How to find an exact solution to the equation $e^{1/x}-2x-1=0$ I know how to find some interval for roots using the intermediate value theorem or getting closer to the roots with the bisection method, but is there a way to actually solve this equation?
 A: I will prove that $x$ is irrational. Add $2x+1$ to both sides:$$e^\frac{1}{x}=2x+1$$Then, raise both sides to the power of $x$: $$e=(2x+1)^x$$Since $e$ is a transcendental number, it couldn't be expressed by radicals and so $x$ is irrational. This means that it doesn't matter if we have a closed form expression or not, and an approximation is best. Turns out that this has two solutions, one approximately $-0.447$ and another approximately $0.943$. Cancel the $0.943$ solution since it $e^{0.943}+0.943>0$.
A: Here is a Lagrange reversion result:
$$\sqrt[x]e-2x=1\implies x=-\frac12+\sum_{n=1}^\infty\frac{1}{2^nn!}\left.\frac{d^{n-1}}{dx^{n-1}}e^\frac nx\right|_{x=-\frac12}$$
tested here. Now to expand $e^y$ as a series and take the $n$th derivative with factorial power $u^{(v)}$:
$$x=-\frac12+\sum_{n=1}^\infty\sum_{k=0}^\infty(-k)^{(n-1)}\frac{(-1)^{n+k}2^{k-1}n^k}{n!k!} $$
Next, interchange sums and use the confluent hypergeometric function $\,_1\text F_1(a;b;z)$:
$$x=\sum_{n=1}^\infty\frac{(-1)^n(-1)^{(n-1)}\,_1\text F_1(n;2;-2n)}{\Gamma(n)}=-0.4466986177721\dots$$
The $(-1)^{(n-1)}=(-1)^n n!,n\in\Bbb N$, so we actually just get:
$$\boxed{\sqrt[x]e-2x=1\implies x=-\sum_{n=1}^\infty\,_1\text F_1(n;2;-2n)=x≈-0.446698617772146595\dots}$$
Shown here. Can one find an expression for the second root at $x=0.9433158\dots$?
A: Your "actually solve" means solving in closed form.
$$e^{\frac{1}{x}}-2x-1=0$$
$x\to\frac{1}{t}$:
$$te^t-2-t=0$$
We see, this equation is solvable neither by elementary functions nor by elementary numbers:
How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?
$$te^t=2+t$$
$$\frac{t}{2+t}e^t=1$$
We see, this equation isn't solvable by elementary functions and Lambert W. But it is solvable by elementary functions together with Generalized Lambert W.
$$t=W\left(^{\ \ 0}_{-2};1\right)$$
$$x=\frac{1}{W\left(^{\ \ 0}_{-2};1\right)}$$
[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
