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Math Help - Sequence inequality

  1. #1
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    Sequence inequality

    Given

    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

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

    -------

    I got it down to:

    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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    x_{n+1}^2=\frac{1}{4}\left(x_n^2+4+\frac{4}{x_n^2}  \right)

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

    =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

    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

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

    -------

    I got it down to:

    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 n=1 and x_2^2=\frac{1}{4}\left(1+\frac{2}{1}\right)^2=\fra  c{9}{4}\geqslant2. Now assume that x_{n}^2\geqslant 2 then 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:
    <br />
\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

    <br />
\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
    <br />
x_{n}^2\geqslant 2<br />
    ?
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  5. #5
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    related question.

    For the same sequence,

    Show that

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

    I did the following:

    Base case: n=2

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

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

    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)

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

    \le \frac{1}{2}\left(\frac{2}{x_{n+1}} - \frac{2}{x_{n}}\right)<br />
= \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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