How show a function in inputs of another function?












0












$begingroup$


I have defined function A as A: D*E --> F and function B as B: S*T --> R.



Now, I want to define function C which gets an instance of A functions and an instance of B functions as inputs and returns a real number.



What is the correct notation for this function?



Is the following notation correct?



C: A(.)*B(.) --> R










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  • $begingroup$
    To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
    $endgroup$
    – David K
    Dec 16 '18 at 1:11


















0












$begingroup$


I have defined function A as A: D*E --> F and function B as B: S*T --> R.



Now, I want to define function C which gets an instance of A functions and an instance of B functions as inputs and returns a real number.



What is the correct notation for this function?



Is the following notation correct?



C: A(.)*B(.) --> R










share|cite|improve this question









$endgroup$












  • $begingroup$
    To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
    $endgroup$
    – David K
    Dec 16 '18 at 1:11
















0












0








0





$begingroup$


I have defined function A as A: D*E --> F and function B as B: S*T --> R.



Now, I want to define function C which gets an instance of A functions and an instance of B functions as inputs and returns a real number.



What is the correct notation for this function?



Is the following notation correct?



C: A(.)*B(.) --> R










share|cite|improve this question









$endgroup$




I have defined function A as A: D*E --> F and function B as B: S*T --> R.



Now, I want to define function C which gets an instance of A functions and an instance of B functions as inputs and returns a real number.



What is the correct notation for this function?



Is the following notation correct?



C: A(.)*B(.) --> R







functions notation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 7:49









Tail of GodzillaTail of Godzilla

535




535












  • $begingroup$
    To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
    $endgroup$
    – David K
    Dec 16 '18 at 1:11




















  • $begingroup$
    To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
    $endgroup$
    – David K
    Dec 16 '18 at 1:11


















$begingroup$
To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 16 '18 at 1:11






$begingroup$
To see how to write your formulas so they look like mathematics rather than like some kind of programming language, you can start here: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 16 '18 at 1:11












2 Answers
2






active

oldest

votes


















0












$begingroup$

Mathematical function notation describes the domain and range of the function in the abstract. So when you have a function $$A: D*E rightarrow {mathbb R}$$ this is a function defined on some product of sets $D*E$ (the domain) taking values in the (ordinary) real numbers (the range).



If your function is taking the output of functions $A$ and $B$ then it is acting on two real numbers, and so $C:{mathbb R}times{mathbb R} rightarrow {mathbb R}$.



If, however, you have an operator $C$ that operates on the domains of functions $A$ and $B$, the domain of $C$ must be jointly the domains of $A$ and $B$. So with your notation,
$$ C: D*E times S*T rightarrow {mathbb R}$$
where $times$ is the Cartesian product of the two domains.



$C:A*B rightarrow {mathbb R}$ isn't right in general because $A$ and $B$ are functions, not sets. ("In general": sometimes notation is abused so that a set is defined by a function, e.g. $[f=alpha]$ to describe a hyperplane.)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since this is a mathematics site, let's use mathematical notation.



    There is no widely accepted mathematical notation that looks like this:
    A: D*E --> F.
    However, there is a notation that looks like
    $A: D times E to F,$
    which says that $A$ is a function to which we give an element of the set $D$ and an element of the set $E,$ upon which $A$ produces an element of the set $F.$



    Note that in this notation, $D,$ $E,$ and $F$ are sets that may contain many different elements, but $A$ is the name of a particular function.
    If you want to describe a function that takes other functions as input, it does not usually make sense to use the name $A$ as part of the "type" of the new function,
    any more than it makes sense to use the name of the number $3$ as part of the "type" of a function. What I mean is, you would not write
    $f: 3 times mathbb R to mathbb R.$
    Technically, you could write $f: {3} times mathbb R to mathbb R,$
    but that's rather silly; it says the first argument to the function must always be $3,$ which kind of defeats the purpose of giving the argument to the function in the first place.



    You certainly can have a function that acts on functions. Here is an example:
    $$ C: (mathbb R times mathbb R to mathbb R) times
    (mathbb R times mathbb R to mathbb R) to mathbb R.$$

    This says that $C$ is a function that takes two arguments, each of which itself is a functions that takes two real numbers arguments, and $C$ returns a real number as its result.
    As a more specific example, we could say that $C$ is the function such that
    $$ C(f,g) = f(0,0) + g(1,1) $$
    for any two functions $f: mathbb R times mathbb R to mathbb R$
    and $g: mathbb R times mathbb R to mathbb R.$






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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      0












      $begingroup$

      Mathematical function notation describes the domain and range of the function in the abstract. So when you have a function $$A: D*E rightarrow {mathbb R}$$ this is a function defined on some product of sets $D*E$ (the domain) taking values in the (ordinary) real numbers (the range).



      If your function is taking the output of functions $A$ and $B$ then it is acting on two real numbers, and so $C:{mathbb R}times{mathbb R} rightarrow {mathbb R}$.



