这是我的序言:
\documentclass[12pt,thmsa]{article}
\usepackage{natbib}
\usepackage{amsmath,amsfonts,amssymb,amsthm}
\usepackage{breqn}
\usepackage{rotating}
\usepackage{threeparttable}
\usepackage{booktabs}
\usepackage[para,multiple]{footmisc}
\usepackage{mathtools}
\begin{document}
第一个方程问题
\begin{flalign}
{w_1} &= \left\{N_1(-d_+(S_{dt},dt)+N_2(d_+(S_{dt},dt)),-d_+(S_{2dt},dt);-\rho_{12})+N_3(d_+(S_{dt},dt),d_+(S_{2dt},2dt),-d_+(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23})...\right\},
\end{flalign}
\begin{flalign}
{w_2} &= \left\{N_1(-d_-(S_{dt},dt)+N_2(d_-(S_{dt},dt)),-d_-(S_{2dt},dt);-\rho_{12})+N_3(d_-(S_{dt},dt),d_-(S_{2dt},2dt),-d_-(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23})+...\right\},
\end{flalign}
困扰我的第二个等式是:
\begin{align}
&\varepsilon {B_t} \left\{ {1 - {e^{ - \delta (T - t)}}N( - {d_ + }(\varepsilon {B_t},K,T - t)) - \delta \int\limits_t^T {{e^{ - \delta (s-t)}}N( - {d_ + }(\varepsilon {B_t},{B_s},s - t))\;ds} } \right\} \\
&= K \left\{ {1 - {e^{ - r(T - t)}}N( - {d_ - }(\varepsilon {B_t},K,T - t)) - r\int\limits_t^T {{e^{ - \delta (s-t)}}N( - {d_ - }(\varepsilon {B_t},{B_s},s - t))} {\kern 1pt} \;ds} \right\}.
\end{align}
我的第三个等式:
\begin{align}
P_3 = &Xe^{-rT/3}N_1[-d_-(B_{T/3},T/3)]-S_{T/3}N_1[-d_+(B_{T/3},T/3)]\nonumber\\
&+Xe^{-2rT/3}N_2[d_-(B_{T/3},T/3), -d_-(B_{2T/3},2T/3); -1/\sqrt{2}]\nonumber\\
&-S_{T/3, 2T/3}N_2[d_+(B_{T/3}, T/3),-d_+(B_{2T/3},2T/3);-1/\sqrt{2}]\nonumber\\
&+Xe^{-rT}N_3[d_+(B_{T/3},T/3), -d_+(B_{2T/3},2T/3), -d_+(X,T);1/\sqrt{2},-1/\sqrt{3},-\sqrt{2/3}]\nonumber\\
&-S_{T/3, 2T/3}N_3[d_-(B_{T/3},T/3),d_-(B_{2T/3},2T/3),-d_-(X,T);1/\sqrt{2},-1/\sqrt{3},-\sqrt{2/3}],
\end{align}
答案1
我可以基于 aligned
或multlined
环境以及medium-sized
来自的公式提出这一点。请注意,如果您加载了,则nccmath
无需加载:amsmath
mathtools
\documentclass[12pt,thmsa]{article}
\usepackage{natbib}
\usepackage{amsfonts,amssymb,amsthm}
%\usepackage{breqn}
\usepackage{rotating}
\usepackage{threeparttable}
\usepackage{booktabs}
\usepackage[para,multiple]{footmisc}
\usepackage{mathtools, nccmath}
\usepackage[showframe]{geometry}
\begin{document}
\begin{align}
w_1 &= \begin{multlined}[t]\Bigl\{N_1(-d_+(S_{dt},dt)+N_2(d_+(S_{dt},dt)),-d_+(S_{2dt},dt);-\rho_{12})\\[-1ex]
+N_3(d_+(S_{dt},dt),d_+(S_{2dt},2dt),-d_+(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23})...\Bigr\},\end{multlined}\\
{w_2} &= \begin{multlined}[t]\Bigl\{N_1(-d_-(S_{dt},dt)+N_2(d_-(S_{dt},dt)),-d_-(S_{2dt},dt);-\rho_{12})\\[-1ex]
+ N_3\bigl(d_-(S_{dt},dt),d_-(S_{2dt},2dt),-d_-(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23}\bigr)+...\Bigr\}\end{multlined}
\end{align}
\vspace{1cm}
\begin{align}
\varepsilon {B_t}&\medmath{\Biggl\{ 1 - {e^{ - \delta (T - t)}}N\bigl( - {d_ + }(\varepsilon {B_t},K,T - t)\bigr) - \delta \int\limits_t^T e^{ - \delta (s-t)} N\bigl( -d_{+}(\varepsilon B_t, B_s, s - t)\bigr)\;ds \Biggr\}} \\
= K &\medmath{\Biggl\{ 1 - e^{ - r(T - t)}N\bigl( -d_{-}(\varepsilon B_t, K, T - t)\bigr) - r\int\limits_t^T e^{ - \delta (s-t)}N\bigl(-d_{-} (\varepsilon B_t, B_s, s - t)\bigr) \;ds \Biggr\}}.
