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Can one find a number $m$ such that $\sin^2 n \geq m > 0 $ for all integers $n$?

By continuity of $\sin x$, it is enough to say that $|n - k \pi| \geq m^{\prime} > 0$ for all integers $n,k$.

Since $\pi$ can be approximated arbitrarily closely by rationals, for any $\epsilon > 0$, we can find integers $n$ and $k$ such that $|n−k\pi| < k \epsilon$. But this is not enough to disprove the existence of the lower bound $m$.

It seems like one needs to use more delicate Diophantine approximation here.

admchrch
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2 Answers2

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Since the set $\{\sin n:n\in\mathbb N\}$ is dense in $[-1,1]$, the set of the squares of its elements is also dense there. In particular, you can approximate zero as much as you want.

Community
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Mariano Suárez-Álvarez
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    A minor correction: the set ${ \sin^2 n,\ n\in \mathbb{N}}$ is not dense in $[-1,1]$ as ${ \sin n,\ n\in \mathbb{N}}$, but it is dense in $[0,1]$. – Pacciu Mar 28 '11 at 08:47
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The set $\{n-2\pi m; m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ so $\sin(n-2\pi m )=\sin(n)$ can get close to any number in $[-1,1].$ Therefore zero is the only lower bound for $\sin^2n$, which is not interesting obviously.

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    Thanks. I accepted the other answer because it came first and provided a link to a proof of the density assertion. – admchrch Mar 28 '11 at 08:05