Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On the unit circle, what does the set $\left \{ n \bmod{2\pi}:n \in \mathbb{N}\right \}$ represent? What is the subsequentieal limits of $\left \{ \sin(n) \right \}_{n\in \mathbb{N}}$?

I am probably really off. But I am thinking that every number will be just a number plus $2\pi$.

share|cite|improve this question
These are the points $(\cos n, \sin n)$, where $n$ ranges over the positive integers. A not very difficult argument shows these are dense in the unit circle. – André Nicolas Oct 1 '12 at 5:39
Okay I know this is a really stupid question. But would the set just be discrete points on the circle? And since we have mod (2pi), the value just plots over itself again and again. – Hawk Oct 1 '12 at 5:46
No, it is not a discrete set, because $\pi$ is irrational. – André Nicolas Oct 1 '12 at 6:34
up vote 2 down vote accepted

Let $P_n$ be the point on the unit circle with coordinates $(\cos n, \sin n)$. The points $P_n$ are all distinct. For if $P_a=P_b$, where $a\ne b$ then $a$ and $b$ differ by a multiple of $2\pi$. This means that $a-b=2k\pi$ for some integer $k$. It follows that $\pi=\frac{a-b}{2k}$, which is impossible, since $\pi$ is irrational.

Now let $\epsilon$ be a (very small) positive real, and let $N\gt \frac{2\pi}{\epsilon}$. Consider the $N$ points $P_1, P_2,\dots,P_N$. They are all distinct, so two of them, say $P_a$ and $P_{a+t}$, must be distance $\lt \epsilon$ from each other, along the unit circle. Let this distance be $\delta$.

Let $Q$ be any point on the unit circle. Consider the sequence of points $P_a,P_{a+t}, P_{a+2t},\dots$. These form a sequence of points that travel clockwise or counterclockwise around the circle, separated by $\delta$. So one of them will be within $\delta$ of $Q$.

We have shown that for any $\epsilon \gt 0$, there are infinitely many points in our sequence $P_1,P_2,P_3,\dots$ that are within distance $\epsilon$ of $Q$. So every point on the unit circle is a limit point of our sequence $(P_n)$.

As a consequence, the set of subsequential limits of the sequences $(\cos n)$ and $(\sin n)$ are each equal to the full interval $[-1,1]$.

Remark: From the wording of the question, it is not clear to me whether you are expected to prove that the sequence $(P_n)$ is dense in the unit circle. Perhaps you are just expected to guess that it is, and draw the appropriate conclusion about subsequential limits of $(\sin n)$.

share|cite|improve this answer
The original question was to plot the set on a computer. But I thought I wanted to understand what it was asking first before I attempted the problem. – Hawk Oct 2 '12 at 3:10
Plotting the points $P_n$ in even small black dots will soon make the entire circumference of the circle look black. Unless of course you use $\pi=22/7$, in which case you get a regular $7$-gon! We need a reasonably good approximation to $\pi$ to make things look like they really are. – André Nicolas Oct 2 '12 at 3:19
Also could you explain what you mean in , the set of subsequential limits of the sequences (cosn) and (sinn) are each equal to the full interval [−1,1]? – Hawk Oct 2 '12 at 3:56
Because the points $P_n$ are dense on the unit circle, their projections onto the $y$-axis (a fancy way of saying their $y$ coordinates) are dense in the interval $[-1,1]$. These projections are the numbers $\sin n$. Informally, we can get close to any $Q=(\cos\theta,\sin\theta)$ with a $(\cos n,\sin n)$. So for any choice of $\theta$, $\sin n$ can be made close to $\sin\theta$. But $\sin\theta$ can be chosen to be any number between $-1$ and $1$. – André Nicolas Oct 2 '12 at 4:07

The value doesn't, as you put it, plot over itself again and again, because the increments are $1$ whereas the modulus is $2\pi$. Since these two numbers are incommensurable, that is, their ratio is irrational, the same point on the circle is never reached twice: If it were, there would be an equation of the form $n_2-n_1=2\pi k$, contradicting the irrationality of $\pi$.

Moreover, it can be shown that this set is dense in the circle, that is, it gets arbitrarily close to every point of the circle.

The second question is ungrammatical since the verb and noun don't agree in number. The plural is correct, since a sequence can have more than one subsequential limit, so it should be "What are the subsequential limits of $\{\sin(n)\}_{n\in\mathbb N}$". Since these are the, say, $x$ coordinates of the points in the first set, which are dense in the circle, they are dense in $[-1,1]$, and thus the sequence has a subsequential limit at every point of $[-1,1]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.