我想像align
这样使用环境:
\begin{itemsize}
\begin{align}
{\color{purple} \mathcal{L} [f'(t)](s) = s \mathcal{L} [f(t)](s) - f(0) } \\
{\color{red} \mathcal{L} [f^{(n)}(t)](s) = s^n \mathcal{L} [f(t)](s) - s^{n-
1}f(0) - s^{n-2}f(0) - \dots - f^{n-1}(0)}
\end{align}
n mal differenzierbar:
\end{itemsize}
输出结果显示"n mal differenzierbar:"
,文本行(本应位于第二个方程后面)位于其下方(我猜是因为长度)。因此,我想将其移至方程上方。
提前谢谢了。
答案1
像这样?
\documentclass{article}
\usepackage{amsmath}
\usepackage{xcolor}
\begin{document}
\begin{align}
{\color{purple} \mathcal{L} [f'(t)](s)\;} & {\color{purple}= s \mathcal{L} [f(t)](s) - f(0) } \\
\intertext{n mal differenzierbar:}
{\color{red} \mathcal{L} [f^{(n)}(t)](s)\;} & {\color{red}= s^n \mathcal{L} [f(t)](s) - s^{n- 1}f(0) - s^{n-2}f(0) - \dots - f^{n-1}(0) }
\end{align}
\end{document}
或这个?
\documentclass{article}
\usepackage{amsmath}
\usepackage{xcolor}
\begin{document}
n mal differenzierbar:
\begin{align}
{\color{purple} \mathcal{L} [f'(t)](s)\;} & {\color{purple}= s \mathcal{L} [f(t)](s) - f(0) } \\
{\color{red} \mathcal{L} [f^{(n)}(t)](s)\;} & {\color{red}= s^n \mathcal{L} [f(t)](s) - s^{n- 1}f(0) - s^{n-2}f(0) - \dots - f^{n-1}(0) }
\end{align}
\end{document}