Working Iterative formula for a system of equations
I have been given a project, I need to show the use of a version of Newton's method to solve these non-linear equations. The version of Newton's method I am required to use is: $ X_{n+1} = X_n - J^{(-1)} F(X_n) $. I have all the values required here, and this works to find the point $ X_1 $. However I need to find a code that inputs the following points $ X_2, X_3, ..., X_n $ automatically. Here are the given values:
f[x_, y_] := x^2 + y^2 - 5;
g[x_, y_] := x^3 - y^3 - 7;
x0 = 2.1;
y0 = 0.9;
f[x0, y0]
g[x0, y0]
0.22
1.532
M = {{2*x0, 2*y0}, {3*x0^2, -3*y0^2}}
{{4.2, 1.8}, {13.23, -2.43}}
J = Inverse[M]
{{0.0714286, 0.0529101}, {0.388889, -0.123457}}
F0 = {{f[x0, y0]}, {g[x0, y0]}}
X0 = {{x0}, {y0}}
X1 = X0 - J.F0
{{0.22}, {1.532}}
{{2.1}, {0.9}}
{{2.00323}, {1.00358}}
Thank you in Advance!
matrix iteration
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
add a comment |
I have been given a project, I need to show the use of a version of Newton's method to solve these non-linear equations. The version of Newton's method I am required to use is: $ X_{n+1} = X_n - J^{(-1)} F(X_n) $. I have all the values required here, and this works to find the point $ X_1 $. However I need to find a code that inputs the following points $ X_2, X_3, ..., X_n $ automatically. Here are the given values:
f[x_, y_] := x^2 + y^2 - 5;
g[x_, y_] := x^3 - y^3 - 7;
x0 = 2.1;
y0 = 0.9;
f[x0, y0]
g[x0, y0]
0.22
1.532
M = {{2*x0, 2*y0}, {3*x0^2, -3*y0^2}}
{{4.2, 1.8}, {13.23, -2.43}}
J = Inverse[M]
{{0.0714286, 0.0529101}, {0.388889, -0.123457}}
F0 = {{f[x0, y0]}, {g[x0, y0]}}
X0 = {{x0}, {y0}}
X1 = X0 - J.F0
{{0.22}, {1.532}}
{{2.1}, {0.9}}
{{2.00323}, {1.00358}}
Thank you in Advance!
matrix iteration
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
3
Have a look atFixedPointandFixedPointListand their optionSameTest.Nest(List)andNestWhile(List)also come to mind. You can also useFindRootwhich has all this built-in.
– Henrik Schumacher
Dec 11 at 12:32
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Oh, this can also be done withFindRoot: TryReap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: UsingLinearSolveinstead ofInverseshould be faster for larger systems and should prevent certain problems with precision loss.
– Henrik Schumacher
Dec 11 at 14:20
add a comment |
I have been given a project, I need to show the use of a version of Newton's method to solve these non-linear equations. The version of Newton's method I am required to use is: $ X_{n+1} = X_n - J^{(-1)} F(X_n) $. I have all the values required here, and this works to find the point $ X_1 $. However I need to find a code that inputs the following points $ X_2, X_3, ..., X_n $ automatically. Here are the given values:
f[x_, y_] := x^2 + y^2 - 5;
g[x_, y_] := x^3 - y^3 - 7;
x0 = 2.1;
y0 = 0.9;
f[x0, y0]
g[x0, y0]
0.22
1.532
M = {{2*x0, 2*y0}, {3*x0^2, -3*y0^2}}
{{4.2, 1.8}, {13.23, -2.43}}
J = Inverse[M]
{{0.0714286, 0.0529101}, {0.388889, -0.123457}}
F0 = {{f[x0, y0]}, {g[x0, y0]}}
X0 = {{x0}, {y0}}
X1 = X0 - J.F0
{{0.22}, {1.532}}
{{2.1}, {0.9}}
{{2.00323}, {1.00358}}
Thank you in Advance!
matrix iteration
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
I have been given a project, I need to show the use of a version of Newton's method to solve these non-linear equations. The version of Newton's method I am required to use is: $ X_{n+1} = X_n - J^{(-1)} F(X_n) $. I have all the values required here, and this works to find the point $ X_1 $. However I need to find a code that inputs the following points $ X_2, X_3, ..., X_n $ automatically. Here are the given values:
f[x_, y_] := x^2 + y^2 - 5;
g[x_, y_] := x^3 - y^3 - 7;
x0 = 2.1;
y0 = 0.9;
f[x0, y0]
g[x0, y0]
0.22
1.532
M = {{2*x0, 2*y0}, {3*x0^2, -3*y0^2}}
{{4.2, 1.8}, {13.23, -2.43}}
J = Inverse[M]
{{0.0714286, 0.0529101}, {0.388889, -0.123457}}
F0 = {{f[x0, y0]}, {g[x0, y0]}}
X0 = {{x0}, {y0}}
X1 = X0 - J.F0
{{0.22}, {1.532}}
{{2.1}, {0.9}}
{{2.00323}, {1.00358}}
Thank you in Advance!
matrix iteration
matrix iteration
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
edited Dec 11 at 13:05
Αλέξανδρος Ζεγγ
4,0421928
4,0421928
asked Dec 11 at 12:17
Andrew Bradley
161
asked Dec 11 at 12:17
Andrew Bradley
161
asked Dec 11 at 12:17
Andrew Bradley
161
161
3
Have a look atFixedPointandFixedPointListand their optionSameTest.Nest(List)andNestWhile(List)also come to mind. You can also useFindRootwhich has all this built-in.
