数学方程式的左对齐

Question

    \begin{Question}{1}
    Part 1) prove $f+g$ is holomorphic at $z_0$.\\
    \\
    We want to show that $(f+g)'(z_0)$ exists:\\
    \\
    $$ (f+g)'(z_0) =\lim_{z \to z_0} \frac{(f+g)(z) - (f+g)(z_0)}{z-z_0}$$\\
    $$ =\lim_{z \to z_0} \frac{f(z)+g(z)- f(z_0)-g(z_0)}{z-z_0}$$\\
    $$=\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0} + \lim_{z \to z_0} 
    \frac{g(z)-g(z_0)}{z-z_0}$$\\
    $$ = f'(z_0) + g'(z_0)$$


    \end{Question}

我尝试使用 \begin{aligned} \end{aligned} 但它只有一行

现在我的测试看起来像这样

我怎样才能将它们左对齐？我不想要水平对齐！

Answer 1

以下是两个使用aligned环境的解决方案。在第一个解决方案中，多行方程的所有四行都是左对齐的，这似乎是您想要的。在第二个解决方案中，对齐是在符号上=。代码块之间的唯一区别在于&第 1 行中对齐说明符的位置。

（就我个人而言，我更喜欢第二种解决方案。）

\documentclass{article} 
\usepackage{amsmath}

\begin{document}
Prove $f+g$ is holomorphic at $z_0$.

We want to show that $(f+g)'(z_0)$ exists:

\smallskip\noindent
$\begin{aligned}
&(f+g)'(z_0) =\lim_{z \to z_0} \frac{(f+g)(z) - (f+g)(z_0)}{z-z_0}\\
&=\lim_{z \to z_0} \frac{f(z)+g(z)- f(z_0)-g(z_0)}{z-z_0}\\
&=\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0} + \lim_{z \to z_0} 
    \frac{g(z)-g(z_0)}{z-z_0}\\
&= f'(z_0) + g'(z_0)
\end{aligned}$

\bigskip\noindent
$\begin{aligned}
(f+g)'(z_0) &=\lim_{z \to z_0} \frac{(f+g)(z) - (f+g)(z_0)}{z-z_0}\\
&=\lim_{z \to z_0} \frac{f(z)+g(z)- f(z_0)-g(z_0)}{z-z_0}\\
&=\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0} + \lim_{z \to z_0} 
    \frac{g(z)-g(z_0)}{z-z_0}\\
&= f'(z_0) + g'(z_0)
\end{aligned}$
\end{document}

Answer 2

我想，您正在寻找类似这样的东西：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
Part 1) prove $f+g$ is holomorphic at $z_0$.

\bigskip
We want to show that $(f+g)'(z_0)$ exists:
    \begin{align*}
(f+g)'(z_0) 
    & = \lim_{z \to z_0} \frac{(f+g)(z) - (f+g)(z_0)}{z-z_0}    \\
    & = \lim_{z \to z_0} \frac{f(z)+g(z)- f(z_0)-g(z_0)}{z-z_0} \\
    & = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0} 
                + \lim_{z \to z_0} \frac{g(z)-g(z_0)}{z-z_0}    \\
    &  = f'(z_0) + g'(z_0)
    \end{align*}
\end{document}

Answer 3

您有一个fleqn由 定义的环境nccmath，其工作原理类似于subequations：此环境内显示的所有方程式都将左对齐。它接受一个可选参数 - 环境开始的左边距的距离（默认为 0）。这里我选择了 \parindent：

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{amsmath, nccmath}

\begin{document}
Part 1) prove $f+g$ is holomorphic at $z_0$.

\bigskip
We want to show that $(f+g)'(z_0)$ exists:
\begin{fleqn}[\parindent]
\begin{align*}
(f+g)'(z_0)
    & = \lim_{z \to z_0} \frac{(f+g)(z) - (f+g)(z_0)}{z-z_0} \\
    & = \lim_{z \to z_0} \frac{f(z)+g(z)- f(z_0)-g(z_0)}{z-z_0} \\
    & = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z-z_0} + \lim_{z \to z_0} \frac{g(z)-g(z_0)}{z-z_0}\\
    & = f'(z_0) + g'(z_0)
    \end{align*}
\end{fleqn}

\end{document}