# Happy Belated Pi Day

I didn’t create this blog until after Pi day, so unfortunately I’m a bit late. I celebrate Pi day by deriving the same series expression for pi every year. I’m not sure who to attribute this particular method to, but I definitely learned in from some old calculus book once upon a time. It’s as simple as calculating $\int_0^1\frac{1}{x^2+1}dx$ two different ways and equating the results. First up, we can make the substitution $x = \tan\theta$ and simplify yielding $\int_0^{\pi/4}d\theta = \frac{\pi}{4}$. On the other hand, if $|x|<1$ then $\sum_{n=0}^\infty(-1)^nx^{2n}$ converges (uniformly on all intervals $|x|\leq r < 1$) to $\frac{1}{x^2+1}$, and so may proceed with

$\int_0^1\frac{1}{1+x^2}dx = \sum_{n=0}^\infty(-1)^n\int_0^1x^{2n}dx$. Each $\int_0^1x^{2n}dx = \frac{1}{2n+1}$, and so putting it all together we’re left with

$\pi = 4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...)$

And there it is!

Advertisements

## Author: ficetea

I'll write more here later!