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Math Help - show that x_n converges to \sqrt{a}

  1. #1
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    show that x_n converges to \sqrt{a}

    Fix a>0 and let x_1>\sqrt{a}. For n >=1, define

    x_{n+1} = 1/2(x_n +a/x_n)

    Show that (x_n) converges and that lim n--> infinity x_n = \sqrt{a}.



    Here is what I have got so far, x_{n+1}= 1/2(x_n +a/x_n) >= \sqrt{a} since \sqrt{ab}<=(a+b)/2
    Last edited by mr fantastic; September 20th 2011 at 08:09 PM. Reason: Deleted tex tags from title (latex does not work in titles!).
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  2. #2
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    Re: show that x_n converges to \sqrt{a}

    Quote Originally Posted by wopashui View Post
    Fix a>0 and let x_1>\sqrt{a}. For n >=1, define

    x_{n+1} = 1/2(x_n +a/x_n)

    Show that (x_n) converges and that lim n--> infinity x_n = \sqrt{a}.



    Here is what I have got so far, x_{n+1}= 1/2(x_n +a/x_n) >= \sqrt{a} since \sqrt{ab}<=(a+b)/2
    1. Use induction to show that x_n < x_{n=1}.

    2. Let the limit be l. Solve l = \frac{1}{2} \left(l + \frac{a}{l}\right).
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    Re: show that x_n converges to \sqrt{a}

    Quote Originally Posted by mr fantastic View Post
    1. Use induction to show that x_n < x_{n=1}.

    2. Let the limit be l. Solve l = \frac{1}{2} \left(l + \frac{a}{l}\right).
    you meant to show x_n > x_{n+1}, since the sequence has to be decreasing to get to \sqrt{a}
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  4. #4
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    Re: show that x_n converges to \sqrt{a}

    Quote Originally Posted by wopashui View Post
    you meant to show x_n > x_{n+1}, since the sequence has to be decreasing to get to \sqrt{a}
    Yes, typo on my part.
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