(02/03/2014, 10:35 PM)tommy1729 Wrote: Im not sure what you meant by reflexion. I noticed the fixpoint is a red x in your plots and there are also green x's. I guess that relates.The green x is the position of the base. With "reflexion" I hope to introduce a meaningful expression for the impression, that the two points on the opposite sides of the "circle" - opposite with respect to the fixpoint- look like mirrored images of each other - and are also just looking like half-circle rotations of each other (which would then be a very nice and significant property).
With the <not-yet-determined-imaginary-value> yes, that 1.3*2*pi is perhaps a good idea. I think it should have a simple solution; I just don't get it at the moment... [update] It should be 2*Pi*I/v where v= log(log(LH(1.3))) where the function t=LH(b) gives the fixpoint t such that b^t=t and v is about 1/v~-1.0503
With the other questions - I do not yet know...
Gottfried
Gottfried Helms, Kassel
