Issue with ContourPlotTrying 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?
Can GPL and BSD licensed applications be used for government work?
What is the purpose of this "red room" in "Stranger Things"?
Company requiring me to let them review research from before I was hired
Is the apartment I want to rent a scam?
how to add 1 milliseconds on a datetime string?
In a script how can I signal who's winning the argument?
Can 々 stand for a duplicated kanji with a different reading?
Why must API keys be kept private?
Monty Hall Problem with a Fallible Monty
How can the artificial womb be made affordable for the common people?
Why is chess failing to attract big name sponsors?
Sextortion with actual password not found in leaks
Why do people say "I am broke" instead of "I am broken"?
My current job follows "worst practices". How can I talk about my experience in an interview without giving off red flags?
Found more old paper shares from broken up companies
Issue with ContourPlot
How may I shorten this shell script?
Reference request: mod 2 cohomology of periodic KO theory
When were "acrobatics" introduced at weddings?
How can I deal with someone that wants to kill something that isn't supposed to be killed?
Why is DC so, so, so Democratic?
Why does the salt in the oceans not sink to the bottom?
Character Frequency in a String
Is there a way to factor age into the mass-luminosity relationship for stars?
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
15 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
15 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
15 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
15 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
15 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 |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f202678%2fissue-with-contourplot%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f202678%2fissue-with-contourplot%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
15 mins ago