Fixed Point Iteration Methods - Convergence We wish to solve the equation $f_a(x) = 0$, where $a ∈ R$ is a parameter and $f_a(x) = (1−a)x+ax^3$. We will study the fixed point iteration $x^{k+1} = φ(x^k)$, where the function $φ(x) = ax(1 − x^2)$.
(i) Show that the fixed point method is consistent and that φ : [0, 1] → [0, 1], provided $0\leqslant a \leqslant \frac{3\sqrt{2}}{2}$.
(ii) Find the positive values of $a$ such that the fixed point iteration converges to the root $α_1 = 0$ for any initial guess $x^0 ∈ [0, 1]$.
(iii) Find a condition on $a$ under which a second zero $α_2 > 0$ exists in the interval [0, 1] and determine for which $a$ the fixed point iteration can approximate $α_2$.
(iv) For what value of $a$ can the fixed point iteration approximate $α_2$ with second order convergence?
Solved part (i): Used the consistency condition $f(\alpha)=0 \iff \phi(\alpha)=\alpha$ which gives LHS: $x=0$, $x=\pm\sqrt\frac{a-1}{a}$. These values both give $\phi(x)=x$.
Solving the RHS gives: $a-a\alpha^2 -1=0$, hence $\alpha=\pm\sqrt\frac{a-1}{a}$ as required.
Then for the next part I calculated the maximum of $\phi(x)$ by finding $\phi'(x)=0$ which gives $x=\pm\sqrt\frac{1}{3}$. Since we are concerned with [0,1], we take $x=+\sqrt\frac{1}{3}$, and find $\phi(\sqrt\frac{1}{3})$ which gives $x=2a\frac{\sqrt3}{9}$. Setting $0\leqslant 2a\frac{\sqrt3}{9} \leqslant 1$ and rearranging achieves the result $0\leqslant a \leqslant \frac{3\sqrt{2}}{2}$.
Attempt at part (ii) Using the local linear convergence theorem, we will try to find the values of a for which the method converges. We first check whether $\phi(x)$ is continuously differentiable or not.
$\phi(x) = ax(1-x^2)$
$\phi'(x) = a - 3ax^2$
Hence we can see that $\phi(x)$ is in fact continuously differentiable.
Then we check that $\alpha_1$ is, in fact, a fixed point of $\phi$:
$\phi(\alpha_1 = 0) = a(0)(1-0) = 0 = \alpha_1$.
Then by the local linear convergence theorem, the fixed point iteration converges if $|\phi'(\alpha_1)| < 1$, for all initial guesses $x^0 \in [\alpha_1-\delta, \alpha_1+\delta]$.
Therefore we have,
$|\phi'(\alpha_1)| = |a-3a(0)^2| = |a|$.
Hence, we have convergence for $|a|<1$.
The following part is the bit that I am unsure about (where you apply the final part of theorem with delta):
I just said, let $\delta = 1$, then for $|a|<1$ the method converges to $\alpha_1 = 0$, for all initial guesses $x^0 \in [-1,1]$
Since we are considering strictly positive values of a:
For $0<a<1$, the fixed point iteration method converges to $\alpha_1 = 0$, for all initial guesses $x^0 \in [0,1]$.
Attempt at part (iii): I am getting $1<a\leqslant a+1$ i.e. $a\in(1,a+1]$ which seems wrong, then I am not sure how to apply part (ii)?
 A: In (ii), you should have learned that you are guaranteed local convergence if $|\varphi'(r)|<1$ and you have global convergence if $|\varphi'(x)|<1$ for all $x$ in the domain. So this is just a matter of messing with some explicit inequalities now.
For (iii), as you say, the other possible zeros are $\pm \sqrt{\frac{a-1}{a}}$. These are real if $a \geq 1$ and the positive one is strictly positive if $a>1$. Is the positive one always in $(0,1]$ for $a>1$, or not? Once you know under what conditions it is in $(0,1]$, you can repeat what you did in (ii). (IMO the prompt is a bit vague, because it says "for which $a$ you can approximate $\alpha_2$?" and it isn't clear whether that means "for some $x^0 \in [0,1]$" or "for all $x^0 \in [0,1]$". If it's the former, then you can just check local convergence instead of global convergence.)
For (iv) you should have learned that the condition for second order convergence is $\varphi'(r)=0$, which is just an algebraic equation you can solve for $a$ since you know the root is $\sqrt{\frac{a-1}{a}}$.
