Lindemann–Weierstrass theorem

Lindemann–Weierstrass theorem Part of a series of articles on the mathematical constant π 3.1415926535897932384626433... Uses Area of a circleCircumferenceUse in other formulae Properties IrrationalityTranscendence Value Less than 22/7ApproximationsMemorization People ArchimedesLiu HuiZu ChongzhiAryabhataMadhavaJamshīd al-KāshīLudolph van CeulenFrançois VièteSeki TakakazuTakebe KenkoWilliam JonesJohn MachinWilliam ShanksSrinivasa RamanujanJohn WrenchChudnovsky brothersYasumasa Kanada History ChronologyA History of Pi In culture LegislationPi Day Related topics Squaring the circleBasel problemSix nines in πOther topics related to π vte Part of a series of articles on the mathematical constant e Properties Natural logarithmExponential function Applications compound interestEuler's identityEuler's formulahalf-lives exponential growth and decay Defining e proof that e is irrationalrepresentations of eLindemann–Weierstrass theorem People John NapierLeonhard Euler Related topics Schanuel's conjecture vte In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states the following: Lindemann–Weierstrass theorem — if α1, ..., αn are algebraic numbers that are linearly independent over the rational numbers {estilo de exibição mathbb {Q} } , then eα1, ..., eαn are algebraically independent over {estilo de exibição mathbb {Q} } .

Em outras palavras, the extension field {estilo de exibição mathbb {Q} (e^{alfa _{1}},pontos ,e^{alfa _{n}})} has transcendence degree n over {estilo de exibição mathbb {Q} } .

An equivalent formulation (Baker 1990, Capítulo 1, Teorema 1.4), é o seguinte: An equivalent formulation — If α1, ..., αn are distinct algebraic numbers, then the exponentials eα1, ..., eαn are linearly independent over the algebraic numbers.

This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over {estilo de exibição mathbb {Q} } by using the fact that a symmetric polynomial whose arguments are all conjugates of one another gives a rational number.

The theorem is named for Ferdinand von Lindemann and Karl Weierstrass. Lindemann proved in 1882 that eα is transcendental for every non-zero algebraic number α, thereby establishing that π is transcendental (Veja abaixo).[1] Weierstrass proved the above more general statement in 1885.[2] O teorema, along with the Gelfond–Schneider theorem, is extended by Baker's theorem, and all of these would be further generalized by Schanuel's conjecture.

Conteúdo 1 Naming convention 2 Transcendence of e and π 3 p-adic conjecture 4 Modular conjecture 5 Lindemann–Weierstrass theorem 5.1 Prova 5.1.1 Preliminary lemmas 5.1.2 Final step 5.1.3 Equivalence of the two statements 6 Veja também 7 Notas 8 Referências 9 Leitura adicional 10 External links Naming convention The theorem is also known variously as the Hermite–Lindemann theorem and the Hermite–Lindemann–Weierstrass theorem. Charles Hermite first proved the simpler theorem where the αi exponents are required to be rational integers and linear independence is only assured over the rational integers,[3][4] a result sometimes referred to as Hermite's theorem.[5] Although apparently a rather special case of the above theorem, the general result can be reduced to this simpler case. Lindemann was the first to allow algebraic numbers into Hermite's work in 1882.[1] Shortly afterwards Weierstrass obtained the full result,[2] and further simplifications have been made by several mathematicians, most notably by David Hilbert[6] and Paul Gordan.[7] Transcendence of e and π The transcendence of e and π are direct corollaries of this theorem.

Suppose α is a non-zero algebraic number; então {uma} is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {} is an algebraically independent set; or in other words eα is transcendental. Em particular, e1 = e is transcendental. (A more elementary proof that e is transcendental is outlined in the article on transcendental numbers.) alternativamente, by the second formulation of the theorem, if α is a non-zero algebraic number, então {0, uma} is a set of distinct algebraic numbers, and so the set {e0, } = {1, } is linearly independent over the algebraic numbers and in particular eα cannot be algebraic and so it is transcendental.

