# Determine $p$, $q$ and $r$ so that the order of the fixed point iteration for computing $a^{1/3}$ becomes as high as possible

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?

• Don't forget to choose $p,q,r$ so that $a^{1/3}$ is in fact a fixed point of $g(x)$. Since $g(a^{1/3})=(p+q+r)a^{1/3}$, this can't converge to $a^{1/3}$ unless $p+q+r=1$. Commented Nov 21, 2014 at 3:55
• Also, your derivative calculations are all wrong. For example, I get $g'(a^{1/3})=p-2q+5r$. If you recalculate them, it will probably make a lot more sense. Commented Nov 21, 2014 at 3:59

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 $$p+q+r=1,\quad\text{and}\quad p-2q-5r=0$$
• This is what I've got so far: $p + q + r = 1, \,\,$ $p - 2q - 5r = 0, \,\,$ $q + 5r = 0, \,\,$ $24q + 210r = 0, \,\,$ I can keep finding the derivatives of g(x) forever. How do I know when I find highest order convergence for $g(x)$ because to do that I would need to explicitly make the $k^{th}$ derivative not equal to 0. How do I know that I have found the highest derivative possible for $g(x)$? Commented Nov 21, 2014 at 16:45
• It's not so much making a derivative not equal to zero, but to see how many derivatives you can keep to be zero. For example, with your 4 equations, you can solve the first 3 to ensure that $g(x)=1$, $g'(x)=0$ and $g''(x)=0$, but when you put the solution for $p$, $q$ and $r$ into $g'''(x)$, you will find it is not zero. Commented Nov 21, 2014 at 22:57