Norm related to diophantine approximation? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T20:37:00Zhttp://mathoverflow.net/feeds/question/86492http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/86492/norm-related-to-diophantine-approximationNorm related to diophantine approximation?Darren Ong2012-01-23T22:36:49Z2012-01-23T23:38:25Z
<p>I'm trying to read this paper:
<a href="http://www.springerlink.com/content/g0046660260825x3/" rel="nofollow">http://www.springerlink.com/content/g0046660260825x3/</a></p>
<p>or <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.20.1813&rep=rep1&type=pdf" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.20.1813&rep=rep1&type=pdf</a></p>
<p>But I don't understand the norm they are using in (1.3). They write down the condition
$\vert\vert k\omega\vert\vert>C(\vert k\vert \log(1+\vert k\vert)^A)^{-1}$ </p>
<p>($\omega$ is an irrational number, and I think $k$ is an integer), and say that this expression is a "strong Diophantine condition" on $\omega$.</p>
<p>Another line in the paper which may be illuminating is </p>
<p>"for $\vert k\vert\leq q/2, \vert kw-ka/q\vert<1/2q$ and hence $\vert\vert k\omega\vert\vert >1/2q$." </p>
<p>Here $a/q$ is a lowest terms fraction with property $\vert \omega-a/q\vert<1/q^2$.</p>
<p>Context suggests to me that the norm should be a related to how easy it is to approximate an irrational number with rationals, but they don't have a definition in the paper, and I don't think I understand Diophantine approximation well enough to guess what it is.</p>