Well, Henryk,
that opended eyes, thanks. With the following parametrization I got the difference.
Starting with the general
x' = x/e-d, x"=(x+d)*e, e*d=fixpoint
and fixing e=1
e= 1 , d0=t0=fixpoint_0, d1=t1=fixpoint_1
and development of the matrices accordingly I got differences, too. So thanks for the hints!
Now I'd like to understand, what is the source of the difference of selection e=1 vs d=1. The most obvious difference was in the given case
f(x) = x^2 + x - 1/16 = x + (x - 1/4) (x + 1/4) = x + (x-t0)*(x-t1)
that I kept the eigenvalues in both developments equal. Maybe my transformation introduces somehow a triviality - but note, that in my tetration-examples I also used d=1, e0=t0 versus e1=t1 for the definition of the triangular matrices, just this selection, that you say is "illegal"
I tried with a general definition of a quadratic function (fixpoints p,q)
f(x) = x + (x - p) (x - q) = x^2 + (1 - (p+q))x + pq
g(z) = a*z^2 + b*z
z = x/e-d
f(x) = (g(z) + d)*e
By comparing coefficients at powers of x in f(x) and g(z) I get conditions for a and b in g(z):
a = e
b = 1 - (p+q) + 2de , b-1 = 2de - (p+q)
and then for d and e the condition occurs
(de - p) (de - q) = 0
where either e or d is free.
It seems then, if b equals for each selection of fixpoints, such that either de=p or de=q and the resulting b's are equal, then the resulting powerseries seem to be compatible and give equal results (I've to check that with more examples)
But again: maybe this introduces some triviality... On the other hand: it is intriguing, that using d=1 , e=fixpoint is the appropriate selection to convert the tetration-matrices into triangular matrices for decremented iterated exponentiation.
Hmmm.
Gottfried
that opended eyes, thanks. With the following parametrization I got the difference.
Starting with the general
x' = x/e-d, x"=(x+d)*e, e*d=fixpoint
and fixing e=1
e= 1 , d0=t0=fixpoint_0, d1=t1=fixpoint_1
and development of the matrices accordingly I got differences, too. So thanks for the hints!
Now I'd like to understand, what is the source of the difference of selection e=1 vs d=1. The most obvious difference was in the given case
f(x) = x^2 + x - 1/16 = x + (x - 1/4) (x + 1/4) = x + (x-t0)*(x-t1)
that I kept the eigenvalues in both developments equal. Maybe my transformation introduces somehow a triviality - but note, that in my tetration-examples I also used d=1, e0=t0 versus e1=t1 for the definition of the triangular matrices, just this selection, that you say is "illegal"
I tried with a general definition of a quadratic function (fixpoints p,q)
f(x) = x + (x - p) (x - q) = x^2 + (1 - (p+q))x + pq
g(z) = a*z^2 + b*z
z = x/e-d
f(x) = (g(z) + d)*e
By comparing coefficients at powers of x in f(x) and g(z) I get conditions for a and b in g(z):
a = e
b = 1 - (p+q) + 2de , b-1 = 2de - (p+q)
and then for d and e the condition occurs
(de - p) (de - q) = 0
where either e or d is free.
It seems then, if b equals for each selection of fixpoints, such that either de=p or de=q and the resulting b's are equal, then the resulting powerseries seem to be compatible and give equal results (I've to check that with more examples)
But again: maybe this introduces some triviality... On the other hand: it is intriguing, that using d=1 , e=fixpoint is the appropriate selection to convert the tetration-matrices into triangular matrices for decremented iterated exponentiation.
Hmmm.
Gottfried
Gottfried Helms, Kassel