Embedded in {$ \sqrt{x^2 + 2xy + y^2} $} text
 Embedded in  \sqrt{x^2 + 2xy + y^2}  text
Centered {$$ \sqrt{x^2 + 2xy + y^2} $$}
 Centered 
\sqrt{x^2 + 2xy + y^2}
{$$ {1\over 2} \quad {n+1\over 3} \quad {n+1 \choose 3} \quad \sum_{n=1}^3 Z_n^2 $$}
 
{1\over 2} \quad {n+1\over 3} \quad {n+1 \choose 3} \quad \sum_{n=1}^3 Z_n^2
{$$ {\textstyle \sum x_n}\quad {\sum x_n} \quad \sum_{n=1}^m x_n^2 $$}
 
{\textstyle \sum x_n}\quad {\sum x_n} \quad \sum_{n=1}^m x_n^2

For some reason the built-in demonstration markup does not interpret correctly the expression under the integral droping dt one line. However, no everything is lost. Outside the demonstration markup the integral comes out right: \int_0^\infty f(t)\,dt . You can see the corresponding LaTex code by clicking on that expression.

{$$\sqrt{2} = \sup \left\{ x\in\mathbb{Q}: x^{2}<2\right\} $$}
\sqrt{2} = \sup \left\{ x\in\mathbb{Q}: x^{2}<2\right\}

Would you guess it:

{$$ 2 = \sqrt{2+\sqrt{2+\sqrt{2+ ....}}} $$}
 
2 = \sqrt{2+\sqrt{2+\sqrt{2+ ....}}}

And this is how you get the Taylor series:

{$$ f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots  + {h^{n+1}\over (n+1)!}f^{(n+1)}(u_0) + o(h^{n+1}). $$}
 
f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots + {h^{n+1}\over (n+1)!}f^(n+1)(u_0) + o(h^{n+1}).