I am trying the following exercise:
Prove that the function $ f(x) = x^3 - 2x - 5 $ has a root, $ r $, at $ [2, \ 2.2] $. If we consider the succesion $ (x_n)_n $ defined by Newton Method starting with $ x_1 = 2 $, prove that $ |x_{n + 1} - r| \leq 0.7 |x_n - r|^2 $.
I have no problem with most of that, but I am not sure about the inequality... I know that if we call $ e_n = x_n - r $, then Taylor Polynomial shows that $$ e_{n + 1} = \frac{e_n f'(x_n) - f(x_n)}{f'(x_n)} = \frac{f''(\xi_n) / 2}{f'(x_n)} e_n^2 = \frac{3 \xi_n}{3x_n^2 - 2} e_n^2 $$ for some $ \xi_n $ between $ r $ and $ x_n $. So, I suppose there is some way to get $$ \frac{3 \xi_n}{3x_n^2 - 2} \leq 0.7 \ \forall n \in \mathbb{N} $$ But I have not succeeded. The reason is I do not know anything about how $ x_n $ behave (it's bounded? maybe $ x_n $ is always between $ 2 $ and $ 2.2 $?). I failed trying to prove these things, so I post it here for help.