convergence rate of newton's method So, I'm currently studying Newton method used for finding the 0's of a function, however my professor has only announced  that the speed of this algorithm can be more than quadratic, however I'm wondering when this happens, since in the demonstration used by him to explain the quadratic case, there is no evidence about the &quot;more than quadratic&quot;
Using Taylor and Lagrange
$f(\xi)=f(x_n)+f'(x_n)(\xi-x_n)+\frac{f''(z_n)}{2}(\xi-x_n)^2$
$-\frac{f(x_n)}{f'(x_n)}=\xi-x_n+\frac{f''(z_n)}{2f'(x_n)}(\xi-x_n)^2$
$x_{n+1}-x_n=\xi-x_n+\frac{f''(z_n)}{2f'(x_n)}(\xi-x_n)^2$
$e_{n+1}=\left|x_{n+1}-\xi\right|=c_ne_n^2 \quad \text{with} \quad c_n=\frac{1}{2}\frac{\left|f''(z_n)\right|}{\left|f'(x_n)\right|}$
Can someone please tell me when (example) the order of the convergence is cubic, and why?=
Since you asked specifically about cubic, consider $f(x)=x+x^3$.
The Newton iteration would be $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n+x_n^3}{1+3x_n^2}=\frac{2x_n^3}{1+3x_n^2}$$
It converges to the root $x=0$. So, for $1-3|x_n|^2&gt;1/2$,
$$\epsilon_{n+1}=|x_{n+1}-0|=\frac{2|x_n|^3}{|1+3x_n^2|}\leq \frac{2|x_n|^3}{1-3|x_n|^2}\leq 4|x_n-0|^3=4\epsilon_n^3$$