Interesting Tommy. I'd like to add some points.
If we consider \( \beta_\lambda(s) \) then,
\(
\exp(\beta_\lambda(s)) = \beta_\lambda(h(s)) = \beta_\lambda(s+1)*(1+ e^{-\lambda s})\\
\)
So we should expect,
\(
h(s) = \beta_\lambda^{-1}(\beta_\lambda(s+1)*(1+ e^{-\lambda s})) = s+1 + \frac{\beta_\lambda(s+1)}{\beta_\lambda'(s+1)} e^{-\lambda s} + ...\text{higher order terms}\\
\)
The value,
\(
\frac{\beta_\lambda(s+1)}{\beta_\lambda'(s+1)} \to 0\,\,\text{as}\,\,\Re(s) \to \infty\\
\)
So, we can shorten this to,
\(
h(s) = s+1 + o(e^{-\lambda s})\\
\)
This seems like a very good way to get the super-logarithm. That's something that's been bugging me a bit, finding a way to limit towards the super-logarithm. Which is to say, this could definitely aid in showing the limit,
\(
A_k = \beta_\lambda^{-1}(\exp^{\circ k}(z)) - k \to F^{-1}_\lambda(z)\\
\)
Where \( F_\lambda \) is the tetration function associated to the multiplier \( \lambda \). Even though the function \( \beta_\lambda^{-1} \) is an absolute monster I can't imagine numerically evaluating.
But using this idea of the \( h \) function, we get that,
\(
A_k = h^{\circ k} (\beta_\lambda^{-1}(z)) - k
\)
And this should be bounded by a nice sum of exponentials which should converge as \( k\to\infty \)... I believe...
Regards, James
Edit:
Please, Tommy, take a look at my Pari-GP code. This isn't a proposed, maybe it works, solution. I can produce accuracy of 200 decimal places.
The function \( \text{Abl_L} \) pretty much produces \( F_\lambda \); minus some caveats. This function is absolutely holomorphic. The trouble I'm having with is only the real tetration \( \text{tet}_\beta \) which is the real tetration we care about. But \( F_\lambda(s) \) is absolutely holomorphic.
If we consider \( \beta_\lambda(s) \) then,
\(
\exp(\beta_\lambda(s)) = \beta_\lambda(h(s)) = \beta_\lambda(s+1)*(1+ e^{-\lambda s})\\
\)
So we should expect,
\(
h(s) = \beta_\lambda^{-1}(\beta_\lambda(s+1)*(1+ e^{-\lambda s})) = s+1 + \frac{\beta_\lambda(s+1)}{\beta_\lambda'(s+1)} e^{-\lambda s} + ...\text{higher order terms}\\
\)
The value,
\(
\frac{\beta_\lambda(s+1)}{\beta_\lambda'(s+1)} \to 0\,\,\text{as}\,\,\Re(s) \to \infty\\
\)
So, we can shorten this to,
\(
h(s) = s+1 + o(e^{-\lambda s})\\
\)
This seems like a very good way to get the super-logarithm. That's something that's been bugging me a bit, finding a way to limit towards the super-logarithm. Which is to say, this could definitely aid in showing the limit,
\(
A_k = \beta_\lambda^{-1}(\exp^{\circ k}(z)) - k \to F^{-1}_\lambda(z)\\
\)
Where \( F_\lambda \) is the tetration function associated to the multiplier \( \lambda \). Even though the function \( \beta_\lambda^{-1} \) is an absolute monster I can't imagine numerically evaluating.
But using this idea of the \( h \) function, we get that,
\(
A_k = h^{\circ k} (\beta_\lambda^{-1}(z)) - k
\)
And this should be bounded by a nice sum of exponentials which should converge as \( k\to\infty \)... I believe...
Regards, James
Edit:
Please, Tommy, take a look at my Pari-GP code. This isn't a proposed, maybe it works, solution. I can produce accuracy of 200 decimal places.
The function \( \text{Abl_L} \) pretty much produces \( F_\lambda \); minus some caveats. This function is absolutely holomorphic. The trouble I'm having with is only the real tetration \( \text{tet}_\beta \) which is the real tetration we care about. But \( F_\lambda(s) \) is absolutely holomorphic.