Alternate solution of tetration for "convergent" bases discovered

[sheldonison]

Now, onto B=1.33+1.28i. The fixed repelling point is
L1 = -1.13500 - 1.98958i
with a corresponding period of -3.6051+2.4470i.
\( \text{period}(B)=2Pi*I/(L*\log(B)+\log(\log(B))) \)

its seems familiar , so forgive me for not having my day , but period of what ? a kind of sexp with base B i assume. which sexp ? computed how ?

i wanted to ask about the formula , but i think it follows when i see the answer to the above.

forgive me if i ask this again or if it has been explained before.

The period for the regular superfunction, is a somewhat complicated mess. As z->-infinity, the behavior of the regular superfunction is approximated by an exponential. This is the formula I use.

\( \text{RegularSuper}_{B}(z) = \lim_{n \to \infty}
B^{[n](L + {(L\times\ln(B))}^{z-n})} \)

The next step, is to figure out what the periodicity is, taking into account because L is a fixed point, then B^L=L.

As z -> -infinity, \( \text{RegularSuper}(z) = L+{(L\times\log(B))}^z \)
\( \text{Period}=2Pi*I/\log(L*\log(B)) \)
\( \text{Period}=2Pi*i/(\log(L) + \log(\log(B))) \)
substitute: \( L=B^L \), \( \log(L)=\log(B^L) \), \( \log(L)=L\times\log(B) \)
\( \text{Period}=2Pi*i/(L\times\log(B) + \log(\log(B))) \)