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Given $x_0 \in [a,b]$, let the sequence $(x_n)$ be defined recursively by: $$ x_n = g( x_{n-1}), n=1,2,... $$ where $g \in C^1 [a,b]$

Assume that $x_n \to c \in [a,b]$, then: $$ c=\lim_{n \to \infty} x_n = \lim_{n \to \infty} g(x_{n-1})= g (\lim_{n \to \infty} (x_{n-1})) = g(c) $$ So the sequence converges to a fixed point of $g$ in $[a,b]$.

Assume also that $g'(c)=0$.

Question: I am trying to prove the following relation: $$ x_{n+1}-x_n = o(x_n - x_{n-1}) $$ i.e., $$ \lim_{n \to \infty} \frac{x_{n+1}-x_n}{x_n - x_{n-1}} = 0 $$

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That relation doesn't seem to be true. Consider the function $f(x)=\frac{1}{2}x$ and $x_0=1$. Now clearly $x_n=2^{-n}$, so $\frac{x_{n+1}-x_n}{x_n - x_{n-1}}=1/2$ for all $n$.

EDIT: When you assume further that $g'(c)=0$, we have $$ \frac{x_{n+1}-x_n}{x_n - x_{n-1}}=\frac{g(x_n) - g(x_{n-1})}{x_n - x_{n-1}}=g'(\xi). $$ where $\xi$ lie between $x_n$ and $x_{n-1}$. Since $x_n\to c$, $\xi\to c$, $g'(\xi) \to g'(c)=0$ because $g'$ is continuous. So your result is proved.

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  • $\begingroup$ I added another assumption that $g'(c) = 0$. $\endgroup$ – A Slow Learner Mar 6 at 5:51
  • $\begingroup$ @ASlowLearner See my edit $\endgroup$ – Holding Arthur Mar 6 at 10:47

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