怎样对方程的推导进行对齐？

Question

方程的推导如下所示。有人能建议如何对齐方程，以便所有方程都包含在同一个方程编号下吗？

\begin{multline}\label{eqn-5.9}
s_{p_m}=\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\\
=\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\\
=\Sigma_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\Sigma_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
\end{multline}

当前获得的输出如下。

Answer 1

嗯，您可以使用align。split如果您希望数字居中并且标签用于整个方程式，也可以使用 （但我不推荐这样做）。并且，\sum对于总和，您有 ，无需使用\Sigma：

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
s_{p_m}&=\sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\nonumber\\
&=\sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\nonumber\\
&=\sum_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\sum_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
\end{align}
\end{document}

或者如果你愿意的话，可以采用更小的总数（内联样式）

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align}
s_{p_m}&=\textstyle\sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\nonumber\\
&=\textstyle\sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\nonumber\\
&=\textstyle\sum_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\sum_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
\end{align}
\end{document}

Answer 2

还有一种可能性：

\documentclass{article}
\usepackage{nccmath}

\begin{document}
    \begin{equation}\label{eqn-5.9}
\begin{split}
s_{p_m}
    & = \sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty}
        f_k\exp\Bigl(jk(\phi_{i_m}-\mfrac{2\pi p_m}{P_m}\Bigr) \\
    & = \sum_{i=1}^{I}\alpha_{i_m}\sum_{k=-\infty}^{\infty}
        f_k\exp\Bigl(jk\phi_{i_m} - \mfrac{jk2\pi p_m}{P_m}\Bigr)\\
    & = \sum_{k=-\infty}^{\infty} f_k\exp\Bigl(-j\mfrac{2\pi k p_m}{P_m}\Bigr)
        \sum_{i=1}^{I}\alpha_{i_m}\exp\Bigl(jk\phi_{i_m}\Bigr)
\end{split}
    \end{equation}
\end{document}

Answer 3

这取决于你想要什么。我更喜欢把方程编号放在最后一行：

\begin{align}
  s_{p_m}&=\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\nonumber\\
  &=\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\nonumber\\
  &=\Sigma_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\Sigma_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
  \label{eqn-5.9}
\end{align}

Answer 4

也许这就是您想要的其中之一？

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\begin{equation}\label{eqn-5.9}
\begin{aligned}
s_{p_m} & =\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\\
 & =\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\\
 & =\Sigma_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\Sigma_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
\end{aligned}
\end{equation}

\begin{equation}
\begin{multlined}[0.9\linewidth]
s_{p_m} =\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk(\phi_{i_m}-(2\pi p_m/P_m))}\\
  =\Sigma_{i=1}^{I}\alpha_{i_m}\Sigma_{k=-\infty}^{\infty} f_k e^{jk\phi_{i_m}-jk2\pi p_m/P_m)}\\
  =\Sigma_{k=-\infty}^{\infty} f_k e^{-j(2\pi k p_m/P_m)}\Sigma_{i=1}^{I}\alpha_{i_m}e^{jk\phi_{i_m}}
\end{multlined}
\end{equation}

\end{document}