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Math Help - Functions and convergence

  1. #1
    Super Member Bacterius's Avatar
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    Functions and convergence

    Hello,
    I want to prove this :

    \mathbb{D}_f is the domain of f, and \mathbb{R}_f is the range of f.

    Let f be some function, and let x \in \mathbb{D}_f \cap \mathbb{R}_f satisfy f(x) = x. Let a series be defined by :

    \left \{<br />
\begin{array}{l}<br />
u_0 \in \mathbb{D}_f \\<br />
u_{n + 1} = f(u_n)<br />
\end{array}<br />
\right.

    Show that if this series converges towards a finite value, then it converges towards x.

    Note : you may use f(x) = \frac{1}{x} + 1, with x = \phi as an example.
    This can be explained because then \frac{1}{x} + 1 = x, yielding x^2 - x - 1 = 0.
    But I've never approached a problem that required to mix functions and series. I guess I could use some notions on series convergence, but could anybody give me a head start or even just tips ? Thanks all
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  2. #2
    A Plied Mathematician
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    This problem is way in the wrong forum. It belongs in the University Math Help/Analysis, Topology and Differential Geometry forum.

    I don't buy the result. You're dealing with fixed points here ( f(x) = x). The problem is, there's no constraints on f, other than the iteration scheme you've set up is known to converge, and there's no constraint on x, either. So, I could easily produce a function f that has two fixed points x and y\not=x. I then could pick u_{0}=y. The resulting sequence is convergent (it converges immediately!), but does not converge to x.
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  3. #3
    MHF Contributor

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    One "constraint" on f that can be used is that f is continuous.

    IF the sequence defined recursively by u_{n+1}= f(u_n}
    converges, to "x", say, then we can take the limit on both sides of the equation: \lim_{n\to\infty} u_{n+1}= \lim_{n\to\infty}f(u_n). If, additionally, f is continuous, that becomes [tex]\lim_{n\to\infty}u_{n+1}= f(\lim_{n\to\infty}u_n). But those are limits of the same sequence, just shifted one place.

    \lim_{n\to\infty}u_{n+1}= x and f(\lim_{n\to\infty}u_n)= f(x)
    Last edited by HallsofIvy; July 21st 2010 at 11:47 PM.
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  4. #4
    A Plied Mathematician
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    Reply to HallsofIvy:

    However, as I have already pointed out, even then you're not guaranteed that the limit of the sequence will be x. A continuous function could have multiple fixed points. Here's a continuous function that has uncountably many fixed points: f(x) = x.

    What you really need is a condition on f that guarantees uniqueness of the fixed point (you have existence already by assumption). Otherwise, this entire question is an exercise in futility.
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  5. #5
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Ackbeet View Post
    This problem is way in the wrong forum. It belongs in the University Math Help/Analysis, Topology and Differential Geometry forum.

    I don't buy the result. You're dealing with fixed points here ( f(x) = x). The problem is, there's no constraints on f, other than the iteration scheme you've set up is known to converge, and there's no constraint on x, either. So, I could easily produce a function f that has two fixed points x and y\not=x. I then could pick u_{0}=y. The resulting sequence is convergent (it converges immediately!), but does not converge to x.
    Really ? Sorry, I haven't done any topology or any of this advanced math ...
    Perhaps I didn't write the question very well, I should've thrown in a "show that there exists a convergent series of the form ... that converges towards x" (this is true because the initial term can be different).
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  6. #6
    A Plied Mathematician
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    Well, given the lack of conditions on f, I'd say the only sequence guaranteed to converge to x is the one that starts right on x and never moves! That is, u_0 = x. It's kind of a boring sequence, but if you start anywhere else, then there might be a different, closer fixed point to which your sequence converges besides x.
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