03/09/2008, 09:30 PM

Gottfried Wrote:I tried to find one e, for which the resulting powerseries are linear scalings of each other, such that

g0(x)=c * g1(x)

This is possible.

Assume again, that, denoting the fixpoints p and q, such that

f(x) = x + (x - p) (x - q)

Denote also the substitution

z = x' = x/e - d

and the backsubstitution

x" = (x + d)*e

Recall, that for a fixpoint p, d and e must be selected that p = e*d

If I select

then the resulting G0 and G1-matrices provide the same powerseries with a linear scaling.Code:`´`

e0 = 1 + (p-q) d0=p/e0 // first fixpoint

e1 = 1 + (q-p) d1=q/e1 // second fixpoint

Ok, short verification:

right. Thats indeed interesting.

Quote:Anyway- this does not (yet?) provide a linear scaling for the half-powers

Yes and thats the critical point here. There is no such law like

but at least we have traced back the regular iteration at different fixed points to the regular iteration of different scalings. Does this also work for higher order polynomials?