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I recently ran into this interesting exercise:

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

Now, define$g(x)=\sum_{n=0}^{\infty}\frac{1}{2^n}h(2^nx).$Consider the sequence $x_m=1/2^m$, where $m\in\mathbb{N}\cup\{0\}$. Show that$\frac{g(x_m)-g(0)}{x_m-0}=m+1$and use this to prove that g'(0) does not exist.

I solved this the following way:$g(x_m)=\sum_{n=0}^{\infty}\frac{1}{2^n}h\left(\frac{2^n}{2^m}\right)=\frac{1}{2^m}(m+1),$$\frac{g(x_m)-g(0)}{x_m-0}=\frac{1/2^m(m+1)}{1/2^m}=m+1,$$g'(0)=\lim_{m\to\infty}\frac{g(x_m)-g(0)}{x_m-0}=\lim_{m\to\infty}m+1=\infty.$Therefore, the equality holds, and g'(0) does not exist.

I also extended the above 'proof' to how that neither g'(1) nor g'(1/2) exist. However, I now want to show that if $x=p/2^k$, where $p\in\mathbb{Z}$ and $k\in\mathbb{N}\cup\{0\}$, then g'(x) does not exist, but I am running into the problem where I am having to consider lots of cases, such as when $k, $k=n$, etc. which seems to be a rather unfeasible way. Do you guys have any ideas about a better approach? Thanks in advance!

Edit 1: I have not been able to progress any further than showing that the summation boils down to$g(x)=\sum_{n=0}^{k}\frac{1}{2^n}h\left(\frac{2^n}{2^k}p\right).$

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Start by arguing that you don't need to worry about the first $k+1$ terms of $g$, because the sum of those terms is linear on the interval $[x,x+\varepsilon]$ for some small finite $\varepsilon$. Therefore you can subtract that linear functions and only consider the remaining part $\sum_{n=k+1}^\infty$ of the series instead -- which can be tackled in the same way as $g(0)$, except starting at a larger $m$.

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    I hope so -- the _point_ of ignoring the first $k+1$ terms is to make the value of the resulting series be $0$ at $x$ and grow in the same way just to the right of $x$ as it does just to the right of $0$. I'm not quite sure that $k+1$ is exactly the right number of terms to drop, though; it's my intuitive guess, but you'll have to do the algebra to check if I've made a fencepost error...2012-03-21