Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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):

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

alt text

$$\sin^{[i]}(x)$$ http://storage8.static.itmages.ru/i/11/0727/h_1311793019_d78445f160.png

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

share|improve this question
1  
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

3 Answers 3

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

http://zakuski.utsa.edu/~jagy/other.html

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 $\lambda = i.$

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.