Finding the maximum real part of rootsPlot the maximum of the real parts of the eigenvalues of a trancendental equationPlot cubic root which includes the negative graphFindRoot gives a wrong solution which obviously should not be thereEvaluating the real part of an expressionStability analysis of transcendental equation (stability crossing curves)A question about ContourPlotFinding Real Roots and Determining RangeThe function is real, while its integral is complexFinding the real solutions using the ones from a complexificationHow to separate the real part from the imaginary part?Finding the real part of a complicated complex expression
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Finding the maximum real part of roots
Plot the maximum of the real parts of the eigenvalues of a trancendental equationPlot cubic root which includes the negative graphFindRoot gives a wrong solution which obviously should not be thereEvaluating the real part of an expressionStability analysis of transcendental equation (stability crossing curves)A question about ContourPlotFinding Real Roots and Determining RangeThe function is real, while its integral is complexFinding the real solutions using the ones from a complexificationHow to separate the real part from the imaginary part?Finding the real part of a complicated complex expression
$begingroup$
Suppose that I have this problem
roots =
Reduce[
Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet;
ListPlot[Re[z], Im[z] /. ToRules[roots],
PlotLabel ->
Style[TraditionalForm[Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]]], 14],
PlotStyle -> Red, AspectRatio -> 1]
Thus as in https://www.wolfram.com/mathematica/newin7/content/TranscendentalRoots/PlotTheRootsOfANestedTranscendentalEquation.html
I get a beautiful solution. Suppose now that this equation depends on quantity a in a range of (1, 2).
roots[a_] :=
Reduce[
Sin[z + Sin[z + Sin[z]]] == a + a Cos[z + a Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet; `
But I don't want the real and imaginary part for each vale of a specified a, rather I would like to have a plot that is a continuous function of a, and the maximum of the real part of the z.
Is there any way to do it?
I have tried this,
Plot[Max[Re[z]] /. ToRules[roots[a_]], a, 1, 2,
PlotLabel -> PlotStyle -> Red, AspectRatio -> 1]
but it has been running for a day and I still have not got any result.
plotting equation-solving complex
edited May 1 at 16:41
m_goldberg
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
$endgroup$
add a comment |
$begingroup$
Suppose that I have this problem
roots =
Reduce[
Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet;
ListPlot[Re[z], Im[z] /. ToRules[roots],
PlotLabel ->
Style[TraditionalForm[Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]]], 14],
PlotStyle -> Red, AspectRatio -> 1]
Thus as in https://www.wolfram.com/mathematica/newin7/content/TranscendentalRoots/PlotTheRootsOfANestedTranscendentalEquation.html
I get a beautiful solution. Suppose now that this equation depends on quantity a in a range of (1, 2).
roots[a_] :=
Reduce[
Sin[z + Sin[z + Sin[z]]] == a + a Cos[z + a Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet; `
But I don't want the real and imaginary part for each vale of a specified a, rather I would like to have a plot that is a continuous function of a, and the maximum of the real part of the z.
Is there any way to do it?
I have tried this,
Plot[Max[Re[z]] /. ToRules[roots[a_]], a, 1, 2,
PlotLabel -> PlotStyle -> Red, AspectRatio -> 1]
but it has been running for a day and I still have not got any result.
plotting equation-solving complex
edited May 1 at 16:41
m_goldberg
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
$endgroup$
1
$begingroup$
Try discretizing theavalue in the relevant range. If you assume the result to be smooth, you'll get a good approximation.
$endgroup$
– Kagaratsch
May 1 at 16:14
add a comment |
$begingroup$
Suppose that I have this problem
roots =
Reduce[
Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet;
ListPlot[Re[z], Im[z] /. ToRules[roots],
PlotLabel ->
Style[TraditionalForm[Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]]], 14],
PlotStyle -> Red, AspectRatio -> 1]
Thus as in https://www.wolfram.com/mathematica/newin7/content/TranscendentalRoots/PlotTheRootsOfANestedTranscendentalEquation.html
I get a beautiful solution. Suppose now that this equation depends on quantity a in a range of (1, 2).
roots[a_] :=
Reduce[
Sin[z + Sin[z + Sin[z]]] == a + a Cos[z + a Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet; `
But I don't want the real and imaginary part for each vale of a specified a, rather I would like to have a plot that is a continuous function of a, and the maximum of the real part of the z.
