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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}. In terms of $(x_{n})$, show $x_{2n-2}.

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

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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    @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?2012-12-06

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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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    @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. The same method can be used to prove that $x_{2n}$ is decreasing.2012-12-06