Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question – Charles Feb 23 '12 at 21:29
up vote 3 down vote accepted

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.

share|cite|improve this answer
Thank you for you answer. – admchrch Mar 28 '11 at 8:06
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 8:47

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.

share|cite|improve this answer
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 8:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.