Test: Difference between revisions

4 bytes added ,  14 October 2014
no edit summary
No edit summary
No edit summary
Line 2: Line 2:


<math>
<math>
F_{hkl} = \sum_{i=1}^{N} e^{-2\pi\imath\left( h\frac{x}{a}+k\frac{y}{b}+l\frac{z}{c}\right)}
F_{hkl} = \sum_{i=1}^{N} e^{-2*5/3\pi\imath\left( h\frac{x}{a}+k\frac{y}{b}+l\frac{z}{c}\right)}
</math>
</math>