So I'm given the following equation for computing $a^{1/3}$ $$x_{k+1}=px_k + \frac{qa}{x_k^2} + \frac{ra^2}{x_k^5}$$ and I have to find the p, q, and r so that this equation converges to $a^{1/3}$ as fast as possible.
We were told to use the Rate of Convergence Theorem which states:
For the fixed point iteration $x_{k+1} = g(x_k)$ converging to $\tilde{x}$, if $g'(\tilde{x}) =g''(\tilde{x}) = ... =g^{(p-1)}(\tilde{x}) = 0 $ but $g^{(p)}(\tilde{x}) \ne 0$ we have $p^{th}$ order convergence.
My work:
$$g(a^{1/3}) = (p + q + r)a^{1/3} $$ $$g'(a^{1/3}) = \frac{1}{3}(p + q + r)a^{-2/3} $$ $$g''(a^{1/3}) = \frac{-2}{9}(p + q + r)a^{-5/3} $$ $$g'''(a^{1/3}) = \frac{10}{27}(p + q + r)a^{-8/3} $$ $$g^{(4)}(a^{1/3}) = \frac{-80}{81}(p + q + r)a^{-11/3} $$
If I select $p + q + r = 0$, then for any $k^{th}$ derivative $g^{(k)} = 0$ which obviously doesn't help us in solving the question. Any hints?
Your derivatives are incorrect, you need to do derivatives with respect to $x$ then substitute $x=a^{1/3}$, rather than substitute then take the derivative with respect to $a$.
Then you'll get $$g(x)=px+q\frac{a}{x}+r\frac{a^2}{x^5}$$ $$g'(x)=p-2q\frac{a}{x^2}-5r\frac{a^2}{x^6}$$ etc., and the first two equations you get for $g(a^{1/3})=a^{1/3}$, $g'(a^{1/3})=0$, etc. are $$p+q+r=1,\quad\text{and}\quad p-2q-5r=0$$
I think you should be able to go on from there.