Finding a function with arbitrary Jacobian determinant everywhere
If we have a function $g: mathbb{R}^n rightarrow mathbb{R}$ and $ forall x, g(x) > 0$, can we always find a function $f: mathbb{R}^n rightarrow mathbb{R}^n$ s.t. $forall x, |det frac{partial f(x)}{partial x}| = g(x)$ under reasonable continuity / differentiability assumptions about $g$? I think it should be true, but I'm not sure how to prove it.
multivariable-calculus determinant jacobian
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
add a comment |
If we have a function $g: mathbb{R}^n rightarrow mathbb{R}$ and $ forall x, g(x) > 0$, can we always find a function $f: mathbb{R}^n rightarrow mathbb{R}^n$ s.t. $forall x, |det frac{partial f(x)}{partial x}| = g(x)$ under reasonable continuity / differentiability assumptions about $g$? I think it should be true, but I'm not sure how to prove it.
multivariable-calculus determinant jacobian
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
add a comment |
If we have a function $g: mathbb{R}^n rightarrow mathbb{R}$ and $ forall x, g(x) > 0$, can we always find a function $f: mathbb{R}^n rightarrow mathbb{R}^n$ s.t. $forall x, |det frac{partial f(x)}{partial x}| = g(x)$ under reasonable continuity / differentiability assumptions about $g$? I think it should be true, but I'm not sure how to prove it.
multivariable-calculus determinant jacobian
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
If we have a function $g: mathbb{R}^n rightarrow mathbb{R}$ and $ forall x, g(x) > 0$, can we always find a function $f: mathbb{R}^n rightarrow mathbb{R}^n$ s.t. $forall x, |det frac{partial f(x)}{partial x}| = g(x)$ under reasonable continuity / differentiability assumptions about $g$? I think it should be true, but I'm not sure how to prove it.
multivariable-calculus determinant jacobian
multivariable-calculus determinant jacobian
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
asked Nov 22 '18 at 23:19
user1825464user1825464
1296
1296
add a comment |
add a comment |
2 Answers
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It will be enough to take $f(x_1,ldots,x_n)=bigl(G(x_1,ldots,x_n),x_2,ldots,x_nbigr)$, where $frac{partial G}{partial x_1}=g$.
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
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If $g$ is continuous then it has an antiderivative $G$ you can simply take
$$ f(x_1, x_2,dots,x_n) = (G(x_1),x_2,dots,x_n). $$
Probably continuity of $g$ is about as general as you'd want to ask this question for.
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
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It will be enough to take $f(x_1,ldots,x_n)=bigl(G(x_1,ldots,x_n),x_2,ldots,x_nbigr)$, where $frac{partial G}{partial x_1}=g$.
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It will be enough to take $f(x_1,ldots,x_n)=bigl(G(x_1,ldots,x_n),x_2,ldots,x_nbigr)$, where $frac{partial G}{partial x_1}=g$.
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
add a comment |
It will be enough to take $f(x_1,ldots,x_n)=bigl(G(x_1,ldots,x_n),x_2,ldots,x_nbigr)$, where $frac{partial G}{partial x_1}=g$.
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
It will be enough to take $f(x_1,ldots,x_n)=bigl(G(x_1,ldots,x_n),x_2,ldots,x_nbigr)$, where $frac{partial G}{partial x_1}=g$.
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 22 '18 at 23:32
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
add a comment |
add a comment |
If $g$ is continuous then it has an antiderivative $G$ you can simply take
$$ f(x_1, x_2,dots,x_n) = (G(x_1),x_2,dots,x_n). $$
Probably continuity of $g$ is about as general as you'd want to ask this question for.
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
add a comment |
If $g$ is continuous then it has an antiderivative $G$ you can simply take
$$ f(x_1, x_2,dots,x_n) = (G(x_1),x_2,dots,x_n). $$
Probably continuity of $g$ is about as general as you'd want to ask this question for.
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
add a comment |
If $g$ is continuous then it has an antiderivative $G$ you can simply take
$$ f(x_1, x_2,dots,x_n) = (G(x_1),x_2,dots,x_n). $$
Probably continuity of $g$ is about as general as you'd want to ask this question for.
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
If $g$ is continuous then it has an antiderivative $G$ you can simply take
$$ f(x_1, x_2,dots,x_n) = (G(x_1),x_2,dots,x_n). $$
Probably continuity of $g$ is about as general as you'd want to ask this question for.
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
answered Nov 22 '18 at 23:32
Trevor GunnTrevor Gunn
14.2k32046
14.2k32046
add a comment |
add a comment |
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