\end{align}
\vspace{1cm}
\begin{equation}
\begin{aligned}
P_3 &= {} Xe^{-rT/3}N_1\Bigl[-d_-\Bigl(B_{\tfrac{T}{3}},\mfrac{T}{3}\Bigr)\Bigr]-S_{T/3}N_1\Bigl[-d_+\Bigl(B_{\tfrac{T}{3}},\mfrac{T}{3}\Bigr)\Bigr] \\
&+Xe^{-2rT/3}N_2\Bigl[d_-\Bigl(B_{\tfrac{T}{3}},\mfrac{T}{3}\Bigr), -d_-\Bigl(B_{\tfrac{2T}{3}},\mfrac{2T}{3}\Bigr); -\mfrac{1}{\sqrt{2}}\Bigr] \\
&-S_{\tfrac{T}{3}, \tfrac{2T}{3}} N_2\Bigl[d_+\Bigl(B_{\tfrac{T}{3}},\mfrac{T}{3}\Bigr),-d_+ \Bigl(B_{\tfrac{2T}{3}},\mfrac{2T}{3}\Bigr);-\mfrac{1}{\sqrt{2}}\Bigr] \\
&+Xe^{-rT}N_3\Biggl[d_+\Bigl(B_{\tfrac{T}{3}},\mfrac{T}{3}\Bigr), -d_+\Bigl(B_{\tfrac{2T}{3}},\mfrac{2T}{3}\Bigr), -d_+(X,T);\mfrac{1}{\sqrt{2}},-\mfrac{1}{\sqrt{3}},-\sqrt{\mfrac{2}{3}}\Biggr] \\
&-S_{\tfrac{T}{3}, \tfrac{2T}{3}}N_3\Biggl[d_-\Bigl(B_{\tfrac{T}{3}}, \mfrac{T}{3}\Bigr),d_-(B_{2T/3}, 2T/3),-d_-(X,T);\mfrac{1}{\sqrt{2}},-\mfrac{1}{\sqrt{3}},-\sqrt{\mfrac{2}{3}}
\Biggr],
\end{aligned}
\end{equation}
\end{document}
答案2
你的公式有点太长了。我尝试将它们分成多行。这是我能做的最好的事情,但请考虑将一些较长的公式部分定义为常量/字母,否则很难跟踪和排版。
哦,请注意,该align
环境旨在同时用不同的数字对齐多个方程。你想要的可能是split
或multline
\documentclass[12pt,thmsa]{article}
\usepackage{natbib}
\usepackage{amsmath,amsfonts,amssymb,amsthm}
\usepackage{breqn}
\usepackage{rotating}
\usepackage{threeparttable}
\usepackage{booktabs}
\usepackage[para,multiple]{footmisc}
\usepackage{mathtools}
\begin{document}
\begin{multline}
w_1 = \{N_1(-d_+(S_{dt},dt)+N_2(d_+(S_{dt},dt)),-d_+(S_{2dt},dt);-\rho_{12})\\
+N_3(d_+(S_{dt},dt),d_+(S_{2dt},2dt),-d_+(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23})\ldots\},
\end{multline}
\begin{multline}
{w_2} = \{N_1(-d_-(S_{dt},dt)+N_2(d_-(S_{dt},dt)),-d_-(S_{2dt},dt);-\rho_{12})\\
