Suppose I have a discrete dynamical system given by: $$ X^{n+1} = f(X^{n}) \qquad X^0 =x , $$ where $f$ is some diffeomorphism on $\mathbb{R}^d$ and $x \in \mathbb{R}^d$. When does there exist a function $\tilde{f}:\mathbb{R}^d\rightarrow \mathbb{R}^d$ such that:

$$ Z^{n+1}= Z^{n} + \tilde{f}(Z^{n}) \qquad Z^0=x, $$

and (either of):

Not Ideally:

and for every $\epsilon>0$ there is some $n_{\epsilon}>0$ satisfying: $$ \|Z^n - X^n\|<\epsilon \qquad (\forall n\geq n_{\epsilon})? $$


Or it possible, the stronger condition holds: $X^n = Z^n$ for all large $n$?

  • $\begingroup$ Should that be $Z^{n+1}= X^{n+1} + \tilde{f}(Z^{n})$? $\endgroup$ – Omnomnomnom Jan 23 at 7:58
  • $\begingroup$ Nope, it should be $Z^{n+1}= Z^{n} + \tilde{f}(Z^{n})$. $\endgroup$ – AnnieTheKatsu Jan 23 at 8:02
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    $\begingroup$ Why not simply define $\tilde f(x) = f(x) - x$ to get $Z^n = X^n$, then? $\endgroup$ – Omnomnomnom Jan 23 at 8:04
  • $\begingroup$ True! I'd accept that as an answer. Also, if you're interested in answering the stochastic version of there question; here it is: mathoverflow.net/questions/350990/… $\endgroup$ – AnnieTheKatsu Jan 23 at 8:35

It suffices to take $\tilde f(x) = f(x) - x$, which leads to $Z^n = X^n$ for all $n$.


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