Question 15.4.3

Using the fact that \[\frac{\dif }{\dif x}\tan^{-1}x = \frac{1}{1+x^2}\] which is the sum of the infinite geometric series \[1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots\] show that \[\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\] This series was discovered by the Indian mathematician-astronomer Madhava of Sangamagrama, and used to compute $\pi$ to astonishing precision (see Question 11.4.17). Show answer

\begin{align*} \tan^{-1} x - \tan^{-1} 0 &= \int_0^x \frac{1}{1 + u^2}\dif u\\ \tan^{-1} x &= \int_0^x \left(1 + (-u^2) + (-u^2)^2 + (-u^2)^3 + \cdots\right)\dif u\\ &= \int_0^x \left(1 - u^2 + u^4 - u^6 + \cdots\right)\dif u\\ &= \left[u - \frac{u^3}{3} + \frac{u^5}{5} - \frac{u^7}{7} + \cdots\right]_0^x\\ &= x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \end{align*}