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    Let x(n+1)= (3*x(n)) / (x(n))^2+2
    n belongs in N, xo>=0 is some starting value

    a) Show that if the sequence converges to a finite limit then this is 0 or 1.
    b) Show that if 0<x(n)<1 then 0<x(n)<x(n+1)<1 and determine the limit behavior of the sequence when 0<x0<1.
    c) Show that if 1<x(n)<2 then 1<x(n+1)<x(n) and determine the limit behavior of the sequence when 1<x0<2
    d) What happens for x0>=2?


    My work and if someone can check it and correct me if I am anywhere wrong.

    a) If x(n) tends to K in R then so does x(n+1) so K=3K/K^2+2 so K=0 or K=1 or K=-1 but we cross out K=-1 since x0>=0 and so ...>x2>x1>x0 (x0>0) or all x(n)=0 (x0=0)

    b) x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) which is >0 if 0<x(n)<1 and so x(n+1)>x(n). So the sequence is an increasing one with 0<x(n)<x(n+1)<1, since the only positive finite limit is 1.

    When 0<x0<1 our sequence is increasing and so 1>...>x2>x1>x0>0. It is bounded above by 1 since is the only positive finite limit.

    c) If 1<x(n)<2 then x(n+1)-x(n)<0 so x(n)>x(n+1) so our sequence is decreasing but >1 since is the only positive finite limit and so 1<x(n+1)<x(n)

    When 1<x0<2 then 1<.....<x2<x1<x0 our sequence is decreasing and bounded below by 1.

    d) If x0=2 then all x(n) are zero so we have a sequence full of zeros.
    If xo>2 we observe that x1<x0 but 1>...>x3>x2>x1 and our sequence is bounded above by 1 since is the only possible finite limit.


    This is my work. Could anyone please check it and confirm my work? Thanks in advance, appreciate it!!
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    No one?
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    Note: I'm somewhat rusty on what's acceptable.

    That doesn't seem very rigorous, with chunks missing. Only one error I can see; bar that, as a sketch it would be fine, but....


    (Original post by Darkprince)

    My work and if someone can check it and correct me if I am anywhere wrong.

    a) If x(n) tends to K in R then so does x(n+1) so K=3K/K^2+2 so K=0 or K=1 or K=-1 but we cross out K=-1 since x0>=0 and so ...>x2>x1>x0 (x0>0) or all x(n)=0 (x0=0)
    Although the full details are not there, it's clear you've worked them out.

    b) x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) which is >0 if 0<x(n)<1 and so x(n+1)>x(n). So the sequence is an increasing one with 0<x(n)<x(n+1)<1, since the only positive finite limit is 1.

    When 0<x0<1 our sequence is increasing and so 1>...>x2>x1>x0>0. It is bounded above by 1 since is the only positive finite limit.

    I'd prefer better justification on x(n+1)<1, as you haven't shown that it goes to a finite limit at this stage.

    c) If 1<x(n)<2 then x(n+1)-x(n)<0 so x(n)>x(n+1) so our sequence is decreasing but >1 since is the only positive finite limit and so 1<x(n+1)<x(n)

    When 1<x0<2 then 1<.....<x2<x1<x0 our sequence is decreasing and bounded below by 1.
    As previous comment but for x(n+1) > 1

    d) If x0=2 then all x(n) are zero so we have a sequence full of zeros.
    Sorry, "nil point" on that one.

    And that's the error that would make me think you've not worked through all the details, and were being a bit sloppy.

    If xo>2 we observe that x1<x0 but 1>...>x3>x2>x1 and our sequence is bounded above by 1 since is the only possible finite limit.
    NOTE: See my opening comment.
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    How should I explain more that the sequence converges to that limit? Thanks for the answer!!! Appreciate it!!
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    You need to show that if x(n) < 1 then x(n+1) is also less than 1. (And similarly for the > 1 case).
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    We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help!
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    (Original post by Darkprince)
    We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help!
    You can argue the wording, but my interpretation would be that for (b) you can't assume that the sequence converges to a finite limit. That is something you need to show. (With some actual calculation).
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    (Original post by Darkprince)
    We know from part a that the only possible finite limits is 1 and 0. So if x(n)>0, the series converges to 1. if x(n)=0 it converges to zero so for my two cases x0 is larger than zero, so the sequence converges to 1. Isn't this enough? What should I say more? Thanks in advance, appreciate all the help!
    The operative part of (a) is IF THE SEQUENCE CONVERGES. At no point have you shown or been told that the sequence does converge.

