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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

3

0

$begingroup$

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

share|improve this question

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

3

0

$begingroup$

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

share|improve this question

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

$begingroup$

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

share|improve this question

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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]

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; ` 

Plot[Max[Re[z]] /. ToRules[roots[a_]], a, 1, 2, 
 PlotLabel -> PlotStyle -> Red, AspectRatio -> 1]

share|improve this question

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited May 1 at 16:41

m_goldberg

89.5k873202

edited May 1 at 16:41

m_goldberg

89.5k873202

asked May 1 at 14:31

vanessavanessa

204

New contributor

vanessa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 1 at 14:31

vanessavanessa

204

1 Answer
1

active

oldest

votes

6

$begingroup$

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.

share|improve this answer

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

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6

$begingroup$

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.

share|improve this answer

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 | 

6

$begingroup$

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.

share|improve this answer

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.

share|improve this answer

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

$endgroup$

Clear["Global`*"]

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]]

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]]")]

share|improve this answer

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598

answered May 1 at 16:16

Bob HanlonBob Hanlon

62.2k33598