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How to distinguish argument parentheses and grouping parentheses?
Which parentheses are implied by $prod$?decimal digit grouping delimitersShould I use parentheses when writing $log$?Notation regarding different derivativesNotations: use of parentheses with “mod” and the “|” symbolNotation question, parenthesesHow do you write that a unique element exists in a collection?Products of factorials and parenthesesAbusing mathematical notation, are these examples of abuse?How do I notate the conditional combination of two sequences?
$begingroup$
$$s(a+b)$$
Does it mean s times a+b, or is s a funtion and take a+b as arguments?
$$(s(a+b))(c+d)$$
Does it mean s(a+b) times (c+d), or is s a high order function and s(a+b) return a function which then take c+d as arguments?
As a reader, how do I distinguish these cases?
As a writer, how can I make my formula super clear when it's needed?
I don't believe I'm the first one who find them confusing. I expect to see references/literatures. Thanks.
notation
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
$endgroup$
add a comment |
$begingroup$
$$s(a+b)$$
Does it mean s times a+b, or is s a funtion and take a+b as arguments?
$$(s(a+b))(c+d)$$
Does it mean s(a+b) times (c+d), or is s a high order function and s(a+b) return a function which then take c+d as arguments?
As a reader, how do I distinguish these cases?
As a writer, how can I make my formula super clear when it's needed?
I don't believe I'm the first one who find them confusing. I expect to see references/literatures. Thanks.
notation
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
$endgroup$
3
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57
add a comment |
$begingroup$
$$s(a+b)$$
Does it mean s times a+b, or is s a funtion and take a+b as arguments?
$$(s(a+b))(c+d)$$
Does it mean s(a+b) times (c+d), or is s a high order function and s(a+b) return a function which then take c+d as arguments?
As a reader, how do I distinguish these cases?
As a writer, how can I make my formula super clear when it's needed?
I don't believe I'm the first one who find them confusing. I expect to see references/literatures. Thanks.
notation
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
$endgroup$
$$s(a+b)$$
Does it mean s times a+b, or is s a funtion and take a+b as arguments?
$$(s(a+b))(c+d)$$
Does it mean s(a+b) times (c+d), or is s a high order function and s(a+b) return a function which then take c+d as arguments?
As a reader, how do I distinguish these cases?
As a writer, how can I make my formula super clear when it's needed?
I don't believe I'm the first one who find them confusing. I expect to see references/literatures. Thanks.
notation
notation
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
asked Mar 14 at 6:54
GqqnbigGqqnbig
275212
275212
3
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57
add a comment |
3
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57
3
3
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The notation for the different things you mention are completely identical, and yes, without context it is certainly ambiguous.
However, such expressions (should) never exist in a vacuum. Before (or possibly immediately after) such an expression is written, it should be made clear whether $s$ is a function or a number (and it should also be made clear what $a$ and $b$ are). And with such a clarification, the expression is no longer ambiguous.
That being said, $(a+b)s$ is less prone to being misunderstood, as most mathematicians apply functions to whatever is to the right, not to the left.
answered Mar 14 at 7:10
ArthurArthur
119k7118202
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
The notation for the different things you mention are completely identical, and yes, without context it is certainly ambiguous.
However, such expressions (should) never exist in a vacuum. Before (or possibly immediately after) such an expression is written, it should be made clear whether $s$ is a function or a number (and it should also be made clear what $a$ and $b$ are). And with such a clarification, the expression is no longer ambiguous.
That being said, $(a+b)s$ is less prone to being misunderstood, as most mathematicians apply functions to whatever is to the right, not to the left.
answered Mar 14 at 7:10
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
The notation for the different things you mention are completely identical, and yes, without context it is certainly ambiguous.
However, such expressions (should) never exist in a vacuum. Before (or possibly immediately after) such an expression is written, it should be made clear whether $s$ is a function or a number (and it should also be made clear what $a$ and $b$ are). And with such a clarification, the expression is no longer ambiguous.
That being said, $(a+b)s$ is less prone to being misunderstood, as most mathematicians apply functions to whatever is to the right, not to the left.
answered Mar 14 at 7:10
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
The notation for the different things you mention are completely identical, and yes, without context it is certainly ambiguous.
However, such expressions (should) never exist in a vacuum. Before (or possibly immediately after) such an expression is written, it should be made clear whether $s$ is a function or a number (and it should also be made clear what $a$ and $b$ are). And with such a clarification, the expression is no longer ambiguous.
That being said, $(a+b)s$ is less prone to being misunderstood, as most mathematicians apply functions to whatever is to the right, not to the left.
answered Mar 14 at 7:10
ArthurArthur
119k7118202
$endgroup$
The notation for the different things you mention are completely identical, and yes, without context it is certainly ambiguous.
However, such expressions (should) never exist in a vacuum. Before (or possibly immediately after) such an expression is written, it should be made clear whether $s$ is a function or a number (and it should also be made clear what $a$ and $b$ are). And with such a clarification, the expression is no longer ambiguous.
That being said, $(a+b)s$ is less prone to being misunderstood, as most mathematicians apply functions to whatever is to the right, not to the left.
answered Mar 14 at 7:10
ArthurArthur
119k7118202
edited Mar 14 at 8:10
answered Mar 14 at 7:10
ArthurArthur
119k7118202
answered Mar 14 at 7:10
ArthurArthur
119k7118202
answered Mar 14 at 7:10
ArthurArthur
119k7118202
119k7118202
add a comment |
add a comment |
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3
$begingroup$
Generally, if $s$ is a function then it would be defined before-hand. Also, only the bracket after the function is considered.
$endgroup$
– Mohammad Zuhair Khan
Mar 14 at 6:57