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.










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  • 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
















0












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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asked Jan 14 at 7:39









BM YoonBM Yoon

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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














  • 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










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