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 1s 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 3s 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.

Advertisements

2 Comments

Filed under Uncategorized

2 responses to “Where Does Pi Come In?

  1. Pingback: Playing With Blocks | The Number Field

  2. coolbeans

    cool beans. Not spam, thought that this was really cool. Beans.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s