## Publication: Test -- Item w/ Markdown + MathJax description

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##### Abstract
#### this is Markdown And this is MathJax $$1+2=3$$ $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ Same thing inline: $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$ When $a \ne 0$, there are two solutions to $ax^2 + bx + c = 0$ and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ $$\begin{array}{c} \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{array}$$ $$A=\begin{bmatrix} a & b\\ c & d\\ e & f \end{bmatrix}$$ $$$$\label{eq:bayes} P(\theta|\textbf{D}) = P(\theta ) \frac{P(\textbf{D} |\theta)}{P(\textbf{D})}$$$$ The following but should be rendered as an error because it defines a duplicate equation label $$$$\label{eq:bayes} P(\theta|\textbf{D}) = P(\theta ) \frac{P(\textbf{D} |\theta)}{P(\textbf{D})}$$$$