Is there any way to do it?
I have tried this,
Plot[Max[Re[z]] /. ToRules[roots[a_]], a, 1, 2,
PlotLabel -> PlotStyle -> Red, AspectRatio -> 1]
but it has been running for a day and I still have not got any result.
plotting equation-solving complex
edited May 1 at 16:41
m_goldberg
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
$endgroup$
Suppose that I have this problem
roots =
Reduce[
Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet;
ListPlot[Re[z], Im[z] /. ToRules[roots],
PlotLabel ->
Style[TraditionalForm[Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]]], 14],
PlotStyle -> Red, AspectRatio -> 1]
Thus as in https://www.wolfram.com/mathematica/newin7/content/TranscendentalRoots/PlotTheRootsOfANestedTranscendentalEquation.html
I get a beautiful solution. Suppose now that this equation depends on quantity a in a range of (1, 2).
roots[a_] :=
Reduce[
Sin[z + Sin[z + Sin[z]]] == a + a Cos[z + a Cos[z + Cos[z]]] &&
-3 < Re[z] < 3 && -3 < Im[z] < 3, z] // Quiet; `
But I don't want the real and imaginary part for each vale of a specified a, rather I would like to have a plot that is a continuous function of a, and the maximum of the real part of the z.
Is there any way to do it?
I have tried this,
Plot[Max[Re[z]] /. ToRules[roots[a_]], a, 1, 2,
PlotLabel -> PlotStyle -> Red, AspectRatio -> 1]
but it has been running for a day and I still have not got any result.
plotting equation-solving complex
plotting equation-solving complex
edited May 1 at 16:41
m_goldberg
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
edited May 1 at 16:41
m_goldberg
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
edited May 1 at 16:41
m_goldberg
89.9k873203
edited May 1 at 16:41
m_goldberg
89.9k873203
edited May 1 at 16:41
m_goldberg
89.9k873203
89.9k873203
asked May 1 at 14:31
vanessavanessa
204
asked May 1 at 14:31
vanessavanessa
204
asked May 1 at 14:31
vanessavanessa
204
204
1
$begingroup$
Try discretizing theavalue in the relevant range. If you assume the result to be smooth, you'll get a good approximation.
$endgroup$
– Kagaratsch
May 1 at 16:14
add a comment |
1
$begingroup$
Try discretizing theavalue in the relevant range. If you assume the result to be smooth, you'll get a good approximation.
$endgroup$
– Kagaratsch
May 1 at 16:14
1
1
$begingroup$
Try discretizing the
a value in the relevant range. If you assume the result to be smooth, you'll get a good approximation.$endgroup$
– Kagaratsch
May 1 at 16:14
$begingroup$
Try discretizing the
a value in the relevant range. If you assume the result to be smooth, you'll get a good approximation.$endgroup$
– Kagaratsch
May 1 at 16:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Clear["Global`*"]
Use a numeric technique, i.e., NSolve.
f[a_?NumericQ] :=
Max@Re[z /.
NSolve[Sin[z + Sin[z + Sin[z]]] ==
a + Cos[z + a*Cos[z + a*Cos[z]]] && -3 < Re[z] < 3 &&
-3 < Im[z] < 3 && 1 <= a <= 2, z]]
Since f uses a numeric technique, its argument is restricted to numeric values by using PatternTest with NumericQ. Even using numeric techniques the calculations are slow.
AbsoluteTiming[data = Table[a, f[a], a, 1, 2, .025];]
(* 803.729, Null *)
ListLinePlot[data,
Frame -> True,
FrameLabel ->
(Style[#, 12, Bold] & /@ "a", "Max[Re[z]]")]
The plot is not smooth so if you were to use Plot its adaptive sampling would further increase the time required.
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
$endgroup$
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Clear["Global`*"]
Use a numeric technique, i.e., NSolve.
f[a_?NumericQ] :=
Max@Re[z /.
