E)   = {n \choose k} p^k (1-p)^{ n-k} \]

<div class="math-header">An Identity of Ramanujan</div>

\[ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} =
1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}}
{1+\frac{e^{-8\pi}} {1+\ldots} } } } \]

<div class="math-header">A Rogers-Ramanujan Identity</div>

\[  1 +  \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|&lt;1$}. \]

<div class="math-header">Maxw