如何在分段函数的条件中添加第二行?我正在使用环境multlined
,但它产生的结果相当丑陋:
理想情况下,我希望这两个条件正好位于彼此之上,并且没有任何缩进。代码相当丑陋,但它基本上只是:
\begin{cases}
..... & \begin{multlined}
.....\\
.....
\end{multlined}\\
..... & \begin{multlined}
.....\\
.....
\end{multlined}
\end{cases}
完整代码如下:
\begin{equation}
\forall m \geq a_{a}, s_{a}(m) = \begin{cases}
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{m-1 \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{multlined}
a\equiv 0 \Mod{4},\\
m\equiv 1 \Mod{2}
\end{multlined}\\
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{\tfrac{m}{2} \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{multlined}
a\equiv 0 \Mod{4},\\
m\equiv 0 \Mod{2}
\end{multlined}\\
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{m-1 \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{multlined}
a\equiv 1 \Mod{4},\\
m\equiv 1 \Mod{2}
\end{multlined}\\
0 &\begin{multlined}
a\equiv 1 \Mod{4},\\
m\equiv 0 \Mod{2}
\end{multlined}\\
0 & a > 1 \Mod{4}
\end{cases}
\end{equation}
答案1
\begin{multlined}...\end{multlined}
我没有使用环境,而是直接使用了\begin{split}...\end{split}
环境。结果如下:
如果有人想看,这里是完整的代码:
\begin{equation}
\forall m \geq a_{a}, s_{a}(m) = \begin{cases}
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{m-1 \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{split}
a\equiv 0 \Mod{4},\\
m\equiv 1 \Mod{2}
\end{split}\\[2em]
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{\tfrac{m}{2} \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{split}
a\equiv 0 \Mod{4},\\
m\equiv 0 \Mod{2}
\end{split}\\[2em]
\mathlarger{\sum\limits_{e=0}^{\left\lfloor\tfrac{a}{8}\right\rfloor}{\tfrac{\left\lfloor\tfrac{m}{2}\right\rfloor^{2}-\left\lfloor\tfrac{m}{2}\right\rfloor}{2} \choose e}{m-1 \choose \tfrac{2\left\lfloor\tfrac{a}{8}\right\rfloor-8e}{4}}} &\begin{split}
a\equiv 1 \Mod{4},\\
m\equiv 1 \Mod{2}
\end{split}\\[2em]
0 &\begin{split}
a\equiv 1 \Mod{4},\\
m\equiv 0 \Mod{2}
\end{split}\\[1.5em]
0 & a > 1 \Mod{4}
\end{cases}
\end{equation}