Solving the eigenvalue system
0
$begingroup$
$Au_{m,n}+Bu_{m-10,n-10}+Cu_{m+10,n+10}+Du_{m,n-10}+Eu_{m-10,n}=lambda u_{m,n}$
where $A=0.01(m^2+n^2),quad B=0.1(m+n),quad C=0.1(m-n),quad D=0.1 m, quad E=0.1 n$. and $m,n$ range from $[-50,50]$. My problem is hhow to transform the equation into a matrix form.
matrices linear-algebra
asked Jan 2 at 6:01
yun shiyun shi
1
$endgroup$
migrated from mathematica.stackexchange.com Jan 8 at 19:14
This question came from our site for users of Wolfram Mathematica.
add a comment |
0
$begingroup$
$Au_{m,n}+Bu_{m-10,n-10}+Cu_{m+10,n+10}+Du_{m,n-10}+Eu_{m-10,n}=lambda u_{m,n}$
where $A=0.01(m^2+n^2),quad B=0.1(m+n),quad C=0.1(m-n),quad D=0.1 m, quad E=0.1 n$. and $m,n$ range from $[-50,50]$. My problem is hhow to transform the equation into a matrix form.
matrices linear-algebra
asked Jan 2 at 6:01
yun shiyun shi
1
$endgroup$
migrated from mathematica.stackexchange.com Jan 8 at 19:14
This question came from our site for users of Wolfram Mathematica.
$begingroup$
This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
$endgroup$
– Lotus
Jan 2 at 9:37
$begingroup$
This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
$endgroup$
– Roman
Jan 2 at 9:37
add a comment |
0
0
0
$begingroup$
$Au_{m,n}+Bu_{m-10,n-10}+Cu_{m+10,n+10}+Du_{m,n-10}+Eu_{m-10,n}=lambda u_{m,n}$
where $A=0.01(m^2+n^2),quad B=0.1(m+n),quad C=0.1(m-n),quad D=0.1 m, quad E=0.1 n$. and $m,n$ range from $[-50,50]$. My problem is hhow to transform the equation into a matrix form.
matrices linear-algebra
asked Jan 2 at 6:01
yun shiyun shi
1
$endgroup$
$Au_{m,n}+Bu_{m-10,n-10}+Cu_{m+10,n+10}+Du_{m,n-10}+Eu_{m-10,n}=lambda u_{m,n}$
where $A=0.01(m^2+n^2),quad B=0.1(m+n),quad C=0.1(m-n),quad D=0.1 m, quad E=0.1 n$. and $m,n$ range from $[-50,50]$. My problem is hhow to transform the equation into a matrix form.
matrices linear-algebra
matrices linear-algebra
asked Jan 2 at 6:01
yun shiyun shi
1
asked Jan 2 at 6:01
yun shiyun shi
1
asked Jan 2 at 6:01
yun shiyun shi
1
asked Jan 2 at 6:01
yun shiyun shi
1
asked Jan 2 at 6:01
yun shiyun shi
1
1
migrated from mathematica.stackexchange.com Jan 8 at 19:14
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com Jan 8 at 19:14
This question came from our site for users of Wolfram Mathematica.
