Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $k<n$, $k=n$, 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$$g(x)=\sum_{n=0}^{k}\frac{1}{2^n}h\left(\frac{2^n}{2^k}p\right).$$

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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 $m$.

share|improve this answer
    
I see. :) But would the series $g(x)=\sum_{n=k+1}^{\infty}\frac{1}{2^n}h(\frac{2^n}{2^k}p)=0$ in that case? –  Josué Molina Mar 21 '12 at 19:40
    
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... –  Henning Makholm Mar 21 '12 at 19:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.