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Show that F'(x) exists for all x [tex]\in[a,b] [/tex]

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider
    F(x) = x2 sin(1/x2) if 0<x[tex]\leq1[/tex]
    and = 0 if x[tex]\leq0[/tex]

    Show that F'(x) exists for all x [tex]\in[a,b] [/tex] but that F':[0,1] [tex]\rightarrow1[/tex] is not integrable.

    2. Relevant equations
    So we have to show we do not have F(1)-F(0) = [tex]\int[/tex] F'(x)dx
    (integral going from 0 to 1)



    3. The attempt at a solution
    I'm having trouble proving this statement.
    Where should I start?
    To show that F'(x) exists, should I just take the derivative or do I have to go under some long theorems of analysis to PROVE?
    Thanks in advance.:shy:
     
  2. jcsd
  3. Feb 10, 2009 #2
    Most likely you'll have to actually *prove* that this derivative exists - it clearly exists for all x other than 0 because [tex]x^{2}sin(\frac{1}{x^{2}})[/tex] is the composition of differentiable functions. The only tricky part is whether F'(0) exists. To show that it does indeed exist you'll need to show that the following limit exists:

    [tex]\lim_{h\to 0}\frac{f(h) - f(0)}{h}[/tex]

    But this just equals

    [tex]\lim_{h\to 0}\frac{h^{2}sin(\frac{1}{h^{2}})}{h} = lim_{h\to 0}hsin(\frac{1}{h^{2}}) = 0[/tex]

    Then by application of the product rule and the chain rule,

    [tex]F'(x) = 2xsin(\frac{1}{x^{2}}) - \frac{2cos(\frac{1}{x^{2}})}{x^{2}}[/tex]

    if x is anything other than 0, and 0 if x=0. So all you need to show is that the above function is unbounded and it will consequently not be integrable.
     
  4. Feb 10, 2009 #3
    right,
    But I think you made a little mistake with differentiation,
    I think it will be:

    F'(x)= 2x sin(1/x2) - 2cos(1/x2)/ x

    Which still, limits exists.

    But How do I prove that F'(x) is unbounded?
     
  5. Feb 10, 2009 #4

    HallsofIvy

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    Take [itex]x= 1/\sqrt{2n\pi}[/itex]. Then [itex]1/x^2= 2n\pi[/itex] so [itex]sin(1/x^2)= 0[/itex] and [itex]cos(1/x^2)= 1[/itex]. [itex]F'(x)= F'(1/\sqrt{2n\pi})= -2\sqrt{2n\pi}[/itex]. x will go to 0 as n goes to infinity. What happens to F'(x)?
     
  6. Feb 10, 2009 #5
    F'(x) will be Undefined as x approaches zero [tex]\rightarrow[/tex] unbounded [tex]\rightarrow [/tex] non integrable ??????
     
  7. Feb 10, 2009 #6
    HallsofIvy just defined F'(x) above, so it *will* be defined; however, as n-->infinity, F'(x) also goes to infinity and this implies that the derivative is unbounded.

    Also, thanks for pointing out my mistake; I was in an Econ class :P
     
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