What is the functional inverse (with respect to $h$ (!)) of $f^{\circ h}(x)={F(h) +x F(h-1) \over F(1+h) +x F(h) }$?

Question

I've considered the fractional iteration of $f(x) = {1 \over 1+x} $ for which the general expression depending on the iteration-height parameter $h$ might be assumed by the formula $$ f^{\circ h}(x) = f(x,h)={F(h) +x \cdot F(h-1) \over F(1+h) +x \cdot F(h) }$$ where
$ \displaystyle \qquad \small F(h)=\operatorname{fibonacci}(h) = {\varphi^h - (1-\varphi)^h \over \sqrt5 }$ and

$ \displaystyle \qquad \small \varphi = {1+\sqrt5 \over 2} \sim 1.618... $

(Note, that for fractional iterates we need $x \in \mathbb C$)

Today I tried to find a formulation for the functional inverse with respect to h, but don't find a good starting point so:

Q: How would look a function $ h = \operatorname{hgh}(x_0,x_h) $ which would indicate the required iteration-height $h$ given $x_0$ and $x_h = f(x_0,h)$ ? Is there even a closed form for it (I'd call it still closed-form if it possibly includes e.g. the Lambert-W function)

[update]: for more background see the older exercise of mine

Answer 1

Fractional linear transformation $f$ can be represented by the $2\times 2$ matrix $\displaystyle \left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)$, which is diagonalized as $$\left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right)= P\left(\begin{array}{cc} \varphi & 0 \\ 0 & 1-\varphi\end{array}\right)P^{-1},\qquad P=\left(\begin{array}{cc} \varphi-1 & -\varphi \\ 1 & 1\end{array}\right).$$ Therefore the matrix corresponding to $h$th iteration is $$P\left(\begin{array}{cc} \varphi^h & 0 \\ 0 & (1-\varphi)^h\end{array}\right)P^{-1}= \frac{1}{\sqrt{5}}\left(\begin{array}{cc} (\varphi-1)\cdot\varphi^h +\varphi\cdot(1-\varphi)^h & \varphi^h -(1-\varphi)^h \\ \varphi^h -(1-\varphi)^h & \varphi\cdot\varphi^h +(\varphi-1)\cdot(1-\varphi)^h \end{array}\right)$$ and we obtain $$\left(\frac{1-\varphi}{\varphi}\right)^{h}=\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}.$$ Taking the logarithm, one finds the expression $$\mathrm{hgh}(x_h,x_0)=\ln\frac{x_hx_0+\varphi x_h-(\varphi-1)x_0-1}{x_hx_0-(\varphi-1)x_h+\varphi x_0-1}\Bigl/\Bigr.\ln\frac{1-\varphi}{\varphi}.$$ Note, however, that the expressions in the logarithms can become negative or even complex (in the denominator logarithm it is always negative), so some care should be taken to choose correct branches.

Answer 2

Ahh, the hints to the matrices and to the superfunction were useful. They forced me to look more seriously in my own exercise - and of course the Abel-function is the log of the Schröder-function and in the exercise I've already defined a very simple series for the Schröder-function.

Let's denote $ \small \beta = \varphi-1 \sim 0.618...$ and $ \small u =-\beta^2$ then according to my own formula 2.7 the (recentered) Schröder function is $$ \sigma (t) = -\sqrt5 \sum_{k=1}^\infty (-{t \over \sqrt5})^k $$ and the inverse schröderfunction $$ \sigma^{-1} (t) = \sqrt5 \sum_{k=1}^\infty ({t \over \sqrt5})^k $$ such that the continuous iterate is $$ f(x,h) = \sigma^{-1} \left(u^h \cdot \sigma( x - \beta) \right) + \beta $$

The height-function $hgh(x_h,x_0)$ can be taken from the logarithm to the base $u$ of the ratio of the schröder-values as $$ \operatorname{hgh}(x_h,x_0) = \log_u \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) $$ and if we use the fact, that the Schröder-function is just a geometric series and can be expressed in a closed form we get finally: $$ \operatorname{hgh}(x_h,x_0) = \log_u \left( {x_h-\beta \over x_0 - \beta } \cdot{ \sqrt5 + x_0-\beta \over \sqrt5 + x_h-\beta } \right) $$

Because $u$ is negative we get complex values for fractional $h$ and this must then be handled by selection of the correct branch of the complex logarithm.

Also it must be handled, that the trajectory of iterations with real heights form a spiral through the complex plane with one round per 2 units in the height and thus the winding-number for heights $h \ge 2 $ must be reflected, say for the example $hgh(2/3,1) = 2$ we need to introduce that correction writing/branching of log $$ \cdots = { \log \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) + 2 \pi i\over \log u} $$ otherwise we get "the shortest distance" expressed by a complex height of $\small 0.171595 + 0.560130 i$ instead.

But that are details which I can now work on by myself. Thanks for the hints - sometimes one needs only an external impulse to recover a unremembered concept ...