13
  • I have an equation extending into the right margin and off the page.
  • I have many equations. For this equation only, I would like to shift
  • it to the left.

    MWE below.

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

Here is an example of an equation I wish to shift left horizontally as it is too far to the right.

\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}
  • 1
    Could you include the actual equation? Is breaking the equation up in multiple lines an option? – Michel Mar 30 '11 at 7:52
  • @Michel,all I want to do do is shift anything wrapped in the equation command to the left. This should be a common requirement. I have 100's of equations but 5 or so overlap off the page of my PhD thesis. I want to shift those equations only to the left. I gave ab example of an equation as requested.NOTE: There are other formatting issues going on,so the example is not off the page but does extend far right. I added the example to a separate section as not to clutter the real question. Which is just to shift anything in the equation or dmath or align command to the right. – Ian Gregory Mar 30 '11 at 8:43
  • Is not the same question ! I modified my answer – Alain Matthes Mar 30 '11 at 8:55
  • Please edit your question if you want to provide more details. The answer section is for solutions only. I moved your answer to the question to keep the site tidy. – Martin Scharrer Mar 30 '11 at 11:12
20
![\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}]

enter image description here

With 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} 

enter image description here

  • 1
    The problem here is that the equation is simply too wide, it's wider than the textblock. A better solution would be to change the equation, a break after the equals sign could be an option. – Michel Mar 30 '11 at 9:07
  • 2
    @Michel I agree with you but the OP wants to shift left horizontally the equation so ... – Alain Matthes Mar 30 '11 at 9:16

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