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Last Edit : 2019-10-08 12:16:0816 / 09 / 2019 à 21h50
$$\sigma_{\psi}(\gamma)=\psi\Big(\gamma, \begin{pmatrix} |\gamma|^{1/2} & 0 \\\ 0 & |\gamma|^{-1/2}\end{pmatrix}\Big)$$
$$\sigma_{\psi}(\gamma)=\psi\Big(\gamma, \begin{pmatrix} |\gamma|^{1/2} & 0 \\\ 0 & |\gamma|^{-1/2}\end{pmatrix}\Big)$$
$$\sin(\alpha)^{\theta}=\sum_{i=0}^{n}(x^i + \cos(f))$$
$$\sin(\alpha)^{\theta}=\sum_{i=0}^{n}(x^i + \cos(f))$$
$$\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi$$
$$f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi$$
$$\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}}
{1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
$$\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
$$\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b
\_k^2 \right)$$
$$\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b \_k^2 \right)$$
16 / 09 / 2019 à 21h50
$$\oint_C x^3\, dx + 4y^2\, dy$$
$$\oint_C x^3\, dx + 4y^2\, dy$$
$$\bigcap_1^n p \bigcup_1^k p$$
$$\bigcap_1^n p \bigcup_1^k p$$
$$e^{i \pi} + 1 = 0$$
$$e^{i \pi} + 1 = 0$$
$$\left ( \frac{1}{2} \right )$$
$$\left ( \frac{1}{2} \right )$$
$$x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}$$
$$x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}$$
$${\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}$$
$${\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}$$
$$\textstyle \sum_{k=1}^N k^2$$
$$\textstyle \sum_{k=1}^N k^2$$
$$\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }{ 1-\tfrac{1}{2} } = s_n$$
$$\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }{ 1-\tfrac{1}{2} } = s_n$$
$$\binom{n}{k}$$
$$\binom{n}{k}$$
$$0+1+2+3+4+5+6+7+\cdots$$
$$0+1+2+3+4+5+6+7+\cdots$$
16 / 09 / 2019 à 21h51
$$
\begin{array}{l}
x(t) = - \lambda {p_0}\left( t \right) + \mu {p_1}\left( t \right),\\
y(t) = \lambda {p_{j - 1}}\left( t \right) - \left( {\lambda + j\mu } \right){p_j}\left( t \right) + \mu \left( {j + 1}
\right){p_{j + 1}}\left( t \right)
\end{array}
$$
$$\begin{array}{l} x(t) = - \lambda {p_0}\left( t \right) + \mu {p_1}\left( t \right),\\ y(t) = \lambda {p_{j - 1}}\left( t \right) - \left( {\lambda + j\mu } \right){p_j}\left( t \right) + \mu \left( {j + 1} \right){p_{j + 1}}\left( t \right) \end{array}$$
$$
f(x) = \int_{-\infty}^\infty
\hat f(\xi)\,e^{2 \pi i \xi x}
\,d\xi
$$
$$f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi$$
$$
\displaystyle
\left( \sum\_{k=1}^n a\_k b\_k \right)^2
\leq
\left( \sum\_{k=1}^n a\_k^2 \right)
\left( \sum\_{k=1}^n b\_k^2 \right)
$$
$$\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b\_k^2 \right)$$
$$
\dfrac{
\tfrac{1}{2}[1-(\tfrac{1}{2})^n] }
{ 1-\tfrac{1}{2} } = s_n
$$
$$\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] } { 1-\tfrac{1}{2} } = s_n$$
$$
\displaystyle
\frac{1}{
\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{
\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {
1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}}
{1+\cdots} }
}
}
$$
$$\displaystyle \frac{1}{ \Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{ \frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} { 1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
$$
f(x) = \int_{-\infty}^\infty
\hat f(\xi)\,e^{2 \pi i \xi x}
\,d\xi
$$
$$f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi$$