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Using a method explained in this answer to How to solve $f(f(x)) = \cos(x)$?, 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 imaginary part):

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

i-th iterate of arctangent

$$\sin^{[i]}(x)$$ i-th iterate of sine

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

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    $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Jul 27, 2011 at 19:20
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    $\begingroup$ 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. $\endgroup$
    – j.c.
    Jul 27, 2011 at 19:42
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    $\begingroup$ That Alexander book is excellent. Lots of historical remarks, too. Section 2.2, "Analytic Iteration". $\endgroup$ Jul 27, 2011 at 20:28
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    $\begingroup$ @Will: It's fairly likely that Professor Alexander does not control the copyright on his book, since it's fairly standard now, and was very standard in 1994, for the author to transfer the copyright to the publisher. And even now, if an author retains the copyright, the publisher often insists on an exclusive right to publish and distribute the material. I'm not saying that this is a good system, but probably from a legal standpoint, you'd need permission from Friedr Vieweg & Sohn Verlagsgesellschaft, who published the book. (It is irrelevant whether the book is for currently available.) $\endgroup$ Jul 27, 2011 at 22:35
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    $\begingroup$ @L Spice: That's interesting. So probably the copyright reverted to the author, which means that the author could legally have someone scan a copy and then put it on the web for free download. $\endgroup$ Jul 28, 2011 at 2:43

5 Answers 5

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Added, May 2023: about half of the intended pdf's are now at zakuski

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.

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 $\lambda = i.$

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  • $\begingroup$ How do we make curly braces here, as is usual for defining a set? $\endgroup$
    – Will Jagy
    Jul 27, 2011 at 20:23
  • $\begingroup$ Interesting material. Of course investigation of the power series version goes back to Caley (1860). $\endgroup$ Jul 27, 2011 at 20:28
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    $\begingroup$ If backslash-brace is not working for whatever reason, you can also use the "\lbrace" and "\rbrace" commands $\lbrace \text{like this} \rbrace$. $\endgroup$ Aug 20, 2012 at 7:52
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    $\begingroup$ @WillJagy (and Tom Copeland) If it is of any use: I might have a full (or partial) download of your url from 2014(?) This are 24 files, but don't know whether this is some mixture from various links you might have given in that years or just from your "other.html". $\endgroup$ May 10, 2023 at 19:12
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    $\begingroup$ Thanks, I hope to return to this topic one day to explore more deeply the connections to renormalization in QFT. Would be great if you could also post a copy from the Academie’s archives of Pincherle's entry for the Grand Prix. The French MO user mathoverflow.net/users/25256/duchamp-g%c3%a9rard-h-e has a fairly keen interest in related topics and the history of mathematics. (Alexander was not aware of the connection of Schroeder's analysis to the beauty of Lagrange inversion, so best to look at the original sources.) $\endgroup$ May 13, 2023 at 22:14
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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 Schröder-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, in some sense, "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^{\small ^-}$ .
  • if you start at $ \small z_0=x=3$ you can approach $ \small z_\infty = 2^{\small ^+}$ or even $ \small z_{-\infty}=4^{\small ^-}$ .
  • if you start at $ \small z_0=x=5$ you can approach $ \small z_{-\infty} = 4^{\small ^+}$ or even $ \small z_{\infty}= \infty$ .

$ \implies $ 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 with height say $h=1$ gives $ \small \sigma°^{-1}( \lambda^1 s)\cdot 2 +2=b=\sqrt 2$ because that is the iteration of height 1 (in the exponent of $ \small \lambda$ ). (Remark: the "circle"-super-postfix $\sigma°^{-1}$ means the functional inverse, not the reciprocal)
  • If we replace that exponent by $ \small h_w = i \cdot {\pi \over \ln \lambda } $ then we get $ \small \sigma°^{-1}( \lambda^{h_w} s) \cdot 2 +2=2.46791405022...$ which is thus, in some sense, "beyond infinity" with respect to the iteration height.

late update I add a picture to illustrate the previous statements.

This is picture, where I studied the application of imaginary heights, using the base for exponentiation $b=\sqrt2$. It has the attracting real fixpoint $t=2$.

