My Math Forum Eulers Identity: e^(i*pi) = -1 vs bellCurve: sqrt( e^(i*x)^(i*x) / circle ) ?

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 April 7th, 2015, 03:15 PM #1 Senior Member   Joined: Jan 2013 Posts: 209 Thanks: 3 Eulers Identity: e^(i*pi) = -1 vs bellCurve: sqrt( e^(i*x)^(i*x) / circle ) ? Eulers identity: e^(i*pi) = -1 Bell curve: e^(-x*x/2)/sqrt(2*pi) (x^y)^z = (x^z)^y = x^(y*z) e^(-x*x/2) = sqrt( e^(-x*x) ) = sqrt( e^((i*x)*(i*x)) ) = sqrt( (e^(i*x))^(i*x) ) Divide by circumference of a circle (which I left off to simplify until now): bell curve height = sqrt( (e^(i*x))^(i*x) / (2*pi) ) Looks alot like e^(i*pi) = -1, doesnt it?
 April 8th, 2015, 08:01 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Nonsense is off topic here.

 Tags bellcurve, circle, eipi, eixix, eulers, identity, sqrt

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