# Approximation of Dynmical System

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: $$\|Z^n - X^n\|<\epsilon \qquad (\forall n\geq n_{\epsilon})?$$

## Ideally:

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

• Should that be $Z^{n+1}= X^{n+1} + \tilde{f}(Z^{n})$? – Omnomnomnom Jan 23 at 7:58
• Nope, it should be $Z^{n+1}= Z^{n} + \tilde{f}(Z^{n})$. – AnnieTheKatsu Jan 23 at 8:02
• Why not simply define $\tilde f(x) = f(x) - x$ to get $Z^n = X^n$, then? – Omnomnomnom Jan 23 at 8:04
• 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/… – 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 $$n$$.