[exercise] fractional iteration of f(z)= 2*sinh (log(z)) ?

[JmsNxn]

Hi,     
 in a thread in math.stackexchange.com (limit of a recursive function) I came over the question of what could be a closed form for   
\(
z_{k+1} = z_k - \frac 1{z_k}\\
z_0 = z \in \mathbb{R}\\
t = f(z_0) = \lim_{n \rightarrow \infty} z_n\\
\)

 I fiddled a bit with it and its reverse operation, finding interesting properties, for instance the existence of periodic points of any order, after another contributor showed that the only 1-periodic point (=fixpoint) would be infinity. (see https://math.stackexchange.com/a/4056192/1714)   

To extend my knowledge about this sequence/function beyond that MSE-discussion I pondered the possibility of fractional iteration (or as one might say: indefinite summation) but couldn't find a promising ansatz to establish such a routine. However, further thinking showed, that the recursive expression could as well be as iteration of the function g(z)= 2*sinh (log(z)) , and since we had discussions here about iteration of 2*sinh(z) there might as well be an idea for the fractional iteration of the form g(z).        

Someone out here with an idea? (Feel free to contribute to the thread in MSE)

Gottfried

Convert the fixed point at infinity to a fixed point at zero via the conjugation \( z \mapsto 1/z \)

So that we have a new sequence,

\(
w_{k+1} = \frac{1}{\frac{1}{w_k} - w_k}\\
\)

Call,

\(
\phi(\xi) = \frac{1}{\frac{1}{\xi} - \xi}\\
\phi'(\xi) = \frac{1}{(\frac{1}{\xi} - \xi)^2} (\frac{1}{\xi^2} - 1)\\
\phi'(0) = \lim_{\xi \to 0} \frac{1}{\xi^2} \frac{1}{\frac{1}{\xi^2} - 1} = 1\\
\)

So we have a neutral fixed point at zero. At this point, the very stable route would be to produce an Abel function using Ecalle's method, which will converge not in a neighborhood of zero but on a petal by zero. This will give a fractional iteration, and furthermore; one which is unique. Then conjugate back to get the original sequence.

Although I haven't published anything on this, there are situations where you can use a different method. It's difficult when the fixed point is neutral, but seems to work in the cases I've tried (for instance for \( \sin(\xi) \) on the real line in a neighborhood of \( 0 \) (real neighborhood)).

That would be to use Ramanujan's master theorem. We write this as,

\(
\Gamma(1-z) \phi^{\circ z}(\xi) = \sum_{n=0}^\infty \phi^{\circ n+1}(\xi) \frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty \phi^{\circ n+1}(\xi) \frac{(-x)^n}{n!})x^{-z}\,dx\\
\)

This form has the added benefit of being the exact mechanism I use to solve the indefinite sum. Now I'd tread carefully here because the fixed point is neutral--I'm only able to do this absolutely rigorously when \( 0 < \phi'(0) < 1 \); and in some cases have managed to use a sequence of \( \phi \) functions with geometric fixed points, to converge towards a neutral solution (as I managed to do with \( \sin \) which largely convinced me of the result). Quite frankly I would be much more confident in this expression if we took a \( 0<q<1 \) and instead talked about the iteration,

\(
w_{k+1} = \frac{q}{\frac{1}{w_k} - w_k}\\
\)

Whereby we can fractionally iterate this using the above method--but taking the limit \( q \to 1 \) proves to be tricky, especially when we talk about where \( \xi \) lives (luckily for the \( \sin \) function we know the real-line will be invariant under this process so just stick to the real line). In the general case it's necessary we discuss the deformation of the basins of attraction as we take \( q \to 1 \); and even more troubling, we can't simply restrict ourselves to a neighborhood of the fixed point, because necessarily it will diverge for some points in that neighborhood (the nature of neutral fixed points, they're not attractive in a neighborhood).

This method is also the basis for constructing bounded tetration for \( 1 \le \alpha \le \eta = e^{1/e} \); where at \( \eta \) we do much the same thing, iterate at a neutral fixed point using Ramanujan's Master Theorem by approximating uniformly using geometric fixed points \( \alpha \to \eta \). Though again, this is a special case because the exponential map is just so pretty (as is the \( \sin \) function).

All in all, I would say the safest bet is to use Ecalle's method of constructing an Abel function on a petal by the fixed point--but the Ramanujan method does work in certain instances; just be careful when discussing domains.