Chebyshev's method: $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}-\frac{f''(x_n)}{2f'(x_n)}\left(\frac{f(x_n)}{f'(x_n)}\right)^2$$ The question might seem simple, but I can't seem to find an answer to it. I have scanned through several numerical analysis/methods books, but all of them just say it has cubic convergence. They don't explain why!
So does anyone know how to show that it has cubic convergence or know a place/book/site where it's explained? It would be much appreciated.
On
Set $s=-\frac{f(x)}{f'(x)}$ as the Newton update. Then we know from the quadratic convergence of Newton's method (or from a Taylor expansion) that $f(x+s)=O(s^2)$. Any further correction $v$ in direction root will have to be smaller than $s$, that is, has size $v=O(s^2)$ in the sufficiently smooth case. Then sorting the Taylor expansion of the correction gives \begin{align} f(x+s+v)&=f(x)+f'(x)(s+v)+\frac12f''(x)(s+v)^2+O(s)^3\\ &=f'(x)v+\frac12f''(x)s^2+O(s^3), \end{align} which means that with $v=-\dfrac{f''(x)}{2f'(x)}s^2$ one gets $$ f(x+s+v)=O(s^3)=O(x-x^*)^3 $$ which establishes the cubic convergence of the Euler-Chebyshev method.
See Second order Taylor method to solve system of equations for the application of the same logic to systems of equations.
The answer can be found here: Метод Чебышева. Some knowledge of the Russian language is needed to understand the explanation.