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How can I add a second line to the condition in my piecewise function? I'm using the multlined environment, but it's producing quite an ugly result:

enter image description here

Ideally, I would want the two conditions right on top of each other, and without any indentation. The code is pretty ugly, but it's pretty much just:

\begin{cases}
..... & \begin{multlined}
    .....\\
    .....
\end{multlined}\\
..... & \begin{multlined}
    .....\\
    .....
\end{multlined}
\end{cases}

Here's the full code:

\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}
Eagle Bound
  • 1
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 29 '21 at 06:38

1 Answers1

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Instead of using the \begin{multlined}...\end{multlined} environment, I just used the \begin{split}...\end{split} environment. That produced this result:

enter image description here

In case anyone wants to see, here's the full code:

\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}
Eagle Bound
  • 1