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Given a strictly positive integer $A$, let $D(A)$ denote the set of all real quadratic algebraic numbers with a continued fraction having almost all coefficients $\leq A$.

Consider the field $Q_A$ generated by all elements of $D(A)$. One has $Q_1=\mathbb Q[\sqrt{5}]\subset Q_2\subset Q_3,\dots$.

The inclusion $Q_1\subset Q_2$ is strict since $Q_2$ contains for example $\sqrt{2}=[1;2,2,2,\dots]$ and $\sqrt{3}=[1;1,2,1,2,\dots]$.

Are there other strict inclusions? Are there cases of equality? (I ignore for example if $Q_2$ is a proper subfield of the field $\mathbb Q[\sqrt{\mathbb N}]$ generated by all real quadratic number-fields.)

A related question: Given a real quadratic algebraic number $\alpha$ with continued fraction expansion $\alpha=[a_0;a_1,a_2,\dots]$ consider the mean value $\mu(\alpha)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^na_j$. Since $[a_0;a_1,\dots]$ is ultimately periodic, this mean value is a well-defined rational number $\geq 1$ if $\alpha$ is irrational.

Are there examples of quadratic number-fields $\mathbb Q[\sqrt{N}]$ such that $\inf_{\alpha\in \mathbb Q[\sqrt{N}]\setminus\mathbb Q}\mu(\alpha)>1$?

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Two questions about the question : (1) I guess "almost all" = "all but finitely many" ? (2) are you sure $Q_A$ will be a number field (ie: finite extension of $\mathbb Q$)? –  Julien Puydt May 17 '12 at 10:20
    
Almost all means indeed all but finitely many (it neglects thus the non-periodic part). Thank you for your observation concerning $Q_A$. It is a field but probably not finitely generated. (I have corrected this mistake). –  Roland Bacher May 17 '12 at 10:28
    
In response to your final question: Are there examples of quadratic number-fields $\mathbb Q[\sqrt{N}]$ other than $N=5$ such that $\inf_{\alpha\in \mathbb Q[\sqrt{N}]\setminus\mathbb Q}\mu(\alpha)=1$? I don't know, but it is not clear to me that there are. –  Aaron Meyerowitz May 17 '12 at 20:52
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2 Answers

up vote 8 down vote accepted

In this paper of McMullen, he asks on p. 842 whether all real quadratic fields contain infinitely many continued fractions with coefficients bounded by 2? In this case, one would have $Q_d=Q_2=\mathbb{Q}(\sqrt{\mathbb{N}})$ for all $d\geq 2$. See also McMullen's slides from his talk at MSRI in February.

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Nothing is known for sure but it seems unlikely that there are any other inclusions. For $d \ge 2$ there are irrational non-quadratic numbers in $Q_d$ (of course they are algebraic). It seems likely that most, or probably all, such algebraic numbers have unbounded partial quotients which are distributed according to the Gauss–Kuzmin distribution: they are $1$ about $41 \%$ of the time and $k$ with probability $$p(k)=-\log_2 \left(1-\frac{1}{(1+k)^2} \right).$$ This is the case with probability $1$ for a uniformly distributed real number. However, there is no proof that there is even one algebraic number with unbounded partial quotients.

I have no idea about your last question. It might be more natural to ask about $\lim_{n\rightarrow\infty}\left(\prod_{j=1}^na_j\right)^{\frac{1}{n}}$.

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Concerning the first question: I do not care for numbers with unbounded coefficients. The question is essentially: Is there a smallest integer $A$ such that every real quadratic number field can is generated by a number having a continued fraction involving only coefficients $\leq A$. Replacing the arithmetic by the geometric sum in the last question changes (essentially) nothing since the question is for quadratic reals which have eventually periodic continued fraction expansions. –  Roland Bacher May 18 '12 at 8:21
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