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Absolute quadratic inequalities.

  1. Sep 26, 2004 #1
    A bit of a newbie question, but I was wondering how does one go about solving these?

    For example: (I was working on a problem posted on another thread on Homework Help)
    [tex]
    |3n-4| < 9\epsilon n^2 + 3 \epsilon
    [/tex]
    Epsilon is a small positive number of course :P

    The tricky part is when I split it up..
    [tex]
    \begin{align*}
    -9\epsilon n^2 - 3n - 3\epsilon + 4 < 0 \\
    9\epsilon n^2 -3n + 3\epsilon + 4 > 0
    \end{align*}
    [/tex]

    Wouldn't the solution for n then be 4 inequalities? That doesn't make sense, does it?
     
  2. jcsd
  3. Sep 26, 2004 #2

    Tide

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    try factoring the quadratic and analyze according to the sign of the factors
     
  4. Sep 26, 2004 #3

    HallsofIvy

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    A general method for solving inequalities is to solve the equation first. The points mark the boundaries of the intervals on which the inequality is true.

    In this case, solve the equation |3n-4|= 9&epsilon;n2+ 3&epsilon;

    If 3n-4> 0 that is equivalent to 3n-4= 9&epsilon;n2+ 3&epsilon; which is the quadratic equation 9&epsilon;n2- 3n+ (4+3&epsilon)= 0.

    If 3n-4< 0 that is equivalent 4- 3n= 9&epsilon;n2+ 3&epsilon; which is the quadratic equation 9&epsilon;n2+ 3n- (4+3&epsilon)= 0.

    Those two equations have 4 solutions which divide all real numbers into 5 intervals. You can choose one point in each interval to determine whether you get ">" or "<".
     
  5. Sep 26, 2004 #4
    Hmm.. I was afraid it would come to this.

    I was trying to solve this for a delta-epsilon proof of a limit at infinity (finding what N of epsilon could be that is < |n|.

    I got my two quadratic equations, so technically, the smallest one could be N? or the largest? Or do all of them work?

    Because there are two quadratic inequalities to solve, this puts a bound on what n could be, so it would be redundant to say that |n| is greater than all N of epsilon.

    Or am I missing something very painfully obvious? :P
     
  6. Sep 27, 2004 #5

    HallsofIvy

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    In that case, you don't really need to solve the inequality, just show that it is true for "sufficiently large" n. Certainly as soon as n> 4/3, 3n- 4 will be positive so you don't need the second of the ways you are "splitting up" the absolute value.
    From [tex] 9\epsilon n^2 + 3n + 3\epsilon - 4 0 0[/tex]
    you can use the quadratic formula to determine where left side is larger than 0. If that is true for all n larger than some number, you are done.
     
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