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Math Help - Problem

  1. #1
    Newbie
    Joined
    Dec 2009
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    Problem

    Hi, I've attached an image of this little snippet from a lecture.

    I just do not get why this equality holds. It's probably something to do with factorising etc... but I cannot see why.

    X bar is the sum of xi over n (i.e. the mean).

    My notation is a bit scruffy, but the summation on the left hand side is on the numerator only (not the entire fraction).

    Cannot see why it equals 1!?
    Attached Thumbnails Attached Thumbnails Problem-imag0424.jpg  
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Re: Problem

    Hello, tahsocks!

    I'll re-write the problem in a more manageable form.


    \text{Show that: }\:\sum^n_{i=1}\!x_i(x_i- \overline{x}) \;=\;\sum^n_{i=1}\!\left(x_i - \overline {x})^2

    Recall that \overline{x} is a constant: . \overline{x} \:=\:\frac{\sum x_i}{n} \quad\Rightarrow\quad \sum\! x_i \:=\:n\!\cdot\!\overline{x}


    The left side is: . \sum\! x_i(x_i - \overline{x}) \;\;=\;\;\sum \!(x_1^2 - \overline{x}\,x_i)

    . . =\;\;\sum\! x_i^2 - \overline{x}\sum x_i \;\;=\;\;\sum\! x_i^2 - \overline{x}\left(n\!\cdot\!\overline{x}\right) \;\;=\;\;\sum\!x_1^2 - n\!\cdot\!\overline{x}^2



    The right side is: . \sum\!(x_1 - \overline{x})^2 \;\;=\;\;\sum\left(x_i^2 - 2\overline{x}x_i + \overline{x}^2\right)

    . . =\;\;\sum\!x_i^2 - 2\overline{x}\sum\!x_i + \overline{x}^2\sum\!1 \;\;=\;\; \sum\!x_i^2 - 2\overline{x}(n\!\cdot\!\overline{x}) + \overline{x}^2\!\cdot\!n

    . . =\;\; \sum\! x_i^2 - 2n\!\cdot\!\overline{x}^2 + n\!\cdot\!\overline{x}^2 \;\;=\;\; \sum\!x_i^2 - n\!\cdot\!\overline{x}^2


    The left side and the right side are equal . . . \tex{ta-}D\!AA!

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