Solving $e^{-x} + x e^{-x} + \frac12x^2e^{-x} + \frac16x^3 e^{-x}=0.99$ Equation:
$$e^{-x} + x e^{-x} + \frac12 x^2 e^{-x} + \frac16x^3e^{-x}=0.99$$
I tried simplifying it and got it down to
$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94$$
although I am not sure how I would solve the $(6+6x+3x^2+x^3)$ portion.
Thanks to Wolfram Alpha I know the roots (solution for $x$), so I am really looking for guidance on how you would approach this equation and how far you can simplify it before having to use a calculator.
 A: 1.)
$$e^{-x} + x e^{-x} + \frac12 x^2 e^{-x} + \frac16x^3e^{-x}=0.99\ \ \ |\cdot 6$$
$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94\ \ \ |\cdot 100e^x$$
$$600+600x+300x^2+100x^3=594\ e^x\ \ \ |-594e^x$$
$$600+600x+300x^2+100x^3-594e^x=0$$
Because your equation is an irreducible algebraic equation of $x$ and $e^x$, it cannot be solved by an $x\neq 0$ by elementary functions or elementary numbers. That's a consequence of the theorems in [Ritt 1925]/[Risch 1979], [Lin 1983], [Chow 1999].
Your equation cannot be solved using Lambert W either.
The usual computer algebra systems cannot solve this equation in closed form either. A generalized Lambert W like in Alex R.'s answer is needed.
[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
2.)
You could split your unwieldy equation into a system of two equations. Take your equation
$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94.$$
Substitute $5.94=ce^{-x}$, where $c\in\mathbb{C}$.
$$e^{-x}\left(6+6x+3x^2+x^3\right)=ce^{-x}$$
$$6+6x+3x^2+x^3=c$$
That's an algebraic equation in $x$. It's of degree $3$ and you could solve it e.g. in closed form first.
A: The roots of the cubic polynomial are distinct, so this requires a generalization of Lambert's W function to a cubic polynomial. You can find an exposition on this here:
Numerics of the Generalized Lambert W Function - Scott, Fee:
https://www.researchgate.net/publication/275483330_Numerics_of_the_Generalized_Lambert_W_function
I'm not sure if there's an easy closed-form solution for every possible root, but you can follow the above guide to do a simple iteration to converge on the roots.
A: To solve $0.99=\frac{1+x+x^2/2+x^3/6}{1+x+x^2/2+x^3/6+x^4/24+\cdots}\in1-x^4/24+o(x^4)$, work numerically (e.g. by Newton–Raphson), starting from $x\approx0.24^{1/4}\approx0.70$. We quickly get a root it sounds like you already know, $0.82$. But I doubt a non-numerical option is available.
A: Beside numerical method, having noticed that
$$f(x)=e^{-x} \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)-\frac{99}{100}$$ is small when $x=1$ (which is normal since the quantity inside parentheses is smaller than $e^x$), we can build the infinite series
$$f(x)=\left(\frac{8}{3 e}-\frac{99}{100}\right) +\frac 1{6e}\sum_{n=1}^\infty (-1)^{n+1}\frac{n^3-9 n^2+23 n-16 }{n!} (x-1)^n$$ Truncated to some order $p$, it is then the Taylor series to $O\left((x-1)^{p+1}\right)$.
Now we can use series reversion to get
$$x=1-t-t^2-\frac{11 }{6}t^3-4 t^4-\frac{383 }{40}t^5-\frac{364 }{15}t^6-\frac{21481
   }{336}t^7+O\left(t^8\right)$$ where $t=\frac{297 e-800}{50} $.
Using this truncated series would lead to the estimate
$x\sim 0.823274$ while the "exact" solution, obtained using Newton method, is $x=0.823249$.
For sure, adding more terms will give better and better approximations.
