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Is there a ternary operator in math
Considering math or computer scienceWhat is a good language to develop in for simple, yet customizable math programs?Translate Programming code to MathMath vocab: operator on $S$ and into $S =$?A set system generated by a closure operator?Evaluating math proofs by computeris all math beyond arithmetic just advanced arithmetic?Are there dictionaries in math?What is a good route for a math student to self study computer science systematically and efficiently?Multiplying two numbers using only the “left shift” operator
$begingroup$
Is there a math equivalent of the programming ternary operator?
a = b + c > 0 ? 1 : 2
The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.
computer-science
asked 2 hours ago
dataphiledataphile
415
$endgroup$
add a comment |
$begingroup$
Is there a math equivalent of the programming ternary operator?
a = b + c > 0 ? 1 : 2
The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.
computer-science
asked 2 hours ago
dataphiledataphile
415
$endgroup$
$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago
add a comment |
$begingroup$
Is there a math equivalent of the programming ternary operator?
a = b + c > 0 ? 1 : 2
The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.
computer-science
asked 2 hours ago
dataphiledataphile
415
$endgroup$
Is there a math equivalent of the programming ternary operator?
a = b + c > 0 ? 1 : 2
The above meaning that if c is greater than 0 then a = b + 1 else a = b + 2.
computer-science
computer-science
asked 2 hours ago
dataphiledataphile
415
asked 2 hours ago
dataphiledataphile
415
asked 2 hours ago
dataphiledataphile
415
asked 2 hours ago
dataphiledataphile
415
asked 2 hours ago
dataphiledataphile
415
415
$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago
add a comment |
$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago
$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered 2 hours ago
Paul ChildsPaul Childs
3247
$endgroup$
add a comment |
$begingroup$
This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered 1 hour ago
FredHFredH
50629
$endgroup$
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 2 hours ago
Siong Thye GohSiong Thye Goh
102k1466118
102k1466118
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
add a comment |
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
$begingroup$
Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
$endgroup$
– dataphile
1 hour ago
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered 2 hours ago
Paul ChildsPaul Childs
3247
$endgroup$
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered 2 hours ago
Paul ChildsPaul Childs
3247
$endgroup$
add a comment |
$begingroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered 2 hours ago
Paul ChildsPaul Childs
3247
$endgroup$
In math, equations are written in piecewise form by having a curly brace enclose multiple lines; each one with a condition excepting the last which has "otherwise".
There are a few custom operators that also occasionally make an appearance. E.g. the Heavyside function mentioned by Alex, the Dirac function, and the cyclical operator $delta_{ijk}$ - all of which can be used to emulate conditional behaviour.
answered 2 hours ago
Paul ChildsPaul Childs
3247
answered 2 hours ago
Paul ChildsPaul Childs
3247
answered 2 hours ago
Paul ChildsPaul Childs
3247
answered 2 hours ago
Paul ChildsPaul Childs
3247
3247
add a comment |
add a comment |
$begingroup$
This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
add a comment |
$begingroup$
This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
$endgroup$
This is not a ternary operator, it is a function of two variables. There is one operation that results in $a$. You can certainly define a function
$$f(b,c)=begin {cases} b+1&c gt 0\
b+2 & c le 0 end {cases}$$
You can also define functions with any number of inputs you want, so you can define $f(a,b,c)=a(b+c^2)$, for example. This is a ternary function.
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
answered 2 hours ago
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
add a comment |
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered 1 hour ago
FredHFredH
50629
$endgroup$
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
add a comment |
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered 1 hour ago
FredHFredH
50629
$endgroup$
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
add a comment |
$begingroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered 1 hour ago
FredHFredH
50629
$endgroup$
In Concrete Mathematics by Graham, Knuth and Patashnik, the authors use the "Iverson bracket" notation: Square brackets around a statement represent $1$ if the statement is true and $0$ otherwise. Using this notation, you could write
$$
a = b + 2 - [c gt 0].
$$
answered 1 hour ago
FredHFredH
50629
edited 51 mins ago
answered 1 hour ago
FredHFredH
50629
answered 1 hour ago
FredHFredH
50629
answered 1 hour ago
FredHFredH
50629
50629
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
add a comment |
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
2
2
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
aka en.wikipedia.org/wiki/Iverson_bracket
$endgroup$
– qwr
56 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
@qwr Thank you! I did not recall the name.
$endgroup$
– FredH
54 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
$begingroup$
I would find it clearer to write $a = b + 1[c>0] + 2[cle0]$.
$endgroup$
– Rahul
46 mins ago
add a comment |
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$begingroup$
It probably depends on the form of the conditions and the results, but the example you gave can be expressed with the unit step $u(x)$ (with an appropriate definition at $x=1$): $a = u(b+c) + 1$
$endgroup$
– Alex
2 hours ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
2 hours ago