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Sequence inequality, epsilon N argument

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data


    I already have the solutions, but I am not sure what the solutions are trying to say.

    [PLAIN]http://img194.imageshack.us/img194/2595/unledlvc.jpg [Broken]

    So in
    I don't understand this, we have

    [tex]n > \frac{1}{\epsilon}[/tex]

    and If (and I am guessing we really want this to hold)

    [tex]N > \frac{1}{\epsilon}[/tex]

    How does that deduce n > N? Wouldn't it make more sense to say that [tex]N < \frac{1}{\epsilon}[/tex] and hence [tex]N < \frac{1}{\epsilon} < n[/tex]?

    One final question from me is, how does N relate to epsilon? I understand that

    [tex]|a_n - L | < \epsilon[/tex] is speaking how close a_n is to our limit L for a very very small epsilon and the goal of the proof-theorem is to find such an epsilon, what role does n > N play with respect to [tex]|a_n - L | < \epsilon[/tex]

    I have accompanied this with a picture

    [PLAIN]http://img213.imageshack.us/img213/5001/unledsy.jpg [Broken]

    I cna see that the epsilon is trying to "squeeze" our limit and how n > N is supporting this too. So is the intuition that they (both inequalities) are saying the same thing but they are both supporting each other?

    I'll clarify if possible
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 7, 2011 #2

    Mark44

    Staff: Mentor

    Suppose ε is given as 1/10. Then we can take N = 10. Then for any integer n for which n > N, 1/n < ε. That's really all the above is trying to say.

    For example:
    n = 11, 1/11 < 1/10
    n = 12, 1/12 < 1/10
    and so on.

    In answer to your question, if we take N < 1/ε, as you suggest, then using my example value of ε = 1/10, we could take N = 8 (or N = 5 are whatever, as long as N < 10). That will cause problems, because 1/9 > 1/10 = ε.
     
  4. Nov 7, 2011 #3
    Now I am even more confused because the conditions were [tex]N > \frac{1}{\epsilon}[/tex] or [tex]N < \frac{1}{\epsilon}[/tex]

    Take ε = 1/10, in either case you get something like this

    [tex]N = 10 > \frac{1}{\frac{1}{10}} = 10[/tex], i.e. 10 > 10 and WLOG, 10 < 10, which is absurd in either case (I don't know why my book uses "absurd" when we do contradictions proofs, it seemed kinda rude)

    Do you actually mean ε = 10 instead?
    So I thinky ou really do mean ε = 10 and where did you get 1/9 from this time?

    thanks
     
  5. Nov 7, 2011 #4

    Mark44

    Staff: Mentor

    No, I don't mean ε = 10. ε is generally a "small" number, typically much smaller than 1. The smaller ε is, the farther out in the sequence you have to go to find terms in the sequence that are smaller than ε.

    You want N > 1/ε. (Some texts will say N >= 1/ε.)

    If ε = .01, then 1/ε = 100, and we can take N = 100.

    Now for any n > N, 1/n < ε.

    You're really overthinking this.

    Where are you getting that the conditions include N < 1/ε? There's nothing in your OP that says this.
     
  6. Nov 7, 2011 #5
    How did you arrive at N < 10 again?

    edit: you got it from n > N right? So n > 10 and n > N, then N <= 10 which is consistent with my initial condition (not from book)

    [strike]No I made that condition up because I thought should work, but I don't follow your counterexample from above at all. [/strike]



    One of the conditions we needed was n > N, so that 1/n < 1/N, can assume there exists this N that n > N? Because if we don't, then it wouldn't work in either case (with more deduction)

    When I suggested that N < 1/ε, then 1/N > ε <=> ε < 1/N

    So we have ε < 1/N and ε > 1/n => 1/n < 1/N => n > N, so now we could confirm with the assumption n > N.

    Do you see where I am going with this?
     
    Last edited: Nov 7, 2011
  7. Nov 8, 2011 #6

    Mark44

    Staff: Mentor

    I didn't. I was trying to show you why what you were doing was incorrect. This is what I said (emphasis added).
    The way this limit process works is as a dialogue. You're trying to prove that [itex]\lim_{n \to \infty} \frac{1}{n} = 0[/itex].

    Let's say I'm skeptical, and don't believe you.

    I say, OK, can you find some point in the sequence beyond which all of the following numbers are < 1/100 (that's ε).

    You do a quick calculation, and say sure. You're thinking, I want to find a number N that works. So, because of this particular sequence, you calculate N = 1/ε = 100.

    And then you show me that all of the terms from a101 on are smaller than ε.

    If I'm still not satisfied, I'll come back, "Can you make them smaller than 1/1000?" Using the same process as before, you tell me that all of the terms from a1001 on are smaller than ε.

    The idea is that you want to find an index N so that on one side, all of the terms are smaller than ε. That means all of the terms whose indexes n are larger than N. On the other side of n, the terms in the sequence are >= ε, so those terms don't serve to prove that the sequence is converging to whatever the limit is supposed to be.

     
    Last edited: Nov 8, 2011
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