# The life of "pi"

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

Note by Elijah Tan
4 years, 11 months ago

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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 $$\{(x,0) : 0 \le x \le 1\}$$ by a square wave. Let $$a$$ be any positive real, and $$n$$ any positive integer. Consider the square wave which:

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

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

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

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

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

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

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

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

If, in particular, we chose $$a=\tfrac{1}{n+1}$$, then we have a sequence of polygonal curves, each of length $$2$$, where the $$n$$th curve is never further than $$\tfrac{1}{n+1}$$ from the target line.

Even worse, if we chose $$a = \tfrac{1}{\sqrt{n}}$$, then we would have a sequence of polygonal curves whose lengths diverged to $$\infty$$, but where the $$n$$th curve is never further than $$\tfrac{1}{\sqrt{n}}$$ 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 $\int_a^b \sqrt{1 + y'(x)^2}\,dx$ Thus if $$y_n$$ is a sequence of curves such that their derivatives $$y_n'$$ converge uniformly, then their lengths converge (if $$a$$ and $$b$$ remain constant and finite). However, even uniform convergence of the sequence $$(y_n)$$ is not enough to guarantee the uniform convergence of the derived sequence $$(y_n')$$. Essentially (we need to tidy up the vertical lines and sharp corners), either in the $$\pi$$ 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 $$\tfrac{1}{4}\pi$$, and there would be no apparent paradox.

- 4 years, 11 months ago

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

- 4 years, 11 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.

- 4 years, 11 months ago

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

- 4 years, 11 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.

- 4 years, 11 months ago