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我有下一份文件:
\documentclass[12pt]{article}
\usepackage[english, ukrainian]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{geometry}
\geometry{
paper=a5paper,
inner=0cm, outer=0cm, top=0cm, bottom=0cm,
bindingoffset=0cm,
textwidth=\textwidth, textheight=\textheight,
includehead=false, includefoot=false, landscape, twoside}
\usepackage{amsmath}
\usepackage{booktabs}
\newcommand{\cheattitle}[1]{%
{\sf \slshape \center \hspace{\fill}#1\hspace{\fill}}%
}
\begin{document}
\smallj
\cheattitle{Основні правила диференціювання}
\begin{align*}
&(u+v)'=u'+v' & & (u-v)'=u'-v' & & (uv)'=u'v+uv' & & (Cu)'=Cu'\\
& (uvw)'=u'vw+uv'w+uvw' & & \left(\frac{u}{v}\right)' =\frac{u'v-uv'}{v^2},\ v\neq 0 & & \left(\frac{C}{v}\right)'=-C\frac{v'}{v^2},\ v\neq 0 & & \left(\frac{u}{C}\right)'=\frac{1}{C}u'\\
\end{align*}
\cheattitle{Таблиця похідних}
\begin{align*}
& C'=0 & & (u^n)'=nu^{n-1}u' & & \left(\sqrt{u}\right)=\frac{1}{2\sqrt{u}}u' & & \left(\frac{1}{u}\right)'=-\frac{1}{u^2}u' & & \left(a^u\right)'=a^u\ln a\cdot u' \\
& (e^u)'=e^uu' & & \left(\log_au\right)'=\frac{1}{u\ln a} u' & & \left(\ln u\right)'=\frac{1}{u}u' & & \left(\sin u\right)'=\cos u\cdot u' & & \left(\cos u\right)'=-\sin u\cdot u'\\
& \left(\tg u\right)' = \frac{1}{\cos^2u}u' & &\left(\ctg u\right)' = \frac{1}{\sin^2u}u' & & \left(\arcsin u\right)' = \frac{1}{\sqrt{1-u^2}}u' & & \left(\arccos u\right)'= -\frac{1}{\sqrt{1-u^2}} & & \left(\arcctg u\right)' = \frac{1}{i+u^2}u'\\
& \left(\arcctg u\right)'=-\frac{1}{1+u^2}u' & & \left(\sh u\right)' = \ch u \cdot u' & & \left(\ch u\right)' = \sh u\cdot u' & & \left(\th u\right)' = \frac{1}{\ch^2u}u' & & \left(\cth u\right)' = -\frac{1}{\sh^2u}u'
\end{align*}
\cheattitle{Формула Тейлора}
\begin{align*}
f(x) = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{n}(0)}{n!}x^n+\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}
\end{align*}
\end{document}
\end{document}
排版后,它会产生:
问题是我无法删除“Таблиця похідних”和“Формула Тейлора”之前的垂直空格。我正在编写数学小抄,希望它尽可能紧凑。
提前致谢。
答案1
不要使用\center
,而是\centering
,并使显示前后的空格为零。由于您使用align
,\noindent
因此是必要的。不要在行后留下空行\cheattitle
。
另外,不要留下尾随\\
,align
这会在显示中产生一个空白行。
我建议amsart
直接支持8pt
。
\documentclass[8pt]{amsart}
\usepackage[T1,T2A]{fontenc}
\usepackage[english, ukrainian]{babel}
\usepackage{geometry}
\geometry{
paper=a5paper,
inner=0cm, outer=0cm, top=0cm, bottom=0cm,
bindingoffset=0cm,
includehead=false,
includefoot=false,
landscape,
twoside
}
\usepackage{amsmath}
\usepackage{booktabs}
\newcommand{\cheattitle}[1]{%
\par{\sffamily\slshape\centering#1\par}\noindent\ignorespaces
}
\begin{document}
\setlength{\abovedisplayskip}{0pt}
\setlength{\abovedisplayshortskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\setlength{\belowdisplayshortskip}{0pt}
\cheattitle{Основні правила диференціювання}
\begin{align*}
&(u+v)'=u'+v' && (u-v)'=u'-v' & & (uv)'=u'v+uv' & & (Cu)'=Cu'\\
& (uvw)'=u'vw+uv'w+uvw' & & \left(\frac{u}{v}\right)' =\frac{u'v-uv'}{v^2},\ v\neq 0 & & \left(\frac{C}{v}\right)'=-C\frac{v'}{v^2},\ v\neq 0 & & \left(\frac{u}{C}\right)'=\frac{1}{C}u'
\end{align*}
\cheattitle{Таблиця похідних}
\begin{align*}
& C'=0 & & (u^n)'=nu^{n-1}u' & & \left(\sqrt{u}\right)=\frac{1}{2\sqrt{u}}u' & & \left(\frac{1}{u}\right)'=-\frac{1}{u^2}u' & & \left(a^u\right)'=a^u\ln a\cdot u' \\
