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How to show a sequence converges

  1. Jun 21, 2010 #1
    Hey guys,

    I have a function y=(x+2)/(x+1) and I have performed iterations to show that for any initial value other than -sqrt{2} the sequence converges to sqrt{2}.

    So I have found that sqrt{2} is a stable fixed point and -sqrt{2} is an unstable fixed point.

    Now I have to prove my findings using a standard test for convergence?

    Can anyone help.
     
  2. jcsd
  3. Jun 21, 2010 #2
    hi there.

    a sufficient condition is that
    the derivative in the fixed point has to be smaller than one in absolute value
     
  4. Jun 21, 2010 #3

    Mark44

    Staff: Mentor

    How are you getting sqrt(2)? As x --> infinity, (x + 2)/(x + 1) --> 1.
     
  5. Jun 21, 2010 #4
    its a fixed point iteration scheme. this specific case you can derive from newtons method to find the zeros of x^2-2.
     
  6. Jun 21, 2010 #5

    Mark44

    Staff: Mentor

    But what do the zeroes of x^2 - 2 have to do with (x + 2)/(x - 1)?
     
  7. Jun 25, 2010 #6
    Hi all,

    I need help to show that the following sequence converges:

    n^n/ ((n + 3)^ (n + 1))

    I can't seem to find any standard limits which can help prove that it is convergent.

    now consider the limit as
     
  8. Jun 25, 2010 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Mark, the iteration is [itex]x_{n+1}= y(x_n)[/itex] or
    [tex]x_{n+1}= \frac{x_n+ 2}{x_n+1}[/itex]

    Assuming that has a limit, L, taking the limit on both sides gives
    [tex]L= \frac{L+2}{L+ 1}[/tex]
    and from that, [itex]L(L+1)= L^2+ L= L+ 2[/itex] so that [itex]L^2= 2[/itex].

    henry22, you should be able to show that
    1) If [itex]x_0> \sqrt{2}[/itex] then [itex]\{x_n\}[/itex] is a decreasing sequence with lower bound.

    2) If [itex]x_0< \sqrt{2}[/itex] then [itex]\{x_n\}[/itex] is an increasing sequence with upper bound.
     
  9. Jun 25, 2010 #8
    Wouldn't an initial value of -1 also be a problem? Also the sequence of starting values -3/2, -7/5, -17/12 ... which reach the value -1 at some stage of iteration.
     
    Last edited: Jun 25, 2010
  10. Jun 25, 2010 #9
    No, if [itex]x_0=-\sqrt{2}[/itex] then [itex]x_n=-\sqrt{2}[/itex]. Also if [itex]x_0=-5/4[/itex] then [itex]x_1=-3<x_0[/itex].

    Neither do positive terms behave as suggested. E.g. from the second example [itex]x_2=\frac{1}{2}[/itex], [itex]x_3=\frac{5}{3}>x_2[/itex], [itex]x_4=\frac{11}{8}<x_3[/itex] and [itex]x_5=\frac{27}{19}>x_4[/itex], so the sequence is neither monotonic increasing from [itex]x_2<\sqrt{2}[/itex] nor monotonic decreasing from [itex]x_3>\sqrt{2}[/itex].
     
    Last edited: Jun 25, 2010
  11. Jun 25, 2010 #10

    Mark44

    Staff: Mentor

    Thanks!
     
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