07/02/2022, 10:37 AM
(06/10/2022, 11:27 PM)JmsNxn Wrote:(06/09/2022, 05:58 PM)bo198214 Wrote: I think it is based on regular iteration of the formal power series of sqrt(2)^x at fixed point 2.
If you look at https://www.researchgate.net/publication..._tetration
chapter 2.3.1 under "Regular C-iteration" you have a similar formula, though the infinite sum is not sorted by powers of s (in your case s=(ln2)^x ) but by powers of z, while in your case z=0 (at the regular iteration of f, not h). I guess after sorting it by powers of s you would arrive at the given formula (and know the further terms).
But this really involves quite some understanding and calculation, would be great if someone has some time to explicit it here.
I'm confused, isn't this just the Fourier expansion of \(F(x) = \sqrt{2} \uparrow \uparrow x\)?
\[
F(x) = 2 + \sum_{k=1}^\infty c_k \log(2)^{kx}\\
\]
At least, this second order expansion agrees as I look at it (you can just massage this to make the formal power series regular iteration method). This isn't much more than:
\[
\Psi^{-1}(\log(2)^x\Psi(1))\\
\]
For the Schroder function \(\Psi\). Am I missing something?
No, you are not missing anything. So can you calculate some more series coefficients, and show Catullus how its done?
