A half-pie ratio

Posted on by Wojciech Aleksander Wołoszyn

Allow me to tell you about an interesting construction I learned from the Stewart’s Calculus book. Start with a square of area one. Next, extend your current rectangle, alternately, on top or beside, by another rectangle of area one. The first 500 steps of this construction are shown below.

animate-3

If you watch the animation carefully, you will notice that, as n increases, two times the ratio of width and height seems to approach a number close to 3.14 – and what else could it be if not \pi.

Let us look closer on what happens here; set a_n to be the width of our figure at step n, and, similarly, b_n the height. Obviously, we start with

\begin{cases} a_1 = 1\\ b_1 = 1. \end{cases}

When we extend the rectangle, we add some number to either a_n or b_n (depending on whether it’s width’s or height’s turn to be extended). The situation is symmetrical so we can assume it is width’s turn, and therefore n=2k-1. Whatever we add, however, say a number c, it must be the case that c \cdot b_{2k-1} is equal to one. Thus, c = \frac{1}{b_{2k-1}}, and so

\begin{cases} a_{2k}=\frac{a_{2k-1} b_{2k-1} + 1}{b_{2k-1}} \\ b_{2k}=b_{2k-1}; \end{cases}

by symmetry we get

\begin{cases} b_{2k+1}=\frac{a_{2k} b_{2k} + 1}{a_{2k}} \\ a_{2k+1}=a_{2k}. \end{cases}

It follows by induction that

\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots.

Now we need to tackle the infinite product. There is a quite known reduction formula that is helpful here:

\int \sin^n x \, dx = -\frac{1}{n} \cos x \sin^{n-1} x + \frac{n-1}{n} \int \sin^{n-2} x \, dx.

To prove the above, set u = \sin^{n-1} x, dv = \sin x and integrate by parts. Note that if we integrate from 0 to \frac{\pi}{2}, the first part of the summand dissapears as \sin 0 = 0 and \cos \frac{\pi}{2} = 0. Therefore, we have

\int_0^{\frac{\pi}{2}}  \sin^n x \, dx = \frac{n-1}{n} \int_0^{\frac{\pi}{2}} \sin^{n-2} x \, dx.

Denote the above integral by I_n. Using foregoing formula, it is easy to see that

\begin{cases} I_{2k}= \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \ldots \cdot \frac{2k-1}{2k} \cdot \frac{\pi}{2} \\ I_{2k+1}= \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{2k}{2k+1}. \end{cases}

Now, use squeeze theorem to get \frac{I_{2k+1}}{I_{2k}} \to 1. But

\frac{I_{2k+1}}{I_{2k}} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{2k}{2k-1} \cdot \frac{2k}{2k+1} \cdot \frac{2}{\pi},

so, by combining everything, indeed

\frac{a_n}{b_n} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \ldots = \frac{\pi}{2},

as the animation suggested!

Leave a Reply

Your email address will not be published. Required fields are marked *

Published by Wojciech Aleksander Wołoszyn

I am a DPhil student in Mathematics at the University of Oxford. My academic interests revolve around the philosophy of foundations of mathematics in the context of V and the multiverse debate. Currently, I am mostly preoccupied with research on modal model theory - an emerging field in mathematical logic. My academic advisers are Professor Joel David Hamkins (University of Notre Dame), Professor Ehud Hrushovski, and Professor W. Hugh Woodin (Harvard University). Previously, I was a Recognised Student in Philosophy at the University of Oxford, under Professor Hamkins's supervision.