How to solve $x e^x + x = 1$?

Question

I have seen a mathematics-related video here, which introduce a Lambert W function to the audience: $$f(x)=xe^x\\ W(x)=f^{-1}(x)$$

Then we can use $W(x)$ to solve some transcendental equation conveniently, such $x^x=4$. But he left homework at the end of the video:

To solve equation $x e^x + x = 1$

Well, this homework made me lose sleep for 2 days. I don't know how to solve this equation with the $W(x)$. Can someone please guide me?

Answer 1

The equation cannot be solved in terms of Lambert W.

To show this, try to rearrange the equation to a form $f(x)e^{f(x)}=c$, $f$ a function in the complex numbers and $c$ a complex constant.

$$xe^x+x=1$$

$$xe^x=1-x$$

$$\frac{x}{1-x}e^x=1$$

We see: the exponent is $x$, a polynomial in $x$, and the factor is not a polynomial in $x$. There is no chance to rearrange the equation by only elementary operations to a form where the factor is equal to the exponent. Therefore your equation is not in a form that can be solved by Lambert W.

But the equation can be solved by generalized Lambert W:

$$x=W(^0_1;-1)=-W(^{-1}_0;-1)$$

$-$ see the references below.
$\ $

[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

[Stoutemyer 2022] Stoutemyer, D. R.: Inverse spherical Bessel functions generalize Lambert W and solve similar equations containing trigonometric or hyperbolic subexpressions or their inverses. 2022

Answer 2

May be, the goal of the exercise is to show that we have an immediate upper bound of the solution $$f(x)=xe^x+x-1 > xe^x-1 \implies x< W(1)=\Omega$$

Making a series expansion $$f(x)=\left(e^{\Omega } \Omega +\Omega -1\right)+\left(e^{\Omega } (\Omega +1)+1\right) (x-\Omega )+\frac{1}{2} e^{\Omega } (\Omega +2) (x-\Omega )^2+O\left((x-\Omega )^3\right)$$ and a series reversion gives as an estimate $$x=\Omega+\frac{1-\left(e^{\Omega }+1\right) \Omega }{1+e^{\Omega } (\Omega +1)}-\frac{e^{\Omega } (\Omega +2) \left(e^{\Omega } \Omega +\Omega -1\right)^2}{2 \left(1+e^{\Omega } (\Omega +1)\right)^3}=0.4028\cdots$$ while the solution is $0.4011\cdots$.