Function inversion

Question

I have an expression of this kind

t = 2 m (Sqrt[4 x^2 (1 - m^2) + m^4 + 4 m^2 x]/(1 - m^2) + 
  m^2/(4 (1 - m^2) Sqrt[m^2 - 1]) ArcSin[(2 (1 - m^2) x + 4 m^2)/(4 m^3)])

that I would like to invert so that to have x[t].

I have tried with

InverseFunction[2 m (Sqrt[4 x^2 (1 - m^2) + m^4 + 4 m^2 x]/(1 - m^2) + 
  m^2/(4 (1 - m^2) Sqrt[m^2 - 1]) ArcSin[(2 (1 - m^2) x + 4 m^2)/(4 m^3)]) - t][0]

but it does not seem to work.

Am I doing something wrong or it is just that the function is not invertible?

Can you say something about the range of m? It looks like it must be less than 1 in order to use real numbers. — Jack LaVigne, May 06 '16 at 01:41
Hey Jack Lavigne, I actually found out that the expression that I wrote has some typos. Here is the corrected version
2 m (Sqrt[4 x^2 (1 - m^2) + m^4 + 4 m^2 x]/(1 - m^2) + m^2/(4 (1 - m^2) Sqrt[m^2 - 1]) ArcSin[(2 (1 - m^2) x + 4 m^2)/(4 m^3)])

that I tried to invert using InverseFunction[], but still cannot be solved. The condition for m is m>1 — Vale, May 06 '16 at 04:59

score 6 · Answer 1 · answered May 06 '16 at 19:40

define a function of two variables,

f[x_, m_] = 2 m (Sqrt[4 x^2 (1 - m^2) + m^4 + 4 m^2 x]/(1 - m^2) +
     m^2/(4 (1 - m^2) Sqrt[m^2 - 1]) ArcSin[(2 (1 - m^2) x + 
          4 m^2)/(4 m^3)]);

then tell InverseFunction to invert w.r.t. the first argument:

inv = InverseFunction[f, 1, 2];

Show[{Plot[f[x, 2], {x, -10, 10}, PlotRange -> All],
  ListPlot[Table[{inv[y, 2], y} // N, {y, -7, -1/2, 1/4}], 
   PlotStyle -> Red]}, PlotRange -> All]

of course this is actually inverted numerically and so is rather slow. Also the inverse is not single valued and you get no control over which solution you get.

Thank you george2079, I will try with this solution – Vale May 09 '16 at 23:47 — Vale, May 09 '16 at 23:47

Function inversion

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