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Convergence of sequences

  1. Mar 15, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Determine the convergence, both pointwise and uniform on [0,1] for the following sequences :

    (i) ##s_n(x) = n^2x^2(1 - cos(\frac{1}{nx})), x≠0; s_n(0) = 0##
    (ii) ##s_n(x) = \frac{nx}{x+n}##
    (iii) ##s_n(x) = nsin(\frac{x}{n})##

    2. Relevant equations

    ##s_n(x) → s(x)## as ##n→∞##

    3. The attempt at a solution

    (i) So for this one, I can split it into a piecewise function since ##s_n(x)## has been defined at the origin.

    Taking the limit as n → ∞, I can observe two limiting functions occuring. s(x) = 0 and s(x) = 1/2.

    Therefore ##s_n(x)## converges pointwise, but not uniformly on [0,1].

    (ii) As n → ∞ ##s_n(x) → x = s(x)## for all x in [0,1]. Hence ##s_n(x)## converges pointwise AND uniformly on [0,1].

    (iii) Exact same limit as (ii) so the same answer will follow.

    Do these look okay? It seems way too straightforward so it has me a bit worried.

    If anyone could confirm it would be great :). Thanks.
     
  2. jcsd
  3. Mar 15, 2013 #2

    Dick

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    Yes, the conclusions about convergence are correct. But I don't see any reason why you think the convergence is uniform. A sequence of continuous functions can converge to a continuous function in a way that is not uniform. You need a better reason than just that the limit is continuous.
     
    Last edited: Mar 15, 2013
  4. Mar 15, 2013 #3

    Zondrina

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    As in for the second and third ones? I think I see what you mean.


    (ii)

    ##s_n(x) = 0 → 0## as ##n→∞## for ##x = 0##.
    ##s_n(x) = n/(n+1) → 1## as ##n→∞## for ##x = 1##.

    So we get only pointwise convergence for (ii) and not uniform for all x in [0,1].

    (iii)

    ##s_n(x) = 0 → 0## as ##n→∞## for ##x = 0##.
    ##s_n(x) = nsin(1/n) → 1## as ##n→∞## for ##x = 1##.

    So as is in (ii) we only get pointwise and not uniform for all x in the interval [0,1].
     
  5. Mar 15, 2013 #4

    Dick

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    No, that's not what I meant at all. You were correct that they are uniformly convergent. You just didn't give a good reason (or any) reason why.
     
  6. Mar 15, 2013 #5

    Zondrina

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    Are you implying I need to go all the way back to the epsilon definition for this?
     
  7. Mar 15, 2013 #6

    Dick

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    No, I wouldn't go back that far. Don't you have any other theorems that will show that convergence is uniform? Here's a hint. In both of those cases ##s_n(x)## is differentiable and the derivative is bounded on [0,1] independent of n. Could that be useful?
     
  8. Mar 15, 2013 #7

    Zondrina

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    Uhm I think I have this theorem I just found :

    Suppose ##\{s_n'(x)\}## converges uniformly on an interval I.

    If each ##s_n'(x)## is continuous and ##\{s_n(x_0)\}## converges for some ##x_0 \in I##, then ##\{s_n(x)\}##
    converges uniformly on I to s(x) and :

    ##lim_{n→∞} s_n'(x) = s'(x)##

    EDIT : So ##s_n'(x) = \frac{n^2}{(n+x)^2}##

    So ##s_n'(x)## is continuous for all x in the interval.

    If we choose x = 0, we see the sequence will converge to 1.

    My only question is the "Suppose {s_n'(x)} converges uniformly on an interval I". Would I say : As n goes to infinity, {s_n'(x)} goes to 1. Which also happens to be the derivative of x just like the theorem is describing.
     
    Last edited: Mar 15, 2013
  9. Mar 15, 2013 #8

    Dick

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    Well that just displaces the problem from showing ##s(x)## converges uniformly to showing ##s_n'(x)## converges uniformly. If you can do that then go ahead. I was thinking of something more like if ##|s'_n(x)|<M## for some constant M then if ##s_n## converges pointwise then it converges uniformly. If you can't find a theorem like that then think how you might prove it yourself. [0,1] is compact.
     
  10. Mar 15, 2013 #9

    Zondrina

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    Wait. So you're saying that if I can bound all the positive terms of my derivative by some constant M_n, then if I have just pointwise convergence of s_n, then it converges uniformly.

    ##|s_n'(x)| = \frac{n^2}{(n+x)^2} ≤ \frac{n^2}{n^2} = 1 = M_n##

    Since we already have shown s_n(x) is pointwise convergent, then because its derivative is bounded we can conclude that s_n(x) is uniformly convergent.
     
