Talk:Lindemann–Weierstrass theorem

message: on most browsers they appear as apostrophes.  —Preceding unsigned comment added by 76.182.194.195 (talk) 08:41, 15 March 2010 (UTC)
thread_title:  font for the exponents of the initial statement of the theorem too tiny 

message: What beta? 07:13, 9 Aug 2004 (UTC)
replies: text: Is this argument valid? The problem I have with it is that simply because {eα} is a linearly independent set over the algebraic numbers doesn't mean that the number eα is transcendental. For instance, {log 2} is linearly independent over the rationals (if a*(log 2) = 0 for rational a, then a = 0.), and also {elog 2 = 2} is linearly independent set over the algebraic numbers (if a*2 = 0 for algebraic a, then a = 0.), yet 2 is not transcendental. I also understand the urge to condense everything to be elegant, but I think expressing things out explicitly in terms of linear combinations is still helpful for people who might not be able to immediately mentally untangle "linearly independent" or "algebraically independent". Revolver 04:01, 2 Nov 2004 (UTC)
replies: text: I see...you're secretly still using the fact that the number is algebraic. I was seeing that ({a} LI over Q) does not imply ({a} AI over A, which is true, but if you add a being alg. to hypothesis, it does go through. Revolver 02:51, 9 September 2005 (UTC)
thread_title:  needs more information 

message: :An equivalent formulation , is the following: If α1,...,αn are distinct algebraic numbers, then the exponentials e^{\alpha_1},\ldots,e^{\alpha_n} are linearly independent over the algebraic numbers.
The part about Baker's theorem needs a fix,  extra hypothesis are required (a non real fifth root of unity provides a counter example)  —Preceding unsigned comment added by 153.90.244.6 (talk) 18:57, 5 December 2007 (UTC)
replies: text: A better counterexample: e^i+e^{-i}=0. Unless Baxter's theorem concerns only real numbers this is definitely false.  Grue  22:42, 5 December 2007 (UTC)
replies: text:  But e^i+e^{-i}= 2cos(1), not zero. JRSP (talk) 23:47, 23 December 2007 (UTC)
replies: text:  OMG, did I say something stupid on Wikipedia again. Indeed, you are right, and if e^x+e^y=0 then x=y+\pi i+2\pi i k, so they cannot be both algebraic. Now we need to figure out if the original counterexample holds (probably not, since roots of 1 have transcendental logarithms), and whether the formulation is actually true.  Grue  11:41, 24 December 2007 (UTC)
replies: text:  According to planetmath.org "An equivalent version of the theorem states that if  α1,...,αn are distinct algebraic numbers over , then  e^{\alpha_1},\ldots,e^{\alpha_n} are linearly independent over \mathbb{Q}." (not over the algebraic numbers). I also found the proof there. (we could add the external link). In the proof they say : "Theorem 3: If  α1,...,αn are algebraic and distinct, and if β1,...,βn  are algebraic and non-zero, then \beta _1 e^{\alpha _1 }  + ... + \beta _n e^{\alpha _n }  \ne 0 " which basically says the exponentials are LI over the algebraic numbers. JRSP (talk) 15:01, 24 December 2007 (UTC)

text:  Let's see anon ip's argument: "Take α to be a non-real fifth root of unity. Then \alpha,\alpha^2,\alpha^3,\alpha^4 are distint, but α satisfies 1+x+x^2+x^3+x^4=0." This only says that 1,\alpha,\alpha^2,\alpha^3,\alpha^4 are linearly dependent but does not show if e,e^{\alpha},\ldots,e^{\alpha^4} are LD. JRSP (talk) 15:24, 24 December 2007 (UTC)
thread_title:  The part about Baker's characterization needs a fix. 

message: Given that the alpha_js are irrational each e^alpha denotes infinitely many numbers. Now what does Q(e^alpha_j) mean? All powers added, or one by one separately? If the latter does Q(e^alpha_1...e^alpha_n) mean any combination of those for all j? Scineram (talk) 17:28, 29 October 2009 (UTC)
thread_title:  Abuse of notation? 

