From charlesreid1

Revision as of 16:41, 29 September 2010 by ::1 (talk)

This is a math test:


$ \operatorname{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2}\,dt = \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n)!}{n!(2x)^{2n}} $

$ y = x + 2 $