NSolve[Sin[z + Sin[z + Sin[z]]] ==
a + Cos[z + a*Cos[z + a*Cos[z]]] && -3 < Re[z] < 3 &&
-3 < Im[z] < 3 && 1 <= a <= 2, z]]
Since f uses a numeric technique, its argument is restricted to numeric values by using PatternTest with NumericQ. Even using numeric techniques the calculations are slow.
AbsoluteTiming[data = Table[a, f[a], a, 1, 2, .025];]
(* 803.729, Null *)
ListLinePlot[data,
Frame -> True,
FrameLabel ->
(Style[#, 12, Bold] & /@ "a", "Max[Re[z]]")]
The plot is not smooth so if you were to use Plot its adaptive sampling would further increase the time required.
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
$endgroup$
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
add a comment |
$begingroup$
Clear["Global`*"]
Use a numeric technique, i.e., NSolve.
f[a_?NumericQ] :=
Max@Re[z /.
NSolve[Sin[z + Sin[z + Sin[z]]] ==
a + Cos[z + a*Cos[z + a*Cos[z]]] && -3 < Re[z] < 3 &&
-3 < Im[z] < 3 && 1 <= a <= 2, z]]
Since f uses a numeric technique, its argument is restricted to numeric values by using PatternTest with NumericQ. Even using numeric techniques the calculations are slow.
AbsoluteTiming[data = Table[a, f[a], a, 1, 2, .025];]
(* 803.729, Null *)
ListLinePlot[data,
Frame -> True,
FrameLabel ->
(Style[#, 12, Bold] & /@ "a", "Max[Re[z]]")]
The plot is not smooth so if you were to use Plot its adaptive sampling would further increase the time required.
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
$endgroup$
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
add a comment |
$begingroup$
Clear["Global`*"]
Use a numeric technique, i.e., NSolve.
f[a_?NumericQ] :=
Max@Re[z /.
NSolve[Sin[z + Sin[z + Sin[z]]] ==
a + Cos[z + a*Cos[z + a*Cos[z]]] && -3 < Re[z] < 3 &&
-3 < Im[z] < 3 && 1 <= a <= 2, z]]
Since f uses a numeric technique, its argument is restricted to numeric values by using PatternTest with NumericQ. Even using numeric techniques the calculations are slow.
AbsoluteTiming[data = Table[a, f[a], a, 1, 2, .025];]
(* 803.729, Null *)
ListLinePlot[data,
Frame -> True,
FrameLabel ->
(Style[#, 12, Bold] & /@ "a", "Max[Re[z]]")]
The plot is not smooth so if you were to use Plot its adaptive sampling would further increase the time required.
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
$endgroup$
Clear["Global`*"]
Use a numeric technique, i.e., NSolve.
f[a_?NumericQ] :=
Max@Re[z /.
NSolve[Sin[z + Sin[z + Sin[z]]] ==
a + Cos[z + a*Cos[z + a*Cos[z]]] && -3 < Re[z] < 3 &&
-3 < Im[z] < 3 && 1 <= a <= 2, z]]
Since f uses a numeric technique, its argument is restricted to numeric values by using PatternTest with NumericQ. Even using numeric techniques the calculations are slow.
AbsoluteTiming[data = Table[a, f[a], a, 1, 2, .025];]
(* 803.729, Null *)
ListLinePlot[data,
Frame -> True,
FrameLabel ->
(Style[#, 12, Bold] & /@ "a", "Max[Re[z]]")]
The plot is not smooth so if you were to use Plot its adaptive sampling would further increase the time required.
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
answered May 1 at 16:16
Bob HanlonBob Hanlon
62.4k33598
62.4k33598
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
add a comment |
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
$begingroup$
It took longer for me (2673.97 seconds), but it work! Thank you so much for your clever answer!
$endgroup$
– vanessa
May 1 at 17:57
add a comment |
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$begingroup$
Try discretizing the
avalue in the relevant range. If you assume the result to be smooth, you'll get a good approximation.$endgroup$
– Kagaratsch
May 1 at 16:14