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Math Help - minusing summations

  1. #1
    Junior Member
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    minusing summations

    Hi, is this right?

    If x_1 = 30 \mu + \sum_j B_j + \sum_j(xB)_{1j} = y_1

    and

    x_2 = 30 \mu + x_2 + \sum_j B_j + \sum_j(xB)_{2j} = y_2

    then is  x_1 - x_2 = - x_2 + (y_1 - y_2)

    is that right?

    coz im confused about how to minus the summations:

     \sum_j(xB)_{1j} - \sum_j(xB)_{2j}

    can someone please help me out

    oh and j goes from 1..3
    Last edited by Dgphru; May 14th 2010 at 05:03 AM. Reason: oops i wrote it wrong
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  2. #2
    Super Member
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    Quote Originally Posted by Dgphru View Post
    Hi, is this right?

    If x_1 = 30 \mu + \sum_j B_j + \sum_j(TB)_{1j} = y_1

    and

    x_2 = 30 \mu + x_2 + \sum_j B_j + \sum_j(TB)_{2j} = y_2

    then is  x_1 - x_2 = - x_2 + (y_1 - y_2)

    is that right?

    coz im confused about how to minus the summations:

     \sum_j(TB)_{1j} - \sum_j(TB)_{2j}

    can someone please help me out

    oh and j goes from 1..3
    What is B? What is \mu? What is T? We will be able to help you much more if you give the whole context of the question, or at least some more of it.

    Regardless, if we assume for a second that x_1-x_2 = -x_2 + y_1 - y_2 then we get x_1 = y_1-y_2 but according to what you wrote, x_1 = y_1 \Rightarrow y_2 = 0... and I'm sure this isn't what you wanted.
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  3. #3
    Junior Member
    Joined
    Sep 2009
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    umm, the whole question is calculate x_1 - x_2

    when

     x_i = 30\mu + 30x_i + \sum_j B_j + \sum_j (xB)_{ij} = \sum y_{i}

    B stands for beta, and i goes from 1 to 3 and j goes from 1 to 3.

    that's actually the whole question..please help me
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