我尝试创建 2 eqnarray
,但这些 2 中的方程式eqnarray
不一致。代码如下:
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{txfonts}
\begin{document}
\begin{eqnarray}
\label{eq1}
&& diag\frac{\partial P}{\partial \theta}=
diag(-diag(V)[Gdiag(V)\sin(Abus)^{T}-Bdiag(V)cos(Abus)^{T}]
\nonumber\\
&& -diag(V).^{2}diag(diag(B)))) \\
\label{eq2}
&& nondiag\frac{\partial P}{\partial \theta}=
VV^{T}.* G .* \sin(Abus) -VV^{T}.*B .* \cos(Abus)
\end{eqnarray}
Replace the diagnol elements of the (\ref{eq2}) with the elements of the
(\ref{eq1}), we will get the complete $\partial P/\partial \theta$
\begin{eqnarray}
\label{eq3}
&& diag\frac{\partial P}{\partial V}=
diag(Gdiag(V)\cos(Abus)^{T}+Bdiag(V)sin(Abus)^{T}
\nonumber\\
&& +diag(V)diag(diag(G)))) \\
\label{eq4}
&& nondiag\frac{\partial P}{\partial V}=
diag(V)G .* \cos(Abus) + diag(V)B .* \sin(Abus)
\end{eqnarray}
Replace the diagnol elements of the (\ref{eq4}) with the elements of the
(\ref{eq3}), we will get the complete $\partial P/\partial V$
\end{document}
结果是: 我们可以看到方程(1)、(2)和方程(3)、(4)的起始位置不同。我该如何解决这个问题?谢谢。
答案1
eqnarray
用结果替换align
并不完全对齐,但我看起来像是,而且我认为比用更好eqnarray
。
根据 Ian 的建议diag
,和nondiag
被声明为数学运算符。括号大小也已调整。
在 Mico 的评论之后,dotstar
已经宣布了一条新命令。我不清楚它的作用是什么。
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{txfonts}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\nondiag}{nondiag}
\newcommand\dotstar{\mathbin{.*}}
\begin{document}
\begin{align}
\label{eq1}
\diag\frac{\partial P}{\partial \theta} & =
\diag\left(-\diag(V)\left[Gdiag(V)\sin(Abus)^{T}-B\diag(V)cos(Abus)^{T}\right]\right.
\nonumber\\
&\quad \left.-\diag(V).^{2}\diag\left(\diag(B)\right)\right) \\
\label{eq2}
\nondiag\frac{\partial P}{\partial \theta} & =
VV^{T}\dotstar G\dotstar\sin(Abus)-VV^{T}\dotstar B\dotstar\cos(Abus)
\end{align}
Replace the diagnol elements of the (\ref{eq2}) with the elements of the
(\ref{eq1}), we will get the complete $\partial P/\partial \theta$
\begin{align}
\label{eq3}
\diag\frac{\partial P}{\partial V} & =
\diag\left(G\diag(V)\cos(Abus)^{T}+B\diag(V)sin(Abus)^{T}\right.
\nonumber\\
&\quad +\Bigl.\diag(V)\diag\left(\diag(G)\right)\Bigr) \\
\label{eq4}
\nondiag\frac{\partial P}{\partial V}&=
\diag(V)G\dotstar\cos(Abus) + \diag(V)B\dotstar\sin(Abus)
\end{align}
Replace the diagnol elements of the (\ref{eq4}) with the elements of the
(\ref{eq3}), we will get the complete $\partial P/\partial V$
\end{document}
但是如果您希望所有方程式都完全对齐,请在命令中插入中间段落\intertext
。
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{txfonts}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\nondiag}{nondiag}
\newcommand\dotstar{\mathbin{.*}}
\begin{document}
\begin{align}
\label{eq1}
\diag\frac{\partial P}{\partial \theta} & =
\diag\left(-\diag(V)\left[Gdiag(V)\sin(Abus)^{T}-B\diag(V)cos(Abus)^{T}\right]\right.
\nonumber\\
&\quad \left.-\diag(V).^{2}\diag\left(\diag(B)\right)\right) \\
\label{eq2}
\nondiag\frac{\partial P}{\partial \theta} & =
VV^{T}\dotstar G\dotstar\sin(Abus)-VV^{T}\dotstar B\dotstar\cos(Abus)
%\end{align}
\intertext{Replace the diagnol elements of the (\ref{eq2}) with the elements of the
(\ref{eq1}), we will get the complete $\partial P/\partial \theta$}
%\begin{align}
\label{eq3}
\diag\frac{\partial P}{\partial V} & =
\diag\left(G\diag(V)\cos(Abus)^{T}+B\diag(V)sin(Abus)^{T}\right.
