(01/22/2021, 08:50 AM)sheldonison Wrote: My understanding is that James' Tetration sould be defined in terms of \( \phi \) as follows:
\( \text{Tet}_\phi(z+k)=\lim_{n\to\infty}\ln^{[\circ n]}(\phi(z+n)) \), where k is chosen so Tet(0)=1
We can write an equation for when there are singularities for \( \ln(\ln(\phi(z))) \), by starting with
\( \phi(z+1)=\exp(\phi(z))\cdot\exp(z+1)=\exp(\phi(z)+z+1) \)
So then taking the logarithm once
\( \ln(\phi(z+1))=\phi(z)+z+1 \)
And taking the logarithm twice there are singularities where \( \phi(z)+z+1=0 \)
There is a trivial version at the real axis, approximately at z=-1.3018, phi(z)=0.3018. But how do we find the values for these singularities in the complex plane, other than the trivial 2pi i periodic examples? Finding these values is complicated... My conjecture is that there are an infinite number of complex values leading to singularities in \( \ln(\ln(\phi))) \) as \( \Re(z) \) increases where phi(z)+z+1~=0 ... More later ...
Yes, that's the gist of the solution; except you've shifted \( \phi \) by one, it should be \( \phi(s+1) = e^{s+\phi(s)} \).
I concur whole heartedly there are singularities in the complex plane. I have no idea how to find them though; but the idea that,
\(
\phi(s) + s= 0\\
\)
has solutions with arbitrarily large \( \Re(s) \) seems like a very smart way of going about it.
I seem to have mistaken my tetrations. I thought Kneser had singularities other than at the negative integers less than one... I must be thinking of a different tetration which had branch cuts all over the place; i.e: the tetration had zeroes...