Benedict Irwin edited Random.tex  over 9 years ago

Commit id: 0107e000a4c9dfa5187afcba60f058350e1fd6e9

deletions | additions      

       

\sum_{i=-\infty}^{\infty} (-1)^i\prod_{j=-\infty}^\infty\frac{x-j\pi}{i\pi-j\pi} = cos(x)  \end{equation}  If \begin{equation}  \sum_{n=0}^\infty n = \frac{-1}{12}  \end{equation}  Does \begin{equation}  \prod_{n=0}^\infty x^n = \frac{1}{\sqrt[12]{x}}  \end{equation}  Is there something special about $12$ then, such that \begin{equation}  \prod_{n=0}^{\infty}\frac{e^{12n}}{n!} = 1  \end{equation}