Proving two subsequences have different limits: x0∈[0,1] and xn+1=1−x2n

[pisrationalhahaha]

\(\displaystyle (x)_n\) is a sequence defined by \(\displaystyle x_0\in [0,1]\) and \(\displaystyle x_{n+1}=1-x_n^2\)
Already proved that \(\displaystyle \forall n\in \mathbb{N},x_n\in [0,1]\) (first part of the exercise)
The second part they asked to show that the two sequences \(\displaystyle (x_{2n})\) and \(\displaystyle (x_{2n+1})\) converge to different limits
The problem is that I never used to find limits of recurrent sequences whose first term is defined by an interval instead of a value
I'm stuck at this point never knowing what to do
I tried to study the function f(x)=1-x^2 which is decreasing on [0,1] and see if f(f(\(\displaystyle (x_{2n})\))) is increasing or not....
Any hints ?

 

[Dr.Peterson]

\(\displaystyle (x)_n\) is a sequence defined by \(\displaystyle x_0\in [0,1]\) and \(\displaystyle x_{n+1}=1-x_n^2\)
Already proved that \(\displaystyle \forall n\in \mathbb{N},x_n\in [0,1]\) (first part of the exercise)
The second part they asked to show that the two sequences \(\displaystyle (x_{2n})\) and \(\displaystyle (x_{2n+1})\) converge to different limits
The problem is that I never used to find limits of recurrent sequences whose first term is defined by an interval instead of a value
I'm stuck at this point never knowing what to do
I tried to study the function f(x)=1-x^2 which is decreasing on [0,1] and see if f(f(\(\displaystyle (x_{2n})\))) is increasing or not....
Any hints ?

I would start by just trying to get familiar with the problem. The idea is that x₀ can be anything in 0<=x<=1, and you create a recursive sequence from it. Try a few different starting numbers, such as 0, .1, .5, .9, 1 and see if it is true that the even terms approach one limit and the odd terms another. (I did that with a quick spreadsheet, and it quickly became clear - especially starting with 0 or 1!)

Having convinced myself that the claim is right, and seen what it means in specific cases, I would then probably consider x_n+2 as a function of x_n, and think about what that transformation preserves -- that is, what will be common to all even, or all odd, terms. Your suggestion (is it an increasing function) may be exactly what you need.

Main idea here: when you have no idea what to do, "play" (i.e. experiment). Then go with whatever you discovered.

 

[Steven G]

\(\displaystyle (x)_n\) is a sequence defined by \(\displaystyle x_0\in [0,1]\) and \(\displaystyle x_{n+1}=1-x_n^2\)
Already proved that \(\displaystyle \forall n\in \mathbb{N},x_n\in [0,1]\) (first part of the exercise)
The second part they asked to show that the two sequences \(\displaystyle (x_{2n})\) and \(\displaystyle (x_{2n+1})\) converge to different limits
The problem is that I never used to find limits of recurrent sequences whose first term is defined by an interval instead of a value
I'm stuck at this point never knowing what to do
I tried to study the function f(x)=1-x^2 which is decreasing on [0,1] and see if f(f(\(\displaystyle (x_{2n})\))) is increasing or not....
Any hints ?

You should certainly play around with real number in [0,1] to see what is going on. After you are satisfied as to what is going on you do your proof by letting x₀ = a, where \(\displaystyle a\in [0,1]\) and go from there.

 

[pisrationalhahaha]

I would start by just trying to get familiar with the problem. The idea is that x₀ can be anything in 0<=x<=1, and you create a recursive sequence from it. Try a few different starting numbers, such as 0, .1, .5, .9, 1 and see if it is true that the even terms approach one limit and the odd terms another. (I did that with a quick spreadsheet, and it quickly became clear - especially starting with 0 or 1!)

Having convinced myself that the claim is right, and seen what it means in specific cases, I would then probably consider x_n+2 as a function of x_n, and think about what that transformation preserves -- that is, what will be common to all even, or all odd, terms. Your suggestion (is it an increasing function) may be exactly what you need.

Main idea here: when you have no idea what to do, "play" (i.e. experiment). Then go with whatever you discovered.

\(\displaystyle x_{n+2}=1-x_{n+1}^2=1-(1-x_n^2)^2=(-x_n^2+2)(x_n^2)\)
Then what ?
We take two cases ? (If n is odd or even)
I can't yet understad how these two subsequences have different limits
Why ?

 

[Dr.Peterson]

\(\displaystyle x_{n+2}=1-x_{n+1}^2=1-(1-x_n^2)^2=(-x_n^2+2)(x_n^2)\)
Then what ?
We take two cases ? (If n is odd or even)
I can't yet understand how these two subsequences have different limits
Why ?

What did you learn from experimentation? You should have at least a guess.

The important thing is that this recurrence produces either the sequence of odd terms or the sequence of even terms. Whether n is odd or even in itself has no effect; it will be a question of what kind of number you start with.

One thing you might do is to find out when x_n+2 = x_n. (You may at first think this can't be solved, but it can.) Then you can consider whether this subsequence is increasing or decreasing.

 

[Steven G]

\(\displaystyle x_{n+2}=1-x_{n+1}^2=1-(1-x_n^2)^2=(-x_n^2+2)(x_n^2)\)
Then what ?
We take two cases ? (If n is odd or even)
I can't yet understad how these two subsequences have different limits
Why ?

Please let x₀ = 1 and then tell us the first 10 terms. Please do the same with x₀ = 0. Then try it with x₀ = .1 (just to the first 5 terms). We just want to make sure you see what is going on!

