09/14/2010, 11:33 PM
(09/14/2010, 10:30 PM)sheldonison Wrote: 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 ...
