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${\displaystyle \delta R_{ij}=\Gamma _{il}^{k}B_{k}dx^{l}}$

Now consider arbitrary contravariant and covariant vectors ${\displaystyle A^{i}}$ and ${\displaystyle B_{i}}$ respectively. Since ${\displaystyle A^{i}B_{i}}$ is a scalar, ${\displaystyle \delta (A^{i}B_{i})=0}$, one arrives at:

${\displaystyle \Rightarrow \delta B_{i}=\Gamma _{il}^{k}B_{k}dx^{l}}$

## Connection Between Covariant And Regular Derivatives

${\displaystyle R_{xyxy}={\frac {1}{2}}\left(frac{\partial ^{2}g_{xx}}{\partial y^{2}}+{\frac {\partial ^{2}g_{xx}}{\partial y^{2}}}\right)}$

${\displaystyle R_{xyxy}={\frac {1}{2}}\left({\frac {\partial ^{2}g_{xx}}{\partial y^{2}}}+{\frac {\partial ^{2}g_{xx}}{\partial y^{2}}}\right)}$

${\displaystyle \Rightarrow \Gamma _{il}^{n}=\delta _{m}^{n}\Gamma _{il}^{m}={\frac {1}{2}}g^{kn}\left(g_{ik,l}+g_{kl,i}-g_{li,k}\right)}$

${\displaystyle R_{x}yxy={\frac {1}{y^{2}}}}$

\sqrt x=3 ${\displaystyle {\sqrt {x}}=3}$

${\displaystyle \int _{-n}^{n}e^{x}\ }$

${\displaystyle \int _{-n}^{n}e^{-x{2}}\mathrm {d} x}$ ${\displaystyle \int _{-n}^{n}e^{-x{2}}\mathrm {d} x}$ // Aide de Julien1311 ${\displaystyle \int _{-n}^{n}e^{-x{2}}\mathrm {d} x}$

${\displaystyle \int _{-n}^{n}e^{-x{2}}\mathrm {d} x}$


${\displaystyle \int _{-n}^{n}e^{x}\,\mathrm {d} x}$

${\displaystyle a^{2}}$

${\displaystyle \sum _{k=1}^{n}k^{2}}$ OK

${\displaystyle \prod _{i=1}^{n}x_{i}}$

${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$

Intégrale double $document} »): {\displaystyle \iint e^x}\, {\rm d}x {\rm d}y[itex] [itex]\iint e^{-\frac{x^2+y^2}{2}\, {\rm d}x {\rm d}y[itex] \begin{document} $\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ \end{document} [itex] x_i$