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Hard calculus proofs!

  1. Feb 24, 2010 #1
    I've got two calculus proofs that I can't seem to get! I was wondering if you guys could help me out a bit...

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

    Suppose [tex]x_{n}[/tex] is the sequence defined recursively by

    [tex]x_{1}=0[/tex] and [tex]x_{n+1}=\sqrt{5 + 2x_{n}}[/tex] for [tex]n=0, 1, 2, 3, ....[/tex]

    Prove that [tex]x_{n}[/tex] converges and find its limit


    The attempt at a solution

    So far, by taking the limit of both sides "[tex] l [/tex]", I've found the limit

    [tex]l=\sqrt{5 + 2l}[/tex]
    [tex]l=\frac{-2 +\sqrt{24}}{2}[/tex]

    but I still have to prove that the sequence converges. I've tried induction hypothesis using

    [tex] x_{n} \leq x_{n+1} \leq \frac{-2 +\sqrt{24}}{2}[/tex]

    and then I attempt to modify [tex]x_{n}[/tex] and [tex]x_{n+1}[/tex] such that the equation becomes (in the end):

    [tex] x_{n+1} \leq x_{n+2} \leq \frac{-2 +\sqrt{24}}{2}[/tex]

    but I can never get it to work out!


    2.
    The problem statement, all variables and given/known data

    The second proof I'm having trouble with is basically one of the same type...

    Let [tex]x_{n}[/tex] be the sequence of real numbers defined recursively by

    [tex]x_{0} = 0 [/tex] and [tex] x_{n+1} = \ln (2 + x_{n}) [/tex] for [tex] n=0, 1, 2, 3,..... [/tex]

    Show that [tex] x_{n} \leq x_{n+1} \leq 2 [/tex]

    The attempt at a solution

    Again, I try to modify it such that i get
    [tex] \ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq 2 [/tex]

    but I keep ending up with
    [tex] \ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq \ln (2 + 2) [/tex]



    Is there a trick to calculus proofs?

    Thanks!
    Steve
     
  2. jcsd
  3. Feb 25, 2010 #2
    In my memory,to prove such thing with induction,you need an induction hypothesis that xn is not only bounded above,but also bounded below

    Add a proper lower bound,and try mathematical induction again
     
  4. Feb 25, 2010 #3

    Mark44

    Staff: Mentor

    You have a sign error in your calculation of L.
    Starting with
    [tex]L=\sqrt{5 + 2L}[/tex]
    [tex]\Rightarrow L^2=5 + 2L[/tex]
    [tex]\Rightarrow L^2 - 2L - 5 = 0[/tex]

    By the Quadratic Formula, this gives
    [tex]L = \frac{2 \pm \sqrt{4 - (-20)}}{2} = 1 + \sqrt{6} \approx 3.4495[/tex]
    The other solution, 1 - sqrt(6), is negative, and can be discarded. The recursive relation won't give negative values.

    Since [tex]L = lim_{n \to \infty} x_{n + 1}[/tex]
    this shows that the sequence converges, and gives its limit.
     
  5. Feb 25, 2010 #4
    I figured them out! Thanks Mark!

    Is a lower bound really necessary? I assumed not, because it will obviously be larger than [tex]x_{1}=0[/tex] since the sequence is increasing
     
  6. Feb 25, 2010 #5
    It is clear that it is bounded below. You can also consider monotone convergence theorem (although, with Mark's remarks, the proof is correct as it is).
     
  7. Feb 25, 2010 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't think so. What it shows is that if the sequence converges, then that value of L is its limit. But you must still argue that it converges. For consider x1 = 1 and xn+1 = 2xn. That argument applied to this would give L = 2L which would imply the limit exists and is 0.
     
  8. Feb 25, 2010 #7

    statdad

    User Avatar
    Homework Helper

    LCKurtz is correct. You've shown what has to happen if it converges; you still must show that it does converge. You can do this by showing
    1) That the sequence is increasing
    2) That the sequence is bounded above (doesn't matter what number you choose, just show that it is bounded above)

    For 1: you already know [tex]x_0 = 0 < x_1 = \sqrt{5 + 2 x_0} = \sqrt{5} [/tex]

    Now use induction: assume [tex] x_k < x_{k+1} [/tex] and show [tex] x_{k+1} < x_{k+2} [/tex]. (Hint: look at simplifying [tex] x_{k+2}^2 - x_{k+1}^2 [/tex]

    For 2: try to show something like [tex] x_n < 5 [/tex] for all n.
     
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