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

Let $S$ be an infinite set of positive integers. Let us say that a "best S-approximation" to a real irrational $r$ is a rational number $p/q$, with $p$ and $q$ integers and $q \in S$, such that for any integer $m\in S$ with $1 \le m<q$, and any integer $n$, it holds that $|p/q-r|<|n/m-r|$.

In the case that $S$ is the positive integers, it is well known that for any $r$ there are infinitely many best $S$-approximations $p/q$ which are less than $r$ and infinitely many that are greater than $r$. This is proved for example in Khintchin's book "Continued Fractions."

What if $S$ is the set of squares of integers? Is it known that for every irrational $r$ there are infinitely many best $S$-approximations less than $r$ and infinitely many greater than $r$?

Experimentation with numerous commonplace irrationals gives the impression that the ratio of the number of under-estimates to the number of over-estimates tends to 1. Could this be true for every irrational $r$? Is this known?

share|improve this question
add comment

2 Answers

I believe that a natural version of the problem is finding how good are rational approximations of the form $p/q^k$ to a given irrational number $\theta$. This classical problem goes back to H. Heilbronn [Quart. J. Math. (Oxford) 19 (1948) 249--256] and I. Danicic [Mathematika 5 (1958) 30--37] and the best known results are due to C. Hooley [in: Analytic number theory, Vol. 2 (Allerton Park, IL, 1995) 471–-486, Progr. Math. 139 Birkhäuser Boston, Boston, MA, 1996] and G. Harman [Glasgow Math. J. 38 (1996) 299–308]. They show that there are infinitely many coprime pairs $p,q$ with the property $$ \left|\theta-\frac p{q^k}\right|<\frac1{q^{k+\rho_k-\varepsilon}} $$ where $\rho_2=\frac25$ and $\rho_k=1/(3\cdot 2^{k-2}-1)$ if $k\geq 3$.

share|improve this answer
    
@Wadim: Thank you for the references. Maybe my question is going to have to wait on a better understanding of how good rational approximations of the form $p/q^k$ can be.... But the numerical experiments I performed show such regular alternations between under and over-estimates that I hoped for a simple explanation. A real number all of whose best approximations of the form $p/q^2$ were overestimates would strike me as a very strange beast! –  SJR Oct 12 '10 at 6:34
    
There is much more in the papers of Hooley and Hartman (for example, they discuss the question of choosing $q$ in the interval $1\le q\le N$). I believe that there is also a discussion (maybe, in Hooley's previous paper) on the expected $\rho_2$. –  Wadim Zudilin Oct 12 '10 at 6:39
1  
And a possible choice for your beast would be the Liouville example $\sum_{n=2}^\infty 2^{-n!}$... –  Wadim Zudilin Oct 12 '10 at 6:46
    
@Wadim: Why??!! –  SJR Oct 12 '10 at 16:14
1  
Aren't all best approximations of the form $\sum_{n=2}^N2^{-n!}=p/q^2$? –  Wadim Zudilin Oct 12 '10 at 20:47
add comment

First of all, I think when you say "any real" you mean "any real irrational." Second, you can find a real irrational such that all the even convergents, and none of the odd convergents, have square denominator. I think such an irrational would have infinitely many best $S$-approximations below but none above.

EDIT: Glyn Harman wrote a series of 4 papers, Metric Diophantine approximation with two restricted variables, which may have some relevance here.

share|improve this answer
1  
Why should best S-approximations be convergents? –  Gjergji Zaimi Sep 10 '10 at 13:23
    
Gerry, I think you are failing to distinguish between best approximations of the "first and second kinds", as described in Khintchine's book, page 24. As Gjergji mentioned, best approximations in the sense of my question needn't be convergents! As for restricting the question to irrationals, I'm not sure that's necessary, but it will avoid trivialities. I'll edit. –  SJR Sep 10 '10 at 13:35
    
Best approximations needn't be convergents, but convergents are best approximations. My construction gives convergents which have square denominators and are underestimates. I'm suggesting this leaves no room for any best approximations which have square denominators and are overestimates. I acknowledge that there is some hand-waving here and many details to work out. –  Gerry Myerson Sep 10 '10 at 23:25
    
More hand-waving; denominators of best approximations to a given real irrational are sparse. You wouldn't expect one to have a lot with square denominator - I bet almost all irrationals have only finitely many best approximations with square denominator. And if you construct one, as you can, with every other convergent having square denominator, then it's hugely unlikely to have many best approximations on the other side with square denominator. –  Gerry Myerson Sep 11 '10 at 11:46
    
@Gerry: Best $S$-approximations in the sense of the question are very different from best approximations with square denominators. I just don't see how best approximations can be used to control best $S$-approximations -- I wish you could tell me! Incidentally, experimentation with various common irrationals (square roots of 2,3,5 etc) suggests that best $S$-approximations are about evenly divided between "bigger" and "smaller." Maybe there is a theorem to this effect, that holds for all irrationals? That's my guess. –  SJR Sep 11 '10 at 19:39
show 4 more comments

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.