03/06/2008, 04:58 PM
Gottfried Wrote:What I've got so far is a good result for the half-iterate, which seems to be exact to more than 20 digits. I've also used a shifting of x->x' to get a triangular matrix with exact terms, where I denote
\( \hspace{24}
\begin{matrix}
x' &=& x/(1/4) - 1 \\
x'' &=& (x+1)*1/4 \\
g(z)&=& 1/4 z^2 + 3/2 z
\end{matrix} \)
and with the above then
\( \hspace{24} f(x) = g(x')'' \)
Wah thats strong tobacco, I just had used a *translation*, for example
\( x' = x-\frac{1}{4} \)
\( x'' = x+\frac{1}{4} \)
this \( g_1(x)=f(x')'' \) is \( f \) with the fixed point \( \frac{1}{4} \) moved to 0. And \( g_2(x)=f(x'')' \) corresponds to moving the fixed point \( -\frac{1}{4} \) to 0.
Computing the half iterate with your triangular matrices corresponds to computing the regular fractional iterate at the corresponding fixed point (to get \( f^{\circ 1/2} \) you of course have to move back the fixed point 0 of \( g_1 \) or \( g_2 \) to its original place. The half iterate at both fixed points is different.
And from here you already see variants:
1. You dont need to use a translation. You can use any bijective (analytic) function \( \tau \) which moves the fixed point to 0 (you used a *linear* transformation). You then take the fractional iterate of \( g=\tau^{-1}\circ f\circ \tau \) and get the fractional iteration of \( f \) as \( f^{\circ t}=\tau\circ g^{\circ t}\circ \tau^{-1} \).
2. The matrix operator method is more general. It does not only work at fixed points but also in between where you use the triangulated finite matrices as approximating sequence. So you can use *any* translation and then apply the matrix operator method. And it is quite sure that all those results differ (by a super small amount).
3. And then you can go even a step further and use any (analytic) bijective \( \tau \) before applying the matrix operator method.