请问怎样在latex如下环境下插入公式编号啊!
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请问怎样在latex如下环境下插入公式编号啊!
\begin{align*}
& \frac{1}{2}\zeta \left( 2 \right)\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}}}} - \ln 2\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} - \ln 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}}}{{{n^3}}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}{L_n}\left( 1 \right)}}{{{n^3}}}} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{H_n^2 + {\zeta _n}\left( 2 \right)}}{{{n^3}}}} {\left( { - 1} \right)^{n - 1}} + \frac{3}{4}\zeta \left( 2 \right)\zeta \left( 3 \right) - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}}}{{{n^4}}}} \\
\end{align*}
\begin{align*}
& \frac{1}{2}\zeta \left( 2 \right)\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}}}} - \ln 2\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} - \ln 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}}}{{{n^3}}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}{L_n}\left( 1 \right)}}{{{n^3}}}} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{H_n^2 + {\zeta _n}\left( 2 \right)}}{{{n^3}}}} {\left( { - 1} \right)^{n - 1}} + \frac{3}{4}\zeta \left( 2 \right)\zeta \left( 3 \right) - \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}{H_n}}}{{{n^4}}}} \\
\end{align*}
![请问怎样在latex如下环境下插入公式编号啊!](/uploads/image/z/8853878-38-8.jpg?t=%E8%AF%B7%E9%97%AE%E6%80%8E%E6%A0%B7%E5%9C%A8latex%E5%A6%82%E4%B8%8B%E7%8E%AF%E5%A2%83%E4%B8%8B%E6%8F%92%E5%85%A5%E5%85%AC%E5%BC%8F%E7%BC%96%E5%8F%B7%E5%95%8A%21)
\begin{align}
\end{align}
不带星号的命令显示公式编号,带星号则不显示.
类似的如\begin{equation}
\end{equation}环境
\end{align}
不带星号的命令显示公式编号,带星号则不显示.
类似的如\begin{equation}
\end{equation}环境