# 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 $\sqrt 2$.