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Consider the function $$g(x)=\frac{x^3+bx}{cx^2+d}$$How can I determine the constants $b,c,d$ so that the succession $x_{n+1}=g(x_n)$ has a cubic convergence towards $\sqrt{z}$?. My professor solved it calculating $$g(\sqrt{z})-\sqrt{z}=\frac{\sqrt{z}N}{D}$$$$g'(x)=\frac{F}{D^2}$$$$g''(x)=\frac{E}{D^3}$$ where $N,D,F,R$ are equations of variables $c,b,d,z$. Finally just solving the system of equations formed by $N,F,E$ he has $b=3z,c=3,d=z$. I'd appreciate if someone could guide me trough this solution.

codingnight
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    Which part(s) of this solution do you fail to get, and why? – Did Mar 28 '18 at 16:14
  • Everything I think. I know when an order of convergence is cubic, that this is a fixed point method and therefore if it has to converge we must have $g'(s)\leq{0}$. Why in the first calculation he didn't take $\sqrt{z}$? Why did he rest $\sqrt{z}$ only on the first calculation? Why is the solution given by the system of equations of the numerators? – codingnight Mar 28 '18 at 16:36
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    "Why in the first calculation he didn't take √z?" Which "first c[omput]ation"? Please add some much needed context. – Did Mar 28 '18 at 17:03
  • Here: $g(\sqrt{z})-\sqrt{z}=\frac{\sqrt{z}N}{D}$. In the other cases, he took all the numerator to build the system of equations, here he takes $N$ instead of $\sqrt{z}N$. I understand this post is vague, sorry for that, it is because I really can't figure out the point of the solution. Maybe it would be better if you could explain how you would solve this problem. – codingnight Mar 28 '18 at 17:14
  • My point is that at present one cannot address your concerns about your professor's solution because one does not know what this solution says. – Did Mar 28 '18 at 17:23
  • Also, this is the Halley method for $f(x)=x^2-z$ which is also the Newton method for $f(x)=x^{1.5}-zx^{-0.5}$. – Lutz Lehmann Mar 29 '18 at 21:08

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You want $$ g(x)-\sqrt{z}=O((x-\sqrt z)^3). $$ On the other hand, $$ g(x)-\sqrt{z}=\frac{x^3+bx-\sqrt{z}(cx^2+d)}{cx^2+d} $$ has a third degree polynomial as denominator. To get the error order, the denominator must be equal to $$ (x-\sqrt z)^3=x^3-3x^2\sqrt{z}+3xz-\sqrt{z}^3=x^3+3zx-\sqrt{z}(3x^2+z) $$ which directly gives the coefficients.

What your lecture probably contained was that as fixed point of $g$, $\sqrt z$ is a root of $$g(x)-x=\frac{(x^3+bx)-(cx^3-dx)}{cx^2+d}=x\frac{(1-c)x^2+(b-d)}{cx^2+d},$$ thus $N(\sqrt z)=(1-c)z+(b-d)=0$. To have cubic convergence, $\sqrt z$ also needs to be a root of $g'(x)$ and $g''(x)$ which have the indicated form with $D=cx^2+d$.

Lutz Lehmann
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