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Prove sequence converges

  1. Nov 1, 2009 #1
    1. The problem statement, all variablproves and given/known data
    X0=1, Xn+1=1/4 (1+2/Xn) for n=0,1,2,3,4....
    prove that this sequence converges ,and find the limit

    2. Relevant equations



    3. The attempt at a solution
    I though that Squeeze law might be used ,and let L=1/4(1+2/L) L=0.8437
    and the sequence is not monotonic
    so , how can i prove it next ?
     
  2. jcsd
  3. Nov 1, 2009 #2

    Mark44

    Staff: Mentor

    I would calculate a few terms in the series to see if I could come up with a non-recursive formula for xn.
     
  4. Nov 1, 2009 #3
    but how can i apply Squeeze law.
     
  5. Nov 1, 2009 #4

    Mark44

    Staff: Mentor

    First things first. Try to find a non-recursive formula for xn, and then worry about the technique to use to see if the sequence converges or not.
     
  6. Nov 1, 2009 #5

    jbunniii

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    Science Advisor
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    Gold Member

    [edit] oops, made a mistake, deleting it so I can fix it
     
  7. Nov 3, 2009 #6
    I'm interested how this is solved. Has anyone done it? If so, how.
     
  8. Nov 3, 2009 #7
    I've just been having a pop at this and decided to turn it into two sequences, one increasing, one decreasing, where:

    X_n+1 = 1/4 + 2/(1+2/X_n)
    With X_0 = 1 or 3/4

    Would it be sufficient to show that one or both of these are bounded and monotonic?
     
  9. Nov 4, 2009 #8

    jbunniii

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    I think I finally found a solution, but it's ugly and hopefully someone knows a much cleaner way.

    Let [itex]L[/itex] be the positive solution to

    [tex]L = \frac{1}{4}\left(1 + \frac{2}{L}\right)[/tex]

    or equivalently

    [tex]4L^2 - L - 2 = 0[/tex]

    Thus, [tex]L \approx0.84307[/tex].

    If [itex](x_n)[/itex] has a limit, then it must be [itex]L[/itex] because it's easy to see that [itex]x_n[/itex] is never negative.

    The hard part is proving that [itex](x_n)[/itex] has a limit. I shall do this by proving that the sequence

    [tex]y_n = x_n - L[/tex]

    has a limit of 0. To do this, I shall find a constant [itex]C < 1[/itex] such that

    [tex]\left|\frac{y_{n+1}}{y_n}\right| < C[/tex]

    for all [itex]n[/itex].

    First, I need a small lemma, namely:

    [tex]|y_n| < 0.2[/tex] for all n

    This will be proved inductively. Clearly it's true for [itex]n = 0[/itex], because

    [tex]|y_0| = |x_0 - L| \approx |1 - 0.84307| \approx 0.15693[/tex]

    Now suppose that [itex]|y_n| < 0.2[/itex]; we must show that this implies [itex]|y_{n+1}| < 0.2[/itex]. Indeed, [itex]|y_n| < 0.2[/itex] implies that [itex]L - 0.2 < x_n < L + 0.2[/itex]. Therefore,

    [tex]\frac{2}{L+0.2} < \frac{2}{x_n} < \frac{2}{L-0.2}[/tex]

    and this means

    [tex]\frac{1}{4}\left(1 + \frac{2}{L+0.2}\right) < \frac{1}{4}\left(1 + \frac{2}{x_n}\right) < \frac{1}{4}\left(1 + \frac{2}{L - 0.2}\right)[/tex]

    This is the same as

    [tex]0.72935 < x_{n+1} < 1.0275[/tex]

    and we also have [itex]L - 0.2 < 0.72935[/itex] and [itex]1.0275 < L + 0.2[/itex], so

    [tex]L - 0.2 < x_{n+1} < L + 0.2[/tex]

    or equivalently

    [tex]|y_{n+1}| < 0.2[/tex], concluding the proof of the lemma.

    For the rest of the proof, we substitute [itex]x_n = y_n + L[/itex] into the original recurrence relation, cross-multiply the denominator, and use the fact that [itex]4L^2 - L - 2 = 0[/itex] to simplify the result to

    [tex]4y_{n+1}y_n + 4L(y_n + y_{n+1}) = y_n[/tex]

    and solving for [itex]y_{n+1}[/itex] we get

    [tex]y_{n+1} = \frac{y_n(1 - 4L)}{4(y_n+L)}[/tex]

    Then, using the fact that

    [tex]\frac{1}{|y_n+L|} = \frac{1}{|x_n|} = \frac{1}{x_n} < \frac{1}{L - 0.2}[/tex]

    we have

    [tex]|y_{n+1}| = \frac{|y_n| \cdot |1 - 4L|}{4|y_n+L|} < \frac{|y_n| \cdot |1 - 4L|}{4|L - 0.2|} = |y_n| \cdot 0.92225[/tex]

    Thus

    [tex]\frac{|y_{n+1}|}{|y_n|} < C[/tex]

    where [itex]C = 0.92225[/itex], concluding the proof.

    It ain't pretty, but it gets the job done, unless I screwed up somewhere.

    By the way, there is nothing magic about the number 0.2. At some point I concluded that I needed a number less than 0.25 but big enough to handle the variation of the sequence about its limit, and I chose 0.2 after "cheating" by plotting the sequence in Matlab.
     
    Last edited: Nov 4, 2009
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