├── .gitignore ├── Chapters ├── chapter0.aux ├── chapter0.tex ├── chapter1.aux ├── chapter1.tex ├── chapter10.aux ├── chapter10.tex ├── chapter11.aux ├── chapter11.tex ├── chapter12.aux ├── chapter12.tex ├── chapter13.tex ├── chapter2.aux ├── chapter2.tex ├── chapter3.aux ├── chapter3.tex ├── chapter4.aux ├── chapter4.tex ├── chapter5.aux ├── chapter5.tex ├── chapter6.aux ├── chapter6.tex ├── chapter7.aux ├── chapter7.tex ├── chapter8.aux ├── chapter8.tex ├── chapter9.aux └── chapter9.tex ├── Fonts └── LXGWWenKai-Regular.ttf ├── LICENSE ├── LouAnalysis.pdf ├── LouAnalysis.tex ├── Pics ├── wx.png └── zfb.jpg ├── README.md ├── clean.pyw └── mbook.cls /.gitignore: -------------------------------------------------------------------------------- 1 | 2 | LouAnalysis.aux 3 | *.log 4 | LouAnalysis.synctex.gz 5 | LouAnalysis.toc 6 | -------------------------------------------------------------------------------- /Chapters/chapter0.aux: -------------------------------------------------------------------------------- 1 | \relax 2 | \providecommand\hyper@newdestlabel[2]{} 3 | \gdef \LT@i {\LT@entry 4 | {1}{61.16116pt}\LT@entry 5 | {1}{377.14047pt}} 6 | \@writefile{toc}{\contentsline {chapter}{\numberline {第0章\hspace {.3em}}闲谈}{0}{chapter.0}\protected@file@percent } 7 | \@writefile{lof}{\addvspace {10.0pt}} 8 | \@writefile{lot}{\addvspace {10.0pt}} 9 | \@writefile{toc}{\contentsline {section}{\numberline {0.1}勘误}{0}{section.0.1}\protected@file@percent } 10 | \@writefile{toc}{\contentsline {section}{\numberline {0.2}符号说明}{0}{section.0.2}\protected@file@percent } 11 | \@setckpt{Chapters/chapter0}{ 12 | \setcounter{page}{3} 13 | \setcounter{equation}{0} 14 | \setcounter{enumi}{0} 15 | \setcounter{enumii}{0} 16 | \setcounter{enumiii}{0} 17 | \setcounter{enumiv}{0} 18 | \setcounter{footnote}{0} 19 | \setcounter{mpfootnote}{0} 20 | \setcounter{part}{0} 21 | \setcounter{chapter}{0} 22 | \setcounter{section}{2} 23 | \setcounter{subsection}{0} 24 | \setcounter{subsubsection}{0} 25 | \setcounter{paragraph}{0} 26 | \setcounter{subparagraph}{0} 27 | \setcounter{figure}{0} 28 | \setcounter{table}{1} 29 | \setcounter{parentequation}{0} 30 | \setcounter{tcbbreakpart}{0} 31 | \setcounter{tcblayer}{0} 32 | \setcounter{tcolorbox@number}{0} 33 | \setcounter{float@type}{16} 34 | \setcounter{caption@flags}{0} 35 | \setcounter{continuedfloat}{0} 36 | \setcounter{nlinenum}{0} 37 | \setcounter{section@level}{0} 38 | \setcounter{Item}{0} 39 | \setcounter{Hfootnote}{0} 40 | \setcounter{Hy@AnnotLevel}{0} 41 | \setcounter{bookmark@seq@number}{0} 42 | \setcounter{lstnumber}{1} 43 | \setcounter{tcblisting}{0} 44 | \setcounter{tcbrastercolumn}{1} 45 | \setcounter{tcbrasterrow}{1} 46 | \setcounter{tcbrasternum}{1} 47 | \setcounter{tcbraster}{0} 48 | \setcounter{mn@abspage}{13} 49 | \setcounter{LT@tables}{1} 50 | \setcounter{LT@chunks}{1} 51 | \setcounter{tcb@cnt@theorem}{0} 52 | \setcounter{tcb@cnt@definition}{0} 53 | \setcounter{tcb@cnt@proposition}{0} 54 | \setcounter{tcb@cnt@lemma}{0} 55 | \setcounter{woes}{0} 56 | \setcounter{lstlisting}{0} 57 | \setcounter{mistake}{7} 58 | } 59 | -------------------------------------------------------------------------------- /Chapters/chapter0.tex: -------------------------------------------------------------------------------- 1 | \begin{titlepage} 2 | \maketitle %首页 3 | \end{titlepage} 4 | \restoregeometry 5 | \begin{center} 6 | \Large 感悟 7 | \end{center} 8 | \noindent \Large 今 9 | \normalsize 10 | 天是2023年5月20日，一个特别而又普通的日子：为什么说特别，因为我是单身，为什么普通？还是这个原因。这一天我读完了这套数学分析的第一遍，并将部分习题抄录至此，但大概有百分之七十的习题并没有弄懂，你可能觉得是时间仓促导致，但事实并非如此：笔者从22年8月开始阅读，至今已十月有余。我听闻难事可让人一夜白头，然这十月后我的头发依旧乌黑(只是掉了几根)，这样比起来楼红卫十分温柔，或者还有一个原因：我还没努力到白头的程度。 11 | 12 | 我每天都乐此不疲的发现让我头疼的题目，并不是我喜欢头疼，若非迫不得已，我宁愿躺在床上睡大觉。但我每天都很苦恼，这让我几乎无法入眠：大佬做题犹如作画，随便一笔我都必须揣摩良久，最后只能承认自己的愚蠢，这是十分值得伤心的事，我也有理由尽情悲伤。但我尽量幽默一点，因为学习与生活永远不可能让我一直保持笑脸，但幽默总能。 13 | 14 | 这本书极不适合初学者，当然你如果天赋异禀，那本书能让你飞黄腾达。笔者大一大二学过一点数学分析等之类的课程，看这套书就像牙齿掉光的老太太吃五仁月饼，不过值得一提的是，许多都是一股脑吞下的，以至于事后问我这什么味道：干巴巴，没什么味儿。 