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December 2nd, 2016, 07:32 AM  #1 
Newbie Joined: Dec 2016 From: Montreal Posts: 4 Thanks: 0  Sensitive Dependence on Initial Condtion
I would like to prove that a function f(x) has sensitive dependence on initial conditions. The theory, as I understand it is: We let J be an interval, and f : J → J. Then f has sensitive dependence on initial conditions at x if there is an ε > 0 such that for each d > 0 there is y ∈ J and a positive integer n such that x − y ≤ d and (f^n)(x) − (f^n)(y) ≥ ε How do I show that f(x) = x^3 has sensitive dependence at 1, and no sensitive dependence at 0. 
December 2nd, 2016, 01:00 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,412 Thanks: 716 
well looking at this what's happening is that the $\delta$ interval about $1$, i.e. $(1\delta, 1+\delta),~\delta>0$ will necessarily contain some values $x:~x+1<\delta \wedge x>1$ $\displaystyle{\lim_{n\to\infty}}\left \left(x^3\right)^n\right = \begin{cases} 0 &x<1 \\ 1 &x = 1 \\ \infty &x>1 \end{cases}$ So for these values with magnitude greater than 1 in the delta interval you can make $\left\left(x^3\right)^n\right$ as large as you like no matter how small $\delta$ is. On the other hand the $\delta$ interval about $x=0$ is such that $x < \delta < 1 \Rightarrow \displaystyle{\lim_{n\to\infty}}\left \left(x^3\right)^n\right = 0$ 
December 2nd, 2016, 01:40 PM  #3 
Newbie Joined: Dec 2016 From: Montreal Posts: 4 Thanks: 0 
Wow thanks! That makes a lot of sense.


Tags 
condtion, dependence, initial, sensitive 
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