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Overlay of two functions leaves gaps


Cannot Plot FunctionStability analysis of transcendental equation (stability crossing curves)Implicitly defined compact complicated surfaceHow to invert an Elliptic function where the elliptic nome is a function of an independent variable?Visualizing the primes with the Riemann Zeta functionGroup delay of a transfer functionHow would one go about plotting this parameterized curve using numerical resources (analitically it's too hard)?How to study the behavior of this series in Mathematica?What am I doing wrong when trying to plot this function?ComplexPlot3D and essential singularities













2












$begingroup$


I have a function defined as:



$rho_mleft(epsilon,mright)=left[-2epsilon rpmleft(4epsilon^2r^2+mlambda r^3right)^frac12right]^frac12$



I want to plot it for some $min mathbbZ$, so I wrote this code:



Clear[r,[Lambda]];
[Lambda]=685*10^-9;
r=25*10^-3;
[Rho]1[[Epsilon]_,m_]=(-2*[Epsilon]*r+(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
[Rho]2[[Epsilon]_,m_]=(-2*[Epsilon]*r-(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
M=Range[-5,5,1];
p1=Show[Plot[[Rho]1[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
p2=Show[Plot[[Rho]2[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
Show[p1,p2]


Which outputs:



enter image description here



However, there are some tiny gaps where the two functiosn meet, but I was expecting them to be continuous. How can I fix that?






plotting graphics







share|improve this question









$endgroup$











  • $begingroup$
    Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
    $endgroup$
    – Bill
    8 hours ago










  • $begingroup$
    I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
    $endgroup$
    – C. E.
    8 hours ago















2












$begingroup$


I have a function defined as:



$rho_mleft(epsilon,mright)=left[-2epsilon rpmleft(4epsilon^2r^2+mlambda r^3right)^frac12right]^frac12$



I want to plot it for some $min mathbbZ$, so I wrote this code:



Clear[r,[Lambda]];
[Lambda]=685*10^-9;
r=25*10^-3;
[Rho]1[[Epsilon]_,m_]=(-2*[Epsilon]*r+(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
[Rho]2[[Epsilon]_,m_]=(-2*[Epsilon]*r-(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
M=Range[-5,5,1];
p1=Show[Plot[[Rho]1[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
p2=Show[Plot[[Rho]2[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
Show[p1,p2]


Which outputs:



enter image description here



However, there are some tiny gaps where the two functiosn meet, but I was expecting them to be continuous. How can I fix that?






plotting graphics







share|improve this question









$endgroup$











  • $begingroup$
    Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
    $endgroup$
    – Bill
    8 hours ago










  • $begingroup$
    I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
    $endgroup$
    – C. E.
    8 hours ago













2












2








2





$begingroup$


I have a function defined as:



$rho_mleft(epsilon,mright)=left[-2epsilon rpmleft(4epsilon^2r^2+mlambda r^3right)^frac12right]^frac12$



I want to plot it for some $min mathbbZ$, so I wrote this code:



Clear[r,[Lambda]];
[Lambda]=685*10^-9;
r=25*10^-3;
[Rho]1[[Epsilon]_,m_]=(-2*[Epsilon]*r+(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
[Rho]2[[Epsilon]_,m_]=(-2*[Epsilon]*r-(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
M=Range[-5,5,1];
p1=Show[Plot[[Rho]1[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
p2=Show[Plot[[Rho]2[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
Show[p1,p2]


Which outputs:



enter image description here



However, there are some tiny gaps where the two functiosn meet, but I was expecting them to be continuous. How can I fix that?






plotting graphics







share|improve this question









$endgroup$




I have a function defined as:



$rho_mleft(epsilon,mright)=left[-2epsilon rpmleft(4epsilon^2r^2+mlambda r^3right)^frac12right]^frac12$



I want to plot it for some $min mathbbZ$, so I wrote this code:



