2
$\begingroup$

Does there exist a parametric smooth curve that visits all integer points $(x,y),\, x,y \in \mathbb{N}$ of the plane?

Something similar to this:

$$\begin{align} x = &\theta \cos(2\sin(\theta\pi))\\ y = &\theta \sin(\theta\pi)\end{align}$$

see the plot on Fooplot ... a sort of (inverted) smooth Cantor pairing function.

Or can it be proven that such a curve cannot exist?

I asked this question on math.stackexchange last week, but didn't get a satisfactory answer: the curve suggested in the answer is the inverse Cantor's pairing function, so it uses the floor and I would like to know if floor/ceil can be avoided.

Improve this question
$\endgroup$
6
  • $\begingroup$ I think it is clear that what you want is not what you are asking for (and hence the votes to close). @BjørnKjos-Hanssen answered your question as asked. If you changed smooth to analytic (as Bjørn Kjos-Hanssen suggested in a comment below), one could still possibly describe such an analytic function by giving something other than a formula. I think what you are looking for is a formula using only the type of continuous functions found in a calculus book (and no piece-wise functions). However, such a question is very difficult since that class of functions is fairly ad-hoc. $\endgroup$
    – Jason Rute
    Oct 30, 2015 at 16:25
  • 2
    $\begingroup$ Having said that, I think the analytic question is interesting. If you do ask that, ask it in a separate question and don't reject an answer just because it is not of the form you were hoping for. $\endgroup$
    – Jason Rute
    Oct 30, 2015 at 16:42
  • $\begingroup$ Perhaps a better question could be "is there a pair of real analytic functions such that $(x(t),y(t))$ visits all integer points in the plane?" Or even "is there a holmorphic function (in a neighborhood of the real axis) such that the image of the real axis contains all Gauss integers?" $\endgroup$
    – Bruno Le Floch
    Oct 30, 2015 at 22:30
  • 1
    $\begingroup$ @JasonRute: These new versions of the question are still trivial. An entire function can take any desired values on any discrete set. See for example my very first MO answer: mathoverflow.net/questions/161473/… $\endgroup$
    – Christian Remling
    Oct 30, 2015 at 23:26
  • $\begingroup$ @ChristianRemling, I stand corrected. $\endgroup$
    – Jason Rute
    Oct 30, 2015 at 23:31

1 Answer 1

Reset to default
4
$\begingroup$

Yes. Start with a compact smooth "bump" curve $$\mathbf r(t)=\langle x(t),y(t)\rangle,\qquad 0\le t\le 1$$ with $$x^{(n)}(0)=y^{(n)}(1)=0\qquad\text{for all }n,$$ $\mathbf r(0)=\langle 0,0\rangle$, and $\mathbf r(1)=\langle 0,1\rangle$.

Then by rotating and translating it, and using straight line segments as well, you can stitch together your smooth function as follows.

enter image description here (Bump curve shown in blue.)

Improve this answer
$\endgroup$
7
  • $\begingroup$ But is it definiable using a compact parametric representation ($x=f(\theta), y=g(\theta)$)? $\endgroup$
    – Marzio De Biasi
    Oct 30, 2015 at 7:32
  • 4
    $\begingroup$ @Marzio De Biasi: what is the definition of "compact parametric representation"? $\endgroup$
    – Alexandre Eremenko
    Oct 30, 2015 at 13:21
  • $\begingroup$ @AlexandreEremenko: sorry for the bad terminology; in this answer Bjorn uses $t$, $n$ and $r$ and it is unclear for me if they can be considered as ONE parameter... what I would like to know is if such a curve can be epxressed as $( f(\theta), g(\theta) ),\; \theta \geq 0$ in which $f,g$ are smooth (i.e. continuously differentiable) .... so floor/ceil are not allowed, and also using extra parameters "over closed intervals" is not allowed. Apparently it seems an easy question (4 close votes at the moment), but I'm not able to figure it out ... or I'm using the wrong terminology. $\endgroup$
    – Marzio De Biasi
    Oct 30, 2015 at 14:28
  • $\begingroup$ @Marzio De Biasi Yes there's only one parameter, $t $. Note that a function defined by cases can still be smooth. OTOH if you had asked for analytic functions it becomes harder. $\endgroup$
    – Bjørn Kjos-Hanssen
    Oct 30, 2015 at 14:42
  • 1
    $\begingroup$ Nevermind. I can't read. (I saw the $x$ and $y$ as a $\mathbf{r}$ for some reason. Now this makes sense.) $\endgroup$
    – Jason Rute
    Oct 30, 2015 at 17:55

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