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2018-02-13T19:33:08Z

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Test Math

2018-02-13T16:50:50Z

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== nowiki Test==
<source lang='html5'>
<math>E=mc^2</math>
</source>
<math>E=mc^2</math>

<source lang='html5'>
<nowiki><math>E=mc^2</math></nowiki>
</source>
<nowiki><math>E=mc^2</math></nowiki>

== Inequality Sign Test ==
<source lang='html5'>
<math>1<2</math>
</source>
<math>1<2</math>

<source lang='html5'>
<math>2>1</math>
</source>
<math>2>1</math>

<source lang='html5'>
<math>1\lt 2</math>
</source>
<math>1\lt 2</math>

<source lang='html5'>
<math>2\gt 1</math>
</source>
<math>2\gt 1</math>

== Inequality Sign Test 2 ==
<source lang='html5'>
<math>a<b</math>
</source>
<math>a<b</math>

<source lang='html5'>
<math>a < b</math>
</source>
<math>a < b</math>

<source lang='html5'>
<math>a>b</math>
</source>
<math>a>b</math>

<source lang='html5'>
<math>a > b</math>
</source>
<math>a > b</math>

== UTF-8 Test ==
<source lang='html5'>
<math>전압 = 전류 \times 저항</math>
</source>
<math>전압 = 전류 \times 저항</math>

<source lang='html5'>
<math>저항 = \frac{전압}{전류}</math>
</source>
<math>저항 = \frac{전압}{전류}</math>

<source lang='html5'>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
</source>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>

<source lang='html5'>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?
</source>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?

== The Lorenz Equations ==
<source lang='html5'>
<math>\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}</math>
</source>

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

== The Cauchy-Schwarz Inequality ==
<source lang='html5'>
<math>\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)</math>
</source>
<math>\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)</math>

== A Cross Product Formula ==
<source lang='html5'>
<math>\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}</math>
</source>

<math>\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}</math>

== The probability of getting k heads when flipping n coins is ==
<source lang='html5'>
<math>P(E) = {n \choose k} p^k (1-p)^{ n-k}</math>
</source>

<math>P(E) = {n \choose k} p^k (1-p)^{ n-k}</math>

== An Identity of Ramanujan ==
<source lang='html5'>
<math>\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+\ldots} } } }</math>
</source>

<math>\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+\ldots} } } }</math>

== A Rogers-Ramanujan Identity ==
<source lang='html5'>
<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad for\,|q|<1.</math>
</source>

<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad for\,|q|<1.</math>

== Maxwell’s Equations ==
<source lang='html5'>
<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>
</source>

<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>

Test Math

2018-02-13T16:49:43Z

Admin: init

= nowiki Test==
<source lang='html5'>
<math>E=mc^2</math>
</source>
<math>E=mc^2</math>

<source lang='html5'>
<nowiki><math>E=mc^2</math></nowiki>
</source>
<nowiki><math>E=mc^2</math></nowiki>

== Inequality Sign Test ==
<source lang='html5'>
<math>1<2</math>
</source>
<math>1<2</math>

<source lang='html5'>
<math>2>1</math>
</source>
<math>2>1</math>

<source lang='html5'>
<math>1\lt 2</math>
</source>
<math>1\lt 2</math>

<source lang='html5'>
<math>2\gt 1</math>
</source>
<math>2\gt 1</math>

== Inequality Sign Test 2 ==
<source lang='html5'>
<math>a<b</math>
</source>
<math>a<b</math>

<source lang='html5'>
<math>a < b</math>
</source>
<math>a < b</math>

<source lang='html5'>
<math>a>b</math>
</source>
<math>a>b</math>

<source lang='html5'>
<math>a > b</math>
</source>
<math>a > b</math>

== UTF-8 Test ==
<source lang='html5'>
<math>전압 = 전류 \times 저항</math>
</source>
<math>전압 = 전류 \times 저항</math>

<source lang='html5'>
<math>저항 = \frac{전압}{전류}</math>
</source>
<math>저항 = \frac{전압}{전류}</math>

<source lang='html5'>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
</source>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>

<source lang='html5'>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?
</source>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?

== The Lorenz Equations ==
<source lang='html5'>
<math>\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}</math>
</source>

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

== The Cauchy-Schwarz Inequality ==
<source lang='html5'>
<math>\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)</math>
</source>
<math>\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)</math>

== A Cross Product Formula ==
<source lang='html5'>
<math>\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}</math>
</source>

<math>\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}</math>

== The probability of getting k heads when flipping n coins is ==
<source lang='html5'>
<math>P(E) = {n \choose k} p^k (1-p)^{ n-k}</math>
</source>

<math>P(E) = {n \choose k} p^k (1-p)^{ n-k}</math>

== An Identity of Ramanujan ==
<source lang='html5'>
<math>\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+\ldots} } } }</math>
</source>

<math>\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+\ldots} } } }</math>

== A Rogers-Ramanujan Identity ==
<source lang='html5'>
<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad for\,|q|<1.</math>
</source>

<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
= \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad for\,|q|<1.</math>

== Maxwell’s Equations ==
<source lang='html5'>
<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>
</source>

<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>

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2016-08-04T12:19:28Z

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