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Let $\{x_n\}_{n\ge 0}$ be a sequence of reals such that $x_{n+1}=g(x_n)$, where $g:\mathbb{R}\to \mathbb{R}$ is a continuous function such that $0$ is a fixed point of $g$. My question is the following:

If for an initial condition $x_0$, we have $x_n\to 0$ as $n\to \infty$, how should one proceed to find the rate of convergence of the sequence $x_n$?

I know that if $g$ is a contraction, then this rate of convergence is linear. However,

how should one proceed if $g$ is not a contraction?

For example, let $g(x) = x^2$. If $x_0<1$, then we have $x_n = x_0^{2^n}\stackrel{n\to \infty}{\to} 0$, and the rate of convergence is quadratic.

Note that if $g$ is Lipschitz, then I can always find an initial condition suitably so that the sequence converges to $0$ linearly. However, this estimated rate would only be an upper bound on the actual rate, as it was seen for the $x^2$ example, the actual rate is quadratic and is thus very fast.

Please provide some references to relevant literature. Thanks in advance.

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    $\begingroup$ But note e.g. that for $g(x)=x-x^2$, that has $g'(0)=1$, starting with $0<x_0\le1$ produces $x_n=1/n +o(1/n)$ (more precisely $x_n=1/n-(\log n)/n^2+o(1/n^2)$ I think) $\endgroup$ Commented Apr 20, 2018 at 8:27
  • $\begingroup$ A Taylor expansion around $0$ of $g$ should give you an answer for most of the practical cases. $\endgroup$
    – RaphaelB4
    Commented Apr 20, 2018 at 8:51
  • $\begingroup$ @PietroMajer, for $g(x) = x - x^2$, I see that the convergence is sublinear. So I guess what I am asking should be applicable for superlinearly converging sequences. $\endgroup$ Commented Apr 23, 2018 at 7:29
  • $\begingroup$ @PietroMajer, also can you kindly tell me how you obtained the expansion for $x_n$? And can that method be applied to any ``sufficiently smooth function'' to get an idea of the convergence rate? $\endgroup$ Commented Apr 23, 2018 at 7:44
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    $\begingroup$ @PietroMajer: see this answer and thank you for your comments here! $\endgroup$ Commented Jun 28, 2019 at 8:22

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