Question

If a_(1)<**<a_(n) find the minimum value of f(x)=sum_(i=1)^(n)(x-a_(i))^(2) .

Answer 1

On each interval $(a_k,a_{k+1})$ :

$f(x)=\displaystyle\sum_{i=1}^k(a_i-x)+\displaystyle\sum_{i=k+1}^k(x-a_i)=(n-2k)x+\displaystyle\sum_{i=1}^ka_i-\displaystyle\sum_{i=k+1}^ka_i$

Thus, $f(x)$ is decreasing on $(a_k,a_{k+1})$ as long as $n-2k>0$, i.e. $k<n/2$, increasing as soon as $k>n/2$.

There are two cases:

If $n$ is odd, $n=2p+1$, there is a unique minimum, attained at the middle point $a_{p+1}$, and its value is:

$f(a_{n+1})=-a_1-...-a_n+a_{n+2}+...+a_{2p+1}$

If $n$ is even, $n=2p$, the function attains its minimum on the middle interval $[a_p,a_{p+1}]$ and its value is:

$-a_1-...-a_p+a_{p+1}+...+a_{2p}$