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Math Help - Summation question - Prove

  1. #1
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    Post Summation question - Prove

    Show that (Sorry about the lack of latex here)

    [Σ from i=1 to n] of (x[i] - x(bar))^2 = [Σ from i=1 to n] of (x^2[i] - n*x^2(bar)
    Last edited by mitch_nufc; March 5th 2009 at 09:48 AM.
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  2. #2
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    Hello, mitch_nufc!

    We need to know the definition of the mean: . \frac{\sum x_i}{n} \:=\: \overline{x}


    Show that .

    \sum^n_{i= 1} \left(x_i^2 - \overline{x}\right)^2 \;=\;\sum^n_{i=1}x_i^2 - n\!\cdot\!\overline{x}^2

    We have: . \sum\left(x_i - \overline{x}\right)^2

    . . . =\:\sum\bigg(x_i^2 - 2\,\overline{x}\,x_i + \overline{x}^2\bigg)

    . . . = \;\sum x_i^2 - \sum 2\,\overline{x}\,x_i + \sum \overline{x}^2

    . . . =\;\sum x_i^2 - 2\,\overline{x}\sum x_i + \overline{x}^2\sum 1 .[1]


    \text{Multiply the middle sum by }\frac{n}{n}\!:\;\;\frac{n}{n}\bigg(-2\,\overline{x}\sum x_i\bigg) \;=\;-2n\,\overline{x}\cdot\overbrace{\left(\frac{\sum x_i}{n}\right)}^{\text{This is }\overline{x}} \;=\;-2n\cdot\overline{x}^2

    The last sum is: . \overline{x}^2\sum 1 \:=\:\overline{x}^2\!\cdot\!n \:=\:n\!\cdot\!\overline{x}^2


    Then [1] becomes: . \sum x_i^2 - 2n\!\cdot\!\overline{x}^2 + n\!\cdot\!\overline{x}^2 \;=\;\sum x_i^2 - n\!\cdot\!\overline{x}^2 . . . There!

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