\documentclass{article}
\usepackage{setspace}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document
\doublespacing{$f(\chi_0)=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]$\\
$f(\chi_1)=\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]$\\
$f(\chi_2)=\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]$\\
$f(\chi_3)=\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]$\\
$f(\chi_4)=\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]$\\}
\smallskip\\
\text{Now we want to calculate $\hat f(a)$ hence:}\\
\begin{align}
\hat f(0)&=f(\chi_0)+f(\chi_1)+f(\chi_2)+f(\chi_3)+f(\chi_4)\\
&=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]+\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]+\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]+\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]+\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]
&=f(0)+\frac{f(1)}{5}[1+a+a^2+a^3+a^4]+\frac{f(2)}{5}[1+a+a^2+a^3+a^4]+\frac{f(3)}{5}[1+a+a^2+a^3+a^4]+\frac{f(4)}{5}[1+a+a^2+a^3+a^4]
\end{align}
我尝试使用这个对齐一组方程来帮助我，但我似乎无法弄清楚哪里出了问题。如能得到帮助我将不胜感激。
我忘了说我想要：
f(\chi_2), f(\chi_3) and f(\chi_4)
出现在下一行，同时还想删除右侧的方程编号
答案1
不要尝试将 LaTeX 当作文字处理器来使用。
\documentclass{article}
\usepackage{amsmath,amssymb}
\begin{document}
\begin{align*}
f(\chi_0)&=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
f(\chi_1)&=\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]\\
f(\chi_2)&=\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]\\
f(\chi_3)&=\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]\\
f(\chi_4)&=\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]
\end{align*}
Now we want to calculate $\hat{f}(a)$:
\begin{align*}
\hat{f}(0)
&= f(\chi_0)+f(\chi_1)+f(\chi_2)+f(\chi_3)+f(\chi_4)\\
&= \begin{aligned}[t]
&\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
&\hphantom{f(0)}+\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]\\
&\hphantom{f(0)}+\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]\\
&\hphantom{f(0)}+\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]\\
&\hphantom{f(0)}+\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]
\end{aligned}
\\
&=\begin{aligned}[t]
f(0) &+ \frac{f(1)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(2)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(3)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(4)}{5}[1+a+a^2+a^3+a^4]
\end{aligned}
\end{align*}
\end{document}
嵌套aligned可align确保间距和对齐。特定情况似乎需要\hphantom{f(0)}对齐块中的 + 符号。
1/5 分数对齐的变体：
\documentclass{article}
\usepackage{amsmath,amssymb}
\begin{document}
\begin{align*}
f(\chi_0)&=\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
f(\chi_1)&=\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]\\
f(\chi_2)&=\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]\\
f(\chi_3)&=\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]\\
f(\chi_4)&=\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]
\end{align*}
Now we want to calculate $\hat{f}(a)$:
\begin{align*}
\hat{f}(0)
&= f(\chi_0)+f(\chi_1)+f(\chi_2)+f(\chi_3)+f(\chi_4)\\
&\begin{aligned}[t]
{}={}&\frac{1}{5}[f(0)+f(1)+f(2)+f(3)+f(4)]\\
{}+{}&\frac{1}{5}[f(0)+af(1)+a^2f(2)+f(3)+f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^2f(1)+a^4f(2)+f(3)+f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^3f(1)+af(2)+f(3)+f(4)]\\
{}+{}&\frac{1}{5}[f(0)+a^4f(1)+a^3f(2)+f(3)+f(4)]
\end{aligned}
\\
&=\begin{aligned}[t]
f(0) &+ \frac{f(1)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(2)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(3)}{5}[1+a+a^2+a^3+a^4]\\
&+ \frac{f(4)}{5}[1+a+a^2+a^3+a^4]
\end{aligned}
\end{align*}
\end{document}
