1. Summation question - Prove

Show that (Sorry about the lack of latex here)

[Σ from i=1 to n] of (x[i] - x(bar))^2 = [Σ from i=1 to n] of (x^2[i] - n*x^2(bar)

2. Hello, mitch_nufc!

We need to know the definition of the mean: . $\frac{\sum x_i}{n} \:=\: \overline{x}$

Show that .

$\sum^n_{i= 1} \left(x_i^2 - \overline{x}\right)^2 \;=\;\sum^n_{i=1}x_i^2 - n\!\cdot\!\overline{x}^2$

We have: . $\sum\left(x_i - \overline{x}\right)^2$

. . . $=\:\sum\bigg(x_i^2 - 2\,\overline{x}\,x_i + \overline{x}^2\bigg)$

. . . $= \;\sum x_i^2 - \sum 2\,\overline{x}\,x_i + \sum \overline{x}^2$

. . . $=\;\sum x_i^2 - 2\,\overline{x}\sum x_i + \overline{x}^2\sum 1$ .[1]

$\text{Multiply the middle sum by }\frac{n}{n}\!:\;\;\frac{n}{n}\bigg(-2\,\overline{x}\sum x_i\bigg) \;=\;-2n\,\overline{x}\cdot\overbrace{\left(\frac{\sum x_i}{n}\right)}^{\text{This is }\overline{x}} \;=\;-2n\cdot\overline{x}^2$

The last sum is: . $\overline{x}^2\sum 1 \:=\:\overline{x}^2\!\cdot\!n \:=\:n\!\cdot\!\overline{x}^2$

Then [1] becomes: . $\sum x_i^2 - 2n\!\cdot\!\overline{x}^2 + n\!\cdot\!\overline{x}^2 \;=\;\sum x_i^2 - n\!\cdot\!\overline{x}^2$ . . . There!