15 | 16 | 我几乎要放弃的时候非常感谢各位群友的帮助，是大家让我间歇性放弃找女朋友的念头而是改为买一只宠物蜈蚣，相信很快就会和大家见面的！另外，明天开始就要开始要第二遍了。届时会完善这些习题。 17 | \pagenumbering{Roman} 18 | \tableofcontents 19 | \newpage 20 | \pagenumbering{arabic} 21 | \setcounter{page}{0} 22 | \setcounter{chapter}{-1} 23 | \newcounter{mistake} 24 | \renewcommand{\themistake}{\arabic{mistake}} 25 | \newcommand{\mist}[1]{\par\textcolor{red}{\refstepcounter{mistake}\textbf{\themistake.} \textbf{Page #1 }}\hspace{0.5em}} 26 | \chapter{闲谈} 27 | \section{勘误} 28 | 29 | \mist{21} 第4题(3)中对于\(R(x)\)的系数为有理数时结论才可被证明, 当\(R(x)\)中包含无理系数时结论不一定成立. 30 | \mist{27} (F)(O)(C)错印为 (F)(D)(C). 31 | \mist{281} 最后一行\(g\)应为\(f\). 32 | \mist{196} \(f^{(n)}(x)\)后\(-1\)的幂次应为\(n\)而非\(n-1\). 33 | \mist{299} 第二个行列式第一行第六列的元素应为24而非0. 34 | \mist{315} 第七题. 35 | \mist{23}定理8.2.7中大括号. 36 | \section{符号说明} 37 | \begin{longtable}{ll} 38 | \hline 39 | \(\bbn\)&自然数集\\\hline 40 | \(\bbz,\bbz_+\)&整数集,正整数集\\\hline 41 | \(\bbq,\bbq_+\)&有理数域, 正有理数集\\\hline 42 | \(\bbr,\bbr_+\)&实数域, 正实数集\\\hline 43 | \(\bbr^n\)&\(n\)维欧氏空间\\\hline 44 | \(\bbc\)&复数域\\\hline 45 | \(\bbc^n\)&\(n\)维复空间\\\hline 46 | \(S^{n-1}\)&\(\bbr^n\)中的单位球面, 即\(\{\boldsymbol{x}\in\bbr^n\big||x|=1\}\)\\\hline 47 | \(\mathbb{S}^n\)&\(n\)阶实对称矩阵全体\\\hline 48 | \(B_r(\boldsymbol{x})\)&半径为\(r\)中心在\(\boldsymbol{x}\in\bbr^n\)中的开球\\\hline 49 | \(\mathring{B}_r(\boldsymbol{x})\)&半径为\(r\)中心在\(\boldsymbol{x}\in\bbr^n\)中的去心开球\\\hline 50 | \(Q_{\delta}(\boldsymbol{x})\)&边长为\(2\delta\)中心在\(\boldsymbol{x}\in\bbr^n\), 且各边平行于坐标轴的开正方体\\\hline 51 | \(\boldsymbol{A}^\top,\boldsymbol{x}^\top\)&矩阵\(\boldsymbol{A}\), 向量\(\boldsymbol{x}\)的转置\\\hline 52 | \(\boldsymbol{x}\cdot\boldsymbol{y}\)&\(\bbr^n\)中向量\(\boldsymbol{x}\)与\(\boldsymbol{y}\)的数量积, 也常常用\(\inp{x,y},\boldsymbol{x\top y}\)表示\\\hline 53 | \(\inp{x,y}\)&内积空间中两个元素\(x,y\)的内积\\\hline 54 | \(\left\|\boldsymbol{x}\right\|_p\)&\(\bbr^n\)中的向量\(\boldsymbol{x}=(x_1,x_2,\cdots,x_n)^\top\)的\(p-\)范数\(\left(\sum_{k=1}^{n}|x_k|^p\right)^p\)\phantom{\huge\(\int\)}\\\hline 55 | \(\left\|\boldsymbol{A}\right\|_p\)&方阵\(\boldsymbol{A}\bbr^{n\times n}\)的诱导范数\(\left\|\boldsymbol{A}\right\|_p=\max_{\left\|\boldsymbol{x}\right\|=1}\left\|\boldsymbol{Ax}\right\|_p\)\\\hline 56 | \(|\boldsymbol{x}|\)&\(\bbr^n\)中向量\(\boldsymbol{x}\)通常的范数, 即\(\left\|\boldsymbol{x}\right\|_2\)\\\hline 57 | \(\left\|\boldsymbol{A}\right\|\)&方阵\(\boldsymbol{A}\bbr^{n\times n}\)通常的诱导范数\(\left\|\boldsymbol{A}\right\|_2\)\\\hline 58 | \(|E|\)&\(\bbr^n\)中集合\(E\)的Lebesgue测度(或Jordan测度)---长度, 面积, 体积\\\hline 59 | \(|E|^*,|E|_*\)&\(\bbr^n\)中集合\(E\)的Jordan外测度, 内测度\\\hline 60 | \(m^*E,m_*E\)&\(\bbr^n\)中集合\(E\)的Lebesgue外测度, 内测度\\\hline 61 | \(a^+.a_-\)&实数\(a\)的正部\(\left(|a|+a\right)/2\)与负部\(\left(|a|-a\right)/2\)\\\hline 62 | \(a\vee b,a\wedge b\)&实数\(a,b\)的最大值和最小值\\\hline 63 | \(\Re z,\Im z\)&复数\(z=a+b\ii\)的实部\(a\)和虚部\(b\), 其中\(a,b\)为实数\\\hline 64 | \(\chi_E\)&集合\(E\)的特征函数, 既在\(E\)上取值为\(1\), 在其余点取值为\(0\)\\\hline 65 | \(\exists\)&存在\\\hline 66 | \(\forall\)&对于任意\\\hline 67 | \(\gg,\ll\)&大大大于,大大小于\\\hline 68 | \(\mathrm{a.e.}\)&几乎处处\\\hline 69 | \(\mathrm{s.t.}\)&满足或使得\\\hline 70 | \(\varnothing\)&空集\\\hline 71 | \(\in,\ni,\notin\)&\(a\in E\)与\(E\ni a\)均表示\(a\)是\(E\)的元素, \(a\notin E\)表\(a\)不是\(E\)中的元素\\\hline 72 | \(\subseteq,\supseteq\)&\(E\subseteq F\)与\(F\supseteq E\)均表示\(E\)包含于集合\(F\), 即\(F\)包含\(E\)\\\hline 73 | \(E\{\varphi\in F\}\)&表示集合\(\{x\in E\big|\varphi(x)\in F\}\). 