0

Let $(p,q)$ be any interval inside $[-1,1]$. I need to find a point $x \in (p,q)$ such that $x = \cos n $ for $n \in \mathbb{N}$

I mean if we look at the graph it is obvious that such an $x$ exists, but we require a rigorous proof here (we try to prove that $\cos n$ is dense on $[-1,1]$)

We know that there is an irrational $i$ inside $p,q$ so if we put $x = \cos([\arccos(i)])$, then $x \in (p,q)$. I was however marked this question as wrong. Is there another way to do it?

Bernard
  • 179,256
Theoneandonly
  • 683
  • Well $\arccos(i)$ needn't equal a natural number, right? – healynr May 08 '20 at 21:21
  • "I mean if we look at the graph it is obvious that such an $x$ exists" Why? If you can answer my question, you will be most of the way to writing down a rigorous proof. – Neal May 08 '20 at 21:25
  • If we travel along a circle one radian at a time, we will never cross the same point twice. – healynr May 08 '20 at 21:31
  • Here's a hint. https://mathoverflow.net/questions/127726/integer-multiples-of-a-irrational-dense-in-r-z – Neal May 08 '20 at 21:33

1 Answers1

1

This question is similar to that your asked recently. Using the notation from your question, we see that $\cos n=\cos 2\pi a''_n$, where $a''_n$ is $a_n$ with $\alpha=1/2\pi$. It is well-known that $\alpha$ is irrational number, so a set $A''=\{a''_n\}$ is dense in $[0,1]$ and thus a set $\{\cos a''_n\}$ is dense in $[-1,1]$ being an image of the set $A''$ by a continuous map $\cos$ onto $[-1,1]$.

Alex Ravsky
  • 106,166