Derivative of an L1 norm of transform of a vector.
$begingroup$
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,
$$sumnolimits|{x_{1}|}^{1}_{1} = sumnolimitssqrt{x^*_{i}x_{i} + epsilon}$$
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
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
$endgroup$
add a comment |
$begingroup$
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,
$$sumnolimits|{x_{1}|}^{1}_{1} = sumnolimitssqrt{x^*_{i}x_{i} + epsilon}$$
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
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
$endgroup$
$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21
add a comment |
$begingroup$
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,
$$sumnolimits|{x_{1}|}^{1}_{1} = sumnolimitssqrt{x^*_{i}x_{i} + epsilon}$$
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
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
$endgroup$
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,
$$sumnolimits|{x_{1}|}^{1}_{1} = sumnolimitssqrt{x^*_{i}x_{i} + epsilon}$$
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
linear-algebra functional-analysis wavelets
edited Feb 20 '15 at 2:45
Mathemagician1234
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
edited Feb 20 '15 at 2:45
Mathemagician1234
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
edited Feb 20 '15 at 2:45
Mathemagician1234
14k24159
edited Feb 20 '15 at 2:45
Mathemagician1234
14k24159
edited Feb 20 '15 at 2:45
Mathemagician1234
14k24159
14k24159
asked Feb 20 '15 at 2:29
user212257user212257
213
asked Feb 20 '15 at 2:29
user212257user212257
213
asked Feb 20 '15 at 2:29
user212257user212257
213
213
$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21
add a comment |
$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21
$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21
$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21
add a comment |
1 Answer
1
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oldest
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$begingroup$
Solving in coordinates, use the formula $frac{partial}{partial x_k} |mathbf{x}|_p = frac{x_k |x_k|^{p-2}}{|mathbf{x}|_{p}^{p-1}}$ 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
Community♦
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
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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
$begingroup$
Solving in coordinates, use the formula $frac{partial}{partial x_k} |mathbf{x}|_p = frac{x_k |x_k|^{p-2}}{|mathbf{x}|_{p}^{p-1}}$ 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
Community♦
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
$endgroup$
add a comment |
$begingroup$
Solving in coordinates, use the formula $frac{partial}{partial x_k} |mathbf{x}|_p = frac{x_k |x_k|^{p-2}}{|mathbf{x}|_{p}^{p-1}}$ 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
Community♦
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
$endgroup$
add a comment |
$begingroup$
Solving in coordinates, use the formula $frac{partial}{partial x_k} |mathbf{x}|_p = frac{x_k |x_k|^{p-2}}{|mathbf{x}|_{p}^{p-1}}$ 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
Community♦
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
$endgroup$
Solving in coordinates, use the formula $frac{partial}{partial x_k} |mathbf{x}|_p = frac{x_k |x_k|^{p-2}}{|mathbf{x}|_{p}^{p-1}}$ 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
Community♦
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Apr 30 '15 at 1:31
rychrych
2,4961717
answered Apr 30 '15 at 1:31
rychrych
2,4961717
answered Apr 30 '15 at 1:31
rychrych
2,4961717
2,4961717
add a comment |
add a comment |
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$begingroup$
Why do you need to do this? If you're solving an optimization problem, there might be a better way.
$endgroup$
– littleO
Feb 20 '15 at 3:21