& (e^u)'=e^uu' & & \left(\log_au\right)'=\frac{1}{u\ln a} u' & & \left(\ln u\right)'=\frac{1}{u}u' & & \left(\sin u\right)'=\cos u\cdot u' & & \left(\cos u\right)'=-\sin u\cdot u'\\
& \left(\tg u\right)' = \frac{1}{\cos^2u}u' & &\left(\ctg u\right)' = \frac{1}{\sin^2u}u' & & \left(\arcsin u\right)' = \frac{1}{\sqrt{1-u^2}}u' & & \left(\arccos u\right)'= -\frac{1}{\sqrt{1-u^2}} & & \left(\arcctg u\right)' = \frac{1}{i+u^2}u'\\
& \left(\arcctg u\right)'=-\frac{1}{1+u^2}u' & & \left(\sh u\right)' = \ch u \cdot u' & & \left(\ch u\right)' = \sh u\cdot u' & & \left(\th u\right)' = \frac{1}{\ch^2u}u' & & \left(\cth u\right)' = -\frac{1}{\sh^2u}u'
\end{align*}
\cheattitle{Формула Тейлора}
\begin{align*}
f(x) = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{n}(0)}{n!}x^n+\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}
\end{align*}
\end{document}
或者,article
使用fontsize
。
\documentclass{article}
\usepackage[T1,T2A]{fontenc}
\usepackage[english, ukrainian]{babel}
\usepackage[fontsize=8pt]{fontsize}
\usepackage{geometry}
\geometry{
paper=a5paper,
inner=0cm, outer=0cm, top=0cm, bottom=0cm,
bindingoffset=0cm,
includehead=false,
includefoot=false,
landscape,
twoside
}
\usepackage{amsmath}
\usepackage{booktabs}
\newcommand{\cheattitle}[1]{%
\par{\sffamily\slshape\centering#1\par}\noindent\ignorespaces
}
\begin{document}
\setlength{\abovedisplayskip}{0pt}
\setlength{\abovedisplayshortskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\setlength{\belowdisplayshortskip}{0pt}
\cheattitle{Основні правила диференціювання}
\begin{align*}
&(u+v)'=u'+v' && (u-v)'=u'-v' & & (uv)'=u'v+uv' & & (Cu)'=Cu'\\
& (uvw)'=u'vw+uv'w+uvw' & & \left(\frac{u}{v}\right)' =\frac{u'v-uv'}{v^2},\ v\neq 0 & & \left(\frac{C}{v}\right)'=-C\frac{v'}{v^2},\ v\neq 0 & & \left(\frac{u}{C}\right)'=\frac{1}{C}u'
\end{align*}
\cheattitle{Таблиця похідних}
\begin{align*}
& C'=0 & & (u^n)'=nu^{n-1}u' & & \left(\sqrt{u}\right)=\frac{1}{2\sqrt{u}}u' & & \left(\frac{1}{u}\right)'=-\frac{1}{u^2}u' & & \left(a^u\right)'=a^u\ln a\cdot u' \\
& (e^u)'=e^uu' & & \left(\log_au\right)'=\frac{1}{u\ln a} u' & & \left(\ln u\right)'=\frac{1}{u}u' & & \left(\sin u\right)'=\cos u\cdot u' & & \left(\cos u\right)'=-\sin u\cdot u'\\
& \left(\tg u\right)' = \frac{1}{\cos^2u}u' & &\left(\ctg u\right)' = \frac{1}{\sin^2u}u' & & \left(\arcsin u\right)' = \frac{1}{\sqrt{1-u^2}}u' & & \left(\arccos u\right)'= -\frac{1}{\sqrt{1-u^2}} & & \left(\arcctg u\right)' = \frac{1}{i+u^2}u'\\
& \left(\arcctg u\right)'=-\frac{1}{1+u^2}u' & & \left(\sh u\right)' = \ch u \cdot u' & & \left(\ch u\right)' = \sh u\cdot u' & & \left(\th u\right)' = \frac{1}{\ch^2u}u' & & \left(\cth u\right)' = -\frac{1}{\sh^2u}u'
\end{align*}
\cheattitle{Формула Тейлора}
\begin{align*}
f(x) = f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\dots+\frac{f^{n}(0)}{n!}x^n+\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}
\end{align*}
\end{document}
不要添加textwidth=\textwidth
和textheight=\textheight
:它们只会导致烦人的警告。
还请注意,这\usepackage[utf8]{inputenc}
不再是必要的,您应该加载(最后)西里尔文的 T2A 编码。
最后,\sf
已经被弃用大约 30 年了。
答案2
你可以使用
\newcommand{\cheattitle}[1]{%
{\sf \slshape \center \vspace{-5mm} \hspace{\fill}#1\hspace{\fill}}%
}
以获得更紧密的间距。调整值,但 -5 毫米似乎是最大值。