Given an even positive integer $n$, I am interested in the behaviour of the finite sequence $C_n=\{\cos(k \pi/n)\},$ where $k=1, \ldots, n/2 -1$. In particular, I am looking for an approximation to the size $s_n$ of $$C_n \cap (a,b)$$ for an interval $(a,b) \subset (0,1)$ and $a,b \approx 1$. If, for instance, $|a-1|=|a-b|=O(1/n^2)$, can one prove something like $s_n=O(n)$ or $s_n=O(n^2)$?
Behaviour of sequence $\cos(k \pi/n)$ in the interval $(0,1)$
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calculus
analysis
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0The size of $C_n$ being $n/2-1$, the fact that $s_n=O(n)$ (and that $s_n=O(n^2)$...) seems straightforward (and in your example, much more holds, namely $s_n=O(1)$). – 2012-09-12
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0I'm definitely missing something because I don't see it... the sequence $C_n$ is not uniformly distributed on the interval $(0,1)$. Actually, to be really specific, I'm trying to show that $s_n$ is $O(n^2)$ for $a=1-k/2n^2$ and $b= \sqrt{\frac{2-k/2n^2}{2+k/2n^2}}$, where $k$ is some large constant. – 2012-09-13
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0I made a mistake before. For my specific case, $|a-b|=O(1/n^4)$ (I got this by a Taylor expansion around 1) – 2012-09-13
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0In your specific case, $b-a=O(1/n^2)$... See my answer. – 2012-09-13