Proving Sigma Notation Equals Zero

[bob1182006]

Homework Statement

Show:
\sum_{i=1}^n (x_i - \overline{x}) = 0

Homework Equations

Sigma notation

The Attempt at a Solution

\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
\sum_{i=1}^n x_i = \frac{1}{n}\sum_{i=1}^n \sum_{i=1}^n x_i

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

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

Is it just:
\sum_{i=1}^n 1 = n?

 

[maverick_starstrider]

consider that \overline{x} is idependent of i therefore \sum_{i=1}^n\overline{x} simply equals n \overline{x}. Also IMPORTANT NOTE: In the second line of your attempted solution you use the fact that \sum_{i=1}^n (x_i - \overline{x}) = 0 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.

 

[bob1182006]

Ah, I get it now thanks a lot!

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!

 

[maverick_starstrider]

Err no it'd be
\sum_{i=1}^n (x_i - \overline{x}) = 0
\sum_{i=1}^n x_i - n\overline{x} = 0
n \sum_{i=1}^n \frac{x_i}{n} - n\overline{x} = 0

n\overline{x} - n\overline{x} = 0

which gives you zero... hope he comes back to read that.

 

[bob1182006]

ah okay, I was expanding the xbar so there would be 2 equal sums being subrated which would just =0.

\sum_{i=1}^n x_i - n \frac{1}{n} \sum_{j=1}^n x_j = 0

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.