I define a function SS[t] as an inverse function:

tt[S_] := 10(2Sqrt[(1-S)S] +14/5 ArcTan[Sqrt[S], Sqrt[1-S]] + 
     2/5 ArcTanh[1/2 Sqrt[(1-S)/S]]);
SS = InverseFunction[tt];
N[SS[53], 60]
(* 0.200580902656825260833989366151287711207279221292307930697890 *)

SS[t] works great up to t=53: it's fast, I get a precision of 60, and I can plug it back into the original (forward) function.


Starting at t=54, it's very slow, and the output is way wrong, complex and wrong magnitude. (It should be real and in [0.2, 0.2006]).


  • The forward function works well for inputs S=0.2+$\epsilon$ ($\epsilon$ very small, like $10^{-10}$), giving tt[$\cdot$] up to 500, no problem.

  • $MinPrecision=0, $MaxPrecision=$\infty$ (defaults).

  • Increasing N[$\cdot$,60] to N[$\cdot$,100] makes no difference.

I use Mathematica 8.0.4 on MacOS, installed... 5 years ago?


I ended up implementing my own iterative inverse, which is easy in this case, as the forward function is well behaved and monotonic. It's very fast, and I can get any precision I like.

Still, I find disturbing the lack of control over InverseFunction, and I get that sense of unease when I cannot fully trust a component. Perhaps there is a way to take control that I'm aware of, so I am still interested in an answer.

  • 1
    $\begingroup$ Version 12.1 Windows 10. Works up to t=60. But from t=61 on it is as you describe. $\endgroup$ Mar 2, 2021 at 16:20
  • 4
    $\begingroup$ A series-expansion of tt[S] around $S=1/5$ can be inverted in a more stable manner. $\endgroup$
    – Roman
    Mar 2, 2021 at 16:23

1 Answer 1


Workaround using: FindRoot.

mm[t_, P_] := (S /. 
FindRoot[(10 (2 Sqrt[(1 - S) S] + 
      14/5 ArcTan[Sqrt[S], Sqrt[1 - S]] + 
      2/5 ArcTanh[1/2 Sqrt[(1 - S)/S]])) == 500, {S, 211/1000}, 
 WorkingPrecision -> P, MaxIterations -> 1000]) // Re

mm[500, 110](*For t=500 and WorkingPrecision= 110*)


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