# Where Does Pi Come In?

Yesterday, I described a simple construction that led to a very interesting result: a sequence of rectangles whose aspect ratios (i.e. width divided by height) approach $\frac{\pi}{2}$. Today, I’ll show why that happens. A disclaimer before proceeding: I want this explanation to be accessible to a wide audience. It won’t be as rigorous, abstract, or terse as some would like, while it will be too technical for others. If you want additional explanation for anything, please ask in the comments.

To make the problem more precise, we need to introduce notation. We’ll call $w_n$ and $h_n$ the width and height, respectively, after $n$ steps. Just to get used to the language, what we know is that $w_1 = h_1 = 1$, and we want to show that $\lim_{n \to \infty} \frac{w_n}{h_n} = \frac{\pi}{2}$. First, we need to find the right expressions for $w_n$ and $h_n$ when $n>1$.

Adding a new rectangle to the side corresponds to forming $w_{n+1}$, which is the width of the existing rectangle plus the width of the new addition. The height of the new addition is $h_n$ and its area is $1$, so its width must be $\frac{1}{h_n}$, which shows that $w_{n+1} = w_n + \frac{1}{h_n}$. Likewise for the height, but since we add to the side before we add to the top, there’s a small difference: $h_{n+1} = h_n + \frac{1}{w_{n+1}}$.

Let’s see how it works:

$w_2 = 1+1 = 2,\ h_2 = 1 + \frac{1}{2} = \frac{3}{2}$
$w_3 = 2 + \frac{1}{3/2} = \frac{8}{3},\ h_3 = \frac{3}{2} + \frac{1}{8/3} = \frac{15}{8}$
$w_4 = \frac{8}{3} + \frac{1}{15/8} = \frac{16}{5},\ h_4 = \frac{15}{8} + \frac{1}{16/5} = \frac{35}{16}$
and so forth.

Sometimes when it doesn’t look like there’s much we can do, we might as well try whatever presents itself. In this case, it looks like we could at least factor all of the numerators and denominators:

$w_2 = 2,\ h_2 = \frac{3}{2}$
$w_3 = \frac{2\cdot 4}{1\cdot 3},\ h_3 = \frac{1\cdot 3\cdot 5}{2\cdot 4}$
$w_4 \frac{2\cdot 4 \cdot 2}{1 \cdot 5},\ h_4 = \frac{1 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 2}$.
I’ve thrown in a few extra $1$s to make it look pretty, because that’s always allowed, but there’s something not quite right with the last row. That’s because we wrote everything in lowest terms, which means that some $3$s cancelled out. If we put them back, it looks much better:
$w_4 = \frac{2\cdot 4 \cdot (2\cdot3)}{1 \cdot 3\cdot 5} = \frac{2\cdot 4 \cdot 6}{1 \cdot 3\cdot 5},\ h_4 = \frac{1 \cdot 3\cdot 5 \cdot 7}{2 \cdot 4 \cdot (2\cdot3)} = \frac{1 \cdot 3\cdot 5 \cdot 7}{2 \cdot 4 \cdot 6}$.

Looking at that, and taking care with the variable $n$, that’s enough to conjecture the general form for every $w_n$ and $h_n$, specifically:

$w_n = \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2(n-1)-1)}$
$h_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2(n-1))}$

If that’s enough evidence for you, just skip the next few paragraphs, where we prove that these formulas work. But you shouldn’t, because this is the best part–we’re about to use induction. Specifically, we’ll assume that $w_n$ and $h_n$ both follow the pattern (which is a good assumption, at least when $n \leq 4$), and show that $w_{n+1}$ and $h_{n+1}$ also follow the pattern. We’ll start with $w_{n+1}$:

$w_{n+1} = w_n + \frac{1}{h_n}$
$w_{n+1} = \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2n-3)} + \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2n-1)}$.

Most mathematicians would not even attempt to do the next step in public, because it’s just arithmetic, so you look dumb if you do it at all and even dumber if you do it wrong. It’s just adding fractions with different denominators! But I’ll press forward. First, we factor out like terms:

$w_{n+1} = \left( \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2n-3)} \right)\left( 1 + \frac{1}{(2n-1)}\right)$.
$w_{n+1} = \left( \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2n-3)} \right)\left( \frac{2n-1}{2n-1} + \frac{1}{2n-1}\right)$.
$w_{n+1} = \left( \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))\cdot(2n)}{1 \cdot 3 \cdot 5 \cdots (2n-3)\cdot(2n-1)} \right)$,

which is exactly what we need, although we do have to remember that $n = n +1 -1$ to make everything work out.

Fortunately, there’s a way to leverage this and simplify the calculation of $h_{n+1}$ by going back to the idea of the ever-growing rectangle. When we get to $h_{n+1}$, we’ve added $2n$ rectangles, each with area $1$, in addition to the starting unit square, so the area of the big rectangle, which is $w_{n+1} \cdot h_{n+1}$, must be $2n +1$. If we divide both sides by $w_{n+1}$, which we just calculated, we find that the equation given for $h_n$ is correct!

Now it’s time for the big finish. With formulas in place for $w_n$ and $h_n$, it’s time to put them together in the ratio

$\frac{w_n}{h_n} = \frac{ \frac{2 \cdot 4 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 5 \cdots (2(n-1)-1)} }{ \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2(n-1))} }$.

When we simplify the fraction and line everything up from smallest to largest, we get:

$\frac{w_n}{h_n} = \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdots (2(n-1))}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdots (2(n-1)-1)}$,

which, in the limit (as $n$ gets larger and larger), is known as the Wallis product. There are two ways of showing that it converges to $\frac{\pi}{2}$–one is slick and short, but the other requires only integral calculus. I originally planned to derive at least one of them here, but the Wikipedia article does a good enough job, and this post has gone on long enough.

So there you go. The explanation is always more work and less exciting than the mystery, isn’t it? I guess from that point of view, Lost got it right. You weren’t expecting that, were you? That’s right, I went there:

L O S T

P.S. Hey redditors, thanks for your interest! I have plans to start a dedicated math blog with lots of content like this at some point in the future, but the pieces for that are not yet in place, and for the time being I post mathematical musings here on my personal blog only rarely. You’re welcome to visit whenever you want, but you’ll probably only find stories about my cute kids. I’ll be sure to let you know when I get the math blog going. If you need help with anything math-related, please leave a comment on this post and I’ll get back to you as soon as I can.