08/13/2010, 12:28 AM
(This post was last modified: 08/13/2010, 02:05 AM by sheldonison.)

(08/11/2010, 04:10 PM)tommy1729 Wrote: .... finding the parametric real and im part of ' kneser ' might be as complicated or even equivalent to riemann mapping.I freely admit that my higher mathematics education has been stretched way beyond its limits. My wife thinks I should go back to school, so I can get an advanced degree and teach at a University. Anyway, all that aside. I don't have a really good background in understanding the Riemann mapping theorem, this despite having stumbled upon an algorithm to generate the Riemann mapping for tetration. I gather from reading on the internet, that the constructive Riemann mapping approaches are matrix based, and very slowly converging in the case of a singularity (which of course, is our case).

and i assume the coo prop of the parametric real and im part is equivalent to the coo prop of ' kneser ' or ' sexp '.

I would assume that except at the singularity, one could continue the function, and that the theta(z) at the real axis would have to be analytic, coo.

By the way, thanks for your comments Tommy!

Quote:....I'm not sure if by Kneser curve you mean sexp(z), which does have one inflection point somewhere near -0.5.

but im not finished...

about that wave again

because of kneser(z+1) = exp(kneser(z))

it is clear that the kneser curve has period 1.

kneser(0) = 0 and kneser(1) = 1.

if we draw a straith line trough them , then i assume the curve must be above or below the line and never cross it.

is that true ?

regards

tommy1729

If you're talking about the theta(z) function, it has a really really bad singularity. As z superexponentially approaches the singularity at integer values, the behavior gets worst and worst, than the plots I posted along with the equations. The imag(z) oscillates, going up and down, and the real(z), which looks like it might converge to a value, instead super-logarithmically oscillates, growing to +real infinity, while oscillating in the imag(z)! I'll plot an updated post some time, along with some comments.

- Sheldon