我怎样才能简化这个等式？

Question

我想在 Latex 中写下以下文字方程式：

我有以下代码：

\begin{align}                  
\mathbf {P_{t} &= \sum_{j=1}^{\rho } \left ( I - G_{0} \right )^{-1} \left [ D_{j} + D_{0}  \left ( I - G_{0} \right )^{-1} B_{j}\right ] Y_{t-j} + \sum_{j=1}^{\rho } \left ( I - G_{0} \right )^{-1} \left [ G_{j} + D_{0}  \left ( I - B_{0} \right )^{-1} C_{j}\right ] P_{t-j} \nonumber \\
               &\hphantom{=} + \sum_{j=1}^{\rho } \left ( I - G_{0} \right )^{-1} \left [ G_{j} + D_{0}  \left ( I - B_{0} \right )^{-1} C_{j}\right ] P_{t-j} \nonumber \\
               &\hphantom{=}  + B_{j} Y_{t-j} + \left ( I - G_{0} \right )^{-1} D_{0} \left ( I - B_{0} \right )^{-1}A^{y} v_{t}^{y} + u_{p}^{t}}, \label{eqn:{2}'}                  
\end{align} 

我尝试使用\nonumber \\和来 分解它&\hphantom{=}，但就是不起作用。我遇到了一些错误。任何帮助都将不胜感激。谢谢。

请注意我的序言：

\documentclass[12pt,reqno]{amsart}

% Packages
\usepackage[onehalfspacing]{setspace}
\usepackage{graphicx}
\usepackage{amssymb,amsthm}
\usepackage{amsmath}
\usepackage{tikz}
\usepackage{float}
\usepackage[utf8]{inputenc}
\makeatletter
\newcommand*\rel@kern[1]{\kern#1\dimexpr\macc@kerna}
\newcommand*\widebar[1]{%
  \begingroup
  \def\mathaccent##1##2{%
    \rel@kern{0.8}%
    \overline{\rel@kern{-0.8}\macc@nucleus\rel@kern{0.2}}%
    \rel@kern{-0.2}%
  }%
  \macc@depth\@ne
  \let\math@bgroup\@empty \let\math@egroup\macc@set@skewchar
  \mathsurround\z@ \frozen@everymath{\mathgroup\macc@group\relax}%
  \macc@set@skewchar\relax
  \let\mathaccentV\macc@nested@a
  \macc@nested@a\relax111{#1}%
  \endgroup
}
\makeatother
\usepackage{cite}
%\usepackage{natbib}
%\bibpunct{(}{)}{;}{;}{,}{,}
\usepackage{xr-hyper}
\usepackage[
  colorlinks=true,
  citecolor=blue,
  urlcolor=blue,
  linkcolor=blue
]{hyperref}
\usepackage{bm}
\usepackage{fullpage}
\usepackage{ amssymb }
\usepackage{caption} 
\usepackage{forest}
\captionsetup{labelfont=normalfont,
              labelsep=colon}
\usepackage{mathabx}
\usepackage{subfig}
\usetikzlibrary{arrows.meta,shapes,positioning}
%
\tikzset{
    full/.style={circle,draw,inner sep=0, minimum size=1mm,fill=black},
    every node/.style={minimum height=5mm,font=\footnotesize}
}
%
\makeatletter
\def\subsection{\@startsection{subsection}{1}%
  \z@{.5\linespacing\@plus.7\linespacing}{-.5em}%
  {\normalfont\itshape}}
\def\@sect#1#2#3#4#5#6[#7]#8{%
  \edef\@toclevel{\ifnum#2=\@m 0\else\number#2\fi}%
  \ifnum #2>\c@secnumdepth \let\@secnumber\@empty
  \else \@xp\let\@xp\@secnumber\csname the#1\endcsname\fi
  \@tempskipa #5\relax
  \ifnum #2>\c@secnumdepth
    \let\@svsec\@empty
  \else
    \refstepcounter{#1}%
    \edef\@secnumpunct{%
      \ifdim\@tempskipa>\z@ % not a run-in section heading
        \@ifnotempty{#8}{.\@nx\enspace}%
      \else
        \@ifempty{#8}{.}{.\@nx\enspace}%
      \fi
    }%
    \@ifempty{#8}{%
      \ifnum #2=\tw@ \def\@secnumfont{\bfseries}\fi}{}%
    \protected@edef\@svsec{%
      \ifnum#2<\@m
        \@ifundefined{#1name}{}{%
          \ignorespaces\csname #1name\endcsname\space
        }%
      \fi
      \@seccntformat{#1}%
    }%
  \fi
  \ifdim \@tempskipa>\z@ % then this is not a run-in section heading
    \begingroup #6\relax
    \@hangfrom{\hskip #3\relax\@svsec}{\interlinepenalty\@M #8\par}%
    \endgroup
    \ifnum#2>\@m \else \@tocwrite{#1}{#8}\fi
  \else
  \def\@svsechd{#6\hskip #3\@svsec
    \@ifnotempty{#8}{\ignorespaces#8\unskip
       %\@addpunct.
       }%
    \ifnum#2>\@m \else \@tocwrite{#1}{#8}\fi
  }%
  \fi
  \global\@nobreaktrue
  \@xsect{#5}}
\makeatother

