Polinomios de Bernoulli Enmatemáticas ospolinomios de Bernoulli B n ( x ) {\displaystyle B_{n}(x)} función xeradora exponencial , tal como se expón a continuación:
e t x t e t − 1 = ∑ n = 0 ∞ B n ( x ) t n n ! {\displaystyle {\frac {et^{xt}}{e^{t}-1}}=\sum _{n=0}^{\infty }B_{n}(x){\frac {t^{n}}{n!}}} Aparecen no estudo de moitasfuncións especiais , en particular dafunción zeta de Riemann e dafunción zeta de Hurwitz . Osnúmeros de Bernoulli b n {\displaystyle b_{n}} B n {\displaystyle B_{n}} b n = B n ( 0 ) {\displaystyle b_{n}=B_{n}(0)}
A identidadeB k + 1 ( x + 1 ) − B k + 1 ( x ) = ( k + 1 ) x k {\displaystyle B_{k+1}(x+1)-B_{k+1}(x)=(k+1)x^{k}\,} forma pechada da suma dosn primeiros números enteiros positivos elevados a unha potenciak ,
∑ i = 1 n i k = 1 k + 2 k + ⋯ + n k = B k + 1 ( n + 1 ) − B k + 1 ( 0 ) k + 1 {\displaystyle \sum _{i=1}^{n}{i^{k}}=1^{k}+2^{k}+\cdots +n^{k}={\frac {B_{k+1}(n+1)-B_{k+1}(0)}{k+1}}} Un conxunto similar de polinomios, baseado nunha función xeradora, é a familia depolinomios de Euler . Neste artigo mencionaremos propiedades e fórmulas para ambas as dúas familias.
A funcións xeradora para os polinomios de Bernoulli é
t e x t e t − 1 = ∑ n = 0 ∞ B n ( x ) t n n ! . {\displaystyle {\frac {te^{xt}}{e^{t}-1}}=\sum _{n=0}^{\infty }B_{n}(x){\frac {t^{n}}{n!}}.} E para os polinomios de Euler é
2 e x t e t + 1 = ∑ n = 0 ∞ E n ( x ) t n n ! . {\displaystyle {\frac {2e^{xt}}{e^{t}+1}}=\sum _{n=0}^{\infty }E_{n}(x){\frac {t^{n}}{n!}}.} Para os polinomios de BernoulliB n ( x ) {\displaystyle B_{n}(x)} E n ( x ) {\displaystyle E_{n}(x)}
B n ( x ) = ∑ k = 0 n ( n k ) B n − k x k , {\displaystyle B_{n}(x)=\sum _{k=0}^{n}{n \choose k}B_{n-k}x^{k},} E m ( x ) = ∑ k = 0 m ( m k ) E k 2 k ( x − 1 2 ) m − k . {\displaystyle E_{m}(x)=\sum _{k=0}^{m}{m \choose k}{\frac {E_{k}}{2^{k}}}\left(x-{\tfrac {1}{2}}\right)^{m-k}.} paran ≥ 0 {\displaystyle n\geq 0} B k {\displaystyle B_{k}} números de Bernoulli , e osE k {\displaystyle E_{k}} números de Euler .