 

[pisrationalhahaha]

Please let x₀ = 1 and then tell us the first 10 terms. Please do the same with x₀ = 0. Then try it with x₀ = .1 (just to the first 5 terms). We just want to make sure you see what is going on!

Okay it's much clearer right now
If \(\displaystyle x_0=1\) then \(\displaystyle x_{2n}=1\) and \(\displaystyle x_{2n+1}=0\)
If \(\displaystyle x_{0}=0\) then \(\displaystyle x_{2n}=0\) and \(\displaystyle x_{2n+1}=1\)
So now it's proved that they approach different limits by counterexample, right ?

 

[Dr.Peterson]

Okay it's much clearer right now
If \(\displaystyle x_0=1\) then \(\displaystyle x_{2n}=1\) and \(\displaystyle x_{2n+1}=0\)
If \(\displaystyle x_{0}=0\) then \(\displaystyle x_{2n}=0\) and \(\displaystyle x_{2n+1}=1\)
So now it's proved that they approach different limits by counterexample, right ?

This is only an example, not a proof of anything (except that the sequence does not always have a single limit).

You still need to show that, for any initial value, each of the two sequences has a limit.

Have you tried other starting values?

 

[pisrationalhahaha]

This is only an example, not a proof of anything (except that the sequence does not always have a single limit).

You still need to show that, for any initial value, each of the two sequences has a limit.

Have you tried other starting values?

Yes I tried for 0.5 and 0.1
It's hard for me to prove it for any initial value

 

[Dr.Peterson]

Yes I tried for 0.5 and 0.1
It's hard for me to prove it for any initial value

You've been given some suggestions that, as far as I can tell, you haven't followed.

For simplicity, let me rewrite the formula using x for x_n and x' for x_n+2. You have

x' = x^2(2 - x^2)

I asked when x' = x, and told you that although you might think this is hard, it can be done.

I also asked about the sequence being monotonic. Can you find under what conditions x' > x or x' < x?

These ideas will help you greatly in showing that the limits exist, and perhaps in finding what they are, for any starting x.

Don't just say it's hard; of course it is, or this wouldn't be an interesting problem! Do something, and show what you did. Sometimes just telling someone else what you have done helps you see the next step.

 

[Ishuda]

\(\displaystyle (x)_n\) is a sequence defined by \(\displaystyle x_0\in [0,1]\) and \(\displaystyle x_{n+1}=1-x_n^2\)
Already proved that \(\displaystyle \forall n\in \mathbb{N},x_n\in [0,1]\) (first part of the exercise)
The second part they asked to show that the two sequences \(\displaystyle (x_{2n})\) and \(\displaystyle (x_{2n+1})\) converge to different limits
The problem is that I never used to find limits of recurrent sequences whose first term is defined by an interval instead of a value
I'm stuck at this point never knowing what to do
I tried to study the function f(x)=1-x^2 which is decreasing on [0,1] and see if f(f(\(\displaystyle (x_{2n})\))) is increasing or not....
Any hints ?

Actually this is not quite true. For example when is x_n+1 = x_n in [0,1]. Otherwise consider starting values below this number and starting values above this number, call it x₌. Assume x₀ is less than x₌, what is the relationship between x₁ and x₌. Assume x₀ is greater than x₌, what is the relationship between x₁ and x₌. Given that, what about x₂, x₃, etc.

 

[Steven G]

Yes I tried for 0.5 and 0.1
It's hard for me to prove it for any initial value

Sometimes just telling someone else what you have done helps you see the next step.
Those are words you should follow.

 

[Steven G]

Okay it's much clearer right now
If \(\displaystyle x_0=1\) then \(\displaystyle x_{2n}=1\) and \(\displaystyle x_{2n+1}=0\)
If \(\displaystyle x_{0}=0\) then \(\displaystyle x_{2n}=0\) and \(\displaystyle x_{2n+1}=1\)
So now it's proved that they approach different limits by counterexample, right ?

You only proved it for x₀ = 1 and x₀ = 1. You are done iff that was what you were asked to prove. So do you think that you are done?

 

[maxwell]

Solution

You are trying to prove the following:

If x is a member of [0,1] and x_n+1 = 1 - (x_n)2 then

1.) x_n is a member of [0,1] for all n <= 0, and

2.) the subsequences {x_2n} and {x_2n+1} must converge to different limits.

Statement 1 can be proven true using an inductive argument (or perhaps by some other means). You've already done this so no problem.

Statement 2 can be proven false if there is one x₀ which is a member of [0,1] and for which the subsequences {x_2n} and {x_2n+1} derived fro x₀ converge to the same limit. Let x₀ = (sqrt(5)-1)/2. Then x_n is a member of [0,1] and x_n = x₀ for all n >= 0 (try it). Then {x_n} is a constant sequence. Likewise {x_2n} and {x_2n+1} are constant sequences. Thus both {x_2n} and {x_2n+1} converge to x₀. Hence statement 2 is false.

Statement 2 is true whenever x₀ is a member of [0,1] but x₀ is not equal to (sqrt(5)-1)/2. This is because the sequences {x_2n} and {x_2n+1} are monotonically opposite. When 0 <= x₀ < (sqrt(5)-1)/2 then {x_2n} is monotone decreasing while {x_2n+1} is monotone increasing. Hence, they must converge to different limits. When (sqrt(5)-1)/2 < x₀ <= 1 the situation is reversed. If anyone is interested in more details let me know and I can outline the proof more completely. Very cool problem.