How did you prove that ? If there is no restriction on the starting point, , then you could easily get an increasing sequence.
This is driving me crazy, because it seems wrong but I haven't found it in any of the errata.
Suppose is twice differentiable on 0, and for all . Let be the unique point in at which .
Prove that , and that
My issue is that I can only show i.e. I can't get a strict inequality. I would think there is a counterexample if we take so that regardless of which you choose, .
In the counter-example you mentioned, I think you'd find that . This, of course, is Newton's Method, and it converges in one step for linear functions (maybe even for quadratic?).
I would focus on the fraction . If you can show that that is positive, you'd be done. You already know the denominator is positive, by assumption.
Maybe you could try showing that .
That, I think, would do the trick.
However, it is certainly true that if you actually attain the zero, you'll get equality for all subsequent iterations.
I'm being really sloppy; in the counterexample was the initial condition, as opposed to .
Thanks for the response. If you allow the inequality to be non-strict it's an easy problem. I thought maybe I had missed something in the problem statement that disallowed my counterexample. If then I think you do get a strict inequality, so it doesn't converge in one iterate for quadratics, and will never land directly on , if and .