Question

Show that the sum of the squares of the derivations of a set of values is minimum when taken about mean.

Answer 1

Let the value be a which brings the minimum value ;

$f\left(x\right)=(x_1-a{)}^2+(x_2-a{)}^2+...\dots +(x_n-a{)}^2$

$⟹f^{\prime }\left(x\right)=2\left[\right(x_1-a)+(x_2-a)+...\dots .+(x_n-a\left)\right]=0⟹(x_1+x_2+....+x_n)=na$

$⟹a=\frac{(x_1+x_2+.....+x_n)}{n}$

$\therefore$ hence proved minimum is from mean.