Polynomial equals exponential?

Question

Is there a closed form for the solution to the equation:

$$e^x = x^2 - x?$$

I already found that $0$ and $1$ aren't solutions of this equation, because we would get the contradictions $1=0$ and $e=0$ respectively.
Algebraic numbers other than $0$ cannot be solutions either because the left-hand side of the equation would be a transcendental number and the right-hand side an algebraic number.

I've tried using the Lambert W function but so far I haven't been able to find the expression.

Are there some Special functions, generalizations of Lambert W or particular numbers that solve the equation?

Answer 1

$$e^x=x^2-x$$

$$e^x-x^2+x=0$$

The equation is an irreducible polynomial equation in $e^x$ and $x$ over the algebraic numbers. [Lin 1983] proves that such kind of equations cannot have solutions in the elementary numbers (means by applying elementary functions to rational numbers).

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

We see, Lambert W cannot be applied. But the equation is solvable by Generalized Lambert W ($W$):

$$x=W\left(^{}_{0,1};1\right)=-W\left(^{0,-1}_{};1\right)$$

$-$ see the references below. Consider that Generalized Lambert W has, like Lambert W, different branches.
$\ $

[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 equa-tion. 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

Unfortunately, this expression is not solvable in terms of the Lambert W function. But for example, the Halley's method can be effectively used to iteratively find the approximation of the root.

\begin{align} x_{n+1}&=F(x_n) ,\\ F(x)&=x-\frac{2\,f(x)\,f'(x)}{2f'(x)^2-f(x)\,f''(x)} , \end{align} where \begin{align} f(x)&=x^2+x-\exp(-x) \quad\text{(the root would be $-x$)} ,\\ f'(x)&=2x+1+\exp(-x) ,\\ f''(x)&=2-\exp(-x) . \end{align}

For example, starting with $x_0=0$, we get

\begin{align} x_1&=0.444444444444444444444444444444\\ x_2&=0.444130228824893210739571781394\\ x_3&=0.444130228823966590585466329491\\ x_4&=0.444130228823966590585466329491 , \end{align}

so we get pretty good approximation of the root of the original equation as $-0.444130228823966590585466329491$.

Here is the python code:

import decimal
decimal.getcontext().prec = 30

lg2 = decimal.Decimal(2).log10()

def f(x):
  return x*x+x-decimal.Decimal(-x).exp()

def df(x):
  return x+x+1+decimal.Decimal(-x).exp()

def ddf(x):
  return 2-decimal.Decimal(-x).exp()

def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2*fx*dfx/(2*dfx**2-fx*ddfx)

x=0
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)

# 0.444444444444444444444444444444
# 0.444130228824893210739571781394
# 0.444130228823966590585466329491
# 0.444130228823966590585466329491