Difference between revisions of "Sandbox"

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(Math test)
(Math test)
Line 4: Line 4:
 
:<math>E=mc^2</math>
 
:<math>E=mc^2</math>
 
:$e^{i\phi} = \cos\phi + i\sin\phi$
 
:$e^{i\phi} = \cos\phi + i\sin\phi$
 +
:<math>전압 = 전류 \times 저항</math>
 +
:<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
 +
 +
<math>\begin{align}
 +
\dot{x} & = \sigma(y-x) \\
 +
\dot{y} & = \rho x - y - xz \\
 +
\dot{z} & = -\beta z + xy
 +
\end{align}</math>
 +
 +
<math>\begin{align}
 +
\nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\  \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
 +
\nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\
 +
\nabla \cdot \vec{\mathbf{B}} & = 0
 +
\end{align}</math>
  
 
==Language test==
 
==Language test==

Revision as of 12:42, 29 June 2019


Math test

[math]E=mc^2[/math]
$e^{i\phi} = \cos\phi + i\sin\phi$
[math]전압 = 전류 \times 저항[/math]
[math]償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}[/math]

[math]\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}[/math]

[math]\begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align}[/math]

Language test