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I have seen a [picture like this] several times:

troll proof

featuring a "troll proof" that $\pi=4$. Obviously the construction does not yield a circle, starting from a square, but how to rigorously and demonstratively prove it?

For reference, we start with a circle inscribed in a square with side length 1. A step consists of reflecting each corner of figure $F_i$ so that it lies precisely on the circle and yielding figure $F_{i+1}$. $F_0$ is the square with side length 1. After infinitely many steps we have a figure $F_\infty$. Prove that it isn't a circle.

Possible ways of thinking:

  1. Since the perimeter of figure $F_i$ indeed does not change during a step, it is invariant. Since it does not equal the perimeter of the circle, $\pi\neq4$, it cannot be a circle.

While it seems to work, I do not find this proof demonstrative enough - it does not show why $F_\infty$ which looks very much like a circle to us, is not one.

  1. Consider one corner of the square $F_0$. Let $t$ be a coordinate along the edge of this corner, $0 \leq t \leq 1$ and $t=0, t=1$ being the points of tangency for this corner of $F_0$ and the circle. By construction, all points $t \in A=\{ \frac{n}{2^m} | (n,m\in \mathbb{N}) \& (n<2^m)\}$ of $F_\infty$ lie on the circle. I think it can be shown that the rest of the points, $\bar{A}=[0;1] \backslash A$, lie in an $\varepsilon$-neighbourhood $U$ of the circle. I also think that in the limit $\varepsilon \to 0$, points $ t\in\bar{A}$ also lie on the circle. Am I wrong in thinking this? Can we get a contradiction from this line of thought?

Any other elucidating proofs and thoughts are also welcomed, of course.


marked as duplicate by Semiclassical, user147263, rogerl, ncmathsadist, Gyu Eun Lee Sep 17 '14 at 1:38

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    $\begingroup$ The crucial point here is that the arc length is $\int_S \sqrt{1+\left(\frac{dy}{dx}\right)^2}\;dx$, and depends not only on the curve $S$ itself but on its derivative $\frac{dy}{dx}$. Although the points of the zigzags do converge to a circle, the slopes of the zigzag segments don't converge to the slopes of the corresponding parts of the circle, and this is why the lengths don't match. $\endgroup$ – MJD Sep 15 '14 at 15:18
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    $\begingroup$ There is some earlier discussion of this question here. $\endgroup$ – MJD Sep 15 '14 at 15:27
  • $\begingroup$ @MJD Ah, now I see. Many thanks for the link, too. This kind of thing turned out to be difficult to search for. $\endgroup$ – Minethlos Sep 15 '14 at 15:32
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    $\begingroup$ math.stackexchange.com/questions/12906/is-value-of-pi-4 $\endgroup$ – Ben Sep 15 '14 at 15:38

You have rigorously defined $F_i$, but how do you define $F_\infty$? You cannot say: "after infinitely many steps...".

In this case you could define $F_\infty = \bigcap_i F_i$ (i.e. the intersection of all $F_i$), since $F_i$ is a decreasing sequence this is a good notion of limit. Notice however that $F_\infty$ is a circle! But this does not mean that the perimeter of $F_i$ should converge to the perimeter of $F_\infty$.

You could also choose a metric on subsets of the plane to define some sort of convergence $F_i \to F_\infty$ as $i\to \infty$. In any case, if you choose any good metric you find that either $F_\infty$ is the circle or that the sequence does not converge.

The point here is that the perimeter is not continuous with respect to the convergence of sets... so even if $F_i\to F_\infty$ (in any decent notion of convergence) you cannot say that $P(F_i)\to P(F_\infty)$ (where $P$ is the perimeter).

  • $\begingroup$ I see your point about $F_\infty$ not being rigorously defined. Could you please show why perimeter is not continuous w.r.t. convergence of sets, choosing the standard Euclidean metric, for example? $\endgroup$ – Minethlos Sep 15 '14 at 14:22
  • $\begingroup$ The example you found is exactly a proof that perimeter is not continuous. And when I speak of "metric" I mean a distance between sets, not points. There is no such a think like Euclidean metric on the family of sets. Look for example at en.wikipedia.org/wiki/Hausdorff_distance $\endgroup$ – Emanuele Paolini Sep 15 '14 at 15:26

Obviously the construction does not yield a circle

This is wrong: the limit of the construction is a bona fide circle.

Suppose we have a sequence of curves $\gamma_n$ ("curve" in the technical sense: in our example, each curve is made up of straight-line segments), which tend to a limiting curve $\gamma$. Suppose further that each $\gamma_n$ has the same length $l$. Then it is natural to assume that $\gamma$ has length $l$ too.

But this is false! For instance, all these 'triangle wave' curves have the same length:

   /\      /\      /\      /\      /\      /\   
  /  \    /  \    /  \    /  \    /  \    /  \  
 /    \  /    \  /    \  /    \  /    \  /    \ 
/      \/      \/      \/      \/      \/      \

  /\    /\    /\    /\    /\    /\    /\    /\  
 /  \  /  \  /  \  /  \  /  \  /  \  /  \  /  \ 
/    \/    \/    \/    \/    \/    \/    \/    \

 /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\ 
/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \/  \


But their limit looks like this:


which is less than half as long.

(For the experts: I have blurred the distinction between a curve $-$ in this context, a function from $[0,1]$ to $\mathbb R^2$ $-$ and its image. But I hope my meaning is clear.)


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