  11. Mar 15, 2013 #10

    Dick

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    No, no subscript n. You've shown ##|s_n'(x)|## is bounded by 1. Independent of n. That's important. Showing each ##|s_n'(x)|## is bounded doesn't help. They all have to be bounded by the same constant. And they are, as you've shown. Can you say how this would show pointwise convergence of ##s_n(x)## would show ##s_n(x)## converges uniformly? Think compactness of [0,1] and epsilon. Wish I could think of the name of a theorem to refer to.
     
  12. Mar 15, 2013 #11

    Zondrina

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    Hmm I just read about what a compact space is, but sadly I don't think I'm allowed to use any of this ( the thing about the limit points made sense ).

    I still think I was correct in an earlier post in saying this ( Slightly modified now ) :

    For ##x = 0, \space s_n(x) = 0 → 0## as ##n→∞##.
    For ##x ≠ 0, \space s_n(x) = nx/(n+x) → x## as ##n→∞##.

    So we have two values for the limiting function s(x) depending on x.

    I don't think this is incorrect at all and would result in pointwise, but not uniform convergence.

    Differentiating, I would get :

    For x = 0. ##s_n'(x) = 0 → 0## as n→∞.
    For x ≠ 0. ##s_n'(x) = \frac{n^2}{(n+x)^2} → 1## as n→∞.

    So the convergence is uniform.

    Since ##\{s_n(0)\}## converges, we have one point where ##\{s_n(x)\}## converges and we know that ##\{s_n'(x)\}## converges uniformly so it follows that ##\{s_n(x)\}## converges uniformly.
     
    Last edited: Mar 15, 2013
  13. Mar 15, 2013 #12

    Dick

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    That. Is. Completely. Incorrect. The convergence is uniform. It's late and I'm running out of ways to convince you. But nx/(n+x) does not converge to 1. It converges to x. You had that part right to begin with. What changed your mind?
     
  14. Mar 15, 2013 #13

    Zondrina

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    I had a moment of doubting myself for some reason. I added a bunch to the prior post that makes more sense than what I had before.
     
  15. Mar 16, 2013 #14

    Dick

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    You just don't seem to be grappling with what uniform means. I just had a great idea. Why don't you graph s_n(x)=nx/(x+n) for some values of n, like 1, 10, 100. Does that give you some idea how you could directly prove the convergence is uniform (by epsilon ideas - not theorems)?
     
  16. Mar 16, 2013 #15

    Zondrina

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    Going back to epsilons would probably make me feel more comfortable anyway.

    ##\forall ε > 0, \exists N \space | \space n>N \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]##

    ##|s_n(x) - s(x)| = |\frac{nx}{n+x} - x| = \frac{x^2}{n+x} ≤ \frac{1}{n+1}##

    So choosing ##n > \frac{1}{ε} - 1 \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]##

    Hence (ii) converges both pointwise and uniformly.
     
  17. Mar 16, 2013 #16

    Dick

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    That's the idea.
     
  18. Mar 16, 2013 #17

    Zondrina

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    See, I understand when I go back to the epsilon definition for some reason, but the theorems he's provided for us are not sufficient for this problem I think.

    (iii)

    ##\forall ε>0, \exists N \space | \space n>N \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]##

    ##|s_n(x) - s(x)| = |nsin(x/n) - x| ≤ nsin(1/n) - 1##

    Having a small nitch with the arithmetic here. I can't isolate for n in terms of epsilon.

    EDIT : Wait, what if :

    ##|s_n(x) - s(x)| = |nsin(x/n) - x| ≤ |n||sin(x/n)| + |x| ≤ n + 1##

    So choosing ##n > ε -1 \Rightarrow |s_n(x) - s(x)| < ε, \forall x \in [0,1]##
     
  19. Mar 16, 2013 #18

    Dick

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    |x| isn't very small on [0,1]. That's not going to work. Look at the function x-n*sin(x/n). Is it increasing or decreasing on [0,1]?
     
  20. Mar 16, 2013 #19

    Zondrina

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    Wouldn't it be increasing? It's derivative is 1 - cos(x/n).

    ##2nπ## seems to be relevant here as it is the root of the derivative.
     
    Last edited: Mar 16, 2013
  21. Mar 16, 2013 #20

    Dick

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    Sure. It's increasing. It's zero at x=0 and increases to 1-n*sin(1/n) at x=1. Does that give you any ideas about how to pick an N corresponding to an epsilon?
     
    Last edited: Mar 16, 2013
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