Fixed Point Iteration And Order Of Convergence Of $x_{n+1}=x_n(2-\pi x_n)$

Question

Let $$x_{n+1}=x_n(2-\pi x_n)$$ a. Find the fixed point

b. Find the order of convergence

a. An iterative function is of the form $x=g(x)$ so the find $x$ we can find the root of $x-g(x)=0$ so:

$$x=x(2-\pi x)\iff x=2x-\pi x^2\iff \pi x^2-x=0$$

So the root are $x_1=0$ and $x_2=\frac{1}{\pi}$

$x_1=0$ can not be a fixed point as $g(0)=2*0-\pi*0^2=0$ whereas $g(\frac{1}{\pi})=2*\frac{1}{\pi}-\pi(\frac{1}{\pi})^2\neq 0$ but $g'(\frac{1}{\pi})=2-2\pi (\frac{1}{\pi})=0$

b. we need to find the first derivative that does not vanish from a. we know that the first derivative vanishes so we take the second derivative

$$g''(\frac{1}{\pi})=2\pi$$ so the order of convergence is $2$

Is the solution correct?

Answer 1

Both $x=0$ and $x=\frac{1}{\pi}$ are fixed points of $g(x)=x(2-\pi x)$.

$g'(0)=2$ implies that $0$ is a repelling fixed point.

$g'(\frac{1}{\pi})=0$ implies that $\frac{1}{\pi}$ is a superattracting fixed point. You're right about the order of convergence being $2$ for this fixed point, except that $g''(\frac{1}{\pi})=-2\pi$.

Answer 2

You can with some small effort see that it is possible to complete the square $$ 1-\pi x_{n+1}=(1-\pi x_n)^2\implies 1-\pi x_n=(1-\pi x_0)^{2^n}. $$ This converges to zero from the interval $|1-\pi x_0|<1$ of initial points and diverges to infinity for $|1-\pi x_0|>1$.

This means that $x_*=\frac1\pi$ is attracting and $x_*=0$ is a repelling fixed point.