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Homework Help: Question from Rudin

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the positive integer p and any positive number [tex]\alpha[/tex] and the formula:

    [tex]x_{n+1}=\frac{p-1}{p}x_n+\frac{\alpha}{p}x_n^{-p+1}[/tex]

    and describe the behavior of the resulting sequences [tex]\{x_n\}[/tex]

    2. Relevant equations

    In the problem statement.

    3. The attempt at a solution

    So in the earlier problem of this chapter where p=2. We first choose [tex]x_1>\sqrt{\alpha}[/tex] and then show that the resulting sequence is monotonically decreasing and is bounded below by [tex]\sqrt{\alpha}[/tex]. Hence a limit exists and letting [tex]x_{n+1}[/tex] and [tex]x_n[/tex] be equal to that limit and solving we get the limit of the sequence.

    Using a similar strategy I wanted to begin with [tex]x_1>\alpha^{\frac{1}{p}}[/tex] and show that the resulting sequence is monotonically decreasing and bounded below by [tex]\alpha^{\frac{1}{p}}[/tex] and then show that the sequence converges to [tex]\alpha^{\frac{1}{p}}[/tex].

    I have proved that [tex]x_{n+1}<x_n[/tex] iff [tex]x_n>\alpha^{\frac{1}{p}}[/tex] so proving inductively that [tex]x_n>\alpha^{\frac{1}{p}}[/tex] for all n should do the job. But I can't prove that. This is where I am stuck.

    I have used excel to plot the sequence for various values of [tex]\alpha[/tex] and p and each time the sequence monotonically decreases and converges to [tex]\alpha^{\frac{1}{p}[/tex] if we choose [tex]x_1>\alpha^{\frac{1}{p}}[/tex].

    I'm just stuck at the point I mentioned earlier.

    Any help would be appreciated. Thank you.
     
  2. jcsd
  3. Jun 24, 2010 #2
    You're close, so I'll just give you a nudge. First, rearrange

    [tex]\frac{p-1}{p}x_n+\frac{\alpha}{p}x_n^{-p+1} =
    \frac{x_n}{p}\Big((p-1)+\alpha x_n^{-p}\Big)[/tex]

    Now, use the assumption [tex]x_n> \alpha^{1/p}[/tex] to conclude that [tex]\alpha x_n^{-p} < 1.[/tex]

    Then, fill this back in the the above. Is that enough?

    By the way, there are 3 Rudin books, and people predominately refer to each a just "Rudin".
     
    Last edited: Jun 24, 2010
  4. Jun 24, 2010 #3
    That only helps me show that [tex]x_{n+1}<x_n[/tex] if and only if [tex]x_n>\alpha^{\frac{1}{p}}[/tex].

    It follows that to show that the sequence is monotonically decreasing I need to prove that [tex]x_n>\alpha^{\frac{1}{P}}[/tex] for all n if we start with [tex]x_1>\alpha^{\frac{1}{p}}[/tex].

    How do I prove that?
     
  5. Jun 24, 2010 #4
    First, you're welcome...

    If [tex] x_{n+1} < x_{n},\;\forall n\in\mathbb{N},[/tex] then the sequence is called monotonically decreasing. What you are looking for is called bounded (from below). If it monotonically decreasing and bounded from below, then you can conclude that it converges.

    You have to do something else to prove that it converges to something in particular. You need to show directly that they converge to [tex]\alpha^{1/p},[/tex] not just show they are bounded by it.

    What have you tried?
     
  6. Jun 24, 2010 #5
    Thanks! :)

    (btw this is from baby rudin chapter 3).

    So If I show that it is bounded from below and is monotonically decreasing then it follows from a theorem in the book that the sequence converges, just as you said.

    Once we know that it converges, and we know that the limit is greater than 0 because the greatest lower bound (which will be the limit) has to be greater than or equal to [tex]\alpha^\=frac{1}{p}[/tex], using a theorem in the same chapter, and letting [tex]x_n\to x[/tex]:

    [tex]x = \frac{p-1}{p}x + \frac{\alpha}{p}x^{1-p}[/tex]

    We know that the limits on the right hand side exist because this follows from another theorem in the same chapter.

    Solving this equation gives us [tex]x = \alpha^\frac{1}{p}[/tex]

    So showing that it converges will also help us show to what it converges.

    As I outlined in my initial attempt, i'm trying to show that it converges by showing that it is a monotonically decreasing sequence that is bounded below. In particular, proving [tex]\forall n\,(\,x_n>\alpha^\frac{1}{p}\,)[/tex] would prove this because [tex]x_{n+1}<x_n[/tex] if and only if [tex]x_n>\alpha^\frac{1}{p}[/tex].

    So the point where i'm stuck is proving: [tex]\forall n\,(\,x_n>\alpha^\frac{1}{p}\,)[/tex] if [tex]x_1>\alpha^\frac{1}{p}[/tex]
     
  7. Jun 24, 2010 #6
    This is kind of old school, but you could consider an old calculus trick, the derivative. The function
    [tex]
    f(x) = \frac{p-1}{p}x + \frac{\alpha}{p}x^{1-p}
    [/tex]
    is increasing for [tex]x > \alpha^{1/p}[/tex] and [tex] f(\alpha^{1/p})=\alpha^{1/p}[/tex]

    That's probably simplest. Hack away at that crappy inequality if this doesn't satisfy you, but that's what I would do.

    By the way, the [tex]0< x < \alpha^{1/p}[/tex] case is handled identically by this approach.
     
  8. Jun 24, 2010 #7
    Yea using calculus, we can can show that [tex]\alpha^\frac{1}{p}[/tex] is the minimum point of this function in the positive domain and since this function is positive for [tex]x>0[/tex], [tex]\forall n\,(\,x_n\geq\alpha^\frac{1}{p}\,)[/tex] and the proof is complete.

    I was just wondering whether we can do it without calculus. Rudin talks about a theorem for cauchy products of series in chapter 3 where he delays the proof until he introduces calculus. Maybe that is why he just asked for a discussion in the question statement and not a proof since calculus is needed for this question.

    Thanks a lot for your help! much appreciated! :)
     
    Last edited: Jun 24, 2010
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