F(x) = x^4 sin(1/x) has derivatives change sign indefinitely

[issacnewton]

Homework Statement

Consider the function $f(x) = x^4 \sin(\frac 1 x)$ for $x \ne 0$ and $f(x) = 0$ for $x =0$. I have to prove that $x=0$ is the critical number of this function and its derivative changes the sign indefinitely.

Homework Equations

Definition of the critical number

The Attempt at a Solution

$x=0$ would be the critical number if $f'(0) = 0$ or if $f'(0)$ does not exist. I have been able to show that $f'(0) = 0$ using the squeeze theorem. Now I want to show that the derivative of $f(x)$ changes sign indefinitely. I am totally stuck at this point. How would I progress from here ?

Thanks $\smallsmile$

 

[Mark44]

Homework Statement

Consider the function $f(x) = x^4 \sin(\frac 1 x)$ for $x \ne 0$ and $f(x) = 0$ for $x =0$. I have to prove that $x=0$ is the critical number of this function and its derivative changes the sign indefinitely.

Homework Equations

Definition of the critical number

The Attempt at a Solution

$x=0$ would be the critical number if $f'(0) = 0$ or if $f'(0)$ does not exist. I have been able to show that $f'(0) = 0$ using the squeeze theorem. Now I want to show that the derivative of $f(x)$ changes sign indefinitely. I am totally stuck at this point. How would I progress from here ?

Thanks $\smallsmile$

As x approaches 0 from the right, look at what happens for $x = \frac 2 {\pi}, \frac 2 {3\pi}, \frac 2 {5\pi}, \dots$. What are the values of 1/x at these numbers? What are the values of $\sin(1/x)$ at these numbers?.

 

[issacnewton]

Thanks Mark for replying. I see that the sign will be altering between positive and negative as $x$ approaches $0$ from the the right at these values. So do I just write this, or is there more formal way of writing it as a proof ?

 

[Mark44]

Thanks Mark for replying. I see that the sign will be altering between positive and negative as $x$ approaches $0$ from the the right at these values. So do I just write this, or is there more formal way of writing it as a proof ?

No, don't write just what I said, as I was trying to steer you in the right direction by asking some questions, and showing a sequence $\{x_n \}$ whose limit is zero, and for which the sequence $\{\sin(\frac 1 {x_n}) \}$ oscillates. Have you written up something that captures this idea?

Note that a similar idea can be used to show that the sequence $\{\sin(\frac 1 {x_n}) \}$ has the same behavior as x approaches zero from the left.

 

[issacnewton]

Ok, So I have to prove an existence of a sequence in $\mathbb{R}$, such that this sequence goes to zero as $n \rightarrow \infty$ but the sequence $\{\sin(\frac{1}{x_n})\}$ oscillates. You have already come up with the sequence. So I need to prove that $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$ such that $\sin(\frac{1}{x_k}) > 0$ and $\sin(\frac{1}{x_p}) < 0$. Would this be correct ?

 

[Mark44]

Ok, So I have to prove an existence of a sequence in $\mathbb{R}$, such that this sequence goes to zero as $n \rightarrow \infty$ but the sequence $\{\sin(\frac{1}{x_n})\}$ oscillates. You have already come up with the sequence. So I need to prove that $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$ such that $\sin(\frac{1}{x_k}) > 0$ and $\sin(\frac{1}{x_p}) < 0$. Would this be correct ?

As you note, you don't have to "prove" the existence of a sequence -- I gave you one that does the trick.This sequence contains one subsequence for which $\sin(x_{n_1}) > 0$ and another subsequence for which $\sin(x_{n_2}) < 0$. IMO, you should focus first on figuring this out, and less on the fancy "mathy" formatting, such as this stuff -- $\forall~n \in \mathbb{N}~\exists k, p \in \mathbb{N}$

 

[issacnewton]

So if $x_n = \frac{2}{\pi(2n-1)}$ is the sequence given by you, then for the subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, we have $\sin(x_{n_1}) > 0$ and for the subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$, we have $\sin(x_{n_2}) < 0$

 

[Mark44]

So if $x_n = \frac{2}{\pi(2n-1)}$ is the sequence given by you, then for the subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, we have $\sin(x_{n_1}) > 0$ and for the subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$, we have $\sin(x_{n_2}) < 0$

Yes, those are the subsequences I was thinking of. The whole sequence should demonstrate that f ' changes sign infinitely often as x approaches zero. "Infinitely often" as a better way to say it than "changes signs indefinitely."

