这是我的代码
\usepackage[eulerchapternumbers,pdfspacing]{classicthesis}
\usepackage{amsmath}
\begin{document}
Let there be 2 points.$z_{0}$ and $z_{0}^{\prime}$ in a \textit{\textbf{ complex plane(contour) }}.We divide the contour from $z_{0}$ to $z_{0}^{\prime}$ as designated path C,in \textbf{n} intervals by picking \textbf{n-1} intermediate points as $z_{1} , z_{2} , z_{3} \dots$ on the contour.\\
so the sum\\
{\Large$s_{n}=\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})$}\\
where$\zeta_{j}$ is a point on the curve between $z_{j} \quad and \quad z_{j-1}$(a small part of the curve).\\
now let's $n \rightarrow \infty \quad and \quad |z_{j} - z_{j-1}| \rightarrow 0$ for all j.\\
if {\Large$\lim_{n \rightarrow \infty} s_{n} \quad exists$}, then \\
{\Large$lim_{n \rightarrow \infty}\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})=\int_{z_{0}}^{z_{0}^{\prime}}f(z)dz = \int_{C}f(z)dz$}\\
It can also be defined as \\
\begin{align*}
\int_{z_{1}}^{z_{2}}f(z)dz &= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y) + iv(x,y)][dx + idy]\\
&= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y)dx - v(x,y)dy] + i \int_{x_{1}y_{1}}^{x_{2}y_{2}}[v(x,y)dx + u(x,y)dy]\\
\end{align*}
\end{document}
我收到了这个错误。
段落第 9-17 行的 \hbox 过满（宽 1.60175pt）。
如何解决？
编辑：现在问题已经解决，并且我已将代码编辑为正确的。
答案1
如果我将其添加\documentclass{book}到代码顶部，我会得到以下输出：
我很遗憾地说，这是一张非常糟糕的排版样本。
第 1 行标点不一致；“2 points”应为“two points”。
第 2 行中的粗体 n 和 n-1 应采用数学斜体。
在第 3 行中，三个点前面应该有一个逗号。
第 5 行的显示非常糟糕：根本没有必要将公式放大。下面的显示也一样。
修复这些点将使过满现象消失。
\documentclass{book}
\usepackage[eulerchapternumbers,pdfspacing]{classicthesis}
\usepackage{amsmath}
\begin{document}
Let there be two points $z_{0}$ and $z'_{0}$ in a \textit{\textbf{complex plane (contour)}}.
We divide the contour from $z_{0}$ to $z'_{0}$ as designated path $C$, in~$n$ intervals
by picking $n-1$ intermediate points $z_{1}$, $z_{2}$, $z_{3}$,\dots\ on the contour so the sum
\[
s_{n}\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})
\]
where $\zeta_{j}$ is a point on the curve between $z_{j}$ and $z_{j-1}$
(a small part of the curve).
Now let $n \rightarrow \infty$ and $|z_{j} - z_{j-1}| \rightarrow 0$, for all~$j$.
If $\lim_{n \rightarrow \infty} s_{n}$ exists, then
\[
\lim_{n \rightarrow \infty}\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})=
\int_{z_{0}}^{z'_{0}} f(z)\,dz = \int_{C}f(z)\,dz
\]
It can also be defined as
\begin{align*}
\int_{z_{1}}^{z_{2}}f(z)\,dz
&= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y) + iv(x,y)][dx + i\,dy] \\
&= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y)\,dx - v(x,y)\,dy]
+ i \int_{x_{1}y_{1}}^{x_{2}y_{2}}[v(x,y)\,dx + u(x,y)\,dy]
\end{align*}
\end{document}
还修复了语言：
\documentclass{book}
\usepackage[eulerchapternumbers,pdfspacing]{classicthesis}
\usepackage{amsmath}
\begin{document}
Let us be given two points two points $z_{0}$ and $z'_{0}$ on a \emph{contour} in
the complex plane. We divide the contour from $z_{0}$ to $z'_{0}$, denoted as the path~$C$,
in~$n$ intervals by picking $n-1$ intermediate points $z_{1}$, $z_{2}$, $z_{3}$,\dots\
on the contour and define the sum
\[
s_{n}\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})
\]
where $\zeta_{j}$ is a point on the curve between $z_{j}$ and $z_{j-1}$
(a small part of the curve).
Now let $n \rightarrow \infty$ and $|z_{j} - z_{j-1}| \rightarrow 0$, for all~$j$.
If $\lim_{n \rightarrow \infty} s_{n}$ exists, then
\[
\lim_{n \rightarrow \infty}\sum_{j=1}^{n}f(\zeta_{j})(z_{j}-z_{j-1})=
\int_{z_{0}}^{z'_{0}} f(z)\,dz = \int_{C}f(z)\,dz
\]
The integral can also be defined as
\begin{align*}
\int_{z_{1}}^{z_{2}}f(z)\,dz
&= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y) + iv(x,y)][dx + i\,dy] \\
&= \int_{x_{1}y_{1}}^{x_{2}y_{2}}[u(x,y)\,dx - v(x,y)\,dy]
+ i \int_{x_{1}y_{1}}^{x_{2}y_{2}}[v(x,y)\,dx + u(x,y)\,dy]
\end{align*}
\end{document}
答案2
解决了
似乎有双重上标导致了此错误。
抱歉，没有仔细检查就提问了。
不会再发生这种情况了。