# Thread: Error in Rudin Chapter 5 Exercise 25?

1. ## Error in Rudin Chapter 5 Exercise 25?

This is driving me crazy, because it seems wrong but I haven't found it in any of the errata.

Problem:
Suppose $f$ is twice differentiable on $[a, b], f(a) < 0, f(b) > 0, f'(x) \ge \delta >$0, and $0 \le f''(x) \le M$ for all $x \in [a, b]$. Let $\xi$ be the unique point in $(a, b)$ at which $f(\xi) = 0$.

Define
$x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Prove that $x_{n + 1} < x_n$, and that
$\lim x_n = \xi$.

EDIT: Assume $x_0 \in (\xi, b)$
----------------------------------------------

My issue is that I can only show $x_n \ge x_{n + 1} \ge \xi$ i.e. I can't get a strict inequality. I would think there is a counterexample if we take $f(x) = x$ so that regardless of which $x_1$ you choose, $x_2 = x_3 = ... = 0$.

2. How did you prove that $x_{n+1}\le x_{n}$? If there is no restriction on the starting point, $x_{0}$, then you could easily get an increasing sequence.

3. Whoops, left that out.

$x_0 \in (\xi, b)$

After looking carefully over the wording of the problem, I don't think I left anything out now.

4. In the counter-example you mentioned, I think you'd find that $x_{1}=\xi$. 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 $\frac{f(x_{n})}{f'(x_{n})}$. 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 $f(x)>0\;\forall\,x\in(\xi,b)$.

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.

5. I'm being really sloppy; in the counterexample $x_1$ was the initial condition, as opposed to $x_0$.

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 $0 < f''(x)$ 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 $0$, if $f(x) = x^2$ and $x_0 = 1$.

6. So, are you good to go?