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如何将一个方程仅向左边距移动

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如何将一个方程仅向左边距移动
  • 我有一个方程式延伸到右边距并超出了页面。
  • 我有很多方程。仅对于这个方程，我想移动

  • 它在左边。

    下面的 MWE。

    \documentclass[twoside]{book}                           
\usepackage{amsmath}                                      
     \begin{document}
        \begin{equation}
                \text{Some big equation}
        \end{equation}
\end{document}

这是一个等式的例子，我希望将其水平向左移动，因为它太靠右了。

\documentclass[a4paper,12pt,twoside]{book}                      

        \newcommand{\MyLeftRoundBracket}{(}                                                                         
        \newcommand{\MyRightRoundBracket}{)}                                                                        
        \newcommand{\MyLeftSquareBracket}{[}                                                                        
        \newcommand{\MyRightSquareBracket}{]} 

        \usepackage{amsmath}                                      

     \begin{document}

                \begin{align}\label{eq:CIR_Expected_Return_Variance_Spread}
     ^{Spread}_{CIR}\sigma^2(X^L,X^U,c) &= -{(c-X^L+X^U)^2 \over {_{CIR}\theta_1}^2} {1 \over \Bigg \MyLeftSquareBracket \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2
   {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)-}  \cdots  \notag \\
&\qquad {1 \over 
        \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
   X^U}{{_{CIR}\theta_3}}\Big) -\frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big) + }  \cdots  \notag        \\
   & \qquad\qquad {1 \over \frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1}
   {_{CIR}\theta_2}}{{_{CIR}\theta_3}}, \frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big) \Bigg \MyRightSquareBracket^3 } 
\times      \notag  \\
&\qquad \Bigg [ \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1}  {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
   X^L}{{_{CIR}\theta_3}}\Big)^2-\frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
   X^U}{{_{CIR}\theta_3}}\Big)^2 + \notag \\
&\qquad   \frac{\partial \Phi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
   X^U}{{_{CIR}\theta_3}}\Big)-\frac{\partial \Phi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)+ \notag \\
&\qquad  \frac{\partial \Psi}{\partial a}\Big(0,\frac{2
   {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)^2-\frac{\partial \Psi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
   X^L}{{_{CIR}\theta_3}}\Big)-         \notag \\
&\qquad   \frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big)^2+\frac{\partial \Psi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1}
   {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big)\Bigg ]
   \end{align}              

\end{document}

答案1

\documentclass[twoside]{book}                           
\usepackage{amsmath}                                      
\begin{document}
\begin{equation}
 \hspace*{-10cm}   \text{Some big equation}
\end{equation}
\begin{equation}
       \text{Some big equation}
\end{equation}
\end{document}]

在此处输入图片描述

和align

\documentclass[a4paper,12pt,twoside]{book}                      

        \newcommand{\MyLeftRoundBracket}{(}                                                                         
        \newcommand{\MyRightRoundBracket}{)}                                                                        
        \newcommand{\MyLeftSquareBracket}{[}                                                                        
        \newcommand{\MyRightSquareBracket}{]} 

        \usepackage{amsmath}                                      

     \begin{document}

  \hspace*{-3cm}\vbox{\begin{align}\label{eq:CIR_Expected_Return_Variance_Spread}
       ^{Spread}_{CIR}\sigma^2(X^L,X^U,c) &= -{(c-X^L+X^U)^2 \over {_{CIR}\theta_1}^2} {1 \over \Bigg \MyLeftSquareBracket \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2
     {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)-}  \cdots  \notag \\
  &\qquad {1 \over 
          \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
     X^U}{{_{CIR}\theta_3}}\Big) -\frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big) + }  \cdots  \notag        \\
     & \qquad\qquad {1 \over \frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1}
     {_{CIR}\theta_2}}{{_{CIR}\theta_3}}, \frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big) \Bigg \MyRightSquareBracket^3 } 
  \times      \notag  \\
  &\qquad \Bigg [ \frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1}  {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
     X^L}{{_{CIR}\theta_3}}\Big)^2-\frac{\partial \Phi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
     X^U}{{_{CIR}\theta_3}}\Big)^2 + \notag \\
  &\qquad   \frac{\partial \Phi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
     X^U}{{_{CIR}\theta_3}}\Big)-\frac{\partial \Phi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)+ \notag \\
  &\qquad  \frac{\partial \Psi}{\partial a}\Big(0,\frac{2
     {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^L}{{_{CIR}\theta_3}}\Big)^2-\frac{\partial \Psi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1}
     X^L}{{_{CIR}\theta_3}}\Big)-         \notag \\
  &\qquad   \frac{\partial \Psi}{\partial a}\Big(0,\frac{2 {_{CIR}\theta_1} {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big)^2+\frac{\partial \Psi^2}{\partial a^2}\Big(0,\frac{2 {_{CIR}\theta_1}
     {_{CIR}\theta_2}}{{_{CIR}\theta_3}},\frac{2 {_{CIR}\theta_1} X^U}{{_{CIR}\theta_3}}\Big)\Bigg ]
     \end{align}  }            

\end{document}

在此处输入图片描述

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