Problem with NSolve
I would like to plot the solutions of an equation, for different values of a parameter. This is my code
cdf[x_] := CDF[NormalDistribution[0, 1], x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p];
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0];
sol = NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]
Plot[x /. sol, {h, 0, 1}]
For $h$ between 0 and 0.6 there should be three solutions, for $h$ above 0.6 just one.
The plot, which runs in 5 minutes, returns only the solution for $h$ bigger than 0.6, while I believe I should get 3 different lines.
Can you help me fix this?
Thank you
equation-solving
asked Dec 16 at 18:05
Api
537
add a comment |
I would like to plot the solutions of an equation, for different values of a parameter. This is my code
cdf[x_] := CDF[NormalDistribution[0, 1], x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p];
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0];
sol = NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]
Plot[x /. sol, {h, 0, 1}]
For $h$ between 0 and 0.6 there should be three solutions, for $h$ above 0.6 just one.
The plot, which runs in 5 minutes, returns only the solution for $h$ bigger than 0.6, while I believe I should get 3 different lines.
Can you help me fix this?
Thank you
equation-solving
asked Dec 16 at 18:05
Api
537
add a comment |
I would like to plot the solutions of an equation, for different values of a parameter. This is my code
cdf[x_] := CDF[NormalDistribution[0, 1], x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p];
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0];
sol = NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]
Plot[x /. sol, {h, 0, 1}]
For $h$ between 0 and 0.6 there should be three solutions, for $h$ above 0.6 just one.
The plot, which runs in 5 minutes, returns only the solution for $h$ bigger than 0.6, while I believe I should get 3 different lines.
Can you help me fix this?
Thank you
equation-solving
asked Dec 16 at 18:05
Api
537
I would like to plot the solutions of an equation, for different values of a parameter. This is my code
cdf[x_] := CDF[NormalDistribution[0, 1], x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p];
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0];
sol = NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]
Plot[x /. sol, {h, 0, 1}]
For $h$ between 0 and 0.6 there should be three solutions, for $h$ above 0.6 just one.
The plot, which runs in 5 minutes, returns only the solution for $h$ bigger than 0.6, while I believe I should get 3 different lines.
Can you help me fix this?
Thank you
equation-solving
equation-solving
asked Dec 16 at 18:05
Api
537
asked Dec 16 at 18:05
Api
537
edited Dec 16 at 19:19
asked Dec 16 at 18:05
Api
537
asked Dec 16 at 18:05
Api
537
asked Dec 16 at 18:05
Api
537
537
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
cdf[x_] := CDF[NormalDistribution, x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p] // Simplify;
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0] // Simplify;
Clear[sol]
sol[h_?NumericQ] := NSolve[ratio[x, h] == 1/2 && (-5 < x < 5), x]
Generate data for a ListLinePlot. This is slow
hValues =
Join[Range[0, 0.1, 0.01], Range[0.15, 0.55, 0.05],
Range[0.551, 0.6, 0.001], Range[0.6, 1, 0.1]];
data = Table[{h, #} & /@ (x /. sol[h]), {h, hValues}];
data2 = GatherBy[data, Length];
ListLinePlot[{
Rest@Flatten[data2[[1]], 1],
Sequence @@ Transpose[data2[[2]]]},
PlotLegends -> Placed[Automatic, {0.75, 0.45}]]
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to useContourPlotas suggested by @UlrichNeumann.
– Bob Hanlon
Dec 16 at 21:16
add a comment |
NSolve cannot solve the equation because the equation isn't numeric(depends on h)
For a first insight of the solution use
ContourPlot[ratio[x, h] == 0.5, {x, -1, 3}, {h, 0, 1},MaxRecursion -> 5, FrameLabel -> {x, h}]
Try
sol[h_?NumericQ] := Values[NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]//Flatten]
to get parameter dependent solutions.
sol[.5]
(*{-0.193146, 1.3878, 1.94506}*)
sol[.75]
(*{-0.193147}*)
Addenum
Unfortunately the solution cannot be plotted, because the number of solutions varies ...
Looking at the Contourplot it is possible to evaluate the Contourline using NDSolve:
First we need one point of the contourline, for example point h==0:
NSolve[{ratio[x, 0] == 1/2, -3 < x < 3}, x, Reals] (*{x -> 2.12759}*)
contourline doesn't change
H = NDSolveValue[{D[ratio[x, h[x]], x] == 0 ,h[2.127591638090098`] == 0}, h,{x, -1, 3}]
The left boundary of the solution range
H["Domain"][[1]] (* {-0.193105, 2.23873} *)
x0=%[[1]]
fullfills ratio[x0,h]==1/2 x==x0 is also a solution .
