Show that the iteration: $x_{k+1}=2x_k-αx_{k}^2$ where $α > 0$ converges quadratically to $\frac{1}{α}$ for any $x_0$ such that $0 < x_0 < \frac{2}{α}$.
I have been able to prove that it converges quadratically to $\frac{1}{α}$ since you can consider the fixed point(s) of $g(x)= 2x-αx^2$ which are $x = 0$ and $x = \frac{1}{α}$ and it's quadratic convergence since:
$g(x) - \frac{1}{α} = 2x-αx^2 - \frac{1}{α} $
$g(x) - \frac{1}{α} = -α(x - \frac{1}{α})^2$
and thus
$\frac{|g(x) - \frac{1}{α}|}{|x - \frac{1}{α}|^2} = α$
and hence
$\lim_{k \to ∞} \frac{|x_{k+1} - \frac{1}{α}|}{|x_k - \frac{1}{α}|^2}= \lim_{k \to ∞}\frac{|g(x_k) - \frac{1}{α}|}{|x_k - \frac{1}{α}|^2} = α$
but I'm having trouble understanding why it converges in that specific interval. I know that for a fixed point method to converge in an interval $[a,b]$:
- g must be continuous
- map $[a,b] \to [a,b]$
- there must exist a constant, $ 0 < k < 1 $, such that $|g'(x)| ≤ k$ for all $a < x < b$
so I tried inspecting $g'(x) = 2-2αx $ and seeing where $|g'(x)| = 1$ which is at $ x = \frac{1}{2α} $ and $x = \frac{3}{2α}$ and examing the behavior of $g'(x)$ but that doesn't seem to be leading me to the right answer. I was wondering what I was doing wrong.
Thank you!
You got $1-αx_{k+1}=(1-αx_k)^2$ and thus per recursion $$ 1-αx_n=(1-αx_0)^{2^n} $$ As subsequence of the geometric sequence this converges to zero exactly if $|1-αx_0|<1$.