Chapter 15.10, Problem 1P

Let $m_1,m_2,\cdots ,m_n$ be a set of measurements, and define the values of $x_i$ by $x_1=$

$m_1-a,x_2=m_2-a,\cdots ,x_n=m_n-a,$ where $a$ is some number (as yet unspecified, but the same for all $x_i$ ). Show that in order to minimize ${\displaystyle \underset{i=1}{\overset{n}{\sum }}{x}_i^2},$ we should choose $a=(1/n){\displaystyle \underset{i=1}{\overset{n}{\sum }}{m}_i}.$ Hint: Differentiate ${\displaystyle \underset{i=1}{\overset{n}{\sum }}{x}_i^2}$ with respect to a. You have shown that the arithmetic mean is the "best" average in the least squares sense, that is, that if the sum of the squares of the deviations of the measurements from their "average" is a minimum, the "average" is the arithmetic mean (rather than, say, the median or mode).