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Schanuel's conjecture states:

  • If $\alpha_1,\alpha_2,...,\alpha_n$ are complex numbers linearly independent over $\mathbb{Q}$, then the transcendence degree of the field $\mathbb{Q}(\alpha_1,e^{\alpha_1},\alpha_2,e^{\alpha_2},...,\alpha_n,e^{\alpha_n})$ over $\mathbb{Q}$ is at least $n$.

In What is a closed-form number (The American Mathematical Monthly, Vol. 106, No. 5. (May, 1999), pp. 440-448; MSN), Timothy Y. Chow states the following conjecture:

A) The real root $R$ of the equation $x+e^x=0$ is not in $\mathbb{E}$ (having defined $\mathbb{E}$ as the field of all ‘logarithmic exponential numbers’, which is the intersection of all subfields in $\mathbb{C}$ that are closed under $\exp$ and $\log$).

He also states:

  • Schanuel's conjecture implies conjecture A).

My question is: if Schanuel's conjecture isn't true, does this mean we can find the solution $R$ of $x+e^x=0$ with $R\in\mathbb{E}$?

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    $\begingroup$ Conjecture A is a consequence of a very special case of Schanuel's conjecture. There is no reason to expect the two to be equivalent. $\endgroup$ – Emil Jeřábek Jul 2 '14 at 10:35
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    $\begingroup$ Moreover, Schanuel’s conjecture actually implies that $R$ is not in the bigger field called $\mathbb L$ in Chow’s paper. I see no a priori reason why $R$ couldn’t be in $\mathbb L\smallsetminus\mathbb E$. $\endgroup$ – Emil Jeřábek Jul 2 '14 at 11:57
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    $\begingroup$ Side remark... This number is $-W(1)$, where $W$ is the Lambert W-function. $\endgroup$ – Gerald Edgar Jul 2 '14 at 15:28
  • $\begingroup$ "$K\subset\mathbf{C}$ is closed under $\log$" means "$x\in\mathbf{C}$, $\exp(x)\in K$ $\Rightarrow$ $x\in K$"? or means the a priori weaker "$y\in K$ $\Rightarrow$ $\exists x\in K:\exp(x)=y$"? $\endgroup$ – YCor Jul 7 at 10:27
  • $\begingroup$ @YCor In the linked article, Chow makes it clear that $\log$ denotes the principal branch of logarithm, that is the one satisfying $-\pi<\operatorname{Im}(\log x)\leq \pi$. Then we demand that $K$ is closed under this operation. Of course, this is easily seen to be equivalent to either of your definitions. $\endgroup$ – Wojowu Jul 7 at 10:36
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My answer is not the whole answer.

$R=-W(1)$, where $W$ is the Lambert W function.

$H(x)=x+e^{x}=0$, $H$ is an elementary function and the inverse of $H$ is $H^{-1}$ with $H^{-1}(x)=-W(e^{x})+x$.

Let us assume Schanuel's conjecture is not true and $-W(1)$ is an elementary number. That does not mean the equation $x+e^{x}=0$ can be solved by rearraning the equation according to $x$ by applying only elementary functions that we can read off from the equation and we get $R$ on this way as an elementary expression of a rational number. Solving an equation on this way means applying the inverse of $H$. Solvability of equations on this way is therefore related to the problem of invertibility of elementary functions by elementary functions. This problem was solved in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759. That Lambert W is a non-elementary function follows already from the theorem of J. F. Ritt and Lindemann-Weierstrass theorem.

One would have to show by other methods that $-W(1)$ is an elementary number.

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    $\begingroup$ All this answer seems to show is that $x+e^x =y$ does not admit a solution where $x$ is an elementary function of $y$, not that $x$ cannot be an elementary number if $y=0$. $\endgroup$ – Will Sawin Jul 23 '17 at 20:42
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    $\begingroup$ Like @WillSawin, I don't understand the step from a function (like LamberW) being complicated to a particular one of its values (like LambertW(0)) being complicated. Could you please add some more information about that? $\endgroup$ – Andreas Blass Jul 24 '17 at 1:35
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    $\begingroup$ The point that @WillSawin and @ AndreasBlass are making is, indeed, explicitly mentioned by Chow (§2, p. 441): "we cannot, for example, simply define an "elementary number" to be any number obtainable by evaluating an elementary function at a point, because all constant functions are elementary, and this definition would make all numbers elementary. Furthermore, even if a function (such as $W$) is not elementary, it is conceivable that each particular value that it takes ($W(1)$, $W(2)$, …) could have an elementary expression, but with different-looking expressions at different points." $\endgroup$ – LSpice Jul 26 '17 at 13:40

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