Sum of Squared Deviations from Mean/Proof 1

Theorem

Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.

Let $\overline x$ denote the arithmetic mean of $S$.

Then:

$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n \paren { {x_i}^2 - {\overline x}^2}$

For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.

Then:

$\ds \sum \paren {x_i - \overline x}^2$

$=$

$\ds \sum \paren {x_i - \overline x} \paren {x_i - \overline x}$

$\ds $

$=$

$\ds \sum x_i \paren {x_i - \overline x} - \overline x \sum \paren {x_i - \overline x}$

$\ds $

$=$

$\ds \sum x_i \paren {x_i - \overline x} - 0$

$\ds $

$=$

$\ds \sum x_i \paren {x_i - \overline x} + 0$

$\ds $

$=$

$\ds \sum x_i \paren {x_i - \overline x} + \overline x \sum \paren {x_i - \overline x}$

$\ds $

$=$

$\ds \sum \paren {x_i + \overline x} \paren {x_i - \overline x}$

$\ds $

$=$

$\ds \sum \paren {x_i^2 - \overline x^2}$

$\blacksquare$