Claryfying few things for myself:
If we have complex number z , we can find second superroot of 1/z ssroot(1/z) = ln(1/z)/W(ln(1/z))= -ln(z)/W(-ln(z)) = 1/h(z)
So h(z)= 1/ssroot(1/z)
and we know that h(z^(1/z)) = z .
So infinite tetration is just an operation that is inverse (in a sense 1/) to finding of second superroot, which is a function of type
ssroot(1/z)^ssroot(1/z) =1/z= 1/ h(z^(1/z)) or
h(z^(1/z))= ssroot(1/z)^(-ssroot(1/z)))
where argument of h is selfroot of z, while on the right side we have selfroot of second superroot of 1/z.
as an example, let z= 1/e, than h ( (1/e)^(e))=1/e, ssroot(e) = 1/Omega, 1/e = (1/Omega)^(-1/Omega)
If we have complex number z/i , we can find second superroot of i/z ssroot(1/(z/i)) = ln(1/(z/i))/W(ln(1/(z/i)))= -ln(z/i)/W(-ln(z/i)) = 1/h(z/i)
So h(z/i)= 1/ssroot(i/z)
and we know that h((z/i)^(i/z)) = z/i .
ssroot(i/z)^ssroot(i/z) =i/z= 1/ h((z/i)^(i/z)) or
h((z/i)^(i/z))= ssroot(i/z)^(-ssroot(i/z))
And finally, if we have complex number z/w-division of 2 complex numbers (this and next I have not checked):
h((z/w)^(w/z))=ssroot(w/z)^(-ssroot(w/z))
h((az+b)/(cz+d))^((cz+d)/(az+b))=ssroot((cz+d)/(az+b))^(-ssroot((cz+d)/(az+b)))
and may be even h(P/Q^(Q/P) where P,Q are complex polynomials of any power, perhaps max power of Q>maz power of P.
and h((S1/S2)^(S2/S1)) where S1, S2 are infinite powerseries, sometimes equivalent to analytic functions, sometimes divergent.
Now selfroots and second super roots are hopefully correctly linked via infinite tetration, and I will try to iterate function h backwards and forward as long as its argument can be expressed as z^(1/z). Just to see what happens.
Ivars
If we have complex number z , we can find second superroot of 1/z ssroot(1/z) = ln(1/z)/W(ln(1/z))= -ln(z)/W(-ln(z)) = 1/h(z)
So h(z)= 1/ssroot(1/z)
and we know that h(z^(1/z)) = z .
So infinite tetration is just an operation that is inverse (in a sense 1/) to finding of second superroot, which is a function of type
ssroot(1/z)^ssroot(1/z) =1/z= 1/ h(z^(1/z)) or
h(z^(1/z))= ssroot(1/z)^(-ssroot(1/z)))
where argument of h is selfroot of z, while on the right side we have selfroot of second superroot of 1/z.
as an example, let z= 1/e, than h ( (1/e)^(e))=1/e, ssroot(e) = 1/Omega, 1/e = (1/Omega)^(-1/Omega)
If we have complex number z/i , we can find second superroot of i/z ssroot(1/(z/i)) = ln(1/(z/i))/W(ln(1/(z/i)))= -ln(z/i)/W(-ln(z/i)) = 1/h(z/i)
So h(z/i)= 1/ssroot(i/z)
and we know that h((z/i)^(i/z)) = z/i .
ssroot(i/z)^ssroot(i/z) =i/z= 1/ h((z/i)^(i/z)) or
h((z/i)^(i/z))= ssroot(i/z)^(-ssroot(i/z))
And finally, if we have complex number z/w-division of 2 complex numbers (this and next I have not checked):
h((z/w)^(w/z))=ssroot(w/z)^(-ssroot(w/z))
h((az+b)/(cz+d))^((cz+d)/(az+b))=ssroot((cz+d)/(az+b))^(-ssroot((cz+d)/(az+b)))
and may be even h(P/Q^(Q/P) where P,Q are complex polynomials of any power, perhaps max power of Q>maz power of P.
and h((S1/S2)^(S2/S1)) where S1, S2 are infinite powerseries, sometimes equivalent to analytic functions, sometimes divergent.
Now selfroots and second super roots are hopefully correctly linked via infinite tetration, and I will try to iterate function h backwards and forward as long as its argument can be expressed as z^(1/z). Just to see what happens.
Ivars