Show Convergence of Contractive Sequence Homework

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In summary: Let me fix that up for you.In summary, the conversation discusses a homework problem involving showing that a given sequence is convergent. The problem provides definitions and attempts at a solution using the concept of contractive sequences and Cauchy sequences. The participants suggest using induction and a numerical example to understand the general case and ultimately prove that the sequence is Cauchy. The conversation ends with a proposed solution involving choosing a constant and using it to show that the sequence is Cauchy.
  • #1
bonfire09
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Homework Statement


If ##x_1 < x_2## are arbitrary real numbers and ##x_n=\frac{1}{2}(x_{n-2}+x_{n-1})## for## n > 2##, show that ##(x_n)## is convergent.

Homework Equations



Definition of Contractive Sequence: We say that a sequence ##X=(x_n)## of real numbers is contractive if there exists a constant ##C##; ##0 < C < 1##, such that ##|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|## for all ##n\in\mathbb{N}##. The number ##C## is called the constant of the contractive sequence.

The Attempt at a Solution


I know this sequence is not monotone so I think I need to use contraction. All I have is that ## |x_{n+2}-x_{n+1}|=| \frac{1}{2}(x_{n}+x_{n+1}-\frac{1}{2}(x_{n-1}+x_n|=|\frac{1}{2}x_{n+1}-\frac{1}{2}x_{n-1}|## and ##|x_{n+1}-x_n|=|\frac{1}{2}(x_{n-1}+x_n)-\frac{1}{2}(x_{n-2}+x_{n-1}|=\frac{1}{2}|x_n-x_{n-2}|##. But I can't seem to show that ##\frac{1}{2}|x_{n+1}-x_n|\leq \frac{1}{2}|x_n-x_{n-2}|## which I think is not possible but I am not sure. I think cauchy might work too but not sure how to apply it here.
 
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  • #2
You've show that |##x_{n+2} -x_{n+1}##| = 1/2|##x_{n+1} -x_{n-1}##| and you wish that last ##x_{n-1}## were ##x_{n}## because then you would have it. I messed around with that a bit and didn't immediately see anything too promising.

I think I would try induction. Show it's true for n =2. Then see if you can do an induction step.
 
  • #3
I think I might have to prove that it is a Cauchy Sequence instead. So what I would have is then for scratchwork is that there exists a ##h\in\mathbb{N}## such that if ##n>m## and ##n,m \geq h## then ##|x_n-x_m|=|\frac{1}{2}x_{n-2}+\frac{1}{2}x_{n-1}-\frac{1}{2}x_{m-2}+\frac{1}{2}x_{m-1}|##. But from here I don't know what to do.
 
  • #4
bonfire09 said:
I think I might have to prove that it is a Cauchy Sequence instead. So what I would have is then for scratchwork is that there exists a ##h\in\mathbb{N}## such that if ##n>m## and ##n,m \geq h## then ##|x_n-x_m|=|\frac{1}{2}x_{n-2}+\frac{1}{2}x_{n-1}-\frac{1}{2}x_{m-2}+\frac{1}{2}x_{m-1}|##. But from here I don't know what to do.

Start with a numerical example to try to see what to do in the general case. Suppose x0=0 and x1=1. So |x1-x0|=1. x2=1/2, so |x2-x1|=1/2. x3=3/4, so |x4-x3|=1/4. Etc etc. Now see if you can figure out what going on and generalize that. And I think you have a typo in your problem statement. You mean ##x_n=\frac{1}{2}(x_{n-2}+x_{n-1})##, right? The induction brmath suggests might be helpful.
 
  • #5
Well this problem is located in the Cauchy Sequence section and I am think this sequence does look Cauchy to me. I'm thinking if I can somehow bound ##|x_n-x_m|## by ##\frac{1}{2^{n-1}}## because of the ##\frac{1}{2}## then I would have it since ##\frac{1}{2^{n-1}}## is convergent. Oh yeah thanks I fixed the problem.
 
