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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?

ephe
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  • For analytical methods compare with this post. Also, see this post. – Dietrich Burde Jan 05 '23 at 16:41
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    The only way to "exactly" solve such an equation is to invent a new function whose definition is the solution. Which is very unsatisfying. Look up Lambert Omega function, e.g. – B. Goddard Jan 05 '23 at 16:41
  • The Lambert-W function was established because it has some merit for a class of equations, but I am surprised that for this kind of equation, another such function was also invented. This does not make much sense. Such equations can only be solved numerically, the Lambert-W function can also be calculated only numerically. – Peter Jan 05 '23 at 16:47
  • @B.Goddard But you have to prove that it isn't possible to solve. – Kamal Saleh Jan 05 '23 at 16:48
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    I think it is not very important to have the EXACT solution for such an equation anyway. For polynomials with degree $3$ and $4$ there are exact formulas for the roots , but they are nevertheless almost never used. – Peter Jan 05 '23 at 16:50

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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 $e^{0.943}+0.943>0$.

Kamal Saleh
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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 $(-1)^{(n-1)}=(-1)^n n!,n\in\Bbb N$. For an integral representation, transform the $_1\text F_1(n,2,-2n)$, convert to a Laguerre $\operatorname L_n(x)$ integral representation, and interchange it with the sum to get:

$$\boxed{\sqrt[x]e-2x=1\implies x=-\sum_{n=1}^\infty\,_1\text F_1(n;2;-2n)=\frac1{2e^2}-\frac12+\frac1{2\pi}\int_0^{2\pi}e^{it-2e^{it}-2}\ln\left(1-e^{-2\left(e^{it}+1\right)}-e^{-2e^{it}-it-2}\right)dt=x≈-0.446698617772146595\dots}$$

Shown here. Can one find an expression for the second root at $x=0.9433158\dots$?

Тyma Gaidash
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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

IV_
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