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Searching for an asymptotic to exp[0.5]
#56
The conclusion is that if | f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be , then there are no nonreal zero's for f(z).

Also the converse : If there are no nonreal zero's for f(z) then
| f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be.

I assume this was already known by sheldon but he did not post the proof.
I assumed him to be correct but double checked it. Now its certain.

Its intresting , f(z) is now unique by the region where | f(f(z)) - exp(z) | << 1. Hmm well probably. Seems like another mission.
It is unique (by that criterion above) if f(z) is unique by the condition : no nonreal zero's and the zero's at the real line are the only possible.

Im already considering pro and contra arguments for that mission.
The majority of pro arguments are based on the periodicity of exp(z).
The contra arguments are based on small perturbations of the zero's.

All proofs are based on arg(f(z)). And NO use is made of ln(f(z)) is close to f^[-1](z).

Generalization ideas and conjectures are underway.


However the initial problem is reduced to an equivalence statement :

| f(f(z)) - exp(z) | << 1 where sheldon's picture claims it to be
<=>
there are no nonreal zero's for f(z).

But a full solution is still not here.
Although Im happy with this new insight.

Not sure how kneser is suppose to be related as sheldon talks about.

We seem to be getting a strong foundation for " fake function theory " Smile

regards

tommy1729
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RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 07/24/2014, 10:47 PM

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