message: I could well be missing something, and indeed I hope I am, but I think Lemma C is invalid.  The proof relies on the betas being non-zero.  However the betas are linear combinations of the alphas, and as far as I can see, there is no reason why some of these linear combinations could not be zero.
If some of the betas are zero, and we reform the equation K'+c(1)(.....) then we obtain an equation K"+c(1)(.....) where all the betas to the right of c(1) are non-zero, but K" may be zero.  This invalidates the application of Lemma B because to apply Lemma B we must give K the value K" and if K is zero, then Lemma B cannot be applied.  This invalidates the whole proof.
The proof sketch closely follows Baker's proof that pi is transcendental (Baker's book is included as a reference to the article) and attempts to apply the same method to produce the generalised result. Baker's proof of the L-W Theorem is highly condensed and therefore difficult to follow (at least for someone of my modest mathematical ability) but in particular, he uses essentially the same sequence of functions f that are used by Hermite in his proof of the transcendental nature of e.  In turn, Hermite's proof was closely followed by Lindemann. These are different from the functions f used in this article, which simply differ by a constant factor.  If it were possible to extend the pi-proof in a simple manner to prove the Lindemann-Weierstrass Theorem, I think Baker would have found this method and used it.  —Preceding unsigned comment added by 121.216.107.117 (talk) 05:21, 9 January 2010 (UTC)
thread_title:  Lemma C 

message: The reference at Note 3 is incorrect.  It should be to Lindemann's Math. Ann. paper of 1882.  There is no paper by Lindemann at the 1888 reference cited at Note 3.  —Preceding unsigned comment added by 58.168.108.203 (talk) 06:42, 24 January 2010 (UTC)
thread_title:  Reference at Note 3 

message: There's something extremely fishy about the proof sketch. It says "We prove that if a1, a2... an are non-zero algebraic numbers, and α1, α2... αn are distinct algebraic numbers, then a_1 e^{\alpha_1} + a_2 e^{\alpha_2} + ... a_n e^{\alpha_n}\ne 0.". The same thing is repeated in the "final step" section. But that's not algebraic independence, that's just plain old linear independence! In the n=1 case, this doesn't say that e^\alpha is transcendental, just that (because \mathbb{C} is an integral domain) e^\alpha \neq 0. Well, duh. I'm not all that comfortable with French (or with this proof), so I don't know exactly what needs fixing, but at least the K term from the reference needs to be added for any of this to make sense. 85.226.204.222 (talk) 07:49, 27 August 2010 (UTC)
replies: text: The sketch proof is of Baker's equivalent reformulation mentioned in the lede, that if 1, …,  are distinct algebraic numbers, then the exponentials α1, …,  are linearly independent over the algebraic numbers.  I've mentioned this in the article now. Chenxlee (talk) 12:06, 26 October 2011 (UTC)
thread_title:  Proof sketch 

message: The previous proof was incorrect or at least it had some unclear points. I posted a new proof of the lemmas. This new proof is taken almost entirely from PlanetMath.  — Preceding unsigned comment added by Kappaenne (talk • contribs) 22:14, 22 October 2013 (UTC)
thread_title:  New proof 

message: This article is incomprehensible not just if you are not a mathematician, but if you are not a particular kind of mathematician. That is not good for an encyclopaedic article.  — Preceding unsigned comment added by 84.243.199.239 (talk) 09:04, 27 May 2015 (UTC)
thread_title:  Assumes too much background knowledge 

message: It seems to me that the condition 'If γ(k)_i ≠ γ(u)_v whenever (k, i) ≠ (u, v), then...' in Lemma A is too weak and that it should be 'If γ(k)_i ≠ γ(u)_v whenever k ≠ i, then...'. The reason being that with the current formulation we could just have all the polynomials $T_k$ being identical but with their roots labeled in a different order. It seems pretty obvious that in that case the lemma would be false. However I am reluctant to change it in the main article because I have never seen this proof before and maybe I am misinterpreting something. I would be happy if someone more expert could take a look at this!  — Preceding unsigned comment added by Octonion (talk • contribs) 09:29, 14 December 2021 (UTC)
thread_title:  Condition in Lemma A 

message: As written the introduction seems to imply that Lindeman-Weierstrass follows from Baker's theorem, but I can't find this claim elsewhere and it doesn't seem true to me. 128.12.123.40 (talk) 20:29, 25 July 2022 (UTC)
thread_title:  Does it really follow from Baker's theorem?