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The following parametric equation recently across my mind recently, $$x = \cos (t)$$ $$y = \sin (\sqrt{2} t)$$ And graphing on my calculator shows it eventually traces out a "square". What I am wondering is if this function actually traces out all the points within this square for $t \in \mathbb{R}$, or some region of it as the corners will never be reached.

yerpderpington
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    Desmos confirms, but I think that it's just the closure which would form the square. – Kenny Lau Sep 12 '17 at 16:40
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    The irrationality of $\sqrt{2}$ is the reason that the iteration is not periodic. The question whether every point is reached is interesting (+1) – Peter Sep 12 '17 at 16:43
  • Ok, i actually got a set of points which can be never included, check my argument – neonpokharkar Sep 12 '17 at 18:13
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    Since the curve is smooth, the image of $\mathbb{R}$ under this curve is a measure-zero subset of $\mathbb{R}^2$. So the curve will miss almost every point of the square $[-1, 1]^2$. Also, this type of curve is called Lissajous curve. – Sangchul Lee Sep 12 '17 at 18:20

2 Answers2

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It only fills the square "densely", but not completely.

Just take any $x$ coordinate you want to achieve. Then calulate one of corresponding values of $t$ for which $\cos$ achieves this value (namely $\arccos(x)$)

Then for any $k$, $t+2\pi k$ will have the same $x$ coordinate while $\sqrt{2}\times(t+2\pi k) \pmod {2\pi}$ will densely fill the $[0,2\pi]$ interval due to irrationality, so the closure is indeed the unit square.

On the other hand the point $\{1,1\}$ is never reached. Indeed for that point to be attained, the value of $t$ should be $2l\pi$ for $x$ to be correct, and $\frac{\pi+2k\pi}{\sqrt{2}}$ for the $y$ coordinate to be correct, so you need $k,l\in\mathbb{N}$ such that $$\frac{\pi+2k\pi}{\sqrt{2}}=2l\pi$$ which can be reformulated to $$\frac{\pi+2k\pi}{2l\pi}=\frac{1+2k}{2l}=\sqrt{2},$$ which would contradict the irrationality of $\sqrt{2}$.

b00n heT
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  • Ok, i actually got a set of points(which your argument is discussion case of) which can be never included, check my argument – neonpokharkar Sep 12 '17 at 18:13
  • Just curious, are there any points in the open square ${(x,y):x,y\in(-1,1)}$ that this function will never reach? – LegionMammal978 Sep 12 '17 at 21:40
  • As Sengchun Lee commented: yes there are infinitely many. To compute a concrete example simply try ${1/2,1/2}$, where the argument to prove it is exactly as the one in my answer – b00n heT Sep 13 '17 at 05:50
  • Shouldn't the graph be no-where dense by Baire category theorem? – user113988 Jun 21 '18 at 12:12
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For a fixed $x \in [-1,1]$, there are countably many $t$ that satisfy the parametric equations. Let $T$ denote such values of $t$. $y$ has period $\sqrt2\pi$.

  1. Convince yourself that $T \pmod{\sqrt2\pi}$ is dense in $[0,\sqrt2\pi]$. (1, 2, 3)
  2. Then, convince yourself that $\sin \sqrt 2T$ is dense in $[-1,1]$. (limit commutes with continuous function)
  3. However, convince yourself that $\sin \sqrt 2T$ only has countably many numbers, while any particular vertical strip of the square has uncountably many numbers, so not all points are reached. (see comment)
  4. Conclude that since $x$ is arbitrary, the set of points $(x,y)$ in the graph is dense in $[-1,1]^2$, but does not contain every point in $[-1,1]^2$.
Kenny Lau
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