$begingroup$
This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
$endgroup$
– Lotus
Jan 2 at 9:37
$begingroup$
This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
$endgroup$
– Roman
Jan 2 at 9:37
add a comment |
$begingroup$
This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
$endgroup$
– Lotus
Jan 2 at 9:37
$begingroup$
This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
$endgroup$
– Roman
Jan 2 at 9:37
$begingroup$
This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
$endgroup$
– Lotus
Jan 2 at 9:37
$begingroup$
This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
$endgroup$
– Lotus
Jan 2 at 9:37
$begingroup$
This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
$endgroup$
– Roman
Jan 2 at 9:37
$begingroup$
This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
$endgroup$
– Roman
Jan 2 at 9:37
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
You can set this equation into a SparseArray with
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n-10->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m+10&&n1==n+10->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m&&n1==n-10->(m-s-1)/10,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}] //ArrayFlatten;
This gets you a list of $(2s+1)^2$ equations:
Thread[X.U == [Lambda] U]
(* list of equations *)
The eigensystem is then
Eigensystem[N[X]]
If you want periodic boundary conditions expressed by the rollover function f:
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
f[a_] = Mod[a, 2s+1, -s];
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==-10 ->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==+10 && f[n1-n]==+10 ->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==0 && f[n1-n]==-10 ->(m-s-1)/10,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==0 ->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}]//ArrayFlatten;
answered Jan 2 at 9:43
RomanRoman
1686
$endgroup$
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
$endgroup$
– yun shi
Jan 3 at 8:23
$begingroup$
So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
$endgroup$
– Roman
Jan 3 at 18:36
$begingroup$
eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
$endgroup$
– yun shi
Jan 4 at 4:59
$begingroup$
@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using theModfunction with offset:f[a_]=Mod[a,2s+1,-s]. Try:s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]
$endgroup$
– Roman
Jan 4 at 7:28
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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2
$begingroup$
You can set this equation into a SparseArray with
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n-10->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m+10&&n1==n+10->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m&&n1==n-10->(m-s-1)/10,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}] //ArrayFlatten;
This gets you a list of $(2s+1)^2$ equations:
Thread[X.U == [Lambda] U]
(* list of equations *)
The eigensystem is then
Eigensystem[N[X]]
If you want periodic boundary conditions expressed by the rollover function f:
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
f[a_] = Mod[a, 2s+1, -s];
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==-10 ->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==+10 && f[n1-n]==+10 ->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==0 && f[n1-n]==-10 ->(m-s-1)/10,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==0 ->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}]//ArrayFlatten;
answered Jan 2 at 9:43
RomanRoman
1686
$endgroup$
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
$endgroup$
– yun shi
Jan 3 at 8:23
$begingroup$
So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
$endgroup$
– Roman
Jan 3 at 18:36
$begingroup$
eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
$endgroup$
– yun shi
Jan 4 at 4:59
$begingroup$
@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using theModfunction with offset:f[a_]=Mod[a,2s+1,-s]. Try:s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]
$endgroup$
– Roman
Jan 4 at 7:28
add a comment |
2
$begingroup$
You can set this equation into a SparseArray with
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n-10->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m+10&&n1==n+10->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m&&n1==n-10->(m-s-1)/10,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}] //ArrayFlatten;
This gets you a list of $(2s+1)^2$ equations:
Thread[X.U == [Lambda] U]
(* list of equations *)
The eigensystem is then
Eigensystem[N[X]]
If you want periodic boundary conditions expressed by the rollover function f:
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
f[a_] = Mod[a, 2s+1, -s];
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==-10 ->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==+10 && f[n1-n]==+10 ->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==0 && f[n1-n]==-10 ->(m-s-1)/10,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==0 ->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}]//ArrayFlatten;
answered Jan 2 at 9:43
RomanRoman
1686
$endgroup$
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
$endgroup$
– yun shi
Jan 3 at 8:23
$begingroup$
So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
$endgroup$
– Roman
Jan 3 at 18:36
$begingroup$
eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
$endgroup$
– yun shi
Jan 4 at 4:59
$begingroup$
@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using theModfunction with offset:f[a_]=Mod[a,2s+1,-s]. Try:s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]
$endgroup$
– Roman
Jan 4 at 7:28
add a comment |
2
2
2
$begingroup$
You can set this equation into a SparseArray with
