement

The irrationality measure of $e$ is $2$.

$\endgroup$
  • $\begingroup$ Thank you, that works for me. $\endgroup$ – cdwe Sep 29 '16 at 17:30
1
$\begingroup$

In continued fractions for $\alpha=[a_0,a_1,a_2,\dots,a_n,\dots]$ we have $$\left|\frac{P_n}{Q_n} -\alpha\right|>\frac{1}{2}\left|\frac{P_n}{Q_n}-\frac{P_{n+1}}{Q_{n+1}}\right|=\frac{1}{2Q_nQ_{n+1}}\geq\frac{1}{2Q_n^2(a_n+1)}$$.

So the fact that $Q_n\geq F_n$ (with $F_n$ the $n$th Fibonacci number) and $a_n=O(n)$ when $\alpha=e$ means that $2Q_n^2(a_n+1)$ can't get bigger than $Q_n^3$ for large $n$.

This provides at least one necessary condition for $\alpha$ being Liouville - that there are for any $k$ infinitely man $n$ with $a_n\geq Q_n^k/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.