在证明中使用 Gig 方程

Question

有人能帮助我使以下证明更专业吗：

Overleaf 模板：

\usepackage{amsmath}
\begin{equation*}
\begin{split}

sup_{z\in \mathrm{R}}|\hat{F}(z)-\widetilde {F}(z)| = sup_{z\in \mathrm{R}}\left|\frac{1}{n} 
\sum_{t=1}^{n}\left\{\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-\hat{z}_{t}}{h}\right) d u
-\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-z_{t}}{h}\right) d u\right\}\right|\\ 

& = sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{\int_{-\infty}^{\frac{z-\hat{z}_{t}} 
{h}} k(v) dv - \int_{-\infty}^{\frac{z-{z}_{t}}{h}} k(v) dv\right\}\right|, \text{By using 
$v=\frac{u-\hat{z}_{t}}{h}$ and $dv=\frac{1}{h}du$} \\

& = sup_{z\in \mathrm{R}}\left|=\frac{1}{n} \sum_{t=1}^{n} \left\{G\left(\frac{z-\hat{z}_{t}} 
{h}\right)-G\left(\frac{z-z_{t}}{h}\right)\right\}\right|,  \text{By using $G(z)=\int_{- 
\infty}^{z}k(v)dv$}\\ 

& = sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{G\left(\frac{z-z_{t}} 
{h}\right)+G\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)  + \frac{1}{2 !} 
G^{\prime \prime}\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}- 
\at{z}_{t}}{h}\right)^{2}  + \frac{1}{3 !} G^{\prime \prime \prime}\left(\frac{z-z_{t}} 
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{3} +R_{t} -G\left(\frac{z-z_{t}} 
{h}\right)\right\}\right|, \text{By using third-order Taylor Expension centered about $\frac{z-z_{t}} 
{h}$} \\

& = sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{G\left(\frac{z-z_{t}} 
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)+\frac{1}{2 !} G^{\prime \prime}\left(\frac{z- 
z_{t}} {h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{2} + \frac{1}{3 !} G^{\prime \prime 
\prime}\left(\frac{z- 
z_{t}} {h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{3}+R_{t} \right\}\right| \\

& \quad \leq sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}G\left(\frac{z-z_{t}} 
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)\right| + sup_{z\in \mathrm{R}}\left|\frac{1}{n} 
\sum_{t=1}^{n}\frac{1}{2 !} G^{\prime\prime}\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}- 
\hat{z}_{t}}{h}\right)^{2} \right|\\

& \quad + sup_{z\in \mathrm{R}}\left| \frac{1}{n}   \sum_{t=1}^{n} \frac{1}{3!} G^{\prime \prime 
\prime}\left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}- \hat{z}_{t}}{h}\right)^{3} \right|  + 
sup_{z\in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} R_{t}\right|\\

\end{split} 
\end{equation*}

Let

\begin{equation*}
I_{1}=sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}G\left(\frac{z- 
z_{t}} 
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)\right| 
\end{equation*}

\begin{equation*}
I_{2}=sup_{z\in \mathrm{R}}\left|\frac{1}{n} 
\sum_{t=1}^{n}\frac{1}{2 !} G^{\prime\prime}\left(\frac{z-z_{t}} 
{h}\right)\left(\frac{z_{t}- 
\hat{z}_{t}}{h}\right)^{2} \right|
\end{equation*}

\begin{equation*}
I_{3}=sup_{z\in \mathrm{R}}\left| \frac{1}{n}   \sum_{t=1}^{n} \frac{1}{3!} 
G^{\prime \prime 
\prime}\left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}- \hat{z}_{t}} 
{h}\right)^{3} \right|  
\end{equation*}

\begin{equation*}
I_{4}= sup_{z\in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} R_{t}\right|
\end{equation*}

Answer 1

我将在长篇展示之前解释替换并宣布泰勒展开的用途；读者将能够看到它们实际在哪里使用。

\documentclass{article}
\usepackage{amsmath,amssymb}

\newcommand{\numberset}[1]{\mathbb{#1}}% or whatever you prefer
\newcommand{\R}{\numberset{R}}
\newcommand{\diff}{\mathop{}\!d}

