與兩個標記對齊

Question

如何在以下alignat環境的第一行中的第一個減號處取得第二個對齊標記？我想用它在兩行上寫第二個表達式，對齊西格瑪符號前面的加號。

\documentclass{amsart}
\usepackage{amsmath}


\begin{document}


\begin{alignat*}{2}
f(x_{4}) &= (x_{4} &&- x_{1}) + (x_{4} - x_{2}) + (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{4}) \\
&= \bigl[(x_{4} - x_{3}) + (x_{3} - x_{1})\bigr] + \bigl[(x_{4} - x_{3}) + (x_{3} - x_{2})\bigr] + (x_{4} - x_{3}) \\
&&+ \sum_{i=5}^{8} \bigl[(x_{i} - x_{3}) - (x_{4} - x_{3})\bigr] \\
&= (x_{3} - x_{1}) + (x_{3} - x_{2}) - (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{3}) \\
&< (x_{3} - x_{1}) + (x_{3} - x_{2}) + \sum_{i=4}^{8} (x_{i} - x_{3}) \\
&< \sum_{i=1}^{8} \vert x_{i} - x_{3} \vert \\
&= f(x_{3}) ,
\end{alignat*}

\end{document}

Answer 1

-我會將第二個表達式中的第一個與環境中的+in +(x_4 - x_3)using對齊。alignedalign*

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

\begin{align*}
f(x_{4}) &= (x_{4}-x_{1}) + (x_{4} - x_{2}) + (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{4}) \\
&=\begin{aligned}[t]\bigl[(x_{4} &- x_{3}) + (x_{3} - x_{1})\bigr] + \bigl[(x_{4} - x_{3}) + (x_{3} - x_{2})\bigr] \\
&+(x_{4} - x_{3})+\sum_{i=5}^{8} \bigl[(x_{i} - x_{3}) - (x_{4} - x_{3})\bigr] \end{aligned} \\
&= (x_{3} - x_{1}) + (x_{3} - x_{2}) - (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{3}) \\
&< (x_{3} - x_{1}) + (x_{3} - x_{2}) + \sum_{i=4}^{8} (x_{i} - x_{3}) \\
&< \sum_{i=1}^{8} \vert x_{i} - x_{3} \vert \\
&= f(x_{3}) ,
\end{align*}

\end{document}

Answer 2

這樣的排列？

\documentclass{amsart}

\begin{document}

\begin{align*}
f(x_4) &= (x_4 - x_1) + (x_4 - x_2) + (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_4) \\
&= \bigl[(x_4 - x_3) + (x_3 - x_1)\bigr] + \bigl[(x_4 - x_3) + (x_3 - x_2)\bigr] + (x_4 - x_3) \\
&\quad+ \sum_{i=5}^8 \bigl[(x_i - x_3) - (x_4 - x_3)\bigr] \\
&= (x_3 - x_1) + (x_3 - x_2) - (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_3) \\
&< (x_3 - x_1) + (x_3 - x_2) + \sum_{i=4}^8 (x_i - x_3) \\
&< \sum_{i=1}^8 \vert x_i - x_3 \vert \\
&= f(x_3)
\end{align*}

\end{document}

---

或依照 Thruston 在評論中的建議，用\qquad代替\quad.

---

或者按照 Enrico 的建議，(x_4-x_3)也移到下一行。

\documentclass{amsart}
\begin{document}

\begin{align*}
f(x_4) &= (x_4 - x_1) + (x_4 - x_2) + (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_4) \\
&= \bigl[(x_4 - x_3) + (x_3 - x_1)\bigr] + \bigl[(x_4 - x_3) + (x_3 - x_2)\bigr] \\
&\quad + (x_4 - x_3)+ \sum_{i=5}^8 \bigl[(x_i - x_3) - (x_4 - x_3)\bigr] \\
&= (x_3 - x_1) + (x_3 - x_2) - (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_3) \\
&< (x_3 - x_1) + (x_3 - x_2) + \sum_{i=4}^8 (x_i - x_3) \\
&< \sum_{i=1}^8 \vert x_i - x_3 \vert \\
&= f(x_3) ,
\end{align*}

