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Is value of $\pi = 4$?

In this question we discussed why the fake proof is wrong.

But, what about the area?

The process converges to the same area of the circle ($\frac{\pi}{4}$)?

What's the area when the process repeats infinitely?

How we can prove, using calculus and limits the result?

Pi is equal to 4

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GarouDan
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  • This question just ask about the area. Not about the problem in the proof. – GarouDan Apr 17 '12 at 23:29
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    have you tried to figure it out? – john w. Apr 17 '12 at 23:34
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    It is an interesting fact that intuition about area (in two dimensions) seems to be much more useful than intuition about arclength. – André Nicolas Apr 17 '12 at 23:39
  • Doesn't this result follow from the one showing that $\sqrt{2} = 2$ by (a) going from $(0,0)$ to $(1,1)$ in a straight line and (b) along a path that is alternately parallel to the $x$ axis and the $y$ axis in ever decreasing segments? – Dilip Sarwate Apr 17 '12 at 23:45
  • @Andre: I don't think that's quite the problem: I think it's more an issue that the victim of such an argument doesn't have the knowledge or appreciation for rigor to be able to confidently reject the assertion that the jagged line converges to the circle in a suitable sense. –  Apr 17 '12 at 23:47
  • @AndréNicolas, I agree. – GarouDan Apr 17 '12 at 23:54
  • @DilipSarwate, well, I didn't understand at all what you're saying. – GarouDan Apr 17 '12 at 23:55
  • @johnw., I tried a little, but a limit to became $\pi$ looks hiden. – GarouDan Apr 17 '12 at 23:56
  • How this is a exaclty duplicate of another question? Well, I really hope this question didn't end because of this missunderstood. – GarouDan Apr 18 '12 at 16:16

1 Answers1

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All of the shapes have area greater than that of the circle, therefore the limit must be at least as large as the area of the circle.

A similar process drawing shapes on the inside of the circle would give you another sequence of areas whose limit must be at most as large as the area of the circle.

However, the two limits are equal. And so, by the squeeze theorem -- known to the ancient Greeks in this case as the method of exhaustion -- both of these limits are equal to the area of the circle.

There is a technical point assumed here: that the notion of area makes sense for circles!

One of the general approaches to the notion of area is to define the "outer area" of a general shape by covering it with a grid and adding up the area of all of the rectangles that partially lie within the shape. If you consider all possible grids, there is a greatest lower bound on the area thus computed, and this bound is called the "outer area" of the shape.

A similar method defines the "inner area". Then, the area of a shape is defined if and only if the inner area equals the outer area.

So, using this method to find the area of a circle is actually perfectly rigorous even by modern standards!

Also, it's pretty easy to see that this method is computing the values of Riemann sums for the integral that computes the area of the circle.

As a final general point, never trust a purported proof with a trollface in it.

  • Hurkyl. I'm searching for a algebra proof. Using calculus and limit. Do you know one? – GarouDan Apr 17 '12 at 23:54
  • @Hurkyl Basically, what we're experiencing is $$\Delta S\neq \Delta y+\Delta x$$ but $$\Delta A = \Delta x \Delta y$$ – Pedro Apr 18 '12 at 00:48