在\(E\)明确的情况下, 简记为\(\{\varphi\in F\}\)\\\hline 74 | \(f(D)\)&当\(f\)是映射, \(D\)是集合时, 表示\(D\)的像集\(\{f(x)\big|x\in D\}\)\\\hline 75 | \(\cap\)&集合的交. \(A\cap B\)表示同时属于\(A\)和\(B\)的所有元素组成的集合\\\hline 76 | \(\cup\)&集合的并. \(A\cup B\)表示属于\(A\)或属于\(B\)的所有元素组成的集合\\\hline 77 | \(\backslash\)&集合的差. \(A\backslash B\)表示属于\(A\)但不属于\(B\)的所有元素组成的集合\\\hline 78 | \(\mathscr{C}\)&集合的补. \(\mathscr{C}E\)表示在全集\(X\)明确的情况下, \(E\)的补集\(X\backslash E\)\\\hline 79 | \(E^{\circ}\)&集合\(E\)的内部, 即\(E\)的全体内点\\\hline 80 | \(E'\)&集合\(E\)的导集, 即\(E\)的极限点(聚点)的全体\\\hline 81 | \(\overline{E}\)&集合\(E\)的闭包\\\hline 82 | \(\partial E\)&集合\(E\)的边界\\\hline 83 | \(\subset\subset\)&集合的紧包含关系, \(E\subset\subset F\)当且仅当\(\overline{E}\)是\(F\)的紧子集\\\hline 84 | \(\alpha E+\beta F\)&线性空间中, 集合的伸缩, 代数和与代数差等, 表示集合\(\{\alpha x+\beta y\big|x\in E,y\in F\}\)\\\hline 85 | \(\sum,\prod\)&连加号, 连乘号\\\hline 86 | \([x],{x}\)&实数\(x\)的正数部分(即不大于\(x\)的最大整数)与小数部分(即\(x-[x]\))\\\hline 87 | \(\varlimsup,\varliminf\)&上极限, 下极限\\\hline 88 | \(\upint,\lowint\)&上积分符号, 下积分符号\phantom{\Large\(\int\)}\\\hline 89 | \(C_n^k\)&在\(n\)个元素中选取\(k\)个的组合数\\\hline 90 | \(C^k(\Omega)\)&在\(\Omega\)上有\(k\)阶连续(偏)导数的函数全体\\\hline 91 | \(C_c^k(\Omega)\)&在\(\Omega\)上有紧支集且有\(k\)阶连续(偏)导数的函数全体\\\hline 92 | \(C^{k,\alpha}(\Omega)\)&在\(\Omega\)上\(k\)阶(偏)导数满足\(\alpha\)次H\''{o}lder条件的函数全体\\\hline 93 | \(\mathscr{S} \)&速降函数全体\\\hline 94 | \(\hat{f}\check{f}\)&函数\(f\)的Fourier变换, Fourier逆变换\\\hline 95 | \(f*g\)&函数\(f\)与\(g\)的卷积\\\hline 96 | \end{longtable} 97 | 98 | 99 | 100 | 101 | 102 | -------------------------------------------------------------------------------- /Chapters/chapter1.aux: -------------------------------------------------------------------------------- 1 | \relax 2 | \providecommand\hyper@newdestlabel[2]{} 3 | \@writefile{toc}{\contentsline {chapter}{\numberline {第1章\hspace {.3em}}实数}{3}{chapter.1}\protected@file@percent } 4 | \@writefile{lof}{\addvspace {10.0pt}} 5 | \@writefile{lot}{\addvspace {10.0pt}} 6 | \@writefile{toc}{\contentsline {section}{\numberline {1.1}集合与映射}{3}{section.1.1}\protected@file@percent } 7 | \@writefile{toc}{{\small 集合,关系与映射,Descartes 乘积,函数,集合的势,可数集,不可数集,代数数,超越数,Bernstein定理\par }} 8 | \newlabel{Th:C11}{{1.1.1}{3}{关于Diophantus逼近的Liouville定理}{tcb@cnt@theorem.1.1.1}{}} 9 | \@writefile{toc}{\contentsline {section}{\numberline {1.2}第一次数学危机}{8}{section.1.2}\protected@file@percent } 10 | \@writefile{toc}{{\small 第一次数学危机,可公度量,比例论\par }} 11 | 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\@writefile{toc}{\contentsline {section}{\numberline {1.5}附录}{14}{section.1.5}\protected@file@percent } 22 | \@writefile{toc}{{\small 构造实数系的其他典型方法,实数系的唯一性,序同构,实数系构造的Cantor方法\par }} 23 | \@setckpt{Chapters/chapter1}{ 24 | \setcounter{page}{15} 25 | \setcounter{equation}{0} 26 | \setcounter{enumi}{6} 27 | \setcounter{enumii}{0} 28 | \setcounter{enumiii}{0} 29 | \setcounter{enumiv}{0} 30 | \setcounter{footnote}{0} 31 | \setcounter{mpfootnote}{0} 32 | \setcounter{part}{0} 33 | \setcounter{chapter}{1} 34 | \setcounter{section}{5} 35 | \setcounter{subsection}{0} 36 | \setcounter{subsubsection}{0} 37 | \setcounter{paragraph}{0} 38 | \setcounter{subparagraph}{0} 39 | \setcounter{figure}{2} 40 | \setcounter{table}{0} 41 | \setcounter{parentequation}{0} 42 | \setcounter{tcbbreakpart}{1} 43 | \setcounter{tcblayer}{0} 44 | \setcounter{tcolorbox@number}{20} 45 | \setcounter{float@type}{16} 46 | \setcounter{caption@flags}{2} 47 | \setcounter{continuedfloat}{0} 48 | \setcounter{nlinenum}{0} 49 | \setcounter{section@level}{0} 50 | \setcounter{Item}{31} 51 | \setcounter{Hfootnote}{0} 52 | \setcounter{Hy@AnnotLevel}{0} 53 | \setcounter{bookmark@seq@number}{0} 54 | \setcounter{lstnumber}{1} 55 | \setcounter{tcblisting}{0} 56 | \setcounter{tcbrastercolumn}{1} 57 | \setcounter{tcbrasterrow}{1} 58 | \setcounter{tcbrasternum}{1} 59 | \setcounter{tcbraster}{0} 60 | \setcounter{mn@abspage}{25} 61 | \setcounter{LT@tables}{1} 62 | \setcounter{LT@chunks}{1} 63 | \setcounter{tcb@cnt@theorem}{0} 64 | \setcounter{tcb@cnt@definition}{0} 65 | \setcounter{tcb@cnt@proposition}{0} 66 | \setcounter{tcb@cnt@lemma}{0} 67 | \setcounter{woes}{1} 68 | \setcounter{lstlisting}{0} 69 | \setcounter{mistake}{7} 70 | } 71 | -------------------------------------------------------------------------------- /Chapters/chapter1.tex: -------------------------------------------------------------------------------- 1 | \chapter{实数} 2 | \section{集合与映射} 3 | \precis{集合,关系与映射,Descartes 乘积,函数,集合的势,可数集,不可数集,代数数,超越数,Bernstein定理} 4 | \begin{tcolorbox} 5 | 列入\(\mathcal{A} \)的习题相对基础, 列入\(\mathcal{B}\)的习题相对困难或属于扩展问题, 个别习题可能超前. 