bo198214 Wrote:However your second transformation
,
,
has as translationthe same translation as in your previous transformation!
Onlychanges to
.
So this second transformation moves also the *greater* fixed point to 0. And we have seen that we can dismiss any multiplicative conjugationwith respect to regular iteration. So it has of course the same result (fractional iteration of
) as your first transformation.
Yepp - I just found the position of the error, grmmpff. It is in
Code:
´
using e1=1/4 e2 = -1/4
d1=1/4 / e1 =1 d2 = 1/4 / e2 = -1
x' = 4x - 1 x´ = -4x + 1
x" = (x+1)/4 x´´= (x-1)/-4
Although I had already that not e or d, but e*d = fixpoint, in the computation of d2, I used the same (1/4) as for d1. For clarity, I should have written a symbol for the selected fixpoint:
Code:
´
involving fixpoint p=1/4 fixpoint q=-1/4
using e1=1/4 e2 = -1/4
d1=p / e1 =1 d2 = q / e2 = 1
x' = 4x - 1 x´ = -4x - 1
x" = (x+1)/4 x´´= (x+1)/-4
Hope this corrects things, didn't check it again yet, but it seems that this reflects just the simple sign-changing in the matrices.
The previous G2-matrix and the subsequent considerations in that posting are on a false assumption...

Gottfried
Gottfried Helms, Kassel