# Is it true that $\sin^2((2n+1) \pi x) \geq c$ occurs often (in precise sense)?

Fix $$x \in [0, 1/2]$$ and consider the function $$f(n, x) = \sin^2((2n +1) \pi x)$$.

Based on plotting, it seems like the following is true

For every $$x \in (0, 1/2)$$, there exists $$c \in (0, 1]$$ and a positive integer $$N$$ such that for every $$k \geq 0$$, there exists an integer $$m \in [kN, kN + N)$$ such that $$f(m, x) \geq c$$.

In other words, we can partition the integers $$\mathbb{N} = \cup_{k \geq 0} I_k$$ where $$I_k = [k N, kN + N) \cap \mathbb{N}$$ and on each interval there is some $$m \in I_k$$ for which the function $$f$$ is bounded away from $$0$$ uniformly.

I tried to prove or disprove this claim using the fact that $$\sin^2((2n+1) \pi x) = \sin^2(\pi \{(2n+1) x\})$$ where $$\{t\}$$ denotes the fractional part of $$t$$, but I failed to make much progress.

Is this claim true?

The claim is true. Indeed, consider the sequence $$(z_n)_{n\in\mathbb N}$$ of points of the unit circle $$\mathbb T=\{(u,v)\in\mathbb R^2:u^2+v^2=1\}$$ such that $$z_n=(\cos (2n+1)\pi x,\sin (2n+1)\pi x)$$ for each natural $$n$$. Then $$z_{n+1}$$ is the point $$z_{n}$$ rotated counterclockwise around the origin $$(0,0)$$ by the angle $$\alpha=2\pi x$$.

Suppose first that $$x$$ is rational. Then $$x=\frac{M}{N}$$ for some integers $$M$$ and $$N$$. Since $$x\in \left(0,\frac 12\right)$$, $$N\ge 3$$. It is easy to show that the points of sequence $$(z_n)_{n\in\mathbb N}$$ are exactly the vertices of the regular $$N$$-gon inscribed into the circle $$\mathbb T$$. Since $$N\ge 3$$, there exists a natural number $$n$$ such that $$f(n,x)>0$$. Put $$c=f(n,x)$$. Then $$f(n+mN,x)=f(n,x)=c$$ for each nonnegative integer $$m$$.

Suppose now that $$x$$ is irrational. Pick any $$c<1$$. Since the function $$\sin \pi t$$ is continuous and $$\sin\frac{\pi}2=1$$, there exists $$\varepsilon>0$$ such that $$\sin^2\pi t>c$$ for each real $$t$$ such that $$\left|t-\frac 12\right|\le \frac{\varepsilon}2$$. Pick any natural $$P\ge\frac 1{\varepsilon}$$. Among $$P+1$$ numbers $$\{\{2px\}:p=0,1,\dots, P\}$$ from the segment $$[0,1]$$ there exist two distinct numbers $$\{2px\}$$ and $$\{2p'x\}$$ such that $$0<\{2p'x\}-\{2px\}\le \frac{1}{P}$$. Put $$M=|p'-p|$$ and $$\delta=\{2p'x\}-\{2px\}$$. Pick any natural $$Q\ge\frac 1{\delta}$$. Then among $$Q+1$$ numbers $$\{\{(2(n+qM)+1)x\}:q=0,1,\dots, Q\}$$ there exists a number $$\{(2(n+qM)+1)x\}$$ such that $$\left|\{(2(n+q)+1)x\}-\frac 12\right|\le\frac \delta{2}\le \frac{\varepsilon}2$$, so $$f(n+qM,x)=\sin^2\pi (2(n+qM)+1)x=\sin^2\pi \{(2(n+qM)+1)x\}>c.$$

• Perhaps I didn't understand, but in the irrational case, in order to conclude your lower bound, don't you need to show that $t = \{(2(n+pM) + 1)x\}$ satisfies $|t-1/2|\leq \epsilon/2$? Also, I am a little confused, are your two choices of the natural $P$ the same? Apr 9 at 19:16
• @DrewBrady Thanks for the attention and sorry for the wrong text, I fixed it. Apr 12 at 7:51

## Context

Your question is very similar to this one: Sine function dense in $[-1,1]$, about the density of the sequence $$sin(n)$$.

The idea of the proof is to prove that $$\mathbb{Z}+\pi \mathbb{Z}$$ is dense in $$\mathbb{R}$$, which is a classic exercise in group theory.

This immediately shows that for values of $$\pi x$$ that are $$\mathbb{Q}$$-linearly independant of $$\pi$$, then, you have $$\overline{sin^2((2n+1)\pi x)} = [0, 1]$$.

## Solution to the problem

If $$x \in \mathbb{Q}$$, then write $$x = \frac{p}{q}$$ with $$p \in \mathbb{N}^\star$$ and $$q\in\mathbb{N}^\star$$. It follows that $$\forall n \in \mathbb{N}, f(nq, x) = \sin^2\left((2 q + 1) \pi \frac{p}{q}\right) = \sin^2\left(\pi\frac{p}{q}\right) = c > 0$$

If $$x \not\in \mathbb{Q}$$, then $$\pi x$$ is $$\mathbb{Q}$$-linearly independant of $$\pi$$, and we can apply the aforementionned result, so $$\overline{sin^2((2n+1)\pi x)} = [0, 1]$$.

For instance, choose $$c = \frac{1}{2}$$, the result proves that there exists a subsequence $$(\phi_n)_\mathbb{N}$$ s. t. $$sin^2((2\phi_n+1)\pi x) \to 1$$, giving that $$\forall \varepsilon > 0, \exists n_0 \in \mathbb{N}, \forall n > n_0, sin^2((2\phi_n+1)\pi x) > 1 - \varepsilon$$.

Choosing $$\varepsilon = \frac{1}{2}$$, this yields $$\forall n > n_0, sin^2((2\phi_n+1)\pi x) > c$$.

In conclusion, your claim is true.

• Of course, this does little to solve my problem. Apr 12 at 7:05