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为了看起来更清楚，怎样写这个等式最好？

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为了看起来更清楚，怎样写这个等式最好？

下列等式的最佳书写方式是什么：

\begin{flalign} 
\begin{aligned}
 &\qquad\qquad\frac{\partial^2 \ell}{\partial \bm{\phi}\partial\bm{\phi}'}=\\
 &\sum_{i=1}^{N-N^*} \Bigg(\bigg[\sum_{j=2}^{d_i-1} -\;\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}{(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}
 \mathbf{\tilde{Y}}_{d_i} \mathbf{\tilde{Y}}'_{d_i}\bigg]\\
 &-\bigg[\sum_{j=2}^{d_i-1} \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j+\tfrac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
  \mathbf{\tilde{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \mathbf{\tilde{Y}}'_{d_i} \bigg]\Bigg)\\
 &-\sum_{i=1}^{N^*} \sum_{j=2}^{n_i}\bigg[\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j\bigg]
 \end{aligned}
 &&
 \end{flalign}

下面是我按照上面的代码编写的图像

在此处输入图片描述

我也尝试使用，multiline但也没有得到很好的效果。

我使用更改代码footnotesize，新代码是

{\footnotesize
    \begin{flalign}  \label{18}
\begin{aligned}
 \frac{\partial^2 \ell}{\partial \bm{\phi}\partial\bm{\phi}'}&=\sum_{i=1}^{N-N^*} \Bigg(\bigg[\sum_{j=2}^{d_i-1} -\;\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}{(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}
 \mathbf{\tilde{Y}}_{d_i} \mathbf{\tilde{Y}}'_{d_i}\bigg]\\
 &\qquad-\bigg[\sum_{j=2}^{d_i-1} \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j+\tfrac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
  \mathbf{\tilde{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \mathbf{\tilde{Y}}'_{d_i} \bigg]\Bigg)\\
 &\qquad-\sum_{i=1}^{N^*} \sum_{j=2}^{n_i}\bigg[\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j\bigg]
 \end{aligned}
 &&
 \end{flalign}
}

现在看起来是这样的： 在此处输入图片描述

答案1

写六行。我将一些括号的位置从求和符号之前移到了之后，并缩小了尺寸。

请注意，它应该是\bigl左分隔符和右分隔符（具有、和 的\bigr明显变化）。BigbiggBigg

\documentclass{article}
\usepackage{amsmath,bm}

\begin{document}

\begin{flalign}
\begin{aligned}
\frac{\partial^2 \ell}{\partial \bm{\phi}\,\partial\bm{\phi}'}={}&
   \begin{aligned}[t]
   \sum_{i=1}^{N-N^*}
   \biggl(\,\sum_{j=2}^{d_i-1}\Bigl[
     & -\frac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}
              {(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}
        \tilde{\mathbf{Y}}_j \tilde{\mathbf{Y}}'_j
     \\
     & +\frac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}
             {(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}
       \tilde{\mathbf{Y}}_{d_i} \tilde{\mathbf{Y}}'_{d_i}\Bigr]
  \\
    -\sum_{j=2}^{d_i-1}\Bigl[
     & \frac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
       \tilde{\mathbf{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \tilde{\mathbf{Y}}'_j
     \\
     &  +\frac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
         \tilde{\mathbf{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \tilde{\mathbf{Y}}'_{d_i} \Bigr]\biggr)
   \end{aligned}
\\
 &\begin{aligned}[t]
    -\sum_{i=1}^{N^*} \sum_{j=2}^{n_i}\Bigl[
    & \frac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}
            {(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}
      \tilde{\mathbf{Y}}_j \tilde{\mathbf{Y}}'_j
    \\
    & + \frac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
        \tilde{\mathbf{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \tilde{\mathbf{Y}}'_j\Bigr]
  \end{aligned}
\end{aligned}
&&
\end{flalign}

\end{document}

在此处输入图片描述

我把所有都改成了这样，\mathbf{\tilde{Y}}这样\tilde{\mathbf{Y}}在概念上更合理。您可能还想使用\widetilde，但我不建议这样做。

答案2

减小文本大小或使其分成几行。

\begin{flalign}
{\footnotesize 
\begin{aligned}
\frac{\partial^2 \ell}{\partial \bm{\phi}\partial\bm{\phi}'}=
&\sum_{i=1}^{N-N^*} \Bigg(\bigg[\sum_{j=2}^{d_i-1} -\;\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}{(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}\\&
\mathbf{\tilde{Y}}_{d_i} \mathbf{\tilde{Y}}'_{d_i}\bigg]
-\bigg[\sum_{j=2}^{d_i-1} \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
\mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j+\tfrac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
\mathbf{\tilde{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \mathbf{\tilde{Y}}'_{d_i} \bigg]\Bigg)\\
&-\sum_{i=1}^{N^*} \sum_{j=2}^{n_i}\bigg[\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
\mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j\bigg]
\end{aligned}}
\end{flalign}

