Talk:Lindemann–Weierstrass theorem

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

needs more information
What beta? 07:13, 9 Aug 2004 (UTC)

Any nonzero algebraic number &#945; gives us a set {&#945;} which is trivially a linearly independent set over the rationals, and hence e&#945; is immediately seen to be transcendental.
 * Is this argument valid? The problem I have with it is that simply because {e&alpha;} is a linearly independent set over the algebraic numbers doesn't mean that the number e&alpha; 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)
 * 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)

needs date
If Lindemann proved pi is trancendental, we should cite the published proof, and at least give a date.

The part about Baker's characterization needs a fix.

 * An equivalent formulation, is the following: If &alpha;1,...,&alpha;n 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)
 * 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)


 * But $$e^i+e^{-i}= 2cos(1)$$, not zero. JRSP (talk) 23:47, 23 December 2007 (UTC)
 * 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)
 * According to planetmath.org "An equivalent version of the theorem states that if &alpha;1,...,&alpha;n 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  &alpha;1,...,&alpha;n are algebraic and distinct, and if &beta;1,...,&beta;n  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)
 * Let's see anon ip's argument: "Take &alpha; to be a non-real fifth root of unity. Then $$\alpha,\alpha^2,\alpha^3,\alpha^4$$ are distint, but &alpha; 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)

This example depends on the "second formulation" so I moved it here: Alternatively, using the second formulation of the theorem, we can argue that if &alpha; is a nonzero algebraic number, then {0, &alpha;} is a set of distinct algebraic numbers, and so the set $$\{e^0,e^\alpha\}=\{1,e^\alpha\}$$ is linearly independent over the algebraic numbers and in particular e&alpha; can't be algebraic and so is transcendental. JRSP (talk) 01:22, 23 December 2007 (UTC)

The reference in the second paragraph of the article should be to Baker not Baxter.Fathead99 (talk) 12:07, 30 January 2008 (UTC)

Abuse of notation?
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)

Lemma C
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)

Reference at Note 3
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)

Equivalent formulation?
From the article: "An equivalent formulation (Baker 1975, Chapter 1, Theorem 1.4), is the following: If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over the algebraic numbers." Now, I don't have that book, and maybe I am missing some important property of the set of algebraic numbers, but it seems to me that proving that eα1, ..., eαn are algebraically independent over the algebraic numbers would actually be a stronger statement than proving that they are algebraically independent over the rationals. Now, I know that things like "equivalent", "stronger", and "weaker" are meaningless in the cases of already proven theorems, because all true statements are equivalent over the same system (if you assume A then prove B, but A is True, you are really just proving B without any need for the assumption on A). So am I missing something here? And even if so, it should be clearer in the article. - Asmeurer  ( talk   ♬  contribs ) 00:58, 9 August 2010 (UTC)


 * Suppose we know Baker's version, that is if αi are n algebraic, distinct numbers then exp(αi) are linearly independent over the algebraic numbers.


 * Now let αi be n algebraic numbers, linearly independent over Q and suppose there is an algebraic relation between the exp(αi), say P(exp(α1),&hellip;,exp(αn)) = 0 for some nonzero polynomial P with coefficients in Q. Each term in P(α1,&hellip;,αn) is something like
 * $$\beta_i \exp(m_1\alpha_1+\ldots+m_n\alpha_n)\,$$
 * for integers mi and rational numbers βi. Because the αi are linearly independent over Q these exponents are all distinct, so by Baker's version of the theorem all the βi must be zero, which proves the original version of the theorem.


 * In the opposite direction: assume the original version is true and let αi be n distinct, algebraic numbers. Note that by sifting, some subset of these n numbers is a basis for the vector space generated by them over Q.  We want to show that if βi are n nonzero algebraic numbers then $$\sum_i\beta_i e^{\alpha_i}\neq0$$.  It obviously suffices to show that this left-hand expression multiplied by something else is nonzero, so we multiply it by the exact same expression except with each βi replaced by all of its conjugates in every possible combination.  The result is a very large polynomial in the exp(αi) but now with rational coefficients.  Since occurrences of the αi that aren't in the basis subset can be replaced by linear combinations of those that are, we can treat this as a polynomial in exp(αi) for just the basis elements αi.  In particular the αi are linearly independent over Q, so by the original version of Lindemann-Weierstrass this polynomial is nonzero, hence $$\sum_i\beta_i e^{\alpha_i}\neq0$$. Chenxlee (talk) 15:41, 25 August 2010 (UTC)


 * Your argument is very nice and would be better in the proof section of the article. After all it is necessary for finishing the proof, and would avoid letting people wondering. —Preceding unsigned comment added by 82.244.80.154 (talk) 18:56, 11 May 2011 (UTC)

Proof sketch
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)
 * The sketch proof is of Baker's equivalent reformulation mentioned in the lede, that if $α$1, &hellip;, $α$$n$ are distinct algebraic numbers, then the exponentials $e$α1, &hellip;, $e$$α$$n$ are linearly independent over the algebraic numbers. I've mentioned this in the article now. Chenxlee (talk) 12:06, 26 October 2011 (UTC)

Lemma B
"Thus we may choose the polynomials in such a way, that the value of $u$($k$)$v$($k$)$m$($k$) &minus; 1 is independent of $k$"

This is not true. Follow some of the known proofs.

New proof
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)

Assumes too much background knowledge
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)

Additionally, the articles in other languages look very different, which doesn't fill me with confidence in this one. — Preceding unsigned comment added by 84.243.199.239 (talk) 09:07, 27 May 2015 (UTC)

Condition in Lemma A
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)

Does it really follow from Baker's theorem?
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)