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Let $f$ be an invertible, real-valued function of a real variable. What are the conditions for us to be able to write down the inverse function $f^{-1}$ as a closed-form expression?

Example. Define $g:[0,\infty)\rightarrow\mathbb{R}$ by $g(x)=x^2$. Then we can write down the inverse function $g^{-1}:[0,\infty)\rightarrow[0,\infty)$ as: $g^{-1}(x)=\sqrt{x}$.

Example (edited). Define $h:\mathbb{R}\rightarrow\mathbb{R}$ by $h(x)=x^5+2x+1$. Then we cannot write down the inverse function $h^{-1}:\mathbb{R}\rightarrow\mathbb{R}$ as a closed-form expression.

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  • Codomain of example $g$ must be $[0,\infty)$ (if it is $\mathbb R$ then there it has no inverse function). – drhab Mar 31 '18 at 11:12
  • In general when $f^{-1}$ exists its graphic is symmetric respect to the diagonal $y=x$. For example for the given below $f(x)=x^5+2x+1$ this is a bijection of $\mathbb R$ so there is a function $f^{-1}$ (it has not an algebraic expression but it exists and even you can calculate the value of $f^{-1}(y)=x$ using the above symmetry). – Piquito Mar 31 '18 at 11:46
  • @drhab: I thought what matters is the range? –  Apr 01 '18 at 01:01
  • Statement "function $f:X\to Y$ is invertible" is supposed to be the same as "a function $g:Y\to X$ exists such that the compositions $f\circ g$ and $g\circ f$ are identity functions". – drhab Apr 01 '18 at 07:01

2 Answers2

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The answer to this depends strongly on what you mean by "closed form": For any invertible function $f$, one can also give a name to $f^{-1}$, so the answer depends on which functions are allowed to appear in "closed-form" expressions.

NB if you allow algebraic expressions, then your counterexample isn't one. The inverse of $x \mapsto x^3 + x$ is \begin{multline} y \mapsto \frac{1}{6} \sqrt[3]{108 y + 12 \sqrt{81 y^2 + 12}} - \frac{2}{\sqrt[3]{108 y + 12 \sqrt{81 y^2 + 12}}} \\ = \frac{2}{\sqrt{3}}\,\sinh\left[\frac{1}{3}\,\text{arcsinh}\left(\frac{3\sqrt{3}y}{2}\right)\right].\end{multline} (For a derivation of second expression on the right, see Jack D'Aurizio's answer in the linked question.)

On the other hand, the Galois group of $p(x) := x^5 + 2 x + 1$ is $S_5$, which is nonsolvable, so there is no algebraic expression for its real root and hence there is no algebraic expression for $p^{-1}$.

Travis Willse
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  • Thanks for pointing out my error. I have now edited my question in response to your NB. –  Mar 31 '18 at 11:36
  • @dtcm840 You're welcome. If you want a more specific answer than mine ("[it] depends"), then (as my answer shows) you'll need to be more specific about what you mean by 'explicit'. – Travis Willse Mar 31 '18 at 14:13
  • I've changed the word "explicitly" to "as a closed-form expression". Hopefully it is now sufficiently specific. –  Apr 01 '18 at 00:57
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The mathematical problem is for functions $D\subseteq\mathbb{C}\to\mathbb{C}$.
Let $f$ denote a function of one variable ($f\colon x\mapsto f(x)$) and $\phi$ the inverse of $f$.
The inverse of a function $f$ exists iff $f$ is bijective. Each non-bijective function can be splitted into different bijective functions by restricting the domain.

Let $f$ be in closed form. That means the function term of $f$ is in closed form, means in finite terms.
Because for the inverse $\phi(f(x))=f(\phi(x))=x$ holds, we have: Existing of the inverse $y=\phi(x)$ of $f$ in closed form means the equation $f(y)=x$ is solvable for $y$ in closed form (means: by applying only allowed operations/functions).

Allowed functions could be e.g. the algebraic functions, the elementary functions and/or special functions.

Your examples are algebraic functions.

