07/25/2021, 11:59 PM
I understand what you are saying Tommy, but it seems like extra work for no reason. The whole point of having an asymptotic solution is that we can write the tetration as,
\(
\text{tet}_{\text{Tom}}(s+x_T) = \text{Tom}_A(s) + \tau(s)\\
\)
Where \( \lim_{\Re(s) \to \infty} \tau(s) = 0 \). In your case, the error will look like \( \mathcal{O}(e^{-s^2/2}) \) at least. It's much easier to just find this small number; than worrying about taking certain branches of log's on large numbers. It's the difference between calculating \( \log(X) \) for large \( X \) and \( \log(1+\Delta) \) for small \( \Delta \). The latter way is much simpler.
I can prove the Tommy method converges on \( \mathbb{C}/(-\infty,-2] \); it's just a slight adaptation of showing that the beta method converges. I'll do a quick write-up; a lot of it is line for line from my paper. I'm just concerned with whether it's the beta method or not. If it is, it's a much more efficient error than the beta method--and a huge quality of life upgrade.
Regards,
\(
\text{tet}_{\text{Tom}}(s+x_T) = \text{Tom}_A(s) + \tau(s)\\
\)
Where \( \lim_{\Re(s) \to \infty} \tau(s) = 0 \). In your case, the error will look like \( \mathcal{O}(e^{-s^2/2}) \) at least. It's much easier to just find this small number; than worrying about taking certain branches of log's on large numbers. It's the difference between calculating \( \log(X) \) for large \( X \) and \( \log(1+\Delta) \) for small \( \Delta \). The latter way is much simpler.
I can prove the Tommy method converges on \( \mathbb{C}/(-\infty,-2] \); it's just a slight adaptation of showing that the beta method converges. I'll do a quick write-up; a lot of it is line for line from my paper. I'm just concerned with whether it's the beta method or not. If it is, it's a much more efficient error than the beta method--and a huge quality of life upgrade.
Regards,