- #1

k3N70n

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I wrote a numerical analysis midterm earlier this week and there was one question I couldn't figure out. I was wondering if anyone had some insight.

What I've been told and what I've read in many many places is that

f(x) will converge to a fixed point on an interval I if

1. f(x) is continuous and differentiable on I

2. |f'(x)|<1 on I

Now the question I was posed was given [itex] x_{n+1}=\frac{1}{3}(x_n^2+2)[/itex] prove that [itex]x_n\rightarrow 1[/itex] as [itex]n\rightarrow\infty [/itex] if [itex] -2<x_0<2[/itex] (that is prove [itex]x_{n+1}[/itex] is between [itex]x_n[/itex] and 1 when [itex]n \geq 1[/itex])

What was most natural to me was to find the interval where |f'(x)|<1 which happens to be on [itex](-\frac{3}{2},\frac{3}{2})[/itex]. It easy to see (but not so easy to prove) that the interval (-2,2) will work as well. How would you go about showing that? Do think that it would be possible to generalize this such that the requirement |f'(x)|<1 would not be needed for fixed point iteration?

Anyway, hope this interest someone else.