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Math Help - Cauchy sequence and its limit......

  1. #1
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    Cauchy sequence and its limit......

    Hi, I had difficulty solving this problem. Actually, I'm stuck.

    Suppose (Xn) in R is defined recursively by X1= 1 and  \displaystyle x_{n+1}=\dfrac{1}{3+x_n} , n= 1, 2, ...
    Show (Xn) converges and find its limit.

    And there's a hint: Notice that | Xn+2 - X n+1| < 1/9 |Xn - Xn+1|. Show (Xn) is Cauchy.

    Thanks for any help.

    I edited my mistake. Sorry again.
    Last edited by truevein; October 28th 2010 at 12:02 PM.
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  2. #2
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    Please reread you sequence definition.
    AS posted, that sequence is not bounded.
    So it must be wrong.
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  3. #3
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    Thank you, but why is it not bounded? I could not understand.
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  4. #4
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    Quote Originally Posted by truevein View Post
    Thank you, but why is it not bounded? I could not understand.
    If we add \frac{1}{3} enough times we get a large number.
    The way you have it every third term of the sequence is a positive integer.
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  5. #5
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    Sorry, it is not (1/3)+ Xn but 1/(3+Xn). It is bounded, isn't it?
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  6. #6
    MHF Contributor Also sprach Zarathustra's Avatar
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    Hint:

    Induction (to the Q. hint)
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  7. #7
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    Quote Originally Posted by truevein View Post
    Sorry, it is not (1/3)+ Xn but 1/(3+Xn). It is bounded, isn't it?
    Why not learn to post in symbols? You can use LaTeX tags
    [tex] \displaystyle x_{n+1}=\dfrac{1}{3+x_n} [/tex] gives  \displaystyle x_{n+1}=\dfrac{1}{3+x_n}
    At least learn to use grouping symbols.
    That wasted time and effort.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by truevein View Post
    Hi, I had difficulty solving this problem. Actually, I'm stuck.

    Suppose (Xn) in R is defined recursively by X1= 1 and  \displaystyle x_{n+1}=\dfrac{1}{3+x_n} , n= 1, 2, ...
    Show (Xn) converges and find its limit.

    And there's a hint: Notice that | Xn+2 - X n+1| < 1/9 |Xn - Xn+1|. Show (Xn) is Cauchy.

    Thanks for any help.

    I edited my mistake. Sorry again.
    In terms of difference equation the sequence is the solution of the recursive equation...

    \displaystyle \Delta_{n} = x_{n+1}-x_{n} = \frac{1-3\ x_{n} - x_{n}^{2}}{3+x_{n}} = f(x_{n}) (1)

    The function f(*) is represented here...



    It exists one attractive fixed point in x_{0} = \frac{-3+\sqrt{13}}{2}= .3027756377199... and for any x_{1} > -3 the sequence will converge to x_{0}. An interesting detail: because is |f(x)|> |x_{0} - x| the sequence will be 'oscillating'. That is confirmed for x_{1}=1...

    x_{1}=1, x_{2}= .25, x_{3}= .307692..., x_{4}= .302325..., x_{5}= .3028169..., ...

    For x_{1}< -3 the behavior of the sequence is stiil to be analysed...

    A very interesting question! ...

    Kind regards

    \chi \sigma
    Last edited by chisigma; October 29th 2010 at 12:24 PM. Reason: bad mistake from me in (1) [but nobody discovered it!]... very sorry!...
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    In this post I will try to prove something more general.

    Let a_n be a sequence. Suppose there exist 1>\theta \in \mathbb{R} so that \left | a_{n+1}-a_n \right |<\theta \left | a_n-a_{n-1} \right | for all  n\geq 2. I will prove that a_n converges.

    Proof:

    We prove via induction that for all n\in \mathbb{N}: \left | a_{n+1}-a_n \right |\leq \left | \theta  \right | ^{n-1}\left | a_2-a_1 \right |.

    Base of induction:

    Of course:  \left | a_2-a_1 \right |\leq \left | \theta  \right |^0\left | a_2-a_1 \right |

    Step of induction:

    Let be  k\in \mathbb{N} such that: \left | a_{k+1}-a_k \right |\leq \left | \theta  \right | ^{k-1}\left | a_2-a_1 \right |, hence:

    \left | a_{k+2}-a_{k+1} \right |<\left | \theta  \right |\left | a_{k+1}-a_k \right |\leq \left | \theta \left |\left | \theta  \right |^{k-1}\left | a_2-a_1 \right |=\left | \theta  \right |^k\left | a_2-a_1 \right |.

    ...hence the statement is true for all n\in \mathbb{N} .


    Now, for all  m,n \in \mathbb{N} that n>m we have:

    \displaystyle\left | a_n-a_m \right |=\left | \sum_{k=m}^{n-1}(a_{k+1}-a_k) \right |\leq \sum_{k=m}^{n-1}\left |a_{k+1}-a_k  \right |< \left | a_2-a_1 \right |\sum_{k=m}^{n-1}\left | \theta  \right |^{k-1}=

    \displaystyle= \left | a_2-a_1 \right |\left | \theta  \right |^{m-1}\sum_{r=0}^{n-1-m}\left | \theta  \right |^r\leq \left | a_2-a_1 \right | \left | \theta  \right |^{m-1}\frac{1-\left |\theta  \right | ^{n-m}}{1-\left ( \theta  \right )}\leq \left | a_2-a_1 \right |\left | \theta  \right |^{m-1}\frac{1}{1-\left | \theta  \right |}\overset{m\rightarrow \infty}{\rightarrow}0

    Hence, a_n is Cauchy sequence, and it converges.
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  10. #10
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    Thanks, chisigma and Also sprach Zarathustra, for your time and effort.
    It looks a bit complicated, I'm trying to figure out and I will. Thanks again!
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