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Issue with ContourPlot
Trying to plot ugly expression, not workingHow can I get an approximate contour plot from a 3D plot?Weird plot with SphericalPlot3DParametricPlot3D etc. with parameters satisfying an implicit relationPlotting a parametrically defined vector fieldHow to invert an Elliptic function where the elliptic nome is a function of an independent variable?How to plot this equation for $x,y$?ContourPlot - how to assign specific colours to level curves?Performing a FindRoot from Numerical integrationHow to control the labeling of axes in ContourPlot?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am trying to Contour plot the solutions "$x$" of an equation versus $k$ for some fixed $T$ values. However, the contour plot is giving incorrect plot perhaps due to some accuracy issues. e.g., at small $T$ ($Tleq 1$) I should get three solutions, but I am not getting three on the contour plot.?
The code:
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] + T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) == 0 /. A -> 1/2, B -> 0.00001, z -> -2.37
ContourPlot[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi]
plotting numerics
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
$endgroup$
add a comment |
$begingroup$
I am trying to Contour plot the solutions "$x$" of an equation versus $k$ for some fixed $T$ values. However, the contour plot is giving incorrect plot perhaps due to some accuracy issues. e.g., at small $T$ ($Tleq 1$) I should get three solutions, but I am not getting three on the contour plot.?
The code:
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] + T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) == 0 /. A -> 1/2, B -> 0.00001, z -> -2.37
ContourPlot[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi]
plotting numerics
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
$endgroup$
$begingroup$
You say three, but I get six:NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Close, but distinct:wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7]---ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]
$endgroup$
– Michael E2
23 mins ago
add a comment |
$begingroup$
I am trying to Contour plot the solutions "$x$" of an equation versus $k$ for some fixed $T$ values. However, the contour plot is giving incorrect plot perhaps due to some accuracy issues. e.g., at small $T$ ($Tleq 1$) I should get three solutions, but I am not getting three on the contour plot.?
The code:
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] + T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) == 0 /. A -> 1/2, B -> 0.00001, z -> -2.37
ContourPlot[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi]
plotting numerics
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
$endgroup$
I am trying to Contour plot the solutions "$x$" of an equation versus $k$ for some fixed $T$ values. However, the contour plot is giving incorrect plot perhaps due to some accuracy issues. e.g., at small $T$ ($Tleq 1$) I should get three solutions, but I am not getting three on the contour plot.?
The code:
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] + T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) == 0 /. A -> 1/2, B -> 0.00001, z -> -2.37
ContourPlot[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi]
plotting numerics
plotting numerics
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
edited 8 hours ago
user64494
4,1472 gold badges14 silver badges23 bronze badges
4,1472 gold badges14 silver badges23 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
asked 8 hours ago
AtoZAtoZ
1606 bronze badges
1606 bronze badges
$begingroup$
You say three, but I get six:NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Close, but distinct:wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7]---ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]
$endgroup$
– Michael E2
23 mins ago
add a comment |
$begingroup$
You say three, but I get six:NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Close, but distinct:wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7]---ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]
$endgroup$
– Michael E2
23 mins ago
$begingroup$
You say three, but I get six:
NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]$endgroup$
– Michael E2
1 hour ago
$begingroup$
You say three, but I get six:
NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]$endgroup$
– Michael E2
1 hour ago
$begingroup$
Close, but distinct:
wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7] --- ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]$endgroup$
– Michael E2
23 mins ago
$begingroup$
Close, but distinct:
wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7] --- ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]$endgroup$
– Michael E2
23 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Not an answer but if you do
Plot3D[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi,
PlotRange -> All]
Division by the sin function has zeros which dominate the plot. What are you hoping for?
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thanks. When employing the numerical solution bysol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1]for example gives solution for $T=1$, $k=0.1$, which is2.64172, 2.94159, 2.98454i.e., three solutions. But on the contour plot, I see only two?
$endgroup$
– AtoZ
8 hours ago
add a comment |
$begingroup$
It appears that the solution of x -> 2.94159 that you mention in a comment in @Hugh 's answer is incorrect (or maybe better said as "inappropriate"). (@Hugh 's answer essentially says it all.)
That particular solution is really at a discontinuity. If you plug in $pi-2/10$ for $x$ (i.e., essentially what x -> 2.94159 means), you'll get
Consider a slight change in your equation (rationalizing the constants and removing the ==0):
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] +
T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) /.
A -> 1/2, B -> 10^-5, z -> -237/100
Then we have
eq[x, 1/10, 1]
If $pi-2/10$ is substituted for $x$, then $csc left(x+frac15right)$ results in ComplexInfinity:
Csc[1/5 + x] /. x -> π - 2/10
So, in short, there are only two contour lines of zero, not three, when $T=0.1$.
