07/21/2010, 03:19 AM
(This post was last modified: 07/21/2010, 05:06 PM by sheldonison.)
(07/20/2010, 08:31 PM)tommy1729 Wrote: how do you know the superfunction of 2sinh is periodic in the imaginary direction ?It is interesting. The period is based on the limiting behavior of 2sinh at its fixed point of zero, where the slope=2. If -real(z) is large enough, then 2^z becomes an excellent approximation of the SuperFunction of 2sinh. The limit equation for the superfunction of 2sinh in the complex plane is:
that is intresting.
\( \operatorname{SuperFunction}(z) = \lim_{n \to \infty}
\operatorname{2sinh}^{[n]}(2^{z-n}) \)
Since 2^z is periodic in the imaginary, with period = i*2pi/ln(2), then the SuperFunction of 2sinh is also periodic with that period. This leads to some interesting behavior (pointed out in earlier posts). The SuperFunction grows superexponentially negative at img(z)=i*pi/ln(2), with img(f)=0. At img(z)=0.5i*pi/ln(2), the SuperFunction has real(f)=0, and converges to an imaginary fixed point.
- Sheldon