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!


Author: ficetea

I'll write more here later!

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