Set notation what does the bar symbol mean?
$begingroup$
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?
Similarly in this equation what does the comma mean?
elementary-set-theory notation
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
$endgroup$
add a comment |
$begingroup$
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?
Similarly in this equation what does the comma mean?
elementary-set-theory notation
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
$endgroup$
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
add a comment |
$begingroup$
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?
Similarly in this equation what does the comma mean?
elementary-set-theory notation
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
$endgroup$
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?
Similarly in this equation what does the comma mean?
elementary-set-theory notation
elementary-set-theory notation
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
edited Dec 26 '18 at 12:53
Mauro ALLEGRANZA
66.5k449115
66.5k449115
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
asked Dec 26 '18 at 12:45
cherry aldicherry aldi
203
203
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
add a comment |
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
3
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
$begingroup$
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
2
$begingroup$
See Set-builder notation.
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 12:53
$begingroup$
See Set-builder notation.
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 12:53
1
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
$begingroup$
@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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $2times2$-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.
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $2times2$-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.
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $2times2$-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.
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $2times2$-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.
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
$endgroup$
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 $2times2$-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.
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
edited Dec 28 '18 at 8:55
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
answered Dec 26 '18 at 13:34
Steven WagterSteven Wagter
1569
1569
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
Where did the $I$ come from? Don't you mean $E$?
$endgroup$
– Shaun
Dec 28 '18 at 5:32
$begingroup$
Where did the $I$ come from? Don't you mean $E$?
$endgroup$
– Shaun
Dec 28 '18 at 5:32
1
1
$begingroup$
@Shaun added some clarification, does it make more sense now?
$endgroup$
– Steven Wagter
Dec 28 '18 at 8:42
$begingroup$
@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
$begingroup$
Yes. Thank you, @StevenWagter :)
$endgroup$
– Shaun
Dec 28 '18 at 8:44
add a comment |
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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