# Thread: Function which is not differentiable at certain points

1. ## Function which is not differentiable at certain points

Hello!

I want to know if the following is correct:
I want to prove that the function
$\displaystyle f_{i} (x) = \int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}$ is differentiable on all open intervals $\displaystyle (2^{-i}k, 2^{-i}(k+1))$ with $\displaystyle k \in \mathds{N} \cup \{0 \}$

Here is what I did:
I know that the function is linear on the given open intervals and each linear function is differentiable.

To show that the function is not differentiable on $\displaystyle 2^{i}k$ I calculated using I'Hospital in the differential quotient :
$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{f_{i} (x) - f_{i} (2^{-i} k^{+})}{x-2^{-i} k^{+}} = \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{\int_{0}^{x} (-1)^{\lfloor 2^{i} t\rfloor}- \int_{0}^{2^{-i} k^{+}} (-1)^{\lfloor 2^{i} t\rfloor}}{x-2^{-i} k^{+}}$

I'Hospital gives:
$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{+}} \frac{(-1)^{\lfloor 2^{i} x\rfloor}}{1} = \lim\limits_{x \rightarrow 2^{-i} k^{+}} (-1)^{\lfloor 2^{i} x\rfloor} = (-1)^{\lfloor k^{+} \rfloor} = (-1)^k \quad (1)$

On the other hand
$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k^{-}} \frac{f_{i} (x) - f_{i} (2^{-i} k^{+})}{x-2^{-i} k^{+}} =(-1)^{\lfloor k^{-} \rfloor} = (-1)^{k-1} \quad (2)$

Now when k is even then (1) is 1 and (2) -1 and
if k is odd, then (1) is -1 and (2) is even.
Given that fact, it means that the function $\displaystyle f_i (x)$ is not differentiable at $\displaystyle 2^{i}k$

I would do the same calculation with the 2nd endpoint of the given interval.

My question is: Is it correct what I am doing here?

Thanks.

Regards

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

The use of l'Hospital rule is "dangerous" here because you don't know if your function are differentiable. Anyway, you can compute the limit without this rule. Fix $\displaystyle k\in\mathbb{N}$, and let $\displaystyle x$ such that $\displaystyle 2^ik\leq x<2^ik+2^{-i}$. We have
$\displaystyle f_i(x) -f_i(2^ik) = \int_{2^ik}^x(-1)^{\lfloor 2^it\rfloor}dt$ and since $\displaystyle 2^ik2^i\leq 2^it<2^i2^ik+1$ we get $\displaystyle f_i(x)-f_i(2^ik) =(x-2^ik)(-1)^{2^i2^ik} =(x-2^ik)$. Finally, the right derivative is $\displaystyle 1$. Now do the same for the left derivative.

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

Hello! Thanks for your response. I just realized that I did a mistake in the given open interval. I made changes to it in my first post. The given interval is $\displaystyle (k 2^{-i}, (k+1) 2^{-i})$. Sorry about that.

What you are saying though is correct since I do not know if my function is differentiable.

I need to find a similar way to calculate the integral now.

How did you come up with the idea of putting x into a certain interval?

I mean I know that $\displaystyle 2^{-i} k \leq t \leq x$ and that is why
$\displaystyle k \leq t \cdot 2^{i} \leq 2^{i}x$

But it does not really tell me now which value $\displaystyle \lfloor t \cdot 2^{i} \rfloor$ will take.

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

Since we will take the limit $\displaystyle x\to 2^{-i}k$, we can only take care of the x such that $\displaystyle 2^{-i}k\leq x<2^{-i}k+2^{-i}$. You will find that $\displaystyle k\leq 2^it<k+1$.

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

Originally Posted by girdav
Since we will take the limit $\displaystyle x\to 2^{-i}k$, we can only take care of the x such that $\displaystyle 2^{-i}k\leq x<2^{-i}k+2^{-i}$. You will find that $\displaystyle k\leq 2^it<k+1$.
Maybe that is a stupid question but how do you come up with $\displaystyle x < 2^{-i}k + 2^{-i}$. Do you choose to have $\displaystyle 2^{-i}k + 2^{-i}$ because then you get exactly $\displaystyle k \leq t 2^{i} \leq k+1$?

