Real Analysis Function defined as sum

[Lazerlike42]

First, apologies for the notation.. the computer I am currently using does not support the various buttons to create proper notation.

Homework Statement

Given h(x)=|x| over [-1,1], extend the definition such that h(x+2)=h(x). That is, make h(x) be a periodic sawtooth function.

Now let h_n(x)=(1/2ⁿ)h(2ⁿx).

Then define g(x)=sigma[h_n(x)]=sigma[(1/2ⁿ)h(2ⁿx)] as n goes from 0 to infinity.

Define the sequence x_m=1/2^m, where m=0, 1, 2, ...

Prove that [g(x_m)-g(0)]/[x_m-0]=m+1 and use this to prove that g'(0) does not exist.

Homework Equations

The Attempt at a Solution

I really have no idea. The best I have been able to do is to write the function as:

g(x)=sigma[(1/2ⁿ)|2ⁿx|] as n goes from 0 to infinity

and then substitute x_m for x to yield:

g(x)=sigma[(1/2ⁿ)|2ⁿ*(1/2^m)|] as n goes from 0 to infinity

and simplify it to:

g(x)=sigma[(1/2)ⁿ|2^n-m|] as n goes from 0 to infinity. Otherwise, I have no idea.

 

[Lazerlike42]

If it's helpful, now that I am home I can post these things with better notation:

h(x) = |x|

h_n(x)=$$\frac{1}{2^{n}}$$h(2ⁿx)

g(x)=$$\sum^{\infty}_{n=0}$$h_n(x)=$$\sum^{\infty}_{n=0}$$$$\frac{1}{2^{n}}$$h(2ⁿx)

Let the sequence x_m=$$\frac{1}{2^{m}}$$, where m=0, 1, 2, ...

Prove that $$\frac{g(x_{m})-g(0)}{x_{m}-0}$$ = m + 1
and use this to show that g'(0) does not exist.

My Attempt at a Solution

Rewrite as:

g(x)=$$\sum^{\infty}_{n=0}\frac{1}{2^{n}}$$|2ⁿx|

Then $$\frac{g(x_{m})-g(0)}{x_{m}-0}$$= $$\frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n}\frac{1}{2^{m}}|}{\frac{1}{2^{m}}}$$=$$\frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}{\frac{1}{2^{m}}}$$=$${2^{m}}{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}$$
That's about it...