若习题中加*号, 表示该习题与后续内容相关性大, 请教师布置习题时, 尽量勾选该题. 6 | \end{tcolorbox} 7 | \begin{theorem}{关于Diophantus逼近的Liouville定理}{C11} 8 | 设\(n\geqslant 2\), 无理数\(x\)是某个\(n\)次整系数多项式的零点, 则存在常数\(A\)使得对于任何整数\(q\)和正整数\(p\),成立\[\left|x-\frac{q}{p}\right|>\frac{A}{p^n}.\] 9 | \end{theorem} 10 | \begin{quiza} 11 | \woe 设\(a_0,a_1,\cdots,a_n\)为复常数, 且对任何\(x\in \mathbb{R}\) 成立 \[a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0.\]证明：\(a_0=a_1=\cdots=a_n=0\). 12 | \begin{proof} 13 | 令\(x_0=0\)代入\(x\)可得\(a_0=0\). 分别令\(x=x_i,(i=1,2,\cdots,n)\)满足\(x_i\ne x_j\)\(j=0,1,\cdots,n\). 可得如下关于\(a_1,a_2,\cdots,a_n\)的方程组. 14 | \[\begin{cases} 15 | a_1x_1+a_2x_1^2+\cdots+a_nx^n_1=0\\ 16 | a_1x_2+a_2x_2^2+\cdots+a_nx^n_2=0\\ 17 | \cdots\\ 18 | a_1x_n+a_2x_n^2+\cdots+a_nx^n_n=0\\ 19 | \end{cases}\] 20 | 其系数矩阵的行列式为 21 | \[\left|\begin{matrix} 22 | x_1&x_1^2&\cdots&x_1^n\\ 23 | x_2&x_2^2&\cdots&x_2^n\\ 24 | \vdots&\vdots&&\vdots\\ 25 | x_n&x_n^2&\cdots&x_n^n\\ 26 | \end{matrix}\right|=\prod_{i=1}^{n}x_i\left|\begin{matrix} 27 | 1&x_1&x_1^2&\cdots&x_1^{n-1}\\ 28 | 1&x_2&x_2^2&\cdots&x_2^{n-1}\\ 29 | \vdots&\vdots&\vdots&&\vdots\\ 30 | 1&x_n&x_n^2&\cdots&x_n^{n-1}\\ 31 | \end{matrix}\right|=\prod_{i=1}^{n}x_i\prod_{1\leqslant in_2\). 此时, 若\(T(q_1)=T(q_2)\)即\[(m_1+n_1)^2+n_1=(m_2+n_2)^2+n_2\]必有\(m_1+n_11,t>n_1\), 于是\[\begin{split} 60 | (m_2+n_2)^2-(m_1+n_1)^2-n_1&=(t+k)^2-t^2-n_1\\ 61 | &=k^2+2kt-n_1>0. 62 | \end{split}\] 63 | 即若\((m_2+n_2)^2>(m_1+n_1)^2\)必有\[(m_2+n_2)^2>(m_1+n_1)^2+n_1\]可见此时不可能有\(T(q_1)=T(q_2)\). \(q<0\)的情况可类似说明. 64 | \end{proof} 65 | \woe 证明代数数集是可列集. 66 | \begin{proof} 67 | 设\(P_n(x)\)为\(n\)次整系数多项式构成的集合. \(U_n\)为所有\(n\)次整系数多项式的复零点, 由于对于一个确定的\(n\)次多项式\[f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n\]它的零点为有限多个. 由于可列个可列集的并为可列集, 接下来我们证明\(P_n(x)\)为可列集. 记\(p(n)\)为第\(n+1\)个素数. 68 | 69 | 定义如下映射\[f(x)\mapsto\prod_{j=0}^{n}\left(p(j)\right)^{a_j}.\] 70 | 这便建立了一个\(P_n(x)\)到\(\mathbb{N}\)的一个双射, 从而\(P_n(x)\)为可列集, 故其元素的复零点的并集为\(U_n\)也为可列集, 从而代数数集\(\underset{n}{\bigcup}\,U_n\)为可列集. 71 | \end{proof} 72 | \woe 证明：存在常数\(A>0\)使得对于任何整数\(q\)和正整数\(p\)成立\[\left|\sqrt{3}-\frac{q}{p}\right|>\frac{A}{p^2}\] 73 | \begin{proof} 74 | \(\sqrt{3}\)为\(x^2-3\)的一个零点, 根据定理\ref{Th:C11}可知结论成立. 75 | \end{proof} 76 | \end{quiza} 77 | \begin{quizb} 78 | \woe 设\(P\)为首项系数不为零的\(n\)次系数多项式, \(x_1\)是它的一个零点, 则存在首项系数不为零的\(n-1\)次多项式\(Q\)使得\[P(x)=(x-x_1)Q(x).\] 79 | \begin{proof} 80 | 由于\[x^k-x_1^k=(x-x_1)\left(x^{k-1}+x_1x^{k-2}+\cdots+x_1^{k-2}x+x_1^{k-1}\right).\]则对\(P(x)\)中的每项应用上述事实便得到了结论. 81 | \end{proof} 82 | \woe 设\(a_0,a_1,\cdots,a_n\)为复常数, \(a_n\ne 0\).证明：多项式\[p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\]至多有\(n\)个复零点(含重数). 83 | \begin{proof} 84 | 这正是代数基本定理, 这个定理的第一个严格证明是1799年由Gauss证明的, 后来他又给出了4个证明, Jordan, Weyl等人也给过这个定理的证明. 它的一般表述是: 次数大于零的复数域上的多项式至少有一复数根. 简单起见, 我们使用复变函数中的Liouville定理进行证明, Liouville定理是说: 有界整函数必是常数. 85 | 86 | 由于\(p(z)\)在\(z\)平面上解析, 若\(p(z)\)在\(z\)平面无零点, 则\(\frac{1}{p(z)}\)在\(z\)平面上也解析, 下面证明后者有界, 从而由Liouville定理推出矛盾. 事实上, 由于\[\lim_{z\rightarrow\infty}p(z)=\lim_{z\rightarrow\infty}z^n\left(a_n+\frac{a_{n-1}}{z}+\cdots+\frac{a_0}{z^n}\right)=\infty,\quad\lim_{z\rightarrow\infty}\frac{1}{p(z)}=0,\]故存在充分大的正数\(R\)使得\(|z|>R\)时, \(\left|\frac{1}{p(z)}\right|<1\). 因\(\frac{1}{p(z)}\)在闭圆\(|z|\leqslant R\)上连续, 故而有界, 从而\(\frac{1}{p(z)}\)在\(z\)平面上有界, 由Liouville定理知\(\frac{1}{p(z)}\)为常数, 这不可能. 87 | \end{proof} 88 | \woe 证明 Vi\`{e}te (韦达) 定理：设\(a_n\ne 0,x_1,x_2,\cdots,x_n\)为\[P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\]的\(n\)个复零点(含重根), 则对于\(1\leqslant k\leqslant n\), 成立 \[\sum_{1\leqslant m_1\frac{K}{p^n}\geqslant\left|x_0-\frac{q}{p}\right| ,\]矛盾, 从而结论得证. 126 | \end{proof} 127 | \woe 证明一个复数是代数数当且仅当它的实部和虚部都是代数数. 128 | \begin{proof} 129 | 设\(x=a+b\ii\), 下证\(x\)是代数数当且仅当\(a\)和\(b\)都是代数数. 130 | 131 | 必要性. 132 | 133 | 充分性. 134 | \end{proof} 135 | \end{quizb} 136 | \section{第一次数学危机} 137 | \precis{第一次数学危机,可公度量,比例论} 138 | \begin{quiza} 139 | 140 | \woe 在有关讲座中, 项武义教授给出了图 \ref{c1t1}. 