\pagestyle{plain}

\setlength{\parskip}{\baselineskip}
\setlength{\parindent}{12pt}

\setcounter{secnumdepth}{2}

\allowdisplaybreaks[4]
% Commenting/debugging
\let\IG\iffalse
\let\ENDIG\fi

%% Shortcuts
\newcommand{\td}[2]{\dfrac{d #1}{d #2}}
\newcommand{\std}[2]{\dfrac{d^2 #1}{d {#2}^2}}
\newcommand{\ctd}[3]{\dfrac{d^2 #1}{d #2 d #3}}

\newcommand{\pd}[2]{\dfrac{\partial #1}{\partial #2}}
\newcommand{\spd}[2]{\dfrac{\partial^2 #1}{\partial {#2}^2}}
\newcommand{\cpd}[3]{\dfrac{\partial^2 #1}{\partial #2 \partial #3}}

\newcommand{\pdi}[2]{\partial #1/\partial #2}

\newcommand{\LR}{\Leftrightarrow}
\newcommand{\Lg}{\mathcal{L}}
\newcommand{\half}{\tfrac{1}{2}}
\newcommand{\eqp}{\phantom{=}}
\newcommand{\eqs}{\buildrel s \over =}
\newcommand{\me}{\mathrm{e}}
%begin codes for footnotes (symbols)
\makeatletter
\newcommand*{\myfnsymbolsingle}[1]{%
    \ensuremath{%
        \ifcase#1% 0
        \or % 1
        \dagger%   
        \or % 2
        1 
        \or % 3  
        2
        \or % 4   
        3
        \or % 5
        4
        \else % >= 6
        \@ctrerr  
        \fi
    }%   
}      
\makeatother

\newcommand*{\myfnsymbol}[1]{%
  \myfnsymbolsingle{\value{#1}}%
}

% remove upper boundary by multiplying the symbols if needed
\usepackage{alphalph}
\newalphalph{\myfnsymbolmult}[mult]{\myfnsymbolsingle}{}

\renewcommand*{\thefootnote}{%
  \myfnsymbolmult{\value{footnote}}%
}
%end codes for footnotes (symbols) 

Answer 1

那这个呢：

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\setcounter{equation}{2}

\begin{align}
  \mathbf{P}_{t} = \sum_{j = 1}^{\rho} (\mathbf{I} &-
  \mathbf{G}_{0})^{-1} [\mathbf{D}_{j} + \mathbf{D}_{0} (\mathbf{I} -
  \mathbf{B}_{0})^{-1} \mathbf{B}_{j}] \mathbf{Y}_{t - j} \nonumber
  \\
  &+ \sum_{j = 1}^{\rho} (\mathbf{I} - \mathbf{G}_{0})^{-1}
  [\mathbf{G}_{j} + \mathbf{D}_{0} (\mathbf{I} - \mathbf{B}_{0})^{-1}
  \mathbf{C}_{j}] \mathbf{P}_{t - j} \tag{\theequation$'$} \\
  &+ (\mathbf{I} - \mathbf{G}_{0})^{-1} \mathbf{D}_{0} (\mathbf{I} -
  \mathbf{B}_{0})^{-1} \mathbf{A}^{y} \mathbf{v}_{t}^{y} +
  \mathbf{u}_{t}^{p}, \nonumber
\end{align}

