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The following is a problem that I can't solve, and I need some tips:


Problem: For $x>-1$, define $f_1(x) = x,\ f_{n+1}=\dfrac{1}{1+f_n(x)}$. Find the limit function $f(x)$ and all the subsets of $A=(-1,\infty)$ where the convergence is uniform.


First, I waste hours trying to find a closed form for $f_{n}(x)$. What I've found is that, for $n\ge 3$,$$\ f_n(x)=\dfrac{a_{n-2} + a_{n-3}x}{a_{n-1}+a_{n-2}x}$$ where $a_n$ is the n-th Fibonacci number (starting with $a_0=a_1=1$). If this is correct, then the limit function is $f(x)=\dfrac{1}{\psi}$ ($\psi$ is the golden number)

Then, I try to find where the convergence is uniform. I use the derivative in case it helps, but $$f_n'(x)= \dfrac{(-1)^n}{(a_{n-1}+a_{n-2}x)^2}$$ Cassini's Identity appears and $f_{2n}$ is increasing but $f_{2n+1}$ is decreasing. So it feels the wrong direction.

As always, any help will be appreciated.

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Hint 1: any limit of a recurrence of the form $f_{n+1} = T(f_n)$ must satisfy $T(f) = f$. In your case that means that any limit of your problem must have range contained in $\{ \phi, 1/\phi \}$, and a continuous limit of your problem must be constantly equal to one of these. And any limit which is uniform on compact subsets must be continuous, since $f_n$ is continuous.

Hint 2: consider $g(x) = \frac{1}{1+x}$. Notice that $g((-1,\infty)) = (0,\infty)$, $g((0,\infty)) = (0,1)$, $g((0,1)) = (1/2,1) \subset (0,1)$. So for any $n \geq 3$, $f_n(x)$ is a bounded, continuous function whose range is contained in $(0,1)$. What does that tell you?

Hint 3: Try using my second hint with the Banach fixed point theorem, applied to the bounded continuous functions on $(-1,\infty)$.

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  • $\begingroup$ I get the idea, seems nice! but (0,1) is not complete, so I can't use Fixed Point Theorem... or bounded + continuous guarantees it? I'm confused =S $\endgroup$
    – FormerMath
    Nov 4 '14 at 1:24
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    $\begingroup$ @Luis $(0,1)$ not being complete is not the problem. $(-1,\infty)$, the domain of the functions, is not complete; herein lies your problem. But you can fix it by restricting to any compact subset of $(-1,\infty)$. $\endgroup$
    – Ian
    Nov 4 '14 at 1:28
  • $\begingroup$ Oh, I get it. Let me work with your hints. Thank you! $\endgroup$
    – FormerMath
    Nov 4 '14 at 1:32
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1) You can do in this way:

If $\lim_{n\to \infty}f_n(x)$ exists, say $S$, then $S=\frac{1}{1+S}$.

Can you proceed futher?

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  • $\begingroup$ For computing the limit function, yes. But I don't see how to proceed for uniform convergence. Any help with that? $\endgroup$
    – FormerMath
    Nov 4 '14 at 1:05
  • $\begingroup$ Sorry, I cannot do. $\endgroup$
    – Paul
    Nov 4 '14 at 1:06
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If the limit exists, then $f = \frac{1}{1 + f} $, that is $f^2 + f - 1 = 0$, so the limit is a golden ratio - or its reciprocal. Problem is to figure out for which $x$ it does converge, and to which one of two possible limits.

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  • $\begingroup$ Seems to be a similar answer than @Paul's. How to proceed with uniform convergence? $\endgroup$
    – FormerMath
    Nov 4 '14 at 1:06

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