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I am trying to find out the inverse of function $f:\mathbb R\to\mathbb R, f(x) = x - \tanh(x),\forall\in\mathbb R.$

What I tried:
Since $f(x)$ is invertible, so using $f(f^{-1}(x)) = x,$ I get $x = f^{-1}(x) - \tanh(f^{-1}(x)).$
Expanding or opening up $\tanh(x)$ leads me nowhere. I also tried the property $f^{-1'}(x) = \frac{1}{f^{'}(f^{-1}(x))}.$ But I couldn't get the inverse equation.

Gary
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Mrityunjay Tripathi
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  • Not sure if this helps, but: $$\tanh^{-1}(x)=\ln\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}},;;x=\ln e^x$$ $$f(x)=\ln\left(e^x\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\right)$$ Look at this:https://math.stackexchange.com/questions/1362270/inverse-of-sum-of-two-functions-in-terms-of-individual-inverse-functions – PinkyWay Jan 27 '20 at 19:55
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    Can you tell us why you want to get the inverse equation? It's possible that you can do what you need without an explicit representation of the inverse. – John Hughes Jan 27 '20 at 20:02
  • @VerkhovtsevaKatya First thing, I doubt that it can be done that way. Because then try finding the inverse of f(x) = sinx + cosx, using both ways i.e. first simplifying and then taking the inverse or first taking the inverse explicitly of each term and then simplifying. Secondly, see the question and your solution again – Mrityunjay Tripathi Jan 28 '20 at 04:27
  • @JohnHughes, I don't know how I read the task. Now, 7n the morning, I see what it actually says. – PinkyWay Jan 28 '20 at 08:18
  • @JohnHughes Actually, I do have another way to solve the problem i.e. in terms of f'(x), whose value I have. But what triggered me to put this on StackExchange is that MATLAB responded with "cannot find inverse" even though the function is invertible. – Mrityunjay Tripathi Jan 29 '20 at 13:05
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    Well...you probably shouldn't expect there to be an inverse formula for every invertible function. Think of "tan" -- there wasn't an inverse formula, until someone decided to say "Hunh. We don't have a formula, but we know the inverse exists, so let's call it $\arctan$." Heck, the same was true for $x \mapsto x^2$ on the nonnegative reals; eventually someone said "Let's write $y \mapsto \sqrt{y}$ for the inverse." – John Hughes Jan 29 '20 at 13:15

2 Answers2

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1.)

The inverse of your function isn't an elementary function:

$$f(x)=x-\tanh(x)$$

$$x-\tanh(x)=y$$

For algebraic values $y=a$, we have

$$x-\tanh(x)=a.\tag{1}$$

Each elementary standard function (like $\tanh$) can be represented in an exp-ln form:

$$x-\frac{(e^x)^2-1}{(e^x)^2+1}=a.$$

$$x(e^x)^2-a(e^x)^2-(e^x)^2+x-a+1=0\tag{2}$$

This is an irreducible polynomial equation over $\overline{\mathbb{Q}}$ in dependence of $x$ and $e^x$. The main theorems in [Lin 1983] and [Chow 1999] state that those kinds of equations don't have solutions except $0$ in the Elementary numbers or Explicit elementary numbers respectively. Because equation (2) and (1) have the same set of solutions, the same conclusion holds for equation (1).

And it follows with the theorem in Proof Check: Non-existence of the inverse function in a given class of functions that your function $f$ cannot have an inverse that is an elementary function.

2.)

$$-\frac{(x-1-y)}{x+1-y}e^{2x}=1$$

We see, your function doesn't have an inverse in terms of Lambert W. But the equation can be solved in terms of Generalized Lambert W:

$$x=\frac{1}{2}W(^{2y+2}_{2y-2};1)=-\frac{1}{2}W(^{-2y+2}_{-2y-2};1)$$

$-$ see [Mező 2017], [Mező/Baricz 2017], [Castle 2018].
$\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)

[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

IV_
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Here is a closed form for the inverse of functions containing a limit with it using the Mathematica function Inversebetaregularized $\text I^{-1}_z(a,b)$ which is a quantile function for the Student T Distribution with the Regularized $\text I_z(a,b)$ Incomplete Beta function $\text B_z(a,b)$. Also, a special case of the Lerch Transcendent is expressible in terms of $\text B_z(a,0)$. Please see the links for information on the below functions:

$$\frac12\text B_{\tanh^2(x)}\left(\frac32,0\right)\mathop=^{z\ge0}x-\tanh(x)=z\implies x=\lim_{a\to0}\tanh^{-1}\left(\sqrt{\text I^{-1}_{az}\left(\frac32,\frac a2\right)}\right)\sim z+1$$

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which works. The limit must be taken because the regularized incomplete beta function cannot have any parameter be $0$ or else there would be division by $0$ problem and the subscript was simplified to $az$ . Please correct me and give me feedback!

Тyma Gaidash
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