Timeline for A Simple Generalization of the Littlewood Conjecture
Current License: CC BY-SA 2.5
6 events
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| Oct 11, 2010 at 15:11 | comment | added | Sidney Raffer | @gowers: Do you have a reference for the "known fact" of your last comment? I can prove it if the ratio $d_{n+1}/d_n$ is at least $2+c$, but "$1+c$" is giving me trouble. | |
| Apr 22, 2010 at 17:26 | comment | added | Sidney Raffer |
But that settles it! As I think Roland was suggesting, if $r=(1+5^{1/2})/2$, then the positive values of $x$ that appear in solutions to the inequality $|x(rx-y)| < 1$ are exactly the Fibonacci numbers, which grow exponentially in the sense you describe. Then the $s$ of your known fact gives a counter-example to my generalization of Littlewood's conjecture. Maybe for ANY irrational $r$ the $x$'s will grow exponentially, but I'm not sure about this. I think I see how to attack the proof of your "known fact". Is there a name attached to it? Thanks.
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| Apr 22, 2010 at 16:01 | comment | added | gowers | I can't quite follow what your argument is, but I would point out that it is a known fact that if a sequence $d_1,d_2,...$ grows geometrically (in the strong sense that there is a positive constant c such that all ratios are at least 1+c) then there must exist a real number s and a positive epsilon such that for every n the distance from $sd_n$ to the nearest integer is at least epsilon. | |
| Apr 22, 2010 at 15:14 | comment | added | Roland Bacher | You are of course right. My attempts to fix the resulting complications were unsuccessful. Perhaps it is more promising to consider for $r$ the golden number for which my claim (without error of my part) is indeed correct. Unfortunately it is to small to give useful error-terms and it is difficult to push through the end of the argument. | |
| Apr 22, 2010 at 6:08 | comment | added | Sidney Raffer |
Roland, It is not quite true that the convergents of the continued fraction expansion of $r$ coincide with the solutions of $x(rx-y)| < 1$ in the way you describe. Non-convergents can satisfy such an inequality and moreover there can be solutions $(x,y)$ with $x$ and $y$ not relatively prime. See Rockett and Szusz, Continued Fractions, Section II.5. Does this fatally break the estimate given in the last two sentences of your argument? I'm not sure.... but it needs another look . (Note: This is a corrected version of my initial comment)
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| Apr 21, 2010 at 17:55 | history | answered | Roland Bacher | CC BY-SA 2.5 |