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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 $2\pi$.

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    No, it is not a discrete set, because $\pi$ is irrational.2012-10-01

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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 $(\sin n)$.

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    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 $1$.2012-10-02
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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 $[-1,1]$.