$$\ln(L)\approx\ln(L_0) + \underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{=\dfrac{n}{\mu}-\large{1}}(\mu-n) + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}$$
从上面来看,我需要在下面再加一个下支架
$${\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}$$
这是我尝试过的:
$$\ln(L)\approx\ln(L_0) + \underbrace{\frac{\mathrm{d}\ln(L)}{\mathrm{d}\mu}\bigg|_n}_{\underbrace{=\dfrac{n}{\mu}-\large{1}}_{\text{from (2)}}(\mu-n) + \frac{\mathrm{d^2}\ln(L)}{\mathrm{d}\mu^2}\bigg|_n\frac{(\mu-n)^2}{2!}$$
有任何想法吗,谢谢
答案1
不太清楚你说的双下支撑是什么意思,但这是可能的双彩虹:
\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}
您只需要在\underbrace
现有的周围(内部或外部)添加一个\underbrace
。