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

  1. Feb 7, 2017 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    Definition of the critical number

    3. 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##
     
  2. jcsd
  3. Feb 7, 2017 #2

    Mark44

    Staff: Mentor

    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?.
     
  4. Feb 8, 2017 #3
    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 ?
     
  5. Feb 8, 2017 #4

    Mark44

    Staff: Mentor

    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.
     
  6. Feb 12, 2017 #5
    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 ?
     
  7. Feb 12, 2017 #6

    Mark44

    Staff: Mentor

    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}##
     
  8. Feb 12, 2017 #7
    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##
     
  9. Feb 12, 2017 #8

    Mark44

    Staff: Mentor

    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."
     
  10. Feb 13, 2017 #9
    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.
     
  11. Feb 13, 2017 #10

    Mark44

    Staff: Mentor

    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?
     
  12. Feb 13, 2017 #11
    Ok, I am putting snapshot of the problem.
     

    Attached Files:

  13. Feb 21, 2017 #12
    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.
     
  14. Feb 21, 2017 #13
    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.
     
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