To prove that π is transcendental, we prove that it is not algebraic. If π were algebraic, πi would be algebraic as well, and then by the Lindemann–Weierstrass theorem eπi = −1 (see Euler's identity) would be transcendental, uma contradição. Therefore π is not algebraic, which means that it is transcendental.

A slight variant on the same proof will show that if α is a non-zero algebraic number then sin(uma), porque(uma), tan(uma) and their hyperbolic counterparts are also transcendental.

p-adic conjecture p-adic Lindemann–Weierstrass Conjecture. — Suppose p is some prime number and α1, ..., αn are p-adic numbers which are algebraic and linearly independent over {estilo de exibição mathbb {Q} } , de tal modo que | αi |p < 1/p for all i; then the p-adic exponentials expp(α1), . . . , expp(αn) are p-adic numbers that are algebraically independent over {displaystyle mathbb {Q} } . Modular conjecture An analogue of the theorem involving the modular function j was conjectured by Daniel Bertrand in 1997, and remains an open problem.[8] Writing q = e2πiτ for the square of the nome and j(τ) = J(q), the conjecture is as follows. Modular conjecture — Let q1, ..., qn be non-zero algebraic numbers in the complex unit disc such that the 3n numbers {displaystyle left{J(q_{1}),J'(q_{1}),J''(q_{1}),ldots ,J(q_{n}),J'(q_{n}),J''(q_{n})right}} are algebraically dependent over {displaystyle mathbb {Q} } . Then there exist two indices 1 ≤ i < j ≤ n such that qi and qj are multiplicatively dependent. Lindemann–Weierstrass theorem Lindemann–Weierstrass Theorem (Baker's reformulation). — If a1, ..., an are algebraic numbers, and α1, ..., αn are distinct algebraic numbers, then[9] {displaystyle a_{1}e^{alpha _{1}}+a_{2}e^{alpha _{2}}+cdots +a_{n}e^{alpha _{n}}=0} has only the trivial solution {displaystyle a_{i}=0} for all {displaystyle i=1,dots ,n.} Proof The proof relies on two preliminary lemmas. Notice that Lemma B itself is already sufficient to deduce the original statement of Lindemann–Weierstrass theorem. Preliminary lemmas Lemma A. — Let c(1), ..., c(r) be integers and, for every k between 1 and r, let {γ(k)1, ..., γ(k)m(k)} be the roots of a non-zero polynomial with integer coefficients {displaystyle T_{k}(x)} . If γ(k)i ≠ γ(u)v whenever (k, i) ≠ (u, v), then {displaystyle c(1)left(e^{gamma (1)_{1}}+cdots +e^{gamma (1)_{m(1)}}right)+cdots +c(r)left(e^{gamma (r)_{1}}+cdots +e^{gamma (r)_{m(r)}}right)=0} has only the trivial solution {displaystyle c(i)=0} for all {displaystyle i=1,dots ,r.} Proof of Lemma A. To simplify the notation set: {displaystyle {begin{aligned}&n_{0}=0,&&\&n_{i}=sum nolimits _{k=1}^{i}m(k),&&i=1,ldots ,r\&n=n_{r},&&\&alpha _{n_{i-1}+j}=gamma (i)_{j},&&1leq ileq r, 1leq jleq m(i)\&beta _{n_{i-1}+j}=c(i).end{aligned}}} Then the statement becomes {displaystyle sum _{k=1}^{n}beta _{k}e^{alpha _{k}}neq 0.} Let p be a prime number and define the following polynomials: {displaystyle f_{i}(x)={frac {ell ^{np}(x-alpha _{1})^{p}cdots (x-alpha _{n})^{p}}{(x-alpha _{i})}},} where ℓ is a non-zero integer such that {displaystyle ell alpha _{1},ldots ,ell alpha _{n}} are all algebraic integers. Define[10] {displaystyle I_{i}(s)=int _{0}^{s}e^{s-x}f_{i}(x),dx.