 

[issacnewton]

Ok.. so I will complete the entire proof when I get the time as I think whole process is with me. But this problem is from James Stewart's Calculus (8 ed.) and this is just a computational book. So I am surprised that they have this problem. When we start introducing subsequences, we are using machinery of real analysis.

 

[Mark44]

Ok.. so I will complete the entire proof when I get the time as I think whole process is with me. But this problem is from James Stewart's Calculus (8 ed.) and this is just a computational book. So I am surprised that they have this problem. When we start introducing subsequences, we are using machinery of real analysis.

Maybe you are misinterpreting what the problem is asking for. If it says "show ..." rather than "prove ..." a less rigorous explanation is called for. Can you post a picture of the problem statement?

 

[issacnewton]

Ok, I am putting snapshot of the problem.

 

[issacnewton]

So I will present the proof here. Consider a sequence $x_n = \frac{2}{\pi(2n-1)}$ where $n\in\mathbb{N}$. Then we have a subsequence $x_{n_1} = \frac{2}{\pi(4n_1 - 3)}$, such that $f'(x_{n_1}) = \frac{32}{\pi^3(4n_1-3)^3} >0$ for all $n_1 \in \mathbb{N}$ and we have another subsequence $x_{n_2} = \frac{2}{\pi(4n_2 - 1)}$ such that $f'(x_{n_2}) = \frac{-32}{\pi^3(4n_2-1)^3} <0$ for all $n_2 \in \mathbb{N}$. So this proves that, the derivative changes sign infinitely often. I hope this is correct.

 

[issacnewton]

I also have to prove that $f$ has neither a local maximum nor a local minimum at $0$. Here I present my proof. Suppose $f$ has a local maxima at $x=0$. Then we have some open interval $(a,b)$ containing $x=0$ such that $f(x)\leqslant f(0)$, for all $x \in (a,b)$. This implies $f(x) \leqslant 0$, for all $x \in (a,b)$. Now $a<0<b$. By Archimedean Property, there exists $n_1 \in \mathbb{N}$ such that $\frac 1 n_1 < \frac{\pi b}{2}$. But we have $4n_1-3 \geqslant n_1 >0$, which implies that $\frac{1}{4n_1-3} < \frac{\pi b}{2}$. Hence we have $0< \frac{2}{\pi(4n_1-3)} < b$. So $\frac{2}{\pi(4n_1-3)} \in (a,b)$. This leads us to $f\left(\frac{2}{\pi(4n_1-3)}\right) \leqslant 0$. But $\forall n_1 \in \mathbb{N}$, we have $f\left(\frac{2}{\pi(4n_1-3)}\right) = \frac{16}{\pi^4(4n_1-3)^4} > 0$. So we reach a contradiction. Hence $f$ doesn't have a local maxima at $x=0$. Now suppose that $f$ has a local minima at $x=0$. Again, there exists an open interval $(a,b)$ containing $x=0$ such that $f(0)\leqslant f(x)$, for all $x \in (a,b)$. Since we have $a<0<b$, we can come up with some $n_2 \in \mathbb{N}$, due to Archimedean property, such that $\frac 1 n_2 < \frac{\pi b}{2}$. But $(4n_2-1) > n_2 > 0$, so we get $\frac{1}{4n_2-1} < \frac{\pi b}{2}$. So $0< \frac{2}{\pi(4n_2-1)} < b$. This means that $\frac{2}{\pi(4n_2-1)} \in (a,b)$. Because of our assumption, this means that $f\left(\frac{2}{\pi(4n_2-1)}\right) \geqslant 0$. But $\forall n_2 \in \mathbb{N}$, we have that $f\left(\frac{2}{\pi(4n_2-1)}\right) = \frac{-16}{\pi^4(4n_2-1)^4} < 0$. So again we reach a contradiction, which means that $f$ can not have a local minima at $x=0$. I hope my proof is correct.