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?


2 Answers 2


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 $$f(n+qM,x)=\sin^2\pi (2(n+qM)+1)x=\sin^2\pi \{(2(n+qM)+1)x\}>c.$$

  • 1
    $\begingroup$ 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? $\endgroup$
    – Drew Brady
    Apr 9 at 19:16
  • $\begingroup$ @DrewBrady Thanks for the attention and sorry for the wrong text, I fixed it. $\endgroup$ Apr 12 at 7:51


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.

  • $\begingroup$ Of course, this does little to solve my problem. $\endgroup$
    – Drew Brady
    Apr 12 at 7:05
  • $\begingroup$ I've added details about the specifics of your problem $\endgroup$
    – Jean
    Apr 12 at 11:51
  • $\begingroup$ Your proof in the irrational case doesn’t help. It just says there’s a subsequence which is bounded from zero. It doesn’t say anything about how regular that subsequence is. $\endgroup$
    – Drew Brady
    Apr 12 at 15:34
  • $\begingroup$ Completely missed the part about the regularity, my bad $\endgroup$
    – Jean
    Apr 12 at 15:46

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