# Intro to LaTeX using WordPress

Why blog with LaTeX? At least two reasons:

1. WordPress and other text editors don’t support complex mathematical notation out-of-the-box.
2. Writing in LaTeX is a standard skill in research. So you can blog and be pro at the same time.

The instructions here allow the use of a subset of LaTeX. Specifically, everything gets dumped into a LaTeX \math environment. Basically, this approach will suit your needs for writing equations, but not for creating whole papers or documents.

Steps:

• Install the Jetpack plugin for WordPress.
• Activate the Beautiful Math with LaTeX module in Jetpack.
• Review the documentation, linked above, and start writing your equations in LaTeX!
• You may also want to play around some with an online LaTeX editor. Like this one.
• You can click here to download a text document containing an example of how to use Beautiful Math. The example contains the same text used to render the text below this bulleted list.
• That text is also the solution to a question found on the Advanced Macro I – Ramirez Final Edition exam!
• Thanks to Brian, Ennio, and Josh for showing me how to solve the related problem.

Given:

$latex Y = C(Y-T) + I(i-\pi^e, Y_{-1}) + \bar{G} + X(\rho, Y, Y*) \\\\ \frac{M}{P} = L(i, Y-T) \\\\ BOP = X(\rho, Y, Y*) + \sigma(i – i*) + K$

We must write the total derivative in matrix form. The result will have a column of endogenous variables $latex dY, di,$ and $latex d\rho$.

Solving for $latex dY$:

$latex Y = C(Y-T) + I(i-\pi^e, Y_{-1}) + \bar{G} + X(\rho, Y, Y*) \\\\ dY = \frac{dC}{dY}(dY – dT) + \frac{dI}{dr}(di – d\pi^e) + \frac{dI}{dY}dY_{-1} + d\bar{G} + \frac{dX}{d\rho}d\rho + \frac{dX}{dY}dY + \frac{dX}{dY^*}dY^* \\\\ dY – dY\frac{dC}{dY} – dY\frac{dX}{dY} – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho} = -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY* \\\\ dY(1 – \frac{dC}{dY} – \frac{dX}{dY}) – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho} = -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY* \\\\$

Solving for $latex dL$:

$latex \frac{M}{P} = L(i, Y-T) \\\\ d\frac{M}{P} = \frac{dL}{di}di + \frac{dL}{dY}(dY – dT) \\\\ d\frac{M}{P} = \frac{dL}{di}di + \frac{dL}{dY}dY – \frac{dL}{dY}dT \\\\ \frac{dL}{dY}dY + \frac{dL}{di}di = d\frac{M}{P} – \frac{dL}{dY}dT \\\\$

Solving for $latex d\rho$ from the BOP function:

$latex BOP = 0 = X(\rho, Y, Y*) + \sigma(i – i*) + K \\\\ 0 = \frac{dX}{d\rho}d\rho + \frac{dX}{dY}dY + \frac{dX}{dY^*}dY^* + \frac{d\sigma}{dr}i – \frac{d\sigma}{dr}i^* + dK \\\\ \frac{dX}{dY}dY + \frac{d\sigma}{dr}i + \frac{dX}{d\rho}d\rho = \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK \\\\$

Expressing the system of equations for $latex dY, di,$ and $latex d\rho$ in matrix form:

$latex dY(1 – \frac{dC}{dY} – \frac{dX}{dY}) – di\frac{dI}{dr} – d\rho\frac{dX}{d\rho} = -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY* \\\\ \frac{dL}{dY}dY + \frac{dL}{di}di = d\frac{M}{P} – \frac{dL}{dY}dT \\\\ \frac{dX}{dY}dY + \frac{d\sigma}{dr}i + \frac{dX}{d\rho}d\rho = \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK \\\\ \\\\ \implies \begin{bmatrix} F_1 & \frac{dI}{dr} & -X_\rho\\ L_Y & \frac{dX}{d\rho} & 0\\ X_Y & \sigma_r & X_\rho \end{bmatrix} \begin{bmatrix} dY\\ di\\ d\rho \end{bmatrix} = \begin{bmatrix} F_2\\ d\frac{M}{P} – \frac{dL}{dY}dT\\ F_3 \end{bmatrix} \\\\ \\\\ F_1 = 1 – \frac{dC}{dY} – \frac{dX}{dY} \\\\ F_2 = -\frac{dC}{dY}dT – \frac{dI}{dr}d\pi^e + \frac{dI}{dY}dY + d\bar{G} + \frac{dX}{dY*}dY* \\\\ F_3 = \frac{d\sigma}{dr}i^* + \frac{dX}{dY^*}dY^* + dK$

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