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Why matlab giving me same values too many times

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Sam
Sam el 15 de Mzo. de 2014
Comentada: Walter Roberson el 16 de Mzo. de 2014
Ok my input function was
r=2;
x(1)=0.3;
N=100
for i=1:N
x(i+1)=r*x(i)*(1-x(i))
end
And, matlab calculated my answer but gave me the same x values for about 6 times. It's very annoying since I don't need the same x values over and over again. Please tell me how can I avoid that? Thanks.
  2 comentarios
Azzi Abdelmalek
Azzi Abdelmalek el 15 de Mzo. de 2014
What are you expecting?
Sam
Sam el 15 de Mzo. de 2014
I am expecting to avoid the same values over and over again.

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Mischa Kim
Mischa Kim el 15 de Mzo. de 2014
Editada: Mischa Kim el 15 de Mzo. de 2014
Sumon, I think you are referring to fact that the series approaches x = 0.5 for increasing i. This is absolutely to be expected. For small x the linear term dominates which leads to an increase in x. However, as x is increasing the quadratic term starts to become more prevalent up to a point, where they will balance each other out. This point is "reached" when
x(i) = r*x(i)*(1-x(i))
or, put aside the indices,
x = r*x*(1 - x)
1 = r* (1 - x)
or, solving for x
x = (r - 1)/r = 0.5
At that point, the values for x(i) remain constant, x(i+1) = x(i).
  2 comentarios
Sam
Sam el 15 de Mzo. de 2014
Oh I get it now. Thanks for clearing things up for me.
Walter Roberson
Walter Roberson el 16 de Mzo. de 2014
In theory you only get x = 1/2 if your start with x = 1/2 . But in practice when you start with x in the range 0 to 1 exclusive then within a few steps the loss of precision will get you to floating point 0.5 after which it will stay there.

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Más respuestas (1)

Walter Roberson
Walter Roberson el 15 de Mzo. de 2014
MATLAB uses finite precision floating point binary numbers. After a few steps of the calculation, all of the answers are the same to within the limited precision. If you require increased precision then you will need to use something like the symbolic toolbox.
x(t) = -1/2*((2/5)^(2^(t-1))-1)

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