4
$\begingroup$

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?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ ArcSin is the correct syntax. $\endgroup$
    – Baran Cimen
    May 5, 2016 at 5:35
  • $\begingroup$ Thanks. Still seems not to be working. $\endgroup$
    – Vale
    May 5, 2016 at 5:49
  • $\begingroup$ look at this $\endgroup$
    – user36273
    May 5, 2016 at 9:32
  • $\begingroup$ Can you say something about the range of m? It looks like it must be less than 1 in order to use real numbers. $\endgroup$
    – Jack LaVigne
    May 6, 2016 at 1:41
  • $\begingroup$ 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 $\endgroup$
    – Vale
    May 6, 2016 at 4:59

1 Answer 1

Reset to default
6
$\begingroup$

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]

enter image description here

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you george2079, I will try with this solution $\endgroup$
    – Vale
    May 9, 2016 at 23:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.