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I tried putting the series $$ |1+1+1|+|1+1-1|\\+|1-1+1|+|1-1-1|\\+|-1+1+1|+|-1+1-1|\\+|-1-1+1|+|-1-1-1| $$ into Wolfram Alpha and typing "in summation notation" but it wouldn't tell me what it is in summation notation. I tried to figure it out on my own but I can't figure out how to put this series into summation notation.

How would I express this sequence in summation notation?

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    $\begingroup$ Well, the easiest way to express this would be $12$. ;) $\endgroup$
    – EKons
    Commented Jul 27, 2017 at 13:44
  • $\begingroup$ $$\sum_{v\in C_3}\left|\sum_{i=1}^3(-1)^{v_i}\right|$$ where $v\in C_3$ runs over the vertices of the 3-dimensional unit cube. This generalizes to $$\sum_{v\in C_n}\left|\sum_{i=1}^n(-1)^{v_i}\right|$$ where $C_n$ is the n-dimensional unit hypercube :) Please don't ever actually do this. $\endgroup$ Commented Jul 27, 2017 at 17:41
  • $\begingroup$ this is not a series, just a (finite) sum. $\endgroup$
    – Masacroso
    Commented Jul 27, 2017 at 17:51

7 Answers 7

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$\sum_{i,j,k=0}^1 | (-1)^i + (-1)^j + (-1)^k |$

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A less formal, but common style, would be $$ \sum_{\epsilon_i = \pm 1} \lvert \epsilon_1+\epsilon_2+\epsilon_3 \rvert $$

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  • $\begingroup$ I'm not sure that I would recognize this expression outside the context of already knowing what it's supposed to mean. Kinda looks like $\Sigma_{\forall\epsilon_i\in \left\{-1,1\right\} }{\left|\epsilon_1+\epsilon_2+\epsilon_3\right|}$, where I'd reason that it'd sum $\epsilon_i$ for each value ${\pm}1$, though that inference would strike me as less obvious here. Does this style have a name, or a rule set? Or is it, as you'd stated, a more informal style that a practicing mathematician would just be liable to recognize? $\endgroup$
    – Nat
    Commented Jul 27, 2017 at 16:59
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    $\begingroup$ $\pm 1$ is one of those funny special cases where there are plenty of ways of writing it (see other answers). But I think most mathematicians would decode the notation with little trouble: subscripts are hard to read, so it is traditional to abbreviate quite significantly. Witness the product for the zeta function, which is often written as $\prod_p (1-p^{-s})^{-1}$ instead of $\prod_{p \text{ prime}} (\dots)$. I would say that I've never seen a $\forall$ sign in the sum: it's completely implicit. $\endgroup$
    – Chappers
    Commented Jul 27, 2017 at 18:00
  • $\begingroup$ (On the other hand, there are certainly plenty of mathematicians whom one would not accuse of writing mathematics that is easy to read...) $\endgroup$
    – Chappers
    Commented Jul 27, 2017 at 18:00
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There are several options. It's good to know many; different ideas are convenient in different situations.

Here is one that hasn't been suggested yet but I would recommend considering: $$ \sum_{a=\pm1}\sum_{b=\pm1}\sum_{c=\pm1}|a+b+c|. $$ This is quite similar to the one by Chappers, but the three sums are more explicit here. You can also consider replacing "$a=\pm1$" with "$a\in\{-1,+1\}$".

Here are some more options for the record: $$ \sum_{a,b,c=\pm1}|a+b+c|,\\ \sum_{a,b,c=-1,+1}|a+b+c|,\\ \sum_{a,b,c\in\{-1,+1\}}|a+b+c|,\\ \sum_{a=-1,+1}\sum_{b=-1,+1}\sum_{c=-1,+1}|a+b+c|. $$ As roger suggested in another answer, you can also let each index $i$ go from 0 to 1 and then sum $(-1)^i$. This leads to many more variations of the formulas above.

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$$\sum_{n=0}^{7}|(-1)^{[n/4]}+(-1)^{[n/2]}+(-1)^{n}|$$

where $[x]$ is the integer that satisfies $[x]\leq x < [x]+1$.

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    $\begingroup$ This is a clever one! It produces the correct sum, but it's a little hard to see it at a glance. I would argue that this is not a summation notation for the desired sum, but instead a sum that turns out to be the same after some manipulation or thought. $\endgroup$ Commented Jul 27, 2017 at 12:13
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    $\begingroup$ I disagree. This is just a binary expansion of @roger's version of the desired sum. $\endgroup$
    – Alex
    Commented Jul 27, 2017 at 15:10
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    $\begingroup$ @Alex This is nothing of no one's solution because this is the very first answer that this question got. There were not even comments around. $\endgroup$
    – Peyton
    Commented Jul 29, 2017 at 2:06
  • $\begingroup$ @Peyton I didn't mean that you stole it from roger-- when I said I disagree, I meant to disagree with Joonas's statement that this is not a summation notation for the desired sum. I meant to point out that this is a summation notation for the desired sum-- in fact, the same summation notation that the number 1 answerer, roger, used. Sorry for the ambiguity. I upvoted your answer btw. $\endgroup$
    – Alex
    Commented Aug 1, 2017 at 15:10
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$$\sum_{a=0}^1 \sum_{b=0}^1 \sum_{c=0}^1 |(-1)^a + (-1)^b + (-1)^c|$$

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$$ \sum_{X\in\{-1, +1\}^3} \vert \sum_{i = 0}^{2} X_i \vert $$

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$$\sum_{i,\ j,\ k\ =\ -1}^1 |ijk\left(i + j + k\right)|$$

$ijk$ vanishes when any of them are zero, allowing the summation to be written in the standard form.

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