Alternate solution of tetration for "convergent" bases discovered

[tommy1729]

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: \( L=B^L \), \( \log(L)=\log(B^L) \), \( \log(L)=L\times\log(B) \)
\( \text{Period}=2Pi*i/(L\times\log(B) + \log(\log(B))) \)

hmm i was looking for the superduper general case of how to find the period of the regular super of any function...

i believe such a formula exists .. perhaps even mentioned before ... perhaps even by you ... perhaps in thread " using sinh " ?

i seem to have such a kind of deja vu ...

[sheldonison]

... Not sure exactly what you'd want to know. It seems to be similar to the "Cauchy integral" for bases outside and in some parts of the STR, but the weird part is that when it is used for \( 1 < b < e^{1/e} \), it gives the attracting regular iteration. ???

This makes sense, in that I'm fairly certain that the attracting regular iteration is a 1-cyclic function of the entire regular superfunction. http://math.eretrandre.org/tetrationforu...hp?tid=515 At least, it seems to be the case for base e^(1/e), and for sqrt(2). I'll half to read up on your post, to try to better understand your algorithm, (and ideally, your numerical approximations as well, and how you iteratively develop the solution).
- Sheldon

[mike3]

This makes sense, in that I'm fairly certain that the attracting regular iteration is a 1-cyclic function of the entire regular superfunction. http://math.eretrandre.org/tetrationforu...hp?tid=515 At least, it seems to be the case for base e^(1/e), and for sqrt(2). I'll half to read up on your post, to try to better understand your algorithm, (and ideally, your numerical approximations as well, and how you iteratively develop the solution).
- Sheldon

I'll post the details of the numerical algorithm in the Computation forum, then.