is there a closed algebraic solution to x(x+a)e^x=b, a,b positive reals?

Question

I am looking at the following equation which is solvable in terms of the Lambert-W function when $a=0$ (but it is strictly positive in my case, i.e. $a>0$):

$x(x+a)e^x=b$ $(a,b>0)$

more generally, one can consider a generalization to the lambert function of the form : $(x-r_1)(x-r_2)e^x=b$ (in my case $r_1=0, r_2=-a$). Can the solution $x$ be expressed using the Lambert function or other special functions?

Using the reference arxiv.org/pdf/1501.00138v3.pdf I was able to obtain the answer: $x=\sum_{n=1}^{\infty} \frac{(-\frac{-nb}{a})^n}{n n!}B_{n-1}(2a^{-1})$ for the equation $x(x+a)e^x=b$. — Gregor Samsa, Mar 19 '16 at 07:25
$B_{n-1}(x)$ in the above is the Bessel polynomial of order $n-1$. — Gregor Samsa, Aug 11 '16 at 06:42

score 0 · Answer 1 · answered Mar 17 '16 at 20:09

There is no closed form in terms of the Lambert W function. What you have to be careful about when dealing with the solving using the Lambert W function is addition and exponents.

It is possible to solve $(x+a)e^x=b$ but $x(x+a)e^x=b$ is not solvable.

It would require you to get the exponent to become $x(x+a)$, which means you'd have to exponentiate everything by $x+a$ to get it into the exponent. And, that leads to problems.

If you really need a solution, numerical methods or something a bit simpler like Fixed-point conversion might do the job.

score 0 · Answer 2 · answered Mar 06 '23 at 17:27

For simplicity, I give an answer for solutions in the reals.

$$x(x+a)e^x=b$$

Your equation is an equation of elementary functions. It's an algebraic equation in dependence of $x$ and $e^x$. Because the terms $x,e^x$ are algebraically independent and the equation is irreducible, we don't know how to rearrange the equation for $x$ by only elementary operations (means elementary functions). A theorem of Lin (1983) proves, if Schanuel's conjecture is true, that irreducible algebraic equations involving both $x$ and $e^x$ don't have solutions in the elementary numbers.

We see, your equation cannot be solved in terms of Lambert W but in terms of Generalized Lambert W.

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

The inverse relation of your kind of equations is what Mezö et al. call $r$-Lambert function. They write: "Depending on the parameter $r$, the $r$-Lambert function has one, two or three real branches and so the above equations can have one, two or three solutions"

So we have a closed form for $x$, and the representations of Generalized Lambert W give some hints for calculating $x$.

[Mező 2017] Mező, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553

[Mező/Baricz 2017] Mező, I.; Baricz, Á.: 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

is there a closed algebraic solution to x(x+a)e^x=b, a,b positive reals?

2 Answers2