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Math Help - Problem involving an iterative sequence?

  1. #1
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    Problem involving an iterative sequence?

    I'm struggling on this but i get the feeling it's easier than i think.
    The sequence is defined by:

    xn+1 = (xn + 2) / 6
    where the initial value is 0.

    and you need to show that

    | xn+1 - xn | </= (5/6) * |xn - xn+1 | </= (1/3) * (5/6)^n-1

    after having shown that
    0< xn <1 (which i was able to do).

    ( by the way </= means 'less that or equal to", i didn't know how to put it on the computer though.)

    can anyone help?
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  2. #2
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    Hello, mongrel73!

    Could you restate the problem?
    There seems to be a serious typo . . .


    \text{The sequence is de{f}ined by: }\;x_{n+1} \:=\:\frac{x_{n+2}}{6}\; \text{ where }x_1 = 0

    \text{and you need to show that: }\;\underbrace{\left|x_{n+1} - x_n\right| \;\leq \; \frac{5}{6}\cdot\left|x_n - x_{n+1}\right|}_{{\color{blue}\text{This can't be true!}}} \;\leq \;\left(\frac{5}{6}\right)^{n-1}

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  3. #3
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    sorry no i didn't use latex so i guess it was a little ambiguous

    that wasn't what the definition of the iteration was. i'll try it in latex:

    ok i tried in latex it didn't work so i'll just try to be clearer:

    xn+1 = [ ( (xn)^5 ) + 2 ] / 6
    is the correct formula

    as opposed to

    xn+1 = [ xn+2 ] / 6
    which is where the misunderstanding came from i think.


    also the right-most term of the double inequality should be:
    (1/3) * [ (5/6)^(n-1) ]

    rather than just:
    (
    5/6)^(n-1)


    I think i just completely forgot to put in the ^5 part!


    Then I made another typo with the actual problem:

    | xn+1 - xn | </= (5/6) * |xn - xn-1 | </= (1/3) * (5/6)^n-1

    Thanks, please help!
    Last edited by mongrel73; November 17th 2007 at 11:38 AM. Reason: i made another (important) typo
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