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Stream: community: our work

Topic: Combinatorial species and dynamics

Daniel Geisler (Oct 31 2022 at 02:13):

I am a hobbyist studying dynamics, tetration and the Ackermann function. Just learning a bit about combinatorial species gave me significant insight into the combinatorics of iterated functions. See A Problem in Statistical Classification Theory, Analytic Combinatorics

Recently I found the following identity for iterated functions involving recursive Bell polynomials, unlabeled hierarchies (recursive integer partitions), and labeled hierachies (total partitions).

Derivative of iterated function $H(n,t)=D^nf^t(L)$
Fixed point: $H(0,t)=L$
Lyapunov multiplier: $H(1,t)=f'(L)^t$
$H(n,t)=\sum_{r=0}^\infty(\sum_{k=1}^n \frac{f^{(k)}(L)}{k!} B_{n,k}(H(1,t-1),\ldots, H(n-k+1,t-1)))^r$
$f^t(x)$=$\sum_{k=0}^\infty\frac{1}{k!} H(k,t) (x - L)^k$

While initially $t \in \mathbb{N}$, once a symmetry is added the result is consistent with $t \in \mathbb{C}$. Note: this is not a model of iterated functions, it is iterated functions in a power series form.

I need to write up proofs for my work, but I suspect that I can either write long ugly "classical" proofs or short beautiful proofs using CT and species. For example, set partitions are associated with the derivatives of composite functions. Consider iterated functions as iterated composition. Then there should be a combinatoric structure associated with recursive set partitions. I know from research that the answer is labeled hierarchies defined by $Hier=Union(Z,Set(Hier,card>1))$. Is it that simple to invoke the definition? It seems like little else would be needed.

Just as labeled hierarchies are recursive set partitions, the derivatives of iterated functions are recursive Bell polynomials. I believe that "apparent" connection is actually categorical and can best be represented in short CT proofs.

Mathematica Code

Flow[f_, t_, x_, L_, order_ : 3] := Module[{},
   H[0] = L;
   H[1] = f'[L]^t ;
   Do[
    H[max] =
     First[r[t] /.
       RSolve[{r[0] == 0,
         r[t] == Sum[
            Derivative[k][f][L] BellY[max, k,
              Table[H[j] /. t -> t - 1, {j, max}]], {k, 2, max}] +
           f'[L] r[t - 1]}, r[t], t]],
    {max, 2, order}];
   Sum[1/k! H[k] (x - L)^k, {k, 0, order}]
   ];

Jean-Baptiste Vienney (Oct 31 2022 at 07:29):

It looks interesting but I don’t understand the notations. What are $H(n,t), L$ and the domain and codomain of $f$ ?

Daniel Geisler (Oct 31 2022 at 08:31):

Derivative of iterated function $H(n,t)=D^nf^t(L)$, $L$ is a fixed point $f(L)=L$. The domain and codomain are $\mathbb{C}$.

Jean-Baptiste Vienney (Oct 31 2022 at 16:49):

I don't understand how $f^{t}(x)$ is defined when $t \in \mathbb{C} \backslash \mathbb{N}$?

Daniel Geisler (Oct 31 2022 at 17:52):

Thank you for your question. I am simply deriving the power series of iteration functions which is in $t \in \mathbb{N}$. But once symmetry is assumed as in Schroeder's Equation in hyperbolic dynamics, (see below) or Abel's Equation in parabolic iteration, the solutions satisfies $f^a(f^b(x))=f^{a+b}(x)$, even when $a$ and $b$ take complex values. The underlying mechanism here is the connection with geometric series when not considering a root of unity. The geometric series consist of discrete sums, but in it's closed form solution can take continuous values.

$L+(x-L) f'(L)^t+\frac{(x-L)^2 f''(L) \left(f'(L)^t-1\right) f'(L)^{t-1}}{2 \left(f'(L)-1\right)}+\frac{(x-L)^3 f'(L)^{t-2} \left(f'(L)^t-1\right) \left(-f^{(3)}(L) f'(L)^{t+1}+f^{(3)}(L) f'(L)^{t+2}+f^{(3)}(L) f'(L)^2-f^{(3)}(L) f'(L)+3 f''(L)^2 f'(L)^t-3 f'(L) f''(L)^2\right)}{6 \left(f'(L)-1\right)^2 \left(f'(L)+1\right)}$

See Bell Polynomials of Iterated functions for more background.

Jean-Baptiste Vienney (Oct 31 2022 at 18:43):

If I understand, you say that for every holomorphic function $f:\mathbb{C} \rightarrow \mathbb{C}$ with a fixed point $L \in \mathbb{C}$, the equation
$H(n,t)=\sum_{r=0}^\infty(\sum_{k=1}^n \frac{f^{(k)}(L)}{k!} B_{n,k}(H(1,t-1),\ldots, H(n-k+1,t-1)))^r$, where $H(n,t) := D^{n}f^{t}(L)$ admits one and only one solution $(f^{t})_{t \in \mathbb{C}}$, where $f^{t}:\mathbb{C} \rightarrow \mathbb{C}$ are holomorphic functions, "once a symmetry is added". And this solution verifies that $f^{t}$ is the the $t^{th}$ iterate of $f$ when $t \in \mathbb{N}$.

Is the added symmetry a condition on $f$ and what is the definition? And how do you prove that the equation admits one and only one solution with this further requirement?

Daniel Geisler (Oct 31 2022 at 20:04):

The added symmetry is given by the Classification of Fixed Points, Complex Dynamics by Carleson and Gamelin. The condition on $f$ is the value of $f'(L)$. If

• Superattracting: $f'(L)=0$
• Hyperbolic: $|f'(L)| \ne 1$
• Rationally Neutral: $f'(L)^n=1$ where $n\in\mathbb{N}$.

And how do you prove that the equation admits one and only one solution with this further requirement?

Good question. The equation should have one and only one solution as the additional requirement is simply dividing the value of $f'(L)$ into cases.

Jean-Baptiste Vienney (Nov 01 2022 at 07:51):

How do you prove it for each case?

Jean-Baptiste Vienney (Nov 01 2022 at 07:54):

Other question: do you have a formula for $H(n,t)$ when $t \in \mathbb{N}$? It requires to iterate the Faà di Bruno rule so it should be a heavy formula.

Daniel Geisler (Nov 01 2022 at 09:15):

See my paper regarding the recursive Faà di Bruno formula on pages four and five. I'm considering how to make a stronger argument for uniqueness.

Jean-Baptiste Vienney (Nov 01 2022 at 11:40):

It seems a very interesting question to me how to define continuous iterations of maps, but you should explain carefully what you do and write rigorous and complete proofs in order than anybody can judge your work! So good luck for this.

Jacques Carette (Nov 01 2022 at 17:13):

I believe the Lattes examples give counter-examples. There is also all sorts of weird stuff that happens for the non-rational neutral cases.

Also the classification should split $|f'(L)|>1$ and $|f'(L)|>1$ (repelling and attracting, respectively) as those behave really quite differently.

Daniel Geisler (Nov 02 2022 at 07:05):

Thanks for the Lattes examples, I hadn't seen them before. But I believe we are looking at a superattracting maps and they usually do not have flows.
If by non-rational neutral cases you are talking about the small divisor problems, I am aware of them. The same for attractors and repellors. See Section 3.2 Dynamics and Hierarchies for a twenty year old attempt to derive the Classification of Fixed Points and their properties solely from a recursive Faà di Bruno's formula.