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n-10->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m+10&&n1==n+10->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m&&n1==n-10->(m-s-1)/10,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}] //ArrayFlatten;
This gets you a list of $(2s+1)^2$ equations:
Thread[X.U == [Lambda] U]
(* list of equations *)
The eigensystem is then
Eigensystem[N[X]]
If you want periodic boundary conditions expressed by the rollover function f:
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
f[a_] = Mod[a, 2s+1, -s];
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==-10 ->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==+10 && f[n1-n]==+10 ->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==0 && f[n1-n]==-10 ->(m-s-1)/10,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==0 ->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}]//ArrayFlatten;
answered Jan 2 at 9:43
RomanRoman
1686
$endgroup$
You can set this equation into a SparseArray with
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n-10->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m+10&&n1==n+10->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/;m1==m&&n1==n-10->(m-s-1)/10,
{m_,m1_,n_,n1_}/;m1==m-10&&n1==n->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}] //ArrayFlatten;
This gets you a list of $(2s+1)^2$ equations:
Thread[X.U == [Lambda] U]
(* list of equations *)
The eigensystem is then
Eigensystem[N[X]]
If you want periodic boundary conditions expressed by the rollover function f:
s = 50;
U = Table[Subscript[u, m, n], {m, -s, s}, {n, -s, s}] // Flatten;
f[a_] = Mod[a, 2s+1, -s];
X = SparseArray[{{m_,m_,n_,n_}->((m-s-1)^2+(n-s-1)^2)/100,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==-10 ->((m-s-1)+(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==+10 && f[n1-n]==+10 ->((m-s-1)-(n-s-1))/10,
{m_,m1_,n_,n1_}/; f[m1-m]==0 && f[n1-n]==-10 ->(m-s-1)/10,
{m_,m1_,n_,n1_}/; f[m1-m]==-10 && f[n1-n]==0 ->(n-s-1)/10},
{2s+1,2s+1,2s+1,2s+1}]//ArrayFlatten;
answered Jan 2 at 9:43
RomanRoman
1686
answered Jan 2 at 9:43
RomanRoman
1686
answered Jan 2 at 9:43
RomanRoman
1686
answered Jan 2 at 9:43
RomanRoman
1686
1686
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
$endgroup$
– yun shi
Jan 3 at 8:23
$begingroup$
So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
$endgroup$
– Roman
Jan 3 at 18:36
$begingroup$
eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
$endgroup$
– yun shi
Jan 4 at 4:59
$begingroup$
@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using theModfunction with offset:f[a_]=Mod[a,2s+1,-s]. Try:s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]
$endgroup$
– Roman
Jan 4 at 7:28
add a comment |
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
$endgroup$
– yun shi
Jan 3 at 8:23
$begingroup$
So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
$endgroup$
– Roman
Jan 3 at 18:36
$begingroup$
eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
$endgroup$
– yun shi
Jan 4 at 4:59
$begingroup$
@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using theModfunction with offset:f[a_]=Mod[a,2s+1,-s]. Try:s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]
$endgroup$
– Roman
Jan 4 at 7:28
$begingroup$
Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
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– yun shi
Jan 3 at 8:23
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Thanks a lot. Also, if I want to consider the periodic boundary condition, how do I improve the code? i.e., if $m-10<-(2s+1)$, we need $m-10rightarrow m-10+(2s+1)$ .
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– yun shi
Jan 3 at 8:23
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So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
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– Roman
Jan 3 at 18:36
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So you want $m=-51$ to be equal to $m=+50$? Or do you rather want $m=-50$ to be equal to $m=+50$? It makes a big difference how precisely you're doing the periodic connection.
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– Roman
Jan 3 at 18:36
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eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
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– yun shi
Jan 4 at 4:59
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eg., for $s=50$, the range $m$ is $[-101,101]$, but for $m=-101$, $m-10=-111$, which has exceeded the range, so I need to consider the periodic boundary condition that make $m-10=-111$ be $-111+202=91$. How should I do?
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– yun shi
Jan 4 at 4:59
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@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using the
Mod function with offset: f[a_]=Mod[a,2s+1,-s]. Try: s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]$endgroup$
– Roman
Jan 4 at 7:28
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@yunshi: for $s=50$, the range is $min[-50,50]$. For $m=-50$ you then find $m-10=-60mapsto-60+101=41$ using the
Mod function with offset: f[a_]=Mod[a,2s+1,-s]. Try: s=50;f[a_]=Mod[a,2s+1,-s];Table[f[a],{a,-60,60}]$endgroup$
– Roman
Jan 4 at 7:28
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This does not seem to be a Mathematica-specific question. How would one solve this outside of Mathematica ?
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– Lotus
Jan 2 at 9:37
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This problem is block-diagonal modulo 10: all indices $m$ that end in "1" are coupled only among each other, and all that end in "2" are coupled only among each other, etc., and the same for $n$. You can therefore set up 100 sub-problems with indices ${m,n}$ all ending the same way. This may simplify the calculation.
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– Roman
Jan 2 at 9:37