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How to show that $\sin(2 \pi nx)$ does not converge as n goes to infinity? $x \in (0,1/2)$

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For irrational $x$, show that for any $N\gt 0$ there exist choices of $n$ that make $2\pi nx$ arbitrarily close to an integer multiple of $\pi$, and choices that make $2\pi nx$ arbitrarily close to an integer multiple of $\frac{\pi}{2}$. For rational $x$, I'll leave it to you. – Arturo Magidin Mar 3 '11 at 22:18
so here you use the Weyl's equidistribution theorem. is this the only possible solution to this problem? – user7762 Mar 3 '11 at 23:51
I'll wager there are more than one solution to the problem. But that seems pretty straightforward to me. – Arturo Magidin Mar 4 '11 at 1:14
You don't have to use equidistribution, it's enough to use simple arguments to show that you can get arbitrarily close to zero, and then shift them to show that you can get arbitrarily close to some other point. – Yuval Filmus Mar 4 '11 at 2:36
Exactly. Why use a big theorem? If the limit exists, then it is unique. Show that if the limit of $\sin (2\pi n x)$ exists ($x\in(0,1/2)$) then two subsequences accumulate around (at least) two different "limits". – Glen Wheeler Mar 4 '11 at 20:37
up vote 2 down vote accepted

Let $\alpha_n:=2\pi n x$. Define for a fixed $x\in \left(0,\frac 12\right)$ $$n_k:=\left\lfloor\frac{k+1/2}x\right\rfloor$$ which represents the greatest integer $\leq \frac{k+1/2}x$. We have \begin{aligned} \frac{k+1/2}x\leq n_k< \frac{k+1/2}x+1\\ k+1/2\leq n_kx<k+1/2+x\\ 2\pi k+\pi\leq \alpha_{n_k}<2k\pi+2\pi \end{aligned} hence $\sin(\alpha_{n_k})\leq 0$ and $$\tag{1}\liminf_n\sin(\alpha_n)\leq 0.$$

Now, define $m_k:=\lfloor\frac kx+\frac 18\rfloor$. We have \begin{aligned} \frac kx+\frac 18&\leq m_k\leq \frac kx+\frac 58\\ 2k\pi+\frac{x\pi}4&\leq 2m_k\pi\leq 2k\pi+\frac{5x\pi}4\leq2k\pi+\frac{5\pi}8. \end{aligned} We deduce that $\sin\left(\frac{\pi x}4\right)\leq \sin(\alpha_{m_k})\leq \sin\left(\frac{5\pi}8\right).$ We deduce that $$\tag{2}\limsup_n\sin(\alpha_n)\geq \sin\left(\frac{\pi x}4\right)>0.$$ With (1) and (2), we can see that $\{\sin(2\pi nx)\}$ has no limit for $0<x<1/2$.

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I think you're still missing an $x$ on the RHS? – TMM Jul 8 '12 at 21:57
I bounded it above by $\frac 12$. – Davide Giraudo Jul 8 '12 at 21:59

If $x$ is irrational, there is a stronger result: $\sin(2\pi x \mathbb{Z})$ is dense in $[0,1]$.

Indeed, $\sin(2\pi x \mathbb{Z})= \sin (2\pi x \mathbb{Z} + 2\pi \mathbb{Z})$ and $2\pi x \mathbb{Z}+ 2 \pi \mathbb{Z}$ is a subgroup of $(\mathbb{R},+)$ not of the forme $\alpha \mathbb{Z}$, so it is dense in $\mathbb{R}$. Hence $\sin(2\pi x \mathbb{Z})$ dense in $[0,1]$ since $\sin$ is continuous.

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$\def\eps{\varepsilon}$Since the statement to be proved is negative (the sequence cannot converge) a fairly elementary proof avoiding any heavy machinery is possible. In spite of the use of $\eps,\delta$ below, the proof involves no limits. In the end a slight computation is needed to arrive at an uncontestable contradiction, but I will try to mention the rather simple line of thought that leads to this argument.

Basically the intuition is that as long as $2\pi x$ stays away from the multiples of $\pi$, the sequnce $n\mapsto\sin(2\pi nx)$ oscillates between $-1$ and $+1$, and can therefore not ultimately stay within a very small neigbourhood of any particular value. To show this rigorously one needs to quantify the fact that $2\pi x$ stays away from multiples of $\pi$; therefore let $\eps=\min(\sin(2\pi x),1-|\cos(2\pi x)|)>0$ so as to ensure that both $\sin(2\pi x)\geq\eps$ and $|\cos(2\pi x)|\leq1-\eps$ (actually one always has $\eps=1-|\cos(2\pi x)|$).

Now imagine that $\lim_{n\to\infty}\sin(2\pi nx)$ exists, then $\sin(2\pi nx)_{n\in\mathbf N}$ would be a Cauchy sequence, which means in particular that for any $\delta>0$ there exists $n$ sufficiently large so that the values $\sin(2\pi(n-1)x),\sin(2\pi nx),\sin(2\pi(n+1)x)$ all lie within $\delta$ of each other. Intuitively oscillating behaviour is not compatible with taking almost the same value thrice in succession (although twice could happen by accident), so I shall deduce a contradiction from this, for which I need $\delta$ to be sufficiently small with respect to $\eps$, in a sense to be made precise. The basic formula used is $$ \sin(\alpha\pm y)=\sin\alpha\cos y\pm\cos\alpha\sin y, $$ with $\alpha=2\pi nx$ and $y=2\pi x$. The final terms have opposite signs for $\sin(2\pi(n-1)x)$ and $\sin(2\pi(n+1)x)$, so if those values are to be within $\delta$ of each other one must have $|\cos\alpha\sin y|<\delta/2$, and given that $\sin y\geq\eps$ this means $|\cos\alpha|<\delta/2\eps$. If we take $\delta$ sufficiently less than $\eps$ this means that $|\cos\alpha|$ must be very small, and $|\sin\alpha|$ therefore very close to $1$. But that means that multiplication by $\cos y<1-\eps$ substantially changes the first term $\sin\alpha\cos y$ with respect to $\sin\alpha=\sin(2\pi nx)$, and because we can have either sign for the second term, this pushes either $\sin(2\pi(n-1)x)$ or $\sin(2\pi(n+1)x)$ even further from $\sin(2\pi nx)$, giving a contradiction.

Concretely, take $\delta=0.8\eps$, then for $\sin(2\pi(n-1)x)$ and $\sin(2\pi(n+1)x)$ to be less than $\delta$ apart one needs $|\cos\alpha|<0.4$, and therefore $|\sin\alpha|>\sqrt{0.84}>0.9$; since we have $|\cos y|\leq1-\eps$ this means $|\sin\alpha\cos y-\sin\alpha|>0.9\eps>\delta$, which implies that either $|\sin(\alpha+y)-\sin\alpha|>\delta$ or $|\sin(\alpha-y)-\sin\alpha|>\delta$, contradicting the fact that we have a Cauchy sequence.

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