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[2014] f(z) = f(exp(z)) = f(exp^[m](z))
#1
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
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[2014] f(z) = f(exp(z)) = f(exp^[m](z)) - by tommy1729 - 07/23/2014, 11:20 PM

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