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How to distinguish argument parentheses and grouping parentheses?

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2

$begingroup$

$$s(a+b)$$

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

share|cite|improve this question

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 | 

2

$begingroup$

$$s(a+b)$$

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

share|cite|improve this question

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

share|cite|improve this question

asked Mar 14 at 6:54

GqqnbigGqqnbig

275212

$endgroup$

share|cite|improve this question

asked Mar 14 at 6:54

GqqnbigGqqnbig

275212

share|cite|improve this question

asked Mar 14 at 6:54

GqqnbigGqqnbig

275212

asked Mar 14 at 6:54

GqqnbigGqqnbig

275212

asked Mar 14 at 6:54

GqqnbigGqqnbig

275212

1 Answer
1

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

edited Mar 14 at 8:10

answered Mar 14 at 7:10

ArthurArthur

119k7118202

$endgroup$

add a comment | 

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2

$begingroup$

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.

share|cite|improve this answer

edited Mar 14 at 8:10

answered Mar 14 at 7:10

ArthurArthur

119k7118202

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

edited Mar 14 at 8:10

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.

share|cite|improve this answer

edited Mar 14 at 8:10

answered Mar 14 at 7:10

ArthurArthur

119k7118202

$endgroup$

share|cite|improve this answer

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