– Henrik Schumacher
Dec 11 at 12:32
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Oh, this can also be done withFindRoot: TryReap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: UsingLinearSolveinstead ofInverseshould be faster for larger systems and should prevent certain problems with precision loss.
– Henrik Schumacher
Dec 11 at 14:20
add a comment |
3
Have a look atFixedPointandFixedPointListand their optionSameTest.Nest(List)andNestWhile(List)also come to mind. You can also useFindRootwhich has all this built-in.
– Henrik Schumacher
Dec 11 at 12:32
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Oh, this can also be done withFindRoot: TryReap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: UsingLinearSolveinstead ofInverseshould be faster for larger systems and should prevent certain problems with precision loss.
– Henrik Schumacher
Dec 11 at 14:20
3
3
Have a look at
FixedPoint and FixedPointList and their option SameTest. Nest(List) and NestWhile(List) also come to mind. You can also use FindRoot which has all this built-in.– Henrik Schumacher
Dec 11 at 12:32
Have a look at
FixedPoint and FixedPointList and their option SameTest. Nest(List) and NestWhile(List) also come to mind. You can also use FindRoot which has all this built-in.– Henrik Schumacher
Dec 11 at 12:32
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Oh, this can also be done with
FindRoot: Try Reap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: Using LinearSolve instead of Inverse should be faster for larger systems and should prevent certain problems with precision loss.– Henrik Schumacher
Dec 11 at 14:20
Oh, this can also be done with
FindRoot: Try Reap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: Using LinearSolve instead of Inverse should be faster for larger systems and should prevent certain problems with precision loss.– Henrik Schumacher
Dec 11 at 14:20
add a comment |
2 Answers
2
active
oldest
votes
jac = D[{f[x, y], g[x, y]}, {{x, y}, 1}];
Xlist = NestList[# - Inverse[jac /. Thread[{x, y} -> #]].{f @@ #, g @@ #} &, {x0, y0}, 5]
{{2.1, 0.9}, {2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}, {2., 1.}}
You can get the same from FindRoot:
{res, {steps}} = Reap[FindRoot[{f[x, y], g[x, y]}, {{x, x0}, {y, y0}},
Method -> "Newton", StepMonitor :> Sow[{x, y}]]]
{{x -> 2., y -> 1.}, {{{2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}}}}
answered Dec 11 at 13:02
Coolwater
14.6k32552
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
add a comment |
I'll show it with FixedPointList
f[x_, y_] = {x^2 + y^2 - 5, x^3 - y^3 - 7};
j[x_, y_] = Grad[f[x, y], {x, y}];
with LinearSolve:
FixedPointList[(# - LinearSolve[j[Sequence @@ #], f[Sequence @@ #]]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
with Inverse:
FixedPointList[(# - Inverse[j[Sequence @@ #]].f[Sequence @@ #]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
answered Dec 11 at 13:46
rmw
2047
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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oldest
votes
jac = D[{f[x, y], g[x, y]}, {{x, y}, 1}];
Xlist = NestList[# - Inverse[jac /. Thread[{x, y} -> #]].{f @@ #, g @@ #} &, {x0, y0}, 5]
{{2.1, 0.9}, {2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}, {2., 1.}}
You can get the same from FindRoot:
{res, {steps}} = Reap[FindRoot[{f[x, y], g[x, y]}, {{x, x0}, {y, y0}},
Method -> "Newton", StepMonitor :> Sow[{x, y}]]]
{{x -> 2., y -> 1.}, {{{2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}}}}
answered Dec 11 at 13:02
Coolwater
14.6k32552
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
add a comment |
jac = D[{f[x, y], g[x, y]}, {{x, y}, 1}];
Xlist = NestList[# - Inverse[jac /. Thread[{x, y} -> #]].{f @@ #, g @@ #} &, {x0, y0}, 5]
{{2.1, 0.9}, {2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}, {2., 1.}}
You can get the same from FindRoot:
{res, {steps}} = Reap[FindRoot[{f[x, y], g[x, y]}, {{x, x0}, {y, y0}},
Method -> "Newton", StepMonitor :> Sow[{x, y}]]]
{{x -> 2., y -> 1.}, {{{2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}}}}
answered Dec 11 at 13:02
Coolwater
