# Do maps have flows?

How can one extend recursive function definitions to continuous numbers? What is the continuous analog of the Ackermann function? The symbolic forms of the Ackermann function with a fixed first argument seem to have obvious interpretations for arbitrary real or complex values of the second argument. But is there a general way to extend these kinds of recursive definitions to continuous cases? Given a way to do this, how does it apply to recursive definitions like those on page 130? ... Stephen Wolfram

The following is an example of a flow of a map from MO f(f(x))=exp(x)-1 and other functions “just in the middle” between linear and exponential. .

Consider $g(x)=e^x-1$. Then $g^n(x)= x+\frac{1}{2!}n x^2+\frac{1}{3!} \left(\frac{3 n^2}{2}-\frac{n}{2}\right) x^3+\frac{1}{4!} \left(3 n^3-\frac{5 n^2}{2}+\frac{n}{2}\right) x^4$ $+\frac{1}{5!} \left(\frac{15 n^4}{2}-\frac{65 n^3}{6}+5 n^2-\frac{2 n}{3}\right) x^5$

$+\frac{1}{6!} \left(\frac{45 n^5}{2}-\frac{385 n^4}{8}+\frac{445 n^3}{12}-\frac{91 n^2}{8}+\frac{11 n}{12}\right) x^6$

$+\frac{1}{7!}\left(\frac{315 n^6}{4}-\frac{1827 n^5}{8}+\frac{6125 n^4}{24}-\frac{1043 n^3}{8}+\frac{637 n^2}{24}-\frac{3 n}{4}\right) x^7 + \cdots$

Note that $g^0(x)=x, g^1(x)=e^x-1$ and that a symbolic mathematical program will also confirm that $g^m(g^n(x))=g^{m+n}(x) +O(x^8)$.

The half-iterate is also computed correctly, $g^\frac{1}{2}(x)=x+\frac{x ^2}{4}+ \frac{x^3}{48} +\frac{x^5}{3840}-\frac{7 x^6}{92160} +\frac{x^7}{645120}$ See MO What’s a natural candidate for an analytic function that interpolates the tower function? for more background.

Questions

1. What evidence is there for believing that maps do not have flows? Is there anything known that would prevent proofs to establish existence, uniqueness, and convergence? References would be nice but an explanation would be better.
2. Consider $f(f(x))=g(x)$ where $g: \mathbb{R} \rightarrow \mathbb{R}$. Can $f: \mathbb{R} \rightarrow \mathbb{C}$ be an appropriate solution or must $f: \mathbb{R} \rightarrow \mathbb{R}$?
3. Likewise, is there any reason beyond aesthetics for believing that maps have flows?
• – Will Jagy Oct 23 '10 at 0:15
• People might take your question more seriously if you removed all mention of Wolfram and his weighty tome. – S. Carnahan Oct 23 '10 at 11:05
• Personally, I don't see why we can't take his question seriously regardless. – Cam McLeman Oct 23 '10 at 14:30
• Updated link to Dr Shalizi's review: bactra.org/reviews/wolfram. – jeq Feb 11 '16 at 14:02
• In a more reduced context, your question has some answers. For example, you can ask when the exponential map for a Lie group is onto, and this has some satisfying answers. In the context of homeomorphisms or diffeomorphisms, you can ask if the corresponding mapping class group is trivial -- even this only has satisfying answers fairly specific cases such as low dimensions. But if you study the techniques used, you see some of the complexities involved. – Ryan Budney Sep 6 '17 at 18:59

In general, there are obstructions for a map being the "time one map" of a flow (see this question).

However, and I am not quite sure this is what you are looking for, there is a general procedure to construct flows out of maps, namely the suspention. For the map $g:M \to M$ you consider the flow in $M\times \mathbb{R}$ given by $\varphi_t((x,s))= (x, s+t)$ and you quotient by $(x,s) \sim (g(x),s+1)$. This gives a flow, whose time one map preserves a set homeomophic to $M$ where the dynamics is $g$.

In the case of a map from $[0,\infty)$ to $[0,\infty)$ fixing $0$ you can work out flow to be defined in $\mathbb{C}$ and such that the time one map is $g$, so this would give a partial answer to $2.$.

• Thanks for your response. I've seen some version of this question asked a number of times and it never seems to be substantially resolved. There are a considerable number of researchers studying tetration, iterated exponentiation, where $^{1}a=a, ^{2}a=a^a, ^{3}a=a^{a^a}$. This leads to questions like "can $^\frac{1}{2} a$ be defined?" and is $f(x)=^x e$ a valid solution if $f: \mathbb{R} \rightarrow \mathbb{C}$. Any further insight into problems with extending tetration to the real and complex numbers would be appreciated. – user37691 Oct 23 '10 at 23:52

Number 2 is answered in the affirmative if we take "functions" to mean "analytic functions". Take $\cos(x)$. It can be shown that there exists no function $f : \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = \cos(x)$. This follows by the fact: $\cos(x_0) = x_0$ where $x_0 = 0.739805...$, and $-1<\cos'(x_0) < 0$. Therefore if there was such an $f$, $\cos'(x_0) = f'(f(x_0))f'(x_0) = f'(x_0)f'(x_0)$ and $f'(x_0)$ cannot be real.

To show there does exists an $f: I \to I$, where $I$ is the immediate basin of attraction about $x_0$ and $\mathbb{R} \subset I$, is a little trickier. You'll have to track down a nice paper constructing it (it's a real hassle to construct); but it is possible. It would take about 10 pages of written work to get it out. Then $f:\mathbb{R} \to \mathbb{C}$. There exists exactly two such $f$ corresponding to $f'(x_0) = i\sqrt{|x_0|}$ or $f'(x_0) = -i\sqrt{|x_0|}$.

If you want a more trivial example just look at $-x$ which has a square root $ix$, but no real analytic square root.