  • #6
bonfire09 said:
Well this problem is located in the Cauchy Sequence section and I am think this sequence does look Cauchy to me. I'm thinking if I can somehow bound ##|x_n-x_m|## by ##\frac{1}{2^{n-1}}## because of the ##\frac{1}{2}## then I would have it since ##\frac{1}{2^{n-1}}## is convergent. Oh yeah thanks I fixed the problem.

Yes, and an extension of that argument would show any contractive sequence is Cauchy and hence convergent, right?
 
  • #7
Yes but I doubt I can get anywhere with it proving its contractive. I think I am stuck pretty badly.
 
  • #8
bonfire09 said:
Yes but I doubt I can get anywhere with it proving its contractive. I think I am stuck pretty badly.

I thought from your last comment in post 5 that you had the solution. What's bugging you? How far did you get?
 
  • #9
The bounding part. Showing that ##|x_n-x_m|= |(\frac{1}{2}x_{n-2}+\frac{1}{2}x_{n-1})-(\frac{1}{2}x_{m-2}+\frac{1}{2}x_{m-1})|\leq \frac{1}{2^{n-1}}##.
 
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  • #10
bonfire09 said:
The bounding part. Showing that ##|x_n-x_m|= |(\frac{1}{2}x_{n-2}+\frac{1}{2}x_{n-1})-(\frac{1}{2}x_{m-2}+\frac{1}{2}x_{m-1})|\leq \frac{1}{2^{n-1}}##.

It's not. The difference |xn-xm| is dependent on what x0 and x1 are. If |x1-x0|=C can you show |x2-x1|=C/2? Extend that. It's going to boil down to a geometric series argument to show it's Cauchy.
 
  • #11
Ahh ok I will try to take it from here and post what I get when I am finished.
 
  • #12
If ##|x_2-x_1|=c## and ##x_1<x_2##. Then ##x_3=\frac{1}{2}(x_1+x_2)##. It follows ##\frac{1}{2}|x_3-x_2|=\frac{c}{2}##. So I from here what I see is that if ##|x_1-x_2|=c## then ##|(x_1-x_2)+(x_2-x_3)+...+(x_{n-1}-x_n)|\leq |x_1-x_2|+|x_2-x_3|+...+|x_{n-1}-x_n|= c+\frac{1}{2}c+...+\frac{1}{2^{n-1}}c=2c(1-\frac{1}{2^n})## which I know that ##2c(1-\frac{1}{2^n})## converges. I hope I got this part right. And for the first part I think I can use induction to prove it.

Proof:
 
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  • #13
bonfire09 said:
If ##|x_2-x_1|=c## and ##x_1<x_2##. Then ##x_3=\frac{1}{2}(x_1+x_2)##. It follows ##\frac{1}{2}|x_3-x_2|=\frac{c}{2}##. So I from here what I see is that if ##|x_1-x_2|=c## then ##|(x_1-x_2)+(x_2-x_3)+...+(x_{n-1}-x_n)|\leq |x_1-x_2|+|x_2-x_3|+...+|x_{n-1}-x_n|= c+\frac{1}{2}c+...+\frac{1}{2^{n-1}}c=2c(1-\frac{1}{2^n})## which I know that ##2c(1-\frac{1}{2^n})## converges. I hope I got this part right. And for the first part I think I can use induction to prove it.

Proving it's Cauchy means you want to show |xn-xm| can be made small by making n and m large. An estimate of |x1-xn| won't help with that. Don't start from x1.
 
  • #14
Proof:
This is what I have so far.
Let ##\epsilon>0## and let ##h\in\mathbb{N}## such that if ##m,n \geq h## and ##m>n## then ##|x_m-x_n|=|(x_m-x_{m-1})+(x_{m-1}-x_{m-2})+...+(x_{n-1}-x_n)| \leq |x_m-x_{m-1}|+...+|x_{n-1}-x_n|##. Suppose ##|x_m-x_{m-1}|=c##. It follows from my previous statement that ##|x_m-x_{m-1}|+...+|x_{n-1}-x_n|=c+\frac{c}{2}+...+\frac{c}{2^{n-1}}=2c(1-\frac{1}{2^n})<\epsilon## since we know ##2c(1-\frac{1}{2^n})## converges. I think this might be right maybe my indexes are off. Not sure.
 
