Polinomios de Bernoulli En matemáticas os polinomios 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 moitas funcións especiais , en particular da función zeta de Riemann e da función zeta de Hurwitz . Os números de Bernoulli
b
n
{\displaystyle b_{n}}
B
n
{\displaystyle B_{n}}
b
n
=
B
n
(
0
)
{\displaystyle b_{n}=B_{n}(0)}
A identidade
B
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 dos n primeiros números enteiros positivos elevados a unha potencia k ,
∑
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 de polinomios 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 Bernoulli
B
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}.}
para
n
≥
0
{\displaystyle n\geq 0}
B
k
{\displaystyle B_{k}}
números de Bernoulli , e os
E
k
{\displaystyle E_{k}}
números de Euler .
Dedúcese logo que
B
n
(
0
)
=
B
n
{\displaystyle B_{n}(0)=B_{n}}
A027641 na OEIS ) e denominadores (secuencia A027642 na OEIS ))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 na OEIS ), tendo en conta que hai quen usa outro criterio usando só os números de índice par, ver números 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 do cálculo sombra usado por Édouard 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 da fórmula 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)}
A serie de Fourier dos polinomios de Bernoulli tamén é unha serie 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 para
0
≤
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 da fórmula de Hurwitz .
Dúas integrais definidas que relacionan os polinomios de Bernoulli e Euler cos números de Bernoulli e Euler son:
[ 2]
∫
0
1
B
n
(
t
)
B
m
(
t
)
d
t
=
(
−
1
)
n
−
1
m
!
n
!
(
m
+
n
)
!
B
n
+
m
for
m
,
n
≥
1
{\displaystyle \int _{0}^{1}B_{n}(t)B_{m}(t)\,dt=(-1)^{n-1}{\frac {m!\,n!}{(m+n)!}}B_{n+m}\quad {\text{for }}m,n\geq 1}
∫
0
1
E
n
(
t
)
E
m
(
t
)
d
t
=
(
−
1
)
n
4
(
2
m
+
n
+
2
−
1
)
m
!
n
!
(
m
+
n
+
2
)
!
B
n
+
m
+
2
{\displaystyle \int _{0}^{1}E_{n}(t)E_{m}(t)\,dt=(-1)^{n}4(2^{m+n+2}-1){\frac {m!\,n!}{(m+n+2)!}}B_{n+m+2}}
Outra integral dános [ 3]
∫
0
1
E
n
(
x
+
y
)
log
(
tan
π
2
x
)
d
x
=
n
!
∑
k
=
1
⌊
n
+
1
2
⌋
(
−
1
)
k
−
1
π
2
k
(
2
−
2
−
2
k
)
ζ
(
2
k
+
1
)
y
n
+
1
−
2
k
(
n
+
1
−
2
k
)
!
{\displaystyle \int _{0}^{1}E_{n}\left(x+y\right)\log(\tan {\frac {\pi }{2}}x)\,dx=n!\sum _{k=1}^{\left\lfloor {\frac {n+1}{2}}\right\rfloor }{\frac {(-1)^{k-1}}{\pi ^{2k}}}\left(2-2^{-2k}\right)\zeta (2k+1){\frac {y^{n+1-2k}}{(n+1-2k)!}}}
e casos particulares sen a variábel
y
{\displaystyle y}
función zeta de Riemann
∫
0
1
E
2
n
−
1
(
x
)
log
(
tan
π
2
x
)
d
x
=
(
−
1
)
n
−
1
(
2
n
−
1
)
!
π
2
n
(
2
−
2
−
2
n
)
ζ
(
2
n
+
1
)
{\displaystyle \int _{0}^{1}E_{2n-1}\left(x\right)\log(\tan {\frac {\pi }{2}}x)\,dx={\frac {(-1)^{n-1}(2n-1)!}{\pi ^{2n}}}\left(2-2^{-2n}\right)\zeta (2n+1)}
∫
0
1
B
2
n
−
1
(
x
)
log
(
tan
π
2
x
)
d
x
=
(
−
1
)
n
−
1
π
2
n
2
2
n
−
2
(
2
n
−
1
)
!
∑
k
=
1
n
(
2
2
k
+
1
−
1
)
ζ
(
2
k
+
1
)
ζ
(
2
n
−
2
k
)
{\displaystyle \int _{0}^{1}B_{2n-1}\left(x\right)\log(\tan {\frac {\pi }{2}}x)\,dx={\frac {(-1)^{n-1}}{\pi ^{2n}}}{\frac {2^{2n-2}}{(2n-1)!}}\sum _{k=1}^{n}(2^{2k+1}-1)\zeta (2k+1)\zeta (2n-2k)}
∫
0
1
E
2
n
(
x
)
log
(
tan
π
2
x
)
d
x
=
∫
0
1
B
2
n
(
x
)
log
(
tan
π
2
x
)
d
x
=
0
{\displaystyle \int _{0}^{1}E_{2n}\left(x\right)\log(\tan {\frac {\pi }{2}}x)\,dx=\int _{0}^{1}B_{2n}\left(x\right)\log(\tan {\frac {\pi }{2}}x)\,dx=0}
∫
0
1
B
2
n
−
1
(
x
)
cot
(
π
x
)
d
x
=
2
(
2
n
−
1
)
!
(
−
1
)
n
−
1
(
2
π
)
2
n
−
1
ζ
(
2
n
−
1
)
{\displaystyle \int _{0}^{1}{{{B}_{2n-1}}\left(x\right)\cot \left(\pi x\right)dx}={\frac {2\left(2n-1\right)!}{{{\left(-1\right)}^{n-1}}{{\left(2\pi \right)}^{2n-1}}}}\zeta \left(2n-1\right)}
↑ 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 .
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eabd8f0560b91e8d9f46e88f5aa019c0eabae98)
