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Thread: 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 $\displaystyle x_1>\sqrt{a}$. For n >=1, define

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

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



    Here is what I have got so far, $\displaystyle x_{n+1}$= $\displaystyle 1/2(x_n +a/x_n)$ $\displaystyle >= \sqrt{a}$ since $\displaystyle \sqrt{ab}<=(a+b)/2$
    Last edited by mr fantastic; Sep 20th 2011 at 07:09 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 $\displaystyle x_1>\sqrt{a}$. For n >=1, define

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

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



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

    2. Let the limit be l. Solve $\displaystyle l = \frac{1}{2} \left(l + \frac{a}{l}\right)$.
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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 $\displaystyle x_n < x_{n=1}$.

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

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