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To solve self consistent equations by simultaneously plotting them
Specifying unlabeled FrameTicks in PlotGraphical Plots of PDF and CDFPlotting the results of a system of equationsHow to solve time dependent Optical Bloch Equations for a three level system?Plotting coupled equations with noise modeling with Ito ProcessAxis label shifted in 3D plottranscendental equation solvingVertical axis as function valueHow to omit extra vertical lines in Plot or ShowHow to solve these equations numerically?
$begingroup$
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 Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
$endgroup$
add a comment |
$begingroup$
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 Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
$endgroup$
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
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.
$endgroup$
– Unbelievable
Mar 27 at 10:38
add a comment |
$begingroup$
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 Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
$endgroup$
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 Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
asked Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
asked Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
asked Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
asked Mar 27 at 9:58
UnbelievableUnbelievable
2,230829
2,230829
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
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.
$endgroup$
– Unbelievable
Mar 27 at 10:38
add a comment |
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
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.
$endgroup$
– Unbelievable
Mar 27 at 10:38
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
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.$endgroup$
– Unbelievable
Mar 27 at 10:38
$begingroup$
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.$endgroup$
– Unbelievable
Mar 27 at 10:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Mar 27 at 10:53
HughHugh
6,67921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is away♦
Mar 27 at 13:36
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Mar 27 at 10:53
HughHugh
6,67921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is away♦
Mar 27 at 13:36
add a comment |
$begingroup$
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 Mar 27 at 10:53
HughHugh
6,67921946
$endgroup$
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is away♦
Mar 27 at 13:36
add a comment |
$begingroup$
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 Mar 27 at 10:53
HughHugh
6,67921946
$endgroup$
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 Mar 27 at 10:53
HughHugh
6,67921946
answered Mar 27 at 10:53
HughHugh
6,67921946
answered Mar 27 at 10:53
HughHugh
6,67921946
answered Mar 27 at 10:53
HughHugh
6,67921946
6,67921946
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is away♦
Mar 27 at 13:36
add a comment |
1
$begingroup$
One can even useMeshFunctionsto mark the intersection point.
$endgroup$
– J. M. is away♦
Mar 27 at 13:36
1
1
$begingroup$
One can even use
MeshFunctions to mark the intersection point.$endgroup$
– J. M. is away♦
Mar 27 at 13:36
$begingroup$
One can even use
MeshFunctions to mark the intersection point.$endgroup$
– J. M. is away♦
Mar 27 at 13:36
add a comment |
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-calculus-and-analysis, equation-solving, plotting
$begingroup$
FindRoot for solving and ContourPlot for plotting
$endgroup$
– Lotus
Mar 27 at 10:06
$begingroup$
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.$endgroup$
– Unbelievable
Mar 27 at 10:38