Addition
In general we have for eq[x,k,T]
The discontinuity occurs when Csc[2 k + x] is ComplexInfinity or whenever $2k+x=pi$ or $2k+x=2pi$ no matter what the value of $T$ happens to be for the ranges of interest of $x$ and $k$. So you could always include those (dotted) lines of discontinuity to your contour plots:
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not an answer but if you do
Plot3D[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi,
PlotRange -> All]
Division by the sin function has zeros which dominate the plot. What are you hoping for?
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thanks. When employing the numerical solution bysol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1]for example gives solution for $T=1$, $k=0.1$, which is2.64172, 2.94159, 2.98454i.e., three solutions. But on the contour plot, I see only two?
$endgroup$
– AtoZ
8 hours ago
add a comment |
$begingroup$
Not an answer but if you do
Plot3D[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi,
PlotRange -> All]
Division by the sin function has zeros which dominate the plot. What are you hoping for?
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
$endgroup$
$begingroup$
Thanks. When employing the numerical solution bysol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1]for example gives solution for $T=1$, $k=0.1$, which is2.64172, 2.94159, 2.98454i.e., three solutions. But on the contour plot, I see only two?
$endgroup$
– AtoZ
8 hours ago
add a comment |
$begingroup$
Not an answer but if you do
Plot3D[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi,
PlotRange -> All]
Division by the sin function has zeros which dominate the plot. What are you hoping for?
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
$endgroup$
Not an answer but if you do
Plot3D[Evaluate[eq[x, k, 0.1]], x, 0, Pi, k, 0, Pi,
PlotRange -> All]
Division by the sin function has zeros which dominate the plot. What are you hoping for?
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
answered 8 hours ago
HughHugh
7,2412 gold badges19 silver badges47 bronze badges
7,2412 gold badges19 silver badges47 bronze badges
$begingroup$
Thanks. When employing the numerical solution bysol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1]for example gives solution for $T=1$, $k=0.1$, which is2.64172, 2.94159, 2.98454i.e., three solutions. But on the contour plot, I see only two?
$endgroup$
– AtoZ
8 hours ago
add a comment |
$begingroup$
Thanks. When employing the numerical solution bysol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1]for example gives solution for $T=1$, $k=0.1$, which is2.64172, 2.94159, 2.98454i.e., three solutions. But on the contour plot, I see only two?
$endgroup$
– AtoZ
8 hours ago
$begingroup$
Thanks. When employing the numerical solution by
sol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1] for example gives solution for $T=1$, $k=0.1$, which is 2.64172, 2.94159, 2.98454 i.e., three solutions. But on the contour plot, I see only two?$endgroup$
– AtoZ
8 hours ago
$begingroup$
Thanks. When employing the numerical solution by
sol[T_?NumericQ, k_ /; 0 <= k <= Pi] := x /. NSolve[eq[x, k, T], 0 <= x <= Pi, x, Reals] sol[1, 0.1] for example gives solution for $T=1$, $k=0.1$, which is 2.64172, 2.94159, 2.98454 i.e., three solutions. But on the contour plot, I see only two?$endgroup$
– AtoZ
8 hours ago
add a comment |
$begingroup$
It appears that the solution of x -> 2.94159 that you mention in a comment in @Hugh 's answer is incorrect (or maybe better said as "inappropriate"). (@Hugh 's answer essentially says it all.)
That particular solution is really at a discontinuity. If you plug in $pi-2/10$ for $x$ (i.e., essentially what x -> 2.94159 means), you'll get
Consider a slight change in your equation (rationalizing the constants and removing the ==0):
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] +
T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) /.
A -> 1/2, B -> 10^-5, z -> -237/100
Then we have
eq[x, 1/10, 1]
If $pi-2/10$ is substituted for $x$, then $csc left(x+frac15right)$ results in ComplexInfinity:
Csc[1/5 + x] /. x -> π - 2/10
So, in short, there are only two contour lines of zero, not three, when $T=0.1$.
Addition
In general we have for eq[x,k,T]
The discontinuity occurs when Csc[2 k + x] is ComplexInfinity or whenever $2k+x=pi$ or $2k+x=2pi$ no matter what the value of $T$ happens to be for the ranges of interest of $x$ and $k$. So you could always include those (dotted) lines of discontinuity to your contour plots:
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
$endgroup$
add a comment |
$begingroup$
It appears that the solution of x -> 2.94159 that you mention in a comment in @Hugh 's answer is incorrect (or maybe better said as "inappropriate"). (@Hugh 's answer essentially says it all.)