Show[{Plot[H[x], {x, -1, 3}, PlotRange -> {0, 1}],ParametricPlot[{x0, h}, {h, 0, 1}]}, AxesLabel -> {x, h[x]}]
That's it, hope I could help!
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
The code isPlot[Evaluate[ sol[h]], {h, 0, 1}]but evaluation doesn't finish...
– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
|
show 2 more comments
Your Answer
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2 Answers
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cdf[x_] := CDF[NormalDistribution, x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p] // Simplify;
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0] // Simplify;
Clear[sol]
sol[h_?NumericQ] := NSolve[ratio[x, h] == 1/2 && (-5 < x < 5), x]
Generate data for a ListLinePlot. This is slow
hValues =
Join[Range[0, 0.1, 0.01], Range[0.15, 0.55, 0.05],
Range[0.551, 0.6, 0.001], Range[0.6, 1, 0.1]];
data = Table[{h, #} & /@ (x /. sol[h]), {h, hValues}];
data2 = GatherBy[data, Length];
ListLinePlot[{
Rest@Flatten[data2[[1]], 1],
Sequence @@ Transpose[data2[[2]]]},
PlotLegends -> Placed[Automatic, {0.75, 0.45}]]
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to useContourPlotas suggested by @UlrichNeumann.
– Bob Hanlon
Dec 16 at 21:16
add a comment |
cdf[x_] := CDF[NormalDistribution, x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p] // Simplify;
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0] // Simplify;
Clear[sol]
sol[h_?NumericQ] := NSolve[ratio[x, h] == 1/2 && (-5 < x < 5), x]
Generate data for a ListLinePlot. This is slow
hValues =
Join[Range[0, 0.1, 0.01], Range[0.15, 0.55, 0.05],
Range[0.551, 0.6, 0.001], Range[0.6, 1, 0.1]];
data = Table[{h, #} & /@ (x /. sol[h]), {h, hValues}];
data2 = GatherBy[data, Length];
ListLinePlot[{
Rest@Flatten[data2[[1]], 1],
Sequence @@ Transpose[data2[[2]]]},
PlotLegends -> Placed[Automatic, {0.75, 0.45}]]
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to useContourPlotas suggested by @UlrichNeumann.
– Bob Hanlon
Dec 16 at 21:16
add a comment |
cdf[x_] := CDF[NormalDistribution, x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p] // Simplify;
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0] // Simplify;
Clear[sol]
sol[h_?NumericQ] := NSolve[ratio[x, h] == 1/2 && (-5 < x < 5), x]
Generate data for a ListLinePlot. This is slow
hValues =
Join[Range[0, 0.1, 0.01], Range[0.15, 0.55, 0.05],
Range[0.551, 0.6, 0.001], Range[0.6, 1, 0.1]];
data = Table[{h, #} & /@ (x /. sol[h]), {h, hValues}];
data2 = GatherBy[data, Length];
ListLinePlot[{
Rest@Flatten[data2[[1]], 1],
Sequence @@ Transpose[data2[[2]]]},
PlotLegends -> Placed[Automatic, {0.75, 0.45}]]
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
cdf[x_] := CDF[NormalDistribution, x];
pdf[x_] := cdf'[x];
cdf2[x_] = cdf[x]^20;
pdf2[x_] = cdf2'[x];
mix[x_, h_, p_] = h pdf[x - p] + (1 - h) pdf2[x - p] // Simplify;
ratio[x_, h_] = mix[x, h, 1]/mix[x, h, 0] // Simplify;
Clear[sol]
sol[h_?NumericQ] := NSolve[ratio[x, h] == 1/2 && (-5 < x < 5), x]
Generate data for a ListLinePlot. This is slow
hValues =
Join[Range[0, 0.1, 0.01], Range[0.15, 0.55, 0.05],
Range[0.551, 0.6, 0.001], Range[0.6, 1, 0.1]];
data = Table[{h, #} & /@ (x /. sol[h]), {h, hValues}];
data2 = GatherBy[data, Length];
ListLinePlot[{
Rest@Flatten[data2[[1]], 1],
Sequence @@ Transpose[data2[[2]]]},
PlotLegends -> Placed[Automatic, {0.75, 0.45}]]
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
edited Dec 17 at 0:09
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
answered Dec 16 at 19:56
Bob Hanlon
58.8k23595
58.8k23595
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to useContourPlotas suggested by @UlrichNeumann.
