Proof using the Contraction Mapping Theorem

Question

How do you identify an interval [f,g] so that the Contraction Mapping Theorem guarantees convergence to the positive fixed point for the following:

a) $\frac{14-x^3}{13}\ $ b) $e^{-x}$

I tried drawing a graph and i see it visually but am not so sure about proving it rigorously.

I believe this helps : find an interval (a,b) where $0<f^{'}(x) < \theta, x \in (a,b)$ ($\theta$ is a fixed number between zero and one.) If you find this interval, you will have a root in this interval. try to do the some for the item b) — math student, May 09 '14 at 20:14

score 3 · Accepted Answer · answered May 09 '14 at 20:24

3

a) Let $f(x) = \frac{14 - x^3}{13}$, then $|f'(x)| < 1 \iff |-3x^2/13| < 1 \iff x \in (-\sqrt{13/3}, \sqrt{13/3}).$

b) Similiarly, let $g(x) = e^{-x}$, then $|g'(x)| < 1 \iff |e^{-x}| < 1 \iff x \in (0, \infty).$

answered May 09 '14 at 20:24

IAmNoOne

You type faster than me – recursive recursion May 09 '14 at 20:25
@recursiverecursion, really? I actually took my time on WolfamAlpha to make sure I didn't do any dumb calculation mistakes. – IAmNoOne May 09 '14 at 20:27
Well I am pretty bad with latex, just never got the hang of it. – recursive recursion May 09 '14 at 20:43
Don't worry keep practicing. – IAmNoOne May 09 '14 at 21:01

1 Answers1