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I read and understood the following definition*:

Define $$h(x)=|x|$$ on the interval $[-1,1]$ and extend the definition of $h$ to all of $\mathbb{R}$ by requiring that $h(x+2)=h(x)$. The result is a periodic "sawtooth" function.

However, I was then promptly asked to sketch the graph of $(1/2)h(2x)$ on $[-2,3]$ and to give a qualitative description of the functions $$h_n(x)=\frac{1}{2^n}h(2^nx)$$ as $n$ gets larger.

However, this is where I am confused: Is it not the case that $$h_n(x)=\frac{1}{2^n}h(2^nx)=\frac{1}{2^n}|2^nx|=\frac{2^n}{2^n}|x|=|x|,$$ and $h_n$ behaves just like $h$ regardless of the size of $n$? Either I am misunderstanding something or the book has made a typo. Could anyone clarify this for me? Thanks in advance!

*This is from Stephen Abbott's Understanding Analysis.

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    The problem is that $h(2^n x) = |2^n x|$ only when $|2^n x|\leq 1$, other wise the fact that $h(x+2) = h(x)$ part comes in and messes with it.2012-03-20
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    That would be true if $h$ were just the absolute value function, but it's not.2012-03-20
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    $h(x) = |x|$ only when $x \in [-1,1]$2012-03-20
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    Didn't mean to steal an answer from your comments. I got up to make tea while answering and didn't refresh my browser when I came back.2012-03-20
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    What is the maximum value of $h(x)$? What is the maximum value of $\frac{1}{2^n} h(2^n x)$?2012-03-20
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    I understand that the function looks like a sawtooth because of the periodicity, but I guess that what I truly wanted to ask is: are the graphs of, say, $h_1$, $h_5$ and $h_n$ for any $n\in\mathbb{N}$ alike? As in, exactly the same?2012-03-20
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    @JonasKibelbek, I believe I see it now.2012-03-20
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    @JosuéMolina You might be interested in [this question of mine.](http://math.stackexchange.com/questions/107237/heuristic-iterated-construction-of-the-weierstrass-nowhere-differentiable-functi)2012-03-21

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Not quite. We have that $h(x)=|x|$ only for $-1\leq x\leq 1$. Let $n=1$. Then, for example,

$$ h_1(1)=\frac{1}{2}h(2)=\frac{1}{2}h(0)=0. $$

This is the periodicity working.