As an example, look at the left side, with $z_0=1 + 0\cdot î$. Using iteration with real heights (here in steps of $1/10$ ) we move rightwards to $z_1=b^{z_0}=b = 1.414...$ and by more iterations more towards the fixpoint $t =2+ 0 î$. This is indicated by the orange arrows.

Note that because $t$ is a fixpoint, we cannot arrive at points on the real axis more to the right hand!

But using imaginary heights, iterations move from $z_0 $ to $z_h$ on the indicated circular curve (computed data are in steps of $0.1 { \pi \over \ln \beta} î$ see legend), which is indicated by the blue arrow.

This iteration does not go towards the fixpoint, but repeats to cycle around it. On that cycling the trajectory crosses the real axis beyond the fixpoint.

(Legend: the circular curves which connect the computed iteration-values of imaginary heights are Excel-cubic-splines and thus only very rough approximations of the true continuous iterations)

image

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  • $\begingroup$ Late p.s.: somehow reminds me this of the "tunneling" effect by which charged particles can "jump through" an isolator, but I don't have an idea whether this could be modeled by this or any related ansatz. $\endgroup$ Oct 27, 2020 at 22:57
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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 $$\sup_{t<0}\lVert t(t-A)^{-1}\rVert$$ 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 $\lVert A^{is}\rVert\leq Ce^{\theta\lvert s\rvert}$ 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.

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  • $\begingroup$ After all the composition of maps that is of interest to the OP is a linear operator $f\mapsto f\circ g$, so the case of linear operators is indeed relevant. $\endgroup$ Oct 28, 2020 at 10:39
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John Milnor, in his book Complex Dynamics in One Variable, has Ernst Schröder at the top of a list of the founders of the field of complex dynamics, and in Schröder's investigations of special cases of functional iteration, or iterated functional composition (IFC), the iteration is reduced to translation of a basic flow function.

If you look at Schröder's 1869/1870 paper "Ueber unendlich viele Algorithmen zur Auflösung der Gleichungen", or the translation (which has a transcription error) by Stewart, "On infinitely many algorithms for solving equations", you'll see that Schröder made use of the iterated derivatives, or iterated infinitesimal generators (IGs), $(\frac{1}{f'(z)}\partial_z)^n=(g(z)\partial_z)^n $, in his exploration of IFC related to generalization of Newton's iterative method of finding zeros of an equation. For an analytic function $f$ and its local compositional inverse $f^{(-1)}$, Schröder constructs the series, in terms of the IGs, for $FL(z,t)=f^{(-1)}[t+f(z)]$ evaluated at $t = -f(z)$, giving the zero $z_1 =FL(z,-f(z))= f^{(-1)}(0)$ of $f(z)$. Although Schröder doesn't directly symbolically couch his analysis this way, equation 21 in Schröder's paper is a truncated series expansion for $FL(z,t) \; |_{t = -f(z)}$ as is clear by comparison with the compositional inversion partition polynomials of OEIS A134685. Alexander, on p. 10 in his book A History of Complex Dynamics, mentioned in the comments to the question, displays equation 21 of Schröder as equation 1.6. In discussing Schröder's later 1871 paper "Ueber iterite Functionen", Alexander displays $FL(z,t)=f^{(-1)}[t+f(z)]$ (mod notation) and its functional iterate on p. 14.

For IFC, Schröder's intuition is more or less as follows;

Given an analytic function $f(z)$ and its local inverse $f^{(-1)}(\omega)$, for $s$ and $t$ small enough in amplitude, the flow function

$$FL(z,t) = f^{(-1)}(f(z)+t)$$

satisfies the translation equation (see

$$FL(FL(z,s),t) = FL(z,s+t).$$

(See the Abel equation for relation to the Abel, Schröder, and Böttcher functional equations of iterated function theory.)

Consequently, the $n-$th IFC in $z$ of the flow function is

$$FL^{(( n ))}(z,t)=f^{(-1)}(f(z)+(n+1)t) = FL(z,(n+1)t) .$$

The obvious formal generalization for any complex number $\alpha$ is

$$FL^{(( \alpha ))}(z,t)=f^{(-1)}(f(z)+(\alpha+1) t)= FL(z,(\alpha+1) t) .$$

The analytic flow function $FL(z,t)$ may be generated by the Graves-Lie op as

$$e^{t g(z)\partial_z} \; z = f^{(-1)}(f(z)+t) =FL(z,t),$$

where $g(z) = \frac{1}{f'(z)} = \frac{1}{\partial_z f(z)}.$

In this case, one can regard a complex iterate as either translation (flow) in the complex plane or complex exponentiation of the op $e^{g(z)\partial_z}$.

