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I view the orbit of $\frac{1}{2}$ of the dynamical system defined by $f:[0,1] \to [0,\frac{3}{4}]$ where $f(x)=3x(1-x)$ as the sequence $(x_{n})$. So

$$x_n = \frac{1}{2}, f\bigg(\frac{1}{2}\bigg), f\bigg(f\bigg(\frac{1}{2}\bigg)\bigg), \dots$$ $(x_n)$ appears to dance around the fixed point $\frac{2}{3}$ while slowly converging to it.

The odd-numbered iterates are increasing and the even-number iterates are decreasing. Let $(a_{k})$ be the sequence of odd terms of $(x_{n})$ where $k = 2n-1$ and $(b_{k})$ be the sequence of even terms of $(x_{n})$ where $k = 2n$.

Show $(a_k)$ is monotonic & bounded. 1. Increasing:

$a_{k}<a_{k+1} \forall \ k \in \ \mathbb{N}$. In terms of $(x_{n})$, show $x_{2n-2}<x_{2n+1} \forall \ n \in \ \mathbb{N}$.

$$x_{2n+1} - x_{2n-1}>0$$ $$(-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+9x_{2n-1})-(x_{2n-1}) > 0$$ $$-27x_{2n-1}^4+54x_{2n-1}^3-36x_{2n-1}^2+8x_{2n-1}> 0$$ $$x_{2n-1}(2-3x_{2n-1})^3 > 0$$ $$0<x_{2n-1}<\frac{2}{3}$$ $$0<a_{k}<\frac{2}{3}$$

Hence, the subsequence $(a_k)$ is strictly increasing and $(a_k)$ is bounded below by the first point of the subsequence $a_1=\frac{1}{2}$.

Question: How do I show $a_{k}$ is bounded above by the fixed point $\frac{2}{3}$?

Maybe, use epsilon delta proof to show convergence, which implies $(x_n)$. Or by induction, assume $x_{2n-1}< \frac{2}{3}$ show $x_{2n+1}=f(f(x_{2n-1}))<\frac{2}{3}$

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Please note that your map is actually $[0,1] \rightarrow [0,\frac{3}{4}]$ –  bonext Dec 6 '12 at 16:57
    
@bonext Thanks for pointing that out. Changes made. –  Joshua Burrow Dec 6 '12 at 17:40
    
@richard Could you help with the induction. I have By induction, we will show all odd term values are less than the fixed point $\frac{2}{3}$. For our basis step, $x_1=\frac{1}{2}<\frac{2}{3}$. For our induction step, assume $x_{2n-1} < \frac{2}{3}$, show $x_{2n+1}=f^2(x_{2n-1}) < \frac{2}{3}$. $$-27_{2n-1}^4+54_{2n-1}^3-36_{2n-1}^2+9_{2n-1}<\frac{2}{3}$$ $$x_{2n-1}(2-3x_{2n-1})^3-x_{2n-1} < \frac{2}{3}$$ Why is this true? –  Joshua Burrow Dec 6 '12 at 19:00

1 Answer 1

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Note that $f(\frac{1}{2})=\frac{3}{4}$(as commented by bonext), $f(\frac{2}{3})=\frac{2}{3}$(as you have noticed) and $f(\frac{3}{4})=\frac{9}{16}\in(\frac{1}{2},\frac{2}{3})$. Also note that $f$ is strictly decreasing on $[\frac{1}{2},1]$. Then we have $$f((\frac{1}{2},\frac{2}{3}))\subset(\frac{2}{3},\frac{3}{4})\quad\mbox{and}\quad f((\frac{2}{3},\frac{3}{4}))\subset(\frac{1}{2},\frac{2}{3}).$$

Now by induction, it is easy to see $x_{2n-1}\in(\frac{1}{2},\frac{2}{3})$ and $x_{2n}\in(\frac{2}{3},\frac{3}{4})$ when $n\ge 2$.

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Could you help with the induction. I have By induction, we will show all odd term values are less than the fixed point $\frac{2}{3}$. For our basis step, $x_1=\frac{1}{2}<\frac{2}{3}$. For our induction step, assume $x_{2n-1} < \frac{2}{3}$, show $x_{2n+1}=f^2(x_{2n-1}) < \frac{2}{3}$. $$-27_{2n-1}^4+54_{2n-1}^3-36_{2n-1}^2+9_{2n-1}<\frac{2}{3}$$ $$x_{2n-1}(2-3x_{2n-1})^3-x_{2n-1} < \frac{2}{3}$$ Why is this true? –  Joshua Burrow Dec 6 '12 at 19:00
    
@JoshuaBurrow: There is something wrong in your typing. By the way, why do not you apply the result in my answer to do the induction? –  23rd Dec 6 '12 at 19:03
    
I understand. After I use induction, I can conclude $x_{2n-1}$ is strictly increasing (from my first argument)and bounded by $(\frac{1}{2},\frac{2}{3})$ and $x_{2n}$ is strictly decreasing (similar to my first argument) and bounded by $(\frac{2}{3},\frac{3}{4})$. Then by Monotone Convergence Theorem $(x_n)$ converges to $\frac{2}{3}$. You have been a big help!! Now I can find the forward limit set of the orbit to be the orbit and it's limit point!! –  Joshua Burrow Dec 6 '12 at 20:28
    
@JoshuaBurrow: To show that $x_{2n-1}$ is increasing, an alternative argument is to notice that $f^2:[\frac{1}{2},\frac{2}{3}]\to[\frac{1}{2},\frac{2}{3}]$ is increasing and $x_1=\frac{1}{2}$, $x_1<x_3<\frac{2}{3}$. The same method can be used to prove that $x_{2n}$ is decreasing. –  23rd Dec 6 '12 at 20:36

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