Gottfried Wrote: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
Hm, I had another look at your transformations (which indeed puzzled me). I came up with this:
Your first transformation:
\( x' = 4 x -1 \), \( x'' = \frac{1}{4} x + \frac{1}{4} \), \( g_0(x)=f(x'')' \)
gives the same result as the regular iteration at the upper fixed point, because:
\( x'' = \tau\circ \mu(x) \) where \( \tau(x)=x+\frac{1}{4} \) and \( \mu(x)=\frac{1}{4}x \)
so \( g_0(x)=\mu^{-1}\circ \tau^{-1}\circ f\circ\tau\circ\mu=\mu^{-1}\circ h_0 \circ \mu \) and \( h_0 \) is the function with upper fixed point moved to 0.
But now it is easily derivable that the regular iteration \( (\mu^{-1}\circ h\circ \mu)^{\circ t}=\mu^{-1}\circ h^{\circ t}\circ \mu \), where \( h^{\circ t} \) is the regular iteration of \( h \). This has to do with that the regular iteration is the only one which posses the limit \( \lim_{x\downarrow 0} (f^{\circ t})'(x)=\left(\lim_{x\downarrow 0} f'(x)\right)^t \).
So we have \( f_0^{\circ t}= \tau\circ \mu\circ g_0^{\circ t}\circ \mu^{-1}\circ \tau^{-1}=\tau\circ h_0^{\circ t}\circ \tau^{-1} \).
However your second transformation
\( x' = -4 x +1 \), \( x'' = -\frac{1}{4} x + \frac{1}{4} \), \( g_1(x)=f(x'')' \)
has as translation \( \tau(x)=x+\frac{1}{4} \) the same translation as in your previous transformation!
Only \( \mu \) changes to \( \mu(x)=-\frac{1}{4}x \).
So this second transformation moves also the *upper* fixed point to 0. And we have seen that we can dismiss any multiplicative conjugation \( \mu^{-1}\circ f\circ \mu \) with respect to regular iteration. So it has of course the same result (fractional iteration of \( f \)) as your first transformation.