+N_3(d_-(S_{dt},dt),d_-(S_{2dt},2dt),-d_-(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23})+\ldots\},
\end{multline}
\begin{equation}
\begin{split}
&\varepsilon {B_t} \biggl\{ 1 - {e^{ - \delta (T - t)}}N( - {d_ + }(\varepsilon {B_t},K,T - t)) \\ &{}-\delta \int\limits_t^T {{e^{ - \delta (s-t)}}N( - {d_ + }(\varepsilon {B_t},{B_s},s - t))\;ds} \biggr\} \\
&= K \biggl\{ 1 - {e^{ - r(T - t)}}N( - {d_ - }(\varepsilon {B_t},K,T - t)) \\
&{}- r\int\limits_t^T {{e^{ - \delta (s-t)}}N( - {d_ - }(\varepsilon {B_t},{B_s},s - t))} {\kern 1pt} \;ds \biggr\}.
\end{split}
\end{equation}
\begin{equation}
\begin{split}
P_3 &= Xe^{-rT/3}N_1[-d_-(B_{T/3},T/3)]-S_{T/3}N_1[-d_+(B_{T/3},T/3)]\\
&{}+Xe^{-2rT/3}N_2[d_-(B_{T/3},T/3), -d_-(B_{2T/3},2T/3); -1/\sqrt{2}]\\
&{}-S_{T/3, 2T/3}N_2[d_+(B_{T/3}, T/3),-d_+(B_{2T/3},2T/3);-1/\sqrt{2}]\\
&{}+Xe^{-rT}N_3[d_+(B_{T/3},T/3), -d_+(B_{2T/3},2T/3),\\
&{} -d_+(X,T);1/\sqrt{2},-1/\sqrt{3},-\sqrt{2/3}]\\
&{}-S_{T/3, 2T/3}N_3[d_-(B_{T/3},T/3),\\
&d_-(B_{2T/3},2T/3),-d_-(X,T);1/\sqrt{2},-1/\sqrt{3},-\sqrt{2/3}],\\
\end{split}
\end{equation}
\end{document}
答案3
正如 Moriambar 所说,你的方程太长了。我认为你在所有方程中都遇到了同样的问题,所以我专注于第一个方程。在这里,我首先在第一个主函数之后进行拆分-
,但即便如此,第二行也太宽了。我尝试使用 对齐括号的内容\phantom
。不过,它并不是真正可读的,所以如果可以找到更短的表达式会有所帮助。
\begin{flalign}
{w_1} &= \Bigl\{N_1\Bigl(-d_+(S_{dt},dt)+N_2(d_+(S_{dt},dt)),-d_+(S_{2dt},dt);-\rho_{12}\Bigr)\nonumber\\
&\phantom{{}={}\Bigl\{}
+N_3\Bigl(d_+(S_{dt},dt),d_+(S_{2dt},2dt),\nonumber\\
&\phantom{{}={}\Bigl\{{}+N_3\Bigl(}
-d_+(S_{3dt},3dt);\rho_{12},-\rho_{13},-\rho_{23}\Bigr)\ldots\Bigr\},
\end{flalign}
在这种情况下,我认为您不应该使用\left
和\right
。而是使用例如\Bigl
和\Bigr
在中定义,amsmath
因为它们不需要在同一行上成对出现。它们有许多其他大小。其次,对齐&
应该在=
tio 之前,以获得正确的间距。您可以使用\phantom
从其他点开始。通常我会尝试在最后一行之后使用方程编号,这意味着使用抑制其他\nonumber
。(还有其他方法可以做到这一点,但我现在忘记了)。