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  f(z) = f(exp(z)) = f(exp^[m](z)) tommy1729 Ultimate Fellow     Posts: 1,370 Threads: 335 Joined: Feb 2009 07/23/2014, 11:20 PM 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 « Next Oldest | Next Newest »

 Messages In This Thread  f(z) = f(exp(z)) = f(exp^[m](z)) - by tommy1729 - 07/23/2014, 11:20 PM

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