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Math Help - Function which is not differentiable at certain points

  1. #16
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    Re: Function which is not differentiable at certain points

    Hello again!

    I want to bring up my issue one more time to finally solve it! I want to understand how it works.

    Here is what I have:

    I want to calculate the left derivate.

    I got from the integral 2^{-i}k \leq t \leq  x which is k \leq 2^{i}t \leq 2^{i} x (1)

    The second inequality I get is from the advice from girdav which was to consider 2^{-i}k - x \leq 2^{-i}
    This inequality can be put to k - 2^{i} x \leq 1 and it leads to 2^{i} x \geq k-1 (2)

    I got (1) and (2). How can I use these two inequalities for the left derivate? If I can replace the 2^{i} x in (1) by something else which is bigger than k I would get the same derivate.
    My conclusions should be correct until here but I donīt see how I can proceed from there.

    I need a hint and I would be very thankful for that little hint to finally finish the work.

    Thanks.
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  2. #17
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    Re: Function which is not differentiable at certain points

    Hello!

    I want to raise my question one more time and hope for some help. Here are my thoughts.

    That is what I still got when it comes to the left derivate:

    \lim\limits_{x \rightarrow 2^{-i} k} \frac{f_{i} (x) - f_{i} (2^{-i} k)}{x-2^{-i} k} = \lim\limits_{x \rightarrow 2^{-i} k} \frac{\int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}- \int_{0}^{2^{-i} k} (-1)^{\lfloor 2^{i} t\rfloor}}{x-2^{-i} k}
    cause x \leq 2^{-i} k.


    Because I look at the derivate from the left site I will get
    \lim\limits_{x \rightarrow 2^{-i} k} \frac{\int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}- \int_{0}^{2^{-i} k} (-1)^{\lfloor 2^{i} t\rfloor}}{x-2^{-i} k} = \lim\limits_{x \rightarrow 2^{-i} k} \frac{\int_{2^{-i}k  - x}^{0} (-1)^{\lfloor 2^{i} t\rfloor}}{x-2^{-i} k}

    Is that correct?
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  3. #18
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    Re: Function which is not differentiable at certain points

    Looks like I really do need some help here please! I am stuck.
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  4. #19
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    Re: Function which is not differentiable at certain points

    I donīt want you to solve me the thing but when you look at a problem for way too long you get more and more stuck and that is why I kindly ask you for some help.

    I want to sum up my current thoughts:
    \lim\limits_{x \rightarrow 2^{-i}k} \frac{f_{i} \left(x \right) - f_{i} \left(2^{-i}k \right)}{x - 2^{-i}k} = \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}

    I justify my last step by the fact that I consider those x with x \leq  2^{-i}k cause I want to calculate the left derivate and there is no minus before the integral cause x \geq 0.
    Now I have got x \leq t \leq 2^{-i}k \Rightarrow 2^{i} x \leq 2^{i}t \leq k

    Again girdav gave me the advice to look at 2^{-i}k - x \leq 2^{-i} but I can not use it here cause when I look at it I will get k - 2^{i} x \leq 1. There has to be a < involved somehow. I would get 2^{i}x \geq k-1 and I can not use that fact with my given inequality.

    These are my thoughts and I am stuck here. Any ideas how I can proceed? Thanks.
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  5. #20
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    Re: Function which is not differentiable at certain points

    Could anyone please tell me if I am right on my thoughts in the last post? I need that assignment until tomorrow
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  6. #21
    Super Member girdav's Avatar
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    Re: Function which is not differentiable at certain points

    Quote Originally Posted by Tahoe View Post
    I donīt want you to solve me the thing but when you look at a problem for way too long you get more and more stuck and that is why I kindly ask you for some help.

    I want to sum up my current thoughts:
    \lim\limits_{x \rightarrow 2^{-i}k} \frac{f_{i} \left(x \right) - f_{i} \left(2^{-i}k \right)}{x - 2^{-i}k} = \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}.
    Put y := x-2^{-i}k. When x goes to 2^{-i}k from the left, y goes to 0 from the left. You have to compute l:=-\lim_{y\to 0^-}\frac{\int_{y+2^{-i}k}^{2^{-i}k}(-1)^{\lfloor 2^it\rfloor}dt}y. Using the substitution u =t-2^{-i}k, we see that l =-\lim_{y\to 0^-}\frac 1y\cdot \int_y^0(-1)^{\lfloor 2^iu+k\rfloor}du=(-1)^{k+1}\lim_{y\to 0^-}\frac 1y\int_y^0(-1)^{\lfloor 2^iu\rfloor}du. If you consider the y such that y\geq -2^{-i}, you will see that the integral is y and finally the left derivative is (-1)^{k+1}.
    Last edited by girdav; July 25th 2011 at 01:33 AM.
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  7. #22
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    Re: Function which is not differentiable at certain points

    Thanks for your answer!

    Could it be that you meant y \geq -2^{-i} in your last step?
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  8. #23
    Super Member girdav's Avatar
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    Re: Function which is not differentiable at certain points

    Quote Originally Posted by Tahoe View Post
    Thanks for your answer!

    Could it be that you meant y \geq -2^{-i} in your last step?
    Yes, it was a typo.
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