the inconsistency depending on fixpoint-selection

[bo198214]

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

Code:

´
    e0 = 1 + (p-q)    d0=p/e0          // first fixpoint
    e1 = 1 + (q-p)    d1=q/e1          // second fixpoint

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

Ok, short verification:
\( \begin{align*}
g_0(x)&=f(x'')'\\
&=(x''+(x''-p)(x''-q))'\\
&=(xe_0+p + (xe_0+p-p)(xe_0+p-q))'\\
&=\frac{xe_0+e_0d + (xe_0)(xe_0+p-q)}{e_0} - d \\
&= x+x(xe_0+p-q)\\
&= x^2e_0 + x(1+p-q) = e_0(x^2+x)
\end{align*} \)
\( g_1(x)=e_1(x^2+x) \)
right. Thats indeed interesting.

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
\( (cf)^{\circ t}=c^tf^{\circ t} \)
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?