# Effective Lindemann–Weierstrass theorem

The Lindemann–Weierstrass theorem states that if $\alpha_1, ..., \alpha_n$ are algebraic numbers which are linearly independent over the rational numbers $ℚ$, then $e^{\alpha_1}, ..., e^{\alpha_n}$ are algebraically independent over ℚ.

An equivalent formulation by Baker is the following: If $\alpha_1, ..., \alpha_n$ are distinct algebraic numbers, then the exponentials $e^{\alpha_1}, ..., e^{\alpha_n}$are linearly independent over the algebraic numbers. This means that $\sum_i \beta_i e^{\alpha_i} \neq 0$.

My question is: suppose $\sum_i \beta_i e^{\alpha_i} \neq 0$, how to give a lower bound of $|\sum_i \beta_i e^{\alpha_i}|$. In other words, I am looking for some analogical result of Baker's theorem. Note that Baker's theorem is about the logarithm, i.e., a lower bound of $\sum_i \beta_i \log(\alpha_i)$, but here we have exponents.

Actually Baker's theorem generalizes Lindemann–Weierstrass, so that alredy gives you an effective bound

$$\bigg|\sum_i \beta_i e^{\alpha_i}\bigg| > Ce^{-(\log H)^k}$$

with $C$ an effectively computable constant, and everything else as in Baker's first paper on linear forms of logarithms.

• I am very curious how this can be derived from Baker's result. – maomao Aug 2 '16 at 13:53
• BTW, it is stated that L-W is a special case of Baker's theorem, but I do not really see this ... – maomao Aug 2 '16 at 19:44
• Do you have any idea of how C depends on the alpha_i? And what are H and k? – Izaak Meckler Mar 18 '17 at 6:08

This seems to be addressed in the paper by Sert (available for free on the interwebs, it seems).

Alain Sert, MR 1688184 Une version effective du théorème de Lindemann-Weierstrass par les déterminants d’interpolation, J. Number Theory 76 (1999), no. 1, 94--119.

• It is in French, but it looks to me that the result in the paper requires that $\alpha_i$'s are linear independent in Q, but here I only assume they are distinct. Hence the result is more akin to the first formulation of the L-W, but my question is akin to the Baker's formulation. I guess one can "translate" the result, but I do not know how (partially because I cannot really read French). – maomao Aug 2 '16 at 13:57