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Cauchy sequences or not

  1. Dec 7, 2015 #1
    Hello evry body let be $(u_{n}) \in \mathbb{C}^{\matbb{N}}$
    with $u_{n}^{2} \rightarrow 1$ and $\forall n \in \mathbb{N} (u_{n+1) - u_{n}) < 1$.
    Why does this sequences converge please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
     
  2. jcsd
  3. Dec 7, 2015 #2

    PeroK

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    Why don't you fix the Latex?
     
  4. Dec 7, 2015 #3

    fresh_42

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    Plus consider please whether the difference ##(u_{n+1} - u_{n}) < 1## shouldn't be better ##|u_{n+1} - u_{n}| < 1##
     
  5. Dec 7, 2015 #4
    It would be better indeed. And how to put the latex here please?

    Thank you in advance and have a good afternoon:oldbiggrin:.
     
  6. Dec 7, 2015 #5

    PeroK

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    You double your dollars!
     
  7. Dec 7, 2015 #6

    fresh_42

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    If it was that easy ....
     
    Last edited: Dec 7, 2015
  8. Dec 10, 2015 #7
    So did anyone has a solution please?
     
  9. Dec 10, 2015 #8

    Erland

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    Ok, I fixed Calabi's TeX-notation. See and learn...

    "Hello evry body let be ##(u_{n}) \in \Bbb{C}^{\Bbb{N}}##
    with ##u_{n}^{2} \rightarrow 1## and ##\forall n \in \Bbb{N}\,(u_{n+1}-u_{n})<1##.
    Why does this sequences converge please?"

    For large ##n##, ##u_n^2## is close to ##1##. What can you then say about ##u_n## for large ##n##? Can the sign of ##u_n## alter for large ##n##, considering that ##u_{n+1}-u_n<1##?
     
  10. Dec 10, 2015 #9

    Krylov

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    It's good to see that the quality of the discussion in the technical math sections has reached a new high.
     
  11. Dec 10, 2015 #10

    mathwonk

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    i know what you mean but, in general, complex numbers have no "sign", so your nice hint needs a modified formulation.
     
  12. Dec 11, 2015 #11
    Hello every one sorry Kyrlov I should give what I search on this exercicse even if it's wrong,
    so as $$u_{n}^{2} \rightarrow 1$$ we can say that $$\forall \epsilon > 0, \exists N \in \mathbb{N} / n > N \Rightarrow |u_{n}^{2} - 1| < \epsilon$$. As Erland said it could be a story of sign but we're in $$\mathbb{C}$$.
     
  13. Dec 11, 2015 #12

    Erland

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    Ah, OK, but in that case, the condition ##u_{n+1}-u_n<1## is meaningless, since order does not exist between complex numbers. Presumably, Calabi meant ##|u_{n+1}-u_n|<1##. If so, let us instead ask if the sign of the real part of ##u_n## can alter for large ##n##...
     
  14. Dec 11, 2015 #13

    Erland

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    Are the ##u_n##:s close to some particular set of numbers for large ##n##?
     
  15. Dec 11, 2015 #14
    Ther's 2 possible of numbers -1 or 1. Like the square of u converge in 1 and as $$|u_{n+1} - u_{n}| < 1$$ the term of $$u_{n}$$
    are in one neighbourhood of 1 or -1 I see it but I can't say it good.
     
  16. Dec 11, 2015 #15

    Erland

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    Yes, if we choose small neighborhoods of ##1## and ##-1##, all ##u_n## is in one of them, not the other, for all sufficiently large ##n##, since otherwise we would have ##|u_{n+1}-u_n|\ge 1## for some ##n##. This holds for arbitrarily small neighborhoods of ##1## or ##-1##, so the sequence converges to ##1## or ##-1##. It just remains to make this argument formal.
     
  17. Dec 11, 2015 #16
    Yeah but it's this formalism I can't wright.
     
  18. Dec 11, 2015 #17

    Erland

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    Try! How would you begin?
     
  19. Dec 11, 2015 #18
    We can do it in the absurd way : let suppose for exemple that $$\exists \epsilon > 0/ \forall N \in \matbb{N} \exists n > N / |u_{n} - 1| > \epsilon (1)$$. For this $$\epsilon > 0, \exists N \in \mathbb{N} / n > N \Rightarow |u_{n}^{2} - 1| < \epsilon $$.
    For this N we can find n as in (1). For this n $$|u_{n+1} - u_{n}| < 1$$.
    Hum.
    I don't see the absurdity.
     
  20. Dec 11, 2015 #19

    Erland

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    Hmm, this might lead somewhere. What if you also consider the corresponding relation as your first one but for -1 instead of 1? Also, what if the ##\epsilon##:s you use are unequal, but perhaps related in some way?
     
  21. Dec 11, 2015 #20

    PeroK

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    Can you see why ##u_n## converges? If not, you'll never prove it using ##\epsilon##.

    I'd draw a diagram for ##u_n^2## and try to see geometrically why ##u_n## must converge. Once you understand that, you can try to translate it into a formal ##\epsilon## proof.
     
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