# Logistic map and real numbers

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Hi,

here is my question : does the logistic sequence for some choosen irrational parameter reach every real number inside a real interval, or is it always just a subset ?

(I hope i'm in the right section)

thanks !

Edited by Edgard Neuman

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nobody answered that question, but I think it's important.. if logistic map sequence never repeat and still stay in the interval [0 ; 1], and if we can prove than any x in [0;1] eventually is in it,  then  it's a way to build a bijection between N and [0;1]..(that wouldn't surprise me, but you probably)
I suppose some real in [0:1] are not in the sequence, but how can we build these numbers ?

Edited by Edgard Neuman

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Since the numbers in the interval are uncountable, no (countable) sequence can generate all the numbers in the interval.

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16 hours ago, Edgard Neuman said:

how can we build these numbers ?

What do you mean by "build these numbers"?

For any sequence we can easily construct an element of the interval $$[0,1]$$ that is not in the given sequence. This is standard, and I can guess that you most likely know the answer already.

Are you asking whether it is somehow possible to construct all the elements of $$[0,1]$$ that are not in the sequence?

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4 hours ago, taeto said:

What do you mean by "build these numbers"?

For any sequence we can easily construct an element of the interval [0,1] that is not in the given sequence. This is standard, and I can guess that you most likely know the answer already.

I don't 😩

For instance, with :

x[n+1]=r * x[n] * (1-x[n])

With r = sqrt(2)*2.6  for instance

X[n] is always in [0 ; 1]
How do you find y in [0 ; 1 ] that is not in  { x[n] } ?

Edited by Edgard Neuman

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On 4/16/2019 at 6:29 PM, Edgard Neuman said:

here is my question : does the logistic sequence for some choosen irrational parameter reach every real number inside a real interval, or is it always just a subset ?

Why would you expect the sequence function to converge to other irrational numbers than your parameter?

Irrational numbers are not, in general algebraic expressions of other irrational numbers.

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58 minutes ago, studiot said:

Why would you expect the sequence function to converge to other irrational numbers than your parameter?

Irrational numbers are not, in general algebraic expressions of other irrational numbers.

in that case you have products of x * (1 - x) that are all irrational
but I understand that you don't reach the sequence starting from another irrational (unless the expression if infinite)

Ok so no bijection !

So you have a family of number (that are irrational) for each irrational and the expressions using them (like  sqrt(2)* p /q  for instance are all irrational of the family of sqrt(2)).
That's the object my other question :

Thanks a lot.

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1 hour ago, Edgard Neuman said:

For instance, with :

x[n+1]=r * x[n] * (1-x[n])

With r = sqrt(2)*2.6  for instance

X[n] is always in [0 ; 1]
How do you find y in [0 ; 1 ] that is not in  { x[n] } ?

Okay then, I will attempt to give the standard answer. I assume $$x(0) \neq 0$$ and $$x(0)\neq 1$$ as otherwise the answer is trivial, say with taking  $$y= 1/2.$$

The key is to realize that every real number in $$[0; 1]$$ has a representation, a "name" if you will, as a binary string preceded by a zero and a "decimal" point, such as the representation of the fraction $$3/4$$ as the string $$0.11,$$ meaning $$1/2$$ for the first digit after the decimal point added to $$1/4$$ for the second digit. The fraction $$1/3$$ has a representation which looks like $$0.10101010\ldots$$ etc., hence the representation is not of finite length for every number.

You say that your $$x(n)$$ is always in $$[0,1],$$ for every $$n.$$  In which case it has a representation as described.

Now we construct a number $$y$$ in $$[0,1]$$ that has a representation which is different from the representation of any one of the numbers $$x(n).$$

For each $$n = 1,2,3,\ldots$$ let the $$n$$'th binary digit of $$y$$ be the one that is different from the $$n$$'th binary digit of your $$x(n).$$

Then the $$y$$ just constructed cannot be equal to any one of the numbers $$x(n),$$ since in the $$n$$'th digit it is different.

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17 minutes ago, taeto said:

Okay then, I will attempt to give the standard answer. I assume x(0)0 and x(0)1 as otherwise the answer is trivial, say with taking  y=1/2.

The key is to realize that every real number in [0;1] has a representation, a "name" if you will, as a binary string preceded by a zero and a "decimal" point, such as the representation of the fraction 3/4 as the string 0.11, meaning 1/2 for the first digit after the decimal point added to 1/4 for the second digit. The fraction 1/3 has a representation which looks like 0.10101010 etc., hence the representation is not of finite length for every number.

You say that your x(n) is always in [0,1], for every n.   In which case it has a representation as described.

Now we construct a number y in [0,1] that has a representation which is different from the representation of any one of the numbers x(n).

For each n=1,2,3, let the n 'th binary digit of y be the one that is different from the n 'th binary digit of your x(n).

Then the y just constructed cannot be equal to any one of the numbers x(n), since in the n 'th digit it is different.

thanks a lot !

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