Latex 排版技巧 1——数学公式对齐

begin{aligned} end{aligned}

&

[code]$ f(x)=2*x^3-x-2 $ \quad unique zero $ \hat{x} \in [1, 2] $
\begin{enumerate}
    \item $ x_{n+1} = 2x_n^3-2 $

    \begin{displaymath} 
	\begin{aligned}	 
	 x_{n+1} -\hat{x} & = (2x_n^3-2) - (2\hat{x}_n^3-2) \\
	 & = 2(x_n^2+x_n\hat{x}+\hat{x}^2)(x_n-\hat{x}) \\
	 \varepsilon_{n+1} & = 2(x_n^2+x_n\hat{x}+\hat{x}^2)\varepsilon_n
	 \end{aligned}
	\end{displaymath}

    \item $ x_{n+1} = \frac{2}{2x_n^2-1} $
    \begin{displaymath} 
	\begin{aligned}	 
	 x_{n+1} -\hat{x} & = \frac{2}{2x_n^2-1} - \frac{2}{2\hat{x}^2-1} \\
	 & = -\frac{4(x_n+\hat{x})(x_n-\hat{x})}{(2x_n^2-1)(2\hat{x}-1)} \\
	 \varepsilon_{n+1} & = -\frac{4(x_n+\hat{x})}{(2x_n^2-1)(2\hat{x}-1)}\varepsilon_n
	 \end{aligned}
	\end{displaymath}
\end{enumerate}

[code]\section{Problem}
\begin{enumerate}
	\item Using Lagrange interpolation formula
	\begin{displaymath} 
	\begin{aligned}	 
		f[x_0] & = x-1\\
		f[x_0,x_1] & = (x-1)^2\\
		f[x_0,x_1,x_2] & = \frac{(x-1)(8-x)}{6}\\
		f[x_0,x_1,x_2,x_3] & = -\frac{(x-1)(x-4)(2x-1)}{6}-\frac{(x-1)(x-2)(6x-31)}{21}
	 \end{aligned}
	\end{displaymath}
\end{enumerate}