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Is Michael Spivak Wrong?

  1. Jun 16, 2010 #1
    Is Michael Spivak Wrong!?

    My text says,

    In general, if ε > 0, to ensure that

    |x²sin(1/x)| < ε,

    we need only require that

    |x| < ε and x ≠ 0, provided that ε ≤ 1. If we are given an ε which is greater than 1 (it might be, even thought it is "small" ε's which are of interest), then it does not suffice to require that |x| < ε, but it certainly suffices to require that |x| < 1 and x ≠ 0.

    I don't see why if ε > 1 then it doesn't work. I've tried finding a counter example, but it always seem to work. Can someone please help me out?

    EDIT:

    x²sin(1/x) ≈ x when x → ∞ so it would seem that when x isn't very large then x²sin(1/x) will always be less then x and hence less then ε, even if ε were greater than 1.

    ANY HELP?? Please?

    BTW, this is Calculus by Spivak page 92-93.
     
  2. jcsd
  3. Jun 16, 2010 #2
    Re: Is Michael Spivak Wrong!?

    Try x = 6/pi.

    Then [itex]sin(1/x) = 1/2[/itex], so [itex]x^2\sin(1/x)=18/\pi^2[/itex].

    The reason why it doesn't necessarily work when [itex]x>1[/itex] is that we may have [itex]|x^2| > \epsilon[/itex] so we need sin to be suifficiently small, but x^2 grows towards infinity much faster than sin(1/x) grows towards 0.

    It is true that we could still works with values of epsilon a little above 1, but we would need a different argument then.
     
  4. Jun 16, 2010 #3
    Re: Is Michael Spivak Wrong!?

    So what would be wrong with this proof then?

    sin x < x for x > 0, so then
    |sin x| < |x| for x ≠ 0, so in particular we have

    |sin x| < |x| for nonzero x in the interval (-1,1)

    This means that

    |sin(1/x)| < |1/x| for |x| > 1

    which implies

    |x^2 sin(1/x)| < |x| for |x| > 1

    This should guarantee then that if ε > 1 and 1 < |x| < ε, then
    |x^2 sin(1/x)| < |x| < ε
     
  5. Jun 16, 2010 #4

    Landau

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    Re: Is Michael Spivak Wrong!?

    Yes, you are totally correct.

    In fact, for all x in R, we have |x^2 sin(1/x)|<=|x|, so we can take delta=epsilon.

    I think what Spivak meant, was that in this chapter you know almost nothing about the sine function. He only uses the very rough estimate |sin x|<= 1 for all x, which everyone knows and which is clear from the picture of the unit circle. If you want to extend the argument for epsilon >1, you will have to come up with a better estimate of |sin x| (or |sin(1/x)|): for example you are using the inequality
    |sin x|<=|x|. I think Spivak does not want to assume you to know this fact, because the sine has not been introduced formally.
     
  6. Jun 16, 2010 #5
    Re: Is Michael Spivak Wrong!?

    I'm confused now. Landau, you're telling me I'm right, but rasmhop has provided the counter-example x = 6/&pi;. So I'm massively confused now...
     
  7. Jun 16, 2010 #6
    Re: Is Michael Spivak Wrong!?

    That isn't really a counterexample I responded too hastily. Taking [itex]x=6/\pi[/itex] we get:
    [tex]x^2\sin(1/x) > 1[/tex]
    but obviously not
    [tex]x^2\sin(1/x) > x[/tex]
    (for instance by the argument you provided). So my counterexample doesn't work.
     
  8. Jun 16, 2010 #7
    Re: Is Michael Spivak Wrong!?

    Wait a second, x = 6/&pi; doesn't work!! lol
     
  9. Jun 16, 2010 #8
    Re: Is Michael Spivak Wrong!?

    Well thanks for helping me guys, this had been bothering me lol So thanks a lot for the help!
     
  10. Jun 16, 2010 #9

    Landau

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    Re: Is Michael Spivak Wrong!?

    Yes, you were right. You're welcome :)
     
  11. Jun 16, 2010 #10

    Office_Shredder

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    Re: Is Michael Spivak Wrong!?

    [tex]|sin(\frac{1}{x})|<\frac{1}{x}[/tex] (since |sin(x)|<|x|) so it's certainly always true that [tex]|x^2sin(\frac{1}{x})|<|x|[/tex]

    It might be that the way he proves it is far more convenient to assume that x is small, since a proof of sin(x)<x is non-trivial from first principles
     
  12. Jun 19, 2010 #11
    Re: Is Michael Spivak Wrong!?

    interesting...
     
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