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Math Help - 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  x_1,...,x_n is a set of real numbers
    such that
    <br />
\sum_{i=1,..N} {x_i=a}<br />
    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
    <br />
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.
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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  x_1,...,x_n is a set of real numbers
    such that
    <br />
\sum_{i=1,..N} {x_i=a}<br />
    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!
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