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Is value of $\pi = 4$?
Can anyone explain how to properly resolve two paradoxes in this YouTube video by James Tanton?
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$\begingroup$ See also the answers to this question. $\endgroup$ – Brian M. Scott Nov 4 '12 at 23:18
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2$\begingroup$ I disagree with closing the question as it is not an "exact duplicate" of the other question. $\endgroup$ – Thomas Nov 4 '12 at 23:31
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11$\begingroup$ I think it would be good if an explanation of the video's argument was added to the question, for better searchability. $\endgroup$ – wj32 Nov 4 '12 at 23:37
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3$\begingroup$ I have to agree that this is an duplicate. First of all, the answer to the second paradox is EXACTLY the top voted, accepted answer to the other question. Second, the second and third highest voting answers of the other question (and say the answer by Mike Jones) clearly explain the answer to both paradoxes. So, is the question EXACTLY the same as the previous question? No, but any one who is going to understand the answer is going to any of these questions is going to be able to transfer that knowledge over to any of the other questions. The word EXACT isn't great, but this is a dup. $\endgroup$ – GeoffDS Nov 5 '12 at 19:02
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First: It is not a paradox: it is just wrong. The reasoning is wrong.
About $\pi = 2$ he says: "Well clearly we are approaching the diameter of the circle". That is a statement that he doesn't prove and which is false.
The same problem arises with the $\sqrt{2} = 2$ when he says: "Well clearly this geometric construction approaches the diagonal of the square". How does he know that?
All that this proves is that we have to be careful when we talk about finding limits from purely looking at pictures.
"Just because the sun sets in the west doesn't mean that it has to rise in the west as well.
Edit: There are plenty of example of proofs that seem right, but turn out to be wrong when we go over them in more detail. Take for example the proof that for complex numbers $$ 1 = \sqrt{1} = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\sqrt{-1} = i\cdot i = -1$$ Here again, the argument is invalid because the rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$ doesn't hold for complex numbers.
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If the limit of a sequence of curves $\{\mathcal{C}_n\}$ is a curve $\mathcal{C}$, that does not mean that the limit of the lengths of the $\mathcal{C}_n$ will be the length of $\mathcal{C}$. The presentation makes the assumption that it will, but what validates that assumption, other than it feels intuitively to be true? This is a great example of how intuition, while usually helpful, can occasionally be hurtful. This example can serve as a poster child to champion mathematical rigor.
Take the circle example. After $N$ iterations, the extremely bumpy curve that you have is still much longer than the diameter.
And the staircase example. After $N$ iterations, the extremely jagged curve that you have is still much longer than the diagonal.
As a function from curves of finite length to $\mathbb{R}$, $\operatorname{length}(\phantom{x})$ is not continuous. It may feel like it should be, until you think about examples like the kind brought up here.
answered Nov 4 '12 at 23:14
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$\begingroup$ So the bumps will remain after infinitely many iterations? $\endgroup$ – glebovg Nov 4 '12 at 23:17
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8$\begingroup$ No, they won't. They will indeed be gone: the limit of the curves will be the flat curve. But the limit of the lengths of the curves will not equal the length of the flat curve. In the transition from the very bumpy curves to the flat curve, the length function will demonstrate a discontinuity. $\endgroup$ – alex.jordan Nov 4 '12 at 23:24
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$\begingroup$ @glebovg - The bumps are no longer in the curve $\mathcal{C}$. A priori, there is no reason to believe that the limit of the length of the curves $\mathcal{C}_n$ has anything to do with the length of the curve $\mathcal{C}$. $\endgroup$ – Braindead Nov 4 '12 at 23:41
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If the semi circle is divided into $2k$ semicircles , the total length is always $$\cfrac 12 \cdot 2\pi\left(\cfrac r{2k}\right)\cdot 2k=\pi r$$ Which means we have a constant sequence $a_{2k}=\pi r\rightarrow \pi r$ when $k\rightarrow +\infty$.
