Polinomios de Bernoulli
En matemáticas os polinomios de Bernoulli
B
n
(
x
)
{\displaystyle B_{n}(x)}
son definidos mediante unha 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}}
(normalmente expresados como
B
n
{\displaystyle B_{n}}
e escritos aquí con minúscula para distinguilos dos polinomios) son os termos independentes dos polinomios correspondentes,
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}\,}
expón unha 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 mais Euler
E
n
(
x
)
{\displaystyle E_{n}(x)}
respectivamente, temos,
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}
, onde os
B
k
{\displaystyle B_{k}}
son os números de Bernoulli , e os
E
k
{\displaystyle E_{k}}
son os números de Euler .
Dedúcese logo que
B
n
(
0
)
=
B
n
{\displaystyle B_{n}(0)=B_{n}}
(numeradores (secuencia 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}}
((secuencia 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}}
ou, máis sinxelamente, a 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}
cando
n
≥
2
{\displaystyle n\geq 2}
e para
0
<
x
<
1
{\displaystyle 0<x<1}
cando
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}
onde aparecen a 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 .