# Simple proof

• Sep 5th 2010, 04:56 PM
Iceflash234
Simple proof
I need help proving that the sum of deviations of a set of measurements about their mean is 0.

$\displaystyle \sum_{j=1}^n (x_j-$http://upload.wikimedia.org/math/6/0...a3cc6a1131.png$\displaystyle ) = 0$

• Sep 5th 2010, 05:49 PM
pickslides
Maybe use the fact that that $\displaystyle \bar{x} = \frac{\sum x_j}{n}$
• Sep 5th 2010, 07:30 PM
Soroban
Hello, Iceflash234!

Quote:

I need help proving that the sum of deviations
of a set of measurements about their mean is 0.

. . $\displaystyle \displaystyle \sum_{j=1}^n (x_j-\overline{x }) \;=\;0$

$\displaystyle \displaystyle \text{We have: }\;S \;\;=\;\;\sum ^n_{j=1}(x_j - \overline{x}) \;\;=\;\;\sum^n_{j=1}x_j - \sum^n_{j=1} \overline{x}$

$\displaystyle \displaystyle \text{Since }\overline{x} \text{ is a constant, we have: }\; S \;\;=\;\;\sum^n_{j=1}x_j - n\overline{x}$

$\displaystyle \displaystyle \text{Multiply the sum by }\dfrac{n}{n}\!:\;\;\;S \;\;=\;\; n\,\cdot$ $\displaystyle \underbrace{\begin{array}{c}_n \;\;\;\; \\ \sum x_j\\ ^{j=1}\;\;\; \\ \hline n \end{array}}_{\text{This is }\overline{x}} -\; n\overline{x}$

Therefore: . $\displaystyle S \;\;=\;\;n\overline{x} - n\overline{x}\;\;=\;\;0$