\end{document}

只需在第一行\nonumber之前使用命令、在第二行之前使用命令以及在最后一行末尾使用最终命令即可完成，使用环境即可。\\\tag\\\nonumberalign

另外，您可能想考虑使用\smash[b]{...}（如推荐的barbara beeton 在评论中）来收紧方程式并消除相当大的差距。无论您认为哪种方式看起来最好：

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\setcounter{equation}{2}

\begin{align}
  \mathbf{P}_{t} = \smash[b]{\sum_{j = 1}^{\rho}} (\mathbf{I} &-
  \mathbf{G}_{0})^{-1} [\mathbf{D}_{j} + \mathbf{D}_{0} (\mathbf{I} -
  \mathbf{B}_{0})^{-1} \mathbf{B}_{j}] \mathbf{Y}_{t - j} \nonumber
  \\
  &+ \sum_{j = 1}^{\rho} (\mathbf{I} - \mathbf{G}_{0})^{-1}
  [\mathbf{G}_{j} + \mathbf{D}_{0} (\mathbf{I} - \mathbf{B}_{0})^{-1}
  \mathbf{C}_{j}] \mathbf{P}_{t - j} \tag{\theequation$'$} \\
  &+ (\mathbf{I} - \mathbf{G}_{0})^{-1} \mathbf{D}_{0} (\mathbf{I} -
  \mathbf{B}_{0})^{-1} \mathbf{A}^{y} \mathbf{v}_{t}^{y} +
  \mathbf{u}_{t}^{p}, \nonumber
\end{align}

\end{document}

另一种方法是使用split，通过打碎求和符号并插入垂直幻影来“均衡”线条：

\documentclass{article}
\usepackage{amsmath}

\newcommand{\bv}[1]{\mathbf{#1}}

\begin{document}

\setcounter{equation}{2}

\begin{equation}
\begin{split}
\vphantom{\Bigg|}
\bv{P}_{t} =
  \smash{\sum_{j = 1}^{\rho}}
  (\bv{I} &- \bv{G}_{0})^{-1}
    [\bv{D}_{j} + \bv{D}_{0} (\bv{I} - \bv{B}_{0})^{-1} \bv{B}_{j}] \bv{Y}_{t - j}
\vphantom{\Bigg|}
  \\
  &+ \smash{\sum_{j = 1}^{\rho}} (\bv{I} - \bv{G}_{0})^{-1}
    [\bv{G}_{j} + \bv{D}_{0} (\bv{I} - \bv{B}_{0})^{-1}
    \bv{C}_{j}] \bv{P}_{t - j}
  \\
\vphantom{\Bigg|}
  &+ (\bv{I} - \bv{G}_{0})^{-1} \bv{D}_{0} (\bv{I} -
    \bv{B}_{0})^{-1} \bv{A}_{\vphantom{t}}^{y} \bv{v}_{t}^{y} + \bv{u}_{t}^{p},
\end{split}
\tag{\theequation$'$}
\end{equation}

