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## Introduction

We should note that arctan(1) = π/4. It is the main idea of the proof. We will find a Taylor series representation for the inverse tangent and the proof will be complete.

## Finding the series representation

Observe these derivatives at of the inverse tangent at x=0:

$\displaystyle \frac{d^{0}}{dx^{0}} \arctan(x)|_{0}=0$
$\displaystyle \frac{d^{1}}{dx^{1}} \arctan(x)|_{0} = 1$
$\displaystyle \frac{d^{2}}{dx^{2}} \arctan(x)|_{0} = 0$
$\displaystyle \frac{d^{3}}{dx^{3}} \arctan(x)|_{0} = -2$

From simple observation or mathematical induction, we obtain that the nth derivative is zero if n is even, and it is this when n is odd:

$\frac{d^{n}}{dx^{n}} \arctan(x)|_{0} = (-1)^{\frac{n-1}{2}}(n-1)!$

So we can strike out the even derivative terms from the Taylor series. Doing that, we obtain the following series for arctan:

$\arctan(x) =\sum_{k=0}^{\infty}(-1)^{k}\frac{(2k)!}{(2k+1)!}x^{2k+1}=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{2k+1}, |x|\leq 1$

Evaluation at x=1 yields

$\frac{\pi}{4} = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}$

Hence, it immediately follows that

$\pi = 4\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1} = 4\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}...\right)$

Proof complete.

The above series is very slow to converge, taking about five billion terms to get pi to 10 decimal places.

## Product representation for pi

From the Basel problem, it follows that the infinite product representation Euler found for sin(x)/x is, in fact, true; despite it relying on the factoring of an infinite polynomial. This formula is

$\frac{\sin(x)}{x} = \prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)$

Since sin(π/2) is equal to 1, it immediately follows that

$\prod_{n=1}^{\infty}\left(1-\frac{1}{(2n)^2}\right)=\frac{\sin(\pi/2)}{\pi/2}=\frac{2}{\pi}$

From here, with a bit of rearranging, we obtain

$\prod_{n=1}^{\infty}\left(\frac{(2n)^2-1}{(2n)^2}\right)=\frac{2}{\pi}$
$\prod_{n=1}^{\infty}\left(\frac{(2n)^2}{(2n)^2-1}\right)=\prod_{n=1}^{\infty}\left(\frac{(2n)(2n)}{(2n-1)(2n+1)}\right)=\frac{\pi}{2}$

which gives, when expanded,

$\displaystyle \frac{\pi}{2}=\frac{2 \cdot 2}{1 \cdot 3}\cdot\frac{4 \cdot 4}{3 \cdot 5}\cdot\frac{6 \cdot 6}{5 \cdot 7}\cdot\frac{8 \cdot 8}{7 \cdot 9}...$

## Other representations

Pi is also equal to the values of some definite integrals:

$\pi=\int_{0}^{1}\frac{4}{1+x^{2}}dx$
$\pi=\int_{1}^{-1}\frac{1}{x^{2}-1}dx$
$\sqrt{\pi}=\Gamma(1/2)=\int_{0}^{\infty}t^{-1/2}e^{-t}dt$

Also this holds:

$\pi=\sqrt{6\zeta(2)} = \sqrt{6\sum_{n=1}^{\infty} \frac{1}{n^2}}$

The first two can be verified via integration, the third one follows from the Weierstrass product for the Gamma function and the fourth one is the result of the Basel problem.

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