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Thread: Limit of a sequence

  1. #1
    MHF Contributor arbolis's Avatar
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    Limit of a sequence

    I'm not able to find a limit of a sequence.
    It's related to the Babylonian's method to approximate a square root. In our case, we have $\displaystyle a>0$, and we want to approximate $\displaystyle \sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $\displaystyle x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $\displaystyle x_0>0$. I also showed that the sequence $\displaystyle x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\displaystyle \sqrt a$. Now I must prove that the sequence $\displaystyle {x_n}$ converges to $\displaystyle \sqrt a$ when $\displaystyle n$ tends to $\displaystyle +\infty$ and this is precisely what I'm asking you to help me.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by arbolis View Post
    I'm not able to find a limit of a sequence.
    It's related to the Babylonian's method to approximate a square root. In our case, we have $\displaystyle a>0$, and we want to approximate $\displaystyle \sqrt a$. Thanks to the Newton's method I've found the Babylonian's method. We have $\displaystyle x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}$. We can start the method for any $\displaystyle x_0>0$. I also showed that the sequence $\displaystyle x_n$ is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by $\displaystyle \sqrt a$. Now I must prove that the sequence $\displaystyle {x_n}$ converges to $\displaystyle \sqrt a$ when $\displaystyle n$ tends to $\displaystyle +\infty$ and this is precisely what I'm asking you to help me.


    $\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

    Let L be the limit of the sequence.
    $\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

    Therefore, from (1), we get :

    $\displaystyle L=\frac{L+\frac aL}{2}$

    ...
    Conclude ?
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks so much!! I got $\displaystyle L^2=a$, so $\displaystyle L=\sqrt a$ because $\displaystyle a>0$. I feel I'm not used to find limit of sequences of this type. You helped me a lot.
    I'll try to demonstrate that $\displaystyle x_n$ is bounded below by $\displaystyle \sqrt a$.
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,





    $\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} \frac{x_n+\frac{a}{x_n}}{2} \ (1)$

    Let L be the limit of the sequence.
    $\displaystyle \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L$

    Therefore, from (1), we get :

    $\displaystyle L=\frac{L+\frac aL}{2}$

    ...
    Conclude ?
    Everything else is perfect, you just need to prove that the sequence is convergent.
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  5. #5
    Moo
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    Quote Originally Posted by ThePerfectHacker View Post
    Everything else is perfect, you just need to prove that the sequence is convergent.
    "Decreasing and bounded below"

    If I read correctly, he said he had to prove the limit after showing that it was decreasing and bounded.
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  6. #6
    MHF Contributor arbolis's Avatar
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    I confused Theperfecthacker because I said " I'll try to demonstrate that is bounded below by ." I assumed the sequence was bounded below, but I had to prove it and I wasn't able. (Now I'll try).
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