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For what values is this function continuous

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]F(x) \ = \ (x^{2})^{\frac{\alpha}{2}} \left[ \frac{\alpha sin \left( \frac{1}{x} \right)}{x} \ - \ \frac{cos \left( \frac{1}{x} \right)}{x^{2}} \right] \ if \ x \neq 0[/itex]

    [itex]F(x) \ = \ 0 \ if \ x \ = \ 0[/itex]

    Question basically says, for what values of alpha is this function continuous

    2. Relevant equations



    3. The attempt at a solution

    My mind is torn between either: α = 0, for all α or some constriction on α being EVEN or ODD


    thanks for any help
     
  2. jcsd
  3. Feb 10, 2013 #2
    You need the ##\displaystyle \lim_{x \to 0} F(x) = F(0) = 0## in order for the function to be continuous.
     
  4. Feb 10, 2013 #3
    How do you do that for functions which have "two parts" to them?

    Also, as x starts getting close to 0 everything is going to go crazy because of all the dividing by x's.

    I remember doing a question a while back which was basically "is f(x) = sin(1/x) continuous at 0 if f(0) = 0?" I believe the answer was no because of sequential continuity?
     
  5. Feb 10, 2013 #4

    haruspex

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    First, concentrate on the two terms inside the bracket. Can you see a reason to say that one of them makes the other irrelevant?
    Next, assuming that ab is defined to mean the positive real value when a > 0 and b > 0, can you see a way to simplify ##\left(x^2\right)^{\frac{\alpha}{2}}## that's valid for α > 0 and all x?
     
  6. Feb 10, 2013 #5
    The main difference between the two terms in the brackets are:
    1) the use of α in the sin term
    2) the cos term is over a higher power of x

    Possibly the cos term is irrelevant due to it not depending on α which is what in looking for, for continuity.

    Or possibly the sin term is irrelevant as it tends to ∞ slower than the cos term?

    The reason as two why it is (x^2)^(α/2) is because it was originally |x|^α
     
  7. Feb 10, 2013 #6

    haruspex

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    What do you get if you collect them into a single fraction? Compare the two terms in the numerator. Can you see that one of them becomes irrelevant when compared to the other as x tends to 0?
    I feel the |x|α form will be the easier to work with.
     
  8. Feb 10, 2013 #7
    Sorry in advance for lack of latex, on a mobile.

    Combining brackets gives us:

    [α(x^2)sin(1/x) - xcos(1/x)]/[x^3]

    I can't see how one term is particularly irrelevant.

    As for |x|^α: by all means use that notation, I only changed it as that particular function [F(x)] actually came from finding the derivative of another function and I found it easier using the other notation.

    If you could explain why that notation would be easier to work with would be helpful :)
     
  9. Feb 10, 2013 #8
    Looking back, as x tends to 0 the cos term must become irrelevant.
    e.g. If x = 0.000001,
    x >> x^2
     
  10. Feb 11, 2013 #9

    haruspex

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    It's a little simpler than that: [αxsin(1/x) - cos(1/x)]/x2
    What does each of the terms in the numerator do as x tends to 0?
     
  11. Feb 11, 2013 #10
    Well the sin term will tend to 0 but the cos term will oscillate like crazy
     
  12. Feb 11, 2013 #11

    haruspex

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    Right, so you can ignore the sin term. And of course x2 = |x|2. So what do can you reduce the whole expression to?
     
  13. Feb 11, 2013 #12
    I believe I got it to reduce to [(x^2)^(α/2 - 1)][-cos(1/x)]
     
  14. Feb 12, 2013 #13

    haruspex

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    Right, but I still think you're better off with the form [|x|α-2][-cos(1/x)].
    You've already described what -cos(1/x) will do as x tends to 0. So what do you need |x|α-2 for the whole to converge to 0?
     
  15. Feb 12, 2013 #14
    Would it not converge to 0 for any α >= 2
     
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