MathJax

提供: Internet Web School

(版間での差分)
(memo, testing...)
(memo, testing...)
 
60 行: 60 行:
== memo, testing... ==
== memo, testing... ==
-
$$f(x,y) = \left\{ \begin{array}{ll} \frac{x^3y}{x^2+y^2} & ((x,y) \neq {\bf 0} \\ 0 & ((x,y)={\bf 0}) \end{array} \right$$
+
関数<br/>
 +
$$
 +
f(x,y) =\left\{
 +
    \begin{array}{ll}
 +
    \frac{x^3y}{x^2+y^2} & ((x,y) \neq {\bf 0}\\
 +
      0 & ((x,y)={\bf 0})  
 +
    \end{array} \right.
 +
$$<br/>
 +
を考える。
$$\int x^a dx = \frac{1}{a+1}x^{a+1}+C. \quad (a \ne -1)$$
$$\int x^a dx = \frac{1}{a+1}x^{a+1}+C. \quad (a \ne -1)$$

2017年12月19日 (火) 11:29 時点における最新版

目次

MathJaxプラグインによる数式表示

inline equation

This is a sample equation : $\sqrt{x^2+y^2} = z$

 $\sqrt{x^2+y^2} = z$

blocked equation

$$\int x^a dx = \frac{1}{a+1}x^{a+1}+C. \quad (a \ne -1)$$

 $$\int x^a dx = \frac{1}{a+1}x^{a+1}+C. \quad (a \ne -1)$$

multiple-line blocked equations

$e^{i \theta} = \cos \theta + i \, \sin \theta\\ e^{\pi i} + 1 = 0$
 $e^{i \theta} = \cos \theta + i \, \sin \theta\\ e^{\pi i} + 1 = 0$

SIZE feature

\small

$\small e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $\small e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

normalsize

$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

\large

$\large e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $\large e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

\Large

$\Large e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $\Large e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

\LARGE

$\LARGE e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $\LARGE e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

\Huge

$\Huge e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$
 $\Huge\blue e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$

MathJax homepage

http://www.mathjax.org/

test page (using .js) -http://cdn.mathjax.org/mathjax/latest/test/sample-tex.html

MediaWiki extensions/MathJax.php plugin

http://www.mediawiki.org/wiki/Extension:MathJax

LaTeX Math. symbols

ファイル:LaTeX-symbols.pdf

memo, testing...

関数
$$ f(x,y) =\left\{ \begin{array}{ll} \frac{x^3y}{x^2+y^2} & ((x,y) \neq {\bf 0}) \\ 0 & ((x,y)={\bf 0}) \end{array} \right. $$
を考える。

$$\int x^a dx = \frac{1}{a+1}x^{a+1}+C. \quad (a \ne -1)$$

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