This is probably a trivial question. But I could not find an answer.

Given the explicit expression for ft in te assignment

ft = InverseFunction[f[#] &][t];

I would like to invert this, i.e. to get the function t[f[x]]. Of course I can copy the expression f[#]& and paste it into the right hand side of the assignment

t[x_] = f[#]& [x];

But I think there should be a valid Mathematica method to do the extraction.

The FullForm[] of ft shows a presumed List[] which, however, is a strange contruct which only lets me extract the argument t as its first element.

Example (I write it in Latex just to save space)

$ft = \text{InverseFunction}\left[-\frac{i \sqrt{\frac{2 \text{$\#$1}^2}{\sqrt{4 w+1}-1}+1} \sqrt{1-\frac{2 \text{$\#$1}^2}{\sqrt{4 w+1}+1}} F\left(i \sinh ^{-1}\left(\sqrt{2} \sqrt{\frac{1}{\sqrt{4 w+1}-1}} \text{$\#$1}\right)|\frac{1-\sqrt{4 w+1}}{\sqrt{4 w+1}+1}\right)}{2 \sqrt{\frac{1}{\sqrt{4 w+1}-1}} \sqrt{-\text{$\#$1}^4+\text{$\#$1}^2+w}}\&\right][t]$

I wish to extract the expression in square brackets.

Taking the compact solution 2 from the answer of xzczd we find

tx = ft[[0, 1]][x]

$-\frac{i \sqrt{\frac{2 x^2}{\sqrt{4 w+1}-1}+1} \sqrt{1-\frac{2 x^2}{\sqrt{4 w+1}+1}} F\left(i \sinh ^{-1}\left(\sqrt{2} \sqrt{\frac{1}{\sqrt{4 w+1}-1}} x\right)|\frac{1-\sqrt{4 w+1}}{\sqrt{4 w+1}+1}\right)}{2 \sqrt{\frac{1}{\sqrt{4 w+1}-1}} \sqrt{w-x^4+x^2}}$

  • $\begingroup$ Do you mean something like: t = InverseFunction@Head@ft, or t = InverseFunction@Function[t, #] &@ft, or t = ft[[0, 1]]? $\endgroup$ – xzczd Nov 17 '15 at 11:21
  • $\begingroup$ Have you noticed what happens when you do InverseFunction[InverseFunction[f]]? $\endgroup$ – J. M.'s technical difficulties Nov 17 '15 at 12:07
  • $\begingroup$ Well, actually your @name syntax is wrong, there shouldn't be any white spaces between @ and the name, or it won't trigger a notification :) $\endgroup$ – xzczd Nov 17 '15 at 13:38
  • $\begingroup$ This question would be more clear if you gave a specific example and the expected result. $\endgroup$ – george2079 Nov 17 '15 at 15:49
  • $\begingroup$ @J. M.: Thanks, I have done this first, of course. But it is returned unevaluated, at least in my case of a complicated function f $\endgroup$ – Dr. Wolfgang Hintze Nov 17 '15 at 16:56

OK, let me extend the comment into an answer. If ft still contains InverseFunction, your goal can be achieved by

(* Solution 1 *)
tf = First@Head@ft
(* Solution 2 *)
tf = ft[[0, 1]]
(* Solution 3 *)
tf = InverseFunction@Head@ft
(* Solution 4 *)
tf = InverseFunction@ft[[0]]
(* Solution 5 *)
tf = InverseFunction@Function[t, #] &@ft

Solution 5 should also work for the cases that ft no longer contains InverseFunction.

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