从中可见, 当\(a,b\)依次是正五边形的边长和对角线长时, \(b-a\)和\(a\)时一个更小的正五边形的边长和对角线. 试以此说明正五边形的边长和对角线长不可公度. 141 | \begin{figure}[H] 142 | \centering 143 | \begin{tikzpicture} 144 | \draw[thick] (0,0)--(2,0);\draw[thick] (2,0)--(2.62,1.90);\draw[thick] (2.62,1.90)--(1,2.98); 145 | \draw[thick] (1,2.98)--(-0.62,1.90);\draw[thick] (0,0)--(-0.62,1.90); 146 | \draw[thick,color=cyan] (0,0)--(2.62,1.90);\draw[thick,color=cyan] (2.62,1.90)--(-0.62,1.90); 147 | \draw[thick,color=cyan] (1,2.98)--(0,0);\draw[thick,color=cyan] (2,0)--(-0.62,1.90); 148 | \draw[thick,color=cyan] (2,0)--(1,2.98); 149 | \draw[thick,color=cyan] (2,0)--(3.236,0);\draw[thick,color=cyan] (3.236,0)--(3.546,1.17); 150 | \draw[thick,color=cyan](3.546,1.17)--(2.62,1.90);\draw[dashed,color=cyan](2.62,1.90)--(3.236,0); 151 | \draw[dashed,color=cyan](3.546,1.17)--(2,0);\draw[dashed,color=cyan](3.546,1.17)--(1.61,1.17); 152 | \draw[dashed,color=cyan](1.61,1.17)--(3.236,0); 153 | \node at (-0.8,1.9) {\footnotesize $B$};\node at (-0.1,-0.2) {\footnotesize$C$}; 154 | \node at (2,-0.2) {\footnotesize$D$};\node at (1,3.2) {\footnotesize$A$}; 155 | \node at (2.8,2) {\footnotesize$E$};\node at (3.8,1.2) {\footnotesize$F$}; 156 | \node at (3.5,0) {\footnotesize$G$};\node at (0.5,2.1) {\footnotesize$D'$}; 157 | \node at (1.5,2.1) {\footnotesize$C'$};\node at (0.2,1) {\footnotesize$E'$}; 158 | \node at (1.,0.5) {\footnotesize$A'$};\node at (1.4,1.2) {\footnotesize$B'$}; 159 | \node at (4,3) {\footnotesize $a=AB$};\node at (4,2.7) {\footnotesize $b=AC$}; 160 | \node at (4.5,2.1) {\footnotesize $a'=DB'=b-a$};\node at (4.2,1.8) {\footnotesize $b'=DE=a$}; 161 | \end{tikzpicture} 162 | \caption{} 163 | \label{c1t1} 164 | \end{figure} 165 | \tcbline 166 | 若正五边形的边和对角线可公度, 则可以把\(a,b\)视为正整数, 此时\(a',b'\)也是正整数, 且\(a'm\)的情况, 当\(n=9,99,999,\cdots\)时, 分数的结果都是循环小数, 且循环节的长度与\(n\)的位数相同.对于一个给定的循环小数\(a=0.a_1a_2\cdots a_na_1a_2\cdots\), (\(a_i\)表示当前位数上的数字), 有\begin{gather*} 197 | 10^na-a=a_1a_2\cdots a_n\\ 198 | a=\frac{a_1a_2\cdots a_n}{10^n-1} 199 | \end{gather*} 200 | \end{quizb} 201 | \section{实数公理系统} 202 | \precis{自然数公理,数学归纳法,实数公理系统,实数系,有序域的性质,三角不等式,Newton二项展开式,杨辉三角,广义实数系,区间,单调函数,复数域,周期函数} 203 | \begin{quiza} 204 | \woe 设\((X,+,\cdot,\leqslant)\)是一个有序域, 并用\(0^*,1^*\)表示\(X\)的零元和单位元. 一般的, \(n^*=n1^*\). 我们可以证明\(X\)中成立\textbf{加法消去律}: 设\(\alpha,\beta,\gamma\in X\), 满足条件\(P:\alpha+\gamma=\beta+\gamma\), 则 205 | \[\begin{split} 206 | \alpha&\ndtstile{{}}{{(A3)}}\alpha+0^*\ndtstile{{}}{{(A4)}}\alpha+\left(\gamma+(-\gamma)\right)\ndtstile{{}}{{(A1)}}(\alpha+\gamma)+(-\gamma)\\ 207 | &\ndtstile{{}}{{P}}(\beta+\gamma)+(-\gamma)\ndtstile{{}}{{(A1)}}\beta+\left(\gamma+(-\gamma)\right)=\beta+0^*=\beta. 208 | \end{split}\]类似的证明: 209 | \begin{quizcs} 210 | \item 若\(\alpha\geqslant \beta\), 则\(\alpha+\gamma\geqslant \beta+\gamma\), 且等号当且仅当\(\alpha=\beta\)时取到. 211 | \item \(\alpha>0^*\)当且仅当\(-\alpha<0^*\). 212 | \item 若\(\alpha\gamma=\beta\gamma\), 且\(\gamma\ne 0^*\), 则\(\alpha=\beta\). 213 | \item \(\alpha^2\geqslant 0^*\), 且等号成立当且仅当\(\alpha=0^*\)时成立. 特别的\(1^*>0^*\). 214 | \item \(\alpha>0^*\)当且仅当\(1/\alpha>0^*\). 215 | \item 设\(\alpha>0,\beta>\gamma\), 则\(\alpha\beta>\alpha\gamma.\) 216 | \end{quizcs} 217 | \begin{proof} 218 | (1) 若\(\alpha>\beta\)有\(\alpha+\gamma=\beta+\gamma\), 依\textbf{加法消去律}可知有\(\alpha=\beta\), 与假设矛盾. 反之若\(\alpha=\beta\), 依上述也有\(\alpha+\gamma=\beta+\gamma\). 219 | 220 | (2) \(\alpha>0^*=\alpha+(-\alpha)\), 依消去律得\(-\alpha<0^*\). 221 | 222 | (3)\(\alpha=\alpha 1^*=\alpha(\gamma\gamma^{-1})=(\alpha\gamma)\gamma^{-1}=\beta\gamma\gamma^{-1}=\beta.\) 223 | 224 | (4)先证 \((-1^*)(-1^*)=1^*\):\[\begin{split} 225 | 1^*+(-1^*)=0^*&\Rightarrow (-1^*)*(1^*+(-1)^*)=0^*\Rightarrow -1^*+(-1^*)(-1^*)=0^*\\ 226 | &\Rightarrow (-1^*)(-1^*)=1^*. 227 | \end{split}\] 228 | 若\(\alpha\geqslant 0\), 依乘法的保序性可知\(\alpha^2\geqslant 0\), 并且若\(\alpha>0\)而\(\alpha^2=0^*\), 则两侧乘以\(\alpha^{-1}\)得\(\alpha=0^*\), 可得矛盾. 从而若\(\alpha^2=0^*\)必有\(\alpha= 0^*\).