– Bob Hanlon
Dec 16 at 21:16
add a comment |
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to useContourPlotas suggested by @UlrichNeumann.
– Bob Hanlon
Dec 16 at 21:16
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
Thank you for your answer, it's just great and very helpful. I've got just one question: why is there a little 'hole' between line 3 and 4? I would like to fix this if it were possible somehow. I tried to change $dh$ at .0005 but this results in an error. Thank you
– Api
Dec 16 at 20:38
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to use
ContourPlot as suggested by @UlrichNeumann.– Bob Hanlon
Dec 16 at 21:16
There is a gap because the gradient is very steep and the steps would need to be smaller as you suggest. The better solution is to use
ContourPlot as suggested by @UlrichNeumann.– Bob Hanlon
Dec 16 at 21:16
add a comment |
NSolve cannot solve the equation because the equation isn't numeric(depends on h)
For a first insight of the solution use
ContourPlot[ratio[x, h] == 0.5, {x, -1, 3}, {h, 0, 1},MaxRecursion -> 5, FrameLabel -> {x, h}]
Try
sol[h_?NumericQ] := Values[NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]//Flatten]
to get parameter dependent solutions.
sol[.5]
(*{-0.193146, 1.3878, 1.94506}*)
sol[.75]
(*{-0.193147}*)
Addenum
Unfortunately the solution cannot be plotted, because the number of solutions varies ...
Looking at the Contourplot it is possible to evaluate the Contourline using NDSolve:
First we need one point of the contourline, for example point h==0:
NSolve[{ratio[x, 0] == 1/2, -3 < x < 3}, x, Reals] (*{x -> 2.12759}*)
contourline doesn't change
H = NDSolveValue[{D[ratio[x, h[x]], x] == 0 ,h[2.127591638090098`] == 0}, h,{x, -1, 3}]
The left boundary of the solution range
H["Domain"][[1]] (* {-0.193105, 2.23873} *)
x0=%[[1]]
fullfills ratio[x0,h]==1/2 x==x0 is also a solution .
Show[{Plot[H[x], {x, -1, 3}, PlotRange -> {0, 1}],ParametricPlot[{x0, h}, {h, 0, 1}]}, AxesLabel -> {x, h[x]}]
That's it, hope I could help!
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
The code isPlot[Evaluate[ sol[h]], {h, 0, 1}]but evaluation doesn't finish...
– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
|
show 2 more comments
NSolve cannot solve the equation because the equation isn't numeric(depends on h)
For a first insight of the solution use
ContourPlot[ratio[x, h] == 0.5, {x, -1, 3}, {h, 0, 1},MaxRecursion -> 5, FrameLabel -> {x, h}]
Try
sol[h_?NumericQ] := Values[NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]//Flatten]
to get parameter dependent solutions.
sol[.5]
(*{-0.193146, 1.3878, 1.94506}*)
sol[.75]
(*{-0.193147}*)
Addenum
Unfortunately the solution cannot be plotted, because the number of solutions varies ...
Looking at the Contourplot it is possible to evaluate the Contourline using NDSolve:
First we need one point of the contourline, for example point h==0:
NSolve[{ratio[x, 0] == 1/2, -3 < x < 3}, x, Reals] (*{x -> 2.12759}*)
contourline doesn't change
H = NDSolveValue[{D[ratio[x, h[x]], x] == 0 ,h[2.127591638090098`] == 0}, h,{x, -1, 3}]
The left boundary of the solution range
H["Domain"][[1]] (* {-0.193105, 2.23873} *)
x0=%[[1]]
fullfills ratio[x0,h]==1/2 x==x0 is also a solution .
Show[{Plot[H[x], {x, -1, 3}, PlotRange -> {0, 1}],ParametricPlot[{x0, h}, {h, 0, 1}]}, AxesLabel -> {x, h[x]}]
That's it, hope I could help!
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
The code isPlot[Evaluate[ sol[h]], {h, 0, 1}]but evaluation doesn't finish...
– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
|
show 2 more comments
NSolve cannot solve the equation because the equation isn't numeric(depends on h)
For a first insight of the solution use
ContourPlot[ratio[x, h] == 0.5, {x, -1, 3}, {h, 0, 1},MaxRecursion -> 5, FrameLabel -> {x, h}]
Try
sol[h_?NumericQ] := Values[NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]//Flatten]
to get parameter dependent solutions.
sol[.5]
(*{-0.193146, 1.3878, 1.94506}*)
sol[.75]
(*{-0.193147}*)
Addenum
Unfortunately the solution cannot be plotted, because the number of solutions varies ...
Looking at the Contourplot it is possible to evaluate the Contourline using NDSolve:
First we need one point of the contourline, for example point h==0:
NSolve[{ratio[x, 0] == 1/2, -3 < x < 3}, x, Reals] (*{x -> 2.12759}*)
contourline doesn't change
H = NDSolveValue[{D[ratio[x, h[x]], x] == 0 ,h[2.127591638090098`] == 0}, h,{x, -1, 3}]
The left boundary of the solution range
H["Domain"][[1]] (* {-0.193105, 2.23873} *)
x0=%[[1]]
fullfills ratio[x0,h]==1/2 x==x0 is also a solution .
Show[{Plot[H[x], {x, -1, 3}, PlotRange -> {0, 1}],ParametricPlot[{x0, h}, {h, 0, 1}]}, AxesLabel -> {x, h[x]}]
That's it, hope I could help!
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
NSolve cannot solve the equation because the equation isn't numeric(depends on h)
For a first insight of the solution use
ContourPlot[ratio[x, h] == 0.5, {x, -1, 3}, {h, 0, 1},MaxRecursion -> 5, FrameLabel -> {x, h}]
Try
sol[h_?NumericQ] := Values[NSolve[ratio[x, h] == 0.5 && (-5 < x < 5), x]//Flatten]
to get parameter dependent solutions.
sol[.5]
(*{-0.193146, 1.3878, 1.94506}*)
sol[.75]
(*{-0.193147}*)
Addenum
Unfortunately the solution cannot be plotted, because the number of solutions varies ...
Looking at the Contourplot it is possible to evaluate the Contourline using NDSolve:
First we need one point of the contourline, for example point h==0:
NSolve[{ratio[x, 0] == 1/2, -3 < x < 3}, x, Reals] (*{x -> 2.12759}*)
contourline doesn't change
H = NDSolveValue[{D[ratio[x, h[x]], x] == 0 ,h[2.127591638090098`] == 0}, h,{x, -1, 3}]
The left boundary of the solution range
H["Domain"][[1]] (* {-0.193105, 2.23873} *)
x0=%[[1]]
fullfills ratio[x0,h]==1/2 x==x0 is also a solution .
Show[{Plot[H[x], {x, -1, 3}, PlotRange -> {0, 1}],ParametricPlot[{x0, h}, {h, 0, 1}]}, AxesLabel -> {x, h[x]}]
That's it, hope I could help!
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
edited Dec 16 at 22:35
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
answered Dec 16 at 18:17
Ulrich Neumann
7,092515
7,092515
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
The code isPlot[Evaluate[ sol[h]], {h, 0, 1}]but evaluation doesn't finish...
– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
|
show 2 more comments
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
The code isPlot[Evaluate[ sol[h]], {h, 0, 1}]but evaluation doesn't finish...
– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
I tried, but then this error was reported: SetDelayed::write: Tag List in {{x->-0.193133},{x->0.774654},{x->2.11263}}[h_?NumericQ] is Protected.
– Api
Dec 16 at 18:20
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
Would you mind to produce the code for plotting it? I am sorry, I am an absolute beginner, I tried some guesses but I always get an empty graph
– Api
Dec 16 at 18:26
1
1
The code is
Plot[Evaluate[ sol[h]], {h, 0, 1}] but evaluation doesn't finish...– Ulrich Neumann
Dec 16 at 18:28
The code is
Plot[Evaluate[ sol[h]], {h, 0, 1}] but evaluation doesn't finish...– Ulrich Neumann
Dec 16 at 18:28
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
It takes 5 minutes and then it returns a graph which again shows just one of the solutions
– Api
Dec 16 at 18:34
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
Sorry, I don't know why the Plot doesn't work as expected.
– Ulrich Neumann
Dec 16 at 18:43
|
show 2 more comments
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