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Can anyone provide a proof of the following inequality? If $n$ is a positive integer, $n\geq2$, then $$\cos(n) \leq 1 - 2^{-n}.$$ This is satisfied if $n$ is not within about $2^{-n/2}$ of a multiple of $2\pi$.

This inequality is sufficient for something else I am trying to prove but I and others have been unable to prove it.

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  • $\begingroup$ Why do you believe this to be true? $\endgroup$
    – Igor Rivin
    Feb 16, 2011 at 16:18
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    $\begingroup$ For the first ten thousand $n$ it's correct. ;-) $\endgroup$
    – Martin Brandenburg
    Feb 16, 2011 at 16:44
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    $\begingroup$ Since the windows around increasing multiples of $n$ get smaller and smaller -- and in particular since $\sum_{n \ge 0} 2^{-n/2}$ converges -- any exceptions are likely to be small. $\endgroup$
    – Michael Lugo
    Feb 17, 2011 at 4:52

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It is well known that $$|\tfrac{n}{k}-\pi|\ge \tfrac1{p(k)}$$ where $p$ is a polynomial of some very finite degree. From this your estimate follows for all sufficiently large $n$.

The polynomial $p$ can be written explicitly; all you need to find it in the literature and check first few values of $n$ by hands...

P.S. According to K.Mahler 1956, one can take $p(k)=k^{42}$; i.e.; it is sufficient to check all $n\le 1000$. So, if you trust Martin Brandenburg, your inequality holds for all $n$. ;-)

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    $\begingroup$ and of course Mahler's exponent 42 has been improved during the succeeding half-century. $\endgroup$
    – Gerald Edgar
    Feb 16, 2011 at 17:40
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    $\begingroup$ This is a very clean and transparent argument, but it relies on something not-quite-elementary. Is there a really simple, elementary argument showing that, say, $$ \left| \pi-\frac{n}{k} \right| > \frac1{2^k} $$ for all integer $k\ge k_0$? $\endgroup$
    – Seva
    Feb 16, 2011 at 19:14

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