Plotting a non-closed form of an equation

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SUMMARY

The discussion focuses on plotting the function defined by the equation (1+z)^{\frac{2}{3}}-(1-z)^{\frac{2}{3}} = B for B values ranging from 0 to 100. Users encountered difficulties with Mathematica and Matlab in calculating the inverse of the function. A solution was proposed using Mathematica's Solve function to derive z and then plot the real parts of the solutions across the specified range. It was emphasized that z must remain within the bounds of -1 to 1 to maintain the validity of the expression.

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Hi,

I need to plot this function z(B) with B=0...100 for an assignment:

[tex](1+z)^{\frac{2}{3}}-(1-z)^{\frac{2}{3}} = B[/tex]

But can't seem to discover how. Mathematica can't calculate the inverse (for rather obvious reasons), and neither can Matlab.

A solution in either program is fine. Thanks!
 
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This is just a start and I have some doubts about it's accuracy. The resulting solution has roots which Mathematica extracts default values which may not be the values you want. Would need to work with it further and back-substitute values both for z and b to make sure your answer is correct. For example, what happens to (1-z)^(2/3) when z becomes greater than one?

Code:
mysols = z /. Solve[(1 + z)^(2/3) - (1 - z)^(2/3) == b, z]
Plot[Re[mysols], {b, 0, 100}]
 
Last edited:
jackmell said:
For example, what happens to (1-z)^(2/3) when z becomes greater than one?

Thanks for the answer, will check on it tomorrow (too late here now). z is defined to be no greater than 1 in absolute value, so the expression is symmetrical in z and there is no z>1.

PS: z is actually [tex]\zeta[/tex]=polarisation)
 

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