      If, however, you have an operator $C$ that operates on the domains of functions $A$ and $B$, the domain of $C$ must be jointly the domains of $A$ and $B$. So with your notation,
      $$ C: D*E times S*T rightarrow {mathbb R}$$
      where $times$ is the Cartesian product of the two domains.



      $C:A*B rightarrow {mathbb R}$ isn't right in general because $A$ and $B$ are functions, not sets. ("In general": sometimes notation is abused so that a set is defined by a function, e.g. $[f=alpha]$ to describe a hyperplane.)






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Mathematical function notation describes the domain and range of the function in the abstract. So when you have a function $$A: D*E rightarrow {mathbb R}$$ this is a function defined on some product of sets $D*E$ (the domain) taking values in the (ordinary) real numbers (the range).



        If your function is taking the output of functions $A$ and $B$ then it is acting on two real numbers, and so $C:{mathbb R}times{mathbb R} rightarrow {mathbb R}$.



        If, however, you have an operator $C$ that operates on the domains of functions $A$ and $B$, the domain of $C$ must be jointly the domains of $A$ and $B$. So with your notation,
        $$ C: D*E times S*T rightarrow {mathbb R}$$
        where $times$ is the Cartesian product of the two domains.



        $C:A*B rightarrow {mathbb R}$ isn't right in general because $A$ and $B$ are functions, not sets. ("In general": sometimes notation is abused so that a set is defined by a function, e.g. $[f=alpha]$ to describe a hyperplane.)






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Mathematical function notation describes the domain and range of the function in the abstract. So when you have a function $$A: D*E rightarrow {mathbb R}$$ this is a function defined on some product of sets $D*E$ (the domain) taking values in the (ordinary) real numbers (the range).



          If your function is taking the output of functions $A$ and $B$ then it is acting on two real numbers, and so $C:{mathbb R}times{mathbb R} rightarrow {mathbb R}$.



          If, however, you have an operator $C$ that operates on the domains of functions $A$ and $B$, the domain of $C$ must be jointly the domains of $A$ and $B$. So with your notation,
          $$ C: D*E times S*T rightarrow {mathbb R}$$
          where $times$ is the Cartesian product of the two domains.



          $C:A*B rightarrow {mathbb R}$ isn't right in general because $A$ and $B$ are functions, not sets. ("In general": sometimes notation is abused so that a set is defined by a function, e.g. $[f=alpha]$ to describe a hyperplane.)






          share|cite|improve this answer









          $endgroup$



          Mathematical function notation describes the domain and range of the function in the abstract. So when you have a function $$A: D*E rightarrow {mathbb R}$$ this is a function defined on some product of sets $D*E$ (the domain) taking values in the (ordinary) real numbers (the range).



          If your function is taking the output of functions $A$ and $B$ then it is acting on two real numbers, and so $C:{mathbb R}times{mathbb R} rightarrow {mathbb R}$.



          If, however, you have an operator $C$ that operates on the domains of functions $A$ and $B$, the domain of $C$ must be jointly the domains of $A$ and $B$. So with your notation,
          $$ C: D*E times S*T rightarrow {mathbb R}$$
          where $times$ is the Cartesian product of the two domains.



          $C:A*B rightarrow {mathbb R}$ isn't right in general because $A$ and $B$ are functions, not sets. ("In general": sometimes notation is abused so that a set is defined by a function, e.g. $[f=alpha]$ to describe a hyperplane.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 15:10









          postmortespostmortes

          2,07531120




          2,07531120























              0












              $begingroup$

              Since this is a mathematics site, let's use mathematical notation.



              There is no widely accepted mathematical notation that looks like this:
              A: D*E --> F.
              However, there is a notation that looks like
              $A: D times E to F,$
              which says that $A$ is a function to which we give an element of the set $D$ and an element of the set $E,$ upon which $A$ produces an element of the set $F.$



              Note that in this notation, $D,$ $E,$ and $F$ are sets that may contain many different elements, but $A$ is the name of a particular function.
              If you want to describe a function that takes other functions as input, it does not usually make sense to use the name $A$ as part of the "type" of the new function,
              any more than it makes sense to use the name of the number $3$ as part of the "type" of a function. What I mean is, you would not write
              $f: 3 times mathbb R to mathbb R.$
              Technically, you could write $f: {3} times mathbb R to mathbb R,$
              but that's rather silly; it says the first argument to the function must always be $3,$ which kind of defeats the purpose of giving the argument to the function in the first place.



              You certainly can have a function that acts on functions. Here is an example:
              $$ C: (mathbb R times mathbb R to mathbb R) times
              (mathbb R times mathbb R to mathbb R) to mathbb R.$$

              This says that $C$ is a function that takes two arguments, each of which itself is a functions that takes two real numbers arguments, and $C$ returns a real number as its result.
              As a more specific example, we could say that $C$ is the function such that
              $$ C(f,g) = f(0,0) + g(1,1) $$
              for any two functions $f: mathbb R times mathbb R to mathbb R$
              and $g: mathbb R times mathbb R to mathbb R.$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since this is a mathematics site, let's use mathematical notation.



