Why matlab giving me same values too many times

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Sam
Sam am 15 Mär. 2014
Kommentiert: Walter Roberson am 16 Mär. 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 Kommentare
Azzi Abdelmalek
Azzi Abdelmalek am 15 Mär. 2014
What are you expecting?
Sam
Sam am 15 Mär. 2014
I am expecting to avoid the same values over and over again.

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Akzeptierte Antwort

Mischa Kim
Mischa Kim am 15 Mär. 2014
Bearbeitet: Mischa Kim am 15 Mär. 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 Kommentare
Sam
Sam am 15 Mär. 2014
Oh I get it now. Thanks for clearing things up for me.
Walter Roberson
Walter Roberson am 16 Mär. 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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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 15 Mär. 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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