Mixing
Partial derivative of coordinates with respect to function
$begingroup$
Let $f : mathbb{R}^n rightarrow mathbb{R}^n$. Then
$$frac{partial f^i}{partial x^j} = (nabla f)^i_j$$
where $nabla f$ is the Jacobian matrix of $f$. When reading this paper I came across the expression
$$frac{partial x^i}{partial f^j}$$
Should I interpret this as
$$frac{partial x^i}{partial f^j} = ((nabla f)^{-1})^i_j$$
matrices multivariable-calculus differential-geometry vector-analysis jacobian
asked Jan 11 at 23:40
user76284user76284
1,2031126
$endgroup$
add a comment |
$begingroup$
Let $f : mathbb{R}^n rightarrow mathbb{R}^n$. Then
$$frac{partial f^i}{partial x^j} = (nabla f)^i_j$$
where $nabla f$ is the Jacobian matrix of $f$. When reading this paper I came across the expression
$$frac{partial x^i}{partial f^j}$$
Should I interpret this as
$$frac{partial x^i}{partial f^j} = ((nabla f)^{-1})^i_j$$
matrices multivariable-calculus differential-geometry vector-analysis jacobian
asked Jan 11 at 23:40
user76284user76284
1,2031126
$endgroup$
add a comment |
$begingroup$
Let $f : mathbb{R}^n rightarrow mathbb{R}^n$. Then
$$frac{partial f^i}{partial x^j} = (nabla f)^i_j$$
where $nabla f$ is the Jacobian matrix of $f$. When reading this paper I came across the expression
$$frac{partial x^i}{partial f^j}$$
Should I interpret this as
$$frac{partial x^i}{partial f^j} = ((nabla f)^{-1})^i_j$$
matrices multivariable-calculus differential-geometry vector-analysis jacobian
asked Jan 11 at 23:40
user76284user76284
1,2031126
$endgroup$
Let $f : mathbb{R}^n rightarrow mathbb{R}^n$. Then
$$frac{partial f^i}{partial x^j} = (nabla f)^i_j$$
where $nabla f$ is the Jacobian matrix of $f$. When reading this paper I came across the expression
$$frac{partial x^i}{partial f^j}$$
Should I interpret this as
$$frac{partial x^i}{partial f^j} = ((nabla f)^{-1})^i_j$$
matrices multivariable-calculus differential-geometry vector-analysis jacobian
matrices multivariable-calculus differential-geometry vector-analysis jacobian
asked Jan 11 at 23:40
user76284user76284
1,2031126
asked Jan 11 at 23:40
user76284user76284
1,2031126
asked Jan 11 at 23:40
user76284user76284
1,2031126
asked Jan 11 at 23:40
user76284user76284
1,2031126
asked Jan 11 at 23:40
user76284user76284
1,2031126
1,2031126
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$
x^i = x^i(f^1(x),cdots,f^n(x))
$$
Now apply the chain rule
$$
frac{partial x^i}{partial x^j} = frac{partial x^i}{partial f^k} frac{partial f^k}{partial x^j} = (nabla_f x)^{i}_{;k}(nabla_x f)^{k}_{;j} = delta^i_j
$$
That means that
$$
mathbb{1} = (nabla_f x) (nabla_x f)
$$
Or in other words
$$
nabla_f x = (nabla_x f)^{-1}
$$
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$
x^i = x^i(f^1(x),cdots,f^n(x))
$$
Now apply the chain rule
$$
frac{partial x^i}{partial x^j} = frac{partial x^i}{partial f^k} frac{partial f^k}{partial x^j} = (nabla_f x)^{i}_{;k}(nabla_x f)^{k}_{;j} = delta^i_j
$$
That means that
$$
mathbb{1} = (nabla_f x) (nabla_x f)
$$
Or in other words
$$
nabla_f x = (nabla_x f)^{-1}
$$
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
add a comment |
$begingroup$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$
x^i = x^i(f^1(x),cdots,f^n(x))
$$
Now apply the chain rule
$$
frac{partial x^i}{partial x^j} = frac{partial x^i}{partial f^k} frac{partial f^k}{partial x^j} = (nabla_f x)^{i}_{;k}(nabla_x f)^{k}_{;j} = delta^i_j
$$
That means that
$$
mathbb{1} = (nabla_f x) (nabla_x f)
$$
Or in other words
$$
nabla_f x = (nabla_x f)^{-1}
$$
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
add a comment |
$begingroup$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$
x^i = x^i(f^1(x),cdots,f^n(x))
$$
Now apply the chain rule
$$
frac{partial x^i}{partial x^j} = frac{partial x^i}{partial f^k} frac{partial f^k}{partial x^j} = (nabla_f x)^{i}_{;k}(nabla_x f)^{k}_{;j} = delta^i_j
$$
That means that
$$
mathbb{1} = (nabla_f x) (nabla_x f)
$$
Or in other words
$$
nabla_f x = (nabla_x f)^{-1}
$$
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
$endgroup$
Imagine you can invert the problem $x^i = x^i(f)$. Clearly
$$
x^i = x^i(f^1(x),cdots,f^n(x))
$$
Now apply the chain rule
$$
frac{partial x^i}{partial x^j} = frac{partial x^i}{partial f^k} frac{partial f^k}{partial x^j} = (nabla_f x)^{i}_{;k}(nabla_x f)^{k}_{;j} = delta^i_j
$$
That means that
$$
mathbb{1} = (nabla_f x) (nabla_x f)
$$
Or in other words
$$
nabla_f x = (nabla_x f)^{-1}
$$
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
edited Jan 12 at 0:23
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
answered Jan 12 at 0:15
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
add a comment |
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
$begingroup$
Should be $delta^i_j$ in your second equation, right? And dot product in the third?
$endgroup$
– user76284
Jan 12 at 0:20
1
1
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
$begingroup$
@user76284 Yes, typo in the indices. But the third eqn is correct, it is the matrix product
$endgroup$
– caverac
Jan 12 at 0:24
1
1
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
$begingroup$
Makes sense. I just always write the matrix product with a dot to make it clear I'm performing tensor contraction, and to distinguish it from the tensor product.
$endgroup$
– user76284
Jan 12 at 0:25
add a comment |
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