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 December 2nd, 2016, 08:32 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, 02:00 PM #2 Senior Member     Joined: Sep 2015 From: CA Posts: 1,206 Thanks: 614 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 $|x| < \delta < 1 \Rightarrow \displaystyle{\lim_{n\to\infty}}\left |\left(x^3\right)^n\right| = 0$ Thanks from zactops
 December 2nd, 2016, 02:40 PM #3 Newbie   Joined: Dec 2016 From: Montreal Posts: 4 Thanks: 0 Wow thanks! That makes a lot of sense.

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