The word cover is rather vague. To be more precise, is it the case that, given any $r,\epsilon\in\mathbb{R} $ with $f(r)$ not equal to $f(r+\epsilon)$, that there exists some $N\in\mathbb{Z}$ such that $$\text{min}(\sin(r),\sin(r+\epsilon))\le \sin(N)\le \max(\sin(r),\sin(r+\epsilon))$$ and $$\min(\cos(r),\cos(r+\epsilon))\le \cos(N)\le \max(\cos(r),\cos(r+\epsilon))$$
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2Yes, the sequence ${e^{in}}_{n\geq 1}$ is dense in the unit circle. This is a classical consequence of the irrationality of $\pi$. Such a sequence is actually equidistributed: https://en.wikipedia.org/wiki/Equidistribution_theorem – Jack D'Aurizio Jun 10 '17 at 21:58
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I'd say that "to cover" would mean that any $;(r,s)\in S^1;$ can be expressed as $;(r,s)=\cos x,,\sin x;$ , and this follows at once from the basic properties of those two trigonometric functions. – DonAntonio Jun 10 '17 at 21:59
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@Jack D'Aurizio What background is required in order to understand what "dense" means? Topology and Analysis? The Wikipedia article on "dense set" references topological spaces. – Sid Jun 10 '17 at 22:11
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@Sid: we say that a set $A$ is dense in a set $B$ if for any $b\in B$ there are points of $A$ arbitrarily close to $b$. For instance, $\mathbb{Q}$ is dense in $\mathbb{R}$. – Jack D'Aurizio Jun 10 '17 at 22:13
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@Jack D'Aurizio Wow so if $(a,b)\subseteq \Bbb R$ and $Y$ is dense in $(a,b)$ then for any $\alpha, \beta$ such that $a<\alpha<\beta<b$ there exists some $y\in Y$ where $y \in (\alpha,\beta)$ . This notion perfectly captures the essence of my question. The link between the density of ${e^{in}}_{n\ge1}$ and the irrationality of $\pi$ intrigues me as well. It's not obvious to me how these are connected. – Sid Jun 10 '17 at 22:53
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@Sid: a starting point is here - https://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 – Jack D'Aurizio Jun 10 '17 at 22:59
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@DonAntonio If that is the definition of cover (and it would have be my first interpretation of the word as well) than the unit circle is obviously not covered as there are uncountably many points on the unit circle but only finitely many f(Z). But the definition for cover as in the OP is equally valid (and is compatible with the idea of an "open cover") and in which case it is exactly as Jack D'aurizio explains. – fleablood Jun 10 '17 at 23:07
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If pi were equal to n/m a rational then eventually you will have an integer k = 2n = 2m*pi and basically f(Z) will be the same set of 2n points repeated over and over again. rational numbers "sync up" eventually. Irrational numbers never do. One thing I like to think about is if to car signals start blinking at the same time at different rates. If the two rates have a rational ratio the will eventually synch up. If the two rates have an irrational ratio they will never blink at the same time ever again. They do it once, and never again. – fleablood Jun 10 '17 at 23:17
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Ok so I think I understand why the density of ${e^{in}}{n\ge 1}$ implies the irrationality of $\pi$. $\frac nm =\pi \rightarrow e^{2ni}=e^{2\pi mi}=1 \rightarrow e^{(2n+k)i}=e^{ki} \rightarrow|{e^{in}}{n\ge1}|\le2n$. Thus ${e^{in}}_{n\ge1}$ clearly can't be dense on the unit circle. But this is a contradiction since it known that this set is dense on the unit circle. I'm still trying to figure out the reverse implication via the link that @Jack D'Aurizio posted. – Sid Jun 11 '17 at 00:30
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The crucial point seems to be, if I'm understanding this correctly, that if some real number $r$ is not a rational multiple of $2\pi$ then the orbit of $r$ is dense on the unit circle. There is a link given to something called irrational rotations in dynamical systems. Can't say that I understand the wiki page at all, words like 'diffeomorphism' and 'topologically conjugate' are a bit over my head at this point. – Sid Jun 11 '17 at 00:30
1 Answers
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Ok, let us prove that $\{e^{in}\}_{n\geq 1}$ is dense in the unit circle.
Step1. We prove first that for any $\varepsilon>0$ there are infinite natural numbers such that $\left|e^{in}-1\right|\leq\varepsilon$. We have that $2\pi$ is an irrational number, hence we may extract from its continued fraction a sequence of rational approximations $\left\{\frac{p_m}{q_m}\right\}_{m\geq 1}$ with the property that
$$\left|2\pi-\frac{p_m}{q_m}\right|\leq \frac{1}{q_m^2} $$
implying that $\left|2\pi q_m-p_m\right|\leq \frac{1}{q_m}$.
This sequence gives that $1=e^{2\pi i q_m}$ and $e^{ip_m}$ are arbitrarily close, since the sequence $\{q_m\}_{m\geq 1}$ is unbounded and for any $\eta\in\left[0,\frac{1}{4}\right]$ we have $\left|1-e^{2\pi i \eta}\right|=2\sin(\pi\eta)\leq 2\pi\eta$.
Step2. The sequence $\{e^{in}\}_{n\geq 1}$ is dense in the unit circle. If we assume the opposite we have that there is some open interval $J\subset[0,2\pi)$ such that no element of $\{e^{in}\}_{n\geq 1}$ belongs to $J$. How large can such interval be? There are not many chances, indeed. We may fix some $\varepsilon>0$ and find, through the previous step, some $n_0$ such that the length of the arc joining $1$ with $n_0$ is $\leq\varepsilon$. If we consider $e^{in_0},e^{i2n_0},e^{i3n_0},\ldots$ until completing a full turn around the circle, such points break the circle into intervals having length $\leq \varepsilon$. In particular $|J|$ cannot exceed $\varepsilon$, but since $\varepsilon$ can be chosen as small as we like, there cannot be open intervals in $[0,2\pi)$ without any element of $\{e^{in}\}_{n\geq 1}$ in them.
Extra. The sequence $\{e^{in}\}_{n\geq 1}$ is dense in itself. Since $e^{in}\cdot e^{im}=e^{i(n+m)}$ this simply follows from Step1: in order to approximate $e^{in_0}$, it is enough to consider some $e^{iM}$ that is close to $1$. Then $e^{i(n_0+M)}$ is an arbitrarily good approximation of $e^{in_0}$.
José Carlos Santos
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Jack D'Aurizio
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2Very clever. The identity $|1-e^{2\pi i\eta}|=2sin(\pi \eta )$ took me like 15 minutes to prove lol. Oh, and the fact that ${e^{in}}_{n \ge 1}$ is dense in itself blew my mind. I should really use this site more often, I've learned a lot today. – Sid Jun 11 '17 at 02:25