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Math Help - Summation proof

  1. #1
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    Summation proof

    Prove that n*\sum_{i=0}^nx_i^2 \geq (\sum_{i=0}^nx_i)^2

    I tried proving it directly using summation identities. I then tired using induction, but I kept getting stuck.

    A proof similar to this may be in the forum, but it is difficult to search for. Thanks.

    -Chops
    Last edited by duchops; February 10th 2010 at 11:27 AM.
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  2. #2
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    You cannot prove this, as it is false. However, you can prove

    n*\sum_{i=1}^nx_i^2 \geq (\sum_{i=1}^nx_i)^2

    by using the following fact repeatedly:

    (x_2 - x_1)^2 \geq 0

    x_1^2 - 2 x_1 x_2 + x_2^2 \geq 0

    x_1^2 + x_2^2 \geq 2x_1 x_2
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  3. #3
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    Quote Originally Posted by duchops View Post
    Prove that n*\sum_{i=0}^nx_i^2 \geq (\sum_{i=0}^nx_i)^2

    I tried proving it directly using summation identities. I then tired using induction, but I kept getting stuck.

    A proof similar to this may be in the forum, but it is difficult to search for. Thanks.

    -Chops
    \sum_{i=0}^nx_i=\frac{n(n+1)}{2}\ and\ \sum_{i=0}^n{x_i}^2=\frac{n(n+1)(2n+1)}{6}

    n\sum_{i=0}^nx_i^2=\frac{n^2(n+1)(2n+1)}{6}\



    \left(\sum_{i=0}^nx_i\right)^2=\frac{n^2(n+1)(n+1)  }{4}

    \frac{n^2(n+1)(2n+1)}{6}\ \ge\ \frac{n^2(n+1)(n+1)}{4} ?

    \frac{n^2(n+1)}{2}\ is\ common

    \frac{2n+1}{3}\ \ge\ \frac{n+1}{2} ?

    \frac{4n+2}{6}\ \ge\ \frac{3n+3}{6} ?

    4n+2\ \ge\ 3n+3 ?

    n\ \ge\ 1 ?

    The inequality is true for all natural numbers n from 1 up, and for all natural numbers i. It does not matter if you sum from i=0 or from i=1, since the first term in both sums is zero.
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  4. #4
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    Ah, sorry. Had the limits of the sum incorrect. Thanks for the tip.
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  5. #5
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    Archie, your analysis is incorrect. Check the sums in the problem again. We are not given the values of the x_i.
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  6. #6
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    Yes, it assumes the terms are consequtive natural numbers from 0 to n,
    nothing more,
    if not, another way is necessary.
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  7. #7
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    No, it says \sum_{i=1}^nx_i not \sum_{i=1}^ni

    Icemanfan is correct.
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  8. #8
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    Yep,
    what i wrote takes the terms as a consequtive series of natural numbers from 0 to n. The question is asking to prove it for n arbitrary terms then ?
    ok, I wrongly assumed it was your way of writing the sequence of natural numbers. I'll take another shot in a while.
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  9. #9
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    \sum_{i=0}^nx_i^2=x_0^2+x_1^2+x_2^2+....+x_n^2

    \left(\sum_{i=0}^nx_i\right)^2=\left(x_0+x_1+x_2+.  ...+x_n\right)\left(x_0+x_1+x_2+....+x_n\right)

    (a+b)(a+b)=a^2+b^2+2ab

    2(a^2+b^2)=a^2+b^2+a^2+b^2

    a^2+b^2\ is\ common

    Is\ a^2+b^2\ \ge\ 2ab\ ?\ Yes,\ since\ (a-b)^2=a^2+b^2-2ab\ \ge\ 0

    (a+b+c)^2=(a+b+c)(a+b+c)=a^2+ab+ac++ba+b^2+bc+ca+c  b+c^2 =a^2+b^2+c^2+2ab+2ac+2bc

    3(a^2+b^2)=a^2+b^2+c^2+[a^2+b^2]+[a^2+c^2]+[b^2+c^2]

    Is\ a^2+b^2\ \ge\ 2ab\ ?\ Yes
    Is\ a^2+c^2\ \ge\ 2ac\ ?\ Yes
    Is\ b^2+c^2\ \ge\ 2ac\ ?\ Yes

    (a+b+c+d)^2=(a+b+c+d)(a+c+b+d)
    =a^2+ab+ac+ad+ba+b^2+bc+bd+ca+cb+c^2+cd+da+db+dc+d  ^2
    =a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd

    4(a^2+b^2+c^2+d^2)=a^2+b^2+c^2+d^2+(a^2+b^2)+(a^2+  c^2)+ (a^2+d^2)+(b^2+c^2)+(b^2+d^2)+(c^2+d^2)

    a^2+b^2\ \ge\ 2ab
    a^2+c^2\ \ge\ 2ac
    a^2+d^2\ \ge\ 2ad
    b^2+c^2\ \ge\ 2bc
    b^2+d^2\ \ge\ 2bd
    c^2+d^2\ \ge\ 2cd

    This pattern is consistent.
    Last edited by Archie Meade; February 11th 2010 at 05:34 AM.
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  10. #10
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    Wait, that's not finished!
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  11. #11
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    Hi duchops,
    there is a general pattern to the inequality.
    The above post is another way to view it.

    Actually, that is the same thing icemanfan said,
    so yes, he's quite right.
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