Ancelmo Graceli - integrais , somas, séries polinômios.

setembro 10, 2022

Esta lista de séries matemáticas contém fórmulas para somas finitas e infinitas. Ela pode ser usada em conjunto com outras ferramentas para avaliar somas.

Aqui, considera-se que $0^{0}$ vale $1$
$B_{n}(x)$ é um polinômio de Bernoulli.
$B_{n}$ é um número de Bernoulli, e aqui, $B_{1}=-{\frac {1}{2}}.$
$E_{n}$ é um número de Euler.
$\zeta (s)$ é a função zeta de Riemann.
$\Gamma (z)$ é a função gama.
$\psi _{n}(z)$ é uma função poligama.
$\operatorname {Li} _{s}(z)$ é um polilogaritmo .
$n \choose k$ é o coeficiente binomial
$\exp(x)$ denota a exponencial de $x$

sendo p = progressão,

k, w, h = números reais.

\sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}

/ [ B / PK/PW] / n =

$\sum _{k=1}^{m}k={\frac {m(m+1)}{2}}$
$\sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}$ / m [pk/ pw] / ph] =
$\sum _{k=1}^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}$ / m [pk/ pw] / ph] =

$\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$ $\zeta (s)$ / Bn [[pk/ pw] / ph] =

Os primeiros valores são:

$\zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}$ (o problema de Basileia) -1 / [ $\zeta (s)$ / Bn [[pk/ pw] / ph] =
$\zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}$
$\zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}$ -1 / [ $\zeta (s)$ / Bn [[pk/ pw] / ph] =

Séries de potências

Polilogaritmos de ordem baixa

Somas com uma quantidade finita de termos:

$\sum _{k=0}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}$ , (série geométrica) -1 / [ $\zeta (s)$ / [[pk/ pw] / ph] =
$\sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}$ -1 / [ $\zeta (s)$ / [[pk/ pw] / ph] =
$\sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}$
$\sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}$

Somas com uma infinidade de termos, válidas para $|z|<1$ (ver polilogaritmo):

$\operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =

A propriedade a seguir é útil para calcular polilogaritmos de ordem inteira baixa recursivamente de forma fechada:

${\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =
$\operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}$ / $\operatorname {Li} _{s}(z)$ / [[pk/ pw] / ph] =

$\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$ $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}$ (ver segundo momento da distribuição de Poisson)
$\sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}$ / $\exp(x)$ [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)$ / $\exp(x)$ [[pk/ pw] / ph] =

$\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\mathrm {sen} \,z$ [[pk/ pw] / ph]
$\sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\mathrm {senh} \,z$ [[pk/ pw] / ph]
$\sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z$ [[pk/ pw] / ph]
$\sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z$ [[pk/ pw] / ph] =

$(1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1$ $n \choose k$ / [[pk/ pw] / ph] =
^[3] $\sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1$
$\sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}$ ,/ $n \choose k$ / [[pk/ pw] / ph] =
$\sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}$ , função geradora dos coeficientes binomiais centrais
$\sum _{k=0}^{\infty }{2k+\alpha \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}}\left({\frac {1-{\sqrt {1-4z}}}{2z}}\right)^{\alpha },|z|<{\frac {1}{4}}$

$\sum _{k=0}^{n}{n \choose k}=2^{n}$ / $n \choose k$ / [[pk/ pw] / ph] =
$\sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ where }}n>0$ / $n \choose k$ / [[pk/ pw] / ph] =
$\sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}$ / $n \choose k$ / [[pk/ pw] / ph] =
$\sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}$ / $n \choose k$ / [[pk/ pw] / ph] =
$\sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}$ / $n \choose k$ / [[pk/ pw] / ph] =

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