若\(\alpha <0^*\), 则\(-\alpha>0^*\),从而\[\alpha\alpha=1^*\cdot \alpha\cdot\alpha=(-1^*)(-1^*)\alpha\alpha=(-\alpha)(-\alpha)>0^*.\] 229 | 230 | (5)设\(\alpha>0,\beta<0\), 则\(-\beta>0\), 从而\(\alpha(-\beta)>0^*\), 于是\(\alpha\beta<0\).若\(\alpha>0\)而\(1/\alpha<0\), 则\[\alpha\cdot 1/\alpha=1^*<0^*\]矛盾. 231 | 232 | (6)由于\(\alpha>0,\beta-\gamma>0\), 从而\(\alpha(\beta-\gamma)>0\). 可得\(\alpha\gamma>\alpha\gamma.\) 233 | \end{proof} 234 | \woe 设\((X,+,\cdot,\leqslant)\)是一个全序域, \(\alpha\in X\)满足\(\alpha\geqslant 0\)以及\(\forall \varepsilon>0\), 成立\(\alpha<\varepsilon\). 证明\(\alpha=0.\) 235 | \begin{proof} 236 | 假设\(\alpha>0\), 则令\(\varepsilon=\alpha\), 则与\(\alpha<\varepsilon\)矛盾. 从而\(\alpha=0\). 237 | \end{proof} 238 | \end{quiza} 239 | \begin{quizb} 240 | \woe 对于有序域\((X,+,\cdot,\leqslant)\), 是否一定具有Archimedes性? 241 | \woe 设\((X,+,\cdot,\leqslant)\)是一个全序域, \(\alpha\in X\)满足\(\alpha\geqslant 0\)以及对\(X\)中任何的正有理数\(p\)成立\(\alpha0\), 且\(p^2<2\), 则\(q>p\)且\(q^2<2\). 266 | \item 若\(p>0\), 且\(p^2>2\), 则\(02\). 267 | \end{compactenum} 268 | \begin{solution} 269 | 由\(p>0\)且\(p^2<2\)易知\(q=p+\frac{2-p^2}{p+2}>p\), 并且\[q^2=\left(\frac{2+2p}{p+2}\right)^2=4\left(1-\frac{1}{p+2}\right)^2<4\left(1-\frac{1}{\sqrt{2}+2}\right)^2=2.\]后者情况类似. 270 | \end{solution} 271 | \woe 试求使变换\(q=p+\frac{2-p^2}{ap+b}\)具有习题\textbf{1}中性质(i)---(ii)的所有有理数对\((a,b)\). 272 | \begin{solution} 273 | 274 | \end{solution} 275 | \woe 证明无理数在实数集中的稠密性: 对于任何\(x\in \mathbb{R}\), 以及\(\varepsilon>0\), 存在无理数\(y\)使得\(|y-x|<\varepsilon\). 276 | \woe 设\(a>0\), 而\(n\)为正整数. 证明: \(\min \{a,1\}\leqslant a^{1/n}\leqslant\max\{a,1\}\). 277 | \begin{proof} 278 | 易见\(a\in (0,1)\)时, 有\(a\leqslant a^{1/n}\leqslant 1\); 若\(a\geqslant1\), 则\(1\leqslant a^{1/n}1\)满足\(\frac{1}{p}+\frac{1}{q}=1,a,b>0\). 利用 281 | \begin{equation}\label{c1e1}\tag{\(\heartsuit\)} 282 | (1+x)^\alpha-1\leqslant \alpha x,\qquad x\geqslant 0,0<\alpha\leqslant1 283 | \end{equation} 284 | 证明Young不等式: \(ab\leqslant\frac{1}{p}a^p+\frac{1}{q}b^q\). 285 | \begin{proof} 286 | 不妨设\(a^p>b^q\), 则令\(x=\frac{a^p}{b^q}-1,\alpha=\frac{1}{p}\)带入式\eqref{c1e1}得到\[\frac{a}{b^{q/p}}-1\leqslant \frac{1}{p}\left(\frac{a^p}{b^q}-1\right),\]即\(ab\leqslant\frac{1}{p}a^p+\frac{1}{q}b^q\). 另外的情况类似. 287 | \end{proof} 288 | \woestar 设\(I\)是一个非空指标集, 对\(I\)中每一个元\(\alpha\), 都对应两个实数\(x_\alpha\)以及\(y_\alpha\). 我们将\(\sup\{x_\alpha|\alpha\in I\}\)和\(\inf\{x_\alpha|\alpha\in I\}\)写成\(\underset{\alpha\in I}{\sup}\,x_\alpha\)和\(\underset{\alpha\in I}{\inf}\,x_\alpha\). 证明: 在广义实数系中, 289 | \begin{quizs} 290 | \item \(\underset{\alpha\in I}{\inf}(-x_\alpha)=-\underset{\alpha\in I}{\sup}\,x_\alpha\). 291 | \item 对以下所列的每一个不等式, 若其两端在广义实数系中都有意义, 则该不等式成立:\[\begin{split} 292 | \underset{\alpha\in I}{\inf}\,x_\alpha+\underset{\alpha\in I}{\inf}\,y_\alpha&\leqslant\underset{\alpha\in I}{\inf}(x_\alpha+y_\alpha)\leqslant\underset{\alpha\in I}{\sup}\,x_\alpha+\underset{\alpha\in I}{\inf}\,y_\alpha\\&\leqslant\underset{\alpha\in I}{\sup}(x_\alpha+y_\alpha)\leqslant\underset{\alpha\in I}{\sup}\,x_\alpha+\underset{\alpha\in I}{\sup}\,y_\alpha.\end{split}\] 293 | \item 若\(x_\alpha>0(\forall \alpha\in I)\), 并在此种情形规定\(\frac{1}{0}=+\infty\), 则有\[ \underset{\alpha\in I}{\sup}\,\frac{1}{x_\alpha}=\frac{1}{\underset{\alpha\in I}{\inf}\,x_\alpha}.\] 294 | \item 设\(x_\alpha>0,y_\alpha>0\,(\forall \alpha\in I)\). 对以下所列的每一个不等式, 若其两端在广义实数系中都有意义, 则该不等式成立: \[\begin{split} 295 | \underset{\alpha\in I}{\inf}\,x_\alpha\underset{\alpha\in I}{\inf}\,y_\alpha&\leqslant \underset{\alpha\in I}{\inf}(x_\alpha y_\alpha)\leqslant\underset{\alpha\in I}{\sup}\,x_\alpha\underset{\alpha\in I}{\inf}\,y_\alpha\\&\leqslant 296 | \underset{\alpha\in I}{\sup}(x_\alpha y_\alpha)\leqslant\underset{\alpha\in I}{\sup}\,x_\alpha\,\underset{\alpha\in I}{\sup}\,y_\alpha. 297 | \end{split}\] 298 | \end{quizs} 299 | \begin{proof} 300 | (1) 301 | 302 | (2) 303 | 304 | (3) 305 | 306 | (4) 307 | \end{proof} 308 | \end{quiza} 309 | \begin{quizb} 310 | \woe 对于\(a>0\)以及有理数\(\frac{m}{n}\), 其中\(m,n\)为既约整数, \(n>0\). 