To plot branch cut of logarithmVisualizing Riemann surface (two branches) of logarithmHow to plot the contour...
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To plot branch cut of logarithm
Visualizing Riemann surface (two branches) of logarithmHow to plot the contour of f[x,y]==0 if always f[x,y]>=0Change Contour Plot Overlap OrderingBranch cuts of sqrtDifferentiate contour color based on different functions rather than the contour valuesBranch cut of $sqrt{x^2-1}$?How to Approximate at Non-differentiable Point (forced Series Expansion around Branch Cut)Visualizing the complex logarithm
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I like to see the branch cut of the function:
$$1 - z ln[(1+z)/z].$$
If I plot it in the complex plane:
Plot3D[Re[1 - (x + I y) Log[(1 + x + I y)/(x + I y)]], {x, -2,
2}, {y, -2, 2}]
The result is:
which shows the brach cut correctly between -1 and 0. How can I get rid of the hole in the picture and have a smooth line as a branch cut rather than a white line discontinuity?
Also the same for contour plot:
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> 40]]
plotting calculus-and-analysis complex
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I like to see the branch cut of the function:
$$1 - z ln[(1+z)/z].$$
If I plot it in the complex plane:
Plot3D[Re[1 - (x + I y) Log[(1 + x + I y)/(x + I y)]], {x, -2,
2}, {y, -2, 2}]
The result is:
which shows the brach cut correctly between -1 and 0. How can I get rid of the hole in the picture and have a smooth line as a branch cut rather than a white line discontinuity?
Also the same for contour plot:
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> 40]]
plotting calculus-and-analysis complex
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
2 days ago
add a comment |
$begingroup$
I like to see the branch cut of the function:
$$1 - z ln[(1+z)/z].$$
If I plot it in the complex plane:
Plot3D[Re[1 - (x + I y) Log[(1 + x + I y)/(x + I y)]], {x, -2,
2}, {y, -2, 2}]
The result is:
which shows the brach cut correctly between -1 and 0. How can I get rid of the hole in the picture and have a smooth line as a branch cut rather than a white line discontinuity?
Also the same for contour plot:
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> 40]]
plotting calculus-and-analysis complex
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I like to see the branch cut of the function:
$$1 - z ln[(1+z)/z].$$
If I plot it in the complex plane:
Plot3D[Re[1 - (x + I y) Log[(1 + x + I y)/(x + I y)]], {x, -2,
2}, {y, -2, 2}]
The result is:
which shows the brach cut correctly between -1 and 0. How can I get rid of the hole in the picture and have a smooth line as a branch cut rather than a white line discontinuity?
Also the same for contour plot:
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> 40]]
plotting calculus-and-analysis complex
plotting calculus-and-analysis complex
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Call me potato.
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
asked 2 days ago
Call me potato.Call me potato.
504 bronze badges
504 bronze badges
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Call me potato. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
2 days ago
add a comment |
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
2 days ago
1
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
2
2
$begingroup$
You can use
Exclusions -> None to get rid of the white line.$endgroup$
– C. E.
2 days ago
$begingroup$
You can use
Exclusions -> None to get rid of the white line.$endgroup$
– C. E.
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Change "Rainbow" with any ColorScheme you prefer, and the rescaling values {-2,1} to obtain different scaling.
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-4, 2, .1],
ColorFunction -> (ColorData["Rainbow"][Rescale[#, {-2, 1}]] &),
ColorFunctionScaling -> False, PlotRange -> All]]
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that you can also use the new (as of Version 12) ComplexPlot function, too:
ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10,
MeshFunctions -> {Re[#2] &, Im[#2] &}]
Or the 3D version:
ComplexPlot3D[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I},
Mesh -> 10, PlotRange -> All]
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Change "Rainbow" with any ColorScheme you prefer, and the rescaling values {-2,1} to obtain different scaling.
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-4, 2, .1],
ColorFunction -> (ColorData["Rainbow"][Rescale[#, {-2, 1}]] &),
ColorFunctionScaling -> False, PlotRange -> All]]
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Change "Rainbow" with any ColorScheme you prefer, and the rescaling values {-2,1} to obtain different scaling.
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-4, 2, .1],
ColorFunction -> (ColorData["Rainbow"][Rescale[#, {-2, 1}]] &),
ColorFunctionScaling -> False, PlotRange -> All]]
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Change "Rainbow" with any ColorScheme you prefer, and the rescaling values {-2,1} to obtain different scaling.
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-4, 2, .1],
ColorFunction -> (ColorData["Rainbow"][Rescale[#, {-2, 1}]] &),
ColorFunctionScaling -> False, PlotRange -> All]]
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
$endgroup$
Change "Rainbow" with any ColorScheme you prefer, and the rescaling values {-2,1} to obtain different scaling.
With[{z = x + I y},
ContourPlot[Re[1 - z Log[(1 + z)/(z)]], {x, -2, 2}, {y, -2, 2},
Contours -> Range[-4, 2, .1],
ColorFunction -> (ColorData["Rainbow"][Rescale[#, {-2, 1}]] &),
ColorFunctionScaling -> False, PlotRange -> All]]
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
answered 2 days ago
FraccaloFraccalo
2,9706 silver badges18 bronze badges
2,9706 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
Note that you can also use the new (as of Version 12) ComplexPlot function, too:
ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10,
MeshFunctions -> {Re[#2] &, Im[#2] &}]
Or the 3D version:
ComplexPlot3D[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I},
Mesh -> 10, PlotRange -> All]
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that you can also use the new (as of Version 12) ComplexPlot function, too:
ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10,
MeshFunctions -> {Re[#2] &, Im[#2] &}]
Or the 3D version:
ComplexPlot3D[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I},
Mesh -> 10, PlotRange -> All]
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
$endgroup$
add a comment |
$begingroup$
Note that you can also use the new (as of Version 12) ComplexPlot function, too:
ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10,
MeshFunctions -> {Re[#2] &, Im[#2] &}]
Or the 3D version:
ComplexPlot3D[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I},
Mesh -> 10, PlotRange -> All]
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
$endgroup$
Note that you can also use the new (as of Version 12) ComplexPlot function, too:
ComplexPlot[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I}, Mesh -> 10,
MeshFunctions -> {Re[#2] &, Im[#2] &}]
Or the 3D version:
ComplexPlot3D[1 - z Log[(1 + z)/z], {z, -2 - 2 I, 2 + 2 I},
Mesh -> 10, PlotRange -> All]
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
answered 2 days ago
murraymurray
6,44319 silver badges36 bronze badges
6,44319 silver badges36 bronze badges
add a comment |
add a comment |
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
Call me potato. is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
2 days ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
2 days ago
$begingroup$
Not sure, but if you have an analytical formula of where the branch cut is, then it would be quite simple to plot it on top of the original 3d plot
$endgroup$
– Fraccalo
2 days ago
2
$begingroup$
You can use
Exclusions -> Noneto get rid of the white line.$endgroup$
– C. E.
2 days ago