Show that the iteration $x_{n+1} = x_n - 2\frac{f(x_n)}{f'(x_n)}$ converges quadratically to $x_*$ provided $x_0$ is sufficiently close to $x_*$

Question

We have the following conditions for the above slightly-modified Newton's method iteration:

• $f$ is a real function of one real variable
• $f''$ is Lipschitz continuous
• $f(x_*) = f'(x_*) = 0$

I also have to show that convergence does not occur if $x_0 = x_*$, but that seems much more straightforward.

So far I've figured out that solving this requires Taylor series expansions and a prior result that states $x^{(k+1)} - x_* = O(|x^{(k)} - x_*|^2)$, but have no idea where to go from there.

Answer 1

I suspect you need $f''(x_*) \neq 0$.

If $f(x) = x^3$, then ${ f(x) \over f'(x) } = {x \over 3}$, and the update becomes $x_{n+1} = x_n - 2{x_n \over 3} = { 1 \over 3} x_n$, which is not quadratic.