# sum of squares minimized

• Aug 7th 2006, 02:10 PM
TFT
Hello,
can any one help me in solving the folowing problem please???

show that if $\displaystyle x_1,...,x_n$ is a set of real numbers
such that
$\displaystyle \sum_{i=1,..N} {x_i=a}$
then the quantity
$\displaystyle \sum_{i=1,..N} {x_i^2}$
is minimized when $\displaystyle x_1=x_2=...=x_n=a/n$

the problem give some hint for solution:
assume that $\displaystyle x_i != x_j$
and show that this contradics the assumption that
$\displaystyle x_1,...,x_n$ minimize the sum of squares.to do this show that replacing $\displaystyle x_1 and x_j$ each by
$\displaystyle (x_i+x_j)/2$ reduces the sum of squares while insuring
$\displaystyle x_1+...+x_i_-_1+(x_i+x_j)/2+x_i_+_1$$\displaystyle +...+x_j_-_1+(x_i+x_j)/2+x_j_+_1+...+x_n=a thanks. • Aug 7th 2006, 02:22 PM ThePerfectHacker Quote: Originally Posted by TFT Hello, can any one help me in solving the folowing problem please??? i'm in hurry please help me if it's possible show that if \displaystyle x_1,...,x_n is a set of real numbers such that \displaystyle \sum_{i=1,..N} {x_i=a} then the quantity \displaystyle \sum_{i=1,..N} {x_i^2} is minimized when \displaystyle x_1=x_2=...=x_n=a/n You want to minimize, \displaystyle f(x_1,...,x_n)=x_1^2+x_2^2+...+x_n^2 With constraint curve, \displaystyle C(x_1,x_2,...,x_n)=x_1+...+x_n=a Use Lagrange multipliers, \displaystyle \nabla f=k\nabla g Thus, \displaystyle 2x_1(1,0,...,0)+2x_2(0,1,...,0)+...+2x_n(0,0,...,1 )$$\displaystyle =k(1,0,...,0)+k(0,1,...,0)+...+k(0,0,...,1)$
From here we see that,
$\displaystyle x_1=x_2=...=x_n=k/2$
Thus,
$\displaystyle x_1=...=x_n=\frac{a}{n}$

Take point $\displaystyle (a,0,...,0)$ and note that $\displaystyle f(a,0,...,0)\geq f(x_1,x_2,...,x_n)$. Therefore, the point we found was the minimum!