对齐环境中的边距是否更宽？

Question

我在对齐环境中有一组方程式。当添加方程式编号时，它太宽了，所以转到下一行。

我想将方程式移到左边距，这样每个方程式就只有一行

\begin{align}
a & =  3.64779052-4.5326958\rho+2.8254675\rho^{1.5}-1.1230973\ln(\rho)/\rho +0.02371867\ln(\rho)/\rho^2 \\
b & =  7.0399098 - 13.686034\rho-0.85088666\rho^3+2.9323443e^{\rho}+6.6465928\sqrt{\rho}\ln(\rho)\\
c & = 5.9382949+3.7429435\rho+1.108728\rho^2\ln(\rho)-0.96469546e^{\rho}+0.19983335\ln(\rho)\\
d & =  1.1350252-1.9259676\rho-1.8835841\rho^2\ln(\rho)+1.0381572\rho^3+0.00007112\ln(\rho)\\
\partial a /\partial \rho & =  2.2578288-2.6419506/\rho-0.57462084/\rho^2-0.49861017/\rho^3+0.038266286/\rho^4\\
\partial b /\partial \rho & =  -1.1150611+1.5019557\rho-1.5365115\rho^{1.5}+0.79801674\rho^3-1.2700068/\rho\\
\partial c /\partial \rho & = 2.5339359-5.0383713\rho+6.0947248\rho^{1.5}-1.3679322\rho^{2.5}+0.20489549/\rho\\
\partial d /\partial \rho & =-18.494398+17.445882\rho-16.716728\sqrt{\rho}\ln(\rho)+0.45062181/\rho^{1.5}-0.12072079/\rho^2 
\end{align}}

Answer 1

您可以使用如下例所示的changepage包和adjustwidth环境。（红线表示边距。）

\documentclass{article}

\usepackage{amsmath}
\usepackage{changepage}
\usepackage{lipsum} % For dummy text. Don't use in real document.


\begin{document}
\lipsum[4]
\begin{adjustwidth}{-3cm}{0pt}
\begin{align}
a & =  3.64779052-4.5326958\rho+2.8254675\rho^{1.5}-1.1230973\ln(\rho)/\rho +0.02371867\ln(\rho)/\rho^2 \\
b & =  7.0399098 - 13.686034\rho-0.85088666\rho^3+2.9323443e^{\rho}+6.6465928\sqrt{\rho}\ln(\rho)\\
c & = 5.9382949+3.7429435\rho+1.108728\rho^2\ln(\rho)-0.96469546e^{\rho}+0.19983335\ln(\rho)\\
d & =  1.1350252-1.9259676\rho-1.8835841\rho^2\ln(\rho)+1.0381572\rho^3+0.00007112\ln(\rho)\\
\partial a /\partial \rho & =  2.2578288-2.6419506/\rho-0.57462084/\rho^2-0.49861017/\rho^3+0.038266286/\rho^4\\
\partial b /\partial \rho & =  -1.1150611+1.5019557\rho-1.5365115\rho^{1.5}+0.79801674\rho^3-1.2700068/\rho\\
\partial c /\partial \rho & = 2.5339359-5.0383713\rho+6.0947248\rho^{1.5}-1.3679322\rho^{2.5}+0.20489549/\rho\\
\partial d /\partial \rho & =-18.494398+17.445882\rho-16.716728\sqrt{\rho}\ln(\rho)+0.45062181/\rho^{1.5}-0.12072079/\rho^2 
\end{align}
\end{adjustwidth}
\lipsum[4]
\end{document}

Answer 2

我建议不要使用边距，而是分割长线。

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{alignat}{3}
a & ={} &   3.64779052 &-4.5326958\rho+2.8254675\rho^{1.5} \notag \\
  &     &              &-1.1230973\ln(\rho)/\rho +0.02371867\ln(\rho)/\rho^2 \\
b & ={} &   7.0399098  &-13.686034\rho-0.85088666\rho^3 \notag \\
  &     &              &+2.9323443e^{\rho}+6.6465928\sqrt{\rho}\ln(\rho)\\
c & ={} &   5.9382949  &+3.7429435\rho+1.108728\rho^2\ln(\rho) \notag \\
  &     &              &-0.96469546e^{\rho}+0.19983335\ln(\rho)\\
d & ={} &   1.1350252  &-1.9259676\rho-1.8835841\rho^2\ln(\rho) \notag \\
  &     &              &+1.0381572\rho^3+0.00007112\ln(\rho)\\
\smash[b]{\frac{\partial a}{\partial \rho}}
  & ={} &   2.2578288  &-2.6419506/\rho-0.57462084/\rho^2 \notag \\
  &     &              &-0.49861017/\rho^3+0.038266286/\rho^4\\
\smash[b]{\frac{\partial b}{\partial \rho}}
  & ={} &  -1.1150611  &+1.5019557\rho-1.5365115\rho^{1.5} \notag \\
  &     &              &+0.79801674\rho^3-1.2700068/\rho\\
\smash[b]{\frac{\partial c}{\partial \rho}}
  & ={} &   2.5339359  &-5.0383713\rho+6.0947248\rho^{1.5} \notag \\
  &     &              &-1.3679322\rho^{2.5}+0.20489549/\rho\\
\smash[b]{\frac{\partial d}{\partial \rho}}
  & ={} & -18.494398   &+17.445882\rho-16.716728\sqrt{\rho}\ln(\rho) \notag \\
  &     &              &+0.45062181/\rho^{1.5}-0.12072079/\rho^2 
\end{alignat}

\end{document}

我们\smash[b]欺骗 LaTeX，让它认为分数不会延伸到基线以下，因此下一条分割线不会向下移动。

替代方法：

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align}
a &= 3.64779052-4.5326958\rho+2.8254675\rho^{1.5} \notag \\
  &\qquad -1.1230973\ln(\rho)/\rho +0.02371867\ln(\rho)/\rho^2 \\
b &= 7.0399098-13.686034\rho-0.85088666\rho^3 \notag \\
  &\qquad +2.9323443e^{\rho}+6.6465928\sqrt{\rho}\ln(\rho)\\
c &= 5.9382949+3.7429435\rho+1.108728\rho^2\ln(\rho) \notag \\
  &\qquad -0.96469546e^{\rho}+0.19983335\ln(\rho)\\
d &= 1.1350252-1.9259676\rho-1.8835841\rho^2\ln(\rho) \notag \\
  &\qquad +1.0381572\rho^3+0.00007112\ln(\rho)\\
\smash[b]{\frac{\partial a}{\partial \rho}}
  &= 2.2578288-2.6419506/\rho-0.57462084/\rho^2 \notag \\
  &\qquad -0.49861017/\rho^3+0.038266286/\rho^4\\
\smash[b]{\frac{\partial b}{\partial \rho}}
  &= -1.1150611+1.5019557\rho-1.5365115\rho^{1.5} \notag \\
  &\qquad +0.79801674\rho^3-1.2700068/\rho\\
\smash[b]{\frac{\partial c}{\partial \rho}}
  &= 2.5339359-5.0383713\rho+6.0947248\rho^{1.5} \notag \\
  &\qquad -1.3679322\rho^{2.5}+0.20489549/\rho\\
\smash[b]{\frac{\partial d}{\partial \rho}}
  &= -18.494398+17.445882\rho-16.716728\sqrt{\rho}\ln(\rho) \notag \\
  &\qquad +0.45062181/\rho^{1.5}-0.12072079/\rho^2 
\end{align}

\end{document}