"TeX 샘플"의 두 판 사이의 차이

 
(사용자 2명의 중간 판 7개는 보이지 않습니다)
2번째 줄: 2번째 줄:
;TeX 샘플
;TeX 샘플


==nowiki 테스트==
==texvc==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>\N</math>
</syntaxhighlight>
<math>\N</math>
 
==nowiki==
<syntaxhighlight lang='tex'>
<math>E=mc^2</math>
<math>E=mc^2</math>
</source>
</syntaxhighlight>
<math>E=mc^2</math>
<math>E=mc^2</math>


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


==부등호 테스트==
==부등호==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>1<2</math>
<math>1<2</math>
</source>
</syntaxhighlight>
<math>1<2</math>
<math>1<2</math>


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


<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>1\lt2</math>
<math>1\lt2</math>
</source>
</syntaxhighlight>
<math>1\lt2</math>
<math>1\lt2</math>


<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>2\gt1</math>
<math>2\gt1</math>
</source>
</syntaxhighlight>
<math>2\gt1</math>
<math>2\gt1</math>


==부등호 테스트 2==
==부등호 테스트 2==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>a<b</math>
<math>a<b</math>
</source>
</syntaxhighlight>
<math>a<b</math>
<math>a<b</math>


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


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


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


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


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


<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
</source>
</syntaxhighlight>
<math>償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}</math>
<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'>
<syntaxhighlight lang='tex'>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?
</source>
</syntaxhighlight>
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?
<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?


== Lorenz 방정식 ==
== Lorenz 방정식 ==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>\begin{align}
<math>\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{x} & = \sigma(y-x) \\
83번째 줄: 89번째 줄:
\dot{z} & = -\beta z + xy
\dot{z} & = -\beta z + xy
\end{align}</math>
\end{align}</math>
</source>
</syntaxhighlight>


<math>\begin{align}
<math>\begin{align}
92번째 줄: 98번째 줄:


== Cauchy-Schwarz 부등식==
== Cauchy-Schwarz 부등식==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<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>
<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>
</syntaxhighlight>
<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>
<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>


== 벡터곱 공식 ==
== 벡터곱 공식 ==
<source lang='text'>
<syntaxhighlight lang='tex'>
<math>\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
<math>\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
104번째 줄: 110번째 줄:
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0
\end{vmatrix}</math>
\end{vmatrix}</math>
</source>
</syntaxhighlight>


<math>\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
<math>\mathbf{V}_1 \times \mathbf{V}_2 =  \begin{vmatrix}
113번째 줄: 119번째 줄:


== <math>n</math>개의 동전을 던져 앞면이 <math>k</math>개 나올 확률 ==
== <math>n</math>개의 동전을 던져 앞면이 <math>k</math>개 나올 확률 ==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>P(E)  = {n \choose k} p^k (1-p)^{ n-k}</math>
<math>P(E)  = {n \choose k} p^k (1-p)^{ n-k}</math>
</source>
</syntaxhighlight>


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


== Ramanujan의 항등식==
== Ramanujan의 항등식==
<source lang='html5'>
<syntaxhighlight lang='tex'>
<math>\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
<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^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } }</math>
{1+\frac{e^{-8\pi}} {1+\ldots} } } }</math>
</source>
</syntaxhighlight>


<math>\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
<math>\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
131번째 줄: 137번째 줄:


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


<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
<math>1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots
142번째 줄: 148번째 줄:


== Maxwell의 방정식 ==
== Maxwell의 방정식 ==
<source lang='text'>
<syntaxhighlight lang='tex'>
<math>\begin{align}
<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{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\  \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\
148번째 줄: 154번째 줄:
\nabla \cdot \vec{\mathbf{B}} & = 0
\nabla \cdot \vec{\mathbf{B}} & = 0
\end{align}</math>
\end{align}</math>
</source>
</syntaxhighlight>


<math>\begin{align}
<math>\begin{align}
166번째 줄: 172번째 줄:
==참고==
==참고==
* 문서샘플 모음 http://wiki.ktug.org/wiki/wiki.php/SampleDocument
* 문서샘플 모음 http://wiki.ktug.org/wiki/wiki.php/SampleDocument
* 영어권 샘플 모음 http://www.latextemplates.com/
* 오버리프 매뉴얼 https://www.overleaf.com/learn/latex/Sections_and_chapters
*https://www.mediawiki.org/wiki/Extension:SimpleMathJax
*https://www.mediawiki.org/wiki/Extension:SimpleMathJax
*http://www.mathjax.org/demos/tex-samples/
*http://www.mathjax.org/demos/tex-samples/


[[분류: TeX]]
[[분류: TeX]]

2020년 10월 23일 (금) 20:09 기준 최신판

TeX Samples
TeX 샘플

1 texvc[ | ]

<math>\N</math>

[math]\displaystyle{ \N }[/math]

2 nowiki[ | ]

<math>E=mc^2</math>

[math]\displaystyle{ E=mc^2 }[/math]

<nowiki><math>E=mc^2</math></nowiki>

<math>E=mc^2</math>

3 부등호[ | ]

<math>1<2</math>

[math]\displaystyle{ 1\lt 2 }[/math]

<math>2>1</math>

[math]\displaystyle{ 2\gt 1 }[/math]

<math>1\lt2</math>

[math]\displaystyle{ 1\lt2 }[/math]

<math>2\gt1</math>

[math]\displaystyle{ 2\gt1 }[/math]

4 부등호 테스트 2[ | ]

<math>a<b</math>

[math]\displaystyle{ a\lt b }[/math]

<math>a < b</math>

[math]\displaystyle{ a \lt b }[/math]

<math>a>b</math>

[math]\displaystyle{ a\gt b }[/math]

<math>a > b</math>

[math]\displaystyle{ a \gt b }[/math]

5 UTF-8 테스트[ | ]

<math>전압 = 전류 \times 저항</math>

[math]\displaystyle{ 전압 = 전류 \times 저항 }[/math]

<math>저항 = \frac{전압}{전류}</math>

[math]\displaystyle{ 저항 = \frac{전압}{전류} }[/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]\displaystyle{ 償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間} }[/math]

<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?

[math]\displaystyle{ n }[/math]개의 동전을 던져 앞면 [math]\displaystyle{ k }[/math]가 나올 확률 [math]\displaystyle{ P(E) }[/math]는?

6 Lorenz 방정식[ | ]

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

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

7 Cauchy-Schwarz 부등식[ | ]

<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>

[math]\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) }[/math]

8 벡터곱 공식[ | ]

<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>

[math]\displaystyle{ \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]

9 [math]\displaystyle{ n }[/math]개의 동전을 던져 앞면이 [math]\displaystyle{ k }[/math]개 나올 확률[ | ]

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

[math]\displaystyle{ P(E) = {n \choose k} p^k (1-p)^{ n-k} }[/math]

10 Ramanujan의 항등식[ | ]

<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>

[math]\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+\ldots} } } } }[/math]

11 Rogers-Ramanujan 항등식[ | ]

<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>

[math]\displaystyle{ 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|\lt 1. }[/math]

12 Maxwell의 방정식[ | ]

<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>

[math]\displaystyle{ \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]

13 같이 보기[ | ]

14 참고[ | ]

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