\begin{document}

We can use the substitutions
\begin{equation*}
G(z)=\int_{-\infty}^{z}k(v) \diff v,
\qquad
v=\frac{u-\hat{z}_{t}}{h},\quad \diff v=\frac{1}{h}\diff u
\end{equation*}
and, by using the third-order Taylor expansion centered about $(z-z_{t})/h$ of $G$,
compute
\begin{equation*}
\begin{split}
&\! \sup_{z\in \R}|\hat{F}(z)-\tilde {F}(z)|
\\
&=\sup_{z\in \R}\biggl|
    \frac{1}{n} \sum_{t=1}^{n}\biggl\{
      \int_{-\infty}^{z} \frac{1}{h} k\biggl(\frac{u-\hat{z}_{t}}{h}\biggr)\diff  u
      -\int_{-\infty}^{z} \frac{1}{h} k\biggl(\frac{u-z_{t}}{h}\biggr)\diff  u
    \biggr\}
  \biggr|
\\ 
&=\sup_{z\in \R}\biggl|
    \frac{1}{n} \sum_{t=1}^{n}\biggl\{
      \int_{-\infty}^{\frac{z-\hat{z}_{t}}{h}} k(v)\diff v
      - \int_{-\infty}^{\frac{z-{z}_{t}}{h}} k(v)\diff v
    \biggr\}
  \biggr|
\\
&=\sup_{z\in \R}\biggl|
    \frac{1}{n} \sum_{t=1}^{n} \biggl\{
      G\biggl(\frac{z-\hat{z}_{t}}{h}\biggr)
      -G\biggl(\frac{z-z_{t}}{h}\biggr)
    \biggr\}
  \biggr|
\\
&=\sup_{z\in \R}\biggl|
    \frac{1}{n} \sum_{t=1}^{n}\biggl\{
      \begin{aligned}[t]
      & G\biggl(\frac{z-z_{t}}{h}\biggr)+
        G\biggl(\frac{z-z_{t}}{h}\biggr)\biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)
      + \frac{1}{2!} G''\biggl(\frac{z-z_{t}}{h}\biggr)
                          \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)^{\!2}
      \\
      & + \frac{1}{3!} G'''\biggl(\frac{z-z_{t}}{h}\biggr)
                            \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)^{\!3}
      + R_{t} -G\biggl(\frac{z-z_{t}}{h}\biggr)
    \biggr\}\biggr|\end{aligned}
\\
& = \sup_{z\in \R}\biggl|
    \frac{1}{n} \sum_{t=1}^{n}\biggl\{
      \begin{aligned}[t]
      & G\biggl(\frac{z-z_{t}}{h}\biggr)\biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)
      + \frac{1}{2!} G''\biggl(\frac{z-z_{t}} {h}\biggr)
                          \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)^{\!2}
      \\
      & + \frac{1}{3 !} G'''\biggl(\frac{z-z_{t}} {h}\biggr)
                            \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)^{\!3}
      +R_{t}
   \biggr\}\biggr|\end{aligned}
\\
&\leq \begin{aligned}[t]
  &\! \sup_{z\in \R}\biggl|
        \frac{1}{n} \sum_{t=1}^{n}G\biggl(\frac{z-z_{t}} {h}\biggr)
                              \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)
      \biggr|
   + \sup_{z\in \R}\biggl|
        \frac{1}{n} \sum_{t=1}^{n}\frac{1}{2!} G''\biggl(\frac{z-z_{t}}{h}\biggr)
                                                  \biggl(\frac{z_{t}-\hat{z}_{t}}{h}\biggr)^{\!2}
    \biggr|
  \\
  & + \sup_{z\in \R}\biggl|
        \frac{1}{n} \sum_{t=1}^{n} \frac{1}{3!} G'''\biggl(\frac{z-z_{t}} {h}\biggr)
                                                    \biggl(\frac{z_{t}- \hat{z}_{t}}{h}\biggr)^{\!3}
    \biggr|
   + \sup_{z\in \R}\biggl| \frac{1}{n} \sum_{t=1}^{n} R_{t}\biggr|
  \end{aligned}
\end{split} 
\end{equation*}

\end{document}

这仍然会产生少量的溢出，但这取决于文档中设置的文本宽度。

需要注意的几点：我把所有的\leftand\right都改成了\biggland \biggr，这样可以更好地控制并减少间距。此外，大括号的指数前面要加上 ，\!以避免它们无处可去。

Answer 2

这只是一个起点：您可以使用\intertext带有公式的文本，并记住，如果公式被破坏，您应该使用\biggl(...\biggr)，或\Biggl(...\Biggr)（例如）等。我的建议是使用geometry包来为整个页面提供更多空间。

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}

\begin{document}

\[
\begin{split}
\sup_{z\in \mathbb{R}}|\hat{F}(z)-\widetilde {F}(z)|& = \sup_{z\in \mathbb{R}}\Biggl|\frac{1}{n}
\sum_{t=1}^{n}\left\{\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-\hat{z}_{t}}{h}\right) d u
-\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-z_{t}}{h}\right) d u\right\}\Biggr|\\ 
& = \sup_{z\in \mathbb{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{\int_{-\infty}^{\frac{z-\hat{z}_{t}} 
{h}} k(v) dv - \int_{-\infty}^{\frac{z-{z}_{t}}{h}} k(v) dv\right\}\right|
\intertext{$v=\frac{u-\hat{z}_{t}}{h}$ and $dv=\frac{1}{h}du$}
\sum & = \sup_{z\in \mathbb{R}}\Biggl|\frac{1}{n}
\sum_{t=1}^{n}\left\{\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-\hat{z}_{t}}{h}\right) d u
-\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-z_{t}}{h}\right) d u\right\}\Biggr|\\ 
& = \sup_{z\in \mathbb{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{\int_{-\infty}^{\frac{z-\hat{z}_{t}} 
{h}} k(v) dv - \int_{-\infty}^{\frac{z-{z}_{t}}{h}} k(v) dv\right\}\right|
\end{split} 
\]

