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Adding LaTeX Equations

We have enabled the LaTeX markdown extensions using MathJax on this site. There are three ways to render eqations:

  1. Inline: simply surround the LaTeX expression with dollar signs
  2. Centered: surround the LeTex expression with '''[''' and ''']'''
  3. Block: place double dollar signs on a separate line before and after the LaTeX expression

Rendering Ohm's Law in Markdown

Here is an inline equation rendering of Ohm's Law: \(V=IR\).

$V=IR$

This example is centered:

\[ V = IR \]
\[ V = IR \]
\[ V=IR \]
$$
V=IR
$$

The Shockley Diode Equation

The Shockley diode equation describes the current–voltage relationship of a diode:

\[ I = I_S \left( e^{\frac{V}{n V_T}} - 1 \right) \]

where:

  • \( I \) is the diode current,
  • \( I_S \) is the reverse saturation current,
  • \( V \) is the voltage across the diode,
  • \( n \) is the ideality factor,
  • \( V_T \) is the thermal voltage, \( V_T = \dfrac{k T}{q} \).
$$
I = I_S \left( e^{\frac{V}{n V_T}} - 1 \right)
$$

where:

- \( I \) is the diode current,
- \( I_S \) is the reverse saturation current,
- \( V \) is the voltage across the diode,
- \( n \) is the ideality factor,
- \( V_T \) is the thermal voltage, \( V_T = \dfrac{k T}{q} \).

Voltage Across a Capacitor

The voltage across a capacitor at any given time is determined by the history of the current that has flowed through it. We can express this as the sum of the initial voltage plus a integral of the current over time.

\[ v(t) = v(t_0) + \frac{1}{C} \int_{t_0}^{t} i(t)dt \]

where:

  • \(v(t)\): Voltage across the capacitor at time \(t\)
  • \(v(t_0)\): Initial voltage across the capacitor at time \(t_0​\)
  • \(C\): Capacitance (in farads)
  • \(i(t)\): Current flowing through the capacitor at time \(t\)
  • \(t_0​\): Initial time (often taken as t=0)
v(t) = v(t_0) + \frac{1}{C} \int_{t_0}^{t} i(t)dt

Other Examples

And here is a display equation that includes an integral, infinity, an exponent, a square, and square root and the symbol for \(\pi\):

\[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \]
\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}
\[ \operatorname{ker} f=\{g\in G:f(g)=e_{H}\}{\mbox{.}} \]

Block

1
\operatorname{ker} f=\{g\in G:f(g)=e_{H}\}{\mbox{.}}