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Notation: By $[a_0;a_1,a_2,\ldots]$ I mean the continued fraction $$a_0+\frac1{a_1+\dfrac1{a_2+\dfrac1{a_3+\ddots}}}$$ where $a_n$ is a positive integer.

Context: Let $\alpha$ be an irrational number and $[a_0;a_1,a_2,\ldots]$ its continued fraction. "Big" terms in the sequence $\{a_n\}$ marks that the previous fraction is a "very good" approximation of the number. For example, the term $a_4$ for the number $\pi$ is $292$, that comes just after the extraordinarily good approximation $\pi\approx\dfrac{355}{113}$.

On the other hand, the sequence is bounded if $\alpha$ is the solution of a quadratic equation, since it is periodic.

Question: Is there a "good" characterization of the set of irrational numbers whose continued fraction has a bounded (but not periodic) sequence of terms? (For example $[1;1,2,1,1,2,1,1,1,2,\ldots]$)

A first thought: The set of these numbers is non-countable.

ajotatxe
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    Possibly related: https://math.stackexchange.com/q/2126551/42969. – Martin R Jul 17 '19 at 14:30
  • I see that it is known very little about this... – ajotatxe Jul 17 '19 at 14:33
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    Not sure what you mean by good characterization. What I know is that the sequence of terms in the continued fraction of $\alpha\notin\mathbb{Q}$ is bounded if and only if there exists a constant $C>0$ such that for all $p,q\in\mathbb{Z} (q\ne 0)$ we have $|\alpha-\frac{p}{q}|>\frac{C}{q^2}$. – Mark Jul 17 '19 at 14:34
  • I mean an "algebraic" characterization. Perhaps the solutions of a certain type of equations or the limit of some type of series. – ajotatxe Jul 17 '19 at 14:37
  • @metamorphy Yeah, I thought too fast... Fixed. – ajotatxe Jul 17 '19 at 14:53
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    There almost certainly isn't a nice "algebraic" characterization, considering the set contains uncomputable numbers. – sbares Jul 17 '19 at 14:53
  • @SBareS I'm no expert in mathematical logic, but this seems to me a proof that there is no "good" characterization with the sense I meant. – ajotatxe Jul 17 '19 at 14:56
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    @ajotatxe it all depends what a "good" characterization is. If it means an algorithm, then you are out of luck. On the other hand, the characterization in terms of irrationality measure mentioned by Mark is "good" in my opinion, probably the "best" you'll get. In particular, it gives the precise sense in which such numbers are hard to approximate by rationals. – sbares Jul 17 '19 at 15:25
  • They are definitely uncountable, since the set ${1,2}^{\mathbb N}$ is uncountable, so just the reals with $a_i\in{1,2}$ are uncountable. – Thomas Andrews Jul 17 '19 at 15:42
  • The only result about the size of the continued fraction coefficients that I know about goes in the other direction - real numbers whose continued fractions have too many "large" coefficients are transcendental. I forget what this theorem is called, or the particulars of the condition, though, and it doesn't mean all transcendental numbers have unreasonably large coefficients - I'd wager some of your bounded continued fractions are transcendental. – Thomas Andrews Jul 17 '19 at 15:46
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    These numbers are very intensely studied, but not exclusively using methods of algebra, more using analysis and geometry. One theorem is that the set of these numbers has Lebesgue measure zero, and its intersection with each open interval has Hausdorff dimension 1. The study of these numbers (and of similar mathematical objects in a wide variety of settings) has its own title: "Diophantine approximation". One of my favorite papers in Diophantine approximation is that of Wolfgang M. Schmidt, "On badly approximable numbers and certain games." Trans. Amer. Math. Soc. 123 (1966), 178–199. – Lee Mosher Jul 17 '19 at 15:50
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    A non-rational real number $\alpha$ has bounded continued fraction coefficients if and only if, for $P,Q\in\mathbb Z$ and $Q\neq 0$ you have that: $$f(P,Q)=\frac{1}{|Q||P-Q\alpha|}$$ is bounded above. – Thomas Andrews Jul 17 '19 at 16:03

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