Do complex iterates of functions have any meaning?

Using a method explained in this answer it is possible to calculate not only integer and real iterates of functions but also complex ones, for example, the $i$-th iterate, where $i=\sqrt{-1}$. Here are graphs of the $i$-th iterates of some common functions (the blue is the real part and the red curve is the imagine part):

$$\arctan^{[i]}(x)$$

$$\sin^{[i]}(x)$$

So the questin is whether there is any intuitive meaning in complex iterates, especially, say, $i$-th iterates of functions?

-
I put a number of original articles at zakuski.utsa.edu/~jagy/other.html of which the obituary of Baker is a good first read. I included an early draft of Milnor on complex dynamics. I left out one very nice book, Daniel Alexander, A History of Complex Dynamics. Meanwhile, Baker and his student Liverpool alter the question, instead of talking about iterates they talk about formal power series that commute with each other. – Will Jagy Jul 27 '11 at 19:20
It may be useful to point out that the term "complex iterates" usually refers to integer iterates of functions of complex variables, which is not the use here. – j.c. Jul 27 '11 at 19:42
That Alexander book is excellent. Lots of historical remarks, too. Section 2.2, "Analytic Iteration". – Gerald Edgar Jul 27 '11 at 20:28
If you consult the "Related" questions automatically found by the software, you will find other interesting things. – Gerald Edgar Jul 27 '11 at 20:31
Hi Gerald. I put the book on the web page. Alexander did not answer his phone, I guess I will email him to ask whether it is alright. My impression is that the book is no longer for sale, and Amazon has a single used copy for about 600 dollars. But that does not guarantee Prof. Alexander will be comfortable with his book being online. – Will Jagy Jul 27 '11 at 20:33

The difficult case is around a fixed point of a function with derivative one. Irvine Noel Baker, 1932-2001, studied these from the viewpoint of formal power series with complex coefficients, beginning with some $f(z) = z + a_{m+1} z^{m+1} + \ldots, \; a_{m+1} \neq 0.$ He changed the question to finding those $$f_\lambda(z) = z + \lambda a_{m+1} z^{m+1} + \sum_{n = m+2}^\infty b_n(\lambda) z^n$$ which commute with $f.$ For a given $f = f_0,$ there may or may not be any other $f_\lambda$ such that the power series is convergent near $z=0.$ The big theorem, with one case taken care of by his student Liverpool, is that the set of $\lambda$ for which $f_\lambda(z)$ converges near $0$ is one of three sets: (a) $\{ 0 \},$ (b) with some fixed $N \in \mathbb Z,$ the fractions $\{m/N, \; \mbox{all} \; m \in \mathbb Z\},$ or $\mathbb C$ itself. In the final case, where any complex $\lambda$ is allowed, Baker called the function $f$ embeddable, saying that the function is embeddable in a continuous group of analytic iterates.

In case (b) there is some minimal $1/N$th iterate which cannot be further, um, divided. So there may be half-iterates of something without there being any one-third iterates.

My summary would be that Baker makes it quite sensible to talk about an $i$ iterate. The conceptual switch from trying to do half iterates to asking what formal power series commute with a given formal power series makes the whole thing tractable.

Oh, original articles and books posted at

EDIT: I found some of my notes from 2010. From what I can make out, the only example that we expect to be really pleasant is the family of linear fractional transformations $$f_\lambda(z) = \frac{z}{1 + \lambda z}$$ which all comute with each other, and nothing worse happens than a pole for each one at $z = -1 / \lambda.$ Note the group law $f_\lambda \circ f_\gamma = f_{\lambda + \gamma}$ I felt that all other embeddable families were essentially that, just take some holomorphic $h(z)$ with $h(0) = 0$ and $h'(0) = 1$ and get the very similar $$f_\lambda(z) = h^{-1} \left( \frac{h(z)}{1 + \lambda h(z)} \right),$$ with Fatou coordinate $$\alpha(z) = \frac{1}{h(z)}.$$ There is a bootstrapping method for solving for the Fatou coordinate $\alpha(z)$ which is probably due to Ecalle. I also noted $\beta(z) = \frac{- h^2(z)}{h'(z)}$ but I forget what $\beta$ was for. No, here we go, it is an explicit description in KCG on solving for the Fatou coordinate, pages 346-352, Iterative functional equations by Marek Kuczma, Bogdan Choczewski and Roman Ger. In general $\beta(z) = 1 / \alpha'(z).$

Note, though, that we have now introduced possible bad behavior when either $h(z)$ or, more likely, $h^{-1}(z)$ are undefined, in short we have probably severely curtailed the region of $\mathbb C$ where things are working well.

