\[
\frac{1}{z^p}\left[ f(z)*\left( \frac{z^p - D z^{p+1}}{(1-z)^{2}} + \frac{\beta e^{i\theta}(D-1)z^{p+1}}{(1-z)^{2}}\right) \right]
\]
\[
=\frac{1}{z^p}\left[ f(z)*\left( z^p + \sum _{k=p+1}^{\infty} (k-p+1) z^k - D \sum _{k=p+1}^{\infty} (k-p) z^k +\beta e^{i\theta}(D-1)\sum _{k=p+1}^{\infty} (k-p) z^k\right) \right]
\]
\[
=\frac{1}{z^p}\left[ f(z)*\left( z^p + \sum _{k=p+1}^{\infty} z^k + (1- \beta e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k z^k +p(1-\beta e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} z^k\right) \right]
\]
\[
=\frac{1}{z^p}\left[ f(z) + (1- \beta e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k \alpha _k z^k +p(1-\beta e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} \alpha _k z^k \right] \]
\[
=\frac{1}{z^p}\left[ zf^{'}(z) (1-D) - \beta e^{i\theta }(1-D)(zf^{'}(z)-pf(z)) + (1+pD-p)f(z) \right]
\]
答案1
另一种 amsmath 可能性:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\begin{split}
\frac{1}{z^p}&\Bigl[ f(z)*\Bigl( \frac{z^p - D z^{p+1}}{(1-z)^{2}} +
\frac{\beta e^{i\theta}(D-1)z^{p+1}}{(1-z)^{2}}\Bigr) \Bigr]\\[\jot]
&{}=\frac{1}{z^p}\Bigl[ f(z)*\Bigl( z^p + \sum _{k=p+1}^{\infty} (k-p+1) z^k
- D \sum _{k=p+1}^{\infty} (k-p) z^k\\
&\qquad
{}+\beta e^{i\theta}(D-1)\sum _{k=p+1}^{\infty} (k-p) z^k\bigr)\Bigr]
\\[\jot]
&{}=\frac{1}{z^p}\Bigl[ f(z)*\Bigl( z^p +\\
& \qquad \sum _{k=p+1}^{\infty} z^k + (1- \beta e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k z^k
+p(1-\beta e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} z^k\Bigr) \Bigr]
\\[\jot]
&=\frac{1}{z^p}\Bigl[ f(z) + (1- \beta e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k \alpha _k z^k +
p(1-\beta e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} \alpha _k z^k \Bigr] \\[\jot]
&=\frac{1}{z^p}\Bigl[ zf'(z) (1-D) - \beta e^{i\theta }(1-D)(zf'(z)-pf(z)) + (1+pD-p)f(z) \Bigr]
\end{split}
\]
\end{document}
答案2
如果您不想对行进行编号,则应在此处使用align
或。这些环境是 包 的一部分。您可以改为加载 包,以便将第一行移到左侧。在我看来这样更好。只需根据您的需要调整即可:align*
amsmath
mathtools
[number]
\MoveEqLeft
% arara: pdflatex
\documentclass{article}
\usepackage{mathtools}
\newcommand*{\e}{\mathrm{e}}
\begin{document}
\begin{align*}
\MoveEqLeft[4]\frac{1}{z^p}\biggl[ f(z)\biggl( \frac{z^p - D z^{p+1}}{(1-z)^{2}} + \frac{\beta\e^{i\theta}(D-1)z^{p+1}}{(1-z)^{2}}\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z)\biggl( z^p + \sum _{k=p+1}^{\infty} (k-p+1) z^k - D \sum _{k=p+1}^{\infty} (k-p) z^k\\
&\hphantom{{}=\frac{1}{z^p}\biggl[}+\beta \e^{i\theta}(D-1)\sum _{k=p+1}^{\infty} (k-p) z^k\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z)\biggl( z^p + \sum _{k=p+1}^{\infty} z^k + (1- \beta\e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k z^k\\
&\hphantom{{}=\frac{1}{z^p}\biggl[}+p(1-\beta\e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} z^k\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z) + (1- \beta e^{i \theta })(1-D) \sum _{k=p+1}^{\infty}k \alpha_k z^k\\