Dedúcese logo que
B n ( 0 ) = B n {\displaystyle B_{n}(0)=B_{n}} A027641 naOEIS ) e denominadores (secuenciaA027642 naOEIS ))e
E m ( 1 2 ) = 1 2 m E m {\displaystyle E_{m}{\big (}{\tfrac {1}{2}}{\big )}={\tfrac {1}{2^{m}}}E_{m}} A122045 naOEIS ), tendo en conta que hai quen usa outro criterio usando só os números de índice par, vernúmeros de Euler ).Os primeiros polinomios de Bernoulli son:
B 0 ( x ) = 1 {\displaystyle B_{0}(x)=1\,} B 1 ( x ) = x − 1 / 2 {\displaystyle B_{1}(x)=x-1/2\,} B 2 ( x ) = x 2 − x + 1 / 6 {\displaystyle B_{2}(x)=x^{2}-x+1/6\,} B 3 ( x ) = x 3 − 3 2 x 2 + 1 2 x {\displaystyle B_{3}(x)=x^{3}-{\frac {3}{2}}x^{2}+{\frac {1}{2}}x\,} B 4 ( x ) = x 4 − 2 x 3 + x 2 − 1 30 {\displaystyle B_{4}(x)=x^{4}-2x^{3}+x^{2}-{\frac {1}{30}}\,} B 5 ( x ) = x 5 − 5 2 x 4 + 5 3 x 3 − 1 6 x {\displaystyle B_{5}(x)=x^{5}-{\frac {5}{2}}x^{4}+{\frac {5}{3}}x^{3}-{\frac {1}{6}}x\,} B 6 ( x ) = x 6 − 3 x 5 + 5 2 x 4 − 1 2 x 2 + 1 42 {\displaystyle B_{6}(x)=x^{6}-3x^{5}+{\frac {5}{2}}x^{4}-{\frac {1}{2}}x^{2}+{\frac {1}{42}}\,} Os primeiros polinomios de Euler son:
E 0 ( x ) = 1 , E 4 ( x ) = x 4 − 2 x 3 + x , E 1 ( x ) = x − 1 2 , E 5 ( x ) = x 5 − 5 2 x 4 + 5 2 x 2 − 1 2 , E 2 ( x ) = x 2 − x , E 6 ( x ) = x 6 − 3 x 5 + 5 x 3 − 3 x , E 3 ( x ) = x 3 − 3 2 x 2 + 1 4 , ⋮ {\displaystyle {\begin{aligned}E_{0}(x)&=1,&E_{4}(x)&=x^{4}-2x^{3}+x,\\[4mu]E_{1}(x)&=x-{\tfrac {1}{2}},&E_{5}(x)&=x^{5}-{\tfrac {5}{2}}x^{4}+{\tfrac {5}{2}}x^{2}-{\tfrac {1}{2}},\\[4mu]E_{2}(x)&=x^{2}-x,&E_{6}(x)&=x^{6}-3x^{5}+5x^{3}-3x,\\[-1mu]E_{3}(x)&=x^{3}-{\tfrac {3}{2}}x^{2}+{\tfrac {1}{4}},\qquad \ \ &&\ \,\,\vdots \end{aligned}}} Os polinomios de Bernoulli e Euler obedecen a moitas relacións docálculo sombra usado porÉdouard Lucas , por exemplo.
B n ( x + 1 ) − B n ( x ) = n x n − 1 {\displaystyle B_{n}(x+1)-B_{n}(x)=nx^{n-1}\,} E n ( x + 1 ) + E n ( x ) = 2 x n {\displaystyle E_{n}(x+1)+E_{n}(x)=2x^{n}\,} B n ′ ( x ) = n B n − 1 ( x ) {\displaystyle B_{n}'(x)=nB_{n-1}(x)\,} E n ′ ( x ) = n E n − 1 ( x ) {\displaystyle E_{n}'(x)=nE_{n-1}(x)\,} B n ( x + y ) = ∑ k = 0 n ( n k ) B k ( x ) y n − k {\displaystyle B_{n}(x+y)=\sum _{k=0}^{n}{n \choose k}B_{k}(x)y^{n-k}} E n ( x + y ) = ∑ k = 0 n ( n k ) E k ( x ) y n − k {\displaystyle E_{n}(x+y)=\sum _{k=0}^{n}{n \choose k}E_{k}(x)y^{n-k}} B n ( 1 − x ) = ( − 1 ) n B n ( x ) {\displaystyle B_{n}(1-x)=(-1)^{n}B_{n}(x)} E n ( 1 − x ) = ( − 1 ) n E n ( x ) {\displaystyle E_{n}(1-x)=(-1)^{n}E_{n}(x)} ( − 1 ) n B n ( − x ) = B n ( x ) + n x n − 1 {\displaystyle (-1)^{n}B_{n}(-x)=B_{n}(x)+nx^{n-1}} ( − 1 ) n E n ( − x ) = − E n ( x ) + 2 x n {\displaystyle (-1)^{n}E_{n}(-x)=-E_{n}(x)+2x^{n}} ∀ n ∈ N , B n ( x ) = 2 n − 1 ( B n ( x 2 ) + B n ( x + 1 2 ) ) {\displaystyle \forall n\in \mathbb {N} ,B_{n}(x)=2^{n-1}\left(B_{n}\left({\frac {x}{2}}\right)+B_{n}\left({\frac {x+1}{2}}\right)\right)} ∀ p ∈ N , ∀ n ∈ N , ∑ k = 0 n k p = B p + 1 ( n + 1 ) − B p + 1 ( 0 ) p + 1 {\displaystyle \forall p\in \mathbb {N} ,\forall n\in \mathbb {N} ,\sum _{k=0}^{n}k^{p}={\frac {B_{p+1}(n+1)-B_{p+1}(0)}{p+1}}} Esta última igualdade, deducida dafórmula de Faulhaber , provén da igualdade:∫ x x + 1 B n ( t ) d t = x n {\displaystyle \int _{x}^{x+1}B_{n}(t)\,\mathrm {d} t=x^{n}} serie telescópica
∑ k = 0 n ( B m ( k + 1 ) − B m ( k ) ) = B m ( n + 1 ) − B m ( 0 ) {\displaystyle \sum _{k=0}^{n}\left(B_{m}(k+1)-B_{m}(k)\right)=B_{m}(n+1)-B_{m}(0)} Aserie de Fourier dos polinomios de Bernoulli tamén é unhaserie de Dirichlet , dada polo desenvolvemento[ 1]
B n ( x ) = − n ! ( 2 π i ) n ∑ k ∈ Z k ≠ 0 e 2 π i k x k n = − n ! ∑ k = 1 ∞ e 2 π i k x + ( − 1 ) n e − 2 π i k x ( 2 π i k ) n = − 2 n ! ∑ k = 1 ∞ cos ( 2 k π x − n π 2 ) ( 2 k π ) n {\displaystyle B_{n}(x)=-{\frac {n!}{(2\pi \mathrm {i} )^{n}}}\sum _{k\in \mathbb {Z} \atop k\neq 0}{\frac {\mathrm {e} ^{2\pi \mathrm {i} kx}}{k^{n}}}=-n!\sum _{k=1}^{\infty }{\frac {\mathrm {e} ^{2\pi \mathrm {i} kx}+(-1)^{n}\mathrm {e} ^{-2\pi \mathrm {i} kx}}{(2\pi \mathrm {i} k)^{n}}}=-2\,n!\sum _{k=1}^{\infty }{\frac {\cos \left(2k\pi x-{\frac {n\pi }{2}}\right)}{(2k\pi )^{n}}}} válido só para0 ≤ x ≤ 1 {\displaystyle 0\leq x\leq 1} n ≥ 2 {\displaystyle n\geq 2} 0 < x < 1 {\displaystyle 0<x<1} n = 1 {\displaystyle n=1}
Este é un caso especial dafórmula de Hurwitz .
Dúas integrais definidas que relacionan os polinomios de Bernoulli e Euler cos números de Bernoulli e Euler son:[ 2]
Outra integral dános[ 3]
e casos particulares sen a variábely {\displaystyle y} función zeta de Riemann
↑ Tsuneo Arakawa; Tomoyoshi Ibukiyama; Masanobu Kaneko (2014).Bernoulli Numbers and Zeta Functions .Springer . p. 61. ↑ Takashi Agoh; Karl Dilcher (2011). "Integrals of products of Bernoulli polynomials".Journal of Mathematical Analysis and Applications 381 : 10–16.doi :10.1016/j.jmaa.2011.03.061 . ↑ Elaissaoui, Lahoucine; Guennoun, Zine El Abidine (2017). "Evaluation of log-tangent integrals by series involving ζ(2n+1)".Integral Transforms and Special Functions 28 (6): 460–475.arXiv :1611.01274 .doi :10.1080/10652469.2017.1312366 . 
