Infinite tetration giving I*Omega costant=I*0.567143...
#1
Please could You check analytically or numerically if this derivation is valid?

h(((e^(pi/2))^(1/Omega))*(e^I)) =-W((-pi/(2*Omega)-I))/(pi/(2*Omega)+I)=(-Omega+(I*(pi/2))/(pi/(2*Omega)+I)= Omega*I=I*0,56714329..

The other branch of W should give -I*Omega?

The argument for h is complex, a product of:

e^(pi/(2*Omega))= e^(pi/2*0.567143..) = e^2,769663952
=15,95327204..

And e^I = cos1+I*sin1= 0,540302306+I*0,841470985

So numerically h( 8,619589667+I*13,42421553)= I*0,56714329

Real counterpart of this is simpler:

h(Omega^(1/Omega))=h(1/e) = -W(-ln(1/e)/(ln(1/e))= W(1)/1=Omega=0,56714329

Square superroot of (Omega^1/Omega) :

ssrt(Omega^(1/Omega) = ln(1/e)/W(ln(1/e))= -1/W(-1)= -1/(-0.318131505204764 + 1.337235701430689*I) = 0.16837688705553+0.707755195958823*I.

Ivars
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Infinite tetration giving I*Omega costant=I*0.567143... - by Ivars - 02/27/2008, 10:09 PM

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