4
$$\ln(L)\approx\ln(L_0) + \underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{=\dfrac{n}{\mu}-\large{1}}(\mu-n) + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}$$

From the above, I need another underbrace beneath

$${\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}$$

This is what I tried:

$$\ln(L)\approx\ln(L_0) + \underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{\underbrace{=\dfrac{n}{\mu}-\large{1}}_{\text{from (2)}}(\mu-n) + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}$$

Any ideas, thanks

8

Not exactly sure what you mean by double underbrace, but it's just as possible as double rainbows:

enter image description here

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\[
  \ln(L) \approx \ln(L_0) + \underbrace{\underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{=\tfrac{n}{\mu} - 1}}_{abc}(\mu-n)
    + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}
\]

\[
  \ln(L) \approx \ln(L_0) + \underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{=\underbrace{\tfrac{n}{\mu} - 1}_{abc}}(\mu-n)
    + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}
\]

\[
  \ln(L) \approx \ln(L_0) + \underbrace{\underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}}_{=\tfrac{n}{\mu} - 1}(\mu-n)
    + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}
\]

\end{document}

You merely have to add an \underbrace around (inside or outside) the existing \underbrace.

  • Excellent answer, precisely what I needed to see – BLAZE Nov 29 '15 at 20:44

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