我有一组很长的方程式，是在array环境中编写的，但我似乎无法将其拆分，以便将其分散到多个页面上。我可以使用环境align，但我似乎找不到左、中或右选项，而我想在环境中使用它array：\begin{array}{llrcl} ... \end{array}。这是 LaTeX 代码：
\[
\begin{array}{llrcl}
g_1=\dfrac{f_1}{\norm{f_1}}\\
h_2=f_2-(f_2,g_1)g_1, & g_2=\dfrac{h_2}{\norm{h_2}}, & (g_1,g_2) & = & \dfrac{(g_1,f_2-(f_2,g_1)g_1)}{\norm{h_2}}\\
&&& = & \dfrac{(f_2,g_1)-(f_2,g_1)(g_1,g_1)}{\norm{h_2}}\\
&&& = & 0\\
h_3^1=f_3-(f_3,g_1)g_1, & g_3^1=\dfrac{h_3^1}{\norm{h_3^1}}, & (g_1,g_3^1) & = & \dfrac{(g_1,f_3-(f_3,g_1)g_1)}{\norm{h_3^1}}\\
&&& = & \dfrac{(f_3,g_1)-(f_3,g_1)(g_1,g_1)}{\norm{h_3^1}}\\
&&& = & 0\\
h_3^2=g_3^1-(g_3^1,g_2)g_2, & g_3=g_3^2=\dfrac{h_3^2}{\norm{h_3^2}}, & (g_1,g_3^2) & = & \dfrac{(g_1,g_3^1-(g_3^1,g_2)g_2)}{\norm{h_3^2}}\\
&&& = & \dfrac{(g_1,g_3^1)-(g_2,g_3^1)(g_1,g_2)}{\norm{h_3^2}}\\
&&& = & 0\\
&& (g_2,g_3^2) & = & \dfrac{(g_2,g_3^1-(g_3^1,g_2)g_2)}{\norm{h_3^2}}\\
&&& = & \dfrac{(g_2,g_3^1)-(g_2,g_3^1)(g_2,g_2)}{\norm{h_3^2}}\\
&&& = & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots\\
h_j^1=f_j-(f_j,g_1)g_1, & g_j^1=\dfrac{h_j^1}{\norm{h_j^1}}, & (g_1,g_j^1) & = & \dfrac{(g_1,f_j-(f_j,g_1)g_1)}{\norm{h_j^1}}\\
&&& = & \dfrac{(f_j,g_1)-(f_j,g_1)(g_1,g_1)}{\norm{h_j^1}}\\
&&& = & 0\\
h_j^2=g_j^1-(g_j^1,g_2)g_2, & g_j^2=\dfrac{h_j^2}{\norm{h_j^2}}, & (g_1,g_j^2) & = & \dfrac{(g_1,g_j^1-(g_j^1,g_2)g_2)}{\norm{h_j^2}}\\
&&& = & \dfrac{(g_1,g_j^1)-(g_j^1,g_2)(g_1,g_2)}{\norm{h_j^2}}\\
&&& = & 0\\
&& (g_2,g_j^2) & = & \dfrac{(g_2,g_j^1-(g_j^1,g_2)g_2)}{\norm{h_j^2}}\\
&&& = & \dfrac{(g_2,g_j^1)-(g_j^1,g_2)(g_2,g_2)}{\norm{h_j^2}}\\
&&& = & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots\\
h_j^{j-1}=g_j^{j-2}-(g_j^{j-2},g_{j-1})g_{j-1}, & g_j=g_j^{j-1}=\dfrac{h_j^{j-1}}{\norm{h_j^{j-1}}}, & (g_k,g_j) & = & \dfrac{(g_k,g_j^{j-2}-(g_j^{j-2},g_{j-1})g_{j-1})}{\norm{h_j^{j-1}}}\\
&&& = & \dfrac{\!\!\!\!\!\overbrace{(g_k,g_j^{j-2})}^{\delta_{k,j-1}(g_k,g_j^{j-2})}\!\!\!\!\!-(g_j^{j-2}\!\!\!\!\!,g_{j-1})\overbrace{(g_k,g_{j-1})}^{=\delta_{k,j-1}}}{\norm{h_j^{j-1}}}\\
&&& = & 0
\end{array}
\]
源自\norm{...}自定义命令
\newcommand{\norm}[1]{\ensuremath{\left|\left|{#1}\right|\right|}} % General norm.
答案1
您可以使用，通过在结尾前align*添加 来分隔。但对于第三列，最好使用内部环境，这样这些部分就不会跨页分隔。\displaybreak\\aligned
\documentclass{article}
\usepackage{amsmath,mathtools}
% this is much better for defining \norm
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\begin{document}
\begin{align*}
g_1&=\dfrac{f_1}{\norm{f_1}}
\\
h_2&=f_2-(f_2,g_1)g_1,
& g_2&=\dfrac{h_2}{\norm{h_2}},
&& \!\begin{aligned}[t]
(g_1,g_2)
&= \dfrac{(g_1,f_2-(f_2,g_1)g_1)}{\norm{h_2}} \\
&= \dfrac{(f_2,g_1)-(f_2,g_1)(g_1,g_1)}{\norm{h_2}}\\
&= 0
\end{aligned}
\\
h_3^1&=f_3-(f_3,g_1)g_1,
& g_3^1&=\dfrac{h_3^1}{\norm{h_3^1}},
&& \!\begin{aligned}[t]
(g_1,g_3^1)
&= \dfrac{(g_1,f_3-(f_3,g_1)g_1)}{\norm{h_3^1}}\\
&= \dfrac{(f_3,g_1)-(f_3,g_1)(g_1,g_1)}{\norm{h_3^1}}\\
&= 0
\end{aligned}
\\
h_3^2&=g_3^1-(g_3^1,g_2)g_2,
& g_3&=g_3^2=\dfrac{h_3^2}{\norm{h_3^2}},
&& \!\begin{aligned}[t]
(g_1,g_3^2)
&= \dfrac{(g_1,g_3^1-(g_3^1,g_2)g_2)}{\norm{h_3^2}}\\
&= \dfrac{(g_1,g_3^1)-(g_2,g_3^1)(g_1,g_2)}{\norm{h_3^2}}\\
&= 0
\end{aligned}
\end{align*}
\end{document}
