mobius/mathjax.html

## mathjax.html
<!DOCTYPE html>
<html>
<title> MathJax Tex Test Page</title>
<head>
	<script type="text/x-mathjax-config">
	MathJax.Hub.Config({
		tex2jax: {inlineMath: [['$','$'], ['\\(', '\\)']]},
		processEscape: true,
		TeX: { equationNumbers: { autoNumber: "AMS"}}
	});
	</script>
	<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
	</script>
</head>
<body>
when $a \ne 0$, there are two solution to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

suppose that $\hat{\mathbb{C}} := {\mathbb{C}} \cup \{ \infty \}$ is the extended complex plane  \( A \perp  -\infty \)

paragraph: $$ a < b $$

inline: if $ a \ne b $ then

\begin{equation}
	E = mc^2
\end{equation}

\begin{equation*}
	e^{\pi i} + 1 = 0
\end{equation*}

In equation \eqref{eq:sample}, we find the value of an interesting integral:

\begin{equation}
	\int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15}
	\label{eq:sample}
\end{equation}

\begin{equation}
	\int_{-\infty}^\infty \frac{1}{x^2}\,dx = \frac{1}{x}
\end{equation}

</body>
</html>
	<!DOCTYPE html>
	<html>
	<title> MathJax Tex Test Page</title>
	<head>
	<script type="text/x-mathjax-config">
	MathJax.Hub.Config({
	tex2jax: {inlineMath: [['$','$'], ['\\(', '\\)']]},
	processEscape: true,
	TeX: { equationNumbers: { autoNumber: "AMS"}}
	});
	</script>
	<script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
	</script>
	</head>
	<body>
	when $a \ne 0$, there are two solution to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

	suppose that $\hat{\mathbb{C}} := {\mathbb{C}} \cup \{ \infty \}$ is the extended complex plane \( A \perp -\infty \)

	paragraph: $$ a < b $$

	inline: if $ a \ne b $ then

	\begin{equation}
	E = mc^2
	\end{equation}

	\begin{equation*}
	e^{\pi i} + 1 = 0
	\end{equation*}

	In equation \eqref{eq:sample}, we find the value of an interesting integral:

	\begin{equation}
	\int_0^\infty \frac{x^3}{e^x-1}\,dx = \frac{\pi^4}{15}
	\label{eq:sample}
	\end{equation}

	\begin{equation}
	\int_{-\infty}^\infty \frac{1}{x^2}\,dx = \frac{1}{x}
	\end{equation}

	</body>
	</html>