Define
$$h(x)=|x|$$
on the interval $[-1,1]$ and extend the defintion of $h$ to all of $\mathbb{R}$ by requiring that $h(x+2)=h(x)$.
Now define the function:
$$h_n (x)=\frac{1}{2^n} h(2^n x)$$
Could you please help me understand why the maximum height of the latter function is $\frac{1}{2^n}$ and the period is $\frac{1}{2^{n-1}}$?
My confusion arises from the fact that I see no reason why the $2^n$ cannot be cancelled based on the definition of $h$. We are left with the original function in this case.
Thanks.
The maximum of $h(x)$ is $1$.
We have $h_n(x)=\frac1{2^n}h(t)$, where $t=2^nx$
As we have maximum of $h$ as 1.
$$h(t)\le1$$ $$h_n(x)=\frac1{2^n}h(t)\le\frac{1}{2^n}$$