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Absolute value removed from equation. Why?

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that [itex]\epsilon[/itex]> 0, why is it, that [itex]\left | x_{n} -L\right |< \epsilon [/itex] implicates that [itex] x_{n} > L-\epsilon [/itex] ?

    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 20, 2012 #2

    Hurkyl

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    Try some examples to get a feel for it.

    Since you're dealing with absolute values, you could always use the usual technique for dealing with them: split into two cases depending on whether the argument is positive or negative.
     
  4. Mar 20, 2012 #3
    Well if: l Xn-L l ≥0 then you can remove the the absolute value.
     
  5. Mar 20, 2012 #4

    Ray Vickson

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    No, you can't. We have |-2| > 0; would you say that -2 > 0? Besides, for ANY real number r we always have |r| ≥ 0.

    RGV
     
  6. Mar 20, 2012 #5

    LCKurtz

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    Remember that an equation like ##|x_n-L|<\epsilon## can always be rewritten ##-\epsilon < x_n-L < \epsilon##. Add ##L## to all three sides and notice you are only using half of the result.
     
  7. Mar 20, 2012 #6
    Yes, i thought of that at first, but it doesnt help in this case since [itex]x_{n}-L[/itex] can be either positive or negative at all points, or alternating, depending on what series [itex]x_{n}[/itex] is a part of. So based on that, i cant exclude one of the cases via the definition.
     
  8. Mar 21, 2012 #7

    Ray Vickson

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    If x >= L then x > L-ε (ε>0). So, just look at the case where x < L (assuming |x-L| < ε).

    RGV
     
  9. Mar 21, 2012 #8

    HallsofIvy

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    As I am sure you know, |x|= x if [itex]x\ge 0[/itex], and |x|= -x if x< 0. So if [itex]x_n- L< 0[/itex], then [itex]|x_n- L|= -(x_n- L)= L- x_n< \epsilon[/itex] so that [itex]-x_n< -L+ \epsilon[/itex] and, multiplying boty sides by -1, [itex]x_n> L- \epsilon[/itex]. Of course, if [itex]x_n- L\ge 0[/itex] then [itex]x_n\ge L[/itex] and, since [itex]\epsilon> 0[/itex], [itex]L> L- \epsilon[/itex] so [itex]x_n> L-\epsilon[/itex].

    In either case, [itex]x_n> L- \epsilon[/itex].
     
    Last edited: Mar 21, 2012
  10. Mar 21, 2012 #9

    Hurkyl

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    It doesn't matter that xn comes from a series; the question you asked is one about the property of the number xn.

    And besides, in a proof by cases you don't exclude a cases: you consider all of them. (in this case, only 2)
     
  11. Mar 22, 2012 #10
    Yea that makes sense.

    That also makes sense! And how creative too.

    Your right, i was confused.

    I guess the problem i had (besides being stupid of course) is that i got hung up on the results from the definition. They seemed to conflict and give 2 different cases, but that wasnt the case, wich i feel that HallsOfIvy showed me.

    Thank you all.
     
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