P(S_t) = \frac{1}{\sqrt{2 \pi t S_t} \sigma} e^\left [ -\frac{1}{2t} \left ( \frac{\log \left ( \frac{St}{S_0} \right ) - \mu t }{\sigma} \right ) ^2 \right ]
日志:
Missing { inserted. ...frac{1}{\sqrt{2 \pi t S_t} \sigma} e^\left
Missing } inserted. ... ) - \mu t }{\sigma} \right ) ^2 \right ]$
Overfull \hbox (7.07501pt too wide) in paragraph
Underfull \vbox (badness 10000) has occurred while \output is active []
Underfull \hbox (badness 10000) in paragraph
Underfull \hbox (badness 1237) in paragraph
Underfull \vbox (badness 10000) has occurred while \output is active []
答案1
我借此机会改进了您的表达方式:内部括号是不必要的,我们可以微调分隔符的大小。或者,我们可以使用\exp
和来自的中等大小的分数nccmath
:
\documentclass[a4paper, 11pt]{book}
\usepackage[utf8]{inputenc}
\usepackage{fourier, heuristica}
\usepackage{mathtools, nccmath}
\begin{document}
\[ P(S_t) = \frac{1}{\sqrt{2 \pi t S_t} \sigma} e^{-\frac{1}{2t}\Bigl( \frac{\log \frac{St}{S_0} - \mu t }{\sigma} \Bigr)^2} \]%
May be better-looking, with \verb+\mfrac+ (from \texttt{nccmath}):
\[ P(S_t) = \frac{1}{\sqrt{2 \pi t S_t} \sigma}\exp{\Biggl [ -\mfrac{1}{2t} \Bigl( \mfrac{\log \frac{St}{S_0} - \mu t }{\sigma} \Bigr) ^2 \Biggr]} \]%
\end{document}
答案2
指数周围缺少括号。因此 TeX 尝试写入e
左括号的幂。
而且 TeX 与 Python 等不同。不要使用虚假的空格。您偶尔会得到奇怪的结果,并且在调试时会让您发疯。
答案3
你想要这个等式吗?
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
P(S_t) = \frac{1}{\sqrt{2 \pi t S_t} \sigma}
\mathrm{e}^{\left [ -\frac{1}{2t} \left ( \frac{\log
\left ( \frac{St}{S_0} \right ) -
\mu t }{\sigma} \right ) ^2 \right]}
\]
\end{document}
答案4
由于通常有很多方法可以在 LaTeX 中获得结果,因此这又是另一种方法。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
P(S_{t}) = \frac{1}{\sqrt{2 \pi t S_{t}} \, \sigma} \cdot \exp{\left[ - \frac{\left( \log \left( \frac{S t}{S_{0}} \right) - \mu t \right)^{2}}{2 \, t \, \sigma^{2}} \right]}
\end{align}
\end{document}