# User:Guardian of Light

User at Wikipedia.

## Major Contributions

Some pages I've invested a lot in:

## Creations

The pages I myself have made (from ye olde scratch) are--in chronological order:

## Problems

To show the probability that two integers chosen at random are relatively prime is ${6\over \pi^2}$.

Proof: It is sufficient to show $\sum_{n=1}^{\infty}{1\over n^2} = {\pi^2\over 6}$. When we have a polynomial with constant term one, we may rewrite it in factored form as follows: If $\alpha_1,\alpha_2,...,\alpha_r\,$ are the roots of a polynomial p(z), then we may write $p(z)=\left(1-{z\over\alpha_1}\right)...\left(1-{z\over \alpha_r}\right)$.

Now examine the power series for the function sin(z)/z. ${\sin z\over z}=1-{z^2\over 3!}+{z^4\over 5!}+...+{(-1)^nz^{2n}\over (2n+1)!}$

Well we also know we can rewrite sin(z)/z in terms of its roots to be:

$\left(1-{z\over \pi}\right)\left(1+{z\over \pi}\right)\left(1-{z\over 2\pi}\right)\left(1+{z\over 2\pi}\right)\left(1-{z\over 3\pi}\right)\left(1+{z\over 3\pi}\right)...\left(1-{z\over k\pi}\right)\left(1+{z\over k\pi}\right)...$

If we examine the quadratic term in each we find that:

${1\over 3!} = {1\over \pi^2}\sum_{n = 1}^{\infty}{1\over n^2}\rightarrow {\pi^2\over 6}=\sum_{n = 1}^{\infty}{1\over n^2}\text{Q.E.D.}$