# Schanuel's conjecture and real root of $x+e^x=0$

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 $R\in\mathbb{E}$?

• 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. – Emil Jeřábek Jul 2 '14 at 10:35
• 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 $\mathbb L\smallsetminus\mathbb E$. – Emil Jeřábek Jul 2 '14 at 11:57
• Side remark... This number is $-W(1)$, where $W$ is the Lambert W-function. – Gerald Edgar Jul 2 '14 at 15:28
• "$K\subset\mathbf{C}$ is closed under $\log$" means "$x\in\mathbf{C}$, $\exp(x)\in K$ $\Rightarrow$ $x\in K$"? or means the a priori weaker "$y\in K$ $\Rightarrow$ $\exists x\in K:\exp(x)=y$"? – YCor Jul 7 at 10:27
• @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. – Wojowu Jul 7 at 10:36

$$R=-W(1)$$, 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.
• 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 $y=0$. – Will Sawin Jul 23 '17 at 20:42
• 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." – LSpice Jul 26 '17 at 13:40