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It is clear that one can obtain a discrete dynamical system from a continuous one, but is the converse possible if the system is "nice"?

Define the discrete-time dynamical system on $\mathbb{R}^d$ by $$ x_{n+1} = f(x_n);\, x_0\triangleq x $$ where $f \in C^2(\mathbb{R}^d;\mathbb{R}^d)$ and $x \in \mathbb{R}^d$.

Fix a (large) positive integer $N$, is there a function $F:\mathbb{R}^d\rightarrow \mathbb{R}^d$ such that the solution to the continuous-time dynamical system $$ \partial_t X_t = F(X_t) ; \qquad X_0=x, $$ and $X_n = x_n$ for every $n \in \left\{1,\dots,N\right\}$?

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    $\begingroup$ If you mean "for all $x$", then no, it is not possible. For every $F$, the Jacobian determinant of the corresponding flow is non-zero, and hence it remains positive. Thus, $f(x) = (-x_1,x_2,x_3,\ldots,x_d)$ provides a simple counter-example. On the other hand, if $x$ is fixed, then I do not see the answer right away. $\endgroup$ – Mateusz Kwaśnicki Jan 9 at 10:10
  • $\begingroup$ @MateuszKwaśnicki No, $x$ should be specified (so the "anti-discritization" is dependent on the choice of x); It would only need to hold for an (arbitrarily large but fixed) number of iterations. $\endgroup$ – AIM_BLB Jan 9 at 10:22
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    $\begingroup$ Then this is essentially a question whether there is a smooth simple curve passing through given $N$ points $x_1, \ldots, x_N$ in a given order. Or, more precisely, we require this curve to pass through $M$ points $x_1, \ldots, x_M$ in a given order and have no self-intersections except possibly at $x_M$, in which case it must be tangent to itself; here $M$ is the least number such that $x_M \in \{x_1, \ldots, x_{M-1}\}$, or $M = N$ if no such number exists. This seems to be fairly straightforward, no? $\endgroup$ – Mateusz Kwaśnicki Jan 9 at 10:40
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I provide a discrete expansion of an iterated function, but a symmetry constraint is also needed to simplify it to a continuous solution. Works for Schroeder's and Abel's Functional Equations. Note that convergence issues make if difficult to provide a complete answer.

Let $f(x)$ and $g(x)$ be functions in Banach space, then the composite $f(g(x))$ can be constructed using Faa Di Bruno's formula.

\begin{eqnarray} D^nf(g(x)) = \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^kf)(g(x)) \left(\frac{Dg(x)}{1!}\right)^{k_1} \cdots \left(\frac{D^ng(x)}{n!}\right)^{k_n} \label{eq:FaaDiBruno} \end{eqnarray} where $\pi(n)$ denotes a partition of $n$, usually denoted by $1^{k_1}2^{k_2}\cdots n^{k_n}$ with $k_1+2k_2+ \cdots nk_n=k$; where $k_i$ is the number of parts of size $i$. The partition function $p(n)$ is a decategorized version of $\pi(n)$, the function $\pi(n)$ enumerates the integer partitions of $n$, while $p(n)$ is the cardinality of the enumeration of $\pi(n)$. [Comtet], [Riordan]

In turn $f(g(x))$ can construct any iterated function at a fixed point by setting $f(0)=0$ and $g(x)=f^{m-1}(x)$.

\begin{equation} D^n f^t(x)= \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^k f)(f^{t-1}(x)) \left(\frac{Df^{t-1}(x)}{1!}\right)^{k_1} \cdots \left(\frac{D^n f^{t-1}(x)}{n!}\right)^{k_n} \end{equation}

\begin{equation} D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0) \end{equation} DynamicalRecurranceEquation

The Taylors series of $f^t(x)$ is derived by evaluating the derivatives of the iterated function at a fixed point $f^t(0)$ by setting $x=0$ and separating out the $k_n$ term of the summation that is dependent on $D^n f^{t-1}(0)$.

The remaining $\pi(n)-1$ terms of the summation are only dependent on $D^k f^{t-1}(0)$, where $0<k<n$. Let this partial summation be written as $\sigma(n)$ with $\sigma(0)=0$ and $\sigma(1) = 1$.

Rewriting the $\pi(n)-1$ terms of the summation as $\sigma(n)$ will help in writing a proof by general induction. For $n>1$,

\begin{equation} D^n f^t(0)=\sigma(n) + D'f(0) D^n f^{t-1}(0) \label{eq:Linear Equation} \end{equation}

Theorem: The Taylor series of an iterated holomorphic function $f^t(x)$ can be constructed given a fixed point where $t \in \mathbb{N}$.

Proof. Assume the given fixed point is at zero. The Taylor series of $f^t(x)$ can be constructed for some positive value of $R$ where $0 < |x| < R$ if and only if $D^n f^t(0)$ can be constructed for every $n \geq 0$. prove by strong induction.

Basis Steps:

Case $n=0$. By definition $D^0 f^t(0) = 0$, so $D^0 f^t(0)$ can be constructed.

Case $n=1$. Let $D^1 f^t(0) = D'f(0)^t$, so $D^1 f^t(0)$ can be constructed.

Induction Step: Case $n=k$. Assume that $D^k f^t(0)$ can be constructed for all $k$ where $0 \leq k < n$.

Using Eq. Dynamical Recurrance Equation, $D^k f^t(0)=\sigma(k) + D'f(0) D^k f^{t-1}(0)$. The function $\sigma(k)$ in only dependent on $D^0 f(0), \ldots, D^k f(0)$ and $D^k f^t(0), \ldots, D^{(k-1)} f^t(0)$. By the strong induction hypothesis, $\sigma(k)$ can be constructed. Therefore Eq. Dynamical Recurrance Equation can be reduced to a geometrical progression based on $D'f(0)$ that can be represented by a summation.

\begin{eqnarray} D^k f^t(0) = \sum_{j=0}^{k-1} \sigma(k) D'f(0)^j \end{eqnarray} This completes the induction step that $D^n f^t(0)$ can be constructed for all whole numbers $n$.

The Taylors series for $f^t(x)$ is

\begin{eqnarray} f^t(x) = \sum_{n=0}^\infty \sum_{j=0}^{n-1} \sigma(n) D'f(0)^j x^n \label{eq:Dynamical Equation} \end{eqnarray}

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  • $\begingroup$ But my function $f$ is of class $C^2$ and not holomorphic? $\endgroup$ – AIM_BLB Jan 9 at 11:50
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    $\begingroup$ My mistake. I do have a question out mathoverflow.net/questions/350030/…, but I expect my work requires $C^\infty$. $\endgroup$ – user37691 Jan 9 at 13:21
  • $\begingroup$ That's alright, I'll take a look and see what I can do with it :) $\endgroup$ – AIM_BLB Jan 9 at 15:22

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