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I am trying to obtain a formula for:

$G = m_1m_2 + m_1m_3 + m_2m_3$ using sigma notation.

I've come up with: $\sum_{i=1}^{3} \sum_{j=2}^{3} m_im_j = \sum_{i=1}^{3}m_im_2 + m_im_3 = m_1m_2 + m_1m_3 + m_2m_2 + m_2m_3 +m_3m_2 + m_3m_3$. My question is, how do I exclude the values $m_2m_2, m_3m_2, m_3m_3$ in order to obtain $G$? That is, how should I exclude the values where $i \geq j$? Also, is there any way that I can express the above sum "without" the double sigma?

3 Answers3

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A common notation is \begin{align*} \color{blue}{G }= m_1m_2 + m_1m_3 + m_2m_3\color{blue}{=\sum_{1\leq i<j\leq 3}m_im_j} \end{align*}

Markus Scheuer
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$\sum_{i=1}^2 \sum_{j=i+1}^3 m_im_j$ works.

coffeemath
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  • Doesn't this still include $m_2m_2$ within the sum (i.e. when i = j)? –  Feb 25 '21 at 02:56
  • If the sum included $m_2m_2$ that would imply $i=2,j=2$ going against $j$ starting at $i+1$ – coffeemath Feb 25 '21 at 02:59
  • @abrakadabra_01 [Forgot to ping you.... did you see my reaction above to your question about the term $m_2m_2$ ? – coffeemath Feb 25 '21 at 03:04
  • yes, thank you so much. I'm not too familiar with double sigma, so I am a bit confused. What does the inside sum $\sum_{j = i + 1}^{3} m_im_j$ evaluate to? My intuition was that $\sum_{j = i + 1}^{3} m_im_j$ = $\sum_{j = 2}^{3} m_im_j$ = $m_im_2 + m_im_3$ –  Feb 25 '21 at 03:06
  • @abrakadabra_01 The inside sum has its lower limit depending on $i$ so is not a definite thing. But one would first be putting $i=1$ when the inner sum would start at $m_1m_2.$ Then one would put $i=2$ and the inner sum would start (and end) when $j=3$ giving $m_2m_3.$ – coffeemath Feb 25 '21 at 03:11
  • In above comment in the $i=1$ case after $m_1m_2$ another term $m_1m_3$ would also be part of the $i=1$ case. – coffeemath Feb 25 '21 at 03:18
  • Right, that makes it more clear, thank you. So would it be correct to say $\sum_{i = 1}^{2}\sum_{j = i + 1}^{3} m_im_j = \sum_{j = 2}^{3} m_1m_j + \sum_{j = 3}^{3} m_2m_j = m_1m_2 + m_1m_3 + m_2m_3$? –  Feb 25 '21 at 03:19
  • Not quite. When evaluating a double sum you first take outer limit as small as it can be and at that point you know what the outer index is. Then go to the inner sum and do all terms for which outer index has its least value. Now we increment the outer index to the next value and so on. In your sum there's only two levels of this iterating. – coffeemath Feb 25 '21 at 03:31
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Without a double sigma, you can write $\sum_{i=1,j>i}^{i<3,j\le3}m_im_j=m_1m_2+m_1m_3+m_2m_3$.

Vons
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  • I think one needs an upper limit on $j$ for this method of single sum. [Easy to fix] – coffeemath Feb 25 '21 at 03:07
  • Is this the correct notation? Am I allowed to write $j > I$ in the lower limit without specifying what $j$ is? –  Feb 25 '21 at 03:10
  • @abrakadabra_01 I don't think it is standard notation, but people can certainly understand what they're trying to say. – Kenta S Feb 25 '21 at 03:15
  • @abrakadabra_01 That's what my (small) criticism was in my comment. To fix it one could put the two inequalities under each other: On top put $1 \le i,j \le 3$ and below that put $i<j.$ – coffeemath Feb 25 '21 at 03:15