&\hphantom{{}=\frac{1}{z^p}\biggl[}+p(1-\beta\e^{i\theta})(D-1)\sum _{k=p+1}^{\infty} \alpha _k z^k \biggr]\\
&=\frac{1}{z^p}\Bigl[ zf'(z) (1-D) - \beta\e^{i\theta}(1-D)(zf'(z)-pf(z))\\
&\hphantom{{}=\frac{1}{z^p}\biggl[}+(1+pD-p)f(z) \Bigr]
\end{align*}
\end{document}
答案3
我建议您做以下调整:
最重要的是,使用
align*
环境而不是连续的\[ ... \]
结构。使用\qquad
缩进每个方程的子行。您在几个等式中都使用了多余的括号。尽量使内容看起来简洁明了 — 您只需要一个大括号,而不是两个。
8 个 [!] 符号下方的下标
\sum
比符号本身更宽,导致求和符号两侧出现大量浪费的(可能也是不受欢迎的)空白。将下标项放在\mathclap{...}
(包的宏mathtools
)中可以避免这种情况发生。不要使用
\left
和\right
来调整括号的大小,因为从印刷上来说,它们会使括号过大。请改用\biggl
和\biggr
。可选:有些人建议(坚持?!)将欧拉数
e
和虚数的字母排版为直立字体。如果您这样做,为此目的i
创建名为\e
和的宏会很方便。\im
\documentclass{article}
\usepackage{mathtools} % for '\mathclap' macro
\newcommand{\e}{\mathrm{e}} % typeset Euler's number in upright font
\newcommand{\im}{\mathrm{i}} % ditto for the square root of -1
\begin{document}
\begin{align*}
\frac{1}{z^p}\, f(z)
&*\biggl[ \frac{z^p - D z^{p+1}}{(1-z)^{2}} + \frac{\beta\e^{\im\theta}(D-1)z^{p+1}}{(1-z)^{2}}\biggr] \\
&=\frac{1}{z^p}\, f(z)*\biggl[ z^p + \sum_{\mathclap{k=p+1}}^{\infty} (k-p+1) z^k - D \sum_{\mathclap{k=p+1}}^{\infty} (k-p) z^k\\
&\qquad +\beta \e^{\im\theta}(D-1)\sum_{\mathclap{k=p+1}}^{\infty} (k-p) z^k\biggr] \\
&=\frac{1}{z^p} \, f(z)*\biggl[ z^p + \sum_{\mathclap{k=p+1}}^{\infty} z^k + (1- \beta\e^{\im\theta})(1-D) \sum_{\mathclap{k=p+1}}^{\infty}k z^k\\
&\qquad+p(1-\beta\e^{\im\theta})(D-1)\sum_{\mathclap{k=p+1}}^{\infty} z^k\biggr] \\
&=\frac{1}{z^p}\biggl[ f(z) + (1- \beta\e^{\im\theta})(1-D) \sum_{\mathclap{k=p+1}}^{\infty}k \alpha_k z^k\\
&\qquad +p(1-\beta\e^{\im\theta})(D-1)\sum_{\mathclap{k=p+1}}^{\infty} \alpha _k z^k \biggr]\\
&=\frac{1}{z^p}\Bigl[ zf'(z) (1-D) - \beta\e^{\im\theta}(1-D)\bigl(zf'(z)-pf(z)\bigr)\\
&\qquad +(1+pD-p)f(z) \Bigr]
\end{align*}
\end{document}
答案4
如果您使用该geometry
包并采用该包的默认(甚至更窄)边距设置,则所有方程式都会适合边距,而无需换行。或者,您可以使用环境medsize
(来自nccmath
,~80% 的 \displaystyle):
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[showframe,nomarginpar]{geometry}
\usepackage{mathtools}
\newcommand*{\e}{\mathrm{e}}
\begin{document}
\begin{align*}
\MoveEqLeft[4]\frac{1}{z^p}\biggl[ f(z)\biggl( \frac{z^p - D z^{p+1}}{(1-z)²} + \frac{β \e^{iθ}(D-1)z^{p+1}}{(1-z)²}\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z)\biggl( z^p + ∑ _{k=p+1}^{∞} (k-p+1) z^k - D ∑ _{k=p+1}^{∞} (k-p) z^k+β \e^{iθ}(D-1)∑ _{k=p+1}^{∞} (k-p) z^k\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z)\biggl( z^p + ∑ _{k=p+1}^{∞} z^k + (1- β \e^{i θ })(1-D) ∑ _{k=p+1}^{∞}k z^k+p(1-β \e^{iθ})(D-1)∑ _{k=p+1}^{∞} z^k\biggr) \biggr]\\
&=\frac{1}{z^p}\biggl[ f(z) + (1- β e^{i θ })(1-D) ∑ _{k=p+1}^{∞}k α_k z^k + p(1-β \e^{iθ})(D-1)∑ _{k=p+1}^{∞} α _k z^k \biggr]\\
&=\frac{1}{z^p}\Bigl[ zf'(z) (1-D) - β \e^{iθ}(1-D)(zf'(z)-pf(z))+(1+pD-p)f(z) \Bigr]
\end{align*}
\end{document}