  • #15
bonfire09 said:
Proof:
This is what I have so far.
Let ##\epsilon>0## and let ##h\in\mathbb{N}## such that if ##m,n \geq h## and ##m>n## then ##|x_m-x_n|=|(x_m-x_{m-1})+(x_{m-1}-x_{m-2})+...+(x_{n-1}-x_n)| \leq |x_m-x_{m-1}|+...+|x_{n-1}-x_n|##. Suppose ##|x_m-x_{m-1}|=c##. It follows from my previous statement that ##|x_m-x_{m-1}|+...+|x_{n-1}-x_n|=c+\frac{c}{2}+...+\frac{c}{2^{n-1}}=2c(1-\frac{1}{2^n})<\epsilon## since we know ##2c(1-\frac{1}{2^n})## converges. I think this might be right maybe my indexes are off. Not sure.

I didn't check the details of the indices yet. The problem with showing that that is less than ε is that ##2c(1-\frac{1}{2^n})## converges alright, but it doesn't converge to 0. Think about it a bit.
 
  • #16
Oh yeah it converges to ##2c## and we need something that converges to ##0## since ##\epsilon## is arbitrary. But I don't think there is another sequence bigger than ##2c(1-\frac{1}{2^n})## that converges to 0 because if it did then some of the terms of that sequence would be smaller than ##2c(1-\frac{1}{2^n})##. Oh could I just do this ##2c(1-\frac{1}{2^n})=(2c-\frac{2c}{2^n})=\frac{1}{2^n}(2c*2^n-2c)## and since ##\frac{1}{2^n}## converges to 0 the whole thing must converge to 0.
 
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  • #17
bonfire09 said:
Oh yeah it converges to ##2c## and we need something that converges to ##0## since ##\epsilon## is arbitrary. But I don't think there is another sequence bigger than ##2c(1-\frac{1}{2^n})## that converges to 0 because if it did then some of the terms of that sequence would be smaller than ##2c(1-\frac{1}{2^n})##. Oh could I just do this ##2c(1-\frac{1}{2^n})=(2c-\frac{2c}{2^n})=\frac{1}{2^n}(2c*2^n-2c)## and since ##\frac{1}{2^n}## converges to 0 the whole thing must converge to 0.

No, no. Define c=|x2-x1| once and for all. What's |x3-x2|? What's ##|x_m-x_{m-1}|##? Remember that inductive proof you were going to do?
 

FAQ: Show Convergence of Contractive Sequence Homework

1. What is a contractive sequence?

A contractive sequence is a sequence of numbers where each term is closer to the previous term than the one before it. In other words, the difference between consecutive terms gets smaller and smaller as the sequence progresses.

2. Why is it important to show convergence of a contractive sequence?

Showing convergence of a contractive sequence is important because it proves that the sequence has a limit. This means that as the sequence progresses, the terms get closer and closer to a specific value, making it easier to understand the behavior of the sequence.

3. How do you show the convergence of a contractive sequence?

To show the convergence of a contractive sequence, you can use the definition of a limit. This involves choosing a limit value and showing that the terms of the sequence get closer and closer to this value as the sequence progresses. Alternatively, you can also use the Cauchy criterion, which states that a sequence is convergent if the difference between consecutive terms gets arbitrarily small as the sequence progresses.

4. What is the importance of contractive sequences in mathematics?

Contractive sequences are important in mathematics because they provide a way to prove the existence of a limit and establish the behavior of a sequence. They are also used in various mathematical fields, such as analysis, calculus, and dynamical systems.

5. Can a non-contractive sequence be convergent?

No, a non-contractive sequence cannot be convergent. This is because a non-contractive sequence does not have a limit, and therefore, the terms do not get closer and closer to a specific value as the sequence progresses. Without this behavior, the sequence cannot be considered convergent.

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