01/18/2011, 01:44 PM

we have studied b^^z and z^^b alot.

they can be expressed by sexp_b and slog_b.

but what about f(z) = a , a^^a = z ?

so apart from superlog and superexp , how about a superlambert ?

what brings us to the core of the question :

is superlambert(z) always defined on C* ?

do we need " new numbers " to compute superlambert(z) ?

to give a nice equation for the superlambert , superlambert solves for 'a' in :

sexp_a(a) = z

and how do we solve that ?

maybe like this :

sexp_a(a) + a = z + a

a = sexp_a(a) + a - z

and take the iteration :

a_(n+1) = sexp_a_n(a_n) + a_n - z

to find 'a' by the limit.

however that seems dubious and/or chaotic to me.

not to mention iteration cycles.

what else can we do ?

is there an easy proof that there is always a complex 'a' ?

i think we can conclude there is always a complex 'a' because of riemann surfaces IF f(z) or z^z is locally holomorphic for each z.

this is because IF z^^z is locally holomorphic and defined for all z , then z^z has a riemann surface with range C* , so we have a mapping from C* to C*

... which can be inverted to arrive at another C* to C* mapping.

however notice the 2 IF's and the fact that we havent managed to work in all bases yet !

regards

tommy1729

they can be expressed by sexp_b and slog_b.

but what about f(z) = a , a^^a = z ?

so apart from superlog and superexp , how about a superlambert ?

what brings us to the core of the question :

is superlambert(z) always defined on C* ?

do we need " new numbers " to compute superlambert(z) ?

to give a nice equation for the superlambert , superlambert solves for 'a' in :

sexp_a(a) = z

and how do we solve that ?

maybe like this :

sexp_a(a) + a = z + a

a = sexp_a(a) + a - z

and take the iteration :

a_(n+1) = sexp_a_n(a_n) + a_n - z

to find 'a' by the limit.

however that seems dubious and/or chaotic to me.

not to mention iteration cycles.

what else can we do ?

is there an easy proof that there is always a complex 'a' ?

i think we can conclude there is always a complex 'a' because of riemann surfaces IF f(z) or z^z is locally holomorphic for each z.

this is because IF z^^z is locally holomorphic and defined for all z , then z^z has a riemann surface with range C* , so we have a mapping from C* to C*

... which can be inverted to arrive at another C* to C* mapping.

however notice the 2 IF's and the fact that we havent managed to work in all bases yet !

regards

tommy1729