Sine at Prime Values (Updated)

Several weeks ago, I wrote a response to a question a student asked me about the value of

$\displaystyle\limsup_{n\rightarrow\infty}\sin(n)$,

which turns out to be $1$. We proved this result by showing that the set $\left\{\sin(n):n\in\mathbb{N}\right\}$ is dense in $[-1,1]$.

Well, the natural numbers are certainly interesting, but the subset of prime numbers are even more interesting–not to mention the engine of much curiosity in mathematics. One user on Math.StackExchange asked if the set

$\displaystyle\left\{\sin p: p \text{ is prime}\right\}$

is dense in $[-1,1]$. The answer is yes. Proving this result will require some more sophisticated tools than before, and I will only provide a slightly more detailed outline that was suggested in the comments to the aforementioned post.

The first tool we will need is that of an equidistributed or uniformly distributed sequence of real number, which measures the proportion of elements of the sequence that a “randomly chosen” interval contains.

Def. We say that a bounded sequence of real numbers $(x_{n})_{n=1}^{\infty}$ is equidistributed on an interval $[a,b]$ if, for any subinterval $[c,d]\subset[a,b]$,

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{\left|\left\{x_{1},\cdots,x_{n}\right\}\cap[c,d]\right|}{n}=\dfrac{d-c}{b-a}$

(Here, the notation $\left|\cdot\right|$ denotes cardinality)

Remember that the sequence $(x_{n})_{n=1}^{\infty}$ is not a random variable, it is determinate.

Constructing equidistributed sequences is not so easy. A simple nonexample is the sequence formed by $x_{n}:=\frac{1}{n}$. Indeed,

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{\left|\left\{1,\frac{1}{2},\cdots,\frac{1}{n}\right\}\cap[\frac{1}{2},1]\right|}{n}=0$

A second, less intuitive nonexample is the sequence formed by the dyadic points:

$\displaystyle0,\frac{1}{2},1,\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\ldots$

To see that this second sequence $(u_{k})_{k=1}^{\infty}$ is not equidistributed, we make use of a theorem called Weyl’s criterion.

Theorem (Weyl) The following are equivalent:

1. The sequence $(x_{k})_{k=1}^{\infty}$ is equidistributed modulo $1$
2. For any integer $j\neq 0$, $\displaystyle\lim_{n\rightarrow\infty}\dfrac{1}{n}\sum_{k=1}^{n}e^{2\pi i \cdot j x_{k}}=0$.
3. For any Riemann integrable function $f:\mathbb{T}\rightarrow\mathbb{C}$, $\displaystyle\dfrac{1}{n}\sum_{k=1}^{n}f(x_{k})=\int_{0}^{1}f(t)dt$.

Applying Weyl’s criterion with $f(x):= x$, it suffices to show that

$\displaystyle\dfrac{1}{n}\sum_{k=1}^{n}x_{k}\not\longrightarrow\int_{0}^{1}xdx=\dfrac{1}{2}$

Consider the subsequence of the sums

$\begin{array}{lcl}\displaystyle\dfrac{1}{3\cdot2^{m-1}+1}\sum_{k=1}^{3\cdot2^{m-1}+1}x_{k}&=&\displaystyle\dfrac{1}{3\cdot2^{m-1}+1}\left[\dfrac{2^{m+1}(2^{m+1}+1)}{2^{m+2}}-2^{m-1}\dfrac{2^{m}+1}{2^{m+1}}-\dfrac{2\cdot2^{m-1}(2^{m-1}+1)}{2^{m+2}}\right]\\&=&\dfrac{1}{3+2^{1-m}}\left[2+\dfrac{1}{2^{m}}-\dfrac{1}{2}-\dfrac{1}{2^{m+1}}-\dfrac{1}{4}-\dfrac{1}{2^{m+1}}\right]\\&\longrightarrow&\displaystyle\dfrac{5}{12}\end{array}$

This result also gives us an example of a sequence which forms a dense subset modulo one but is not equidistributed modulo one.

The second tool we will need is a theorem in analytic number theory of I.M. Vinogradov. Let $P = \left\{p_{n}:n\in\mathbb{N}\right\}$ be the set of the primes enumerated by ascending order. In a 1935 paper, the reference for which I cannot find, Vinogradov proved that, for any irrational number $\alpha$, the sequence

$\displaystyle(p_{n}\alpha)_{n=1}^{\infty} = 2\alpha, 3\alpha, 5\alpha,7\alpha,11\alpha,\ldots$

is equidistributed modulo $1$, or, for an interval $[a,b]\subset[0,1]$,

$\displaystyle\lim_{n\rightarrow\infty}\dfrac{\left|\left\{p_{1}\alpha,\ldots,p_{n}\alpha\pmod{1}\right\}\cap[a,b]\right|}{n}=b-a$

Take $\alpha=\frac{1}{2\pi}$. Fix $x\in[-1,1]$ and $\varepsilon>0$. By the intermediate value theorem applied to the function $\sin()$, there exists $\theta \in [0,1]$ such that $x=\sin 2\pi\theta$. Since, for any $\gamma>0$ given,

$\displaystyle \lim_{n\rightarrow\infty}\dfrac{\left|\left\{p_{1}\alpha,\ldots,p_{n}\alpha \pmod{1}\right\}\cap[\theta-\gamma,\theta+\gamma]\right|}{n}=2\gamma>0$,

there exists a prime $p$ and an integer $k$ satisfying

$\displaystyle\left|\frac{p}{2\pi}+k-\theta\right|<\frac{\delta}{2\pi}\Longrightarrow\left|p+2\pi k-2\pi\theta\right|<\delta$,

for any $\delta>0$ given. The continuity of the sine function implies that

$\displaystyle\left|\sin(p+2\pi k)-\sin(2\pi\theta)\right|=\left|\sin(p)-\sin(2\pi\theta)\right|=\left|\sin(p)-x\right|<\varepsilon$