Derivative of an L1 norm of transform of a vector.
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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$
where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that
$$C = psi F$$
then my l1 norm is defined as:
$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,
$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$
My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?
linear-algebra functional-analysis wavelets
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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$
where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that
$$C = psi F$$
then my l1 norm is defined as:
$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,
$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$
My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?
linear-algebra functional-analysis wavelets
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21
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up vote
1
down vote
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up vote
1
down vote
favorite
I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$
where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that
$$C = psi F$$
then my l1 norm is defined as:
$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,
$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$
My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?
linear-algebra functional-analysis wavelets
I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$
where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that
$$C = psi F$$
then my l1 norm is defined as:
$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,
$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$
My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?
linear-algebra functional-analysis wavelets
edited Feb 20 '15 at 2:45
Mathemagician1234
13.6k24054
13.6k24054
asked Feb 20 '15 at 2:29
user212257
112
112
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21
add a comment |Â
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21
add a comment |Â
1 Answer
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Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.
See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.
See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.
add a comment |Â
up vote
0
down vote
Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.
See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.
See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.
Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.
See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.
edited Apr 13 '17 at 12:20
Communityâ¦
1
1
answered Apr 30 '15 at 1:31
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rych
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2,3081717
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Why do you need to do this? If you're solving an optimization problem, there might be a better way.
â littleO
Feb 20 '15 at 3:21