Defining multiple functions
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I have a matrix with 4 elements that I want to turn into functions without explicitely defininig all the functions "by hand":
matrix={{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(
3 u2 + 1. v2), (-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)}, {(
5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(
3 u2 + 1. v2), (-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(
3 u2 + 1. v2)}}
in other words, in order to define a function for each element I have to manually specify the functions:
eq1[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,1]];
eq2[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,2]];
eq3[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,1]];
eq4[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,2]];
(Although the arguments do not all appear in the RHS, they might in different scenarios, so I chose to specify all arguments).
I tried:
Table[eq[i, j][u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[i, j]], {i, 2}, {j, 2}]
but that does not seem to work. Or at least I don't get how to access eq now:
eq[1, 1]
does not work. Any help appreciated.
functions function-construction
asked 2 hours ago
holistic
1,204620
add a comment |
up vote
4
down vote
favorite
I have a matrix with 4 elements that I want to turn into functions without explicitely defininig all the functions "by hand":
matrix={{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(
3 u2 + 1. v2), (-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)}, {(
5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(
3 u2 + 1. v2), (-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(
3 u2 + 1. v2)}}
in other words, in order to define a function for each element I have to manually specify the functions:
eq1[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,1]];
eq2[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,2]];
eq3[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,1]];
eq4[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,2]];
(Although the arguments do not all appear in the RHS, they might in different scenarios, so I chose to specify all arguments).
I tried:
Table[eq[i, j][u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[i, j]], {i, 2}, {j, 2}]
but that does not seem to work. Or at least I don't get how to access eq now:
eq[1, 1]
does not work. Any help appreciated.
functions function-construction
asked 2 hours ago
holistic
1,204620
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have a matrix with 4 elements that I want to turn into functions without explicitely defininig all the functions "by hand":
matrix={{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(
3 u2 + 1. v2), (-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)}, {(
5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(
3 u2 + 1. v2), (-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(
3 u2 + 1. v2)}}
in other words, in order to define a function for each element I have to manually specify the functions:
eq1[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,1]];
eq2[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,2]];
eq3[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,1]];
eq4[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,2]];
(Although the arguments do not all appear in the RHS, they might in different scenarios, so I chose to specify all arguments).
I tried:
Table[eq[i, j][u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[i, j]], {i, 2}, {j, 2}]
but that does not seem to work. Or at least I don't get how to access eq now:
eq[1, 1]
does not work. Any help appreciated.
functions function-construction
asked 2 hours ago
holistic
1,204620
I have a matrix with 4 elements that I want to turn into functions without explicitely defininig all the functions "by hand":
matrix={{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(
3 u2 + 1. v2), (-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)}, {(
5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(
3 u2 + 1. v2), (-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(
3 u2 + 1. v2)}}
in other words, in order to define a function for each element I have to manually specify the functions:
eq1[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,1]];
eq2[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[1,2]];
eq3[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,1]];
eq4[u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[2,2]];
(Although the arguments do not all appear in the RHS, they might in different scenarios, so I chose to specify all arguments).
I tried:
Table[eq[i, j][u1_,u2_, v1_,v2_, x11_,x12_,x21_,x22_] := matrix[[i, j]], {i, 2}, {j, 2}]
but that does not seem to work. Or at least I don't get how to access eq now:
eq[1, 1]
does not work. Any help appreciated.
