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You've probably seen this fallacious proof that $\pi = 4$:

enter image description here

The answers to this question provide a variety of explanations for why it's fallacious. But I'd like to try a different approach. Let the circle be the unit circle (so bigger than the circle shown in the picture) and $\gamma_1(t)$ be a parametrization of the initial square, let $\gamma_2(t)$ be a parametrization of the square with the corners removed, etc. My question is, what is the limit of $\gamma_n(t)$ as $n$ goes to infinity? Is it a standard parametrization of the circle like $(cos(t),(sin(t))$, or is it a weirder parametrization?

I'm wondering whether it's a weirder parametrization of the circle, because there's a lot of weird parametrizations, like ones that run around the circle twice, ones that's go back and forth, etc. So it's possible that in this case the circle just happens to be parametrized in some weird way that really does generate an arc length of $4$.

Keshav Srinivasan
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    @XanderHenderson Yes, it is the circle, but that's wasn't my question. My question is what parametrization of the circle is it? – Keshav Srinivasan Mar 31 '18 at 22:58
  • To be clear, your parametrizations $\gamma_1,\gamma_2,\dots$ are parametrizations by arc length? – Eric Wofsey Mar 31 '18 at 23:01
  • That doesn't seem to reflect what the OP is asking, @Eric. Arc length applies to the circle, not the removal of the corner squares, which don't change the circle. – amWhy Mar 31 '18 at 23:03
  • @EricWofsey Not necessarily, any parametrization of the square and other shapes is fine with me. – Keshav Srinivasan Mar 31 '18 at 23:04
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    Well then the question doesn't have a well-defined answer. If you can use whatever parametrizations you want, they could have all sorts of limits, or no limit at all. – Eric Wofsey Mar 31 '18 at 23:05
  • @XanderHenderson "No matter how you parameterize that curve, it will have length $2\pi$." That's not true in the slightest. What if you go around the circle twice? What's if you go back and forth on the circle? – Keshav Srinivasan Mar 31 '18 at 23:07
  • @EricWofsey I'm just trying to get a sense of what kind of parametrization the limit would yield, if you use a straightforward parametrization of the square and other shapes. – Keshav Srinivasan Mar 31 '18 at 23:13
  • @XanderHenderson Yes, the graphs of those arcs are still the same. To use your terminology, I'm trying to find out whether $(cos(t),sin(t))$ and the limit of $\gamma_n$ as $n$ goes to infinity parametrize the same arcs or parametrize different arcs that have the same graph, namely the unit circle. – Keshav Srinivasan Mar 31 '18 at 23:20
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    @XanderHenderson "In this case, I think that it is reasonably clear that the parameterization that you get at the end of the day is 1-1 (though it may be "faster" or "slower" in some places than the usual parameterization)." OK, that's what's not clear to me. If you can post an answer proving that, that would answer my question. – Keshav Srinivasan Mar 31 '18 at 23:22
  • @XanderHenderson All I would like to do is take the pointwise limit of $\gamma_n(t)$. – Keshav Srinivasan Mar 31 '18 at 23:27
  • @XanderHenderson Is there a reason you deleted all your comments? – Keshav Srinivasan Mar 31 '18 at 23:35

2 Answers2

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This question is not meaningful unless you fix some specific parametrizations $\gamma_n$. Different parametrizations give you different limits, or no limit at all.

It is certainly possible to choose parametrizations such that the limit will be an injective parametrization of the circle, though. For instance, if you let each $\gamma_n$ be the counterclockwise parametrization by arc length starting from $(1/2,0)$ (assuming our circle is centered at the origin), the functions $\gamma_n:[0,4)\to\mathbb{R}^2$ will converge uniformly to a continuous injection $\gamma:[0,4)\to\mathbb{R}^2$ whose image is the circle. Explicitly, for instance, for $t\in [0,1]$, $\gamma(t)=(\frac{1}{2}\cos\theta,\frac{1}{2}\sin\theta)$ where $\theta\in[0,\pi/2]$ is the unique angle such that $\frac{1}{2}(1-\cos\theta)+\frac{1}{2}\sin\theta=t$. This is just because $\gamma_n(t)$ is some point in the first quadrant whose $L^1$ distance from $(1/2,0)$ is $t$, and these points are approaching the circle, so $\gamma(t)$ is some point on the circle whose $L^1$ distance from $(1/2,0)$ is $t$. The point $(\frac{1}{2}\cos\theta,\frac{1}{2}\sin\theta)$ for $\theta$ as described above is the unique such point on the circle which is in the first quadrant, and so it is the only possible value of $\gamma(t)$.

In other words, $\gamma$ is the parametrization of the circle "by arc length" except that the arc length is computed in the $L^1$ metric instead of in the usual Euclidean metric. By choosing different parametrizations $\gamma_n$, you could get $\gamma$ to be the usual parametrization of the circle by Euclidean arc length (just reparametrize the $\gamma_n$ by the same change of variable needed to turn this parametrization $\gamma$ into the usual one).

