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Math Help - 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 a>0, and we want to approximate \sqrt a. Thanks to the Newton's method I've found the Babylonian's method. We have x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}. We can start the method for any x_0>0. I also showed that the sequence x_n is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by \sqrt a. Now I must prove that the sequence {x_n} converges to \sqrt a when n tends to +\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 a>0, and we want to approximate \sqrt a. Thanks to the Newton's method I've found the Babylonian's method. We have x_{n+1}=\frac{x_n+\frac {a}{x_n}} {2}. We can start the method for any x_0>0. I also showed that the sequence x_n is decreasing and I must assume (since I wasn't able to demonstrate it) that it is bounded below by \sqrt a. Now I must prove that the sequence {x_n} converges to \sqrt a when n tends to +\infty and this is precisely what I'm asking you to help me.


    \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.
    \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L

    Therefore, from (1), we get :

    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 L^2=a, so L=\sqrt a because 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 x_n is bounded below by \sqrt a.
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    Quote Originally Posted by Moo View Post
    Hello,





    \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.
    \lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} x_n=L

    Therefore, from (1), we get :

    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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