The associated evolution p.d.e., coding the tangent vector to the field, is

$$\partial_t FL(z,t) = g(z)\partial_z FL(z,t),$$

and all the apparatus for dealing with flow fields can be brought to bear on the problem. (Links and notes on that in OEIS A145271 and A139605.)

The associated autonomous o.d.e. is

$$ \partial_z \;f^{(-1)}(z) = g(f^{(-1)}(z)),$$

which provides a link to the beta function of renormalization group flow.


Some more refs:

"A survey of the theory of functional equations" by Kuczma, eqn. 111 on p. 35.

"A survey on the hypertranscendence of the solutions of the Schröder’s, Böttcher’s and Abel’s equations" by Gwladys Fernandes, eqn. 11 on p. 5.

“Variational aspects of the Abel and Schroeder functional equations” by McKiernan.

“Some differential equations related to iteration theory” by Azcel and Gronau.

Another logician, Frege, also considered the infinitesimal generator approach. See “Gottlob Frege, A Pioneer in Iteration” by Granau.

“Eri Jabotinsky, mathematician and politician: a short biography” by Gronau.

“Analytic iteration” by Jabotinsky.

“On analytic iteration” by Erdos and Jabotinsky.

(Added on 9/6/2022: For a flow function for real functions, "Note on the iterations of functions of one variable" by M. Ward and "The continuous iteration of real functions" by Ward and Fuller.)


Edit, Mar. 6, 2022:

Reading through Anixx's answer to the linked question, I realize I have not directly addressed a complex iterate in the way he has; however, there is a purely formal connection of my answer to his.

He considers the Newton-Gregory interpolation of the positive integer IFCs of a function. In my case that amounts purely formally to

$$\sum_{n = 0}^{\infty} (-1)^n \binom{\alpha}{n} \sum_{k=0}^n (-1)^k \binom{n}{k} FL(z,k)$$

$$ = \sum_{n = 0}^{\infty} (-1)^n \binom{\alpha}{n} \sum_{k=0}^n (-1)^k \binom{n}{k}e^{kg(z)\partial_z}\; z$$

$$ = (1-(1-e^{g(z)\partial_z})^{\alpha} \; z =e^{\alpha g(z)\partial_z} \; z = FL(z,\alpha). $$

This is also intimately related to the formal Mellin transform interpolation

$$\int_0^{\infty} \sum_{n \geq 0} (-1)^n FL(z,n) \frac{u^n}{n!} \frac{u^{s-1}}{{(s-1)}!} du $$

$$ = \int_0^{\infty} \sum_{n \geq 0} (-1)^n e^{ng(z)\partial_z} \frac{u^n}{n!} \frac{u^{s-1}}{{(s-1)}!} du \; z$$

$$ = \int_0^{\infty} e^{-ue^{g(z)\partial_z} } \frac{u^{s-1}}{{(s-1)}!} du \; z = (e^{g(z)\partial_z})^{-s} \; z = e^{-sg(z)\partial_z}z = FL(z,-s)$$

$$ = \int_0^{\infty} e^{-u} e^{(1-e^{g(z)\partial_z)} u} \frac{u^{s-1}}{{(s-1)}!} du \; z$$

$$ = \sum_{n\ge 0} \binom{n+s-1}{s-1}(1-e^{g(z)\partial_z})^n \; z$$

$$= \sum_{n= 0}^{\infty} (-1)^n\binom{-s}{n}\sum_{k=0}^n (-1)^k \binom{n}{k}e^{kg(z)\partial_z} \; z = e^{-sg(z)\partial_z}z .$$

Choosing $e^{-sg(z)\partial_z}z=FL(z,-s)$ for small amplitude/modulus $s$ and then analytically continuing to larger $|s|$ provides an interpretation, or summation method, for the formal maneuvers above.


