方程的多重对齐

Question 1

我认为实现您想要的最简单的方法实际上是使用单一环境。然后，您可以在切换行时align*在的可选参数中的三个表达式之间手动添加一些额外的间距。\\

\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}

\begin{document}
\begin{align*}
\frac{\partial F(\phi)}{\partial \beta_1}  &=  \sum_{i: x_i =0} y_i .... \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab \\[4mm]
\frac{\partial F(\phi)}{\partial \beta_2} &=  \sum_{i: x_i =0} y_i + 2y_{i}^{2} \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab + 2x \\[4mm]
\frac{\partial F(\phi)}{\partial \beta_3} &=  \sum_{i: x_i =0} y_i + (y+1)^2 - (y-1)^2 \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab
\end{align*}
\end{document}

Answer

我认为实现您想要的最简单的方法实际上是使用单一环境。然后，您可以在切换行时align*在的可选参数中的三个表达式之间手动添加一些额外的间距。\\

\documentclass{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}

\begin{document}
\begin{align*}
\frac{\partial F(\phi)}{\partial \beta_1}  &=  \sum_{i: x_i =0} y_i .... \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab \\[4mm]
\frac{\partial F(\phi)}{\partial \beta_2} &=  \sum_{i: x_i =0} y_i + 2y_{i}^{2} \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab + 2x \\[4mm]
\frac{\partial F(\phi)}{\partial \beta_3} &=  \sum_{i: x_i =0} y_i + (y+1)^2 - (y-1)^2 \\
&\mathrel{\phantom{=}} + \sum_{i: y_i > 0} 2y_i + ab
\end{align*}
\end{document}

Question 2

除了使用单一align*环境之外，我还建议您将所有六个\sum项都放入\smashoperator“包装器”中，以使方程看起来更紧凑。（基本上，这些\smashoperator指令让求和下标项向左和向右突出，而不会弄乱符号左右两侧的间距\sum。）我还将&\mathrel{\phantom{=}}用&\quad或替换的所有实例&\qquad。

\documentclass{article}
\usepackage{amssymb,amsthm}
\usepackage{mathtools} % 'mathtools' loads 'amsmath' automatically

\begin{document}
\begin{align*}
\frac{\partial F(\phi)}{\partial \beta_1}  
  &=       \smashoperator{\sum_{i: x_i =0}} y_i \dots \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab \\[\jot]
\frac{\partial F(\phi)}{\partial \beta_2} 
  &=       \smashoperator{\sum_{i: x_i =0}} y_i + 2y_{i}^{2} \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab + 2x \\[\jot]
\frac{\partial F(\phi)}{\partial \beta_3} 
  &=       \smashoperator{\sum_{i: x_i =0}} y_i + (y+1)^2 - (y-1)^2 \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab
\end{align*}
\end{document}

Answer

除了使用单一align*环境之外，我还建议您将所有六个\sum项都放入\smashoperator“包装器”中，以使方程看起来更紧凑。（基本上，这些\smashoperator指令让求和下标项向左和向右突出，而不会弄乱符号左右两侧的间距\sum。）我还将&\mathrel{\phantom{=}}用&\quad或替换的所有实例&\qquad。

\documentclass{article}
\usepackage{amssymb,amsthm}
\usepackage{mathtools} % 'mathtools' loads 'amsmath' automatically

\begin{document}
\begin{align*}
\frac{\partial F(\phi)}{\partial \beta_1}  
  &=       \smashoperator{\sum_{i: x_i =0}} y_i \dots \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab \\[\jot]
\frac{\partial F(\phi)}{\partial \beta_2} 
  &=       \smashoperator{\sum_{i: x_i =0}} y_i + 2y_{i}^{2} \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab + 2x \\[\jot]
\frac{\partial F(\phi)}{\partial \beta_3} 
  &=       \smashoperator{\sum_{i: x_i =0}} y_i + (y+1)^2 - (y-1)^2 \\
  &\qquad+ \smashoperator{\sum_{i: y_i > 0}} 2y_i + ab
\end{align*}
\end{document}

