To plot branch cut of logarithmVisualizing 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 $sqrtx^2-1$?How to Approximate at Non-differentiable Point (forced Series Expansion around Branch Cut)Visualizing the complex logarithm
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To plot branch cut of logarithm
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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 $sqrtx^2-1$?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 10 hours ago
Call me potato.Call me potato.
304 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 10 hours ago
Call me potato.Call me potato.
304 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
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours 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
10 hours ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
10 hours 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 10 hours ago
Call me potato.Call me potato.
304 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 10 hours ago
Call me potato.Call me potato.
304 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 10 hours ago
Call me potato.Call me potato.
304 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 10 hours ago
Call me potato.
asked 10 hours ago
Call me potato.Call me potato.
304 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 10 hours ago
Call me potato.Call me potato.
304 bronze badges
asked 10 hours ago
Call me potato.Call me potato.
304 bronze badges
304 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
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours 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
10 hours ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
10 hours ago
add a comment |
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours 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
10 hours ago
2
$begingroup$
You can useExclusions -> Noneto get rid of the white line.
$endgroup$
– C. E.
10 hours ago
1
1
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
10 hours ago
$begingroup$
Add the plot option: PlotRange -> All
$endgroup$
– Fraccalo
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours 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
10 hours 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
10 hours ago
2
2
$begingroup$
You can use
Exclusions -> None to get rid of the white line.$endgroup$
– C. E.
10 hours ago
$begingroup$
You can use
Exclusions -> None to get rid of the white line.$endgroup$
– C. E.
10 hours 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 9 hours ago
FraccaloFraccalo
2,9306 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 1 hour ago
murraymurray
6,42319 silver badges36 bronze badges
$endgroup$
add a comment |
Your Answer
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2 Answers
2
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 9 hours ago
FraccaloFraccalo
2,9306 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 9 hours ago
FraccaloFraccalo
2,9306 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 9 hours ago
FraccaloFraccalo
2,9306 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 9 hours ago
FraccaloFraccalo
2,9306 silver badges18 bronze badges
answered 9 hours ago
FraccaloFraccalo
2,9306 silver badges18 bronze badges
answered 9 hours ago
FraccaloFraccalo
2,9306 silver badges18 bronze badges
answered 9 hours ago
FraccaloFraccalo
2,9306 silver badges18 bronze badges
2,9306 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 1 hour ago
murraymurray
6,42319 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 1 hour ago
murraymurray
6,42319 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 1 hour ago
murraymurray
6,42319 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 1 hour ago
murraymurray
6,42319 silver badges36 bronze badges
answered 1 hour ago
murraymurray
6,42319 silver badges36 bronze badges
answered 1 hour ago
murraymurray
6,42319 silver badges36 bronze badges
answered 1 hour ago
murraymurray
6,42319 silver badges36 bronze badges
6,42319 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
10 hours ago
$begingroup$
@Fraccalo Many thanks! Can we have a smooth line for the brach cut rather than "white line" discontinuity?
$endgroup$
– Call me potato.
10 hours 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
10 hours ago
2
$begingroup$
You can use
Exclusions -> Noneto get rid of the white line.$endgroup$
– C. E.
10 hours ago