# How do we know if a function has an elementary inverse?

There are certain elementary functions where the inverse (or the branches of the inverse in a non-injective function, or the inverse over its range for a non-surjective function) is non-elementary.

For example, the function

$$y=xe^x$$

does not have an inverse that is elementary. Instead we call its inverse the Lambert W function.

Another example is, in general,

$$y=ax^5+bx^4+cx^3+dx^2+ex+f$$

because an inverse function would imply

$$ax^5+bx^4+cx^3+dx^2+ex+f-y=0$$

has an elementary solution, which contradicts the unsolvability of the quintic in general.

However, in general, given a function $$f(x)$$, how will I determine if the function has an elementary inverse/inverses?

(note: an elementary function is a function that can be represented as the finite sum, product and composition of rational functions, radicals, $$\exp$$, $$\ln$$ and trigonometric functions)

• You could use the inverse function theorem which says that if an inverse existed, then you would have $\frac{d}{dx}(f^{-1}) = \frac{1}{\frac{d}{dx}f}$. Therefore, you have $f^{-1} = \int \frac{1}{\frac{d}{dx} f}$. The Risch algorithm tells you whether you can express the integral in terms of elementary functions. Also, the insolvability of the quintic means it cannot be solved in terms of radical expressions of $a,b,c,d,e,f$. I'm not sure if I can be done in terms of elementary transcendental functions, but the distinction is important
– Rdrr
Commented Aug 12, 2021 at 16:28
• It seems possible that solution of some quintic is possible by elementary functions even if its solution is not possible by radicals. Commented Aug 13, 2021 at 0:53
• @Rdrr Let $\phi$ denote the inverse function of $f$. The correct inverse rule of differentiation is: $\phi'(x)=\frac{1}{f'(\phi(x))}$. Does your method work then? You need $\phi$, but it is unknown in our problem.
– IV_
Commented Aug 13, 2021 at 18:29
• @GEdgar Algebraic functions that cannot be inverted in terms of radicals cannot be inverted in the Explicit elementary functions. Algebraic equations that cannot be solved in radicals cannot be solved in terms of explicit elementary functions. See my answer below.
– IV_
Commented Mar 12, 2022 at 15:40

You aren't asking only about the elementary functions that have an elementary inverse function, but more general about the functions that have an elementary partial inverse.

I want to summarize here what I've found out so far in the last years.

Let's call your elementary functions the explicit elementary functions.

In the following, the inverse means the inverse function for bijective functions and an appropriate partial inverse otherwise.

Let $$^{-1}$$ denote the inverse.

The explicit elementary functions are generated from their complex function variable by applying finite numbers of $$\exp$$, $$\ln$$ and/or (unary or multiary) radicals.

Each elementary standard function (i.e. the trigonometric functions, the hyperbolic functions, the arcus functions, the inverse hyperbolic functions) can be represented in the above form. So "trigonometric functions" in your definition of elementary functions isn't necessary.
The radicals contain the rational expressions. So "rational functions" in your definition of elementary functions isn't necessary.

Theorems and statements about inverse functions can be extended to nonbijective functions because we can decompose a nonbijective function into bijective restrictions by decomposing the domain of the function. So we get the individual partial inverses of the function.

The problem of existence of elementary inverses is related to the problem of solving two-variable equations by elementary functions because the definition of the inverse of a function $$F$$ implies

$$F(y)=z\tag{1}$$

wherein $$y=F^{-1}(z)$$.
For a given equation 1, you can take $$F$$ as function and $$F^{-1}$$ as inverse of $$F$$ and vice versa because a bijective function is the inverse function of a function. $$\$$

a) Algebraic functions

For the algebraic functions, the question is solved in [Ritt 1922]. You can use also Galois theory because the algebraic functions are defined as solutions of an irreducible algebraic equation.

The symbol $$\wp$$ ("\wp") in [Ritt 1922] means the Weierstrass Elliptic Function.
$$\wp u$$ means $$\wp(u)$$.

$$a,b,c,d\in\overline{\mathbb{Q}}$$
$$m\in\mathbb{N}_+$$

Ritt lists i.a. the following functions (represented here by its function term in dependence of its complex function variable $$z$$).

$$a(z+b)^n+c$$

$$2^{m-1}a(bz+c)^m+a\sum_{i=1}^{m-1}\left({\frac{(-1)^im\prod_{l=0}^{i-2}(m-i-1-l)2^{m-2i-1}(bz+c)^{m-2i}}{i!}}\right)+d$$

Perhaps someone will find a better formula by including the first summand under the summation sign.

Algebraic functions that cannot be inverted in radicals cannot be inverted in the Elementary functions. See e.g. my answer to Solvability in radicals, elementary functions and monodromy/Galois groups.

b) Generalized elementary functions

For the generalized elementary functions ([Khovanskii 2014]), the question is partly answered by Ritt's theorem in [Ritt 1925] that's proved also in [Risch 1979]. I adapt the theorem here to explicit elementary inverses.

Theorem 1 - Ritt's theorem, adapted to explicit elementary inverses:
If $$F$$ is a generalized elementary function with an explicit elementary inverse, then $$F(z)=(f_n\circ\ ...\ \circ f_1)(z),$$ where each $$f_i$$ ($$i\in\{1,...,n\}$$) is either an algebraic function that has a radical as inverse, or else $$\exp$$ or $$\ln$$.

So the generalized elementary functions having an explicit elementary inverse are generated from their complex function variable by applying finite numbers of $$\exp$$, $$\ln$$ and/or unary algebraic functions having a radical as inverse.
See e.g. [Ritt 1922] for the corresponding algebraic functions.

It's easy to prove that

$$F^{-1}(z)=(f^{-1}_1\circ\ ...\ \circ f_n^{-1})(z),$$

wherein for $$i\in\{1,...n\}$$ $$f_i^{-1}$$ is the inverse relation of $$f_i$$.

But theorem 1 doesn't prove if an explicit elementary inverse exists. The theorem proves only that an explicit elementary inverse does exist iff a representation of the form of the theorem does exist for $$F$$. It doesn't help in deciding whether such representation exists.

A proof of the existence of generalized elementary numbers and explicit elementary numbers as solutions of irreducible polynomial equations in terms of simultaneously $$z$$ and $$e^z$$ is available by the methods of [Lin 1983] and [Chow 1999] respectively (assuming Schanuel's conjecture is true). They can be used as proof for non-existence of elementary inverses.

If a theorem like Ritt's can be found for other classes of functions is an open question. I made a proposal at How to extend Ritt's theorem on elementary invertible bijective elementary functions?

c) Compositions of Lambert W and generalized elementary functions

Consider the interchangeability between the function $$F$$ and its inverse in the two-variable equation 1 above and correspondingly between the function variable and the solution of the equation and see e.g. the equations in [Edwards 2020] or in section 6 of my answer at Example equation which does not have a closed-form solution.
$$\$$

How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?