Question: Develop three terms of the perturbation solutions to the real roots of $$(x^3 + 2x^2 + x)e^{-x} = \epsilon,$$ identifying the scalings in the expansion sequence $\delta_0(\epsilon)x_0 + \delta_1(\epsilon)x_1 + \delta_2(\epsilon)x_2 + \cdots$.
My approach: The equation can be rewritten as follows: $$x(x+1)^2e^{-x} = \epsilon.$$ Taking the natural log of both sides of the equation, we arrive at the following equation: $$\ln(x) + 2\ln(x+1) -x = \ln(\epsilon),$$ which implies that $$x = \ln(x) + 2\ln(x+1) - \ln(\epsilon).$$
This leads us to the following iteration scheme to find out the two real perturbed roots: $$x_{n+1} = \ln(x_n) + 2\ln(x_n+1) - \ln(\epsilon),$$ for all $n\ge\mathbb{Z}_{\ge 0}$.
To find the smaller root, we can choose the starting point $x_0 = \epsilon$, and for the larger root, a good starting point can be $x_0 = -3\ln(\epsilon)$ or $x_0 = -4\ln(\epsilon)$.
However, I am not being able to find a stable expansion sequence for both roots. Can someone please help me out with this?
For the small root, why not to use Taylor expansion followed by power series reversion to get $$x=\epsilon -\epsilon ^2+\frac{5 }{2}\epsilon ^3-\frac{22}{3}\epsilon ^4+O\left(\epsilon ^5\right)$$ which also gives some bounds.
For the large root, start Newton method with $x_0=-k\log(\epsilon)$ such that $$f(x_0) \times f''(x_0) >0$$ with $$f(x)=-x+\log(x) + 2\log(x+1) - \log(\epsilon)$$ This ensures, by Darboux theorem, that the solution will be reached without any overshoot..
Since $$f''(x)=-\frac{1}{x^2}-\frac{2}{(x+1)^2} ~<~0 \quad \forall x>0$$ it suffices the select $k$ such that $f(x_0)<0$
If $\epsilon=e^{-n}$, at least for $1\leq n \leq 10$ $$k = \left\lceil \frac{6}{n}+\frac{3}{2}\right\rceil$$ seems to be a good choice.
Trying for $n=18$, $k=2$; Newton iterates will be $$\left( \begin{array}{cc} p & x_p \\ 0 & 36.000000 \\ 1 & 28.164131 \\ 2 & 28.074611 \\ 3 & 28.074595 \\ \end{array} \right)$$
Edit
If the equation had been $$f(x)=-x+\log(x) + 2\log(x+\color{red}{0}) - \log(\epsilon)$$ with $\epsilon=e^{n}$ and $x=-k\log(\epsilon)$, we should have $$k=-\frac 3 n \,W_{-1}\left(-\frac{1}{3} e^{-n/3}\right)$$ but, for this value $f(x_0)>0$.
But $$\color{blue}{k=\frac 12-\frac 3 n \,W_{-1}\left(-\frac{1}{3} e^{-n/3}\right)}$$ is fine $\forall n \geq 1$.