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Thread: monotone

  1. #1
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    monotone convergence theorem

    a) X1=8 and Xn+1=.5Xn +2 for n within naturals. Show that Xn is bounded and monotone and find the limit?

    b) X1> and Xn+1=2-1/Xn for n within naturals. Show that Xn is bounded and monotone and find the limit

    for part a i can say x1>x2, thus the sequence is decreasing.
    Last edited by bigb; Sep 25th 2008 at 03:19 PM.
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  2. #2
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    Firstly lets find the limit by solving

    $\displaystyle
    x = x/2 + 2
    $

    gives x = 4, so limit is 4

    lets rewrite
    $\displaystyle
    x_n = 4 + \delta_n
    $
    $\displaystyle
    x_n+1 = (4 + \delta_n) /2 + 2 = 4 + \delta_n / 2
    $
    we can see that if $\displaystyle x_n > 4 $ then $\displaystyle x_n+1 > 4 $ and also that $\displaystyle x_n+1 < x_n$

    so we can conclude that the sequence decreases and has a lower bound of 4.

    hope this helps and makes sense!

    cheers Nobby
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  3. #3
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    Quote Originally Posted by Nobby View Post
    Firstly lets find the limit by solving

    $\displaystyle
    x = x/2 + 2
    $

    gives x = 4, so limit is 4

    lets rewrite
    $\displaystyle
    x_n = 4 + \delta_n
    $
    $\displaystyle
    x_n+1 = (4 + \delta_n) /2 + 2 = 4 + \delta_n / 2
    $
    we can see that if $\displaystyle x_n > 4 $ then $\displaystyle x_n+1 > 4 $ and also that $\displaystyle x_n+1 < x_n$

    so we can conclude that the sequence decreases and has a lower bound of 4.

    hope this helps and makes sense!

    cheers Nobby
    can you explain to me how to find out the sequence is decreasing. I know how to solve for the limit.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bigb View Post
    can you explain to me how to find out the sequence is decreasing. I know how to solve for the limit.
    by induction.

    can you continue?
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  5. #5
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    I guess the formal argument is by induction.

    $\displaystyle x_2< x_1$ you can calculate these first.

    then

    $\displaystyle 4 + \delta/2 < 4 + \delta$

    which we showed was the same as

    $\displaystyle x_{n+1}< x_n$

    Half of something positive is less then the original

    clear?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    by induction.

    can you continue?
    i think soo...
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bigb View Post
    i think soo...
    here's the outline to a more straightforward induction proof. i leave it to you to formalize it. we use induction to show that the sequence is monotonically decreasing (that is $\displaystyle x_{n + 1} < x_n$ for all $\displaystyle n$), and hence is bounded above by the first term, namely 8

    Note that $\displaystyle x_2 = \frac 12 \cdot 8 + 2 = 6$ so that $\displaystyle x_2 < x_1$

    Now assume $\displaystyle x_{n + 1} < x_n$

    then $\displaystyle \frac 12 x_{n + 1} < \frac 12 x_n$

    adding 2 to both sides we get

    $\displaystyle \underbrace{\frac 12 x_{n + 1} + 2}_{\text{note this is }x_{n + 2}} < \underbrace{\frac 12 x_n + 2}_{\text{note this is }x_{n + 1}}$

    but that means

    $\displaystyle x_{n + 2} < x_{n + 1}$
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