\nonumber\\
&\quad +\Bigl.\diag(V)\diag\left(\diag(G)\right)\Bigr) \\
\label{eq4}
\nondiag\frac{\partial P}{\partial V}&=
\diag(V)G\dotstar\cos(Abus) + \diag(V)B\dotstar\sin(Abus)
\end{align}
Replace the diagnol elements of the (\ref{eq4}) with the elements of the
(\ref{eq3}), we will get the complete $\partial P/\partial V$
\end{document}
答案2
切勿使用eqnarray
(见eqnarray 与 align)
align
这里与的结合split
似乎是解决问题的最佳方法。
\documentclass{article}
\usepackage{mathtools}
\usepackage{newtxtext,newtxmath}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\nondiag}{nondiag}
\newcommand{\dotstar}{\mathbin{.{*}}}
\newcommand{\Abus}{\mathit{Abus}}
\begin{document}
\begin{align}
\begin{split}
\diag\frac{\partial P}{\partial \theta} & =
\diag\bigl(-\diag(V)[G \diag(V)\sin(\Abus)^{T}-B\diag(V)\cos(\Abus)^{T}]\\
&\qquad -\diag(V)^{2}\diag(\diag(B))\bigr)
\end{split}
\label{eq1} \\
\nondiag\frac{\partial P}{\partial \theta} & =
VV^{T}\dotstar G\dotstar \sin(\Abus)-VV^{T}\dotstar B\cos(\Abus)
\label{eq2}\\
\intertext{%
By replacing the diagonal elements of the (\ref{eq2}) with the elements of
equation~\eqref{eq1}, we will get the complete $\partial P/\partial \theta$}
\begin{split}
\diag\frac{\partial P}{\partial V} & =
\diag\bigl(G\diag(V)\cos(\Abus)^{T}+B\diag(V)\sin(\Abus)^{T} \\
&\qquad +\diag(V)\diag(\diag(G))\bigr)
\end{split}
\label{eq3}\\
\nondiag\frac{\partial P}{\partial V}&=
\diag(V)G\dotstar \cos(\Abus) + \diag(V)B\dotstar \sin(\Abus)
\label{eq4}
\end{align}
By replacing the diagonal elements of equation~\eqref{eq4} with the elements of
equation~\eqref{eq3}, we will get the complete $\partial P/\partial V$
\end{document}
我txfonts
用newtxtext
和替换了newtxmath
(数学字体更好,并且包得到积极维护)。
如果将tbtags
选项添加到amsmath
方程式中,则编号将与零件的底线对齐split
。
我删除了所有内容\left
,但\right
这样做毫无用处(它们只是增加了不必要的水平空间)。只\big
需要几对。
答案3
在一个对齐环境中使用 intertext 可获得对齐:
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{txfonts}
\begin{document}
\begin{align}
\label{eq1}
& diag\frac{\partial P}{\partial \theta}=
diag(-diag(V)[Gdiag(V)\sin(Abus)^{T}-Bdiag(V)cos(Abus)^{T}]
\nonumber\\
& -diag(V).^{2}diag(diag(B)))) \\
\label{eq2}
& nondiag\frac{\partial P}{\partial \theta}=
VV^{T}.* G .* \sin(Abus) -VV^{T}.*B .* \cos(Abus)
\intertext{Replace the diagnol elements of the (\ref{eq2}) with the elements of the
(\ref{eq1}), we will get the complete $\partial P/\partial \theta$}
\label{eq3}
&diag\frac{\partial P}{\partial V}=
diag(Gdiag(V)\cos(Abus)^{T}+Bdiag(V)sin(Abus)^{T}
\nonumber\\
& +diag(V)diag(diag(G)))) \\
\label{eq4}
& nondiag\frac{\partial P}{\partial V}=
diag(V)G .* \cos(Abus) + diag(V)B .* \sin(Abus)
\end{align}
Replace the diagnol elements of the (\ref{eq4}) with the elements of the
(\ref{eq3}), we will get the complete $\partial P/\partial V$
\end{document}