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Homework Help: Mean value theorem & Cauchy sequence

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let a0=0 and an+1=cos(an) for n≥0.
    a) prove that a2n≤a2n+2≤a2n+3≤a2n+1 for all n≥0.
    b) use mean value theorem to find a number r<1 such that |an+2-an+1| ≤ r|an-an+1| for all n≥0. Using this, prove that the sequence {an} is Cauchy.


    2. Relevant equations
    N/A

    3. The attempt at a solution
    a) I proved part (a) using induction. Furthermore, I showed that all terms of the sequence is bounded between 0 and 1.

    b) Claim: |an+2-an+1| ≤ 0.85|an+1-an|

    Assuming the claim is true, I have that
    |am-an|=|am-am-1|+...+|an+1-an|≤0.85m-1|a1-a0|+...+0.85n|a1-a0|≤0.85n/(1-0.85) ->0 as n->∞. Thus, by squeeze theorem, the sequence is Cauchy.

    But I am not sure how to prove the claim. If I consider a special case an+1≥an (which is clearly not true in our case), i.e. assuming the sequence is nondecreasing, then by mean value theorem, an+2-an+1=-sin(c)(an+1-an) where c must be somewhere between 0 and 1, so |sin(c)|<0.85, and the claim is proved in this special case. But the trouble is that our sequence is actually not even monotone (as shown in part a), so how can we prove the claim in our case?

    Any help is much appreciated! :)
     
  2. jcsd
  3. Jan 21, 2010 #2
    If I rewrite [tex]|a_{n+2} - a_{n+1}| \leq r|a_{n+1} - a_n|[/tex] as [tex]\left| \frac{\cos a_{n+1} - \cos a_n}{a_{n+1} - a_n}\right| \leq r[/tex] what does that remind you of?
     
  4. Jan 21, 2010 #3
    ???

    Mean value theorem: if f is differentiable in (a,b) and continuous on [a,b], then there exists c E (a,b) s.t. f(b)-f(a)=c(b-a)
    The problem is in the theorem, b must be greater than a. So I have to assume an+1≥an in order to say that an+2-an+1=-sin(c)(an+1-an) where c must be somewhere between 0 and 1, and |sin(c)|<0.85. How can I remove the assumption an+1≥an, and arrive at the same conclusion?

    I am not sure how to prove in our case that |an+2-an+1| ≤ 0.85|an+1-an|, which is needed in the proof.

    Does anybody have any idea?
     
    Last edited: Jan 21, 2010
  5. Jan 21, 2010 #4
    There are places you can slap absolute value signs in the statement of the mean value theorem to remove that little problem :)
     
  6. Jan 22, 2010 #5
    I think the claim can be proved if we assume either one of the following:
    (i) an+1≥an for all n, i.e. {an} nondecreasing
    or (ii) an+1≤an for all n, i.e. {an} nonincreasing

    But here in our case{a_n} is not even monotone. a0=0, a1=1, a2=0.54, a3=0.86,...
    So it belongs to neither case (i) nor case (ii). How should I prove the claim?

    Could somebody help, please?
    thanks!!!!
     
  7. Jan 22, 2010 #6
    Look again. You will find that you can apply the mean value theorem separately to each case arising from each index [tex]n[/tex]; therefore the direction of the inequality [tex]a_{n+1} \leq a_n[/tex] or [tex]a_{n+1} \geq a_n[/tex] doesn't have to be the same for all [tex]n[/tex].
     
  8. Jan 22, 2010 #7
    OK, you're right! I got it now :)
    Actually I got confused before about the "for ALL n" part in "a(n+1)>a(n) for ALL n" . I thought we need montone sequences, but acutally we don't.

    Given any two consecutive terms a(n) and a(n+1), either a(n)<=a(n_1) or a(n)>a(n+1), and in either case the MVT gives the result. Done.
     
  9. Jan 22, 2010 #8
    "What is the maximum of |sin(c)| in [0,1]?"

    Actually I used my calculator to get the bound 0.85. Is it possible to answer this question without using a calculator?? I don't think there is any way to tell what sin(1) is equal to...
     
  10. Jan 22, 2010 #9
    You don't need a precise value, only a proven bound less than 1. Try estimating it using the power series.
     
  11. Jan 24, 2010 #10
    I have a question about the end when we're trying to prove that the sequence is Cauchy.

    Case 1: If m≥n, then
    |am-an|≤|am-am-1|+...+|an+1-an|
    ≤0.85m-1|a1-a0|+...+0.85n|a1-a0|
    ≤0.85n/(1-0.85) ->0 as n->∞ (and m->∞).
    Thus, by squeeze theorem, |am-an|->0 as n,m->∞
    (here we are assuming that as n,m->infinity, it is in a manner such that m is always greater than or equal to n)

    Case 2: If m<n, then similarly, we can show that |am-an|->0 as n,m->∞.
    (here we are assuming that as n,m->infinity, it is in a manner such that m is always less than n)

    Is this enough to show that the sequence is Cauchy?
    If so, why? What if n,m->∞ while n and m keep on crossing over each other? (i.e. n,m->∞, but sometimes m≥n and sometimes m<n. How can we handle this case?)
    I don't understand why proving only case 1 and case 2 would be enough to show that the sequence is Cauchy.

    Can someone explain this, please?
    Thank you!
     
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