Arguments for the beta method not being Kneser's method

[sheldonison]

Hi James,
Sounds like you're making progress in understanding the Beta function!  One of the problems with iterating logarithms is knowing how many 2pi i multiple's are required.  This can often be resolved by comparing the logarithm with the function with one less iteration, and picking the 2n*Pi*I branch which is closest.  

The key question seems to be whether the resultant Tetration function converges ... especially near where Beta(z+2)=1, |Beta(z+1)| is small.

But unfortunately I found a singularitiy.  Consider z=5.3136167434369 + 0.80386188968627*I; where beta(z+1,1)=1, and beta(z) is small.  So this location is a logarithmic singularity for your Tetration function.   I found this singularity by looking nearby the zeros of \( \beta(z-1,1)+z=0 \), since looking directly for zeros of \( \beta(z)-\ln(1+\exp(z)) \) gets many nearly zero results that are false positives, where beta(z-1) has a negative real part.

Ember: beautiful plots!

James,
The singularity I was looking for is where beta(z+1)=1; ln(beta(z+1))=0; and beta(z) is small.
\( \ln(\beta(z+1))=\beta(z)-\ln(1+\exp(-z))=0 \)
approximation:
\( \ln(\beta(z+1))\approx\beta(z)-\exp(-z)\approx0 \)
but \( \beta(z)\approx\exp(\beta(z-1))\;\;\; \) also a good approximation
so we get 
\( \exp(\beta(z-1))-\exp(-z)\approx0 \)
\( \exp(\beta(z-1))\approx\exp(-z) \)
So I found a case for n=0 where:
\( \beta(z-1)+z\approx 2n\pi i \)
and then used Newton's to find where beta(z+1)=1 near this result; z=5.3136167434369 + 0.80386188968627*I