# Real Analysis Function defined as sum

1. Dec 4, 2008

### Lazerlike42

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

1. The problem statement, all variables and given/known data

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 hn(x)=(1/2n)h(2nx).

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

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

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

2. Relevant equations

3. 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 xm for x to yield:

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

and simplify it to:

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

2. Dec 4, 2008

### Lazerlike42

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

h(x) = |x|

hn(x)=$$\frac{1}{2^{n}}$$h(2nx)

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

Let the sequence xm=$$\frac{1}{2^{m}}$$, 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 $$\frac{g(x_{m})-g(0)}{x_{m}-0}$$= $$\frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n}\frac{1}{2^{m}}|}{\frac{1}{2^{m}}}$$=$$\frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}{\frac{1}{2^{m}}}$$=$${2^{m}}{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}$$