Proof that no analytical solution exists for $x+e^{-x}=2$

Question

I think, there's no (preferrably real) analytical solution to the equation $$x+e^{-x}=2,$$ however I have no idea how I'd proof that guess. How does one generally proof the nonexistence of an analytical solution?

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A Closed-form expression is defined by Wikipedia as

a mathematical expression expressed using a finite number of standard operations.

I assume an analytical expression doesn't necessarily have to include only a finite number of operations - i.e. an infinite series $\sum_{K=0}^\infty a_k$ is also an analytical solution, if it is convergent? I didn't find much information on a precise definition of analytical expressions.

If that definition is in fact correct, one could argue though that there is an analytical solution. Because the equation is clearly numerically solvable, hence there exists a sequence $x_k$ that approximates the solution $x$, so $x$ can be written as $$x=x_0+\sum_{k=1}^\infty(x_k-x_{k-1})$$ So maybe my question should be how to proof that there is not Closed-form expression that solves above equation.

This seems similar to the Abel-Ruffini theorem, which states that no algebraic solution (a subset of the closed form expressions if I understand it correctly) to polynomials of degree 5 or higher.

However, I don't want to limit the allowed solutions too much. Simply said I want to proof that you can't solve the equation exactly with just a pen and paper and a finite number of transformation steps, like subracting $x$ from both sides.

Answer 1

There are two real solutions:

$$x_1=W\left(-\frac{1}{e^2}\right)+2,$$

$$x_2=W_{-1}\left(-\frac{1}{e^2}\right)+2,$$

where $W(x)$ is the Lambert W-function.

Answer 2

Let's try to solve the equation by hand...

$$x+e^{-x}=2$$

• Let $e^{-x}=t$

$$\ln t-t=-2$$

$$\ln (te^{-t})=-2$$

$$-te^{-t}=-e^{-2}$$

$$W\left(-te^{-t}\right)=W\left(-e^{-2}\right)$$

$$-t=W\left(-e^{-2}\right)$$

$$t=-W\left(-e^{-2}\right)$$

• Remember that $e^{-x}=t \Longleftrightarrow x=- \ln t$, then we have

$$\begin{align}\color {gold}{\boxed {\color{black}{x_1=-\ln \left(-W\left(-\frac{1}{e^2}\right)\right) \\ x_2=-\ln \left(-W_{-1}\left(-\frac{1}{e^2}\right)\right)}}}\end{align}$$

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Additional explanation:

If we write $x$, as $x=-e^{-x}+2$, then, we will get the same solutions: ( but, in a slightly different representation)

$$\begin{align}\color {gold}{\boxed {\color{black}{x_1=W\left(-\frac{1}{e^2}\right)+2 \\ x_2=W_{-1}\left(-\frac{1}{e^2}\right)+2. }}}\end{align}$$

Answer 3

The answer of Anixx is correct. This is definitively the analytical solution of the equation (Well complited by "lone student" ). $$x+e^{-x}=2$$ No need of WolframAlpha to find it :

$$x-2=-e^{-x}$$ $$(x-2)e^x=-1$$ $$(x-2)e^{x-2}=-e^{-2}$$ Let $X=x-2$ and $C=-e^{-2}$ $$Xe^X=C$$ $$X=W(C)\quad\implies\quad x-2=W\left( -e^{-2}\right)$$

$W$ denotes the Lambert W function which is the inverse function of $Xe^X$. This is a multivalued function. The two real branches are noted $W_0(X)$ and $W_{-1}(X)$. So the two real roots of the original equation are : $$x_1=2+W_0\left( -e^{-2}\right)\simeq 1.84140566$$ $$x_2=2+W_{-1}\left( -e^{-2}\right)\simeq -1.14619322$$ We won't discuss about complex branches of the function $W_n(X)$.

I suppose that is not what the OP expected because the question is not well-posed. If the OP was thinking of elementary functions only, this should be specified in the wording of the question.

The Lambert W function is a standard function just like arcos, arccsin, etc. This is a so called special function just like the Bessel functions, the Jacobi functions, or many other well known functions and the inverse functions.

A "closed form" expression can involve all whose standard functions, elementary and/or special. If this is not what is expected, the OP must specify the restricted list of functions to which the question refers.

For people not familiar with the special function, this popularization paper : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales .

Answer 4

This is another answer, based on the paper, linked in a comment to the original post by user metamorphy.