    Other than that, note DFranklin's comments; he's more one the ball with the methodology etc. than I am (I haven't used this stuff in over 30 years!).
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    Thank you guys, what would be the actual calculation in my case?
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    (Original post by Darkprince)
    Thank you guys, what would be the actual calculation in my case?
    You have the basic formula for x(n+1) in terms of x(n). Have a go!
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    Yes i understand but when doing some calculations then it is obvious that the sequence converges! You meant numerical calculations?
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    (Original post by Darkprince)
    Yes i understand but when doing some calculations then it is obvious that the sequence converges! You meant numerical calculations?
    Algebraic, rather than numerical. What you need to show is that if x(n) < 1, then using the recurrance formula you've been given, then x(n+1) < 1.
    This together with the fact that you've already proved that x(n+1) > x(n) is what shows convergence.

    You're not interested in a specific numerical value as such.
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    Oh, so I just have to say that if x(n)<1 then x(n+1)<(3*1)/(1^2+2)<1!
    That's correct since I used the recurrance formula, right?
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    (Original post by Darkprince)
    Oh, so I just have to say that if x(n)<1 then x(n+1)<(3*1)/(1^2+2)<1!
    That's correct since I used the recurrance formula, right?
    Not quite. Just substituting "1" for x(n) into the formula for x(n+1) is only going to work if you know the maximum value of 3x/(x^2+1) on the interval (0,1) is less than one, and in fact on the closed interval it occurs at x=1 with a value of 1, but you've not shown that.

    Since this is dragging on, here's my thinking:

    x(n+1) = 3x(n)/[(x(n))^2+2] = 1 - [(x(n))^2-3x(n)+2]/[(x(n))^2+2]

    = 1 - [(x(n))-2)(x(n)-1)]/[(x(n))^2+2]

    Then in the interval (0,1) your fraction works out as postive, and since it's being subtracted the whole thing is less than 1.
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    Thank you very much! Appreciate it!
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    Ghostwalker sorry for bothering again! I transformed my recursive relation to the one you guided me to do but as you said before it is 1- something positive! Shouldn't I also prove that that something positive is less than one? So 1 - (something smaller than one) so the whole thing would be larger than one??
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    If I want to prove something is smaller than 1, then showing it is 1 - "something positive" is *exactly* what I need to do...
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    Yes correct! I don't know what was I thinking! I am gonna have a final look at my solution and if anything appears I'll let you know. Thanks for all the help, heavily appreciate it!!
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    Ok just reviewed my solution. I was able to prove that if 0<x(n)<1 then 0<x(n)<x(n+1)<1 and that if 1<x(n)<2 then 1<x(n+1)<x(n).

    Then for determining the limit behavior of the sequence when 0<x(0)<1 I said that we have an increasing sequence bounded above by 1 (since 0<x(n)<x(n+1)<1 ), so the series converges to 1 by Monotone Sequence Theorem and 1 is the only possible finite limit for this case.

    For determining the limit behavior of the sequence when 1<x(0)<2 I said that we have a decreasing sequence bounded below by 1 ( since 1<x(n+1)<x(n)) so it converges to 1 by Monotone Sequence Theorem and 1 is the only possible finite limit for this case.

    This sounds better and more mathematically correct? Do you have anything to add? Thanks again for all the help!!
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    You seem to be saying that if x_n is monotonically increasing and bounded above by 1, then x_n must tend to 1. This isn't true. (e.g. x_n = -1/n). All you know is that x_n tends to a limit. You then need to argue (using the first part) that that limit must be 1.
 
 
 
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