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Equivalent definitions of convergence

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\mathbf{D1:}\forall\varepsilon>0,\exists K\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n\geq K\Longrightarrow|x_{n}-x|<\varepsilon[/itex]

    [itex]\mathbf{D2:}\forall\rho>0,\exists M\epsilon\mathbb{N},\forall n\epsilon\mathbb{N},n>M\Longrightarrow|x_{n}-x|\leq\rho[/itex]

    Show these two definitions of convergence (of a sequence xn) are equivalent by showing both D1 implies D2, and D2 implies D1.

    2. Relevant equations

    --

    3. The attempt at a solution

    To show D1 ==> D2, i just took M = K - 1, where K was from D1. But it didn't seem to get me anywhere since i can show

    [itex]n>M\Longrightarrow|x_{n}-x|< \rho[/itex],

    but this does not encompass the possibility of ≤ ρ. So basically i have no idea how to do this.

    Similarly, i need something to get me started on D2 ==> D1. How do I go about this?

    I'm sure i'm overthinking it as I've done much more complicated proofs before, and I've done all other questions on this assignment without help. I just honestly don't know how to do this one. Thanks in advance!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 11, 2013 #2

    vela

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    If ##\lvert x_n - x \rvert \le \rho##, then surely ##\lvert x_n - x \rvert < \rho## is satisfied, right?
     
  4. Sep 11, 2013 #3
    That information is helpful to show D2 ==> D1 right?
     
    Last edited: Sep 11, 2013
  5. Sep 13, 2013 #4
    These are definitions of convergence as [tex]n \rightarrow \infty[/tex] of the sequence [tex]x_n[/tex] to a particular value x. The concept is that the bigger n gets, the closer [tex]x_n[/tex] gets to x.

    Now to state it formally we have to choose an [tex]\epsilon,[/tex] which supposedly is a small number like 1/n or [tex]1/n^2[/tex] and show that if n is big enough

    (1)... [tex]|x - x_n| < \epsilon[/tex].

    When we say n is big enough we mean that the inequality (1) is true for n and every higher integer -- n+1, n+2,... etc. In other words, once we pick the [tex]\epsilon[/tex] there is some point from M on where the inequality holds.

    Of course, we said any [tex] \epsilon [/tex] so if you make it smaller you'll need a bigger M (or K or whatever). The important thing is that once you pick the [tex]\epsilon[/tex] you can always find some M that works.

    Note we said there is some point M, not that M is necessarily the smallest cutoff that will work. If you're not sure M is adequately large, you can always pick a bigger one. So the difference between saying there is an M such that [tex] n \ge M[/tex] makes (1) work, and there is a K such that n> K makes (1) work is almost non-existent. Just pick an N which is bigger than both M and K and let n be larger than that, and it won't matter whether you write < or less than or =.

    Once you understand why convergence is defined in terms of n > N, it really doesn't matter whether you have strict inequality or not.

    I hope you can use these thoughts to construct a proof your teacher will like.

    If I may editorialize, I think this is a poor choice of problem for a student. The strict or not strict inequality is essentially meaningless, and takes your eye off the ball, which is what convergence really means. However, do what you have to to get a good grade.
     
    Last edited: Sep 13, 2013
  6. Sep 14, 2013 #5
    Thanks brmath. I do understand this on an intuitive level. I'm not particularly great at this stuff but I've taken a few junior and even a senior analysis/proof oriented course before, and I have proved much more complex stuff. That's why I'm surprised I couldn't get started here. But I definitely understand and your explanation helped.

    Because I asked this question the night before it was due, I didn't really have time to do much so i just threw down an "explanation". Hopefully I'll get something out of it, but the remainder of the assignment was completed, so I hope it will not affect my grade too much.

    Thanks!
     
  7. Sep 14, 2013 #6
    I remember being a student and doing things the night before they were due. Computer programming cured me of that (missed deadlines where money was involved). My theory became that if I start early and get something done early, this could not possibly hurt. If I start early and something goes wrong (which is practically always) then there is time to recover. I hope your explanation will be satisfactory to your teacher.
     
  8. Sep 14, 2013 #7
    Normally i too start my assignments late (i.e., i start my assignments the night before). this one, however, i started early :) hopefully my habits change and continue this way this semester.
     
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