##### Question 11.4.17

Using Madhava’s series expansion for the arctangent: $\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ (see Question 15.4.3), $\pi$ can be found faster than with Archimedes’ method.

1. Show that $\pi = 4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$ Show hint

2. Unfortunately you would need to compute a lot of terms in that series to find $\pi$ accurately. The smaller $x$ is, the faster the series will converge, so let’s find a series for $\pi$ using $x=1/\sqrt3$. Madhava used $21$ terms of this series to calculate $\pi$ to $11$ decimal places – the most precise estimate of $\pi$ for centuries. Show answer