It is true that $\pi$ is an unknown constant $>0$. From the first semicircle, it is quite obvious that $\pi r > 2r$ which implies that $\pi r \ne 2r$ no matter the value of $\pi$ as long as $r$ stays positive. If you draw semi circles to infinity, we do approach the diameter. In this case, the geometric figure approaches the diameter while the total length remains constant. But no matter the number of semi circles drawn, there are always curves around the diameter and it is the total length of these curves that we are interested in, not the diameter so when this number gets large enough that we have no curves to measure, one simply measures the diameter and assume that the total length is that of the diameter without needing $\pi$ at all. If one now says $\pi r = 2r$, this contradicts the first assertion that $\pi r > 2r$
The geometric constructions might have approached the diameter, but the total lengths stays constant and doesn't approach the length of the diameter.
Applying the same logic to the one of $\sqrt 2$, you find out that his argument is also invalid. The semicircles don't converge to the diameter and the stairs don't converge to the diagonal.
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$\begingroup$ I don't understand. All I see in your answer is, essentially, "It is invalid because it yields the wrong answer." The question is, what incorrect mathematical assumption is being used to get the wrong answer. I don't see anything in your answer that even remotely hints at this. Am I missing something? $\endgroup$ – GeoffDS Nov 5 '12 at 18:30
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$\begingroup$ @Graphth when a sequence is constant, what is its limit? $\endgroup$ – user31280 Nov 5 '12 at 18:32
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$\begingroup$ Obviously it's the constant. What does that have to do with anything? I'm asking, where did you explain a mathematical assumption that was invalid? $\endgroup$ – GeoffDS Nov 5 '12 at 18:34
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$\begingroup$ he assumed that as the number of semi circles drawn tends to infinity, the total length tends to the length of the diameter which is untrue because the sequence is a constant. The first semi circle drawn is obviously longer than the diameter and also equal to $\pi r$ so how can that same $\pi r$ equal to the length of the diameter. He was unable to differentiate the geometrical limit of the semicircles from the limit of the total lengths. $\endgroup$ – user31280 Nov 5 '12 at 18:41
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1$\begingroup$ Yes, the point is, let's say I'm working a similar problem but this time the answer is not obvious. So, I go through and use similar steps and I find an answer and I think it's right because everything I did makes sense. Now, you come along and say, actually the answer is so and so. Okay, help me see where I went wrong. That is what this question is asking. $\endgroup$ – GeoffDS Nov 5 '12 at 19:05
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At 1:00 in the video, it is stated that $2(\frac{1}{2}\times\frac{\pi r}{2})=\pi r$, which is incorrect; the former expression simplifies to $\frac{\pi r}{2}$, not $\pi r$. The $\pi r$ line of reasoning is used to prove that $\pi=2$.
As for the proof that $\sqrt{2}=2$, bear this in mind.
(Great, now $2=4$. We're reinventing even more things than were anticipated now.)
It doesn't actually approach anything; it just looks like it does. Maybe it would help if the staircase were less crude; like, actually a reasonable approximation of the diagonal of the square, in which the median averages relatively near the diagonal.
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3$\begingroup$ At the 1 minute mark, he does NOT have the expression you say he has. What he has actually is $\pi r$. If it were $(\pi r)/2$, then the limit of the lengths would be 0, which would make absolutely no sense as we know the limit must be at least $2r$ because each curve clearly has a longer length than the diameter itself. As in Fola's answer, I don't really see anything that actually answers the question. You just presented another "paradox". The question is, what invalid mathematical assumption is being made? $\endgroup$ – GeoffDS Nov 5 '12 at 18:35
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$\begingroup$ @Graphth AHH, I see what he was doing there. The way that he convincingly misspoke and miswrote the expression confused me. $\endgroup$ – Tortoise Nov 5 '12 at 23:32
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$\begingroup$ That's incorrect. Pointwise, the sequence of curves certainly does approach the circle. But the length of the curve does not approach the length of the circle. $\endgroup$ – Mike Rosoft Sep 28 '19 at 9:41
