MathTest: Difference between revisions
From charlesreid1
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\frac{2}{\sqrt{\pi}} \int_x^{\infty} e^{-t^2}\,dt = | \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}} | \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n)!}{n!(2x)^{2n}} | ||
</math> | |||
<math> | |||
y = x + 2 | |||
</math> | |||
Revision as of 16:41, 29 September 2010
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 $