This article demonstrates one way to write complex equations using LaTeX on your WordPress blog.

Why blog with LaTeX? At least two reasons:

- WordPress and other text editors don’t support complex mathematical notation out-of-the-box.
- 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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