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Absolute value result

  1. Apr 13, 2012 #1
    Just a general question with absolute values:

    Is it possible to have an absolute value of this form:

    | f(x) - L | < -L (the minus sign is meant to be there) and if so how can I convert it into expanded form? i.e: -L < f(x) - L < -L or something of that form.
     
  2. jcsd
  3. Apr 13, 2012 #2

    LCKurtz

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    If L is meant to signify a positive number, then the inequality can't hold because no absolute value is negative. If L might be negative, then there is no problem. An inequality ##|f(x) - a|<b## is always equivalent to ##-b < f(x)-a<b##.
     
  4. Apr 13, 2012 #3
    If L is negative, then your problem reduces to (substitute K = -L, K positive) [tex] |f(x) + K| < K[/tex] which holds if [tex] -2K < f(x) < 0 [/tex].
     
  5. Apr 13, 2012 #4



    So the fact that there is a 2 with the K does not affect the inequality? Wouldn't I have to get rid of the 2 to make it a standard statement?
     
  6. Apr 13, 2012 #5
    Let me write it out more explicitly:
    [tex] |f(x) + K| < K[/tex] means
    [tex] -K < f(x) + K < K.[/tex]
    (compare with the first reply).
    Now substract K from all three terms to get:
    [tex] -2K < f(x) < 0 [/tex].

    Hence the 2 is simply the result of the calculation.
     
  7. Apr 13, 2012 #6
    Ok, I was just confirming if that's all it meant. Thanks.
     
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