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Proving Cauchy Sequences

  1. May 16, 2010 #1
    The problem statement, all variables and given/known data

    Well, my problem is proving that sequences are in fact Cauchy sequences. I know all the conditions that need to be satisfied yet I cannot seem to apply it to questions. (Well, only the easy ones!)

    My question is, prove that [tex]X_{n}[/tex] is a Cauchy sequence, given that [tex]X_{n+1}= f(X_{n})[/tex] where f(x)=cos(x).

    I have been told that [tex]|X_{n+1}-X_{n}| \leq 2^{(1-n)}[/tex]

    The attempt at a solution

    Well basically, I tried using m,n>N and N>0 and m>n>N to say that

    [tex]|X_m-X_n| = cos(X_{m-1}) - cos(X_{n-1})[/tex]

    because this is the way i was doing it for simpler sequences.

    However, I havent got a clue where I would go from here which leads me to thinking I have gone the wrong way about it!
    Any help would be greatly appreciated.
    Thanks
     
    Last edited: May 16, 2010
  2. jcsd
  3. May 16, 2010 #2
    OK, how do we prove that something is a Cauchy sequence?

    First, a very mean man who hates undergrad analysis students is going to choose an [itex]\epsilon > 0[/itex]. What we have to do is, for whatever [itex]\epsilon [/itex] he chooses, be able provide a number N such that for any two numbers m and n that are both greater than N,

    [tex] |{X_n - X_m}| < \epsilon [/tex].

    If we can do this, then it's a Cauchy sequence.

    So now it would be nice if we were able to do cool things with [itex] cos X_{m-1} [/itex], but I don't really see how that's going to help us. However, the inequality they told you about looks promising. Try to figure out a way to provide a number N in response to any [itex] \epsilon [/itex] using that inequality.
     
  4. May 17, 2010 #3
    Hello, thanks for the reply.

    Here is what I have done:

    Take m,n>N
    N>0
    [tex]\epsilon > 0[/tex]

    Take m = n+1 then

    [tex]|X_m-X_n| \leq 2^{1-n} \leq 2^{-n}[/tex]

    But then i don't know where to fit epsilon into it?

    I actually have the answer in front of me, but it doesnt use epsilon at all, or N, which is what I have been used to using. So I wanted to see if there was a way to do it with them included.
     
  5. May 17, 2010 #4
    What definition of Cauchy sequences are you using that doesn't contain an epsilon? Maybe there is some other variable used, like a lot of calculus textbooks now use "h" instead of [itex] \DELTA x [/itex] in the definition of a derivative?

    In any event, you're not allowed to choose m and n; the inequality has to be true for ANY m and n which are both greater than N. Otherwise, you'd have the following (FLAWED) proof:

    Let [itex] \{ x_n \} = \ln x [/itex]. Let [itex] m = n+1 [/itex]. Then [itex] |X_m - X_n| = | \ln (n+1) - \ln n | = \ln \frac{n+1}{n} [/itex].

    Now [itex] \lim_{n \rightarrow \infty} \ln \frac{n+1}{n} = 0. [/itex] Hence for any [itex] \epsilon [/itex], there is an N such that for n > N [itex] \ln \frac{n+1}{n} [/itex] < [itex] \epsilon [/itex]. So [itex] \{ x_n \}[/itex] is Cauchy.

    Of course, this isn't true; ln x doesn't converge at all.
     
  6. May 17, 2010 #5
    Yes I thought that was odd. But it is a follow on question so perhaps they had proved it before, I'm not sure.
    But I want to be able to do it just based on this.

    So I can't take m = n+1, that makes sense. I was just clutching at straws really, trying to use the inequality provided.

    So could we take [tex]\epsilon > 0[/tex] and then we have [tex]|X_m-X_n| \leq 2^{1-n}[/tex]
    but m,n>N, so if we take N > 1

    [tex]|X_m-X_n| \leq 2^{1-n} \leq 2^{1-N} \leq 2^{-N}[/tex]

    Is that wrong to just assume that?
    Sorry, I am not very good at these at all! Thanks for your patience :)
     
  7. May 17, 2010 #6
    You might try doing this.

    Let [itex] \epsilon = .1 [/itex].

    Now give me N such that that no matter what m and n I choose greater than N, [itex] | X_m - X_n | < \epsilon [/itex].

    Then try it for [itex] \epsilon = .001 [/itex]. Then .00001, etc, until you get the method down.

    Then generalize so that you can give me an N for any [itex] \epsilon [/itex] I choose.

    Then in your proof, just write down the method of generating Ns in response to an [itex] \epsilon [/itex] and show that it makes [itex] | X_m - X_n | < \epsilon [/itex].

    There will be a DIFFERENT N for each [itex] \epsilon [/itex], by the way. As [itex] \epsilon [/itex] gets smaller, N gets bigger.
     
  8. May 17, 2010 #7
    you don't have to go back to the cauchy definition, perhaps it was proven somewhere that either

    [tex] | a_{n} - a_{n+1} | \leq c^{n} [/tex] where |c| < 1
    or
    [tex] | a_{n} - a_{n+1} | \leq c^{-n} [/tex]
    implies that the sequence is cauchy. In both cases, you can obviously see that as n gets large, c^n (where 0 < c < 1) or c^(-n) can be made arbitrarily small.

    even if it was not proven formally anywhere, you can still use this
     
  9. May 18, 2010 #8
    Okay that sort of makes sense, i have C defined in the set [-1,1] in a previous question. (if it is the same C !)
    but what about the [tex]2^{n-1}[/tex]
    where does that come in in terms of C ?
     
  10. May 20, 2010 #9
    [tex] 2^{1-n} [/tex] becomes small as n becomes large
     
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