Mon 2020-Jul-20

Math test post

Tagged: About / Math / Obscurantism

Just seeing if I can make the math typesetting work. Nothing to see here, kid… move along.

Inline equation: $f(x) = \frac{1}{sqrt(2\pi)} e^{-x^2 / 2}$

A longer example: $\max\limits_\theta L_{\theta_0}(\theta)$, subject to $D_{KL}^{\rho_{\theta_0}}(\theta_0,\theta)\le\delta$, where $D_{KL}^\rho(\theta_1,\theta_2)=\mathbb{E}_{s\sim\rho}[D_{KL}(\pi_{\theta_1}(\cdot\vert s)\mid\mid\pi_{\theta_2}(\cdot\vert s))]$

Display equation:

\[f(x) = \frac{1}{sqrt(2\pi)} e^{-x^2 / 2}\]


\[\begin{multline} \shoveleft \begin{aligned} G_t&=R_{t+1}+\gamma R_{t+2}+\gamma^2 R_{t+3}+\gamma^3 R_{t+4}+...\\ &=R_{t+1}+\gamma(R_{t+2}+\gamma R_{t+3}+\gamma^2 R_{t+4})+...\\ &=R_{t+1}+\gamma G_{t+1}\\ \end{aligned} \end{multline}\]

Oh, Maxwell…

\[\begin{align} \nabla\times\mathbf{B}-\frac{1}{c}\frac{\partial\mathbf{E}}{\partial t} &= \frac{4\pi}{c}\mathbf{j} \\ \nabla\cdot\mathbf{E} &= 4\pi\rho \\ \nabla\times\mathbf{E}+\frac{1}{c}\frac{\partial\mathbf{B}}{\partial t} &= \mathbf{0} \\ \nabla\cdot\mathbf{B} &= 0 \end{align}\] \[\begin{align*} \dot{x} &= \sigma(y - x) \\ \dot{y} &= \rho x - y - xz \\ \dot{z} &= -\beta z + xy \end{align*}\]


Published Mon 2020-Jul-20

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