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Thread: Sequence inequality

  1. #1
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    Sequence inequality

    Given

    $\displaystyle x_1=1,\quad x_{n+1} = \frac{1}{2}\left( x_n + \frac{2}{x_n}\right),\quad n = 1, 2, 3, \dots$

    Show that

    $\displaystyle x_{n+1}^2 -2 \ge 0$

    -------

    I got it down to:

    $\displaystyle x_{n+1}^2 -2=\frac{x_n^2}{4}+\frac{1}{x_n^2}-1$

    but not sure what to do now...
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  2. #2
    Member Black's Avatar
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    $\displaystyle x_{n+1}^2=\frac{1}{4}\left(x_n^2+4+\frac{4}{x_n^2} \right)$

    $\displaystyle =2+\frac{1}{4}\left(x_n^2-4+\frac{4}{x_n^2}\right)$

    $\displaystyle =2+\frac{1}{4}\left(x_n-\frac{2}{x_n}\right)^2>2.$
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Given

    $\displaystyle x_1=1,\quad x_{n+1} = \frac{1}{2}\left( x_n + \frac{2}{x_n}\right),\quad n = 1, 2, 3, \dots$

    Show that

    $\displaystyle x_{n+1}^2 -2 \ge 0$

    -------

    I got it down to:

    $\displaystyle x_{n+1}^2 -2=\frac{x_n^2}{4}+\frac{1}{x_n^2}-1$

    but not sure what to do now...
    Induction.

    So our base case is obviously $\displaystyle n=1$ and $\displaystyle x_2^2=\frac{1}{4}\left(1+\frac{2}{1}\right)^2=\fra c{9}{4}\geqslant2$. Now assume that $\displaystyle x_{n}^2\geqslant 2$ then $\displaystyle x_{n+1}^2=\frac{1}{4}\left(x_n+\frac{2}{x_n}\right )^2=\frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right )\geqslant\frac{1}{4}\left(4+4+\frac{1}{4}\right)= 2+\frac{1}{16}\geqslant 2$
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  4. #4
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    Thanks.

    I follow everything except this bit:
    $\displaystyle
    \frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right)\ge qslant\frac{1}{4}\left(4+4+\frac{1}{4}\right)$

    shouldn't it be

    $\displaystyle
    \frac{1}{4}\left(x_n^2+4+\frac{1}{x_n^2}\right)\ge qslant\frac{1}{4}\left({\color{red}2}+4+\frac{1}{\ color{red}2}\right)$

    since we have that
    $\displaystyle
    x_{n}^2\geqslant 2
    $
    ?
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  5. #5
    Senior Member
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    related question.

    For the same sequence,

    Show that

    $\displaystyle x_{n+1}-x_n\le 0$ for all $\displaystyle n \ge 2$

    I did the following:

    Base case: n=2

    $\displaystyle x_3 -x_2=\frac{17}{12}-\frac{3}{2}=-\frac{1}{12}\le 0$

    Induction:
    If $\displaystyle x_{n+1}-x_n \le 0$ then

    $\displaystyle x_{n+2}-x_{n+1} = \frac{1}{2}\left(x_{n+1} + \frac{2}{x_{n+1}}\right) - \frac{1}{2}\left(x_{n} + \frac{2}{x_{n}}\right)$

    $\displaystyle = \frac{1}{2}\left(\underbrace{x_{n+1}-x_n}_{\le 0} +\frac{2}{x_{n+1}} - \frac{2}{x_{n}}\right)$

    $\displaystyle \le \frac{1}{2}\left(\frac{2}{x_{n+1}} - \frac{2}{x_{n}}\right)
    = \frac{1}{x_{n+1}} - \frac{1}{x_{n}}$

    but now what? I need to show that that expressions is <= 0.
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