Maybe someone can give me a hint here:
question 1: Given a sequence {$x_n$} which is the Newton-Raphson sequence on some $f(x)$ s.t. $f(x)$ is strictly convex and $f'>0$.
Let $\alpha$ be the unique root of $f$ (i.e $f(\alpha$) = 0).
For any $x_0$ initial guess,
prove that $\alpha < x_n$ and $x_n > x_{n+1}$.
I know that I should use the fact that for any given $x_0,x_1$ s.t. $x_0 < x_1, f'(x_0) < \frac{f(x_1) – f(x_0)}{x_1 – x_0}$, and although I can see the geometrical correctness instantly, I am having a hard time to prove it formally.
Appreciate your guidance,
Yotam.
Best Answer
Because of convexity, the tangent $f(a)+f'(a)(x-a)$ is a supporting hyper"plane" and thus completely below the function. Which means that at the root of the tangent, the function value is above zero, positive. As $f$ is strictly increasing, this means that $x_1$ is right of the root with $f(x_1)>0$, independent of the position of $x_0$, if that is not already the root $α$. This property then also holds for all further Newton iterates.
As the tangent in $x_n$ is also monotonically increasing, the positive value $f(x_n)>0$ implies that the next iterate $x_{n+1}$ has to be left of $x_n$.