# Sigma Notation

bob1182006

## Homework Statement

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

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:
$$\sum_{i=1}^n 1 = n?$$

## Answers and Replies

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 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!

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.