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Not differentiable function

  1. Jul 23, 2011 #1
    Hello!

    I got the following function:

    [itex]\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt, \quad x \in \left[0,1 \right][/itex]

    I want to show it is not differentiable at [itex] x= 2^{-i} k [/itex] where k is a natural number greater equal 0.

    I already calculated the right derivate by considering those x with [itex]2^{-i}k \leq x < 2^{-i}k + 2^{-i}[/itex]. What I got was [itex](-1)^{k} [/itex].

    Now when it comes to the left hand derivate I have the following:
    [itex]\lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt - \int_{0}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt}{x - 2^{-i}k} [/itex]

    But since I consider those x approaching from the left side [itex]x \leq 2^{-i} k[/itex] and [itex]x \geq 0[/itex].
    That is why the above integral will be
    [itex]\lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt - \int_{0}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt}{x - 2^{-i}k} = \lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{x}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt }{x - 2^{-i}k} [/itex]

    I know that for the last integral [itex]x \leq t \leq 2^{-i}k[/itex] which is equivalent to [itex]2^{i}x \leq 2^{i}t \leq k [/itex].
    But that doesn´t help me when it comes to calculating [itex]\lfloor 2^{i} \cdot t \rfloor [/itex]

    How can I proceed from there or is there any mistake in my way of thinking I don´t see at this moment?

    Thanks.
     
  2. jcsd
  3. Jul 24, 2011 #2
    Hello!

    Don´t want to bother but does anybody has an idea how to proceed with the given problem? It makes me crazy not solving it right now. Thank you for any help.
     
  4. Jul 25, 2011 #3
    I'm not sure, just an idea: I think one can prove a generalized form of the first fundamental theorem of calculus that is, if [itex]F(x)=\int_{a}^{x}f(t)dt[/itex] then for some conditions on f, we have [tex]F'_+(x_0)=\lim_{c\rightarrow x_0+}f(c)[/tex] and [tex]F'_-(x_0)=\lim_{c\rightarrow x_0-}f(c)[/tex]

    Where [itex]F'_+(x_0)[/itex] and [itex]F'_-(x_0)[/itex] represent right and left derivatives at x0, respectively. [itex]\lim_{c\rightarrow x_0+}[/itex] and [itex]\lim_{c\rightarrow x_0-}[/itex] represent right sided and left sided limits at x0, respectively.
     
    Last edited: Jul 25, 2011
  5. Jul 31, 2011 #4
    Which type of integral is implied in your o.p.?
     
  6. Jul 31, 2011 #5

    chiro

    User Avatar
    Science Advisor

    Hey Tahoe and welcome to the forums.

    I would recommend the same approach as asmani gave above. If you show that the appropriate limits on either side of the point are unequal, then you have showed that the derivative does not exist at that point.

    (When I mean limits, I mean them in the context of the derivative as you would see in first principles).
     
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