Infinite tetration giving I*Omega costant=I*0.567143...
#8
Interestingly, if we take

z=I*ln(phi)= I* ln(1.6180399..) and

z =I*logomega (phi)= I* (ln(1.6180399..)/ln(Omega))= -0.8484829....,

using:

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega))),
cos (z) = (1/2)*((Omega^(z/(I*Omega))+Omega^(-z/(I*Omega)))

sin(I*logomega (phi))= (-I/2)
cos (I*logomega (phi)) = (1/2) *(sqrt(5))= phi-1/2=1.6180399-0.5=1.1180399


but (I/2)=sin(I*ln(phi), so

sin(I*ln(phi)*sin(I*logomega (phi)) = 1/4
sin(I*ln(phi)+sin(I*logomega (phi)) =0
sin(I*ln(phi)/sin(I*logomega (phi)) =-1
sin(I*ln(phi)-sin(I*logomega (phi)) =-I



Ivars
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Messages In This Thread
RE: Infinite tetration giving I*Omega costant=I*0.567143... - by Ivars - 03/07/2008, 11:02 AM

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