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Question Number 120037 by floor(10²Eta[1]) last updated on 28/Oct/20

Suppose that R > 0, x_{0} > 0, and

x_{n + 1} = \frac{1}{2} (\frac{R}{x_{n}} + x_{n}), n ⩾ 0

Prove : For n ⩾ 1, x_{n} > x_{n + 1} > \sqrt{R} and

x_{n} - \sqrt{R} ⩽ \frac{1}{2^{n}} \frac{{(x_{0} - \sqrt{R})}^{2}}{x_{0}}

Answered by floor(10²Eta[1]) last updated on 29/Oct/20

{2 x}_{n + 1} x_{n} = R + x_{n}^{2}

{2 x}_{n + 1} x_{n} - 2 \sqrt{R} x_{n} = {(x_{n} - \sqrt{R})}^{2} ⩾ 0

{2 x}_{n + 1} x_{n} - 2 \sqrt{R} x_{n} ⩾ 0

x_{n + 1} ⩾ \sqrt{R}

\Rightarrow x_{n + 1} > \sqrt{R}

x_{n}^{2} - {2 x}_{n + 1} x_{n} + R = 0

x_{n} = x_{n + 1} \pm \sqrt{x_{n + 1}^{2} - R} > \sqrt{R}

suppose that x_{n} < x_{n + 1}

x_{n} < \frac{1}{2} (\frac{R}{x_{n}} + x_{n})

{2 x}_{n}^{2} < R + x_{n}^{2} \Rightarrow x_{n} < \sqrt{R} (imp .)

\Rightarrow x_{n} > x_{n + 1} > \sqrt{R}

{2 x}_{n + 1} x_{n} - 2 \sqrt{R} x_{n} = {(x_{n} - \sqrt{R})}^{2}

2 (x_{n} - \sqrt{R}) > 2 (x_{n + 1} - \sqrt{R}) = \frac{{(x_{n} - \sqrt{R})}^{2}}{x_{n}}

x_{n} - \sqrt{R} > \frac{{(x_{n} - \sqrt{R})}^{2}}{{2 x}_{n}} < \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}}

(i) : x_{n} - \sqrt{R} ⩾ \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}}

x_{0} - \sqrt{R} > x_{n} - \sqrt{R} ⩾ \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}}

{2 x}_{0}^{2} - 2 \sqrt{R} x_{0} > x_{0}^{2} - 2 \sqrt{R} x_{0} + R

x_{0}^{2} > R \Rightarrow x_{0} > \sqrt{R}

(ii) :

x_{n} - \sqrt{R} ⩽ \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}}

x_{0} - \sqrt{R} < \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}}

x_{0} < - \sqrt{R}

\Rightarrow x_{n} - \sqrt{R} ⩾ \frac{{(x_{0} - \sqrt{R})}^{2}}{{2 x}_{0}} ⩾ \frac{{(x_{0} - \sqrt{R})}^{2}}{2^{n} x_{0}}