当\(m\ne 1\)时, 定义\[a^{\frac{m}{n}}=\left(a^m\right)^\frac{1}{n}\] 311 | 自然, 当\(m=1\)时, 上式也成立. 进一步, 当\(n=2k+1\)为奇数时, 定义\[(-a)^{m/(2k+1)}=(-1)^ma^{m/2k+1}.\]下设\(a,b>0\), 且\(p,q\)为有理数, 证明: 312 | \begin{compactenum}[\quad (1)] 313 | \item \((a^p)^q=a^{pq}\). 314 | \item \(a^pa^q=a^{p+q}\). 315 | \item \(a^pb^p=(ab)^p\). 316 | \item 若\(a>1,p>0\), 则\(a^p>1\). 317 | \item 当\(a>1\)时, \(a^p\)关于\(p\in\mathbb{Q}\)严格单调递增. 318 | \item 设\(a>1\), 证明: 对于任何\(b\in\mathbb{R}\), 成立\(\mathrm{sup}\{a^p|pb.\}\) 319 | \end{compactenum} 320 | \woe 设\(a\geqslant 1\), 且\(b\)为无理数, 定义\[a^b=\sup\{a^p|p\leqslant b \text{ 且 }p\text{ 为有理数}\}.\] 321 | 证明: 当\(p\)为有理数时上式也成立. 322 | 323 | 进一步, 当\(00\), 且\(x,y\in \mathbb{R}\), 证明: 325 | \begin{quizcs} 326 | \item \(\left(a^x\right)^y=a^{xy}\). 327 | \item \(a^xa^y=a^{x+y}\). 328 | \item \(a^xb^x=(ab)^x\). 329 | \item 若\(a>1,x>0\), 则\(a^x>1\). 330 | \item 若\(a>1\), 则\(a^x\)关于\(x\in \mathbb{R}\)严格单增. 331 | \item 若\(00,a\ne 1,b>0\), 证明有唯一实数\(x\)满足\(a^x=b\). 334 | 335 | 该实数称为以\(a\)为底以\(b\)为真数的对数. 记作\(\log_ab\), 在底\(a\)明确的情况下, 可以简写为\(\log b\). \(log_{10}b\)记作\(\lg b\), \(\log_eb\)记作\(\ln b\), 称为自然对数. 336 | \woe 设\(a,b,x>0,a\ne 1,b\ne 1\). 证明: \(\log_ax=\frac{\log_bx}{\log_ba}\). 337 | \woe 证明: 当\(a>1\)时, \(\log_ab\)关于\(x>0\)严格单增. 当\(00\)严格单减. 338 | \woe 设\(a,x,y>0,a\ne 1\), 证明: \(\log_a(xy)=\log_ax+\log_ay\). 339 | \end{quizb} 340 | \section{附录} 341 | \precis{构造实数系的其他典型方法,实数系的唯一性,序同构,实数系构造的Cantor方法} 342 | \begin{quiza} 343 | \woe 证明: 具有最小上界性的有序域一定具有Archimedes性. 344 | \begin{proof} 345 | 令集合\[P=\left\lbrace n\big|\frac{y}{n}\geqslant x \right\rbrace,\]若\(P=\varnothing\), 则对\(\forall n\in\mathbb{Z}_+\), 都有\(nx>y\), 否则\(P\)有最小上界, 不妨记为\(n_0\), 这时有\[\frac{y}{n_0}\geqslant x>\frac{y}{n_0+1},\]右侧的不等号等价于Archimeds性. 346 | \end{proof} 347 | \end{quiza} 348 | \begin{quizb} 349 | \woe 试在Cantor用Cauchy列构造实数的基础上, 建立上确界存在定理, Cauchy准则等(参见第二章内容). 350 | \end{quizb} -------------------------------------------------------------------------------- /Chapters/chapter10.aux: 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其中\(C\)为空间曲线\(\begin{cases} 8 | x(t)=\cos t,\\y(t)=\sin t,\\z(t)=t 9 | \end{cases}t\in[0,2\pi]\). 10 | \begin{solution} 11 | 由参数方程得 12 | \[\int_Cxyz\dd s=\int_{0}^{2\pi}\cos t\cdot\sin t\cdot t\sqrt{\sin^2 t+\cos^2 t+1}\dd t=-\frac{\sqrt{2}\pi}{2}.\qedhere\] 13 | \end{solution} 14 | \item \(\int_Cx^2\dd s\), 其中\(C\)为平面曲线\(x^{2/3}+ y^{2/3}=1.\) 15 | \begin{solution} 16 | 置\(x=\cos^3t,y=\sin^3t,t\in[0,2\pi]\), 从而\[\int_Cx^2\dd s=\int_{0}^{2\pi}\cos^6t\sqrt{9\sin^2t\cos^2t}\dd t=3\int_{0}^{2\pi}\cos^6t\left|\sin t\cos t\right|\dd t=\frac{3}{2}.\qedhere\] 17 | \end{solution} 18 | \item \(\int_C|xy|\dd s\), 其中\(C\)为平面曲线\((x^2+y^2)^2=x^2-y^2\). 19 | \begin{solution} 20 | 置\(x=\sqrt{\cos 2t}\cos t,y=\sqrt{\cos 2t}\sin t,t\in\left[0,\frac{\pi}{4}\right]\), 利用对称性则有\[\int_C|xy|\dd s=4\int_{0}^{\pi/4}\frac{\cos 2t\cdot\cos t\cdot\sin t}{\sqrt{\cos 2t}}\dd t=\int_{0}^{\pi/2}\sin x\cos^{1/2}x\dd x=\frac{1}{2}B\left(1,\frac{3}{4}\right)=\frac{2}{3}.\qedhere\] 21 | \end{solution} 22 | \end{quizcs} 23 | \end{quiza} 24 | \begin{quizb} 25 | \woe 试将定积分的一些性质移植到第一型曲线积分. 26 | \begin{solution} 27 | 28 | \end{solution} 29 | \end{quizb} 30 | \section{第一型曲面积分} 31 | \precis{曲面,同胚,\(k\)维\(C^m\)曲面,Schwarz的例子,集合的面积,分片\(C^m\)曲面,\(\mathbb{R}^n\)中子集的\(k\)维体积(面积),第一型曲面积分,余面积公式,楔积} 32 | \begin{quiza} 33 | \woe 设\(a,b>0\), 计算椭圆柱面\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)在\(z\geqslant 0,y\geqslant 0\)中夹在平面\(y=z\)和平面\(z=0\)之间部分的侧面积. 34 | \begin{solution} 35 | 记\(L\)为在\(xOy\)平面上的曲线\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), 结合题设有\(\frac{x^2}{a^2}+\frac{z^2}{b^2}=1\), 即\(z(x)=b\sqrt{1-\frac{x^2}{a^2}}\), 于是面积可以表成以下第一型曲线积分\[\int_Lz(x)\dd s=\int_{0}^{\pi}b\sqrt{1-\cos^2\theta}\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}\dd\theta.\] 36 | 37 | \tcbline 38 | 选取\(x,z\)为参数, 则由\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)与\(y=z\)得到题设区域到\(xOz\)面的投影\[D=\left\lbrace(x,z)\Big| \frac{x^2}{a^2}+\frac{z^2}{b^2}\leqslant 1,z\geqslant 0 \right\rbrace .