Then it follows that $\displaystyle \lfloor 2^{i} \cdot t \rfloor = k$
Then $\displaystyle (-1)^{\lfloor 2^{i} \cdot t \rfloor } = (-1)^{k}$
It will switch its value between +1 and -1 now which will give me a hard time calculating the integral.

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

Originally Posted by Tahoe
Maybe that is a stupid question but how do you come up with $\displaystyle x < 2^{-i}k + 2^{-i}$. Do you choose to have $\displaystyle 2^{-i}k + 2^{-i}$ because then you get exactly $\displaystyle k \leq t 2^{i} \leq k+1$?
Yes. (the second inequality should be $\displaystyle <$)

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

Originally Posted by girdav
Yes. (the second inequality should be $\displaystyle <$)
Yes, correct. Got that as well on my paper!
And it leads to $\displaystyle \lfloor 2^{i} \cdot t \rfloor = k$ and then
$\displaystyle (-1)^{\lfloor 2^{i} \cdot t \rfloor} = (-1)^k$

How can I proceed from there?

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

You conclude that the right derivative in $\displaystyle 2^{-i}k$ is $\displaystyle (-1)^k$. Now, by the same way, compute the left derivative.

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

We still got $\displaystyle 2^{-i} \cdot k \leq t \leq x$ which means
$\displaystyle k \leq 2^{i} \cdot t \leq 2^{i} \cdot x$

Since x approaches $\displaystyle 2^{-i} k$ from the left side we got
$\displaystyle 2^{-i}k - 2^{-i} < x \leq 2^{-i} k$ and that is why
$\displaystyle 2^{i} \cdot 2^{-i}k - 2^{i} \cdot 2^{-i} < 2^{i} \cdot x \leq 2^{i} \cdot 2^{-i} k$ which is equivalent to $\displaystyle k - 1 < 2^{i} \cdot x \leq k$

Then it follows that $\displaystyle 2^{i} \cdot t = k$

Something is wrong here I guess but on the other hand I think that my conclusions are correct until here.

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

What I did in my last post is not correct, isnīt it?

Regards

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

Originally Posted by Tahoe
We still got $\displaystyle 2^{-i} \cdot k \leq t \leq x$

Since x approaches $\displaystyle 2^{-i} k$ from the left side
There is a contradiction between these two sentences.

You can take $\displaystyle x$ such that $\displaystyle 2^{-i}k-x\leq 2^{-i}$ and see what happens.

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

Originally Posted by girdav
There is a contradiction between these two sentences.

You can take $\displaystyle x$ such that $\displaystyle 2^{-i}k-x\leq 2^{-i}$ and see what happens.
Thanks!
Then I get:
$\displaystyle 2^{i} 2^{-i} k - 2^{i} x \leq 2^{i} 2^{-i}$ which is
equivalent to
$\displaystyle k - 2^{i} x \leq 1$ which is equivalent to
$\displaystyle 2^{i}x \geq k-1$ which means $\displaystyle 2^{i}t \geq k-1$
$\displaystyle \Leftrightarrow \quad \lfloor 2^{i} t \rfloor = k-1$ and that is why we get
$\displaystyle \lim\limits_{x \rightarrow 2^{-i} k } \frac{f_{i} \left(x \right) - f_{i} \left(2^{-i}k \right)}{x-2^{-i}k }= (-1)^{k-1}$

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

What I did in my last post was not fully correct:
What I get is $\displaystyle k-1 \leq 2^{i} x$ and
$\displaystyle k \leq 2^{i}t \leq 2^{i} x$

I need to get these two inequalities into one piece now.

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

Hello girdav!

Are you sure you got to choose x the way you did?
Since I approach $\displaystyle 2^{-i} k$ from the left side all I know is that $\displaystyle x \leq 2^{-i} k$ And I can choose $\displaystyle x - 2^{-i} < x \leq 2^{-i} k$, but it does not help me neither.

When I take x the way you chose it, I get stuck. Can you tell me how you would proceed?

Thanks.

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

Hello!

I donīt get to calculate the left derivative. I get contradictions. Can anyone please give me a hint on how to calculate the left derivative for $\displaystyle x=2^{-i} k$?Thanks.

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