14.6k32552
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
add a comment |
jac = D[{f[x, y], g[x, y]}, {{x, y}, 1}];
Xlist = NestList[# - Inverse[jac /. Thread[{x, y} -> #]].{f @@ #, g @@ #} &, {x0, y0}, 5]
{{2.1, 0.9}, {2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}, {2., 1.}}
You can get the same from FindRoot:
{res, {steps}} = Reap[FindRoot[{f[x, y], g[x, y]}, {{x, x0}, {y, y0}},
Method -> "Newton", StepMonitor :> Sow[{x, y}]]]
{{x -> 2., y -> 1.}, {{{2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}}}}
answered Dec 11 at 13:02
Coolwater
14.6k32552
jac = D[{f[x, y], g[x, y]}, {{x, y}, 1}];
Xlist = NestList[# - Inverse[jac /. Thread[{x, y} -> #]].{f @@ #, g @@ #} &, {x0, y0}, 5]
{{2.1, 0.9}, {2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}, {2., 1.}}
You can get the same from FindRoot:
{res, {steps}} = Reap[FindRoot[{f[x, y], g[x, y]}, {{x, x0}, {y, y0}},
Method -> "Newton", StepMonitor :> Sow[{x, y}]]]
{{x -> 2., y -> 1.}, {{{2.0032275, 1.0035802}, {2.0000033, 1.0000051}, {2., 1.}, {2., 1.}}}}
answered Dec 11 at 13:02
Coolwater
14.6k32552
edited Dec 11 at 13:17
answered Dec 11 at 13:02
Coolwater
14.6k32552
answered Dec 11 at 13:02
Coolwater
14.6k32552
answered Dec 11 at 13:02
Coolwater
14.6k32552
14.6k32552
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
add a comment |
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
Thank you so much, you are my hero!
– Andrew Bradley
Dec 11 at 13:48
add a comment |
I'll show it with FixedPointList
f[x_, y_] = {x^2 + y^2 - 5, x^3 - y^3 - 7};
j[x_, y_] = Grad[f[x, y], {x, y}];
with LinearSolve:
FixedPointList[(# - LinearSolve[j[Sequence @@ #], f[Sequence @@ #]]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
with Inverse:
FixedPointList[(# - Inverse[j[Sequence @@ #]].f[Sequence @@ #]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
answered Dec 11 at 13:46
rmw
2047
add a comment |
I'll show it with FixedPointList
f[x_, y_] = {x^2 + y^2 - 5, x^3 - y^3 - 7};
j[x_, y_] = Grad[f[x, y], {x, y}];
with LinearSolve:
FixedPointList[(# - LinearSolve[j[Sequence @@ #], f[Sequence @@ #]]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
with Inverse:
FixedPointList[(# - Inverse[j[Sequence @@ #]].f[Sequence @@ #]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
answered Dec 11 at 13:46
rmw
2047
add a comment |
I'll show it with FixedPointList
f[x_, y_] = {x^2 + y^2 - 5, x^3 - y^3 - 7};
j[x_, y_] = Grad[f[x, y], {x, y}];
with LinearSolve:
FixedPointList[(# - LinearSolve[j[Sequence @@ #], f[Sequence @@ #]]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
with Inverse:
FixedPointList[(# - Inverse[j[Sequence @@ #]].f[Sequence @@ #]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
answered Dec 11 at 13:46
rmw
2047
I'll show it with FixedPointList
f[x_, y_] = {x^2 + y^2 - 5, x^3 - y^3 - 7};
j[x_, y_] = Grad[f[x, y], {x, y}];
with LinearSolve:
FixedPointList[(# - LinearSolve[j[Sequence @@ #], f[Sequence @@ #]]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
with Inverse:
FixedPointList[(# - Inverse[j[Sequence @@ #]].f[Sequence @@ #]) &, {2.1, 0.9}, 3]
{{2.1, 0.9}, {2.00323, 1.00358}, {2., 1.00001}, {2., 1.}}
answered Dec 11 at 13:46
rmw
2047
answered Dec 11 at 13:46
rmw
2047
answered Dec 11 at 13:46
rmw
2047
answered Dec 11 at 13:46
rmw
2047
2047
add a comment |
add a comment |
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Have a look at
FixedPointandFixedPointListand their optionSameTest.Nest(List)andNestWhile(List)also come to mind. You can also useFindRootwhich has all this built-in.– Henrik Schumacher
Dec 11 at 12:32
Ok Thank you, I will try some of the commands you have listed. Unfortunately, FindRoot does not help me in my situation as I need to show all the iterations leading up to the point, rather than just find the point itself.
– Andrew Bradley
Dec 11 at 13:02
Oh, this can also be done with
FindRoot: TryReap[FindRoot[Sin[x] == 0.2, {x, Pi/42}, EvaluationMonitor :> Sow[x]]]. Btw.: UsingLinearSolveinstead ofInverseshould be faster for larger systems and should prevent certain problems with precision loss.– Henrik Schumacher
Dec 11 at 14:20