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07/22/2023, 05:09 PM
(This post was last modified: 07/28/2023, 08:14 AM by Daniel.)
Given the logarithm has a helix shaped Riemann surface, the iterated logarithms related to tetration assure that the Riemann surface is quite complex as well as fractal. Has anyone worked with the Riemann surface of tetration? If so, what did you determine?

Edit: removed idea about tetration periods being assicated with Riemann surfaces.

Daniel

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How about just discussing the Riemann surface of iterated logarithms? For example what is the Riemann surface for \(\log{(\log{(z))}}\)?

Daniel

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The fact* see what I did there) that Tiemannian sums are hyper-tetraion some times tell us that the method of tetrating can be less convergent than divergent or also divergent Riemann sums. This means the topology of the surface sphe the surface can Schroeder equate to two Riemann sums. If this is true basically you can iutderive a rieman sum

Hope this helps

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(07/28/2023, 05:42 PM)Daniel Wrote: How about just discussing the Riemann surface of iterated logarithms? For example what is the Riemann surface for \(\log{(\log{(z))}}\)?

(07/28/2023, 05:42 PM)Daniel Wrote: How about just discussing the Riemann surface of iterated logarithms? For example what is the Riemann surface for \(\log{(\log{(z))}}\)?

I think he means just that the Riemann surfaces that parameterize tetration don't parameterize signed area enough for vector valued functions like Gauss groups to hyper-operate with two variables like a system of Schroeder equations or Schroeder equations and a variable

For logarithms the tetration took care of that, but that's good too the iterated log is plenty big and might be too big long enough to show that

this means the topology of the Riemann surfaces can equate to a set of Schroeder equations for two Riemann summs is such that you can outderive a Riemann sum.

Then its all tetration