Answer to Question #168940 in Calculus for Vishal

Question

If a1< ...... <an . Find the minimum value of f(x) = (x - ai)² summation from i=1 to n

Answer 1

Suppose $a_1<a_2<⋯<a_n$ . Then on each interval $(a_k,a_{k+1}),$

$f(x)=∑_{i=1}^k(a_i−x)+∑_{i=k+1}^n(x−a_i)=(n−2k)x+∑_{i=1}^kai−∑_{i=k+1}^na_i$

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

We see 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:

$−a_1−⋯−a_p+a_{p+1}+⋯+a_{2p}.$