(07/21/2022, 11:39 AM)MphLee Wrote: How the previous rank' omega constant pops out in this forumla?Hmm, no not exactly
Quote:\(\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\
\)
Can be this extended to
Quote:\(\alpha \uparrow^{n+1} z = \Psi_{n,\alpha}^{-1}\left(\Psi_{n,\alpha}(1) (\log(\alpha)\omega_n)^z\right)\\
\)
Where the \(\Psi_{n,\alpha}\) is the Schroeder of \(\alpha \uparrow^n-\).
Where you have \(\log(\alpha)\omega_n\) you should have \(\lambda(\alpha) = \frac{d}{dz}\Big{|}_{z=\omega_n} \alpha \uparrow^n z\).
We just get lucky with the exponential fixed point, the derivative formula satisfies \(\frac{d}{dz}\Big{|}_{z=\omega_1} \alpha^z = \log(\alpha) \omega_1\).

