Composition of two multivariate functions
$begingroup$
Lets say I have 2 multivariate functions:
f(x,y) = x - y
g(x,y) = x + y
How do I get the composition of these 2 functions $g(f(x,y))$ ?
functions
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
$endgroup$
add a comment |
$begingroup$
Lets say I have 2 multivariate functions:
f(x,y) = x - y
g(x,y) = x + y
How do I get the composition of these 2 functions $g(f(x,y))$ ?
functions
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
$endgroup$
1
$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
1
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36
add a comment |
$begingroup$
Lets say I have 2 multivariate functions:
f(x,y) = x - y
g(x,y) = x + y
How do I get the composition of these 2 functions $g(f(x,y))$ ?
functions
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
$endgroup$
Lets say I have 2 multivariate functions:
f(x,y) = x - y
g(x,y) = x + y
How do I get the composition of these 2 functions $g(f(x,y))$ ?
functions
functions
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
edited Dec 10 '18 at 20:32
Key Flex
8,28261233
8,28261233
asked Dec 10 '18 at 20:31
KemalKemal
32
asked Dec 10 '18 at 20:31
KemalKemal
32
asked Dec 10 '18 at 20:31
KemalKemal
32
32
1
$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
1
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36
add a comment |
1
$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
1
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36
1
1
$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
1
1
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36
add a comment |
2 Answers
2
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oldest
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$begingroup$
$g(x,y)$ takes in $2$ arguments: $x$ and $y$. $f(x,y)$ is just one value, equal to $x-y$. Therefore, $g(f(x,y))$ doesn't make sense. However, $g(f(x,y,),0), g(0,f(x,y))$ etc. are fine.
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
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add a comment |
$begingroup$
The composition $g(f(x,y))$ as you have written it is not well-defined, since $f:mathbb R^2mapstomathbb R$ and $g:mathbb R^2mapsto mathbb R$.
Suppose we want to compose the functions $a:Amapsto A'$ and $b:Bmapsto B'$. The composition $acirc b$ is defined iff $B'subset A$, and the composition $bcirc a$ is defined iff $A'subset B$. Thus, you cannot compose two functions mapping $mathbb R^2$ to $mathbb R$.
However, you might define composition in a different way to suit your needs; for example, you might consider $g(f(x,y),f(x,y))$.
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$g(x,y)$ takes in $2$ arguments: $x$ and $y$. $f(x,y)$ is just one value, equal to $x-y$. Therefore, $g(f(x,y))$ doesn't make sense. However, $g(f(x,y,),0), g(0,f(x,y))$ etc. are fine.
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
$g(x,y)$ takes in $2$ arguments: $x$ and $y$. $f(x,y)$ is just one value, equal to $x-y$. Therefore, $g(f(x,y))$ doesn't make sense. However, $g(f(x,y,),0), g(0,f(x,y))$ etc. are fine.
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
$endgroup$
add a comment |
$begingroup$
$g(x,y)$ takes in $2$ arguments: $x$ and $y$. $f(x,y)$ is just one value, equal to $x-y$. Therefore, $g(f(x,y))$ doesn't make sense. However, $g(f(x,y,),0), g(0,f(x,y))$ etc. are fine.
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
$endgroup$
$g(x,y)$ takes in $2$ arguments: $x$ and $y$. $f(x,y)$ is just one value, equal to $x-y$. Therefore, $g(f(x,y))$ doesn't make sense. However, $g(f(x,y,),0), g(0,f(x,y))$ etc. are fine.
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
answered Dec 10 '18 at 20:41
Shubham JohriShubham Johri
5,192717
5,192717
add a comment |
add a comment |
$begingroup$
The composition $g(f(x,y))$ as you have written it is not well-defined, since $f:mathbb R^2mapstomathbb R$ and $g:mathbb R^2mapsto mathbb R$.
Suppose we want to compose the functions $a:Amapsto A'$ and $b:Bmapsto B'$. The composition $acirc b$ is defined iff $B'subset A$, and the composition $bcirc a$ is defined iff $A'subset B$. Thus, you cannot compose two functions mapping $mathbb R^2$ to $mathbb R$.
However, you might define composition in a different way to suit your needs; for example, you might consider $g(f(x,y),f(x,y))$.
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
$endgroup$
add a comment |
$begingroup$
The composition $g(f(x,y))$ as you have written it is not well-defined, since $f:mathbb R^2mapstomathbb R$ and $g:mathbb R^2mapsto mathbb R$.
Suppose we want to compose the functions $a:Amapsto A'$ and $b:Bmapsto B'$. The composition $acirc b$ is defined iff $B'subset A$, and the composition $bcirc a$ is defined iff $A'subset B$. Thus, you cannot compose two functions mapping $mathbb R^2$ to $mathbb R$.
However, you might define composition in a different way to suit your needs; for example, you might consider $g(f(x,y),f(x,y))$.
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
$endgroup$
add a comment |
$begingroup$
The composition $g(f(x,y))$ as you have written it is not well-defined, since $f:mathbb R^2mapstomathbb R$ and $g:mathbb R^2mapsto mathbb R$.
Suppose we want to compose the functions $a:Amapsto A'$ and $b:Bmapsto B'$. The composition $acirc b$ is defined iff $B'subset A$, and the composition $bcirc a$ is defined iff $A'subset B$. Thus, you cannot compose two functions mapping $mathbb R^2$ to $mathbb R$.
However, you might define composition in a different way to suit your needs; for example, you might consider $g(f(x,y),f(x,y))$.
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
$endgroup$
The composition $g(f(x,y))$ as you have written it is not well-defined, since $f:mathbb R^2mapstomathbb R$ and $g:mathbb R^2mapsto mathbb R$.
Suppose we want to compose the functions $a:Amapsto A'$ and $b:Bmapsto B'$. The composition $acirc b$ is defined iff $B'subset A$, and the composition $bcirc a$ is defined iff $A'subset B$. Thus, you cannot compose two functions mapping $mathbb R^2$ to $mathbb R$.
However, you might define composition in a different way to suit your needs; for example, you might consider $g(f(x,y),f(x,y))$.
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
answered Dec 10 '18 at 20:36
FrpzzdFrpzzd
23k841109
23k841109
add a comment |
add a comment |
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$begingroup$
Both of these functions map $mathbb R^2$ into $mathbb R$ (I would guess) , so $g circ f$ does not make sense, since the domain of $g$ is incompatible with the codomain of $f$.
$endgroup$
– MisterRiemann
Dec 10 '18 at 20:33
1
$begingroup$
$g(f(x,y))$ doesn’t make sense as the image of $f$ is $mathbb R$ while the domain of $g$ is $mathbb R^2$ providing your variables are real numbers.
$endgroup$
– mathcounterexamples.net
Dec 10 '18 at 20:36