That particular solution is really at a discontinuity. If you plug in $pi-2/10$ for $x$ (i.e., essentially what x -> 2.94159 means), you'll get
Consider a slight change in your equation (rationalizing the constants and removing the ==0):
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] +
T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) /.
A -> 1/2, B -> 10^-5, z -> -237/100
Then we have
eq[x, 1/10, 1]
If $pi-2/10$ is substituted for $x$, then $csc left(x+frac15right)$ results in ComplexInfinity:
Csc[1/5 + x] /. x -> π - 2/10
So, in short, there are only two contour lines of zero, not three, when $T=0.1$.
Addition
In general we have for eq[x,k,T]
The discontinuity occurs when Csc[2 k + x] is ComplexInfinity or whenever $2k+x=pi$ or $2k+x=2pi$ no matter what the value of $T$ happens to be for the ranges of interest of $x$ and $k$. So you could always include those (dotted) lines of discontinuity to your contour plots:
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
$endgroup$
add a comment |
$begingroup$
It appears that the solution of x -> 2.94159 that you mention in a comment in @Hugh 's answer is incorrect (or maybe better said as "inappropriate"). (@Hugh 's answer essentially says it all.)
That particular solution is really at a discontinuity. If you plug in $pi-2/10$ for $x$ (i.e., essentially what x -> 2.94159 means), you'll get
Consider a slight change in your equation (rationalizing the constants and removing the ==0):
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] +
T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) /.
A -> 1/2, B -> 10^-5, z -> -237/100
Then we have
eq[x, 1/10, 1]
If $pi-2/10$ is substituted for $x$, then $csc left(x+frac15right)$ results in ComplexInfinity:
Csc[1/5 + x] /. x -> π - 2/10
So, in short, there are only two contour lines of zero, not three, when $T=0.1$.
Addition
In general we have for eq[x,k,T]
The discontinuity occurs when Csc[2 k + x] is ComplexInfinity or whenever $2k+x=pi$ or $2k+x=2pi$ no matter what the value of $T$ happens to be for the ranges of interest of $x$ and $k$. So you could always include those (dotted) lines of discontinuity to your contour plots:
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
$endgroup$
It appears that the solution of x -> 2.94159 that you mention in a comment in @Hugh 's answer is incorrect (or maybe better said as "inappropriate"). (@Hugh 's answer essentially says it all.)
That particular solution is really at a discontinuity. If you plug in $pi-2/10$ for $x$ (i.e., essentially what x -> 2.94159 means), you'll get
Consider a slight change in your equation (rationalizing the constants and removing the ==0):
eq[x_, k_, T_] := -Sin[3*k + x]/Sin[2*k + x] + z + 2*Cos[k] +
T^2 + (A*T^2*Sin[k]^2)/(Sin[2*k + x]^2 + B*T^4*Sin[k]^2) /.
A -> 1/2, B -> 10^-5, z -> -237/100
Then we have
eq[x, 1/10, 1]
If $pi-2/10$ is substituted for $x$, then $csc left(x+frac15right)$ results in ComplexInfinity:
Csc[1/5 + x] /. x -> π - 2/10
So, in short, there are only two contour lines of zero, not three, when $T=0.1$.
Addition
In general we have for eq[x,k,T]
The discontinuity occurs when Csc[2 k + x] is ComplexInfinity or whenever $2k+x=pi$ or $2k+x=2pi$ no matter what the value of $T$ happens to be for the ranges of interest of $x$ and $k$. So you could always include those (dotted) lines of discontinuity to your contour plots:
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
edited 3 hours ago
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
answered 6 hours ago
JimBJimB
20k1 gold badge28 silver badges65 bronze badges
20k1 gold badge28 silver badges65 bronze badges
add a comment |
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$begingroup$
You say three, but I get six:
NSolve[eq[x, k, 1/10] /. x -> 1.5, 0 < k < Pi, k, WorkingPrecision -> 16]$endgroup$
– Michael E2
1 hour ago
$begingroup$
Close, but distinct:
wp = 100; k0 = k /. NSolve[eq[x, k, 1/10] == 0 /. x -> x0, 0 < k < Pi, k, WorkingPrecision -> wp]; ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[2]] - 1*^-7, k0[[2]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[2]] - 1*^-7, k0[[2]] + 1*^-7]---ListLinePlot[ Table[ eq[1, k, 1/10], k, k0[[3]] - 1*^-7, k0[[3]] + 1*^-7, 1*^-9], GridLines -> k0, None, DataRange -> k0[[3]] - 1*^-7, k0[[3]] + 1*^-7]$endgroup$
– Michael E2
23 mins ago