我有两个方程式。一个大,一个小。我想将它们对齐,并希望这两个方程式只有一个方程式编号。另一个问题是:有没有什么方法可以让这些方程式看起来更吸引人?
\documentclass{article}
\usepackage{amsmath,mathrsfs,xcolor,mathtools}
\begin{document}
\begin{equation}\label{eq:lmlt}
\begin{gathered}
t_0 = 0, \quad s_0 = t_0+t_1, \quad t_{n+1} = s_n + \dfrac{M(1+M(s_n-t_n))
(s_n-t_n)^2}{2(1-M_0t_n)^5},\\
s_{n+m} = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} +
\dfrac{13L(s_n-t_n)^4}{108} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}\\
+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big]
\end{gathered}
\end{equation}
\end{document}
答案1
也许这是你第一个问题的答案
它使用aligned
环境 - 请注意,使用\big[
没有给出正确的大小[]
;我们将在下面的下一个解决方案中解决这个问题。
\subsection*{Original}
\begin{equation}\label{eq:lmlt}
\begin{gathered}
t_0 = 0, \quad s_0 = t_0+t_1\\
\begin{aligned}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108} \\
& \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big]
\end{aligned}
\end{gathered}
\end{equation}
有很多有多种不同的方法来呈现这些方程式,因此这将非常主观。这里有一个替代方案,它并没有太大的不同,只是将初始迭代和后续迭代分开,这样可能更容易阅读。
请注意,此解决方案使用\left[ ... \right.
和\left. ... \right]
来获得正确的尺寸[ ]
;您应该在所使用的任何解决方案中实现这一点。
\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
t_0 = 0, \qquad s_0 = t_0+t_1
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right. \\
& \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right]
\end{align*}
这是完整的 MWE-请注意,mathtools
包已加载amsmath
,因此amsmath
如果您加载mathtools
\documentclass{article}
\usepackage{xcolor,mathtools}
\begin{document}
\subsection*{Original}
\begin{equation}\label{eq:lmlt}
\begin{gathered}
t_0 = 0, \quad s_0 = t_0+t_1\\
\begin{aligned}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\big[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108} \\
& \phantom{=} +\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)} +\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\big]
\end{aligned}
\end{gathered}
\end{equation}
\subsection*{Alternative}
The intial values of $t$ and $s$ are defined by
\begin{equation*}
t_0 = 0, \qquad s_0 = t_0+t_1
\end{equation*}
with subsequent iterations following the formulas
\begin{align*}
t_{n+1} & = s_n + \dfrac{M(1+M(s_n-t_n)) (s_n-t_n)^2}{2(1-M_0t_n)^5}, \\
s_{n+m} & = t_{n+m-1}+ \dfrac{1}{1-M_0t_{n+1}}\left[\dfrac{M(t_{n+1}-s_n)^2}{2} + \dfrac{13L(s_n-t_n)^4}{108}\right. \\
& \phantom{=}+\left.\dfrac{{\color{red}N}M(s_n-t_n)^4}{9(1-M_0t_n)}+\dfrac{M^3(s_n-t_n)^4}{3(1-M_0t_n)^2}\right]
\end{align*}
\end{document}