05/08/2021, 07:58 AM
Again, Leo.
I agree with everything you're saying. What you are saying is fantastically intelligent. But trying to just use "multi-valued" functions will get you nowhere. You have to describe it as a Riemann surface. This is something rarely talked about on this forum or anywhere about tetration. But Tetration defines a Riemann surface, (I forget how to classify it off hand). And the multivalued choice of log; which makes tetration multivalued; is as much a riemann surface, as log is.
Again, I'm not saying you are wrong. You are fantastically right. And what you are saying is highly fascinating. But just slow down. I guarantee you, you need Riemann surfaces to explain what you're talking about.
Imagine I throw three more variables at you,
\(
\zeta(f(z,x,y)|h(z,x,y))
\)
Can you describe how you're multivalued branches behave with these two extra variables? And that they're holomorphic. And that they're well behaved. If you can do that without Riemann surfaces that's incredible. But I guarantee, if you wanna get what you wanna get out of this, read up on Riemann surfaces.
This is really fascinating though, Leo. I really appreciate your dialogue.
Regards.
I agree with everything you're saying. What you are saying is fantastically intelligent. But trying to just use "multi-valued" functions will get you nowhere. You have to describe it as a Riemann surface. This is something rarely talked about on this forum or anywhere about tetration. But Tetration defines a Riemann surface, (I forget how to classify it off hand). And the multivalued choice of log; which makes tetration multivalued; is as much a riemann surface, as log is.
Again, I'm not saying you are wrong. You are fantastically right. And what you are saying is highly fascinating. But just slow down. I guarantee you, you need Riemann surfaces to explain what you're talking about.
Imagine I throw three more variables at you,
\(
\zeta(f(z,x,y)|h(z,x,y))
\)
Can you describe how you're multivalued branches behave with these two extra variables? And that they're holomorphic. And that they're well behaved. If you can do that without Riemann surfaces that's incredible. But I guarantee, if you wanna get what you wanna get out of this, read up on Riemann surfaces.
This is really fascinating though, Leo. I really appreciate your dialogue.
Regards.