09/29/2021, 11:55 AM
(09/29/2021, 12:33 AM)JmsNxn Wrote: EDIT:
I updated the other thread and handled the case where \( b = e^{-e} \)--no obvious errors, as expected. It runs slower than b = 1/2, but seems fine so far. I'm making a complex plane graph at the moment, and I'll see how it looks. I'm still working with the toy model case which is 2pi i periodic solutions.
I graphed some Taylor series for \( b= e^{-e} \) and there are no errors. The infinite composition method works fine here. Again, I'll say that for \( b > 0 \) we don't fall into the same traps we fall into when talking about Schroder functions. We're solving a schroder equation in the neighborhood of infinity; not a fixed point. So the neutral, attracting, repelling paradigm doesn't matter for us. We don't care about fixed points. All we care about is that \( b^z \) and \( \log_b(z) \) are well enough behaved.
We can always find an asymptotic solution, and we're just trying to solve for an error between the asymptotic and the actual tetration. Again, ember, I don't see anything glaringly wrong. This avoids all the problems that the theta mapping method has.
Anyway, that's enough for tonight. I have a zoom with sheldon tomorrow, to talk about \( b = e \) with \( 2 \pi i \)-period case. I'm focusing on \( b = e \) for now; if I can get this to work perfectly, I'll move on to \( b > 0 \); then if I dare \( b \in \mathbb{C} \).
Keep posting challenges though; let's try to break this method together. Find everything that could break it.
10^-3 is not enough to prove the reliability of your method at the singularity 0/1, I continue to request <10^-5, but base=0.066 test pass is indeed enough to suggest that your method may have avoided the problem for Shell-Thron-region.
If things go in the best direction, this could indeed be the strongest contender for Kneser's method.
Because you may incidentally solve the numerical approximation problem of the super-root function, which is very difficult to solve by Kneser's method for the same reason.
