The life of "pi"

Main post link -> http://s254.photobucket.com/user/balthamossa2b/media/1290457745312.jpg.html

I'm very curious about this as to why pi = 3.141..... instead of pi = 4 as shown in the link. Could anyone enlighten me about this? The difference of 4 and 3.141 is too big! Thanks :D

http://s254.photobucket.com/user/balthamossa2b/media/1290457745312.jpg.html

Note by Elijah Tan
7 years, 9 months ago

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2 votes

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Comments

Looking at the lengths of curves that converge to other curves can bring about counterintuitive results.

As an example, consider trying to "approximate" the line {(x,0):0x1}\{(x,0) : 0 \le x \le 1\} by a square wave. Let aa be any positive real, and nn any positive integer. Consider the square wave which:

  • jumps up from (0,0)(0,0) to (0,a)(0,a),

  • runs at y=ay=a for 0x1n0 \le x \le \tfrac{1}{n},

  • drops down from (1n,a)(\tfrac{1}{n},a) to (1n,0)(\tfrac{1}{n},0),

  • runs at y=0y=0 for 1nx2n\tfrac{1}{n} \le x \le \tfrac{2}{n},

  • jumps up from (2n,0)(\tfrac{2}{n},0) to (2n,a)(\tfrac{2}{n},a),

  • runs at y=ay=a for 2nx3n\tfrac{2}{n} \le x \le \tfrac{3}{n},

and so on finally ending up with a vertical line finishing at (1,0)(1,0).

This line is always within aa of the original straight line, but its total length is 1+(n+1)a1 + (n+1)a, since in addition to the nn horizontal lines each of length 1n\tfrac{1}{n}, there are n+1n+1 vertical lines of length aa.

If, in particular, we chose a=1n+1a=\tfrac{1}{n+1}, then we have a sequence of polygonal curves, each of length 22, where the nnth curve is never further than 1n+1\tfrac{1}{n+1} from the target line.

Even worse, if we chose a=1na = \tfrac{1}{\sqrt{n}}, then we would have a sequence of polygonal curves whose lengths diverged to \infty, but where the nnth curve is never further than 1n\tfrac{1}{\sqrt{n}} from the target curve!

Thus we have a sequence of curves which converge uniformly to the line, but whose lengths either converge to something other than the length of the line, or else diverge altogether.

To see the sort of area where the problem lies, consider that the length of a continuously differentiable curve is ab1+y(x)2dx \int_a^b \sqrt{1 + y'(x)^2}\,dx Thus if yny_n is a sequence of curves such that their derivatives yny_n' converge uniformly, then their lengths converge (if aa and bb remain constant and finite). However, even uniform convergence of the sequence (yn)(y_n) is not enough to guarantee the uniform convergence of the derived sequence (yn)(y_n'). Essentially (we need to tidy up the vertical lines and sharp corners), either in the π\pi case or the simpler example above, we have a sequence of continuously differentiable functions which converge uniformly, but whose derivatives do not.

On the other hand, if a sequence of continuous functions converges uniformly, the sequence of their integrals (over a fixed finite interval) does converge. Consequently if we considered the areas of the polygonal regions in your link, they would converge to 14π\tfrac{1}{4}\pi, and there would be no apparent paradox.

Mark Hennings - 7 years, 9 months ago

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wow thanks :O haha. but its abit too much for my small knowledge in maths :P

Elijah Tan - 7 years, 9 months ago

A shorter and less rigorous version of the argument given: The figure seems to approach a circle, but doesn't actually. It has infinitely many "bumps" and "grooves" in the lines.

Michael Tang - 7 years, 9 months ago

There is an interesting anecdote about a certain Dr.Goodwin who "prove" pi=4 by using the area of a square with the same circumference. It actually almost became a law in Indiana.

Jeremy Shuler - 7 years, 9 months ago

With the curve convergence. One of my lecturers used it to "prove" 5=7. Yes, Mark H, it's interesting, this.

Jeremy Shuler - 7 years, 9 months ago
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