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Thread: sum of squares minimized

  1. #1
    TFT
    TFT is offline
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    Help me please

    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 $

    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.
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  2. #2
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    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!
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