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

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

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

EDIT: Assume $\displaystyle x_0 \in (\xi, b)$
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My issue is that I can only show $\displaystyle x_n \ge x_{n + 1} \ge \xi$ i.e. I can't get a strict inequality. I would think there is a counterexample if we take $\displaystyle f(x) = x$ so that regardless of which $\displaystyle x_1$ you choose, $\displaystyle x_2 = x_3 = ... = 0$.

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

3. Whoops, left that out.

$\displaystyle x_0 \in (\xi, b)$

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 $\displaystyle x_{1}=\xi$. 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 $\displaystyle \frac{f(x_{n})}{f'(x_{n})}$. 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 $\displaystyle f(x)>0\;\forall\,x\in(\xi,b)$.

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 $\displaystyle x_1$ was the initial condition, as opposed to $\displaystyle x_0$.

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

6. So, are you good to go?