The original poster asked for a closed form of a number. The problem here is in that the concept of "closed form" of a number is usually not well defined.

On the other hand, the author of the linked paper had proposed a plausible definition of such concept, building it analogously to the concept of elementary function (author admits, "elementary number" would be a better name for his proposal, but the term is already occupied).

So, the author defines a set of $\mathbb{E}$ of "EL numbers" which stands for "elementary" and well as "exponentially-logarithmic".

He defines the set as any numbers that can be produced by applying finite number of field operations, exponential and logarithmic functions to the number $0$.

For instance, in his system

$$1=\exp(0)$$

$$e=\exp(\exp(0))$$

$$i=\exp\left(\frac{\log(-1)}2\right)=\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right)$$

$$\pi=-i\log(-1)=-\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right)\log(0-\exp(0))$$

It turns out that any root of a polynomial with rational coefficients, expressible in radicals, is also in $\mathbb{E}$.

Finally, the author explicitly comes to a similar question as the asker: whether the root of the equation $x+e^x=0$ belongs to $\mathbb{E}$ (the root is expressible via Lambert W-function). The author conjectures, no. But the conjecture is still open.

Answer 5

A closed-form function has a function term that is a closed-form expression. That means, the function term contains only finite numbers of variables, allowed constants, allowed functions and/or allowed operations. It means something like "in allowed finite terms".

The elementary functions (Elementary functions - Differential algebra), if they are allowed, are a subset of the closed-form functions.

For the references below, see:
Prove that an equation has no elementary solution
Why is it that the Lambert W relation cannot be expressed in terms of elementary functions?

Your equation

$$x+e^{-x}=2,\ \ -2+x+e^{-x}=0$$

is an algebraic equation in dependencce of $x$ and $e^{-x}$. With Hermite-Lindemann theorem, we get that the equation cannot have an algebraic solution $x$.

$-2+x+e^{-x}$ is the function term of an elementary function. If it is a composition of allowed functions, we can calculate its inverse relation (the partial inverse functions) by applying the inverse relations of the individual members of that composition in opposite order. [Ritt 1925] and [Risch 1979] prove that compositions of elementary functions are the only elementary functions that are invertible by elementary functions.

The problem is that addition ("+") is not a unary but a binary operation and that $x$ and $e^{-x}$ are algebraically independent.

[Lin 1983] proves, assuming Schanuel's conjecture is true, that algebraic equations of both $x$ and $e^x$ cannot have solutions except $0$ that are elementary numbers. [Chow 1999] presents a method for proving this, assuming Schanuel's conjecture is true, for solutions that are explicit elementary numbers. (Someone should write an article that generalizes Chow's method to larger classes of equations.)

With Lin, Chow and Schanuel's conjecture, we get that your equation cannot have explicit or implicit elementary numbers as solutions.

The proofs come from number theory and differential algebra, because the elementary functions are closed regarding differentiation. But differentiation shouldn't be necessary for this.
$\ $

Lambert W ist the inverse relation of the elementary function $f\colon x\mapsto xe^x$ in the complex numbers. $xe^x$ is an algebraic expression in dependence of $x$ and $e^x$. Using the theorem of [Ritt 1925], we can guess that $f$ cannot have partial inverses that are elementary functions. It is proved in [Bronstein/Corless/Davenport/Jeffrey 2008] by the methods of [Rosenlicht 1969].

Your equation has the form

$$A(x)+B(x)e^{C(x)}=u,$$

with $A,B,C$ functions in the complex numbers, $u$ a complex constant and $x$ a complex variable. Under certain circumstances, equations of this form can be solved by applying Lambert W. See the real solutions $W_0\left(-e^{-2}\right)+2$ and $W_{-1}\left(-e^{-2}\right)+2$ of your equation in the other answers.

If you can bring the equation to the form

$$af(x)^ce^{b+cf(x)}=d,$$

with $a,b,d$ complex constants, $c$ a real constant and $f$ a function in the complex numbers, you can apply Lambert W and the solutions are, obeying the calculation rules for complex numbers,

$$x=f^{-1}\left(W_{k}(e^{\frac{\ln\left(\frac{d}{a}\right)-b+2j\pi i}{c}})\right)\ \ \ (j,k\in\mathbb{Z}),$$

with $f^{-1}$ the inverse relation of $f$.

If you allow $f^{-1}$ and Lambert W, the solutions are not elementary, but in closed form.