Infinite tetration giving I*Omega costant=I*0.567143...
#5
Thanks for proving my little formula.

So nice. Of course, its 4D, that is needed. So we either need to know where to look for projection to 3D, or have some mapping that reduces dimensionality ( if there is one).

I need to add few more formulae and check before we can explain(? ) oscillations related to Omega and W(1),I mentioned in another thread :

Omega^(1/(I*Omega) = e^I

Omega^(-1/(I*Omega)=e^-I

sin (z) = (-I/2)* (Omega^(z/(I*Omega))-Omega^(-z/(I*Omega)))

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

(I*Omega)^(1/Omega) =-0.342726848178+I*0.13369214926..

Module ((I*Omega)^(1/Omega)) = (1/e) = Omega^(-1/Omega)


Arg ((I*Omega)^(1/Omega)) = atan(-2,5632)=-1,198826..rad = -68,6876759..grad

An Interesting complex number with module 1/e.

The angle between these 2 formula values is 2,1988261.. rad =125,983.. degrees.

Ivars
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RE: Infinite tetration giving I*Omega costant=I*0.567143... - by Ivars - 03/06/2008, 09:51 PM

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