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Real Analysis Function defined as sum

  1. Dec 4, 2008 #1
    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. jcsd
  3. Dec 4, 2008 #2
    If it's helpful, now that I am home I can post these things with better notation:

    h(x) = |x|

    hn(x)=[tex]\frac{1}{2^{n}}[/tex]h(2nx)

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

    Let the sequence xm=[tex]\frac{1}{2^{m}}[/tex], where m=0, 1, 2, ...

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

    My Attempt at a Solution

    Rewrite as:

    g(x)=[tex]\sum^{\infty}_{n=0}\frac{1}{2^{n}}[/tex]|2nx|

    Then [tex]\frac{g(x_{m})-g(0)}{x_{m}-0}[/tex]= [tex] \frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n}\frac{1}{2^{m}}|}{\frac{1}{2^{m}}}[/tex]=[tex] \frac{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}{\frac{1}{2^{m}}}[/tex]=[tex] {2^{m}}{\sum^{\infty}_{n=0}\frac{1}{2^{n}}|2^{n-m}|}[/tex]
    That's about it...
     
    Last edited: Dec 4, 2008
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