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Math Help - graphs and points

  1. #1
    Senior Member Sampras's Avatar
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    graphs and points

    Suppose  S = (a,b) with  0 < b < a . We generate points as follows:  x_0 = a, y_0 = b ,  x_{n+1} = \frac{x_n+y_n}{2} and  y_{n+1} = \frac{2x_{n}y_{n}}{x_n+y_n} .

    So  0 < x_{n+1}-y_{n+1} = \frac{x_n-y_n}{x_n+y_n} \cdot \frac{x_n-y_n}{2} < \frac{x_n-y_n}{2} .

    Does the  \frac{x_n-y_n}{2} act similarly to  \epsilon/2 ? And so  \lim x_n = \lim y_n = x ?
    Last edited by Sampras; June 19th 2009 at 04:15 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sampras View Post
    Suppose  S = (a,b) with  0 < b < a . We generate points as follows:  x_0 = a, y_0 = b ,  x_{n+1} = \frac{x_n+y_n}{2} and  y_{n+1} = \frac{2x_{n}y_{n}}{x_n+y_n} .

    So  0 < x_{n+1}-y_{n+1} = \frac{x_n-y_n}{x_n+y_n} \cdot \frac{x_n-y_n}{2} < \frac{x_n-y_n}{2} .

    Does the  \frac{x_n-y_n}{2} act similarly to  \epsilon/2 ?
    the later is a constant, the former varies depending on n.
    And so  \lim x_n = \lim y_n = x ?
    how did you get x here?
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  3. #3
    Senior Member Sampras's Avatar
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    Quote Originally Posted by Jhevon View Post
    the later is a constant, the former varies depending on n. how did you get x here?
    because  x_{n+1}y_{n+1} = x_ny_n .
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