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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.

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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
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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).

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  • $\begingroup$ 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
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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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