# Notation - Summation Over Multiple Elements in a Set

If $E$ is some set $E=\{x_1,x_2,\cdots,x_n,\cdots\}$ and $f:E\times \cdots \times E\to\mathbb{R}$ is a real valued function on $E$, and we have the sum,

$$\sum_{x_1,x_2,\cdots,x_n\in E}f(x_1,x_2,\cdots,x_n)$$

How would the above sum be computed? I'm not sure what it means to sum over multiple elements in a set at the same time.

• If $f: E \to \mathbb{R}$, images of $f$ look like $f(x)$ for $x \in E$, and something like $f(x_1, ..., x_n)$ has no defined meaning. Given the available information, the correct way to sum the images of $f$ over $E$ would be to write $\sum_{x \in E} f(x)$. The summation you ask about is meaningless and is most probably a typo.
– user
May 18, 2018 at 2:06
• Sorry I meant $f:E\times E \times \cdots \times E \to \mathbb{R}$. I've fixed that now. May 18, 2018 at 2:09
• In this case, elements of the set $E^n := E \times ... \times E$ would be tuples of the form $x:= (x_1, ..., x_n)$, your $f$ would know how to produce a single real number for each such tuple $x$, and the summation would just be $\sum_{x \in E^n} f(x)$ as usual. The notation, as it currently stands, is still incorrect.
– user
May 18, 2018 at 2:17

This notation means that the sum should be taken over all $n$-subsets $\{x_1,x_2,\dots,x_n\}$ in $E$.

For example, $\sum_{a,b \in E} ab$ is the sum of all products of pairs of elements from $E$.

If you like, you can rewrite the sum in the question as $$\frac{1}{n!} \sum_{x_1 \in E} \sum_{x_2 \in E\setminus\{x_1\}} \cdots \sum_{x_n \in E \setminus \{x_1,\dots,x_{n-1}\}} f(x_1,x_2,\dots,x_n).$$ (The $1/n!$ is to prevent counting $n$-tuples more than once, and we have to play with the sum bounds to prevent adding any $x_i$s to themselves.)

If you want to take the sum over each ordering of the subsets, then one would just remove the $1/n!$ factor - then each distinct ordering of each subset will be run through $f$ as well. This is probably what you want if $f$ isn't symmetric in its arguments.

• If you want to go from tuples to subsets, you would have to factor out $n!$ different permutations, not $n$. Also, the way you've written it, you aren't excluding when $x_i=x_j$ and $i\neq j$, so you will also get subsets of size less than $n$. May 18, 2018 at 2:21
• Usually, when you write $\sum_{x_1,x_2,\dots,x_n\in E} f(\dots)$, repeats are counted more than once. To prevent repeats, assuming $E$ had an order, you would write $\sum_{x_1\le x_2\le \dots \le x_n} f(\dots)$. As Kevin noted, $1/n$ does not correctly get rid of over-counting. May 18, 2018 at 3:43
• I've fixed it, not too sure about whether repeats should be counted or not.
– user562871
May 18, 2018 at 4:33