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Question of "min" function from Spivak

  1. Oct 10, 2014 #1
    Hi,

    Suppose you want to prove [itex]|x - a||x + a| < \epsilon[/itex]

    You know

    [itex]|x - a| < (2|a| + 1)[/itex]
    You need to prove

    [itex]|x + a| < \frac{\epsilon}{2|a| + 1}[/itex]

    So that

    [itex]|x - a||x + a| < \epsilon[/itex]

    Why does Michael Spivak do this:

    He says you have to prove --> [itex]|x + a| < min(1, \frac{\epsilon}{2|a| + 1})[/itex] in order to finally prove, [itex]|x + a||x - a| < \epsilon[/itex]

    Why do we need the "min" function there?
    Amad27 - The closing itex tag starts with /, not \. I fixed them all for you. - Mark44

    Thanks!
     
    Last edited by a moderator: Oct 10, 2014
  2. jcsd
  3. Oct 10, 2014 #2

    pasmith

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    Homework Helper

    If we assume [itex]|x - a| < 1[/itex] then [itex]a - 1 < x < a + 1[/itex] so that [itex]2a - 1 < x + a < 2a + 1 \leq 2|a| + 1[/itex]. That in turn means that [tex]
    |x - a||x + a| < |x - a|(2|a| + 1)[/tex] and we can ensure [itex]|x - a||x + a| < \epsilon[/itex] by requiring that [tex]
    |x - a|(2|a| + 1) < \epsilon,[/tex] so that [tex]
    |x - a| < \frac{\epsilon}{2|a| + 1}.[/tex] However the preceding argument assumed that [itex]|x - a| < 1[/itex], so [itex]|x - a|[/itex] must satisfy that constraint as well. Hence we must have [tex]
    |x - a| < \min\left\{1, \frac{\epsilon}{2|a| + 1}\right\}.[/tex]
     
  4. Oct 10, 2014 #3
    Hello @pasmith, I dont understand;

    On WHAT basis and WHY are you assuming [itex]|x - a| < 1[/itex]??
     
  5. Oct 10, 2014 #4

    Mark44

    Staff: Mentor

    Amad, be careful when you quote someone. @pasmith did not write "|xa|<1|x - a| < 1" - what he wrote was "If we assume |x−a| < 1 then ..."

    He made this assumption, and if it turns out that this assumption is incorrect, all that is accounted for in the last line of his post where he says that |s - a| will be less than the minimum of 1 and the expression involving ##\epsilon##.
     
  6. Oct 11, 2014 #5
    Hi,

    What does [itex] min(1, \frac{\epsilon}{2|a| + 1|})[/itex]

    Actually mean? I dont understand this "min" idea??
     
  7. Oct 11, 2014 #6

    Mark44

    Staff: Mentor

    It's very simple - it means the minimum, or smaller, of the two values.
     
  8. Oct 11, 2014 #7
    @Mark44, why cant we directly prove:

    [itex] |x - a| < \frac{\epsilon}{2|a| + 1|}[/itex]

    Why is that 1 required.

    Why does it matter if |x - a| < min of something?

    Thanks!
     
  9. Oct 11, 2014 #8

    Mark44

    Staff: Mentor

    The goal is to prove that |x - a||x + a| < ##\epsilon##

    You need to get bounds on the value of |x + a|. He could just as easily assumed that |x - a| < 1/2 or |x - a| < 3 or whatever.
     
  10. Oct 11, 2014 #9
    Oh,

    So its we know that

    [itex] \frac{\epsilon}{2|a| + 1} < 1[/itex]?

    If [itex] \frac{\epsilon}{2|a| + 1} < 1[/itex] is minimum of course.
     
  11. Oct 11, 2014 #10

    Mark44

    Staff: Mentor

    It all depends on the value of ##\epsilon##. For some values ##\frac{\epsilon}{2|a| + 1}## will be smaller than 1, but for suitably large values of ##\epsilon##, 1 will be larger.
     
  12. Oct 12, 2014 #11
    Right, since it depends for every [itex]\epsilon[/itex]

    But why would he randomly choose one (1) as a part of the minimum function? Why not two(2) or three(3)?

    So, if 1 is the minimum then how does it prove

    [itex]|x-a||x+a| < \epsilon[/itex]?
     
  13. Oct 12, 2014 #12

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The "[itex]\delta[/itex]" was derived under the assumption that |a- 1|< 1 (so that -1< a- 1< 1 and then 1< a+ 1< 3). In order for the proof to work, both [tex]|a- 1|< \delta[/tex] and [tex]|a- 1|< 1[/tex] must be true. That will be the case if it is less than the smaller of the two.
     
  14. Oct 12, 2014 #13
    But it has work for all real numbers,

    Not just |x-a| < 1 what about |x-a| > 1 ???
     
  15. Oct 12, 2014 #14

    Mark44

    Staff: Mentor

    We don't care about that. You're missing the big picture here. The goal is to see how close x needs to be to a so that |x2 - a2| < ##\epsilon##.
    We don't care about any x values that are relatively distant from a.
     
  16. Oct 13, 2014 #15
    I really dont understand this concept.

    The definition says you have to prove there is some [itex]\delta[/itex] such that [itex] |x - a| < \delta[/itex] not that [itex]|x - a| < \delta < 1[/itex]

    Also, there is no proper definition of "relatively distant." Relatively close could mean 10000000000000. Why are you making assumptions anyway?
     
  17. Oct 13, 2014 #16

    Char. Limit

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    Gold Member

    Because if you don't make assumptions, you'll never get anywhere.
     
  18. Oct 13, 2014 #17
    I dont believe so; The goal I think is to bound the number [itex]|x - a|[/itex] so that it doesnt go overboard. Because when looking at limits you generally look at values close to x = a.
     
  19. Oct 13, 2014 #18

    Char. Limit

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    Gold Member

    And that right there, that's the assumption. We assume that x is close to a - namely, that |x - a| < 1. We can safely assume this, so we do, and the rest of the proof follows from such.
     
  20. Oct 13, 2014 #19

    pasmith

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    Homework Helper

    We don't care about [itex]|x - a | \geq 1[/itex].

    To conclude that [itex]x^2 - a^2[/itex] is continuous at [itex]a[/itex] you need to prove:

    Statement 1
    For all [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for all [itex]x[/itex], if [itex]|x - a| < \delta[/itex] then [itex]|x^2 - a^2| < \epsilon[/itex].

    What Spivak invites you to prove is a stronger statement:

    Statement 2
    For all [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that [itex]\delta < 1[/itex] and for all [itex]x[/itex], if [itex]|x - a| < \delta[/itex] then [itex]|x^2 - a^2| < \epsilon[/itex].

    Note that Statement 2 implies Statement 1. Of course one could substitute any [itex]R > 0[/itex] for '1' in Statement 2 and the result would still imply Statement 1, but 1 is the most natural choice.
     
  21. Oct 13, 2014 #20

    Mark44

    Staff: Mentor

    Really? In what context could 10 trillion be considered relatively close?

    Saying that "relatively close" could mean 10 trillion comes off in my mind as trolling. Since the question has been asked and answered, I am closing this thread.
     
    Last edited: Oct 13, 2014
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