Addressing the comment by Gottfried Helms:

As I've shown in numerous MO-Q&As, e.g., this MO_A, this one, and this one, for points $(\omega,z)$ for which $\omega = f(z)$ and $z = f^{(-1)}(\omega)$,

$$e^{t g(z) \partial_z} \; z = \exp[t \frac{\partial}{\partial f(z)}] \; z = \exp[t \frac{\partial}{\partial \omega}] \; f^{(-1)}(\omega) $$

$$ =f^{(-1)}(\omega+t) = f^{(-1)}(f(z)+t)= FL(z,t),$$

and we can see that this is equivalent to a Taylor series expansion, which is the approach Schroeder and, as he acknowledges, Theremin before him took. One has to be careful that $t$ is small enough that $\omega+t$ remains within the disc of analyticity about $\omega$.

Parse $t$ into $u+v=t$ for which $\omega+u$ and $\omega+u+v$ lie in the disc of analyticity of $f^{(-1)}$. Then

$$FL(z,t) = e^{t g(z) \partial_z} \; z = e^{vg(z) \partial_z} e^{ug(z) \partial_z}\; z$$

$$=\exp[v \frac{\partial}{\partial \omega}] \exp[u \frac{\partial}{\partial \omega}] \; f^{(-1)}(\omega) $$

$$=\exp[v \frac{\partial}{\partial \omega}] f^{(-1)}(\omega+u) =f^{(-1)}(\omega+u+v) $$

$$= f^{(-1)}(f(z)+u+v) = FL(z, u+v).$$

Reframing the intermediate steps,

$$FL(z,t) = e^{t g(z) \partial_z} \; z = e^{vg(z) \partial_z} e^{ug(z) \partial_z}\; z$$

$$=\exp[v \frac{\partial}{\partial \omega}] \exp[u \frac{\partial}{\partial \omega}] \; f^{(-1)}(\omega) $$

$$=\exp[v \frac{\partial}{\partial \omega}] f^{(-1)}(\omega+u)= e^{vg(z) \partial_z} f^{(-1)}(f(z)+u) =e^{vg(z) \partial_z} FL(z,u) $$

$$ =f^{(-1)}[f[f^{(-1)}(f(z)+v)]+u] = FL(f^{(-1)}(f(z)+v),u)= FL(FL(z,v),u) $$

$$= f^{(-1)}(f(z)+v+u) = FL(z, v+u).$$

For the substitution and reduction in the first equalities of the fourth and fifth lines, i.e.,

$f[f^{(-1)}(f(z)+v)] = f(z)+v = \omega +v,$

to be valid, either $\omega +v$ has to lie in the domain of $f^{(-1)}$ such that it remains the local inverse of $f$ or $f^{(-1)}$ as an analytic expression has to be analytically continued to the local inverse at that point.

An example:

For $m \neq 0$, with $\omega=\frac{z^{-m}}{m}=f(z)$ and $z=(m \cdot \omega)^{\frac{-1}{m}}=f^{(-1)}(\omega)$, the exponential mapping gives

$\exp[-t\cdot z^{m+1}\frac{\partial }{\partial z}] z=\exp[t\cdot \frac{\partial }{\partial \omega}](m \cdot \omega)^{\frac{-1}{m}}=(m \cdot (\omega+t))^{\frac{-1}{m}}=\frac{z}{(1+\ m \cdot t \cdot z^m)^{\frac{1}{m}}} =FL_m(z,t).$

Note upon direct substitution, $FL_{m}(FL_{m}(z,s),t)=FL_{m}(z,s+t)$ for any $s$ and $t$ for $|z|$ small enough regardless of the derivation of $FL_m(z,t)$. (This is satisfied in the limiting case for $m \to 0$ as well.)

Edit 3/10/22:

Consider the basic Möbius, or linear fractional, transformations:

1) Translation

With $f(z) = z$, then $f^{(-1)}(z)=z$, and

$$FL(z,t) = f^{(-1)}(f(z)+t) = z+t $$

with

$$FL(FL(z,s),t) = z+t \; |_{z \to z+s} = z+s+t = FL(z,s+t).$$

Self-composition in $z$ of $h(z) = FL(z,t)$ once gives

$$h^{((1))}(z) = h(h(z)) = FL(FL(z,t),t) = ((z+t) +t) = z+2t = FL(z,2t).$$

Then iterating $n=1,2,...$ times,

$$h^{((n))}(z) = FL(FL(FL(...,t),t),t) = ((z+t)+ t \cdots +t) = z +(n+1)t = FL(z,(n+1)t).$$

