Is there a ternary operator in math












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    40 mins ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    23 mins ago


















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    40 mins ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    23 mins ago
















1












1








1


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 45 mins ago









dataphiledataphile

315




315












  • $begingroup$
    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
    40 mins ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    23 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
    40 mins ago












  • $begingroup$
    @Alex $a = b + 2 - u(c)$
    $endgroup$
    – eyeballfrog
    23 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
40 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
40 mins ago














$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
23 mins ago






$begingroup$
@Alex $a = b + 2 - u(c)$
$endgroup$
– eyeballfrog
23 mins ago












3 Answers
3






active

oldest

votes


















4












$begingroup$

Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indicator is definitely the way to go, since the conditional can define an arbitrary set.
    $endgroup$
    – eyeballfrog
    21 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
    16 mins ago



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        21 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
        16 mins ago
















      4












      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        21 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
        16 mins ago














      4












      4








      4





      $begingroup$

      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$






      share|cite|improve this answer









      $endgroup$



      Using the indicator function notation:$$a=b+1+1_{(-infty, 0]}(c)$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 35 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
        21 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
        16 mins ago


















      • $begingroup$
        Indicator is definitely the way to go, since the conditional can define an arbitrary set.
        $endgroup$
        – eyeballfrog
        21 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
        16 mins ago
















      $begingroup$
      Indicator is definitely the way to go, since the conditional can define an arbitrary set.
      $endgroup$
      – eyeballfrog
      21 mins ago




      $begingroup$
      Indicator is definitely the way to go, since the conditional can define an arbitrary set.
      $endgroup$
      – eyeballfrog
      21 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
      16 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
      16 mins ago











      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 34 mins ago









          Paul ChildsPaul Childs

          3047




          3047























              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 28 mins ago









                  Ross MillikanRoss Millikan

                  298k23198371




                  298k23198371






























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