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This question relates to this one and that one.

Some background

In the setting of discrete holomorphic dynamics (say, Julia sets) an irrational $\lambda$ is said to be well approximated by rational numbers when \begin{eqnarray*} \sum_{n=0}^{+\infty}\frac{\ln q_{n+1}}{q_{n}} & = & +\infty\, \end{eqnarray*} where $\left(\frac{p_{n}}{q_{n}}\right)_{n\in\mathbb{N}}$ is its sequence of convergents (see continued fractions). This "arithmetic" condition was an improvement of Cremer condition (replace $\Sigma$ by $\limsup$) by Brjuno to tackle the (difficult) problem of linearization of germs of a biholomorphism \begin{eqnarray*} \Delta\left(z\right) & = & e^{2\mathtt{i}\pi\lambda}z+o\left(z\right)\. \end{eqnarray*} The theorem of Siegel-Brjuno asserts that if the condition does not hold then any such germ $\Delta$ is locally conformally conjugate to its linear part (the irrational rotation). The converse is a (very clever, needless to say) construction produced by Yoccoz: if the above condition holds then there exists some germs $\Delta$ which are not locally linearizable. His construction boils down to building a $\Delta$ with periodic orbits accumulating on $0$. To this day the moduli space of conjugacy of such germs is not known, and describing it remains an important open question in the field. Somehow all this is also connected to the recent (beautiful) work of Buff and Chéritat, where they build a Julia set of full Lebesgue measure.

It also relates to more "conventional" dynamics, as Perez-Marco exhibited other "arithmetic" conditions involved in the problem of classification of circle diffeomorphisms.

It is well known that irrational numbers well approximated by rational numbers, in the above sense, is a $PSL_{2}\left(\mathbb{Z}\right)$-invariant set (through the action by homographies on the real line) with zero Lebesgue measure. The fact that such numbers are so rare is related to the fact that, for almost every irrational number, the geometric mean of the integers $(a_{n})_{n\in\mathbb{N}}$ appearing in the continued fraction expansion converges to the Khinchin's constant $K_{0}\simeq2,68$.

My question

It is easy to produce theoretical examples of such numbers, just start from a sequence $\left(q_{n}\right)_{n\in\mathbb{N}}$ satisfying the condition and find an adequate sequence $\left(p_{n}\right)_{n\in\mathbb{N}}$, for instance by following a walk in the Stern-Brocot tree, such that $\lim\frac{p_{n}}{q_{n}}\notin\mathbb{Q}$.

Now, does anyone know about an "explicit" (for a reasonable notion of explicit) number which is well approximated by rational numbers?

To the best of my knowledge this question should be answered as "no", but the limits of my knowledge are not that far away, even from my point of view ;) Thank you in advance for any comment (better: answer!) to this wishful question.

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I think the devil will be in the detail of what you mean by explicit here. $\sum 10^{-a_n}$ will certainly satisfy your criterion if $a_n$ is a sufficiently fast growing sequence of integers (i.e. $a_n=2\uparrow\uparrow(2n)$: a tower of twos of height $2n$). An alternative would be to prescribe the sequence of partial quotients (again $a_n=2\uparrow\uparrow (2n)$ should suffice). –  Anthony Quas Jan 16 '13 at 22:14
You're right Anthony, I should precise this. By explicit I mean something of finite presentation, a "conventional" computable number (a "closed-form") obtained from the usual constants of practical-world mathematics. Obviously an infinite series does not match that criterion, as it is in the same vein as producing a sequence of denominators, like I explained in my question. –  Loïc Teyssier Jan 16 '13 at 22:34
Related to your question, how many "explicit" continued fraction expansions do we know? As far as I understand, your definition of being well-approximated depends heavily on this. For instance, is it obvious that Anthony Quass' example (forgetting non-expliciteness) is well-appoximated? –  Filippo Alberto Edoardo Jan 17 '13 at 0:31
@Filippo: you do have a point here. Yet the construction of Anthony does work. Let $x$ be the number $\sum{10^{-a_n}}$ with, say, $a_0=1$. Using the property that if $|x-\frac{p}{q}|<\frac{1}{2q^2}$ then $\frac{p}{q}$ is one of the convergents, we see that the quotients of the convergents are $q_n=10^{-a_n}$ provided $a_{n+1}>\log{2}+2a_n$. Hence it suffices to have $a_{n+1}>10^{a_n}$ to ensure that $x$ is well approximated by rational numbers. –  Loïc Teyssier Jan 17 '13 at 16:28
edit to my previous comment: read $q_n=10^{a_n}$, and $a_{n+1}\geq10^{a_n}$. –  Loïc Teyssier Jan 17 '13 at 16:54
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1 Answer

Your question seems somewhat ill-posed. What type of construction exactly do you allow? I suppose that you might ask for the number to be obtained by basic algebraic operations and perhaps some elementary functions from rational numbers, and perhaps some standard transcendental constants such as $\pi$ and $e$. However, to make it a precise question would require some work, and in some sense whatever you come up with would likely be arbitrary.

I strongly believe that no such number is known, though I do not have a definitive reference stating this, and there are people who know more about this. However, it seems that, where the continued fraction expansions are explicitly known (such as for $e$), the growth of the expansion tends to be at most linear, a far cry from what you would need to violate Bryuno's condition.

The next number you might try is probably $\pi$. However, as far as I understand, it hasn't even been proved that its continued fraction expansion is unbounded! Based on experimental data of the initial part of this expansion, it certainly doesn't seem likely that $\pi$ violates Bryuno condition.

Indeed, given that almost every rational number is Bryuno, it seems rather unlikely that any specific one that we might try will fail the condition, unless there is a specific reason for it.

Moreover, imagine yourself undertaking some work in an unrelated field of mathematics, and you come across a new significant constant $\alpha$. Unbeknownst to you, $\alpha$ is Bryuno, which means it likely looks an awful lot like a rational number $\rho$. Unless you know for some other reason that $\alpha$ must be irrational (which would be rather lucky indeed), how likely are you to be able to tell that $\alpha\neq\rho$?

However, most importantly in my view, it is not clear that you are asking the right question. Suppose that by some miracle, you knew that $x:=\sqrt[5]{e+\pi+\zeta(5)}$ is a non-Bryuno number, how would that help you? (I suppose it would say interesting things about the number from a number-theoretic point of view, but I am assuming that is not really what you are after.)

It seems rather difficult to argue that the number $x$ is much more natural than any other (let's say efficiently) computable real number. And the number in Anthony's comment will certainly be very efficiently computable.

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Thank you Lasse for your answer. My question was asked out of sheer curiosity. This condition has been hanging around for some time now, and so far I've never stumbled on an "explicit" example (again with big quotes). I'm willing to agree the notion of explicit, that you describe at the biginning of your answer, is arbitrary and was the one I was implicetly referring to. Yet most people will find this defintion intuitive enough, the constants $\pi$ and $e$ appearing so naturally and enduringly in maths courses that their genuine transcendance is forgotten. –  Loïc Teyssier Jan 23 '13 at 11:05
That being said, I also agree that it would not be much more helpful to know that some specific number is not Bryuno. Not knowing a "natural" (not better...) example is just a little frustating and I really wanted to know if someone knew better than me. Only out of curiosity... –  Loïc Teyssier Jan 23 '13 at 11:12
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