\end{document}

最后，一些建议，请避免使用\left和\right 除非你需要它们，例如：

• 问题 1，

• 和问题2。

Answer 2

以下是一个例子：

\documentclass[12pt,reqno]{amsart}

\usepackage{newtxmath}

\begin{document}

\setcounter{equation}{1}
\renewcommand\theequation{\arabic{equation}$'$}
\begin{align}
\mathbf {\mathbf{P}}_{t} =& \sum_{j=1}^{\rho } \left ( {\mathbf{I}} - {\mathbf{G}}_{0} \right )^{-1} \left [ {\mathbf{D}}_{j} + {\mathbf{D}}_{0}  \left ( {\mathbf{I}} - {\mathbf{B}}_{0} \right )^{-1} {\mathbf{B}}_{j}\right ] Y_{t-j}\nonumber\\
               &\quad+ \sum_{j=1}^{\rho } \left ( {\mathbf{I}} - {\mathbf{G}}_{0} \right )^{-1} \left [ {\mathbf{G}}_{j} + {\mathbf{D}}_{0}  \left ( {\mathbf{I}} - {\mathbf{B}}_{0} \right )^{-1} {\mathbf{C}}_{j}\right ] {\mathbf{P}}_{t-j} \nonumber \\
               &\quad+ \left ( {\mathbf{I}} - {\mathbf{G}}_{0} \right )^{-1} {\mathbf{D}}_{0} \left ( {\mathbf{I}} - {\mathbf{B}}_{0} \right )^{-1}{\mathbf{A}}^{y} {\mathbf{v}}_{t}^{y} + {\mathbf{u}}^{p}_{t}, \label{eqn:{2}'}
\end{align} 

\end{document}

预览：

Answer 3

那么使用宏来帮助自己怎么样：

\newcomand{\foo}[2][]{\ensuremath{\mathbf{#2}_{#1}}}

用于@Jagath AR 的答案

\begin{align}
  \foo P[t]=\sum_{j=1}^\rho (\foo I-\foo G[0])^{-1} ... \nonumber
  & ... \nonumber
  & ... \foo A^y\foo v[t]^y+\foo u[t]^p
\end{align}

Answer 4

我提出了另外两个解决方案：一个是multline，另一个是align和方程数的对齐之间第二行也是最后一行，如问题图片所示：

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{array, mathtools}

\newtagform{prime}{(}{$'$)}

\begin{document}

\setcounter{equation}{1}
\usetagform{prime}
\begin{multline}
  \mathbf{P}_{t} ={∑_{j = 1}^{ρ}} (\mathbf{I}- \mathbf{G}_{0})⁻¹ [\mathbf{D}_{j} + \mathbf{D}_{0} (\mathbf{I} -
  \mathbf{B}_{0})⁻¹ \mathbf{B}_{j}] \mathbf{Y}_{t - j}
  \\[-3ex]
  + {∑_{j = 1}^{ρ}} (\mathbf{I} - \mathbf{G}_{0})⁻¹
  [\mathbf{G}_{j} + \mathbf{D}_{0} (\mathbf{I} - \mathbf{B}_{0})⁻¹
    \mathbf{C}_{j}] \mathbf{P}_{t - j} \\[-1.5ex]
    + {}(\mathbf{I} - \mathbf{G}_{0})⁻¹ \mathbf{D}_{0} (\mathbf{I} -
    \mathbf{B}_{0})⁻¹ \mathbf{A}^{y} \mathbf{v}_{t}^{y} +   \mathbf{u}_{t}^{p},
  \end{multline}


  \begin{align}
    \mathbf{P}_{t} ={∑_{j = 1}^{ρ}}(\mathbf{I} & - \mathbf{G}_{0})⁻¹ [\mathbf{D}_{j} + \mathbf{D}_{0} (\mathbf{I} -
    \mathbf{B}_{0})⁻¹ \mathbf{B}_{j}] \mathbf{Y}_{t - j} \notag
    \\[-0.5ex]
                                                     & \begin{array}{@{} >{\displaystyle}l@{}}
    {}+ \smash[t]{∑_{j = 1}^{ρ}} (\mathbf{I} - \mathbf{G}_{0})⁻¹
    [\mathbf{G}_{j} + \mathbf{D}_{0} (\mathbf{I} - \mathbf{B}_{0})⁻¹
    \mathbf{C}_{j}] \mathbf{P}_{t - j} \\
    \phantom{\sum}  +  (\mathbf{I} - \mathbf{G}_{0})⁻¹ \mathbf{D}_{0} (\mathbf{I} -
    \mathbf{B}_{0})⁻¹ \mathbf{A}^{y} \mathbf{v}_{t}^{y} +
    \mathbf{u}_{t}^{p},
    \end{array}
  \end{align}

\end{document}