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Math Help - Error in Rudin Chapter 5 Exercise 25?

  1. #1
    Guy
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    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.
    Last edited by Guy; June 9th 2010 at 11:46 AM.
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  2. #2
    A Plied Mathematician
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    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.
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  3. #3
    Guy
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    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.
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  4. #4
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    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.
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  5. #5
    Guy
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    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.
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  6. #6
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    So, are you good to go?
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