To solve self consistent equations by simultaneously plotting them
1
$begingroup$
How can we solve a pair of equations that have to be solved self-consistently (even by plotting them simultaneously) .
y = 1/2 + 1/[Pi] ArcTan[c (0.5 - x)];
x = 1/2 + 1/[Pi] ArcTan[c (0.5 - y)];
For c=0.5 we must have a plot as below one in which the vertical axis is y and horizontal one is devoted to x.
For c=Pi we must have
plotting equation-solving calculus-and-analysis
asked 2 days ago
IrreversibleIrreversible
1,675723
$endgroup$
add a comment |
1
$begingroup$
How can we solve a pair of equations that have to be solved self-consistently (even by plotting them simultaneously) .
y = 1/2 + 1/[Pi] ArcTan[c (0.5 - x)];
x = 1/2 + 1/[Pi] ArcTan[c (0.5 - y)];
For c=0.5 we must have a plot as below one in which the vertical axis is y and horizontal one is devoted to x.
For c=Pi we must have
plotting equation-solving calculus-and-analysis
asked 2 days ago
IrreversibleIrreversible
1,675723
$endgroup$
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and byInverseFunction[f][y]I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.
$endgroup$
– Irreversible
2 days ago
add a comment |
1
1
1
1
$begingroup$
How can we solve a pair of equations that have to be solved self-consistently (even by plotting them simultaneously) .
y = 1/2 + 1/[Pi] ArcTan[c (0.5 - x)];
x = 1/2 + 1/[Pi] ArcTan[c (0.5 - y)];
For c=0.5 we must have a plot as below one in which the vertical axis is y and horizontal one is devoted to x.
For c=Pi we must have
plotting equation-solving calculus-and-analysis
asked 2 days ago
IrreversibleIrreversible
1,675723
$endgroup$
How can we solve a pair of equations that have to be solved self-consistently (even by plotting them simultaneously) .
y = 1/2 + 1/[Pi] ArcTan[c (0.5 - x)];
x = 1/2 + 1/[Pi] ArcTan[c (0.5 - y)];
For c=0.5 we must have a plot as below one in which the vertical axis is y and horizontal one is devoted to x.
For c=Pi we must have
plotting equation-solving calculus-and-analysis
plotting equation-solving calculus-and-analysis
asked 2 days ago
IrreversibleIrreversible
1,675723
asked 2 days ago
IrreversibleIrreversible
1,675723
asked 2 days ago
IrreversibleIrreversible
1,675723
asked 2 days ago
IrreversibleIrreversible
1,675723
asked 2 days ago
IrreversibleIrreversible
1,675723
1,675723
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and byInverseFunction[f][y]I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.
$endgroup$
– Irreversible
2 days ago
add a comment |
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and byInverseFunction[f][y]I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.
$endgroup$
– Irreversible
2 days ago
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and by
InverseFunction[f][y] I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.$endgroup$
– Irreversible
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and by
InverseFunction[f][y] I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.$endgroup$
– Irreversible
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
With[{c = 0.5},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
With[{c = π},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
answered 2 days ago
HughHugh
6,66921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is slightly pensive♦
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
With[{c = 0.5},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
With[{c = π},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
answered 2 days ago
HughHugh
6,66921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is slightly pensive♦
2 days ago
add a comment |
3
$begingroup$
With[{c = 0.5},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
With[{c = π},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
answered 2 days ago
HughHugh
6,66921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is slightly pensive♦
2 days ago
add a comment |
3
3
3
$begingroup$
With[{c = 0.5},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
With[{c = π},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
answered 2 days ago
HughHugh
6,66921946
$endgroup$
With[{c = 0.5},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
With[{c = π},
ContourPlot[{
y == 1/2 + 1/π ArcTan[c (0.5 - x)],
x == 1/2 + 1/π ArcTan[c (0.5 - y)]
},
{x, 0, 1}, {y, 0, 1}]
]
answered 2 days ago
HughHugh
6,66921946
answered 2 days ago
HughHugh
6,66921946
answered 2 days ago
HughHugh
6,66921946
answered 2 days ago
HughHugh
6,66921946
6,66921946
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is slightly pensive♦
2 days ago
add a comment |
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is slightly pensive♦
2 days ago
1
1
$begingroup$
One can even use
MeshFunctions to mark the intersection point.$endgroup$
– J. M. is slightly pensive♦
2 days ago
$begingroup$
One can even use
MeshFunctions to mark the intersection point.$endgroup$
– J. M. is slightly pensive♦
2 days ago
add a comment |
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$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
2 days ago
$begingroup$
I wanted to plot the first function y vs x ordinary and by
InverseFunction[f][y]I wanted to plot the second one (x vs y). For the second one the output plot is not symmetric relative to bisectrix of the first quarter of the coordination plate.$endgroup$
– Irreversible
2 days ago