} Using integration by parts we arrive at {displaystyle I_{i}(s)=e^{s}sum _{j=0}^{np-1}f_{i}^{(j)}(0)-sum _{j=0}^{np-1}f_{i}^{(j)}(s),} where {displaystyle np-1} is the degree of {displaystyle f_{i}} , and {displaystyle f_{i}^{(j)}} is the j-th derivative of {displaystyle f_{i}} . This also holds for s complex (in this case the integral has to be intended as a contour integral, for example along the straight segment from 0 to s) because {displaystyle -e^{s-x}sum _{j=0}^{np-1}f_{i}^{(j)}(x)} is a primitive of {displaystyle e^{s-x}f_{i}(x)} . Consider the following sum: {displaystyle {begin{aligned}J_{i}&=sum _{k=1}^{n}beta _{k}I_{i}(alpha _{k})\[5pt]&=sum _{k=1}^{n}beta _{k}left(e^{alpha _{k}}sum _{j=0}^{np-1}f_{i}^{(j)}(0)-sum _{j=0}^{np-1}f_{i}^{(j)}(alpha _{k})right)\[5pt]&=left(sum _{j=0}^{np-1}f_{i}^{(j)}(0)right)left(sum _{k=1}^{n}beta _{k}e^{alpha _{k}}right)-sum _{k=1}^{n}sum _{j=0}^{np-1}beta _{k}f_{i}^{(j)}(alpha _{k})\[5pt]&=-sum _{k=1}^{n}sum _{j=0}^{np-1}beta _{k}f_{i}^{(j)}(alpha _{k})end{aligned}}} In the last line we assumed that the conclusion of the Lemma is false. In order to complete the proof we need to reach a contradiction. We will do so by estimating {displaystyle |J_{1}cdots J_{n}|} in two different ways. First {displaystyle f_{i}^{(j)}(alpha _{k})} is an algebraic integer which is divisible by p! for {displaystyle jgeq p} and vanishes for {displaystyle j1} , e deixar {estilo de exibição a(1),ldots ,uma(n)} be non-zero algebraic numbers, e {alfa de estilo de exibição (1),ldots ,alfa (n)} distinct algebraic numbers such that: {estilo de exibição a(1)e^{alfa (1)}+cdots +a(n)e^{alfa (n)}=0.} As seen in the previous section, and with the same notation used there, the value of the polynomial {estilo de exibição Q(x_{11},ldots ,x_{nd(n)},s_{1},pontos ,s_{n})=prod nolimits _{sigma in S}deixei(x_{1sigma (1)}s_{1}+pontos +x_{Eu sinto Muito (n)}s_{n}certo),} no {estilo de exibição à esquerda(uma(1)_{1},ldots ,uma(n)_{d(n)},e^{alfa (1)},ldots ,e^{alfa (n)}certo)} has an expression of the form {estilo de exibição b(1)e^{beta (1)}+b(2)e^{beta (2)}+cdots +b(M)e^{beta (M)}=0,} where we have grouped the exponentials having the same exponent. Aqui, as proved above, {estilo de exibição b(1),ldots ,b(M)} are rational numbers, not all equal to zero, and each exponent {beta de estilo de exibição (m)} is a linear combination of {alfa de estilo de exibição (eu)} com coeficientes inteiros. Então, desde {displaystyle n>1} e {alfa de estilo de exibição (1),ldots ,alfa (n)} are pairwise distinct, a {estilo de exibição mathbb {Q} } -vector subspace {estilo de exibição V} do {estilo de exibição mathbb {C} } generated by {alfa de estilo de exibição (1),ldots ,alfa (n)} is not trivial and we can pick {alfa de estilo de exibição (eu_{1}),ldots ,alfa (eu_{k})} to form a basis for {displaystyle V.} For each {displaystyle m=1,dots ,M} , temos {estilo de exibição {começar{alinhado}beta (m)=q_{m,1}alfa (eu_{1})+cdots +q_{m,k}alfa (eu_{k}),&&q_{m,j}={fratura {c_{m,j}}{d_{m,j}}};qquad c_{m,j},d_{m,j}em matemática {Z} .fim{alinhado}}} For each {displaystyle j=1,ldots ,k,} deixar {estilo de exibição d_{j}} be the least common multiple of all the {estilo de exibição d_{m,j}} por {displaystyle m=1,ldots ,M} , and put {estilo de exibição v_{j}={tfrac {1}{d_{j}}}alfa (eu_{j})} . Então {estilo de exibição v_{1},ldots ,v_{k}} are algebraic numbers, they form a basis of {estilo de exibição V} , e cada {beta de estilo de exibição (m)} is a linear combination of the {estilo de exibição v_{j}} com coeficientes inteiros. By multiplying the relation {estilo de exibição b(1)e^{beta (1)}+b(2)e^{beta (2)}+cdots +b(M)e^{beta (M)}=0,} por {estilo de exibição e^{N(v_{1}+cdots +v_{k})}} , Onde {estilo de exibição N} is a large enough positive integer, we get a non-trivial algebraic relation with rational coefficients connecting {estilo de exibição e^{v_{1}},cdots ,e^{v_{k}}} , against the first formulation of the theorem.

