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Sequence and Series, finding relationship

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data


    in general

    2. Relevant equations


    3. The attempt at a solution

    I have no idea
    I tried to solve for Xn and substituting that into another equation...
    however, I don't know how to simplify it down to one single variable....

    How do I take the limit of the sequence, so that I can find the relationship
  2. jcsd
  3. Nov 25, 2007 #2


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    Please state the problem as it is given.
  4. Nov 25, 2007 #3
    the problem was given that way, we are trying to find the relationship between X and y
  5. Nov 25, 2007 #4


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    take the general equations and sub one into the other is all i can suggest
  6. Nov 26, 2007 #5


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    Staff Emeritus
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    Since there was NO X or Y in what you gave, I don't see how that can be true.
  7. Nov 26, 2007 #6


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    Hmm, I think I got what you mean.

    Given 0 < y0 < x0, the sequence (xn), and (yn) are defined as follow:

    [tex]x_{n + 1} = \frac{x_n + y_n}{2}, n \in \mathbb{N}[/tex] (1)
    [tex]y_{n + 1} = \sqrt{x_n y_n}, n \in \mathbb{N}[/tex] (2)

    Now find the limit of xn, and yn (Or does it tell you to prove that the limit of the two sequences are the same?).

    Is that the correct problem?


    Ok, now you must at least have a vision of how xn, and yn behave. Draw a pictures like this:


    Now, we have: x1 = (x0 + y0) / 2, that means, x1 lies exactly at the middle of x0, and y0.

    [tex]y_1 = \sqrt{x_0 y_0} > \sqrt{y_0 ^ 2} = y_0[/tex]

    You'll also have: [tex]x_1 - y_1 = \frac{x_0 - 2\sqrt{x_0 y_0} + y_0}{2} = \frac{(\sqrt{x_0} - \sqrt{y_0}) ^ 2}{2} > 0[/tex], so x1 > y1.


    Now, where do x2, and y2 lie?


    So, in conclusion, let's answer some questions:

    1. xn, and yn, which is greater?

    2. Is (xn) an increase sequence, or a decrease sequence? Can you prove it?

    3. Is (yn) an increase sequence, or a decrease sequence? Can you prove it?

    4. Are they bounded? Do they have limit?

    5. Do they both have the same limit?

    Ok, I think you can take it from here. Can you? :)
    Last edited: Nov 26, 2007
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