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I am trying to understand a particular version of the "Peano curve." Although definitions will vary from person to person, for the purposes of this question I am talking about the one obtained from the process in the picture below.

enter image description here

More specificically, I am concerned with a curve which maps the unit interval surjectively onto the unit square $[0,1] \times [0,1]$.

Now, I find the above figure a little confusing because there are no axes or points labeled, thus it is hard to tell what points the curves touch.

For the first curve pictured, I would describe it by taking a straight path from $(0,0)$ to $(0,1)$, then from $(0,1)$ to $(1/2,1)$, then from $(1/2,1)$ to $(1/2,0)$, then from $(1/2,0)$ to $(1,0)$, and finally from $(1,0)$ to $(1,1).$

I am having trouble describing the second curve, however. I would start by taking a straight paths from $(0,0) \to (0,1/3) \to(1/6,1/3) \to(1/6,0) \to(1/3,0) \to (1/3,1/3)$. So far so good? Now, I think I'm supposed to continue by going $(1/3,1/3) \to (1/3,2/3) \to(1/6,2/3) \to (1/6,1/3) \to (0,1/3) \to (0,2/3)$. But wait a second! This curve overlaps! This is different from the picture!

So, if the picture is wrong, how does one correct it? I'm also unsure where I'm supposed to go once I reach $(1/3,1)$. Do I move back down (even though I just came that way) or do I move to the right?

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Yeah, your picture is wrong. If you draw curve $n$ (the first being number $1$) on a piece of engineering paper such that it fit on the smallest scale possible, the first graph would be $2\times2$, the second $8\times8$ and the third $26\times26$ and so on. You need to fit $3$ graphs of curve $n$ plus $2$ units of spacing so that if $a_n$ is the width and height of graph $n$ then $$a_{n+1}=3a_n+2$$ The solution to this difference equation is $a_n=3^n-1$ so graph $n$ must be shrunk by a factor of $$\frac{a_n}{a_{n+1}}=\frac{3^n-1}{3^{n+1}-1}$$ before inserting it or its mirror image into graph $n+1$.

Thus the first graph should have been shrunk by a factor of $2/8=1/4$ so that its corners would be $(0,0),(0,1/4),(1/8,1/4),(1/8,0),(1/4,0),(1/4,1/4)$. Then leave a spacing of $1/8$ and draw the mirror image above. Repeat until you get to the top, the leave a spacing of $1/8$ of the right and draw the mirror image of the last subgraph on the right and so on until all $9$ subgraphs have been draw. Also draw those little connecting segments.

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  • $\begingroup$ Thank you! I think I now understand how to construct each iteration of the Peano curve. $\endgroup$ – Pascal's Wager Jan 27 at 11:47
  • $\begingroup$ I know this is a bit beyond the scope of the question, but would you mind explaining how one goes about seeing uniform convergence? Any hints? $\endgroup$ – Pascal's Wager Jan 27 at 12:21
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    $\begingroup$ I'm afraid I don't really get the uniform convergence business myself so I wouldn't be the right person to try to explain it. $\endgroup$ – user5713492 Feb 5 at 1:44

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