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Homework Help: Solving an inequation

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    I was doing a certain problem, and in the end, I got this inequation which I can't solve. Probably due to lack of skill, can someone give me a hand on it?

    [itex]|n|<\frac{\epsilon(n^3 + 1)}{2}[/itex]

    3. The attempt at a solution

    [itex]|n|(1-\frac{\epsilon*n^2}{2})< \frac{\epsilon}{2}[/itex]

    I tried rearranging in this way, but still, no dice. I know that n must be positive, and I actually plan on plugging different values of epsilon, but still, I can't solve this inequation.
  2. jcsd
  3. Jan 28, 2013 #2


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    Homework Helper

    What is the problem exactly?
  4. Jan 28, 2013 #3
    The number 27 on this list.
  5. Jan 28, 2013 #4


    Staff: Mentor

    Since n > 0, you don't need the absolute values.

    Are you trying to solve for n or for ##\epsilon##? If you're trying to solve for ##\epsilon##, that's pretty simple, but if you're trying to solve for n, you have a messy third-degree polynomial that makes factoring difficult, if not impossible. That would leave only numerical approximations or graphical solutions as potential approaches.

    BTW, we call them inequalities in English.
  6. Jan 28, 2013 #5
    Oh, I apologize, I'm not from an English speaking country, so Inequality, got it.

    I'm trying to solve for ϵ.
  7. Jan 28, 2013 #6


    Staff: Mentor

    We typically use letters such as M or N to represent large numbers, and Greek letters ##\epsilon## and ##\delta## for small (close to zero) numbers.

    You want to find a (large) number N such that 2n/(n^3 + 1) > N. For any given value of N you can use trial and error to find the smallest value of n that satisfies the inequality.
  8. Jan 28, 2013 #7
    Wait, wait, I thought that :

    \\a_n = \frac{2n}{n^3 + 1}\\
    |{a_n-L}| < \epsilon \\
    L =\lim_{n->\infty} a_n = 0\\
    |{a_n}| < \epsilon \\

    And thus, my problem was:

    [itex]|\frac{2n}{n^3 + 1}|<\epsilon\\[/itex]

    And N would be an particular "n" for a particular epsilon, isn't that the case?
  9. Jan 28, 2013 #8


    Staff: Mentor

    I'm sorry, I didn't look close enough at your sequence, and mistakenly thought that you needed to show that it diverged. Your use of ##\epsilon## was correct. I'm sorry for steering you the wrong direction.

    An ##\epsilon## would normally be given, so your goal would be to find n so that the inequality was true. The usual way would be to use trial and error.

    Note that, since n > 0, you don't need absolute values in the inequality.
  10. Jan 28, 2013 #9
    Ah, that's fine, don't worry!

    While I agree that I don't need absolute values, my problem still is with the inequality :/ How would I go about solving it?
  11. Jan 28, 2013 #10


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    So applying the definition :

    [itex]\forall \epsilon > 0, \exists N \space | \space n>N \Rightarrow |a_n -L| < \epsilon[/itex]

    So now :

    [itex]|a_n - L| =|\frac{2n}{n^3 + 1} - 0| = |\frac{2n}{n^3 + 1}| = \frac{2n}{n^3 + 1}[/itex] since n > 0.

    Lets say we take the case where ε=1 as per your question. Okay, now that we have something plausible to work with consider :

    [itex]\frac{2n}{n^3 + 1} < 1[/itex]
    [itex]\frac{n}{n^3 + 1} < \frac{1}{2}[/itex]
    [itex]n^2 + \frac{1}{n} > 2[/itex]

    Can you continue from here?
  12. Jan 28, 2013 #11
    Thanks for the input Zondrina!

    I can't, that's exactly the point I made in OP, I don't know how to solve inequalities in this form. I usually try to factor them out, but I can't really find a factorization in this case. I can't apply the Rational Roots Theorem either, so, in short, I don't know how to proceed.
  13. Jan 28, 2013 #12

    Ray Vickson

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    Science Advisor
    Homework Helper

    In the context of the original problem you posted, you do NOT need to find the entire range of n values; you just need to find an N = N(ε) such that for all n ≥ N the inequality holds. Finding the *smallest* such N is hard; finding some N that works is a lot easier.

    So, for n > 0 and ε > 0, we have ## \epsilon \, n^3/2 < \epsilon \, (n^3+1)/2,##, so if we satisfy ##n \leq \epsilon \, n^3/2,## that value of n will also satisfy the original inequality. Solving ##n \leq \epsilon \, n^3/2## is a much easier problem.
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