Proving Non-Differentiability of a Function at a Specific Point

In summary, the conversation discusses a function and its differentiability at a specific point. The participants consider the right and left derivatives and suggest using a generalized form of the first fundamental theorem of calculus to prove the non-existence of the derivative at the given point.
  • #1
Tahoe
2
0
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.
 
Physics news on Phys.org
  • #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.
 
  • #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:
  • #4
Which type of integral is implied in your o.p.?
 
  • #5
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).
 

What is a "Not differentiable function"?

A "not differentiable function" is a mathematical function that does not have a derivative at one or more points in its domain. In other words, the function is not smooth and has a sharp corner or cusp at some point.

What causes a function to be not differentiable?

A function can be not differentiable for a few reasons. One common cause is that the function has a sharp point or corner, which means it has a discontinuity at that point. Another reason could be that the function has a vertical tangent, meaning the slope of the function is undefined at that point.

What is the significance of a function being not differentiable?

A not differentiable function can have important implications in calculus and other areas of mathematics. It means that the function is not continuously changing and therefore cannot be described by a derivative. This can affect calculations and interpretations of the function's behavior.

Can a not differentiable function still be continuous?

Yes, a not differentiable function can still be continuous. This means that the function is still defined at all points in its domain and there are no breaks or holes in its graph. However, the function is not smooth and does not have a well-defined slope at certain points.

How can you determine if a function is not differentiable?

To determine if a function is not differentiable, you can look for any sharp points, corners, or vertical tangents on its graph. You can also calculate the derivative of the function and see if it is undefined at any point. If so, the function is not differentiable at that point.

Similar threads

Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
852
Replies
4
Views
144
Replies
2
Views
752
Replies
3
Views
1K
Replies
3
Views
173
Replies
1
Views
2K
Replies
5
Views
943
Replies
3
Views
1K
Back
Top