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Does tetration take the right half plane to itself?
Your second response confuses me. That's VERY close to what I am asking, but it is slightly off. It is DEFINITELY NOT 1-1, first of all. Tetration is periodic, so there's no hope in hell of that, I wouldn't dare say that. Tetration IS injective modulo a period (which is easy to prove). I do know that tetration does live in the right half plane, using experimental arguments. And I can prove it if the Schroder function of is fully monotone about zero, then tetration lives in a small disk about the fixed point of the exponential (that small disk residing in the right half plane). 

Yes, I reread a bunch of your posts regarding it. I read them very slow, and I now know why the inverse Schroder function is fully monotone. It is much simpler than I thought! MUCH simpler than what I was thinking. We just have to use that the composition of two fully monotone functions is fully monotone, and that the uniform limit of fully monotone functions is fully monotone. Where "fully monotone functions" are defined as having "completely monotone first derivative."

It easily follows from this that is fully monotone, it uniformly converges to the inverse Schroder function, and therefore the inverse Schroder function is a fully monotone function.

Therefore, by the above arguments , it actually sends the right half plane to a small disk in the right half plane (but that's extraneous). Therefore
If perhaps you are confused. I'm betting when you read the paper it'll be a lot clearer. I'll be sure to make a head nod and an accreditation to you. I'll PM you to ask how to reference you. I'm rebuilding a paper I wrote two years ago, and adding grout to everything, so there are no leaks and no holes.

Messages In This Thread
RE: Does tetration take the right half plane to itself? - by JmsNxn - 05/16/2017, 04:09 AM

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