答案3

在此处输入图片描述

我使代码更加清晰易读，并将 \Bigg( 更改为 \Biggl( 等。另外，我使用了 amsmath 的多行环境，它在这里很适用。

\documentclass{article}

\usepackage{amsmath}
\usepackage{bm}
\usepackage[margin=1in]{geometry}

\newcommand{\bP}{\bm{P}}
\newcommand{\tY}{\mathbf{\tilde Y}}
\begin{document}
\begin{multline} 
\frac{\partial^2 \ell}{\partial \bm{\phi}\partial\bm{\phi}'}=
 \sum_{i=1}^{N-N^*} \Biggl(\biggl[\sum_{j=2}^{d_i-1} -\;\frac{\bP_{ij}'\bP_{ij} (1-\bP_{ij})'}{(1-\bP_{ij})'(1-\bP_{ij})+\bP_{ij}'\bP_{ij}}\tY_j \tY'_j + \frac{\bP_{id_i}'(1-\bP_{id_i})'(1-\bP_{id_i})}{(1-\bP_{id_i})'(1-\bP_{id_i})+\bP_{id_i}'\bP_{id_i}}
 \tY_{d_i} \tY'_{d_i}\biggr]\\
 -\biggl[\sum_{j=2}^{d_i-1} \frac{\bP_{ij}}{(1-\bP_{ij})'}
  \tY_j \bP_{ij}(1-\bP_{ij}) \tY'_j+\frac{(1-\bP_{id_i})}{\bP'_{id_i}}
  \tY_{d_i} \bP_{id_i}(1-\bP_{id_i}) \tY'_{d_i} \biggr]\Biggr)\\
 -\sum_{i=1}^{N^*} \sum_{j=2}^{n_i}\biggl[\frac{\bP_{ij}'\bP_{ij} (1-\bP_{ij})'}{(1-\bP_{ij})'(1-\bP_{ij})+\bP_{ij}'\bP_{ij}}\tY_j \tY'_j + \frac{\bP_{ij}}{(1-\bP_{ij})'}
  \tY_j \bP_{ij}(1-\bP_{ij}) \tY'_j\biggr]
\end{multline}

\end{document}

答案4

我建议使用包geometry。如果您有具有大边距的默认布局，则可以将方程内容嵌套在环境中medsize（nccmath约为以前大小的 80%）：

\documentclass{article}
\usepackage{geometry} 
\usepackage{showframe}
\renewcommand{\ShowFrameLinethickness}{0.3pt}
\usepackage{amsmath, bm, nccmath}
\usepackage{diffcoeff} 

 \begin{document}

 \begin{equation}
\begin{aligned}\diffp[1,1]{\ell}{\bm{\phi},\bm{\phi}'} =
 \sum_{i=1}^{N-N^*} \Bigg( & \bigg[\begin{aligned}[t]\sum_{j=2}^{d_i-1} -\;\tfrac{\bm{P}_{ij}'
\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}{(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}
 \mathbf{\tilde{Y}}_{d_i} \mathbf{\tilde{Y}}'_{d_i} & \bigg]\\
 %&
 - \bigg[\sum_{j=2}^{d_i-1} \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j+\tfrac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
  \mathbf{\tilde{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \mathbf{\tilde{Y}}'_{d_i} &\bigg]\Bigg)
  \end{aligned}\\
% &
-\sum_{i=1}^{N^*} \sum_{j=2}^{n_i} & \bigg[\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j\bigg]
\end{aligned}
 \end{equation}
 \bigskip

 \begin{equation}
\begin{medsize}
\begin{aligned}\diffp[1,1]{\ell}{\bm{\phi},\bm{\phi}'} =
 \sum_{i=1}^{N-N^*} \Biggl( & \Biggl[\begin{aligned}[t]\sum_{j=2}^{d_i-1} -\;\tfrac{\bm{P}_{ij}'
\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{id_i}'(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})}{(1-\bm{P}_{id_i})'(1-\bm{P}_{id_i})+\bm{P}_{id_i}'\bm{P}_{id_i}}
 \mathbf{\tilde{Y}}_{d_i} \mathbf{\tilde{Y}}'_{d_i} & \Biggr]\\
 %&
 - \Biggl[\sum_{j=2}^{d_i-1} \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j+\tfrac{(1-\bm{P}_{id_i})}{\bm{P}'_{id_i}}
  \mathbf{\tilde{Y}}_{d_i} \bm{P}_{id_i}(1-\bm{P}_{id_i}) \mathbf{\tilde{Y}}'_{d_i} &\Biggr]\Biggr)
  \end{aligned}\\
% &
-\sum_{i=1}^{N^*} \sum_{j=2}^{n_i} & \biggl[\tfrac{\bm{P}_{ij}'\bm{P}_{ij} (1-\bm{P}_{ij})'}{(1-\bm{P}_{ij})'(1-\bm{P}_{ij})+\bm{P}_{ij}'\bm{P}_{ij}}\mathbf{\tilde{Y}}_j \mathbf{\tilde{Y}}'_j + \tfrac{\bm{P}_{ij}}{(1-\bm{P}_{ij})'}
  \mathbf{\tilde{Y}}_j \bm{P}_{ij}(1-\bm{P}_{ij}) \mathbf{\tilde{Y}}'_j\biggr]
\end{aligned}
\end{medsize}
 \end{equation}

\end{document}

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