Wavelet Analysis: A notation problem in this paper
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Because of my research work, I am reading this paper: PDE net. But I ran into a notation problem: In definition 2.1, I do not understand how $textbf{q}[ textbf{k} ]$ yields a real number. Therefore, I could not verify the authors' claims about the order of sum rules of the Haar wavelet frame filter bank and also the filter
begin{equation*}
textbf{q} = begin{pmatrix}
1 & 0 & -1 \
2 & 0 & -2 \
1 & 0 & -1
end{pmatrix}
end{equation*}
I checked their citations and some other literatures, but most of them just said ${textbf{q}[textbf{k}]}$ is a sequence of real numbers, which does not help me to understand the order of sum rules at all. I am grateful if anyone could give me some hint about this notation.
wavelets
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add a comment |
$begingroup$
Because of my research work, I am reading this paper: PDE net. But I ran into a notation problem: In definition 2.1, I do not understand how $textbf{q}[ textbf{k} ]$ yields a real number. Therefore, I could not verify the authors' claims about the order of sum rules of the Haar wavelet frame filter bank and also the filter
begin{equation*}
textbf{q} = begin{pmatrix}
1 & 0 & -1 \
2 & 0 & -2 \
1 & 0 & -1
end{pmatrix}
end{equation*}
I checked their citations and some other literatures, but most of them just said ${textbf{q}[textbf{k}]}$ is a sequence of real numbers, which does not help me to understand the order of sum rules at all. I am grateful if anyone could give me some hint about this notation.
wavelets
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1
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Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
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– Ted
Jan 14 at 8:14
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It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29
add a comment |
$begingroup$
Because of my research work, I am reading this paper: PDE net. But I ran into a notation problem: In definition 2.1, I do not understand how $textbf{q}[ textbf{k} ]$ yields a real number. Therefore, I could not verify the authors' claims about the order of sum rules of the Haar wavelet frame filter bank and also the filter
begin{equation*}
textbf{q} = begin{pmatrix}
1 & 0 & -1 \
2 & 0 & -2 \
1 & 0 & -1
end{pmatrix}
end{equation*}
I checked their citations and some other literatures, but most of them just said ${textbf{q}[textbf{k}]}$ is a sequence of real numbers, which does not help me to understand the order of sum rules at all. I am grateful if anyone could give me some hint about this notation.
wavelets
$endgroup$
Because of my research work, I am reading this paper: PDE net. But I ran into a notation problem: In definition 2.1, I do not understand how $textbf{q}[ textbf{k} ]$ yields a real number. Therefore, I could not verify the authors' claims about the order of sum rules of the Haar wavelet frame filter bank and also the filter
begin{equation*}
textbf{q} = begin{pmatrix}
1 & 0 & -1 \
2 & 0 & -2 \
1 & 0 & -1
end{pmatrix}
end{equation*}
I checked their citations and some other literatures, but most of them just said ${textbf{q}[textbf{k}]}$ is a sequence of real numbers, which does not help me to understand the order of sum rules at all. I am grateful if anyone could give me some hint about this notation.
wavelets
wavelets
asked Jan 14 at 7:39
BM YoonBM Yoon
283
283
1
$begingroup$
Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
$endgroup$
– Ted
Jan 14 at 8:14
$begingroup$
It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29
add a comment |
1
$begingroup$
Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
$endgroup$
– Ted
Jan 14 at 8:14
$begingroup$
It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29
1
1
$begingroup$
Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
$endgroup$
– Ted
Jan 14 at 8:14
$begingroup$
Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
$endgroup$
– Ted
Jan 14 at 8:14
$begingroup$
It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29
$begingroup$
It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29
add a comment |
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$begingroup$
Since $k in mathbb{Z}^2$, the obvious guess is to take $k$ as a pair of indices into the matrix, and 0 if it's outside the bounds of the matrix. Does not that work?
$endgroup$
– Ted
Jan 14 at 8:14
$begingroup$
It works! I can't believe the authors put it in this way! I have been guessing it is about convolution at $textbf{k}$. Thank you so much.
$endgroup$
– BM Yoon
Jan 14 at 8:29