\end{document}

---

在評論中澄清後再次嘗試：

\documentclass{amsart}
\begin{document}

\begin{align*}
f(x_4) &= (x_4 - x_1) + (x_4 - x_2) + (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_4) \\
&= \bigl[(x_4 - x_3) + (x_3 - x_1)\bigr] + \bigl[(x_4 - x_3) + (x_3 - x_2)\bigr]+ (x_4 - x_3) \\
&\phantom{{}=(x_4} + \sum_{i=5}^8 \bigl[(x_i - x_3) - (x_4 - x_3)\bigr] \\
&= (x_3 - x_1) + (x_3 - x_2) - (x_4 - x_3) + \sum_{i=5}^8 (x_i - x_3) \\
&< (x_3 - x_1) + (x_3 - x_2) + \sum_{i=4}^8 (x_i - x_3) \\
&< \sum_{i=1}^8 \lvert x_i - x_3 \rvert \\
&= f(x_3)
\end{align*}

\end{document}

Answer 3

你是這個意思嗎？

% arara: pdflatex

\documentclass{amsart}

\begin{document}
\begin{align*}
    f(x_{4})&= (x_{4} - x_{1}) + (x_{4} - x_{2}) + (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{4}) \\
            &= \bigl[(x_{4} - x_{3}) + (x_{3} - x_{1})\bigr] + \bigl[(x_{4} - x_{3}) + (x_{3} - x_{2})\bigr] + (x_{4} - x_{3}) \\
            &\hphantom{{}= (x_{4}} + \mathop{\smash[b]{\sum_{i=5}^{8}}} \bigl[(x_{i} - x_{3}) - (x_{4} - x_{3})\bigr] \\
            &= (x_{3} - x_{1}) + (x_{3} - x_{2}) - (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{3}) \\
            &< (x_{3} - x_{1}) + (x_{3} - x_{2}) + \mathop{\smash{\sum_{i=4}^{8}}} (x_{i} - x_{3}) \\
            &< \sum_{i=1}^{8} \lvert x_{i} - x_{3} \rvert \\
            &= f(x_{3}),
\end{align*}
\end{document}

不過，我不建議這樣做，因為我看不出有任何理由說明為什麼要與這一點保持一致。但這看起來就像您在上面嘗試做的那樣。我傾向於對齊第二行的第一個，或者更好的第二個，甚至更好的第三個標誌。或按照上面的建議使用一些\quad或類似的公式，以便在所有公式中保持一致。

Answer 4

我會使用包包multlined中的mathtools：

\documentclass{amsart}
\usepackage{mathtools}

\begin{document}
    \begin{align*}
f(x_{4})
    & = (x_{4} - x_{1}) + (x_{4} - x_{2}) + (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{4}) \\
    & = \!\begin{multlined}[t][0.5\linewidth]
        \bigl[(x_{4} - x_{3}) + (x_{3} - x_{1})\bigr]  
        + \bigl[(x_{4} - x_{3}) + (x_{3} - x_{2})\bigr] \\
        + (x_{4} - x_{3})
        + \sum_{i=5}^{8} \bigl[(x_{i} - x_{3}) - (x_{4} - x_{3})\bigr]
        \end{multlined}   \\
    & = (x_{3} - x_{1}) + (x_{3} - x_{2}) - (x_{4} - x_{3}) + \sum_{i=5}^{8} (x_{i} - x_{3}) \\
    & < (x_{3} - x_{1}) + (x_{3} - x_{2}) + \sum_{i=4}^{8} (x_{i} - x_{3}) \\
    & < \sum_{i=1}^{8} \vert x_{i} - x_{3} \vert \\
    & = f(x_{3}) ,
    \end{align*}
\end{document}