I have a problem with proving the following question:
Prove that there exists $x_0\in (0,\pi)$ and $s_0>0$ such that $|\sin(kx_0)|\geq s_0 k^{-1}$ for all $k \in \mathbb{Z^+}$.
Here is my solution:
First, I showed that for all $\alpha \in (0,1)$ irrational number, there exists $c_0>0$ such that $d(\alpha, \frac{m-1}{m})\geq c_0m^{-2}$. I showed that letting $\alpha$ be any irrational number in $(0,1)$, and then fixing $c_0 \leq \alpha$.Then I used induction on $k$, and I got what I want.
Second, I fixed $x_0$ to be $x_0:=\alpha \pi$ such that $x_0 \in (0,\pi)$, and I conclude by the first step that for any $k \in \mathbb{Z^+}$, we have that $$d(kx_0,k\pi)=d(k(\alpha \pi),k\pi )) \geq c_0(k\pi)k^{-2}=c_0\pi k^{-1}.$$
Third step is to show what I want. I am getting stuck here, I tried induction on $k$ by letting $s_0 \in (0,1)$ such that $|\sin (x_0)|\geq s_0$ but I could not go along with that. Also, I tried proving by a contradiction but does not work with me, as well. So I really appreciate any hint or help with that or any other ways or ideas to prove that.
Thank you.
The range of the absolute value of the sin function is [0,1].
Note, $\forall k\in \mathbb{Z^+}$:
$$|\sin(kx_0)|\geq s_0 k^{-1}\rightarrow k\cdot |\sin(kx_0)|\geq s_0$$
So you are left trying to prove that some fraction of $k$ is bigger than some $s_0$ of your choosing.
Hints:
- pick $s_0 < 1$
- You can make $k\cdot |\sin(kx_0)| = k$ by choosing $x_0$ correctly. Specifically, you need $kx_0 = \pi n + \frac{\pi}{2}$