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Suppose I have a Tent Transformation which is defined by:

\begin{align*}T(x)=\begin{cases}3x&\text{if $x\le\dfrac12$,}\\3(1-x)&\text{if $x\ge\dfrac12$.}\end{cases}\end{align*}

After noticing that $T(x)$ is continuous, although not differentiable at $x=\frac12$, if we let s $\in$ $\mathbb{R}$, we can then iterate over $T(x)$.

Like such:

$$ s, T(s), T(T(s)), T(T(T(s)))...$$

Well, in any case, I have been playing around with this problem and plugging in different values for $s$, and well... I'm not sure how to go about the following question.

After finding the pattern through the iterations, use this pattern to find all the points that remain bounded after infinitely many iterations.

EDIT: For which real numbers $s$ is this sequence bounded, and for which real numbers $s$ does this sequence diverge?

Since it says infinitely many iterations am I suppose to find

$$ \lim_{s\to \infty} Z(s), $$ with $Z(s)$ being some recursive function used to iterate over $T(s)$.

I'm not sure. Am I going about this completely wrong ?

Also, how would I graph for example $$y=T^2(x)$$ or $$y=T^3(x)$$

Did
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Marco Neves
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  • For your last question, just write explicit expressions for the second and third iterations, its not that hard – Yuriy S May 02 '16 at 08:46
  • @YuriyS, would you care to write an answer explaining what you mean by that? If it's not asking too much – Marco Neves May 02 '16 at 08:53
  • Ok wait. Is this the question: Consider the sequence of numbers given by $x, T(x), T(T(x)), T(T(T(x))),...$. The limit of this sequence exists and tends to the fixed point of $T$ i.e. a point such that $T(y)=y$, under certain conditions. If this is not the question, then I don't understand it, you will have to explain it. – Sarvesh Ravichandran Iyer May 02 '16 at 09:25
  • @астонвіллаолофмэллбэрг you didn't really write a question? What I want to know is how to find all the points for which $F(x)$ remains bounded, meaning it still lies on [0,1]. Hope that explains it. – Marco Neves May 02 '16 at 09:29
  • @астонвіллаолофмэллбэрг I'll edit the question to make it clear. – Marco Neves May 02 '16 at 09:30

1 Answers1

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If $0\le x\le1/3$ or $2/3\le x\le1$, then $0\le T(x)\le1$; if $1/3<x<2/3$, then $T(x)>1$. Let's think now about $T^2(x)$. When $0\le x\le1$, since $t(x)$ varies between $0$ and $1$, $T^2(x)$ wiill be a tent over the interval $[0,1/3]$. Similarly on $[2/3,1]$. On $(1/3,2/3)$ $T^2$ is negative. So, the graph of $T^2$ has two tents on the extremens and an inverted tent under the $x$ axis in the middle. Iterating the argument, we see that the graph of $T^3$ will have $4$ tents over the intervals $[0,1/9]$, $[2/9,1/3]$, $[6/9,6/9]$ and $[8/9,1]$, and it will be below the $x$-axis on the complement of those intervals. In general, $T^n$ will have $2^{n-1}$ tents over intervals of length $3^{-n+1}$.

It is also easy to see that if $x$ $T^k(x)<0$, then $T^n(x)\to-\infty$ as $n\to\infty$. It follows that the set of $x$ for which $T^n(x)$ remains bounded is the intersection of those intervals, which is Cantor's middle third set.

Julián Aguirre
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