# Getting closer to $\pi$

A lovely curiosity came my way via @mikeandallie and @divbyzero:

Isn't that neat? If I use an estimate $p = 3.142$, then this method gives $\pi \approx p + \sin(p) = 3.141\ 592\ 653\ 6$, which is off by about $10^{-12}$ -- even better than Shanks suggests.

So, why does it work?

It's a two-step chain of reasoning: a trig identity and a Maclaurin series.

The trig identity is that $\sin(\pi - p) = \sin(p)$, by symmetry. We've picked $p$ so that $\pi - p$ is small -- in fact, smaller than $5 \times 10^{-n-1}$.

When an angle $x$ is small, $\sin(x)$ can be approximated as $\sin(x) \approx x - \frac 13 x^3$. In particular, $\sin(p) = \sin(\pi - p) \approx (\pi-p) - \frac 13 (\pi - p)^3$.

Looking at Shanks's original setup, we have $p + \sin(p) \approx p + (\pi - p) - \frac 13 (\pi - p)^3$, which simplifies to $\pi - \frac13 (\pi - p)^3$.

We know that $\pi - p$ is at most $5 \times 10^{-n-1}$, so $\left|\frac 13 (\pi - p)^3 \right| \lt \frac {125}{3} \times 10^{-3n - 3} < 5\times 10^{-3n-2}$. I reckon the approximation is generally correct to $3n+1$ decimal places1. ## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. Perhaps this breaks down in some cases when p's last digit is 4 or 5, but that is left as an exercise. []

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