The why of pi, part three

This isn’t about theology or computer programming, really.

A few years back I had a number of sleepness nights, literally, wondering why the constant $\pi$ is so damned weird. So I wrote to Mark-Jason Dominus, who knows mathematics, and he wrote a long blog post back with lots of good answers. The crux of his answer was

I think the essence of the problem with π is that the Euclidean distance function is nonlinear in the two spatial coordinates x and y.

Which was a good answer (as it turns out, a spectacular answer) but wasn’t particularly mentally satisfying. But often these things aren’t until you see them for yourself. I also wondered why $\pi$ was just a bit more than 3, rather than 5 or 57. MJD replied with a good list of things, but again, I didn’t quite get it. “Circles are a bit bigger than hexagons” seemed just as arbitrary as “pi is a bit more than 3”.

A few nights ago, I started thinking about it again and managed to satisfy myself fairly quickly. I also realised that the tau day people have a point. The one thing that made it all drop into place in my head was to start thinking in terms of $\tau$, not $\pi$. “Why is $\tau$ a little bit more than 1.5?” turned out to be a much more fruitful question than “Why is $\pi$ a little bit more than 3?”

Here is the picture that appeared in my head:

The lower and upper bounds for tau are the distances \(|B C|\) and $|D F|$. With a unit circle, \(|B C|\) is obviously $\sqrt 2$ by Pythagoras’ theorem. “Why is $\tau$ a little bit more than 1.5?” is now a simple question: it’s because it’s actually a little bit more than $\sqrt 2$.

In a sense this is exactly equivalent to Archimedes’ approximation of inscribing a hexagon in a circle, which is precisely what MJD had suggested as an explanation why $\pi$ was around 3, but I found it easier to think in terms of triangles than hexagons. (And yes, I know that triangles are just as arbitrary as hexagons. But obviously not in my mind…) Archimedes did it by measuring but we can do it by geometry. As we’ve mentioned \(|B C|\) is $\sqrt 2$, and \(|D F|\) is (slightly less obviously) 2. So a first approximation of $\tau$ is the average of these two distances, $ 2 + \sqrt 2 \over 2 $.

(Incidentally, we could rather arbitrarily decide that $\tau$ is a quarter of the way between $|B C|$ and $|D F|$ and get an approximation of $ {1 \over 2} + {3 \over 4} \sqrt 2$, which is fairly accurate - within 0.6%)

We can get a better approximation by dividing the circle further. (and thanks to the GeoGebra software for the diagrams.) Again, essentially what Archimedes did, but this time we have to use more trigonometry. $|A E|$ is 1, so $|E F|$ is $\tan 22.5$, which by the half-angle identity is $\sqrt 2 - 1$, so $|D F|$ is $2 \sqrt 2 - 2$. Similarly $|B C|$ is $2 \sin 22.5$, which is $\sqrt{2-\sqrt2}$.

Notice how the $\sqrt 2$s are starting to pile up. Taking the average of the two lengths, and doubling (because we divided the circle in half), we get a second approximation of $\tau$ as $2 \sqrt 2 - 2 + \sqrt{2-\sqrt2}$. Continued halving the circle and applying the half-angle identities
leads to a better approximation, and also a proliferation of 2 and $\sqrt 2$s.

The repeated application of $\sqrt{2-\sqrt{2-\dots}}$ suggests a family resemblance with Viète’s formula. Again, I’m not suggesting that I’ve gone up with anything new here at all, just that I’ve found a way of explaining things to myself that makes sense.

And which proves MJD’s intuition: the weirdness of $\pi$ is all about the weirdness of $\sqrt 2$.