Question 3

根据@Mico 的很好的回答（+1），但使用\pdv包derivative和 multlined环境，\smashoperator以及\mathrlap包mathtools：

\documentclass{article}

\usepackage{amssymb, amsthm}
\usepackage{mathtools}  % it load amsmath too
\usepackage{derivative}

\begin{document}
    \begin{align*}
\pdv{F(\phi)}{\beta_1}  
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i \dots} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab
        \end{multlined}     \\[\jot]
%
\pdv{F(\phi)}{\beta_2}
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i + 2y_{i}^{2}} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab + 2x
        \end{multlined}     \\[\jot]
%
\pdv{F(\phi)}{\beta_3}
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i + (y+1)^2 - (y-1)^2} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab
        \end{multlined}
    \end{align*}
\end{document}

Answer

根据@Mico 的很好的回答（+1），但使用\pdv包derivative和 multlined环境，\smashoperator以及\mathrlap包mathtools：

\documentclass{article}

\usepackage{amssymb, amsthm}
\usepackage{mathtools}  % it load amsmath too
\usepackage{derivative}

\begin{document}
    \begin{align*}
\pdv{F(\phi)}{\beta_1}  
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i \dots} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab
        \end{multlined}     \\[\jot]
%
\pdv{F(\phi)}{\beta_2}
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i + 2y_{i}^{2}} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab + 2x
        \end{multlined}     \\[\jot]
%
\pdv{F(\phi)}{\beta_3}
    & = \begin{multlined}[t]
        \smashoperator{\sum_{i:x_i=0}}  \mathrlap{y_i + (y+1)^2 - (y-1)^2} \\
            + \smashoperator{\sum_{i:y_i>0}}    2y_i + ab
        \end{multlined}
    \end{align*}
\end{document}

Question 4

使用 IEEEtrantools 包。轻松对齐多个复杂方程。使用 \\ 分隔多个方程并使用 \begin{IEEEeqnarray}{} 和 \end{IEEEeqnarray}。开始括号中有三种对齐选择。

\documentclass[letterpaper, 10pt]{memoir}  
\usepackage{graphicx}     
\usepackage{amsmath}   
\usepackage[retainorgcmds]{IEEEtrantools} %for aligning eqns with multiple lines  

\begin{document}

\begin{IEEEeqnarray}{rCl}
  
\overline{C_2B} &=& r = \dfrac{\sqrt {9\Phi^{2}+1} }{2}\,tis.\nonumber \\\  

\overline{BX} &=& \mbox{midface of pentagon to a vertex of pentagon} = \dfrac{\Phi}{\sqrt {\Phi^{2}+1} }\,tis. \nonumber \\\

h^{2} &=& \overline {C_{2} B}^{2}-\overline {BX} 
^{2}=\dfrac{9\Phi^{2}+1}{4}-\dfrac{\Phi^{2}}{\Phi^{2}+1}\nonumber \\\ 

&=& \dfrac{(\Phi^{2}+1)(9\Phi^{2}+1)-4\Phi^{2}}{4(\Phi 
^{2}+1)}\nonumber \\\

&=&\dfrac{9\Phi^{4}+6\Phi^{2}+1}{4(\Phi 
^{2}+1)}=\dfrac{(3\Phi^{2}+1)^{2}}{4(\Phi^{2}+1)}\nonumber \\\ 

h &=& \dfrac{3\Phi^{2}+1}{2\sqrt {\Phi^{2}+1} }\,tis=2.327438437\,tis.
  
\end{IEEEeqnarray}

看起来像这样：

相同的方程，但第一个参数从 {rCl} 变为 {cCl}，每个方程的左侧居中

\begin{IEEEeqnarray}{cCl}

第一个参数从 {lCl} 变为 {lCl} 的相同方程式将每个方程式的左侧一直移到左侧

\begin{IEEEeqnarray}{lCl}

该系统非常灵活，允许您将方程式排列在您想要的位置。

Answer