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Sigma Notation

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Show:
    [tex]\sum_{i=1}^n (x_i - \overline{x}) = 0[/tex]


    2. Relevant equations
    Sigma notation


    3. The attempt at a solution

    [tex]\sum_{i=1}^n x_i - \sum_{i=1}^n \overline{x} = \sum_{i=1}^n x_i - \frac{1}{n}\sum_{i=1}^n \sum_{i=1}^n x_i = 0[/tex]
    [tex]\sum_{i=1}^n x_i = \frac{1}{n}\sum_{i=1}^n \sum_{i=1}^n x_i [/tex]

    By Inspection I know i need to show that:
    [tex]\sum_{i=1}^n \frac{1}{n}=1[/tex]

    Since the LHS has no [itex]x_i[/itex] how can i show that the sum will result in n/n =1?

    Is it just:
    [tex]\sum_{i=1}^n 1 = n?[/tex]
     
  2. jcsd
  3. Jun 11, 2008 #2
    consider that [tex]\overline{x}[/tex] is idependent of [tex]i[/tex] therefore [tex]\sum_{i=1}^n\overline{x}[/tex] simply equals [tex]n \overline{x}[/tex]. Also IMPORTANT NOTE: In the second line of your attempted solution you use the fact that [tex]\sum_{i=1}^n (x_i - \overline{x}) = 0[/tex] how ever that's what you're trying to SHOW so you can't assume it's true. You have to work entirely on the LHS and get it equal to 0.
     
  4. Jun 11, 2008 #3
    Ah, I get it now thanks alot!

    So on the first step it'd just be the n*xbar/n leaving only sum of x - sum of x which =0.

    Thank You!
     
  5. Jun 11, 2008 #4
    Err no it'd be
    [tex]\sum_{i=1}^n (x_i - \overline{x}) = 0[/tex]
    [tex]\sum_{i=1}^n x_i - n\overline{x} = 0[/tex]
    [tex]n \sum_{i=1}^n \frac{x_i}{n} - n\overline{x} = 0[/tex]

    [tex]n\overline{x} - n\overline{x} = 0[/tex]

    which gives you zero.... hope he comes back to read that.
     
  6. Jun 11, 2008 #5
    ah okay, I was expanding the xbar so there would be 2 equal sums being subrated which would just =0.

    [tex]\sum_{i=1}^n x_i - n \frac{1}{n} \sum_{j=1}^n x_j = 0[/tex]

    since n/n =1, and both sums start and end at the same place they are both equal and thus will give 0 as the answer.
     
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