Curve dense inside the unit circle

Question

For $\alpha$ a real, irrational number, I have been to prove that any point $(x,y)$ such that $x^2 + y^2 \leq 2$ can be written as $$(x,y) = (\cos( u) + \cos( \alpha u), \ \sin( u) + \sin(\alpha u)) $$ for some $u \in \mathrm{R}$.

It is easy to show that this is not true if $\alpha$ is rational. In that case, the vector function is periodic, and the curve it traces is closed. It is also easy to see that the function is not periodic if $\alpha$ is rational.

What is less trivial to argue, however, is that the curve fills the full circle.

Answer 1

For any $z\in \mathbb C,$ let $C(z)$ denote the circle of radius $1$ centered at $z.$ Note that for every $z$ in the closed unit disc, there exists $t\in [0,2\pi)$ such that $z\in C(e^{it}).$

Let $a$ be irrational. Well known result: $\{e^{ina\pi}:n\in \mathbb Z\}$ is dense in the unit circle. It follows that if $t\in [0,2\pi),$ then

$$\tag 1 e^{i(t+2n\pi)} + e^{ia(t+2n\pi)} = e^{it} + e^{iat}\cdot e^{ia2n\pi},\, n \in \mathbb Z$$

is dense in $C(e^{it}).$

So given $z$ in the closed unit disc, we have $z$ in some $C(e^{it}),$ and then $(1)$ shows $z$ is the limit of points of the form $e^{i(t+2n\pi)} + e^{ia(t+2n\pi)}.$ This is the desired conclusion.

Answer 2

Given some $(x,y)\in\mathbb{R}^2$ such that $x^2+y^2\leq 1$, let $\rho=\sqrt{x^2+y^2}$. Assuming $$ (x,y)=\left(\cos(u)+\cos(\alpha u),\sin(u)+\sin(\alpha u)\right) \tag{1}$$ and $\alpha>1$ we have $$ \rho^2 = 2+ 2\cos(u)\cos(\alpha u)+2\sin(u)\sin(\alpha u) = 2+2\cos((\alpha-1)u)=4\cos^2\left(\frac{\alpha-1}{2}u\right) \tag{2}$$ hence $u$ is constrained to belong to a discrete subset of $\mathbb{R}$, which depends on $\rho$:

$$ u\in \pm\frac{1}{\alpha-1}\arccos\left(\frac{\rho^2-2}{2}\right)+\frac{2\pi}{\alpha-1}\mathbb{Z}=E_{\rho}.\tag{3} $$ If $\alpha\not\in\mathbb{Q}$ then neither $\cos(u)+\cos(\alpha u)$ or $\sin(u)+\sin(\alpha u)$ are periodic functions, but they are smooth functions with range $[-2,2]$ and they are given by products of smooth periodic functions. In particular $(1)$ implies $$ (x,y) = \left( 2\cos\left(\frac{\alpha-1}{2}u\right)\cos\left(\frac{\alpha+1}{2}u\right),2\cos\left(\frac{\alpha-1}{2}u\right)\sin\left(\frac{\alpha+1}{2}u\right)\right)\tag{4}$$ where for any $u\in E_\rho$ we have $2\cos\left(\frac{\alpha-1}{2}u\right)\in\{-\rho,\rho\}$.
The range of the RHS $(4)$ over $E_\rho$ equals the range of $$ \rho\cdot\left( \cos\left(\frac{\alpha+1}{2(\alpha-1)}(\theta_\rho+2k\pi)\right),\sin\left(\frac{\alpha+1}{2(\alpha-1)}(\theta_\rho+2k\pi)\right)\right).\tag{5}$$ Now we may recall a standard result: if $\beta\not\in\mathbb{Q}$, the range of $e^{2\pi i \beta k}$ for $k\in\mathbb{Z}$ is dense in $S^1$. Density is not affected by the multiplication by the constant $e^{2\pi i\theta_\rho}$. In particular it is not granted that $(1)$ holds as an equality for some $u\in\mathbb{R}$, but for sure $(1)$ holds as an approximate equality (with an arbitrary small error) for infinite $u\in\mathbb{R}$.

Answer 3

Let us identify $\mathbb{R}^2$ with $\mathbb{C}$ and write for the mentioned function $$g(u)= e^{iu}+e^{i\alpha u}.$$ First note that any point $r e^{i \gamma}$ with $0 \leq r \leq 1$ and $\gamma \in [0,2\pi]$ can be written as $$\tag{1}r e^{i \gamma} = e^{ia} + e^{ib}.$$ Indeed, we have so solve $$r^2 = |re^{i\gamma}|^2 = |e^{i(a-b)}+1|^2 = 4 \cos((a-b)/2)^2$$ Thus $$\tag{2}a-b = 2 \arccos(r/2)$$ Now, we have $$r e^{i(\gamma-b)} = 1 + e^{i(a-b))} = \frac{1}{2} r (r+i \sqrt{4-r^2}).$$ After dividing by $r$ we may check that the absolute value of the resulting term is one. So we may choose $b$ suitable and then corresponding $a$ satisfying (2).

If $\alpha$ is irrational, we can choose $u_1 = 2\pi k$ with $k \in \mathbb{N}$ such that $$\tag{3}|e^{i\alpha u} - e^{ia}| < \varepsilon.$$ Now if we take $u_2 = u_1 + 2 \pi\alpha^{-1} n$ with $n \in \mathbb{Z}$ we see that (3) doesn't change!. On the other hand, we have $e^{iu_2} = e^{2\pi \alpha^{-1} n}$. Again, because $\alpha^{-1}$ is irrational, we can take $n \in \mathbb{N}$ with $|e^{2 \pi \alpha^{-1} n} - e^{ib}| < \varepsilon$. This shows that $|g(u_2)-r e^{i \gamma}| \leq 2 \varepsilon$.

Answer 4

Identify the plane with complex plane. The curve at hand becomes $$\mathbb{R} \ni u \quad\mapsto\quad \gamma(u) = e^{iu} + e^{i\alpha u} = e^{iu}(1 + e^{i(\alpha-1)u}) \in \mathbb{C}$$ For simplicity of discussion, we will only consider the case $\beta = \alpha - 1 > 0$.

For any $\eta \in \mathbb{C}$ with $|\eta| \le 2$, pick an $u_0 \in [0, \frac{\pi}{\beta}]$ such that $|\eta| = |\gamma(u_0)| = 2 \left|\cos\frac{\beta u_0}{2}\right|$.

For any $n \in \mathbb{Z}$, let $u_n = u_0 + \frac{2\pi}{\beta} n$. It is easy to see $$\gamma(u_n) = \gamma(u_0) e^{i\frac{2\pi}{\beta}n}$$

Since $\alpha$ is irrational, so does $\frac{1}{\beta}$. The fractional parts of $\frac{n}{\beta}$ will be dense in $[0,1)$.
As a result, the set $\{ \gamma(u_n) : n \in \mathbb{Z} \}$ will be dense in the circle $|z| = |\gamma(u_0)| = |\eta|$.

Since $|\eta|$ can be any number between $0$ and $2$. The image of the curve is dense in the closed disk $|\eta| \le 2$.