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$\sqrt{3}$ 的数值求解

Posted by haifeng on 2011-06-29 16:45:15 last update 2011-06-29 16:47:27 | Edit | Answers (1)

设数列 ${x_{n}}_{n = 1}^{\infty}$ 满足递推关系

x_{n} = \frac{1}{2} (x_{n - 1} + \frac{3}{x_{n - 1}}), n = 1, 2, \dots,

给定初值

x_{0} > 0

. 证明

lim_{n \to \infty} x_{n} = \sqrt{3}

Posted by haifeng on 2011-06-29 16:58:42

根据递推关系, 容易证明

从而 $lim_{n \to \infty} x_{n} = \sqrt{3}$ . 这给出了计算 $\sqrt{3}$ 的一个算法.