# 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 $x_1,...,x_n$ is a set of real numbers
such that
$
\sum_{i=1,..N} {x_i=a}
$

then the quantity
$\sum_{i=1,..N} {x_i^2}$
is minimized when $x_1=x_2=...=x_n=a/n$

the problem give some hint for solution:
assume that $x_i != x_j$
and show that this contradics the assumption that
$x_1,...,x_n$ minimize the sum of squares.to do this show that replacing $x_1 and x_j$ each by
$(x_i+x_j)/2$ reduces the sum of squares while insuring
$
x_1+...+x_i_-_1+(x_i+x_j)/2+x_i_+_1$
$+...+x_j_-_1+(x_i+x_j)/2+x_j_+_1+...+x_n=a$
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???

show that if $x_1,...,x_n$ is a set of real numbers
such that
$
\sum_{i=1,..N} {x_i=a}
$

then the quantity
$\sum_{i=1,..N} {x_i^2}$
is minimized when $x_1=x_2=...=x_n=a/n$

You want to minimize,
$f(x_1,...,x_n)=x_1^2+x_2^2+...+x_n^2$
With constraint curve,
$C(x_1,x_2,...,x_n)=x_1+...+x_n=a$
Use Lagrange multipliers,
$\nabla f=k\nabla g$
Thus,
$2x_1(1,0,...,0)+2x_2(0,1,...,0)+...+2x_n(0,0,...,1 )$ $=k(1,0,...,0)+k(0,1,...,0)+...+k(0,0,...,1)$
From here we see that,
$x_1=x_2=...=x_n=k/2$
Thus,
$x_1=...=x_n=\frac{a}{n}$

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