Is there an easy way to compute the jacobian of a normalized vector?
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If $x in mathbb{R}^3$ I want to compute the jacobian of the following function
$$
f(x) = frac{x}{lVert x rVert }
$$
If I proceed I get a matrix whose elements are
$$
a_{ij} = begin{cases}
frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
-frac{x_i x_j}{lVert x rVert^3} &i neq j
end{cases}
$$
Is this the most compact form?
The derivation is based on the product rule componentwise.
multivariable-calculus jacobian
asked Aug 14 at 18:09
user8469759
1,3361616
add a comment |
up vote
0
down vote
favorite
If $x in mathbb{R}^3$ I want to compute the jacobian of the following function
$$
f(x) = frac{x}{lVert x rVert }
$$
If I proceed I get a matrix whose elements are
$$
a_{ij} = begin{cases}
frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
-frac{x_i x_j}{lVert x rVert^3} &i neq j
end{cases}
$$
Is this the most compact form?
The derivation is based on the product rule componentwise.
multivariable-calculus jacobian
asked Aug 14 at 18:09
user8469759
1,3361616
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $x in mathbb{R}^3$ I want to compute the jacobian of the following function
$$
f(x) = frac{x}{lVert x rVert }
$$
If I proceed I get a matrix whose elements are
$$
a_{ij} = begin{cases}
frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
-frac{x_i x_j}{lVert x rVert^3} &i neq j
end{cases}
$$
Is this the most compact form?
The derivation is based on the product rule componentwise.
multivariable-calculus jacobian
asked Aug 14 at 18:09
user8469759
1,3361616
If $x in mathbb{R}^3$ I want to compute the jacobian of the following function
$$
f(x) = frac{x}{lVert x rVert }
$$
If I proceed I get a matrix whose elements are
$$
a_{ij} = begin{cases}
frac{1}{lVert x rVert} - frac{x_i^2}{lVert x rVert^3} & i = j \
-frac{x_i x_j}{lVert x rVert^3} &i neq j
end{cases}
$$
Is this the most compact form?
The derivation is based on the product rule componentwise.
multivariable-calculus jacobian
multivariable-calculus jacobian
asked Aug 14 at 18:09
user8469759
1,3361616
asked Aug 14 at 18:09
user8469759
1,3361616
edited Nov 21 at 9:08
asked Aug 14 at 18:09
user8469759
1,3361616
asked Aug 14 at 18:09
user8469759
1,3361616
asked Aug 14 at 18:09
user8469759
1,3361616
1,3361616
add a comment |
add a comment |
1 Answer
1
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0
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I get the same result:
$$
begin{align}
J_{ij}
&= partial_j left( frac{x}{lVert x rVert} right)_i \
&= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
&= frac{delta_{ij}}{lVert x rVert} +
x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
&= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
end{align}
$$
answered Aug 14 at 18:20
mvw
31.3k22252
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I get the same result:
$$
begin{align}
J_{ij}
&= partial_j left( frac{x}{lVert x rVert} right)_i \
&= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
&= frac{delta_{ij}}{lVert x rVert} +
x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
&= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
end{align}
$$
answered Aug 14 at 18:20
mvw
31.3k22252
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
add a comment |
up vote
0
down vote
I get the same result:
$$
begin{align}
J_{ij}
&= partial_j left( frac{x}{lVert x rVert} right)_i \
&= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
&= frac{delta_{ij}}{lVert x rVert} +
x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
&= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
end{align}
$$
answered Aug 14 at 18:20
mvw
31.3k22252
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
add a comment |
up vote
0
down vote
up vote
0
down vote
I get the same result:
$$
begin{align}
J_{ij}
&= partial_j left( frac{x}{lVert x rVert} right)_i \
&= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
&= frac{delta_{ij}}{lVert x rVert} +
x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
&= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
end{align}
$$
answered Aug 14 at 18:20
mvw
31.3k22252
I get the same result:
$$
begin{align}
J_{ij}
&= partial_j left( frac{x}{lVert x rVert} right)_i \
&= partial_j x_i left(sum_k x_k^2right)^{-1/2} \
&= frac{delta_{ij}}{lVert x rVert} +
x_i left(-frac{1}{2}right) left(sum_k x_k^2right)^{-3/2}(2 x_k delta_{kj}) \
&= frac{delta_{ij}}{lVert x rVert} - frac{x_i x_j}{lVert x rVert^3}
end{align}
$$
answered Aug 14 at 18:20
mvw
31.3k22252
answered Aug 14 at 18:20
mvw
31.3k22252
answered Aug 14 at 18:20
mvw
31.3k22252
answered Aug 14 at 18:20
mvw
31.3k22252
31.3k22252
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
add a comment |
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
Maybe an equivalent would be $$frac{1}{lVert x rVert}left(I - frac{x cdot x^T}{lVert x rVert^2} right)$$
– user8469759
Aug 14 at 18:55
add a comment |
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