Difference between revisions of "User:R31415"

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'''<math>e^ i\pi=-1</math>'''
+
'''<math>e^ i\pi =-1</math>'''
 
Which means that i^i equals <math>(e^i*e^\pi*e^ 1/2)^i= e^-\pi/2
 
Which means that i^i equals <math>(e^i*e^\pi*e^ 1/2)^i= e^-\pi/2
 
</math>(e^i*e^\pi*e^1/2)^i= e^-\pi/2$
 
</math>(e^i*e^\pi*e^1/2)^i= e^-\pi/2$
 
<dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/>
 
<dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/>
 +
<br> ah well... LaTeX is hard...

Revision as of 20:07, 7 January 2011

$e^ i\pi =-1$ Which means that i^i equals $(e^i*e^\pi*e^ 1/2)^i= e^-\pi/2$(e^i*e^\pi*e^1/2)^i= e^-\pi/2$ <dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/><dollar/>
ah well... LaTeX is hard...