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Math Help - Explain this summation notation please

  1. #1
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    Explain this summation notation please

    Hi, I'm reading the preliminary algebra section of 'Mathematical Methods for Physics and Engineering' and the book explain that for a polynomial

    <br />
a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0<br />

    then

    <br />
\sum\limits_{j = 1}^n {\sum\limits_{k>j}^n {\alpha_j\alpha_k }}=\frac{a_{n-2}}{a_n}<br />

    where

    <br />
\alpha_n,\alpha_{n-1},...,\alpha_2,\alpha_1<br />

    are the zeroes of the polynomial

    Please can someone explain the summation term to me, I thjnk I have an idea what it means.

    Can someone write out the summation explicitly in the case of n=3

    Thanks
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  2. #2
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    I believe what is wanted is the sum of all pairs \alpha_j\alpha_k where j\neq k:

    n=3 \implies \sum_{j=1}^n \sum_{k>j}^n \alpha_j \alpha_k = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3.

    There is a name for this, but it escapes me at the moment....
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  3. #3
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    (The "k> j" doesn't makes sense as a lower bound on a sum. I assume it was k= j+1 so that k goes from j+ 1 to n, always being larger than j.)

    <br />
\sum\limits_{j = 1}^n {\sum\limits_{k=j+1}^n {\alpha_j\alpha_k }}=\frac{a_{n-2}}{a_n}<br />
    means \sum_{j=1}^n \left[a_j \left(\sum_{k=j+1}^n a_n\right)\right]. Notice that the "inner sum" is in every term of the "outer sum".

    If n= 5, that would be a_1(a_2+ a_3+ a_4+ a_5)+ a_2(a_3+ a_4+ a_5)+ a_3(a_4+ a_5)+ a_4(a_5).
    If n= 3, a_1(a_3+ a_3)+ a_2(a_3)= a_1a_2+ a_1a_3+ a_2a_3.

    (I just noticed that my " a_i" is you " \alpha_i". Sorry about that.

    For example, if \alpha_1, \alpha_2, and \alpha_3 are the zeros of a cubic polynomial a_3x^3+ a_2x^2+ a_1x+ a_0, then the polynomial can be factored as a_n(x- \alpha_1)(x- \alpha_2)(x- \alpha_3) and multiplying that out gives a_nx^3- a_n(\alpha_1+ \alpha_2+ \alpha_3)x^2+ a_n(\alpha_1\alpha_2+ \alpha_1\alpha_3+ \alpha_2\alpha_3)x+ a_n\alpha_1\alpha_2\alpha_3. Notice that the coefficient of x is the sum of the products of two \alphas and that coefficient was a_1. That is, a_n(\alpha_1\alpha_2+ \alpha_1\alpha_3+ \alpha_2\alpha_3)= a_1 so \alpha_1\alpha_2+ \alpha_1\alpha_3+ \alpha_2\alpha_3 = \frac{a_1}{a_3}.

    You should be able to see that by multiplying a_n(x- \alpha_1)(x- \alpha_2)(x- \alpha_3)\cdot\cdot\cdot(x- \alpha_n), "sum of products of two \alphas" is the coefficient of x^{n-2}. That's why that sum is \frac{a_{n-2}}{a_n}.
    Last edited by HallsofIvy; July 13th 2009 at 10:25 AM.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    (The "k> j" doesn't makes sense as a lower bound on a sum. I assume it was k= j+1 so that k goes from j+ 1 to n, always being larger than j.)
    What about \sum_{j=1}^n\sum_{k>j}\alpha_j\alpha_k? This notation is commonly used. That's how I arrived at my original answer, which I believe is consistent with yours.

    I think this is called a Kronecker sum?
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  5. #5
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    Quote Originally Posted by AlephZero View Post
    What about \sum_{j=1}^n\sum_{k>j}\alpha_j\alpha_k? This notation is commonly used. That's how I arrived at my original answer, which I believe is consistent with yours.

    I think this is called a Kronecker sum?
    Actually, I take back what I said about that notation. It is perfectly valid. Don't know what I was thinking!
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