04/22/2022, 06:14 PM
(This post was last modified: 04/22/2022, 06:49 PM by sheldonison.)
(03/24/2022, 09:35 AM)JmsNxn Wrote: ... Again, I haven't worked out the details, but the final expression should look something like this (remember, \(b = e^\mu\)):Hi James, Mphlee,
\[
x<s>\omega = \exp_b^{\circ s + \theta(s,x,\omega)}(\log_b^{\circ s + \theta(s,x,\omega)}(x) + \omega)
\]
.... Hopefully this makes sense, I'm kind of just shooting at the wall and seeing what sticks, lol. But I do believe this holds some weight. Think of it as taking all the Bennet hyperoperations, and finding a path along them where the typical Goodstein functional equation works.
I was trying to make sense of what has been posted. In Kneser's original paper, he talks about the half iterate o \( e^z \), which might be written as:
\[f(z)=e^{[0.5]}(z)=\text{tet}_e(\text{slog}_e(z)+0.5);\;\;\;f(f(z))=e^z\]
The base e could be any real or complex base for which Kneser's tetration is defined, and the fractional iterate of 0.5 could be replaced by any arbitrary fractional iterate like \( \frac{1}{n} \). Such a function would also be defined for non-integer values of n, and would be straightforward to compute with fatou.gp
\[f(z)=b^{[\frac{1}{n}]}(z)=\text{tet}_b(\text{slog}_b(z)+\frac{1}{n});\;\;\;f^{[\circ n]}(z)=b^z\]
Then perhaps one might want to think about the super function of f(z), which could be generated surprisingly simply as the following:
\[f^{[\circ z]}=\text{tet}_b(\frac{z}{n})\]
And would about the function g whose iterated superfunction is f(z)? Could we define that as well? How about the following, which should also be reasonably easy to compute with fatou.gp
\[g(z)=f(f^{-1}(z)+1);\;\;\;f(z)=g^{[\circ z]}\]
\[f^{-1}(z)=\text{tet}_b(\text{slog}_b(z)-\frac{1}{n})\]
I hope this is somewhat relevant to the current discussion; this equation might even be closely related to the equation in James PDF which I just downloaded. But my equations here don't lead to a generalized Goodstein/Ackermann functions for non integer values of n.
- Sheldon