More generally for any complex $\alpha$, define the complex iterate as

$$h^{((\alpha))}(z) = FL(FL(z,\alpha t),t) = FL(z, (\alpha +1) t) = z + (\alpha +1) t.$$

From the perspective of the Lie IGs, consistently, $g(z)= 1/f'(z) = 1$, and, for $t$ any complex number,

$$e^{t g(z) \partial_z}\; z =e^{t \partial_z}\; z = z+t = f^{(-1)}(f(z)+t) = FL(z,t).$$

For $u$ and $v$ any complex numbers,

$$e^{v g(z) \partial_z} \; e^{u g(z) \partial_z}\; z =e^{v \partial_z}\; (z+u) = z+u+v = e^{v \partial_z}\; FL(z,u) = FL(FL(z,v),u) = FL(z,u+v).$$

2) Dilation

With $f(z) = \ln(z)$, then $f^{(-1)}(z) = e^z$, and

$$FL(z,t) = f^{(-1)}(f(z)+t) = e^{\ln(z)+t} = e^t\; z $$

with

$$FL(FL(z,s),t) = e^t z \; |_{z \to e^s z} = e^t e^s z = e^{t+s} z = FL(z,s+t).$$

Self-composition once of $h(z) = FL(z,t)$ gives

$$h^{((1))}(z) = h(h(z)) = FL(FL(z,t),t) = e^t (e^t z) = e^{2t} z = FL(z,t+t).$$

Then iterating $n=1,2,...$ times,

$$h^{((n))}(z) = e^t (e^t (...e^t z) = e^{(n+1)t} z =FL(FL(z,nt),t) =FL(z,(n+1)t).$$

More generally for any complex $\alpha$, define

$$h^{((\alpha))}(z) = FL(FL(z,\alpha t),t) = FL(z, (\alpha +1) t) = e^{(\alpha + 1)t}\; z.$$

Consistently, $g(z) = 1/f'(z) = z$, and

$$e^{tg(z)\partial_z} \; z = e^{tz\partial_z} \;z = \sum_{n\ge 0}\; \frac{t^n}{n!} (z\partial_z)^n \:z = e^t \; z = FL(z,t). $$

For $u$ and $v$ any complex numbers,

$$e^{v g(z) \partial_z} \; e^{u g(z) \partial_z}\; z =e^{v z\partial_z}\; e^u z = e^u e^v z = e^{u+v} z$$

$$ = e^{v \partial_z}\; FL(z,u) = FL(FL(z,v),u) = FL(z,u+v).$$

3) Special linear fractional transformation

With $f(z) = \frac{1}{z}$, then $f^{(-1)}(z) = \frac{1}{z}$, and

$$FL(z,t) = f^{(-1)}(f(z)+t) = \frac{1}{z} \; |_{z \to f(z)+t = \frac{1+tz}{z}} = \frac{z}{1+tz} $$

with

$$FL(FL(z,s),t) = \frac{z}{1+tz} \; |_{z \to \frac{z}{1+sz}} = \frac{z}{1+(s+t)z} = FL(z,s+t).$$

One functional self-composition of $h(z) = FL(z,t)$ gives

$$h^{((1))}(z) = h(h(z)) = FL(FL(z,t),t) = \frac{z}{1+(t+t)z} = FL(z,t+t).$$

Then iterating $n=1,2,...$ times,

$$h^{((n))}(z) = \frac{z}{1+(n+1)tz} =FL(FL(z,nt),t) =FL(z,(n+1)t).$$

More generally for any complex $\alpha$ define

$$h^{((\alpha))}(z) = FL(FL(z,\alpha t),t) = FL(z, (\alpha +1) t) = \frac{z}{1+(\alpha +1)tz}.$$

Consistently, $g(z) = 1/f'(z) = -z^2$, and, for $|tz| < 1$,

$$e^{tg(z)\partial_z} \; z = e^{-tz^2\partial_z} \;z = \sum_{n\ge 0}\; \frac{t^n}{n!} (-z^2\partial_z)^n \;z = \sum_{n\ge 0}\; (-tz)^n\; z = \frac{z}{1+tz} = FL(z,t). $$