functions function-construction
functions function-construction
asked 2 hours ago
holistic
1,204620
asked 2 hours ago
holistic
1,204620
asked 2 hours ago
holistic
1,204620
asked 2 hours ago
holistic
1,204620
asked 2 hours ago
holistic
1,204620
1,204620
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
5
down vote
accepted
matrix = {{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(3 u2 + 1. v2),
(-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)},
{(5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(3 u2 + 1. v2),
(-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(3 u2 + 1. v2)}};
Table[eq[i, j][u1_, u2_, v1_, v2_, x11_, x12_, x21_, x22_] :=
Evaluate[matrix[[i, j]]], {i, 2}, {j, 2}];
?eq
(definitions displayed)
Evaluation examples
eq[1, 1][1, 2, 3, 4, 5, 6, 7, 8]
0.5
Table[eq[i, j][1, 2, 3, 4, 5, 6, 7, 8], {i, 2}, {j, 2}]
{{0.5, -1.5}, {12.4, -5.8}}
answered 1 hour ago
Chris Degnen
21.8k23584
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
matrix = {{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(3 u2 + 1. v2),
(-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)},
{(5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(3 u2 + 1. v2),
(-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(3 u2 + 1. v2)}};
Table[eq[i, j][u1_, u2_, v1_, v2_, x11_, x12_, x21_, x22_] :=
Evaluate[matrix[[i, j]]], {i, 2}, {j, 2}];
?eq
(definitions displayed)
Evaluation examples
eq[1, 1][1, 2, 3, 4, 5, 6, 7, 8]
0.5
Table[eq[i, j][1, 2, 3, 4, 5, 6, 7, 8], {i, 2}, {j, 2}]
{{0.5, -1.5}, {12.4, -5.8}}
answered 1 hour ago
Chris Degnen
21.8k23584
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
add a comment |
up vote
5
down vote
accepted
matrix = {{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(3 u2 + 1. v2),
(-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)},
{(5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(3 u2 + 1. v2),
(-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(3 u2 + 1. v2)}};
Table[eq[i, j][u1_, u2_, v1_, v2_, x11_, x12_, x21_, x22_] :=
Evaluate[matrix[[i, j]]], {i, 2}, {j, 2}];
?eq
(definitions displayed)
Evaluation examples
eq[1, 1][1, 2, 3, 4, 5, 6, 7, 8]
0.5
Table[eq[i, j][1, 2, 3, 4, 5, 6, 7, 8], {i, 2}, {j, 2}]
{{0.5, -1.5}, {12.4, -5.8}}
answered 1 hour ago
Chris Degnen
21.8k23584
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
matrix = {{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(3 u2 + 1. v2),
(-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)},
{(5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(3 u2 + 1. v2),
(-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(3 u2 + 1. v2)}};
Table[eq[i, j][u1_, u2_, v1_, v2_, x11_, x12_, x21_, x22_] :=
Evaluate[matrix[[i, j]]], {i, 2}, {j, 2}];
?eq
(definitions displayed)
Evaluation examples
eq[1, 1][1, 2, 3, 4, 5, 6, 7, 8]
0.5
Table[eq[i, j][1, 2, 3, 4, 5, 6, 7, 8], {i, 2}, {j, 2}]
{{0.5, -1.5}, {12.4, -5.8}}
answered 1 hour ago
Chris Degnen
21.8k23584
matrix = {{(6. u2 - 3 v2 + 0.3 u2 x11 + 0.1 v2 x11)/(3 u2 + 1. v2),
(-15. + 1.5 u2 x12 + 0.5 v2 x12)/(-3. u2 - v2)},
{(5. u2 v2 + 3.6 u2 x21 + 1.2 v2 x21)/(3 u2 + 1. v2),
(-9. u2 + 2 v2 - 1.8 u2 x22 - 0.6 v2 x22)/(3 u2 + 1. v2)}};
Table[eq[i, j][u1_, u2_, v1_, v2_, x11_, x12_, x21_, x22_] :=
Evaluate[matrix[[i, j]]], {i, 2}, {j, 2}];
?eq
(definitions displayed)
Evaluation examples
eq[1, 1][1, 2, 3, 4, 5, 6, 7, 8]
0.5
Table[eq[i, j][1, 2, 3, 4, 5, 6, 7, 8], {i, 2}, {j, 2}]
{{0.5, -1.5}, {12.4, -5.8}}
answered 1 hour ago
Chris Degnen
21.8k23584
edited 1 hour ago
answered 1 hour ago
Chris Degnen
21.8k23584
answered 1 hour ago
Chris Degnen
21.8k23584
answered 1 hour ago
Chris Degnen
21.8k23584
21.8k23584
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
add a comment |
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
Ah I see..well I was close :). Thank you!
– holistic
1 hour ago
add a comment |
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