# Fixed-point iteration: little-oh relation between consecutive pair of elements

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., $$\lim_{n \to \infty} \frac{x_{n+1}-x_n}{x_n - x_{n-1}} = 0$$

## 1 Answer

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.

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