- last post: 01.01.0001 12:00 AM PDT
Posted by: un gato
Posted by: Superbxen
\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}
What I get from this is
divided by the variable (greek for "z") equals divided by the sum (of?) ((the variable n which equals one) to infinity which still equals one) divided by the fraction (1 times the (variable n to the variable s power) which equals the variable n to the variable s power)
So the simplified load of crap is
/z(s)=/sum_1/frac(n^s)
Which is still a load of bs.
Oww. Don't make my head hurt like that showoff.