As a part of finding square roots, I am asked to use Halley's method on $f(x)=x^2-a$. Using this I derived the iteration formula:
$$ p_{n+1} =\frac{p_n^3 + 3a p_n}{3p_n^2 +a}$$
Now I am asked to investigate whether the convergence process converges to a unique fixed point $p$. As a hint I should consider the intitial point $p_0$ and the unique fixed point $p$. More precisely:
Hint: consider the cases $p_0 > p$ and $0 < p_0 < p$ separately.
I am not sure how to use this hint, I think it boils down to manipulating inequalities, but my problem is that I have an equation in terms of $p_n$ and not $p$ and $p_0$. I guess an assumption would be that if we get a fixed point then $p_{n+1}= p_n$. Any ideas on this?
Best Answer
You can show that $$ q_{n+1}=\frac{p_{n+1}-\sqrt{a}}{p_{n+1}+\sqrt{a}}=\left(\frac{p_{n}-\sqrt{a}}{p_{n}+\sqrt{a}}\right)^3=q_n^3. $$ The convergence behavior directly follows from $|q_0|=\left|\frac{p_{0}-\sqrt{a}}{p_{0}+\sqrt{a}}\right|<1$ for $p_0>0$, as then $$p_n=\frac{1+q_0^{3^n}}{1-q_0^{3^n}}\sqrt{a}.$$