Edit toooo: the Fatou coordinate may be defined on only a sector out of the origin, anyway $$\alpha(f(z)) = \alpha(z) + 1.$$ Then we get a family (but maybe only in a sector) by $$f_\lambda(z) = \alpha^{-1}( \lambda + \alpha(z) ),$$ where $f_1 = f$ in this recipe. So once again, as in the linear fractional transformations, we can plug in $\lambda = i.$

-
How do we make curly braces here, as is usual for defining a set? – Will Jagy Jul 27 '11 at 20:23
Interesting material. Of course investigation of the power series version goes back to Caley (1860). – Gerald Edgar Jul 27 '11 at 20:28
@Will: backtick dollar backslash{ a, b, c backslash} dollar backtick. – Joseph O'Rourke Jul 27 '11 at 20:30
Joseph, it worked. – Will Jagy Jul 27 '11 at 20:38
backtick is the British name for this character ` – Gerald Edgar Jul 27 '11 at 21:33

Complex iterates of linear operators on Banach spaces, in particular imaginary iterates, have quite a lot of meaning in operator theory and they have applications to, among others, abstract parabolic equations.

Given a sectorial operator $A$, i.e. a linear closed injective densely defined operator $A$ on a Banach space $X$ such that $(-\infty,0)$ is contained in the resolvent set of $A$ and $$\sup_{t<0}\|t(t-A)^{-1}\|$$ is finite, we say that $A$ admits bounded imaginary powers if the operators $(A^{is})_{s\in\mathbb{R}}$ form a $C_0$-group of bounded operators on $X$ where $A^{is}$ is defined via a suitable functional calculus.

As far as I know there is no reasonable partial differential operator on $L^p(\Omega)$ with $1<p<\infty$ known not to admit bounded imaginary powers (at least after a suitable translation along the real axis); the situation changes once we pass to $\Psi$DOs, though.

If an operator $A$ admits bounded imaginary powers this has remarkable consequences:

1. If $X$ is a $UMD$-space and there is $\theta\in (0,\frac{\pi}{2})$ such that the group $(A^{is})_{s\in\mathbb{R}}$ satisfies $\|A^{is}\|\leq Ce^{\theta |s|}$ for all $s\in\mathbb{R}$ then the operator $A$ has the maximal regularity property by a result of Dore and Venni.
2. The domain of the complex powers $A^z$ of $A$ for $\Re z\geq 0$ can be obtained using complex interpolation: $$D(A^z)=\left[X,D(A^k)\right]_{\frac{\Re z}{k}}$$ for $k\in\mathbb{N}$ with $k>\Re z$.
3. If $X$ is a Hilbert space then the functional calculus $f\mapsto f(A)$ for bounded holomorphic $f$ is continuous with respect to the norm topology.

A good source for this and related aspects of operator theory is the book Functional calculus for sectorial operators by Markus Haase.

-

I'm discussing this from the view of iterated exponentiation (although the technical process should be the same with other functions as well).

If you can use the Schroeder-function for the continuous iteration, then the iteration-height-parameter (say "h") goes into the exponent of some basis (the log of the fixpoint, often denoted as $\small \lambda$ ). Imaginary heights h then switch the value of the schröder-function to the negative; this allows then to extend the iteration beyond "infinite height".

For instance, use base $\small b = \sqrt 2$ for iterated exponentiation, $\small z_0=x, z_1=b^x , z_2=b^{b^x}, \ldots$. Then if you begin at, say, $\small z_0=x=1$ you can iterate to infinite height to approach the limit at $\small z_\infty = 2$ . If you start at $\small z_0=x=3$ you can approach $\small z_\infty = 2$ or even $\small z_{-\infty}=4$ . But you cannot iterate from a value $\small z_m<2$ to a value $\small 2 < z_w < 4$ using real heights, even when infinite.

But if you use the imaginary unit height you iterate directly from $\small z_m=1$ to something like $\small z_{m+i}=2.4$.

Assume again $\small z_0=1$. Then the value of the schröder-function (which is assumed to be normed to have the powerseries $\small \sigma(x)= 1x+\sum_{k>1} a_k x^k$ ) is about $\small s=-0.316049330525$. Then $\small \sigma^{o-1}( \lambda^1 s)\cdot 2 +2=b=\sqrt 2$ because that is the iteration of height 1 (in the exponent of $\small \lambda$ ).
If we replace that exponent by $\small h_w = i \cdot {\pi \over \ln \lambda }$ then we get $\small \sigma^{o-1}( \lambda^{h_w} s) \cdot 2 +2=2.46791405022...$ which is, in some sense "beyond infinity" with respect to the iteration height.

-