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Homework Help: Need help understanding this equality

  1. Apr 4, 2007 #1
    Hello, I am really bothered by this because I can't seem to be able to prove the following is true.

    [tex]\sum_{i=1}^n (x_i - u)^2 = \sum_{i=1}^n (x_i - x_{avg})^2 + n(x_{avg} - u)^2[/tex]

    for [tex]u = constant[/tex]
    and [tex]x_{avg} = \frac{\sum x_i}{n}[/tex]

    I just can't seem to grasp how I can prove that the right-side equals the left-side. Any push in the right direction would be extremely helpful.

    Thanks in advance.

    Last edited: Apr 4, 2007
  2. jcsd
  3. Apr 4, 2007 #2


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    What've you tried doing? Have you tried expanding the RHS and using the definition of xavg? What do you get after doing this?
  4. Apr 4, 2007 #3


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    Your definition of "xavg" is wrong. You need
    [tex]x_{avg}= \frac{\sum x_i}{n}[/tex]
    so [itex]\sum x_i= nx_{avg}[/itex]

    Multiply it out on the left: [itex]\sum (x_i^2- 2ux_i+ u^2)= \sum x_i^2- 2u\sum x_i+ u^2 \sum 1[/itex]
    Of course, [itex]\sum x_i= nx_{avg}[/itex] and [itex]\sum 1[/itex]= n so that is [itex]\sum x_i^2- 2nux_{avg}+ nu^2[/itex]

    Now do the same on the right and you should be able to cancel some things and arrive at the same answer.
  5. Apr 4, 2007 #4
    oh that helped a lot. Thx, cristo and halls, and thx for the correction. I got to the conclusion:

    [tex]n(u^2-2ux_{avg}) == n(u^2-2ux_{avg})[/tex]


    I had one more question. I need to know this relation to use in a proof in my 'probability' class. Is there an intuitive way to look at the equality to make it seem more obvious, and/or easier to memorize? Any takes on that would be helpful. Thanks again.

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