See also Gelfond–Schneider theorem Baker's theorem; an extension of Gelfond–Schneider theorem Schanuel's conjecture; if proven, it would imply both the Gelfond–Schneider theorem and the Lindemann–Weierstrass theorem Notes ^ Jump up to: a b Lindemann 1882a, Lindemann 1882b. ^ Saltar para: a b Weierstrass 1885, pp. 1067–1086, ^ Hermite 1873, pp. 18–24. ^ Hermite 1874 ^ Gelfond 2015. ^ Hilbert 1893, pp. 216-219. ^ Gordan 1893, pp. 222-224. ^ Bertrand 1997, pp. 339–350. ^ (em francês) french Proof's Lindemann-Weierstrass (pdf)[link morto] ^ Up to a factor, this is the same integral appearing in the proof that e is a transcendental number, where β1 = 1, ..., βm = m. The rest of the proof of the Lemma is analog to that proof. References Gordan, P. (1893), "Transcendenz von e und π.", Anais Matemáticos, 43 (2-3): 222-224, doi:10.1007/bf01443647, S2CID 123203471 Hermite, C. (1873), "Sur la fonction exponentielle.", Comptes rendus de l'Académie des Sciences de Paris, 77: 18–24 Hermite, C. (1874), Sur la fonction exponentielle., Paris: Gauthier-Villars Hilbert, D. (1893), "Ueber die Transcendenz der Zahlen e und π.", Anais Matemáticos, 43 (2-3): 216-219, doi:10.1007/bf01443645, S2CID 179177945, arquivado a partir do original em 2017-10-06, recuperado 2018-12-24 Lindemann, F. (1882), "Über die Ludolph'sche Zahl.", Relatórios da sessão da Royal PREUSSIAN Academy of Sciences em Berlim, 2: 679–682 Lindemann, F. (1882), "Über die Zahl π.", Anais Matemáticos, 20: 213–225, doi:10.1007/bf01446522, S2CID 120469397, arquivado a partir do original em 2017-10-06, recuperado 2018-12-24 Weierstrass, K. (1885), "Zu Lindemann's Abhandlung. "Über die Ludolph'sche Zahl".", Sitzungsberichte der Königlich Preussischen Akademie der Wissen-schaften zu Berlin, 5: 1067–1085 Further reading Baker, Alan (1990), Transcendental number theory, Cambridge Mathematical Library (2ª edição), Cambridge University Press, ISBN 978-0-521-39791-9, SENHOR 0422171 Bertrand, D. (1997), "Theta functions and transcendence", The Ramanujan Journal, 1 (4): 339–350, doi:10.1023/UMA:1009749608672, S2CID 118628723 Gelfond, A.O. (2015) [1960], Transcendental and Algebraic Numbers, Dover Books on Mathematics, translated by Boron, Leo F., Nova york: Publicações de Dover, ISBN 978-0-486-49526-2, SENHOR 0057921 Jacobson, Natan (2009) [1985], Basic Algebra, volume. EU (2ª edição), Publicações de Dover, ISBN 978-0-486-47189-1 Weissstein esquerdo externo, Eric W. "Hermite-Lindemann Theorem". MathWorld. Weisstein, Eric W. "Lindemann-Weierstrass Theorem". MathWorld. Categorias: E (mathematical constant)ExponentialsPiTheorems in number theoryTranscendental numbers

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