# A half-pie ratio

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.

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!