在单独的 \begin[split] 函数内对齐方程

Question

我正在尝试在对齐环境内包含几个拆分环境，我想知道是否有任何方法可以将拆分环境内的方程与它们外部的方程对齐？

\begin{align}
    y_{1}^{'} &= -\mu x^{\lambda - 1}sin(\mu \ln(x))+ x^{\lambda - 1} cos(\mu \ln(x))\\
    y_{2}^{'} &= \mu x^{\lambda - 1}cos(\mu \ln(x))+ x^{\lambda - 1} sin(\mu \ln(x)) \\ 
\begin{split}
    W(y_{1},y_{2}) &= (\mu x^{2\lambda - 1} cos^{2}(\mu \ln(x)) + x^{2\lambda -1} sin(\mu 
\ln(x))cos(\mu \ln(x))) \\
    & - (-\mu x^{2\lambda - 1}sin^{2}(\mu \ln(x))+x^{2\lambda - 1}cos(\mu \ln(x))sin(\mu \ln(x))) \\
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[(\mu  cos^{2}(\mu \ln(x)) + sin(\mu \ln(x))cos(\mu \ln(x))) \\
    & - (-\mu sin^{2}(\mu \ln(x))+cos(\mu \ln(x))sin(\mu \ln(x)))] \\
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[\mu  cos^{2}(\mu \ln(x)) + sin(\mu \ln(x))cos(\mu \ln(x)) \\
    & + \mu sin^{2}(\mu \ln(x)) - cos(\mu \ln(x))sin(\mu \ln(x)))] \\
\end{split}\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu sin^{2}(\mu \ln(x)) + \mu cos^{2}(\mu \ln(x))]\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu (sin^{2}(\mu \ln(x)) + cos^{2}(\mu \ln(x)))]\\
    W(y_{1},y_{2}) &= \mu x^{2\lambda - 1}
\end{align}

Answer 1

无需在环境split中嵌入环境align。只需一个align环境以及合理放置的\notag说明\quad即可完成格式化工作。

请注意，我还分别将 (a) 的两个实例替换^{'}为'和 (b) 和 的所有实例替换为sin和。此外，无需在等式 (4) 至 (8) 的左侧重复。cos\sin\cosW(y_{1},y_{2})

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{align}
    y_{1}' &= -\mu x^{\lambda-1}\sin(\mu \ln(x))+ x^{\lambda-1} \cos(\mu \ln(x))\\
    y_{2}' &=  \mu x^{\lambda-1}\cos(\mu \ln(x))+ x^{\lambda-1} \sin(\mu \ln(x)) \\ 
    W(y_{1},y_{2}) 
    &= (\mu x^{2\lambda-1} \cos^{2}(\mu \ln(x)) + x^{2\lambda-1} \sin(\mu \ln(x))\cos(\mu \ln(x))) \notag\\
    &\quad- (-\mu x^{2\lambda-1}\sin^{2}(\mu \ln(x))+x^{2\lambda-1}\cos(\mu \ln(x))\sin(\mu \ln(x))) \\
    %W(y_{1},y_{2}) 
    &= x^{2\lambda-1}[(\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x))) \notag \\
    &\quad-(-\mu \sin^{2}(\mu \ln(x))+cos(\mu \ln(x))\sin(\mu \ln(x)))]\\
    %W(y_{1},y_{2}) 
    &= x^{2\lambda-1}[\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x)) \notag\\
    &\quad + \mu \sin^{2}(\mu \ln(x))-\cos(\mu \ln(x))\sin(\mu \ln(x)))] \\
    %W(y_{1},y_{2}) 
    &= x^{2\lambda-1} [\mu \sin^{2}(\mu \ln(x)) + \mu \cos^{2}(\mu \ln(x))]\\
    %W(y_{1},y_{2}) 
    &= x^{2\lambda-1} [\mu (\sin^{2}(\mu \ln(x)) + \cos^{2}(\mu \ln(x)))]\\
    %W(y_{1},y_{2}) 
    &= \mu x^{2\lambda-1} \,.
\end{align}

\end{document}

Answer 2

\\在amsmath对齐环境中切勿出现尾随。

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align}
    y_{1}' &= -\mu x^{\lambda - 1}\sin(\mu \ln(x))+ x^{\lambda - 1} \cos(\mu \ln(x))\\
    y_{2}' &= \mu x^{\lambda - 1}\cos(\mu \ln(x))+ x^{\lambda - 1} \sin(\mu \ln(x)) \\
\begin{split}
    W(y_{1},y_{2}) &= (\mu x^{2\lambda - 1} \cos^{2}(\mu \ln(x)) + x^{2\lambda -1} \sin(\mu 
\ln(x))\cos(\mu \ln(x))) \\
    & - (-\mu x^{2\lambda - 1}\sin^{2}(\mu \ln(x))+x^{2\lambda - 1}\cos(\mu \ln(x))\sin(\mu \ln(x)))
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[(\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x))) \\
    & - (-\mu \sin^{2}(\mu \ln(x))+\cos(\mu \ln(x))\sin(\mu \ln(x)))]
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x)) \\
    & + \mu \sin^{2}(\mu \ln(x)) - \cos(\mu \ln(x))\sin(\mu \ln(x)))]
\end{split}\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu \sin^{2}(\mu \ln(x)) + \mu \cos^{2}(\mu \ln(x))]\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu (\sin^{2}(\mu \ln(x)) + \cos^{2}(\mu \ln(x)))]\\
    W(y_{1},y_{2}) &= \mu x^{2\lambda - 1}
\end{align}

\end{document}

我固定一切sin和cos为\sin和\cos；也^{'}应该是公正的'。

Answer 3

split总是与外部对齐对齐，但虚假的尾随\\使其混淆

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{align}
    y_{1}' &= -\mu x^{\lambda - 1}\sin(\mu \ln(x))+ x^{\lambda - 1} \cos(\mu \ln(x))\\
    y_{2}' &= \mu x^{\lambda - 1}\cos(\mu \ln(x))+ x^{\lambda - 1} \sin(\mu \ln(x)) \\ 
\begin{split}
    W(y_{1},y_{2}) &= (\mu x^{2\lambda - 1} \cos^{2}(\mu \ln(x)) + x^{2\lambda -1} \sin(\mu 
\ln(x))\cos(\mu \ln(x))) \\
    & - (-\mu x^{2\lambda - 1}\sin^{2}(\mu \ln(x))+x^{2\lambda - 1}\cos(\mu \ln(x))\sin(\mu \ln(x)))
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[(\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x))) \\
    & - (-\mu \sin^{2}(\mu \ln(x))+\cos(\mu \ln(x))\sin(\mu \ln(x)))]
\end{split}\\
\begin{split}
    W(y_{1},y_{2}) &= x^{2\lambda - 1}[\mu  \cos^{2}(\mu \ln(x)) + \sin(\mu \ln(x))\cos(\mu \ln(x)) \\
    & + \mu \sin^{2}(\mu \ln(x)) - \cos(\mu \ln(x))\sin(\mu \ln(x)))]
\end{split}\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu \sin^{2}(\mu \ln(x)) + \mu \cos^{2}(\mu \ln(x))]\\
    W(y_{1},y_{2}) &= x^{2\lambda - 1} [\mu (\sin^{2}(\mu \ln(x)) + \cos^{2}(\mu \ln(x)))]\\
    W(y_{1},y_{2}) &= \mu x^{2\lambda - 1}
\end{align}
\end{document}