CÁLCULO GRACELI SINTÉTICO [8]

d / dt y [t] [ $\zeta (s)$ ] = f [y [t] ,t] ,[ $\zeta (s)$ ]

d / dt y [t] [ $\Gamma (z)$ ] = f [y [t] ,t] ,[ $\Gamma (z)$ ]

d / dt y [t] [ $\psi _{n}(z)$ = f [y [t] ,t] . [ $\psi _{n}(z)$ ]

d / dt y [t] [ $\operatorname {Li} _{s}(z)$ ] = f [y [t] ,t] , [ $\operatorname {Li} _{s}(z)$ ]

d / dt y [t] [ $n \choose k$ ] = f [y [t] ,t] , [ $n \choose k$ ]

d / dt y [t] [ $\sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}$ ] = f [y [t] ,t] , [ $\sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}$ ]

d / dt y [t] [ $\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$ ] = f [y [t] ,t] ,[ $\zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}$ ]

d / dt y [t] [ $\operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}$ ] = f [y [t] ,t] , [ $\operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}$ ]

d / dt y [t] [ $\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$ ] = f [y [t] ,t], [ $\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$ ]

d / dt y [t] [ ${\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}}$ )] = f [y [t] ,t] , [ ${\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}}$ )]

d / dt y [t] [ $\sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}$ ] = f [y [t] ,t] , [ $\sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}$ ]

d / dt y [t] [ $\displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)$ ] = f [y [t] ,t] [ $\displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)$ ]

d / dt y [t] [ $\displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}$ ] = f [y [t] ,t] , [ $\displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}$ ]

d / dt y [t] [ $\sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}$ ] = f [y [t] ,t] , [ $\sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}$ ]

d / dt y [t] = $\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1+a\pi \coth(a\pi )}{8a^{4}}}$ ~= f [y [t] ,t] , $\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1+a\pi \coth(a\pi )}{8a^{4}}}$ ]

d / dt y [t] [ $\psi ^{*}$ = f [y [t] ,t] , [ $\psi ^{*}$

d / dt y [t] [ $\psi ^{*}$ ${\boldsymbol {\mu }}$ [ $\psi ^{*}$ ${\boldsymbol {\mu }}$

d / dt y [t] [ $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$ = f [y [t] ,t],[] $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$

d / dt y [t] [ $\lambda ={\frac {h}{p}}$ / ${\boldsymbol {\mu }}=-g_{s}\mu _{B}({\boldsymbol {s}}/\hbar )$ / $\Delta \mathrm {E} =hf=hc/\lambda$ ] = f [y [t] ,t], $\lambda ={\frac {h}{p}}$ / ${\boldsymbol {\mu }}=-g_{s}\mu _{B}({\boldsymbol {s}}/\hbar )$ / $\Delta \mathrm {E} =hf=hc/\lambda$ =]

d / dt y [t] [ $\psi ^{*}$ [ $\psi ^{*}$ ${\boldsymbol {\mu }}$ ] $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$ $-i\hbar {\frac {\partial }{\partial x_{n}}}$ [x,t] ] = ] = f [y [t] ,t] , [ $\psi ^{*}$ [ $\psi ^{*}$ ${\boldsymbol {\mu }}$ ] $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$ $-i\hbar {\frac {\partial }{\partial x_{n}}}$ [x,t] ] = ]

d / dt y [t] [ $\psi ^{*}$ = f [y [t] ,t] , [ $\psi ^{*}$

d / dt y [t] [ $\psi ^{*}$ ${\boldsymbol {\mu }}$ [ $\psi ^{*}$ ${\boldsymbol {\mu }}$

d / dt y [t] [ $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$ = f [y [t] ,t],[] $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$

Pesquisar este blog

CÁLCULO GRACELI SINTÉTICO [8]

Comentários

Postar um comentário