                There is no widely accepted mathematical notation that looks like this:
                A: D*E --> F.
                However, there is a notation that looks like
                $A: D times E to F,$
                which says that $A$ is a function to which we give an element of the set $D$ and an element of the set $E,$ upon which $A$ produces an element of the set $F.$



                Note that in this notation, $D,$ $E,$ and $F$ are sets that may contain many different elements, but $A$ is the name of a particular function.
                If you want to describe a function that takes other functions as input, it does not usually make sense to use the name $A$ as part of the "type" of the new function,
                any more than it makes sense to use the name of the number $3$ as part of the "type" of a function. What I mean is, you would not write
                $f: 3 times mathbb R to mathbb R.$
                Technically, you could write $f: {3} times mathbb R to mathbb R,$
                but that's rather silly; it says the first argument to the function must always be $3,$ which kind of defeats the purpose of giving the argument to the function in the first place.



                You certainly can have a function that acts on functions. Here is an example:
                $$ C: (mathbb R times mathbb R to mathbb R) times
                (mathbb R times mathbb R to mathbb R) to mathbb R.$$

                This says that $C$ is a function that takes two arguments, each of which itself is a functions that takes two real numbers arguments, and $C$ returns a real number as its result.
                As a more specific example, we could say that $C$ is the function such that
                $$ C(f,g) = f(0,0) + g(1,1) $$
                for any two functions $f: mathbb R times mathbb R to mathbb R$
                and $g: mathbb R times mathbb R to mathbb R.$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since this is a mathematics site, let's use mathematical notation.



                  There is no widely accepted mathematical notation that looks like this:
                  A: D*E --> F.
                  However, there is a notation that looks like
                  $A: D times E to F,$
                  which says that $A$ is a function to which we give an element of the set $D$ and an element of the set $E,$ upon which $A$ produces an element of the set $F.$



                  Note that in this notation, $D,$ $E,$ and $F$ are sets that may contain many different elements, but $A$ is the name of a particular function.
                  If you want to describe a function that takes other functions as input, it does not usually make sense to use the name $A$ as part of the "type" of the new function,
                  any more than it makes sense to use the name of the number $3$ as part of the "type" of a function. What I mean is, you would not write
                  $f: 3 times mathbb R to mathbb R.$
                  Technically, you could write $f: {3} times mathbb R to mathbb R,$
                  but that's rather silly; it says the first argument to the function must always be $3,$ which kind of defeats the purpose of giving the argument to the function in the first place.



                  You certainly can have a function that acts on functions. Here is an example:
                  $$ C: (mathbb R times mathbb R to mathbb R) times
                  (mathbb R times mathbb R to mathbb R) to mathbb R.$$

                  This says that $C$ is a function that takes two arguments, each of which itself is a functions that takes two real numbers arguments, and $C$ returns a real number as its result.
                  As a more specific example, we could say that $C$ is the function such that
                  $$ C(f,g) = f(0,0) + g(1,1) $$
                  for any two functions $f: mathbb R times mathbb R to mathbb R$
                  and $g: mathbb R times mathbb R to mathbb R.$






                  share|cite|improve this answer









                  $endgroup$



                  Since this is a mathematics site, let's use mathematical notation.



                  There is no widely accepted mathematical notation that looks like this:
                  A: D*E --> F.
                  However, there is a notation that looks like
                  $A: D times E to F,$
                  which says that $A$ is a function to which we give an element of the set $D$ and an element of the set $E,$ upon which $A$ produces an element of the set $F.$



                  Note that in this notation, $D,$ $E,$ and $F$ are sets that may contain many different elements, but $A$ is the name of a particular function.
                  If you want to describe a function that takes other functions as input, it does not usually make sense to use the name $A$ as part of the "type" of the new function,
                  any more than it makes sense to use the name of the number $3$ as part of the "type" of a function. What I mean is, you would not write
                  $f: 3 times mathbb R to mathbb R.$
                  Technically, you could write $f: {3} times mathbb R to mathbb R,$
                  but that's rather silly; it says the first argument to the function must always be $3,$ which kind of defeats the purpose of giving the argument to the function in the first place.



                  You certainly can have a function that acts on functions. Here is an example:
                  $$ C: (mathbb R times mathbb R to mathbb R) times
                  (mathbb R times mathbb R to mathbb R) to mathbb R.$$

                  This says that $C$ is a function that takes two arguments, each of which itself is a functions that takes two real numbers arguments, and $C$ returns a real number as its result.
                  As a more specific example, we could say that $C$ is the function such that
                  $$ C(f,g) = f(0,0) + g(1,1) $$
                  for any two functions $f: mathbb R times mathbb R to mathbb R$
                  and $g: mathbb R times mathbb R to mathbb R.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 1:09









                  David KDavid K

                  54.9k343120




                  54.9k343120






























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