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Well, I have a basic question about double map function, let's go to it...

The double map $D:[0,1) \to [0,1)$ is define by $$ D(x) = 2x\pmod 1 = \begin{cases} 2x & \quad \text{if } 0 \leq x < {1\over 2}\\ 2x-1 & \quad \text{if } {1\over 2}\le x < 1 \end{cases}$$

My doubt is that the Devaney's book "Differential Equations, Dynamics Systems, and An Introduction to Chaos" says "An easy computation shows that $D^n(x) = 2^n x \pmod 1$".

Firstilly, I was open that first equation like

\begin{align} D(x) & \equiv 2x\pmod 1\\ 1 | (2x -D(x)) &\Rightarrow k = 2x -y \end{align} where $y = D(x)$ and $k \in \mathbb{Z}$. So $$ \begin{align} &\text{If } k=0 \text{ then} \quad 2x -y=0 \Rightarrow y=2x \begin{cases} y=0 & \Rightarrow 2x=0 \Rightarrow x=0\\ y=1 & \Rightarrow 2x=1 \Rightarrow x=1/2 \end{cases}\\ &\text{So } 0 \leq x < 1/2. \end{align} $$

$$ \begin{align} &\text{If } k=1 \text{ then} \quad 2x -y=1 \Rightarrow y=2x -1 \begin{cases} y=0 & \Rightarrow 2x-1=0 \Rightarrow x=1/2\\ y=1 & \Rightarrow 2x-1=1 \Rightarrow x=1 \end{cases}\\ &\text{So } 1/2 \leq x < 1. \end{align} $$

Now, I want to proof that $D^2(x) = D(2x) = 2^2 x \pmod 1$. But I do not undestant how. What did I do

$$D(D(x)) = 2(2 x \pmod 1) \pmod 1 \Rightarrow 2(2x-y)-y=k$$

And It is not effective...

So, how can I work with the second and next $D$ iterations, to proof $D^k(x)$ for $k \in \mathbb{N}$?

Thanks for help!

Mostafa Ayaz
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    You can do induction over $k$. Then we have (where all the equalities are as element in $\mathbb{R}/\mathbb{Z}$, which is exactly what mod 1 means) $$D^{k+1}(x)= DD^k(x)= 2 D^k(x)=2^{k+1} x$$ – Severin Schraven Aug 24 '19 at 23:05

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Write $\{x\}$ for the fractional part of $x$, and $[x]$ for the integer part of $x$, so $\{x\}=x-[x]$ and $[x]$ is an integer and $0\le\{x\}<1$. Note that $x\bmod 1$ is just another way to say $\{x\}$. You want to know why $2(2x\bmod1)\bmod1=4x\bmod1$. We'll prove that if $a,b$ are integers then $\{abx\}=\{a\{bx\}\}$. Your question is just the case $a=b=2$.

$a\{bx\}=a(bx-[bx])=abx-a[bx]\equiv abx\bmod1$, so $\{a\{bx\}\}=\{abx\}$.

Gerry Myerson
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