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Math Help - Sequence behavior

  1. #1
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    Sequence behavior

    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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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Darkprince View Post
    Let x(n+1)= (3*x(n)) / (x(n))^2+2 n belongs in N, xo>=0 is some starting value...
    There is an essential detail: the recursive relation is...

    a) \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} +2}

    b) \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} +2n}

    c) somewhat else...

    Kind regards

    \chi \sigma
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  3. #3
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    What do you mean? My approach is completely wrong? Why in b the recursive relation is 3xn/(x(n)^2+2n)?
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  4. #4
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    Oh sorry, now I understand. I meant n belongs in N in my first sentence. So the sequence is just x(n+1)= (3*x(n)) / (x(n))^2+2
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Darkprince View Post
    Oh sorry, now I understand. I meant n belongs in N in my first sentence. So the sequence is just x(n+1)= (3*x(n)) / (x(n))^2+2
    ... so the recursive relation is...

    \displaystyle x_{n+1}= \frac{3\ x_{n}}{x^{2}_{n} + 2} (1)

    ... and that can be written as...

    \displaystyle \Delta_{n}= x_{n+1}-x_{n} = \frac{x_{n}- x^{3}_{n}}{x^{2}_{n}+2}= f(x_{n}) (2)

    The function f(x) is represented here...



    If we define 'attractive fixed point' a value of x for which f(x) crosses the x axis with negative slope, then f(x) has two attractive fixed points in x= \pm 1. Because f(x) is an 'odd function' we can analyse what does happen for x_{0}>0. The point x=0 isn't an attractive fixed point and the initial value x_{0}=0 will produce the 'all zero sequence'. Setting x^{*}=1 the positive attractive fixed point, any x_{0} which satisfies the condition |f(x_{0})|< |x_{0}-x^{*}| [the 'red line' in the figure...] will produce a sequence monotonically convergent at x=1. If x_{0}>2 the first iteration will produce an x_{1} between 0 and 1 and after that the sequence converges monotonically at x=1...

    Kind regards

    \chi \sigma
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  6. #6
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    Thank you very much for the help
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  7. #7
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    Sorry Chi Sigma for bothering again. I understood your logic. My solution was correct? I mean I prefer to work on my approach, so if you can tell me if it was correct or if I have to add anything I would greatly appreciate it! Thank you!
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Darkprince View Post
    Sorry Chi Sigma for bothering again. I understood your logic. My solution was correct? I mean I prefer to work on my approach, so if you can tell me if it was correct or if I have to add anything I would greatly appreciate it! Thank you!
    Your work was correct!... a minor detail is in the point a)...if the sequence converges to a finite limit then this is 0 or 1...

    ... more exact is to say that if x_{0}>0 the sequence converges to 1, if x_{0}<0 the sequence converges to -1 and if x_{0}=0 the sequence is 'all zeroes'...

    Kind regards

    \chi \sigma
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  9. #9
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    Thank you very much Chi Sigma, appreciate all your help
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  10. #10
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    Hello Chi Sigma, sorry for bothering again. I want to ask if there is anything more to add to my answers for part b and c so I can deduce that my series is converging in that cases. I mean the pattern is obvious but is there something to add up to my answers to explain them better mathematically?
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