Find a value which minimizes a given sum

Let $a_1, \ldots, a_n \in \mathbb{R}$ . Prove

$\sum_{k=1}^n (x-a_k)^2$

is minimal when $x$ is the arithmetic mean of $a_1, \ldots, a_n$ .

Proof. Recall that the arithmetic mean of $a_1, \ldots, a_n \in \mathbb{R}$ is given by

$x = \frac{a_1 + \cdots + a_n}{n}.$

Now, consider the given sum

$\sum_{k=1}^n (x-a_k)^2.$

To find the minimum we first take the derivative, using the linearity of the derivative over this (finite) sum, (i.e., using that $(f+g)' = f' + g'$ ),

$\begin{align*} \left( \sum_{k=1}^n (x-a_k)^2 \right)' &= \sum_{k=1}^n ((x - a_k)^2)' \\ &= \sum_{k=1}^n 2(x-a_k). \end{align*}$

Setting this equal to 0 and solving,

$\begin{align*} \sum_{k=1}^n 2(x - a_k) = 0 && \implies && 2 \left(\sum_{k=1}^n x - \sum_{k=1}^n a_k \right) &= 0 \\ && \implies && nx - \sum_{k=1}^n a_k &= 0 \\ && \implies && nx &= \sum_{k=1}^n a_k \\ && \implies && x &= \frac{a_1 + \cdots + a_n}{n}. \qquad \blacksquare \end{align*}$

Stumbling Robot

Find a value which minimizes a given sum

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