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I am new to the Mathematica and I have a question about the plotting. I am trying to plot the transcendental equation

$$\tan{(z)}=\sqrt{\frac{1}{2m_0}\left (\frac{z_0}{z}\right)^2-1} $$

where the parameter $z_0=4.8$. I am just wondering can I keep the parameter $\frac{1}{2m_0}$ without specifying an exact value to it? I have searched some similar questions but still don't figure this out. Could anyone give some suggestions? Thanks!

thorimur
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user81752
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1 Answers1

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In this particular case, you can view the solution set in the $z,m_0$ plane (i.e. allowing $m_0$ to vary) with ContourPlot:

z0 = 4.8;

ContourPlot[ Tan[z] == Sqrt[(1/(2 m0)) (z0/z)^2 - 1], {z, -5, 5}, {m0, 0, 2}, PlotPoints -> 50]

enter image description here (PlotPoints -> 50 just makes the resolution a bit better.)


Note: as @Artes mentions in a comment on the original question, these sorts of equations are difficult to handle. As such, the above plot might not reveal the full solution set. However, by plotting both sides of the equation in 3D and viewing the result from directly above to see where the intersections lie, it does seem we've got them all in this range.

Plot3D[{Tan[z], Sqrt[(1/(2 m0)) (z0/z)^2 - 1]}, {z, -5, 5}, {m0, 0, 2},
 PlotPoints -> 50, ViewPoint -> {0, 0, Infinity}, 
 ViewAngle -> All, Ticks -> False, AxesLabel -> Automatic, 
 PlotRange -> {-100, 100}]
thorimur
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