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Strictly decreasing sequence (analysis)

  1. Dec 22, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = (3 sin(x))/(2 + cos(x))

    let x_0 be a number in (0, (2/3)*pi], and define a sequence recursively by setting
    x_n+1 = f(x_n)

    (1) prove that the sequence {x_n} is strictly decreasing sequence in (0, (2/3)*pi] and that lim x_n =0

    (2) Find an integer k greater or equal to 1 such at {n^(1/k) x_n} is convergent to a finite, non-zero real number and evaluate lim n^(1/k) x_n

    2. Relevant equations



    3. The attempt at a solution

    I tried the first problem by induction but I don't know how to prove it for P(n+1) (assuming P(n) is true). Any help is appreciated thank you.
     
  2. jcsd
  3. Dec 22, 2009 #2
    Don't think about this as an inductive argument at the bottom, but as an inequality.

    Because your sequence is defined by iterating the function [tex]f[/tex], the simplest way to prove that [tex]x_{n+1} = f(x_n) < x_n[/tex] for all [tex]n[/tex] is to prove that [tex]f(x) < x[/tex] for all [tex]0 < x \leq \frac{2\pi}3[/tex].

    (You also need to show that [tex]f(x)[/tex] still lies in [tex]\left(0, \frac{2\pi}3\right][/tex]; otherwise you can't conclude [tex]f(x_{n+1}) < x_{n+1}[/tex], because you don't know that [tex]x_{n+1}[/tex] is in the domain to which your inequality applies. This part of the argument is an induction.)

    To show that [tex]x_n \to 0[/tex], you need a little more than that. When you prove the fact above, try to show not just that [tex]f(x) < x[/tex], but that [tex]f(x) < x[/tex] in some kind of systematic or predictable way. Think about examples of simple sequences that converge to zero, and try to establish a comparison with one of those.
     
    Last edited: Dec 22, 2009
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