Chapter 8.1, Problem 142E

To determine

To prove: the give expression.

Explanation of Solution

Given information:

The expression which needs to be prove is as follows

  ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$ .

Formula used:

  $\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$ .

Proof:

The expression which needs to be prove is as follows

  ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$

Given that $\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$

  $n\overline{x}={\displaystyle \sum _{i=1}^n{x}_i}$

Consider the left hand side of expression needs to be proved

  ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^n\left(x_i{}^2+{\left(\overline{x}\right)}^2-2.x_i.\overline{x}\right)}$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}+{\displaystyle \sum _{i=1}^n{\left(\overline{x}\right)}^2}-2{\displaystyle \sum _{i=1}^n{x}_i}\overline{x}$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}+{\left(\overline{x}\right)}^2{\displaystyle \sum _{i=1}^{n}1}-2\overline{x}{\displaystyle \sum _{i=1}^n{x}_i}$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}+{\left(\overline{x}\right)}^2{\displaystyle \sum _{i=1}^{n}1}-2\overline{x}\times n\overline{x}$ [using $n\overline{x}={\displaystyle \sum _{i=1}^n{x}_i}$ ]

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}+n-2n{\left(\overline{x}\right)}^2$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}-n{\left(\overline{x}\right)}^2$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}-n{\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$ [using $\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$ ]

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}-n\times \frac{1}{n^2}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$

  $={\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$

Hence proved that ${\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}={\displaystyle \sum _{i=1}^n{x}_i{}^2}-\frac{1}{n}{\left({\displaystyle \sum _{i=1}^n{x}_i}\right)}^2$ .