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S May 18 at 8:14 history suggested The Amplitwist CC BY-SA 4.0
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S May 18 at 8:14
Apr 13, 2017 at 12:58 history edited CommunityBot
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Oct 14, 2016 at 18:58 comment added Charles @GeorgeLowther Is there a standard reference for the case when we know the irrationality measure of θ?
Oct 14, 2010 at 2:31 comment added George Lowther Well, my guess of $S_n=o(n^x)$ for every x > 1/2 and almost every $\theta$ was correct. We can't do much better than this, as Fedor's answer shows that $\limsup_{n\to\infty}|S_n|/n^x=\infty$ for every x < 1/2 and almost every $\theta$.
Oct 14, 2010 at 1:42 history edited George Lowther CC BY-SA 2.5
answered follow-up question
Oct 13, 2010 at 20:52 comment added George Lowther ...although, trying a couple of plots with randomly chosen $\theta$, $S_n$ appears to be bounded, but is probably growing very slowly.
Oct 13, 2010 at 20:20 comment added George Lowther I expect that $S_n=o(n^x)$ for every x > 1/2 and almost every $\theta$, by thinking of $S_n$ as similar to a random walk. I'll have to think about that a bit more though.
Oct 13, 2010 at 18:40 vote accept Portland
Oct 13, 2010 at 18:39 comment added Portland George, this is very nice. "There will be an uncountable dense set of irrational $\theta$ for which we can rule out bounds such as $S_n = O(n^x)$ with $x<1$." But is there any $\theta \in \mathbb{R}\setminus \mathbb{Q}$ s.t. there exists $x<1$ and $S_n = O(n^x)$, or can we rule that out as well?
Oct 13, 2010 at 13:10 history edited George Lowther CC BY-SA 2.5
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Oct 13, 2010 at 13:04 history answered George Lowther CC BY-SA 2.5