Determine the constants $b,c,d$ so that the succession $x_{n+1}=g(x_n)$ has a cubic convergence towards $\sqrt{z}$

Question

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.

Answer 1

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.

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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 $D=cx^2+d$.