I have this exercise in my book:
Let $f(x)=x^3-5$ and we are looking for the solution $f(x^*)=0$, that is $x^*=5^{1/3}$, with Newton's method. So $x_{n+1}=x_n-\frac{x_n^3-5}{3x_n^2}$ and let $e_n=x_n-x^*$ be the $n$-th error. If we take $x_0>x^*$, is it true that $e_{n+1}\le L\cdot e_n$ for any $n\ge 0$ and constant $L<1$?
And I don't know what is wrong with it. I read that Newton's method order of convergence is $2$ that is $e_{n+1}\le L\cdot e_n^2$ for some constant $L<1$. Isn't this inequality in this problem weaker? Anyway, I am missing something probably. I will be very grateful for any explanation and how to answer above question.
