My profesor said that (if $\alpha$ is the root of the function) then $$|x_{k+1}-\alpha|\leq \displaystyle\frac{M_2}{2m_1}|x_{k+1}-x_{k}|,$$ where $M_2$ is the maximum of the second derivative and $m_1$ is the minimum of the First derivative. He said it's proven using Taylor Expansion between $x_k$ and $x_{k+1}$, but I can't see why.
Any help?
One has, assuming the inequality is about a scalar function $f$, $$ m_1|x_{k+1}-α|\le |f(x_{k+1})| $$ by applying the mean-value formula and then using the lower bound on the derivative.
In the next step, apply the linear Taylor expansion with quadratic remainder term and apply the upper bound for the second derivative, $$ f(x_{k+1})=\underbrace{f(x_k)+f'(x_k)(x_{k+1}-x_k)}_{=0}+\frac12f''(c_k)(x_{k+1}-x_k)^2. $$ Combined this results indeed in something very similar to the claimed inequality $$ |x_{k+1}-α|\le \frac{M_2}{2m_1}|x_{k+1}-x_k|^2. $$ Try to find out where the square got lost. Without the square the inequality is only valid if $|x_{k+1}-x_k|\le1$
As a possible variation on the topic, a simplified version on the Newton-Kantorovich theorem says that if $M_2|f'(x_k)^{-1}|\,|x_{k+1}-x_k|\le\frac12$, then there is exactly one root in the ball or interval given as center and radius $B(x_{k+1},|x_{k+1}-x_k|)$, that is, $|x_{k+1}-α|\le|x_{k+1}-x_k|$ or $α\in[x_k,2x_{k+1}-x_k]$.