\end{document}

Answer 3

这是一种可能性，在环境中的相关位置使用multlined& 。我还使用中等大小的分数 ( )作为数值分数系数：alignedsplit\mfracnccmath

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

    \begin{document}

\begin{equation*}
\begin{split}
\MoveEqLeft[-1]\sup_{z\in \mathrm{R}}|\hat{F}(z)-\widetilde{F}(z)|
 = \sup_{z\in \mathrm{R}}\left|\frac{1}{n}
\sum_{t=1}^{n}\left\{\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-\hat{z}_{t}}{h}\right) d u
-\int_{-\infty}^{z} \frac{1}{h} k\left(\frac{u-z_{t}}{h}\right) d u\right\}\right|\\
& = \sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}\left\{\int_{-\infty}^{\frac{z-\hat{z}_{t}}
{h}} k(v) dv - \int_{-\infty}^{\frac{z-{z}_{t}}{h}} k(v) dv\right\}\right|,\qquad
 \text{by using $v=\frac{u-\hat{z}_{t}}{h}$ and $dv=\frac{1}{h}du$}\\
& = \sup_{z\in \mathrm{R}}\left|=\frac{1}{n} \sum_{t=1}^{n} \left\{G\left(\frac{z-\hat{z}_{t}}
{h}\right)-G\left(\frac{z-z_{t}}{h}\right)\right\}\right|, \qquad\enspace
\text{setting $G(z)=\int_{- \infty}^{z}k(v)dv$}\\
\intertext{and by using third-order Taylor Expansion centered about $\frac{z-z_{t}}{h}$\vskip 2ex}
& \begin{multlined}[t][0.95\linewidth]= \sup_{z\in \mathrm{R}}\biggl|\frac{1}{n} \sum_{t=1}^{n}\biggl\{G\left(\frac{z-z_{t}}
{h}\right) +G\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)+ \mfrac{1}{2!}
G''\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}- \hat{z}_{t}}{h}\right)^{2}\\[-1.5ex]
+ \mfrac{1}{3!} G'''\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{3} +R_{t} -G\left(\frac{z-z_{t}}
{h}\right)\biggr\}\biggr|,\end{multlined} %
 \\
&\begin{multlined}[t][0.95\linewidth] = \sup_{z\in \mathrm{R}}\biggl|\frac{1}{n} \sum_{t=1}^{n}\biggl\{G\left(\frac{z-z_{t}}
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)+\mfrac{1}{2 !} G''\left(\frac{z-
z_{t}} {h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{2}\\[-1.5ex]
-\mfrac{1}{3 !} G'''\left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)^{3}+R_{t} \biggr\}\biggr| ,\end{multlined} %
\\
 & \leq \begin{aligned}[t]\sup_{z\in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}G\left(\frac{z-z_{t}}
{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)\right| & + \sup_{z\in \mathrm{R}}\left|\frac{1}{n}
\sum_{t=1}^{n}\mfrac{1}{2 !} G''\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}-
\hat{z}_{t}}{h}\right)^{2} \right|\\
 & + \sup_{z\in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} \mfrac{1}{3!} G''' \left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}- \hat{z}_{t}}{h}\right)^{3} \right| +
\sup_{z\in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} R_{t}\right| \end{aligned}%
\end{split}
\end{equation*}

    \end{document} 

至于第二部分，我将以稍有不同的方式排列，分为两行：

Let%
\begin{align*}
I_{1} & =\sup_{z \in \mathrm{R}}\left|\frac{1}{n} \sum_{t=1}^{n}G\left(\frac{z-z_{t}}{h}\right)\left(\frac{z_{t}-\hat{z}_{t}}{h}\right)\right|, &
I_{2} & =\sup_{z \in \mathrm{R}}\left|\frac{1}{n}
\sum_{t=1}^{n}\mfrac{1}{2 !} G^{\prime\prime}\left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}- \hat{z}_{t}}{h}\right)^{\!2} \right|, \\[1ex]
I_{3} & =\sup_{z \in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} \mfrac{1}{3!}G'''\left(\frac{z-z_{t}} {h}\right)\left(\frac{z_{t}- \hat{z}_{t}}{h}\right)^{\!3} \right| ,&
I_{4} & = \sup_{z \in \mathrm{R}}\left| \frac{1}{n} \sum_{t=1}^{n} R_{t}\right|.
\end{align*}