Is there a ternary operator in math
$begingroup$
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 53 mins ago
dataphiledataphile
365
$endgroup$
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 53 mins ago
dataphiledataphile
365
$endgroup$
$begingroup$
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
48 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins 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 53 mins ago
dataphiledataphile
365
$endgroup$
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 53 mins ago
dataphiledataphile
365
asked 53 mins ago
dataphiledataphile
365
asked 53 mins ago
dataphiledataphile
365
asked 53 mins ago
dataphiledataphile
365
asked 53 mins ago
dataphiledataphile
365
365
$begingroup$
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
48 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins 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
48 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins 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
48 mins 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
48 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 42 mins 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
29 mins 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
24 mins 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 41 mins ago
Paul ChildsPaul Childs
3147
$endgroup$
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 35 mins ago
Ross MillikanRoss Millikan
298k23198371
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 42 mins 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
29 mins 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
24 mins ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 42 mins 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
29 mins 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
24 mins ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 42 mins ago
Siong Thye GohSiong Thye Goh
102k1466118
$endgroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 42 mins ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 42 mins ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 42 mins ago
Siong Thye GohSiong Thye Goh
102k1466118
answered 42 mins 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
29 mins 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
24 mins ago
add a comment |
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
29 mins 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
24 mins ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
29 mins ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
29 mins 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
24 mins 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
24 mins 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 41 mins ago
Paul ChildsPaul Childs
3147
$endgroup$
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 41 mins ago
Paul ChildsPaul Childs
3147
$endgroup$
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 41 mins ago
Paul ChildsPaul Childs
3147
$endgroup$
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 41 mins ago
Paul ChildsPaul Childs
3147
answered 41 mins ago
Paul ChildsPaul Childs
3147
answered 41 mins ago
Paul ChildsPaul Childs
3147
answered 41 mins ago
Paul ChildsPaul Childs
3147
3147
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 35 mins 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 35 mins 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 35 mins 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 35 mins ago
Ross MillikanRoss Millikan
298k23198371
answered 35 mins ago
Ross MillikanRoss Millikan
298k23198371
answered 35 mins ago
Ross MillikanRoss Millikan
298k23198371
answered 35 mins ago
Ross MillikanRoss Millikan
298k23198371
298k23198371
add a comment |
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
48 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
31 mins ago