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Math Help - Real sequence

  1. #1
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    Real sequence

    Hello, I want some help writing a matlab script.
    x(n+1) = (x(n))^3-a*x(n) is my real sequence
    The script should read in the value of a and the value x1.
    I achieved to do that.
    It should then carry out a large number of iterations, without printing the results, to bypass the short-term, transient behaviour, and then print out on the screen a moderate number of further iterations.
    I don't know how to do this part.
    The script should also display the results in a graphical way. This I can do it after finding a way to do the above part.

    Any help is appreciated, thank you very much!
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  2. #2
    MHF Contributor chisigma's Avatar
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    The 'recursive relation' defining the sequence can be written as...

    \displaystyle \Delta_{n} = x_{n+1}-x_{n} = x_{n}^{3} - (1+a)\ x_{n}= f(x_{n}) (1)

    The behavior of the sequence depends from the 'initial value' x_{0} and from the parameter a. We have three possibilities...

    a) a<-1. In this case any x_{0}<0 will produce a sequence that tends to - \infty, any x_{0}>0 a sequence that tends to  + \infty and x_{0}=0 the 'all zeroes sequence'...

    b) -1 < a < 1. In this case any x_{0}<- \sqrt{1+a} will produce a sequence that tends to - \infty, any x_{0}> + \sqrt{1+a} a sequence that tends to  + \infty and any - \sqrt{1+a} < x_{0}< + \sqrt{1+a} a sequence that tends to 0...

    c) a>1. In this case any x_{0}<- \sqrt{1+a} will produce a sequence that tends to - \infty, any x_{0}> + \sqrt{1+a} a sequence that tends to  + \infty, any - \sqrt{1+a} < x_{0}< + \sqrt{1+a}, x_{0}\ne 0 an 'oscillating sequence' and x_{0}=0 the 'all zeroes sequence'...

    Kind regards

    \chi \sigma
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  3. #3
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    Thank you very much for your help! So there is no case that the sequence converges to a number and not only diverges? And do you know how is it possible to do this in a Matlab script using iterations? I firstly thought something like

    for i=1:200
    x(i+1)=x(i)^3-a*x(i);
    if abs(x(i+1)-x(i))<0.0000001 (Using Cauchy's convergence test)
    fprintf('Sequence converge to some number');
    break;
    else if (x(i+1)==inf) | (x(i+1)==-inf)
    fprintf('The sequence diverges);
    break;
    end;
    end;
    end;


    I have no other idea...Is it possible to help me a bit? Thanks again!! Appreciate it!
    Last edited by Darkprince; March 15th 2011 at 04:11 PM. Reason: Used j by mistake instead of i
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