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I am unclear about what the following summation means given that $\lambda_i: \forall i \in \{1,2,\ldots n\}$:

$\mu_{4:4} = \sum\limits_{i=1}^{4} \lambda_i + \mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2}) + \mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})$

I understand how this term expands:

$\sum\limits_{i=1}^{4} \lambda_i = \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4$.

But, I don't understand what how this term expands

$\mathop{\sum\sum}_{\substack{1\leq i_1 < i_2 \leq 4}}(\lambda_{i_1} + \lambda_{i_2})$

Nor do I understand how this term expands

$\mathop{\sum\sum\sum}_{\substack{1\leq i_1 < i_2 <i_3 \leq 4}}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3}) $

Any help in these matters would be appreciated.

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    $\begingroup$ Whenever I've seen summations like this, they just used one summation sign. $\endgroup$ – user2357112 supports Monica May 14 '14 at 5:55
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I understand how this term expands

$\sum\limits_{i=1}^{4} \lambda_i = \lambda_1 + \lambda_2 + \lambda_3 + \lambda_4$.

But, I don't understand what how this term expands

$\mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2})$

The subscript is just another way of indicating the domain of the indices.

Like so: $\displaystyle\quad\quad \sum\limits_{i=1}^{4} \lambda_i = \sum\limits_{1\leq i \leq 4} \lambda_i$

Thus, $1\leq i_1 < i_2 \leq 4$ means: $i_1\in[1\,..\,(i_2-1)], i_2\in[(i_1+1)\,..\,4]$

Hence: $$\mathop{\sum\sum}_{1\leq i_1 < i_2 \leq 4}(\lambda_{i_1} + \lambda_{i_2}) \\ = \sum\limits_{i_1=1}^{3}\left(\sum\limits_{i_2=i_1+1}^4(\lambda_{i_1} + \lambda_{i_2})\right) \\ = ((\lambda_1+\lambda_2)+(\lambda_1+\lambda_3)+(\lambda_1+\lambda_4))+((\lambda_2+\lambda_3)+(\lambda_2+\lambda_4))+((\lambda_3+\lambda_4)) \\ = 3(\lambda_1 +\lambda_2+\lambda_3+\lambda_4) \\ = 3\sum_{i=1}^4 \lambda_i$$

Nor do I understand how this term expands

$\mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})$

$$\mathop{\sum\sum\sum}_{1\leq i_1 < i_2 <i_3 \leq 4}(\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3}) \\ = \sum_{i_1=1}^2\left(\sum_{i_2=i_1+1}^{3}\left(\sum_{i_3=i_2+1}^{4} (\lambda_{i_1} + \lambda_{i_2} + \lambda_{i_3})\right)\right) \\ = (\lambda_1\!+\!\lambda_2\!+\!\lambda_3)\!+\!(\lambda_1\!+\!\lambda_2\!+\!\lambda_4)\!+\!(\lambda_1\!+\!\lambda_3\!+\!\lambda_4)\!+\!(\lambda_2\!+\!\lambda_3\!+\!\lambda_4) \\ = 3( \lambda_1 + \lambda_2+\lambda_3+\lambda_4)\\ = 3\sum_{i=1}^4 \lambda_i$$

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    $\begingroup$ Doesn't explain the summation very well but does add clever insight into simplification of the expression. $\endgroup$ – Zhuli May 14 '14 at 5:37
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The $\sum\sum_{1\le i_1\lt i_2\le 4} \ldots $ means that you take the sum over all pairs of $i_1$ and $i_2$ that satisfy the specified condition, namely that $$1\le i_1\lt i_2\le 4.$$ In this case that means that $i_1$ and $i_2$ should take the following pairs of values: $$\begin{array}{cc} i_1&i_2\\\hline 1&2\\1&3\\1&4\\2&3\\2&4\\3&4\end{array}$$ so the summation consists of 6 terms.

Similarly the triple sum is the sum over all triples $(i_1, i_2, i_3)$ for which all of $$1\le i_1\\ i_1\lt i_2\\ i_2\lt i_3\\i_3\le 4$$ all hold, so there are 4 terms in the sum:

$$\begin{array}{ccc} i_1 & i_2&i_3\\\hline 1 & 2 & 3 \\ 1 & 2 & 4 \\ 1 & 3 & 4 \\ 2 & 3 & 4 \end{array}$$

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    $\begingroup$ Thanks for your help - I understand the notation now. $\endgroup$ – PiE May 14 '14 at 3:05
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I would interpret ${\sum\sum}_{\substack{1\leq i_1<i_2\leq 4}}(\lambda_{i_1}+\lambda_{i_2})$ as $(\lambda_1+\lambda_2)+(\lambda_1+\lambda_3)+(\lambda_1+\lambda_4)+(\lambda_2+\lambda_3)+(\lambda_2+\lambda_4)+(\lambda_3+\lambda_4)$

I would interpret ${\sum\sum\sum}_{\substack{1\leq i_1 < i_2 <i_3 \leq 4}}(\lambda_{i_1}+\lambda_{i_2}+\lambda_{i_3})$ as $(\lambda_1+\lambda_2+\lambda_3)+(\lambda_1+\lambda_2+\lambda_4)+(\lambda_1+\lambda_3+\lambda_4)+(\lambda_2+\lambda_3+\lambda_4)$

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  • $\begingroup$ Usually I don't use so many $\Sigma$ notation but only one. $\endgroup$ – Y.H. Chan May 14 '14 at 2:54
  • $\begingroup$ In the first expression, the summation is done over two variables, hence there are two summation symbols. For the second expression, there are three variables, hence three symbols. $\endgroup$ – Joel Reyes Noche May 14 '14 at 2:55
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    $\begingroup$ Ok - I got this now - Thanks! $\endgroup$ – PiE May 14 '14 at 3:05
  • $\begingroup$ @UnemChan: The multiple sigmas indicates multiple summation indices are used. The wider space can also provide more room to list their criteria underneath them. $\endgroup$ – Graham Kemp May 14 '14 at 3:42

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