左辺にたくさんの方程式を入れる方法

Question

次の方程式をテキストの左側に配置したいのですが、どのコマンドが役に立つかわかりません。 align を使用しましたが、すべての方程式が紙の右端に配置されました。より正確には、すべての方程式を互いの下に配置し、それらの位置を紙の左側にしたいのです。

\begin{document}
\begin{align*}
(f,f)_{[x^3]_{1}}  = (f,f)_{[x^3]_{1}}=0 \\
    (f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0\\
    (f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0\\
    (f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0\\
    (f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
    (f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    (g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
\end{align*}
\end{document}

Answer 1

このパッケージは、式をテキストの左の境界に押し出すnncmath環境を提供します。fleqn

\documentclass{article}
\usepackage{nccmath}

%---------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%

\begin{document}

\begin{fleqn}
    \begin{gather*}
(f,f)_{[x^3]_{1}}  = (f,f)_{[x^3]_{1}}=0 \\
(f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0 \\
(f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0 \\
(f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0 \\
(f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
(f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
(g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
    \end{gather*}
\end{fleqn}

\end{document}

（赤い線はテキストの境界線を示します）

Answer 2

以下に、3 つの異なる配置を含む可能性のある提案を示します (赤い線は余白を示します)。

\documentclass{article}
\usepackage{amsmath}


\begin{document}
\begin{flalign*}
(f,f)_{[x^3]_{1}}  &= (f,f)_{[x^3]_{1}}=0 &&\\ 
    (f,g)_{[x^3]_{1}} &=(f,g)_{[x^3]_{3}}=0 \\
    (f,h)_{[xxy]_{1}}&=(f,h)_{[xxy]_{3}}=0 \\
    (f,p)_{[xxy]_{1}}&=(f,p)_{[xxy]_{3}}=0 \\
    (f,m)_{[xxt]_{1}}&=[xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
    (f,m)_{[xxt]_{3}}&=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    (g,f)_{[x^3]_{1}}&=(g,f)_{[x^3]_{3}}=0\\
\end{flalign*}

\begin{align*}
(f,f)_{[x^3]_{1}}  &= (f,f)_{[x^3]_{1}}=0 \\ 
    (f,g)_{[x^3]_{1}} &=(f,g)_{[x^3]_{3}}=0 \\
    (f,h)_{[xxy]_{1}}&=(f,h)_{[xxy]_{3}}=0 \\
    (f,p)_{[xxy]_{1}}&=(f,p)_{[xxy]_{3}}=0 \\
    (f,m)_{[xxt]_{1}}&=[xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
    (f,m)_{[xxt]_{3}}&=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    (g,f)_{[x^3]_{1}}&=(g,f)_{[x^3]_{3}}=0\\
\end{align*}

\begin{flalign*}
    &(f,f)_{[x^3]_{1}}  = (f,f)_{[x^3]_{1}}=0 &&\\ 
    &(f,g)_{[x^3]_{1}} =(f,g)_{[x^3]_{3}}=0 \\
    &(f,h)_{[xxy]_{1}}=(f,h)_{[xxy]_{3}}=0 \\
    &(f,p)_{[xxy]_{1}}=(f,p)_{[xxy]_{3}}=0 \\
    &(f,m)_{[xxt]_{1}}=[xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2})= [xtx]_{1}\\
    &(f,m)_{[xxt]_{3}}=[xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    &(g,f)_{[x^3]_{1}}=(g,f)_{[x^3]_{3}}=0\\
\end{flalign*}

\end{document}

Answer 3

いくつかの方程式を中央揃えにし、他の方程式を左揃えにするちょっとしたバリエーションを追加するだけです。もちろん、@leandriis の回答ほどエレガントではありませんが、役に立つかもしれません。

\documentclass{article}
\usepackage{amsmath}
\usepackage{blindtext}

\begin{document}

\begin{align*}
    & (f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\ 
    & (f,g)_{[x^3]_{1}} = (f,g)_{[x^3]_{3}}=0 \\
    & (f,h)_{[xxy]_{1}} = (f,h)_{[xxy]_{3}}=0 \\
    & (f,p)_{[xxy]_{1}} = (f,p)_{[xxy]_{3}}=0 \\
    & (f,m)_{[xxt]_{1}} = [xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2}) = [xtx]_{1}\\
    & (f,m)_{[xxt]_{3}} = [xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    & (g,f)_{[x^3]_{1}} =(g,f)_{[x^3]_{3}}=0\\
\end{align*}

%some blind text to check that margins are correct
\blindtext

\begin{align*}
    & (f,f)_{[x^3]_{1}} = (f,f)_{[x^3]_{1}}=0 \\ 
    & (f,g)_{[x^3]_{1}} = (f,g)_{[x^3]_{3}}=0 \\
    \noalign{\hspace{-\parindent}$(f,h)_{[xxy]_{1}} = (f,h)_{[xxy]_{3}}=0$}
    \noalign{\hspace{-\parindent}$(f,p)_{[xxy]_{1}} = (f,p)_{[xxy]_{3}}=0$}
    & (f,m)_{[xxt]_{1}} = [xx]_{1} \dashv t  - x \dashv ([xt]_{1} - [tx]_{2}) = [xtx]_{1}\\
    & (f,m)_{[xxt]_{3}} = [xx]_{1}\vdash t - x \vdash ([xt]_{1} - [tx]_{2})=[xxt]_3-[xxt]_2+[xtx]_3\\
    & (g,f)_{[x^3]_{1}} =(g,f)_{[x^3]_{3}}=0\\
\end{align*}

\end{document}

出力は次のようになります。