1. ## Functions and convergence

Hello,
I want to prove this :

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

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

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

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

Note : you may use $\displaystyle f(x) = \frac{1}{x} + 1$, with $\displaystyle x = \phi$ as an example.
This can be explained because then $\displaystyle \frac{1}{x} + 1 = x$, yielding $\displaystyle x^2 - x - 1 = 0$.
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

2. 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 ($\displaystyle f(x) = x$). The problem is, there's no constraints on $\displaystyle f$, other than the iteration scheme you've set up is known to converge, and there's no constraint on $\displaystyle x$, either. So, I could easily produce a function $\displaystyle f$ that has two fixed points $\displaystyle x$ and $\displaystyle y\not=x$. I then could pick $\displaystyle u_{0}=y.$ The resulting sequence is convergent (it converges immediately!), but does not converge to $\displaystyle x$.

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

IF the sequence defined recursively by $\displaystyle u_{n+1}= f(u_n}$
converges, to "x", say, then we can take the limit on both sides of the equation:$\displaystyle \lim_{n\to\infty} u_{n+1}= \lim_{n\to\infty}f(u_n)$. 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.

$\displaystyle \lim_{n\to\infty}u_{n+1}= x$ and $\displaystyle f(\lim_{n\to\infty}u_n)= f(x)$

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.

5. Originally Posted by Ackbeet
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 ($\displaystyle f(x) = x$). The problem is, there's no constraints on $\displaystyle f$, other than the iteration scheme you've set up is known to converge, and there's no constraint on $\displaystyle x$, either. So, I could easily produce a function $\displaystyle f$ that has two fixed points $\displaystyle x$ and $\displaystyle y\not=x$. I then could pick $\displaystyle u_{0}=y.$ The resulting sequence is convergent (it converges immediately!), but does not converge to $\displaystyle x$.
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).

6. 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.