CÁLCULO GRACELI SINTÉTICO [112]

NÚMEROS E SEQUÊNCIA DE GRACELI .

1 / P =

SENDO P = PROGRESSÃO DE 3.

$\pi$ = PI / [1.1] . =

Gδ(x) = e−x2 $\delta _{a}(x)={\tfrac {1}{{\sqrt {\pi }}a}}\cdot e^{-{\frac {x^{2}}{a^{2}}}}$ e−x2 =

$G (x) =$ e−x2 = = $f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}\;\;\mathrm {e} ^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}$ e−x2 =

$\zeta (s)=\sum _{k=1}^{\infty }k^{-s}.$ e−x2 =

$\sum {\frac {1}{n}}$ e−x2 = =

$\sum {\frac {1}{n^{2}}}$ , e−x2 =

$y'=p(x)y+q(x)y^{n}.$ e−x2 =

$\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ e−x2 =

$y'+P(x)y=Q(x)y^{n}.$ e−x2 =

${\bar {X}}_{n}:={\frac {1}{n}}\sum _{i=1}^{n}X_{i}$ e−x2 =

$N(k,n)={n \choose k}={\frac {n!}{k!(n-k)!}}$ e−x2 =

$\mathbb {P} ([S_{n}=k])={n \choose k}p^{k}(1-p)^{n-k}$ e−x2 =

$n!={\sqrt {2\pi n}}\;n^{n}e^{-n}\left(1+{\mathcal {O}}\left({\frac {1}{n}}\right)\right)$ e−x2 =

$H=-p\log _{2}p-(1-p)\log _{2}(1-p)$ e−x2 =

NÚMEROS E SEQUÊNCIA DE GRACELI .

1 / P =

SENDO P = PROGRESSÃO DE 3.

$\pi$ = PI / [1.1] . =

Gδ(x) = e−x2 $\delta _{a}(x)={\tfrac {1}{{\sqrt {\pi }}a}}\cdot e^{-{\frac {x^{2}}{a^{2}}}}$ e−x2 / S =

$G (x) =$ e−x2 = = $f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}\;\;\mathrm {e} ^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}$ e−x2 / S

$\zeta (s)=\sum _{k=1}^{\infty }k^{-s}.$ e−x2 / S =

$\sum {\frac {1}{n}}$ e−x2 / S =

$\sum {\frac {1}{n^{2}}}$ , e−x2 / S =

$y'=p(x)y+q(x)y^{n}.$ e−x2 / S =

$\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ e−x2 /S =

$y'+P(x)y=Q(x)y^{n}.$ e−x2/ S =

${\bar {X}}_{n}:={\frac {1}{n}}\sum _{i=1}^{n}X_{i}$ e−x2 / S=

$N(k,n)={n \choose k}={\frac {n!}{k!(n-k)!}}$ e−x2 / S =

$\mathbb {P} ([S_{n}=k])={n \choose k}p^{k}(1-p)^{n-k}$ e−x2 / =

$n!={\sqrt {2\pi n}}\;n^{n}e^{-n}\left(1+{\mathcal {O}}\left({\frac {1}{n}}\right)\right)$ e−x2 / S=

$H=-p\log _{2}p-(1-p)\log _{2}(1-p)$ e−x2 / S =

x2 =