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I understand multiple summations, but I do not know how to do this summation. Could someone break this down for me?

For example: $$\sum_{ 1 \leq i < j \leq 3 }(2i+j). $$

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  • $\begingroup$ You take $i$ and $j$ to be all $i$ and $j$ that satisfy the inequality. $\endgroup$ – 3x89g2 Aug 25 '16 at 1:17
  • $\begingroup$ So like i=1 j=2 you plug that in get 4. Then, i=2 j=3 you plug that in get 7. The whole summation is equal to 4+7=11 then. $\endgroup$ – Will Aug 25 '16 at 1:24
  • $\begingroup$ Think about the double integral $\iint_{0 \leq x < y \leq 1} (2x+y)\,dx\,dy$. Does that bother you, or do you understand what it means? $\endgroup$ – KCd Aug 25 '16 at 1:52
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With two parameters $i$ and $j$ under the sum, this is a double sum. It means you have to find all $i$ and $j$ that are $1$, $2$, or $3$, and also $i < j$. Thus, the possible pairs are $$(i, j) \in \{(1, 2), (1, 3), (2, 3)\}.$$ Then, you plug in each pair into the expression $2i + j$ and add it all up: $$[2(1) + 2] + [2(1) + 3] + [2(2) + 3] = 4 + 5 + 7 = 16.$$

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  • $\begingroup$ Intuitively, I read i and j to be integers, but is this somewhere specified? Could the sum alternativly sum over the (infinite) i, j pairs where i < j and i and j are floats? $\endgroup$ – René Nyffenegger Aug 25 '16 at 12:28
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    $\begingroup$ @RenéNyffenegger I think at that point it would have been written as an integral. $\endgroup$ – Kyle Aug 25 '16 at 13:18
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    $\begingroup$ @RenéNyffenegger: "Float" isn't the correct term here, it's a CS term referring to a specific way to represent numbers. "Real numbers" would be more appropriate. And yes, in the most abstract notion of summation you are correct - the fact that $i,j \in\mathbb{Z}$ should be explicitly mentioned. But in common usage it is assumed implicitly, and in any case summation isn't defined for uncountably many nonzero summands - integrals and measure theory are used instead for the continuous case. $\endgroup$ – Meni Rosenfeld Aug 25 '16 at 14:39
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I think it is clearer to call $a_{ij} = 2i+j$ to explain this. We must sum over all pairs $(i,j)$ satisfying $1\leq i < j \leq 3$, that is: $$\sum_{ 1\leq i < j \leq 3 } a_{ij} = a_{12} + a_{13} + a_{23} = 4 + 5 + 7 = 16. $$

In general you can rewrite it as a double summation: $$\sum_{1 \leq i < j\leq n} a_{ij} = \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_{ij}.$$

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This notation is telling you to take the sum over all terms $2i + j$ where $1 \leq i \lt j \leq 3$. That being said, let's just write it out.

$$ \sum_{1 \leq i \lt j \leq 3}2i + j = (2\cdot 1 + 2) + (2\cdot 1 + 3) + (2\cdot 2 + 3). $$

In this case, there's only three terms to sum. Note that we take the sum over all ordered pairs $(i, j)$ where $1 \leq i \lt j \leq 3$.

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As has been stated, this is a double sum.

If we let the upper limit be $n$ instead of $3$, it can be explicitly converted into a double sum and then evaluated like this:

$\begin{array}\\ \sum_{ 1 \leq i < j \leq n }(2i+j) &=\sum_{1 < j \le n}\sum_{1\le i < j}(2i+j)\\ &=\sum_{j=2}^{ n}\sum_{i=1}^{j-1}(2i+j)\\ &=\sum_{j=2}^{ n}\sum_{i=1}^{j-1}(2i)+\sum_{j=2}^{n}\sum_{i=1}^{j-1}(j)\\ &=\sum_{j=2}^{ n}2\dfrac{(j-1)j}{2}+\sum_{j=2}^{n}j(j-1)\\ &=2\sum_{j=2}^{n}j(j-1)\\ \end{array} $

and this you should be able to evaluate.

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    $\begingroup$ Why do you need to convert the upper limit to n? Can't you do the exact same thing when the upper limit is 3? $\endgroup$ – Chris Aug 25 '16 at 11:18

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