Solve $\frac d{dx}W_{-1}(-1/x)=-1$

Question

This problem arose when I tried to find the maximum of $f(x)=\ln(x\ln(x\ln(x\cdots)))-x$. This can be written as $f(x)=\exp(-W_{-1}(-1/x))/x -x$ by substituting the recursion into $f$. The negative branch of $W$ is taken as $-1/x$ is negative for all positive real $x$. Now $$f'(x)=(x-x^2\exp(W_{-1}(-1/x)))^{-1}-1$$ and equating to zero yields $x-x^2\exp(W_{-1}(-1/x))=1$ for stationary points. Notice that $\exp(W_{-1}(-1/x))=-\frac1{xW_{-1}(-1/x)}$ so this is equivalent to $$x+\frac x{W_{-1}(-1/x)}=1\implies-\frac{W_{-1}(-1/x)}{x(W_{-1}(-1/x)+1)}=-1$$ LHS is the derivative of $W_{-1}(-1/x)$ so this leaves us with $\frac d{dx}W_{-1}(-1/x)=-1$. How to continue?

Answer 1

Via Lagrange reversion:

$\def\W{\operatorname W}$ $$\frac d{dx}\left(-\W_{-1}\left(-\frac1x\right)-x\right)=0\implies x=-\frac{e^{-u}}u,e^uu^2+u+1=0,u=-1+\sum_{n=1}^\infty\frac{(-1)^n}{n!}\left.\frac{d^{n-1}e^{nu}u^{2n}}{du^{n-1}}\right|_{-1}$$

then General Leibniz rule uses Laguerre $\operatorname L_v^a(x)$:

$$\boxed{x=-\frac{e^{-u}}u,u=-1-\sum_{n=1}^\infty\frac{\operatorname L_{n-1}^{n+1}(n)}{ne^n}}$$

giving your maximum at $x=2.9902\dots$

Answer 2

$$x=-\frac{e^{-u}}{u}=-\frac{1}{ue^u},\ \ u^2e^u+u+1=0$$ $$u^2e^u+u+1=0\tag{1}$$

We see, equation (1) is a polynomial equation of more than one algebraically independent monomials ($u,e^u$). The main theorem in [Ritt 1925] cannot help in these cases: We cannot read an elementary inverse function over a non-discrete domain directly from the equation because we don't know how to rearrange the equation for $u$ by applying only finite numbers of elementary functions (operations) we can read from the equation.

We also see, equation (1) is an irreducible algebraic equation of both $u$ and $e^u$. According to the theorems in [Lin 1983] and [Chow 1999], assuming Schanuel's conjecture is true, such kind of equations cannot have solutions except $0$ that are elementary numbers or explicit elementary numbers respectively.

$$\frac{u^2}{u+1}e^u=-1$$

We see, this equation cannot be solved in terms of Lambert W but in terms of Generalized Lambert W:

$$\frac{u^2}{u-(-1)}e^u=-1$$ $$u=W\left(^{0\ 0}_{-1};-1\right)$$ $$x=-\frac{1}{W\left(^{0\ 0}_{-1};-1\right)e^{W\left(^{0\ 0}_{-1};-1\right)}}$$

So we have a closed form for $x$, and the series representations of Generalized Lambert W give some hints for calculating $x$.
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[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90

[Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[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