Explain this summation notation please

**AAKhan07** · July 12th 2009, 04:35 PM

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

$ a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 $

then

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

where

$ \alpha_n,\alpha_{n-1},...,\alpha_2,\alpha_1 $

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

**AlephZero** · July 12th 2009, 07:56 PM

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....

**HallsofIvy** · July 13th 2009, 09:02 AM

(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.)

$ \sum\limits_{j = 1}^n {\sum\limits_{k=j+1}^n {\alpha_j\alpha_k }}=\frac{a_{n-2}}{a_n} $
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 $\alpha$ s 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 $\alpha$ s" is the coefficient of $x^{n-2}$ . That's why that sum is $\frac{a_{n-2}}{a_n}$ .

**AlephZero** · July 13th 2009, 02:53 PM

Originally Posted by HallsofIvy

(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?

**HallsofIvy** · July 15th 2009, 04:10 AM

Originally Posted by AlephZero

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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