I would like to find the exact order of convergence of the sequence $\{x_n\}_{n=0}^{\infty }$ given by $x_{n+1}=2x_n-\pi {x_n}^2$ with $x_0=1/3$.
I found that the limit of the sequence is $\frac{1}{\pi }$.
But, I'm not sure if this method of finding the order of convergence is right.
I gave it a try as follows.
At first, I let $F(h)=2h-\pi h^2+\frac{1}{\pi }$ so that $\lim_{h\rightarrow 0}F(h)=\frac{1}{\pi }$.
Get a Taylor expansion of $F(h)$ at $0$, then
$$F(h)=F(0)+F'(0)h+F''(0)\frac{h^2}{2 } + O(h^3) \\ =\frac{1}{\pi }+2h-\pi h^2 + O(h^3) = F(h)+O(h^3) $$
Thus, the order of convergence is $O(h^3)$.
Is this the right method?
Any help would be appreciated.
Best Answer
The iteration $x_{n+1}=2x_n-ax_n^2$ can be quadratically completed to $$ (1-ax_{n+1})=(1-ax_n)^2, $$ and from that one can directly conclude the order of convergence.
If you are examining the convergence at $\frac1\pi$, then the basis point for the Taylor expansion has to be $\frac1\pi$, not zero. Or you need to correctly insert $x_n=\frac1\pi+h$. If $x_{n+1}=g(x_n)$ and $x_*$ is the limit, then you need to find the $p$ in $g(x_*+h)=x_*+O(h^p)$.