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+--- Thread: [2014] f(z) = f(exp(z)) = f(exp^[m](z)) (/showthread.php?tid=906)
[2014] f(z) = f(exp(z)) = f(exp^[m](z)) - tommy1729 - 07/23/2014
An unavoidable question.
We are intrested in the function f(z) = f(exp(z)).
However , we know that by substitution that implies f(z) = f(exp^[m](z)) for all integer m , if we want the function and equation to be defined everywhere.
But iterations of exp are chaotic !
So a nontrivial f(z) analytic on a large segment of the complex plane that satisfies this cannot exist !
But we have functions like sin(sexp(z)).
So what gives ?
The question becomes how many substitutions we can take and where the function is analytic.
So for instance if we expand a Taylor series at x = 100 + 25 i and have a radius that does not touch the real line then satisfying the equation is probably no problem. ( I try to stay away from the fixpoints )
But that is just a solution within a circle , what happens with continuation ?
A general theory seems to be missing.
One could conjecture that f(z) = sin(sexp(z+theta(z))) but even if that is true , that does not answer all questions.
There seems to be work to be done here.
Consider perhaps :
for every z , such that F(z) is analytic :
F(z) = F(exp(z)) or F(z) = F(ln(z))
However that seems inconsistant.
( F(z) - F(exp(z) must be analytic too )
Not sure how to proceed.
It seems that the number of singularities within a radius must grow kinda like the number of iterations of exp within that radius.
Another untraditional idea from tetration it seems.
regards
tommy1729