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