1
$\begingroup$

In set notation could somebody explain the meaning of $\mid$ in the equation below please? How does it read?

I read it as $s$ and $j$ are an element of $E$ but what does the $j \mid$ mean?

summation

Similarly in this equation what does the comma mean?

flow

$\endgroup$
  • 3
    $\begingroup$ Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$. $\endgroup$ – Steven Wagter Dec 26 '18 at 12:48
  • 2
    $\begingroup$ See Set-builder notation. $\endgroup$ – Mauro ALLEGRANZA Dec 26 '18 at 12:53
  • 1
    $\begingroup$ @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.) $\endgroup$ – John Bentin Dec 26 '18 at 13:03
2
$\begingroup$

Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) \subset I \subset \Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then

$$Inputs(i) = \sum_{j|(j,i) \in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) \in E$.

Here $E$ is some subset of $\Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) \in E$.

In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.

EDIT - in response to Shaun:

$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2\times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = \{ (1,1),(1,2),(2,1),(2,2)\}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.

For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = \{(1,1),(2,1),(2,2)\}$, then $Inputs(2) = m_{2,2}$.

Another example of indexing sets: $$1/2+1/4+1/8 +... = \sum_{n \in \Bbb{N} } \frac{1}{2^n}$$

Here $\Bbb{N}$ is the indexing set.

$\endgroup$
  • $\begingroup$ Where did the $I$ come from? Don't you mean $E$? $\endgroup$ – Shaun Dec 28 '18 at 5:32
  • 1
    $\begingroup$ @Shaun added some clarification, does it make more sense now? $\endgroup$ – Steven Wagter Dec 28 '18 at 8:42
  • $\begingroup$ Yes. Thank you, @StevenWagter :) $\endgroup$ – Shaun Dec 28 '18 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.