With $\alpha= u+v$ for $u$ and $v$ any complex numbers, $|uz|<1$, and $|\frac{uz}{1+vz}|<1$,

$$e^{\alpha g(z) \partial_z}\; z = e^{v g(z) \partial_z} \; e^{u g(z) \partial_z}\; z =e^{-v z^2\partial_z}\; \frac{z}{1+uz}$$

$$ = e^{-v z^2\partial_z}\; \sum_{n\ge 0}\; (-uz)^n\; z =\sum_{n\ge 0}\; (-u)^n\; \sum_{k\ge 0}\; (-v)^k \frac{(n+k)!}{n!} z^{n+k+1} $$

$$ = \sum_{n\ge 0}\; (-u)^n\; (\frac{z}{1+vz})^{n+1} = \frac{\frac{z}{1+vz}}{1+\frac{uz}{1+vz}}= \frac{z}{1+(u+v)z}$$

$$ = e^{-vz^2 \partial_z}\; FL(z,u) = FL(FL(z,v),u) = FL(z,u+v) = FL(z,\alpha) .$$

In Appendix VIII of "Renormalization Group Functional Equations", Curtright and Zachos discuss the relation between these diff op equations (Eqn. 90 in C & S) and Schröder's functional conjugacy equation of iterated function theory. In their intro, they state that the renormalization group of Gell-Mann and Low and of Stueckelberg and Petermann has an elegant mathematical expression in terms of the functional conjugation methods of Schröder.

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  • $\begingroup$ Hmm, concerning the formula $FL(FL(z,s),t) = FL(z,s+t)$: in my question mathoverflow.net/q/391772/7710 I've discussed the observation on periodic points, by which $FL(FL(z,s),t) \ne FL(FL(z,t),s) $ and thus $FL(FL(z,s),t) \ne FL(z,s+t)$ if $z$ is a n-periodic point and $s=n$ is the period-length. For me this problem/special case is still not clear/not resolved with the given answers, and it would be instructive, if some required restrictions on "for $s$ and $t$ small enough" would be expressed more explicite. (Hope I'm not asking for complete trivia and/or nonsense here ...) $\endgroup$ Mar 7, 2022 at 10:00
  • $\begingroup$ @GottfriedHelms, hope I've addressed you comments correctly in the edit as to my meaning of small for the domains of $s$ and $t$ although, as my example indicates, the domains could extend far beyond that allowed in the differential-shift characterization. $\endgroup$ Mar 7, 2022 at 21:19
  • $\begingroup$ TomCopeland - thank you very much for your detailed answer! Unfortunately it contains too much material which I am still unfamiliar with, and I don't know, whether I'll ever get the spirit again to involve appropriately. If I can and find some way through I'll come back to this another day - this comment is just to tell you today I'm yet thankful for your kind workout w.r. to my specific question. $\endgroup$ Mar 9, 2022 at 17:07
  • $\begingroup$ @GottfriedHelms, I included some basic examples in more detail to illustrate the analysis. $\endgroup$ Mar 11, 2022 at 8:02
  • $\begingroup$ See also p. 46 and 47 of Iterative Functional Equations by Kuczma, Choczewki. and Ger. $\endgroup$ Mar 11, 2022 at 14:12
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I'm kind of embarrassed to give such a simple answer, but since I don't think any of the answers are in this direction: there is a whole world of study in dynamical systems associated to acting groups beyond $\mathbb{Z}$. What I mean is that the "traditional" case of a single invertible map $T$ acting on a space $X$ can be thought of as an entire $(\mathbb{Z},+)$-action $\{T^n\}_{n \in \mathbb{Z}}$ (which just happens to be determined by the single map $T$).

If you view this way, then it's easy to picture generalizing to other groups; for any group $(G,\cdot)$, you can define a $(G,\cdot)$-dynamical system as coming from a $(G,\cdot)$-action $\{T^g\}_{g \in G}$ of self-maps of $X$. Then, if $G = (\mathbb{C},+)$, it's easy to talk of complex iterates.

Your maps are not assumed invertible, but everything above works for semigroup actions too, so you could work with the upper-right quadrant of $\mathbb{C}$.

Maybe what you want is more computational (and maybe other answers are doing this in a more direct way), but this is the first thing that came to mind as a dynamicist.

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