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帮助将特定方程组刷新到左边距

tag-icon • tag-icon vertical-alignment amsmath
帮助将特定方程组刷新到左边距

所以我有这个代码

\begin{align}
{P_1} = {p_0}
\end{align}

\begin{align}
{P_2} = {p_0} + {\frac{rKT}{2}}{e^{ - {\frac{rT}{2}{rT}}}}N( - {d_-}({S_0},{B_{{\frac{T}{2}}}},T/2)) - \frac{\delta S_0 T}{2}e^{-\frac{\delta T}{2}}N( - {d_+}({S_0},{B_{{\frac{T}{2}}}},T/2))
\end{align}

\begin{align}
{P_3} = &{p_0} + {\frac{rKT}{3}}\left[ { e^{ - {\frac{rT}{3}}}N( - {d_-}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2rT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))}\right] \nonumber\\
&- {\frac{{S_0}T}{3}}\left[ {     e^{ - {\frac{\delta T}{3}}}N( - {d_+}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2aT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))} \right]
\end{align}

它产生以下内容（带有展示框）在此处输入图片描述

遗憾的是，我希望 p1 也能被刷新到左侧

答案1

使用multlined来自mathtools（并清理方程中的杂乱之后）：

在此处输入图片描述

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{mathtools}

\begin{document}
\begin{align}
P_{1}   & = p_{0}                                                               \\
P_{2}   & = p_{0} + \frac{rKT}{2}\mathrm{e}^{- \frac{rT}{2}{rT}}
                N\bigl(- d_{-}(S_{0},B_{\frac{T}{2}},T/2)\bigr) 
                - \frac{\delta S_{0}T}{2}\mathrm{e}^{-\frac{\delta T}{2}}
                 N\bigl(- d_{+}(S_{0},B_{\frac{T}{2}}T/2)\bigr)                           \\
P_{3} & =  \begin{multlined}[t]
    p_{0} + \frac{rKT}{3}\Bigl[\mathrm{e}^{-\frac{rT}{3}}
        N\bigl(-d_{-}(S_{0},B_{\frac{T}{3}},T/3)\bigr)     
        + \mathrm{e}^{-\frac{2rT}{3}} 
            N(-d_{+}(S_{0},B_{\frac{2T}{3}},2T/3))\Bigr]        \\[1ex]
        - \frac{{S_0}T}{3}\Bigl[\mathrm{e}^{-\frac{\delta T}{3}}
        N\bigl(-d_{+}(S_{0},B_{\frac{ T}{3}},T/3)\bigr)
        + \mathrm{e}^{- \frac{2aT}{3}}
        N\bigl(- d_{+}(S_{0},B_{\frac{2T}{3}},2T/3)\bigr) \Bigr]
            \end{multlined}
\end{align}
\end{document}

或使用split：

在此处输入图片描述

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{amsmath}

\begin{document}
\begin{align}
P_{1}   & = p_{0}                                                               \\
P_{2}   & = p_{0} + \frac{rKT}{2}\mathrm{e}^{- \frac{rT}{2}{rT}}
                N\bigl(- d_{-}(S_{0},B_{\frac{T}{2}},T/2)\bigr)
                - \frac{\delta S_{0}T}{2}\mathrm{e}^{-\frac{\delta T}{2}}
                 N\bigl(- d_{+}(S_{0},B_{\frac{T}{2}}T/2)\bigr)                           \\
    \begin{split}
P_{3} & =  p_{0} + \frac{rKT}{3}\Bigl[\mathrm{e}^{-\frac{rT}{3}}
        N\bigl(-d_{-}(S_{0},B_{\frac{T}{3}},T/3)\bigr)
        + \mathrm{e}^{-\frac{2rT}{3}}
            N(-d_{+}(S_{0},B_{\frac{2T}{3}},2T/3))\Bigr]        \\[1ex]
      &\qquad
       - \frac{{S_0}T}{3}\Bigl[\mathrm{e}^{-\frac{\delta T}{3}}
        N\bigl(-d_{+}(S_{0},B_{\frac{ T}{3}},T/3)\bigr)
        + \mathrm{e}^{- \frac{2aT}{3}}
        N\bigl(- d_{+}(S_{0},B_{\frac{2T}{3}},2T/3)\bigr) \Bigr]
    \end{split}
\end{align}
\end{document}

如果您希望所有方程式都左对齐（缩进一些），那么请使用以下fleqn选项amsmath：

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage[fleqn]{amsmath}
...

答案2

您可以使用flalignas in，但应避免使用如此长的（不间断的）公式。也许您还应该考虑仅使用一个flalign具有换行符和正确对齐的环境。

对齐

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{flalign}
{P_1} &= {p_0}\\
{P_2} &= {p_0} + {\frac{rKT}{2}}{e^{ - {\frac{rT}{2}{rT}}}}N( - {d_-}({S_0},{B_{{\frac{T}{2}}}},T/2)) - \frac{\delta S_0 T}{2}e^{-\frac{\delta T}{2}}N( - {d_+}({S_0},{B_{{\frac{T}{2}}}},T/2))\\
{P_3} &= {p_0} + {\frac{rKT}{3}}\left[ { e^{ - {\frac{rT}{3}}}N( - {d_-}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2rT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))}\right] \nonumber\\
&- {\frac{{S_0}T}{3}}\left[ {     e^{ - {\frac{\delta T}{3}}}N( - {d_+}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2aT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))} \right]
\end{flalign}
\end{document}

答案3

我将使用flalign提供的环境amsmath

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\begin{flalign}
P_{1}&=p_{0}\\
P_{2}&=p_{0} + {\frac{rKT}{2}}{e^{ - {\frac{rT}{2}{rT}}}}N( - {d_-}({S_0},{B_{{\frac{T}{2}}}},T/2))- \frac{\delta S_0 T}{2}e^{-\frac{\delta T}{2}}N( - {d_+}({S_0},{B_{{\frac{T}{2}}}},T/2))\\
P_{3}&={p_0} + {\frac{rKT}{3}}\left[ { e^{ - {\frac{rT}{3}}}N( - {d_-}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2rT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))}\right] \nonumber\\
&- {\frac{{S_0}T}{3}}\left[ {     e^{ - {\frac{\delta T}{3}}}N( - {d_+}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2aT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))} \right]
\end{flalign}
\end{document}

答案4

我将使用alignat并在符号处拆分第一个等式-：

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{alignat}{2}
{P_1} &= {p_0} && \\
{P_2} &= {p_0} &&+ {\frac{rKT}{2}}{e^{-{\frac{rT}{2}{rT}}}}N(-{d_-}({S_0},{B_{{\frac{T}{2}}}},T/2))\notag\\
      &        &&- \frac{\delta S_0 T}{2}e^{-\frac{\delta T}{2}}N( - {d_+}({S_0},{B_{{\frac{T}{2}}}},T/2))\\
{P_3} &= {p_0} &&+ {\frac{rKT}{3}}\left[ { e^{ - {\frac{rT}{3}}}N( - {d_-}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2rT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))}\right] \nonumber\\
      &        &&- {\frac{{S_0}T}{3}}\left[ {     e^{ - {\frac{\delta T}{3}}}N( - {d_+}({S_0},B_{\frac{T}{3}},T/3)) + e^{ - {\frac{2aT}{3}}}N( - {d_+}({S_0},B_{\frac{2T}{3}},2T/3))} \right]
\end{alignat}

\end{document}

在此处输入图片描述

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