## Hurry up Archimedes!

Archimedes would have liked the following function in two real variables

$f(x, y) = \dfrac{y (x + 14)}{6 x + 9}$

Can you think of a reason why, given the following expressions?

$2 f(0, 1) = \dfrac{28}{9} \approx 3.11111\cdots$

$3 f(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}) = \dfrac{29 \sqrt{3}}{16} \approx 3.13934\cdots$

$4 f(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}) = \dfrac{82 \sqrt{2} - 50}{21} \approx 3.14121\cdots$

$6 f(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}) = \dfrac{81 - 25 \sqrt{3}}{12} \approx 3.14156\cdots$

Here’s the pattern. When the point $(x, y)$ lies on the unit circle, say in the first quadrant, then $f(x, y)$ is a pretty good approximation of the arc length from $(1, 0)$ to this point. I like the first example above where a quarter circle already gives a very decent approximation to $\pi$. The last example above estimates $\pi$ with one twelfth of a circle. The result is close enough to have the famous approximation

$\pi \approx \dfrac{355}{113}$

in its partial fraction expansion. Where did this formula come from? The idea was to choose parameters a, b, c such that

$\dfrac{ \sin(x) (\cos(x) + a)}{b \cos(x) + c} \approx x$

The choices 14, 6 and 9 give the following approximation

$\dfrac{\sin(x) (\cos(x) + 14)}{6 \cos(x) + 9} = x - \dfrac{x^7}{2100} -O(x^9)$

These are also optimal in the sense that all other choices approximate to a lower order. Archimedes started with one sixth of a circle and halved the arc in each subsequent step. With this order seven approximation the error in the approximation of $\pi$ is reduced by a factor of 64 in each step. Using the function $f$ instead of chord lengths (as Archimedes did) requires only a third of the number of steps for the same accuracy.