1.) Algebraic functions

An algebraic function (over $\mathbb{Q}$) of $n$ variables is a solution of an irreducible algebraic equation over $\mathbb{Q}[x_1,...,x_n]$. An algebraic function is in closed form if its function term is a radical expression. That are the explicit algebraic functions.

The inverse of an algebraic function is an algebraic function. To see if $\phi(x)$ is in closed form, you have to check if the equation $f(y)=x$ is solvable for $y$ in closed form.
Galois theory says which algebraic equations have solutions that are radical expressions and which not.

[Ritt 1922] lists all types of rational functions whose inverses are expressible in terms of radicals.

Some solutions of algebraic equations of higher degree can be represented e.g. with Bring radicals and/or theta functions.

2.) Transcendental elementary functions

The elementary functions of Liouville, Ritt, Risch and Lin contain also the implicit algebraic functions.

For the elementary functions, [Ritt 1925] and [Risch 1979] prove the

Theorem:
"If $F(z)$ and its inverse are both elementary, there exist $n$ functions $$\phi_{1}(z),\ \phi_{2}(z),\ ...,\ \phi_{n}(z),$$ where each $\phi(z)$ with an odd index is algebraic, and each $\phi(z)$ with an even index is either $e^{z}$ or $\log z$, such that $$F(z)=\phi_{n}\ \phi_{n-1}\ ...\ \phi_{2}\ \phi_{1}(z)$$ each $\phi_{i}(z)$ $(i<n)$ being substituted for $z$ in $\phi_{i+1}(z)$."

Because $F$ has to be a composition, the algebraic functions have to be unary.

$EL$ are the elementary numbers:
http://mathworld.wolfram.com/LiouvillianNumber.html, https://en.wikipedia.org/wiki/Elementary_number, http://mathworld.wolfram.com/ElementaryNumber.html, [Chow 1999].
The explict elementary numbers are used in [Chow 1999].

[Lin 1983] proves the

Theorem:
"If Schanuel's conjecture is true and $f(X,Y)\in\overline{\mathbb{Q}}[X,Y]$ is an irreducible polynomial involving both $X$ and $Y$ and $f\left(\alpha,exp(\alpha)\right)=0$ for some non-zero $\alpha$ in $\mathbb{C}$, then $\alpha$ is not in $EL$."

That also means, the inverse of a bijective function $F\colon x\mapsto f(x,e^x)$ cannot be an elementary function.

3.) Functions in general

One can easily prove the

Theorem:
Let $n \in \mathbb{N}_0$,
$f_{1},...,f_{n}$ bijective functions,
$f=f_{n}\circ f_{n-1}\circ\ ...\ \circ f_{2}\circ f_{1}$ a bijective function,
$\phi$ the inverse of $f$.
Then $\phi=\phi_{1}\ \circ\ \phi_{2}\ \circ\ ...\ \circ\ \phi_{n-1}\ \circ\ \phi_{n}$, wherein for all $i$ with $1\leq i\leq n$, $\phi_{i}$ is the inverse of $f_{i}$.

For inverses in the elementary functions and LambertW, I could apply this theorem for my answer at Equations solvable by Lambert Function.

Khovanskii and Burda give another method: see [Burda/Khovanskii 2011], [Khovanskii 2014], [Belov-Kanel/Malistov/Zaytsev 2020].
$\ $

[Belov-Kanel/Malistov/Zaytsev 2020] Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205

[Burda/Khovanskii 2011] Burda, Y.; Khovanskii, A.: Branching Data for Algebraic Functions and Representability by Radicals. 2011

[Chow 1999] Chow, T. Y.: What is a Closed-Form Number? Amer. Math. Monthly 106 (1999) (5) 440-448 or https://arxiv.org/abs/math/9805045

[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory. Solvability and Unsolvability of Equations in Finite Terms. Springer 2014

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

[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759

[Ritt 1922] Ritt, J. F.: On algebraic functions which can be expressed in terms of radicals. Trans. Amer. Math. Soc. 24 (1922) (1) 21-30

[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

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