- 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}
Best Answer
With
align