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Why do I get the right answer?

  1. Oct 14, 2004 #1

    quasar987

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    I have this problem that I could not solve, so I looked up the answer. It more of less said how we had to start to get to the answer. So I try that and get the right answer, but I still don't understand what this first step MEANS.

    The question is about sequences and it says: Find an interval which contains all terms but the first 20 of the sequence defined by

    [tex]x_n =4+\frac{(-1)^n}{n}[/tex]

    If we start by saying "we are looking for an [itex]\epsilon>0[/itex] such that [itex]\forall n>20[/itex], [itex]|x_n - 4|<\epsilon[/itex]", everything follows very smoothly and we find that this epsilon is 1/20 and that the interval is (3+19/20 , 4+1/20).

    But this "we are looking for an [itex]\epsilon>0[/itex] such that..." begining is not the real begining, because what we are looking for really is an INTERVAL. An interval [itex]I[/itex] such that [itex]\forall n>20[/itex], [itex]x_n \in I[/itex]. So how do the "we are looking for an [itex]\epsilon>0[/itex] such that..." follows from the real question that is asked?
     
  2. jcsd
  3. Oct 14, 2004 #2
    I don't understand your confusion. You already have the answer. What's the problem?
     
  4. Oct 14, 2004 #3

    Gokul43201

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    Define y(n) = x(n) - 4 and find the smallest interval that contains all y(n>20).
    Also see that y(n+1) = -(n/n+1)*y(n) => |y(n+1)| < |y(n)|
     
  5. Oct 15, 2004 #4

    quasar987

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    Ok so the line of reasoning I was looking for is: Since 4 is a constant term in every term of the sequence, finding the interval [itex]I[/itex] that contains all but the 20 first terms is the same as finding an interval [itex](-\epsilon, \epsilon)[/itex] that contains all but the first 20 [itex]y_n[/itex] (as defined by Gokul) because our interval [itex]I[/itex] will then be simply [itex](4-\epsilon, 4+\epsilon)[/itex].

    There is still something that is unclear in my head about this method, though. Supposing this is the general method of solving all problems of the "find an interval that contains all but the first /whatever terms" kind, I am bothered by this special case: Suppose the sequence converges towards 0 and looks like this:

    [tex]\{-100, 100, -50, 97, -23, 92, -11, 84, -2 ...\}[/tex]

    i.e. the negative terms converge much faster than the positive ones. Now if we try to find the interval [itex](-\epsilon, \epsilon)[/itex] that contain all but the first 5 terms, we cannot set [itex]\epsilon=23[/itex] because this does not include the sixth term. And we cannot set [itex]\epsilon=93[/itex] because it would include the third and the fifth terms. So there is no such [itex]\epsilon[/itex].

    There is always the posibility that there exists no [itex]x_n[/itex] that generates a sequence that behaves in such a way. That would be nice. :smile:
     
  6. Oct 15, 2004 #5

    shmoe

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    You needn't be bothered as problems of this type are going to be impossible for some sequences, hence there is no general method that works on all sequences.

    Your original sequence is nice in the following way, the sequence [tex]|y_n|[/tex] is a strictly decreasing sequence. That is [tex]|y_1|>|y_2|>|y_3|>\ldots[/tex]. In a case like this you can always find an interval that excludes the first N terms, for any value of N. (note [tex]y_n=x_n-\lim_{n\rightarrow\infty}x_n[/tex])
     
  7. Oct 15, 2004 #6
    The confusion seems to be because there isn't a unique answer to the problem. It seems that any interval contained in (3+18/19,4+1/20) and containing [3+20/21,4+1/22] would do the trick. The answer given is the largest interval which is symmetric about 4.
     
  8. Oct 15, 2004 #7

    HallsofIvy

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    xn satisfying [itex]|x_n - 4|<\epsilon[/itex]
    IS an interval:
    This is the same as
    [tex]-\epsilon< x_n- 4<\epsilon[/tex]
    or
    [tex]4-\epsilon< x_n< 4+\epsilon[/tex]
     
  9. Oct 15, 2004 #8

    quasar987

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    I realised that. It's why I was able to solve the problem. But why 4±epsilon and not something else is what I was wondering. I know now.
     
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