I'm trying to find the accumulation points for the sequence $c_n = \lfloor \cos(\sqrt{n}) \rfloor, n \in \mathbb{N}_0$. I know what the points are, but I'm having trouble coming up with an explicit proof. Here's what I have so far:
Since $\cos(x) \in [-1,1] \; \forall x$ there are only three cases for $\lfloor \cos(\sqrt{n}) \rfloor$ and we can rewrite $c_n$ as:
$$ c_n = \left \lfloor{\cos(\sqrt{n})}\right \rfloor = \begin{cases}
1 & \cos(\sqrt{n}) = 1 \\
0 & 0 \leq \cos(\sqrt{n}) < 1 \\
-1 & -1 \leq \cos(\sqrt{n}) < 0
\end{cases} $$
Thus, the only candidates for accumulation points are $\{-1,0,1\}$, since that's the only values $c_n$ can assume.
Regarding 1, we have:
\begin{align*}
c_n &= 1\\
\Leftrightarrow \cos(\sqrt{n}) &= 1\\
\Leftrightarrow \sqrt{n} &= 2k \pi \text{ for et $k \in \mathbb{N}_0$} \\
\Leftrightarrow n &= 4k^2 \pi^2 \text{ for et $k \in \mathbb{N}_0$}
\end{align*}
We see that this equality only holds for $n=0$, since $\pi^2$ is irrational and $n \in \mathbb{N}_0$. So 1 is not an accumulation point, since the sequence only assumes the value 1 once.
I know both 0 and -1 are accumulation points, corresponding to $\sqrt{n} \in (0, \frac{\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi)$ and $\sqrt{n} \in (\frac{\pi}{2}, \frac{3\pi}{2})$ respectively. But I'm not sure how to prove that $\sqrt{n}$ "visits" both of these regions infinitely often.
This is my first question on here, so I'm sorry if there is any formatting weirdness. Thanks in advance for your help!
Best Answer
Both $0$ and $-1$ are accumulation points.
We have to show that $\cos \sqrt n \ge 0$ and $\cos \sqrt n < 0$ for infinitely many $n$.
$\cos \sqrt n \ge 0$ means $\sqrt n \in I_k := [-\frac \pi 2 + 2k\pi,\frac \pi 2 + 2k\pi]$ for some integer $k \ge 0$ and $\cos \sqrt n < 0$ means $\sqrt n \in J_k := (\frac \pi 2 + 2k\pi,\frac{3\pi}{2} + 2k\pi)$ for some integer $k \ge 0$.
Since $(\sqrt{n+1}^2 = n + 1 < n + 1 + 2\sqrt n = (\sqrt n + 1)^2$, we see that $\sqrt{n+1} < \sqrt n + 1$.
This implies that if $\sqrt n \in I_k$, then $\sqrt m \in J_k$ for some $m > n$. In fact, since $(\sqrt n)$ is unbounded and strictly increasing, there exists a unique $r \ge n$ such that $\sqrt r \le \frac \pi 2 + 2k\pi$ and $\sqrt{r+1} > \frac \pi 2 + 2k\pi$. Hence $$\frac \pi 2 + 2k\pi < \sqrt{r+1} < \sqrt r + 1 \le \frac \pi 2 + 2k\pi + 1 < \frac \pi 2 + 2k\pi + \pi = \frac{3\pi}{2} + 2k\pi .$$ Thus $\sqrt{r+1} \in J_k$. A similar argument shows that $\sqrt n \in J_k$, then $\sqrt m \in I_{k+1}$ for some $m > n$.
This proves that $\sqrt n \in \bigcup_{k=0}^\infty I_k$ for infinitely many $n$ and $\sqrt n \in \bigcup_{k=0}^\infty J_k$ for infinitely many $n$ .