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

   $\tan\frac{\pi}{4} = 1$

   Show answer

   Using $x = 1$, we get \begin{align*} \tan^{-1} 1 &= 1 - \frac{1^3}{3} + \frac{1^5}{5} - \frac{1^7}{7} + \cdots\\ \frac{\pi}{4} &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\\ \pi &= 4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right) \end{align*}

   

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

   \begin{align*} \tan^{-1}\frac{1}{\sqrt3} &= \frac{1}{\sqrt3} - \frac{1}{\sqrt3^3 \times 3} + \frac{1}{\sqrt3^5 \times 5} - \frac{1}{\sqrt3^7 \times 7} + \cdots\\ \frac{\pi}{6} &= \frac{1}{\sqrt3} - \frac{1}{3\sqrt3 \times 3} + \frac{1}{3^2\sqrt3 \times 5} - \frac{1}{3^3\sqrt3 \times 7} + \cdots\\ &= \frac{1}{\sqrt3}\left(1 - \frac{1}{3 \times 3} + \frac{1}{3^2 \times 5} - \frac{1}{3^3 \times 7} + \cdots\right)\\ \pi &= \frac{6}{\sqrt3}\left(1 - \frac{1}{3 \times 3} + \frac{1}{3^2 \times 5} - \frac{1}{3^3 \times 7} + \cdots\right)\\ &= \frac{6\sqrt3}{3}\left(1 - \frac{1}{3 \times 3} + \frac{1}{3^2 \times 5} - \frac{1}{3^3 \times 7} + \cdots\right)\\ &= 2\sqrt{3}\left(1 - \frac{1}{3 \times 3} + \frac{1}{3^2 \times 5} - \frac{1}{3^3 \times 7} + \cdots\right) \end{align*}