The moral here, as the good answers to the original question (which is not all of the top answers!) mention, is that the length of a (parametrized) curve is not continuous with respect to pointwise or even uniform limits of parametrizations. In particular, a limit of curves parametrized by arc length need not be parametrized by arc length. This has absolutely nothing to do with the parametrizations failing to be injective, as you seem to be proposing.

Eric Wofsey
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  • "In particular, a limit of curves parametrized by arc length need not be parametrized by arc length." This is interesting. Can you give me the speed function for your parametrization? Is it a non-uniform speed? – Keshav Srinivasan Mar 31 '18 at 23:33
  • It is non-uniform. If I'm not mistaken, the speed while at a given point $(\frac{1}{2}\cos\theta,\frac{1}{2}\sin\theta)$ is $\frac{1}{|\sin\theta|+|\cos\theta|}$. – Eric Wofsey Mar 31 '18 at 23:38
  • (This comes from the equation $\frac{1}{2}(1-\cos\theta)+\frac{1}{2}\sin\theta=t$ for $t\in[0,1]$. This equation writes $t$ as a function of $\theta$, so $\theta$ as a function of $t$ is its inverse. The derivative of $\theta$ with respect to $t$ can thus be obtained by differentiating the left side with respect to $\theta$ and taking the reciprocal.) – Eric Wofsey Mar 31 '18 at 23:40
  • This should make sense intuitively: when you are away from the axes and the circle is going diagonally rather than vertically or horizontally, the staircase path is less efficient (it is going around the circle slower than just going diagonal would). So the speed of $\gamma$ should be slower away from the axes, and indeed $|\sin\theta|+|\cos\theta|$ is larger away from the axes. – Eric Wofsey Mar 31 '18 at 23:45
  • Interestingly, if you instead used staircase paths to approximate a diagonal straight line rather than a circle, then the limit would still have constant speed--just a slower constant speed than the staircase paths. – Eric Wofsey Mar 31 '18 at 23:46
  • "In other words, $\gamma$ is the parametrization of the circle "by arc length" except that the arc length is computed in the $L^1$ metric instead of in the usual Euclidean metric." This is another really interesting point in your answer. It's a point that even a lot of people who know why the proof is fallacious probably aren't aware of. – Keshav Srinivasan Apr 01 '18 at 00:04
  • "Interestingly, if you instead used staircase paths to approximate a diagonal straight line rather than a circle, then the limit would still have constant speed--just a slower constant speed than the staircase paths." Oh, do you know how much slower it is off the top of your head? – Keshav Srinivasan Apr 01 '18 at 00:06
  • Well, just compare their actual lengths! A staircase path from $(a,b)$ to $(c,d)$ has length $|c-a|+|d-b|$ (assuming it has no silly backtracking). So, if you take a limit of staircase paths with uniform speed $1$, you will get a diagonal path whose domain is an interval of that length. So its speed is the actual diagonal length $\sqrt{(c-a)^2+(d-b)^2}$ divided by $|c-a|+|d-b|$. – Eric Wofsey Apr 01 '18 at 00:09
  • Also, if it's not too much trouble could you prove the statement "for $t\in [0,1]$, $\gamma(t)=(\frac{1}{2}\cos\theta,\frac{1}{2}\sin\theta)$ where $\theta\in[0,\pi/2]$ is the unique angle such that $\frac{1}{2}(1-\cos\theta)+\frac{1}{2}\sin\theta=t$"? – Keshav Srinivasan Apr 01 '18 at 00:10
  • That would be rather messy to actually prove, since it would require actually precisely defining what the staircase paths used are. But this should intuitively be pretty obvious: after travelling for a time $t$ along a staircase path, you will have travelled to a point $(a,b)$ near the circle whose $L^1$ distance from $(1/2,0)$ is $t$. If the point were actually on the circle, then it would have to be the point I described as $\gamma(t)$. – Eric Wofsey Apr 01 '18 at 00:11
  • So, I guess, assuming you know that $\gamma_n(t)$ does converge to some point on the circle for all $t$, that proves it must converge to the point I described, since the $L^1$ distance function is continuous. – Eric Wofsey Apr 01 '18 at 00:15
  • OK, thanks for all your help! – Keshav Srinivasan Apr 01 '18 at 00:37
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There are two different limits here.

One is the limit of the arc-lengths of the polygons and that is $4$ because it is the limit of a constant sequence of $ 4$

The other one is the arc-length of the limits of the polygons and that is $\pi $ because the limit of polygons is the circle.

The arc-length is not a continuous measure, so limit of the arc-length is not necessarily the same as are-length of the limit.

Can we reparametrize the circle to have an arc-length of $4$, the answer in no , because the arc-length is independent of parametrization.

Mohammad Riazi-Kermani
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