Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $I_{\lambda},$ $\lambda>0$ be a subset of all irrational numbers $\rho=[a_{1},a_{2},...,a_{n},...]\in(0,1)$ such that $a_{n}\leq \text{const}\cdot n^{\lambda}.$

Here, $[a_{1},a_{2},...,a_{n},...]$ is the continued fraction with partial quotients $a_1,a_2,\dots$.

My question is: under what values of $\lambda$ the set $I_{\lambda}$ has a positive Lebesgue measure?

share|improve this question
2  
Your notation seems to imply you want $I_\lambda$ to be countable, which would make your question trivial. Could you please clarify if this is really what you want? –  Yemon Choi Jul 16 at 22:41
3  
@Yemon Choi: It looks like the question is about the set of numbers whose simple continued fraction coefficients don't have early large values. –  Douglas Zare Jul 16 at 22:56
    
@DouglasZare Ah, that makes much more sense. Thanks. –  Yemon Choi Jul 16 at 22:59
    
I don't think @EmilJeřábek is right here. The fact that the geometric mean of the $a_i$'s is something doesn't tell you how big the largest one is. –  Anthony Quas Jul 17 at 1:28
3  
See my answer to mathoverflow.net/questions/161441/… which gives a reference to Khinchin's book, and a Theorem of Khinchin that discusses general such problems. –  Lucia Jul 17 at 5:04

2 Answers 2

up vote 10 down vote accepted

The condition $\lambda>1$ is sufficient and, at least almost, necessary:

To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to $\omega^\omega$ under the map that sends $\frac{1}{a_1+\frac{1}{a_2+\cdots}}$ to a function $f$ satisfying $f(n)=a_{n+1}-1$. (As is well known.)

This way Lebesgue measure on $(0,1)$ induces a "Gaussian" measure on $\omega^\omega$.

Under this measure $\mu$, I claim that the following set has measure one: $$\{f:f(n)\text{ is eventually bounded by $n^t$ for $t>1$, but not for $t=1$}\}$$

Proof: With $\delta_n=\sum_{k\le n^t}\epsilon_{nk}$ and $\Delta_n=\delta_n\log 2$ and $\log=\log_e$, we have

$$\log\prod_{n=1}^\infty \sum_{k\le n^t}\frac{\log\left(1+\frac{1}{k(k+2)}\right)}{\log 2}+\epsilon_{nk}=$$ $$\log\prod_{n=1}^\infty \frac{ \log 2+(\log (n^t+1)-\log(n^t+2))}{\log 2}+\delta_n =$$ $$\sum_{n=1}^\infty \log\left[ \log 2+(\log (n^t+1)-\log(n^t+2)) +\Delta_n \right]-\log\log 2$$

Here $\epsilon_{nk}=\pm\frac{A}{k(k+1)}e^{-\lambda\sqrt{n-1}}$ for certain constants $A,\lambda$.

By comparison of the negative of this series with $\sum \frac{1}{n^t}$, where we get a constant limit, the series converges iff $t>1$:

$$\lim_{n\to\infty}\frac{\log\log 2-\log(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))}{n^{-t}}=$$ $$\lim_{n\to\infty} \frac{(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))^{-1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]} {-tn^{-t-1}}=$$

$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}{t(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))}=$$

$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{n^{t-1}}{n^t+1}-\frac{n^{t-1}}{n^t+2}+\Delta'(n)/t]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$

$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{n^t+1}-\frac{1}{n^t+2}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$

$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{(n^t+1)(n^t+2)}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$ $$\frac{1}{\log 2}$$

provided $n^{t+1}\Delta'(n)\to 0$.

And indeed

$$n^{t+1}\Delta'(n)=(\pm A)(\log 2)n^{t+1}\frac{d}{dn}[e^{-\lambda\sqrt{n-1}}\sum_{k\le n^t}\frac{1}{k(k+1)}]$$

Ignoring the constant and using $f(n)=\sum_{k\le n^t}g(k)\approx \int_1^{n^t}g(k)dk$, so $f'(n)\approx g(n^t)-g(1)$, we have

$$ n^{t+1} \left[ e^{-\lambda\sqrt{n-1}}[-\lambda\frac{1}{2\sqrt{n-1}}]f(n)+e^{-\lambda\sqrt{n-1}}f'(n) \right]= $$

$$ e^{-\lambda\sqrt{n-1}} n^{t+1} \left[ \left(-\lambda\frac{1}{2\sqrt{n-1}}\right)f(n)+f'(n) \right] $$

and this clearly goes to 0 as $n\to\infty$.

So $\log\prod=-\infty$ and hence $\prod=0$. This completes the proof.

share|improve this answer
    
Thank you very much. Could you suggest me books regarding these facts. –  sokho Jul 16 at 23:07
    
Actually my coauthor Bonnie S. Huggins and I just proved it from scratch. But there should be some books about continued fractions and the Gaussian measure on the irrational numbers. –  Bjørn Kjos-Hanssen Jul 16 at 23:12
2  
Thank you. Say my regards to Bonnies S. Huggins also. –  sokho Jul 16 at 23:16

A precise solution to this problem is known. In Khintchine's book on continued fractions it says:

$\mathbf{Theorem~30}$ Suppose that $\varphi(n)$ is an arbitrary positive function with natural argument $n$. The inequality $$ a_n = a_n(\alpha) \geq \varphi(n) $$ is, for almost all $\alpha$, satisfied by an infinite number of values $n$ if the series $\sum_n 1/\varphi(n)$ diverges. On the other hand, this inequality is, for almost all $\alpha$, satisfied by only a finite number of values of $n$ if the series $\sum_n 1 /\varphi(n)$ converges.

(quoting from: A. Ya. Khintchine. Continued Fractions. University of Chicago Press, 1964)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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