A double summation
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 '18 at 18:56
Zilch
63
add a comment |
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 '18 at 18:56
Zilch
63
add a comment |
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
asked Nov 29 '18 at 18:56
Zilch
63
While going through the article given here (page 23), it seems that the following doublesummation is carried out
$sum_{i=1,j=1}^{N^2} B_{ij} = Big( sum_{i,j=1}^{N^2-1} + sum_{i=1,j=N^2}^{N^2} + sum_{i=N^2,j=1}^{N^2} - sum_{i=N^2,j=N^2}^{N^2} Big) B_{ij}$. Where $B_{ij} = c_{ij} F_i rho_S F_j^dagger$, is an operator.
Could anyone help me to understand how the double sum on the left breaks into the four sums on the right hand side?
sequences-and-series summation
sequences-and-series summation
asked Nov 29 '18 at 18:56
Zilch
63
asked Nov 29 '18 at 18:56
Zilch
63
asked Nov 29 '18 at 18:56
Zilch
63
asked Nov 29 '18 at 18:56
Zilch
63
asked Nov 29 '18 at 18:56
Zilch
63
63
add a comment |
add a comment |
1 Answer
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When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
add a comment |
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1 Answer
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When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
add a comment |
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
add a comment |
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
When you write $sum_{i,j=1}^{N^2-1} B_{ij}$, to get to $sum_{i,j}^{N^2}B_{ij}$ you are missing all the terms that have either $i=N^2$, or $j=N^2$. The authors are dividing these in two groups:
$i=1,ldots,N^2$; $j=N^2$
$i=N^2$; $j=1,ldots,N^2$
But now they are counting $i=N^2$, $j=N^2$ twice, so they subtract it once.
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
answered Nov 29 '18 at 23:27
Martin Argerami
124k1176175
124k1176175
add a comment |
add a comment |
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