I found an interesting expression of the fibonacci sequence as iteration of the function

$$f(z) = \frac{1}{1+z}$$

if we look at it:

\begin{align}

f^{\circ 2}(z) &= \frac{1}{1+\frac{1}{z+1}} = \frac{1+z}{2+z}\\

f^{\circ 3}(z) &= \frac{1}{1+\frac{1+z}{2+z}} = \frac{2+z}{3+2z}\\

f^{\circ 4}{z} &= \frac{1}{1+\frac{2+z}{3+2z}} = \frac{3+2z}{5+3z}\\

\end{align}

So we see already the scheme

$$ f^{\circ n}(z) = \frac{\phi_n + \phi_{n-1}z}{\phi_{n+1} + \phi_n z}$$

where \(\phi_n\) is the Fibonacci sequence.

There is a well known continuous formula for the fibonacci sequence:

$$\phi_t = \frac{\Phi^t - \Psi^t}{\Phi -\Psi}, \Phi,\Psi=\frac{1\pm \sqrt{5}}{2}$$

which we can use in our iteration formula.

The linear fraction \(f\) has the fixed points \(\Phi\) and \(\Psi\) (solution of \(1=z+z^2\)).

Unfortunately it is not a nice continuous iteration, because for non-integer t we have an imaginary part (due to \(\Psi\) being negative). This has to do with the pole in the middle between the two fixed points which would not allow for e.g. a real half iterate.

Nonetheless we can look at the complex behaviour (here we look at the function \(f^{\circ t}(z) - z\) to better see the fixed points but still see the poles):

We see how the pole of the iterate moves from one fixed point to the other in an S-spiral (I love S-spirals! lol).

The t-iterates are analytic for each t at both fixed points.

Hence we talk about *regular* iteration. The continuous version of the Fibonacci sequence can be obtained / is regular iteration.

And again we have the case that we can continue the regular iteration from one fixed point to the other (though not through the pole but beside the pole) which is not the case for e.g. the regular iteration of \(b^z\) and probably most other functions (and for which JmsNxn is still owing a proof ). And again JmsNxn would say that we don't have a vicinity around both fixed points where each iterate is holomorphic

$$f(z) = \frac{1}{1+z}$$

if we look at it:

\begin{align}

f^{\circ 2}(z) &= \frac{1}{1+\frac{1}{z+1}} = \frac{1+z}{2+z}\\

f^{\circ 3}(z) &= \frac{1}{1+\frac{1+z}{2+z}} = \frac{2+z}{3+2z}\\

f^{\circ 4}{z} &= \frac{1}{1+\frac{2+z}{3+2z}} = \frac{3+2z}{5+3z}\\

\end{align}

So we see already the scheme

$$ f^{\circ n}(z) = \frac{\phi_n + \phi_{n-1}z}{\phi_{n+1} + \phi_n z}$$

where \(\phi_n\) is the Fibonacci sequence.

There is a well known continuous formula for the fibonacci sequence:

$$\phi_t = \frac{\Phi^t - \Psi^t}{\Phi -\Psi}, \Phi,\Psi=\frac{1\pm \sqrt{5}}{2}$$

which we can use in our iteration formula.

The linear fraction \(f\) has the fixed points \(\Phi\) and \(\Psi\) (solution of \(1=z+z^2\)).

Unfortunately it is not a nice continuous iteration, because for non-integer t we have an imaginary part (due to \(\Psi\) being negative). This has to do with the pole in the middle between the two fixed points which would not allow for e.g. a real half iterate.

Nonetheless we can look at the complex behaviour (here we look at the function \(f^{\circ t}(z) - z\) to better see the fixed points but still see the poles):

We see how the pole of the iterate moves from one fixed point to the other in an S-spiral (I love S-spirals! lol).

The t-iterates are analytic for each t at both fixed points.

Hence we talk about *regular* iteration. The continuous version of the Fibonacci sequence can be obtained / is regular iteration.

And again we have the case that we can continue the regular iteration from one fixed point to the other (though not through the pole but beside the pole) which is not the case for e.g. the regular iteration of \(b^z\) and probably most other functions (and for which JmsNxn is still owing a proof ). And again JmsNxn would say that we don't have a vicinity around both fixed points where each iterate is holomorphic