# Thread: Function which is not differentiable at certain points

1. ## 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 don´t 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.

2. ## 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?

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

Looks like I really do need some help here please! I am stuck.

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

I don´t 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.

5. ## 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

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

Originally Posted by Tahoe
I don´t 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 $(-1)^{k+1}$.

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

Could it be that you meant $y \geq -2^{-i}$ in your last step?

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

Originally Posted by Tahoe
Could it be that you meant $y \geq -2^{-i}$ in your last step?