Is there a ternary operator in mathConsidering math or computer scienceWhat is a good language to develop in...
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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
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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
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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
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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 $x=1$): $a = u(b+c) + 1$
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– Alex
52 mins ago
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@Alex $a = b + 2 - u(c)$
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– eyeballfrog
35 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
$endgroup$
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 57 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
52 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
35 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
52 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
35 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
52 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
52 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
35 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
35 mins ago
add a comment |
3 Answers
3
active
oldest
votes
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Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
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Indicator is definitely the way to go, since the conditional can define an arbitrary set.
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– eyeballfrog
33 mins ago
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Thank you, that is very elegant. I will definitely try out all of the answers as I often use this in various scenarios.
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– dataphile
28 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.
$endgroup$
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.
$endgroup$
add a comment |
Your Answer
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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:$$a=b+1+1_{(-infty, 0]}(c)$$
$endgroup$
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
33 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
28 mins ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
$endgroup$
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
33 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
28 mins ago
add a comment |
$begingroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
$endgroup$
Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$
answered 47 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
33 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
28 mins ago
add a comment |
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
33 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
28 mins ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
33 mins ago
$begingroup$
Indicator is definitely the way to go, since the conditional can define an arbitrary set.
$endgroup$
– eyeballfrog
33 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
28 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
28 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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 45 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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 40 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
52 mins ago
$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
35 mins ago