Clear[r,[Lambda]];
[Lambda]=685*10^-9;
r=25*10^-3;
[Rho]1[[Epsilon]_,m_]=(-2*[Epsilon]*r+(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
[Rho]2[[Epsilon]_,m_]=(-2*[Epsilon]*r-(4*[Epsilon]^2*r^2+m*[Lambda]*r^3)^(1/2))^(1/2);
M=Range[-5,5,1];
p1=Show[Plot[[Rho]1[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
p2=Show[Plot[[Rho]2[[Epsilon]*10^-3,#]*10^3, [Epsilon],-0.5,0.5, PlotRange -> -0.5,0.5,0, 5,AxesOrigin->-0.5,0,PlotTheme->"Monochrome"] & /@ M];
Show[p1,p2]


Which outputs:



enter image description here



However, there are some tiny gaps where the two functiosn meet, but I was expecting them to be continuous. How can I fix that?






plotting graphics




plotting graphics






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









RodrigoRodrigo

1056




1056











  • $begingroup$
    Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
    $endgroup$
    – Bill
    8 hours ago










  • $begingroup$
    I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
    $endgroup$
    – C. E.
    8 hours ago
















  • $begingroup$
    Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
    $endgroup$
    – Bill
    8 hours ago










  • $begingroup$
    I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
    $endgroup$
    – C. E.
    8 hours ago















$begingroup$
Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
$endgroup$
– Bill
8 hours ago




$begingroup$
Adding the option PlotPoints->1000 to both your Plots will make those gaps much less visible.
$endgroup$
– Bill
8 hours ago












$begingroup$
I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
$endgroup$
– C. E.
8 hours ago




$begingroup$
I think that the problem may be that the functions become imaginary at $epsilon = 0$. Plot doesn't plot anything at all when the value is imaginary. When it happens precisely at the point where they're supposed to meet I guess it becomes a numerical issue, hence why PlotPoints may help.
$endgroup$
– C. E.
8 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

If you turn the equation around and plot $epsilon$ as a function of $rho$, then there are no gaps and no branches:



λ = 685*10^-9;
r = 25*10^-3;
ParametricPlot[Table[10^3 (m r^3 λ - ρ^4)/(4 r ρ^2), ρ, m, -5, 5],
 ρ, 0, 5*10^-3, AspectRatio -> 1/GoldenRatio]


enter image description here






share|improve this answer









$endgroup$













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    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    If you turn the equation around and plot $epsilon$ as a function of $rho$, then there are no gaps and no branches:



    λ = 685*10^-9;
    r = 25*10^-3;
    ParametricPlot[Table[10^3 (m r^3 λ - ρ^4)/(4 r ρ^2), ρ, m, -5, 5],
     ρ, 0, 5*10^-3, AspectRatio -> 1/GoldenRatio]


    enter image description here






    share|improve this answer









    $endgroup$

















      5












      $begingroup$

      If you turn the equation around and plot $epsilon$ as a function of $rho$, then there are no gaps and no branches:



      λ = 685*10^-9;
      r = 25*10^-3;
      ParametricPlot[Table[10^3 (m r^3 λ - ρ^4)/(4 r ρ^2), ρ, m, -5, 5],
       ρ, 0, 5*10^-3, AspectRatio -> 1/GoldenRatio]


      enter image description here






      share|improve this answer









      $endgroup$















        5












        5








        5





        $begingroup$

        If you turn the equation around and plot $epsilon$ as a function of $rho$, then there are no gaps and no branches:



        λ = 685*10^-9;
        r = 25*10^-3;
        ParametricPlot[Table[10^3 (m r^3 λ - ρ^4)/(4 r ρ^2), ρ, m, -5, 5],
         ρ, 0, 5*10^-3, AspectRatio -> 1/GoldenRatio]


        enter image description here






        share|improve this answer









        $endgroup$



        If you turn the equation around and plot $epsilon$ as a function of $rho$, then there are no gaps and no branches:



        λ = 685*10^-9;
        r = 25*10^-3;
        ParametricPlot[Table[10^3 (m r^3 λ - ρ^4)/(4 r ρ^2), ρ, m, -5, 5],
         ρ, 0, 5*10^-3, AspectRatio -> 1/GoldenRatio]


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 8 hours ago









        RomanRoman

        6,29611132




        6,29611132



























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