\]又记\(F=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0\), 则有\(y_x=-\frac{F_x}{F_y}=-\frac{b^2x}{a^2y}\), 于是题设区域的面积\[S=\iint_D\sqrt{1+y_x^2}\dd x\dd z,\]置\(x=ar\cos\theta,z=br\sin\theta\), 结合\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), 代入\(y_x\), 则有\[\begin{split} 39 | S=\iint_D\sqrt{1+y_x^2}\dd x\dd z=\int_{0}^{\pi}\dd\theta\int_{0}^{1}\sqrt{1+\frac{b^2r^2\cos^2\theta}{a^2\left(1-r^2\cos^2\theta\right)}}\cdot abr\dd r 40 | \end{split}\] 41 | \end{solution} 42 | \woe 计算螺旋面\(\begin{cases} 43 | x=r\cos\theta,\\y=r\sin\theta,\\z=a\theta 44 | \end{cases}\left(r\in[0,R],\theta\in[0,2\pi]\right)\)的面积. 45 | \tcbline 46 | 记\(\boldsymbol{r}(r,\theta)=\left(r\cos\theta,r\sin\theta,a\theta\right)\), 从而\(\boldsymbol{r}_r=\left(\cos\theta,\sin\theta,0\right),\boldsymbol{r}_{\theta}=\left(-r\sin\theta,r\cos\theta,a\right)\), 于是题设曲面的面积\(\left(D=\left\lbrace(r,\theta)\big|r\in[0,R],\theta\in[0,2\pi]\right\rbrace\right)\)\[\begin{split} 47 | S&=\iint_{D}\left|\boldsymbol{r}_u\times\boldsymbol{r}_{\theta}\right|\dd r\dd\theta=\int_{0}^{2\pi}\dd\theta\int_{0}^{R}\sqrt{a^2+r^2}\,\dd r\\&=\pi R\sqrt{a^2+R^2}+\pi a^2\ln\left(\frac{R+\sqrt{a^2+R^2}}{a}\right). 48 | \end{split}\] 49 | \tcbline 50 | \woe 设\(\varSigma\)为三角形\(x+y+z=1(x,y,z\geqslant 0)\), 计算第一型曲面积分\(\iint_{\varSigma}\frac{1}{x+2y+3z}\dd S\). 计算曲面积分\(\int_{\varSigma}x^2\dd S\). 51 | \tcbline 52 | \begin{gather*} 53 | \iint_{\varSigma}\frac{1}{x+2y+3z}\dd S=\int_{0}^{1}\dd x\int_{0}^{1-x}\frac{\sqrt{3}\,\dd y}{3-2x-y}=\sqrt{3}\int_{0}^{1}\ln\left(\frac{3-x}{2-x}\right)\dd x=\sqrt{3}\ln\frac{27}{16}.\\ 54 | \iint_{\varSigma}x^2\dd S=\int_{0}^{1}\dd x\int_{0}^{1-x}x^2\dd y=\int_{0}^{1}x^2(1-x)\dd x=\frac{1}{12}. 55 | \end{gather*} 56 | \tcbline 57 | \woe 设球面\(x^2+y^2+z^2=R^2\)上均匀分布着单位面积质量为\(\rho\)的物质. 某质点质量为\(m\), 位于点\((0,0,r),r\ne R\). 试求球面上的物质的质量以及球面对于该质点的引力. 58 | \begin{solution} 59 | 60 | \end{solution} 61 | \woe 设\(\ell\)是长为\(L\)且足够光滑的平面闭曲线. 对于\(\delta>0\), 设\(W_{\delta}\)为所有以\(\ell\)上的点为球心, \(\delta\)为半径的球的并. 证明: 当\(\delta\)足够小时, \(W_{\delta}\)的表面积为\(2\pi\delta L\). 62 | \tcbline 63 | 当\(\delta\)足够小时, \(W_\delta\)暴露在外的面积为\[\int_{0}^{L}\int_{0}^{2\pi}\theta\delta\dd\theta\dd L=2\pi\delta L.\] 64 | \tcbline 65 | \woe 设\(\varphi,\psi\)在\([a,b]\)上连续可导. 证明: 曲线\(\begin{cases} 66 | x=\varphi(t),\\y=\psi(t) 67 | \end{cases}(t\in[a,b])\)绕\(x\)轴旋转一周所得到的旋转面的面积为\(2\pi\int_{a}^{b}|\psi(t)|\sqrt{|\varphi'(t)|^2+|\psi'(t)|^2}\,\dd t\). 68 | \tcbline 69 | 曲线绕\(x\)轴旋转一周得到旋转面的面积\(S\)正是\[\begin{split} 70 | \sup\left\lbrace\sum_{j=1}^{N}2\pi \alpha_j\left|E_j\right|\Big|\sum_{j=1}^{N}\alpha_j\chi_{E_j}\text{为}[a,b]\text{上的简单函数, }0\leqslant\sum_{j=1}^{N}\alpha_j\chi_{E_j}\leqslant|y|\right\rbrace, 71 | \end{split}\]记题设曲线为\(L\), 于是\(S=2\pi\int_L|y|\dd s=2\pi\int_{a}^{b}|\psi(t)|\sqrt{|\varphi'(t)|^2+|\psi'(t)|^2}\,\dd t.\) 72 | \end{quiza} 73 | 74 | \begin{quizb} 75 | \woe 设\(a,b,c\)为实数, \(A=\sqrt{a^2+b^2+c^2}, f\in L^1[-A,A]\), 证明Poisson(泊松)公式\[\iint\limits_{x^2+y^2+z^2=1}f(ax+by+cz)\dd S=2\pi\int_{-1}^{1}f(Au)\dd u.\] 76 | \begin{proof} 77 | 我们对坐标系\(Oxyz\)进行旋转, 记新坐标系为\(Ouvw\), 平面\(Ovw\)即为\(ax+by+cz=0\), \(u\)轴垂直于该平面, 于是有 \[u=\frac{ax+by+cz}{\sqrt{a^2+b^2+z^2}}.\]则有\[\iint\limits_{x^2+y^2+z^2=1}f(ax+by+cz)\dd S=\iint\limits_{u^2+v^2+w^2=1}f(Au)\dd S,\]于是有\(v^2+w^2=\left(\sqrt{1-u^2}\right)^2\), 则令\[u=u,\quad v=\sqrt{1-u^2}\cos\theta,\quad w=\sqrt{1-u^2}\sin\theta,\] 78 | \end{proof} 79 | 80 | \end{quizb} 81 | \section{第二型曲线积分} 82 | \precis{第二型曲线积分,第一、二型曲线积分的关系,曲线的方向,Jordan闭曲线定理} 83 | 84 | \begin{quiza} 85 | \woe 计算以下第二型曲线积分: 86 | \begin{quizs} 87 | \item \(\int_{C}y\ee^{xy}\cos z\dd x+x\ee^{xy}\cos z\dd y-\ee^{xy}\sin z\dd z,\) 其中\(C\)为曲线\(\begin{cases} 88 | x(t)=\cos t,\\y(t)=\sin t,\\z(t)=t 89 | \end{cases}\)对应于\(t\)从\(0\)到\(2\pi\)的那一段. 90 | \item \(\int_C (3x+2y)\dd x+(x^2-y^2)\dd y\), 其中\(C\)为平面闭曲线\(x^{2/3}+y^{2/3}=1\)的逆时针方向. 91 | \end{quizs} 92 | \begin{solution} 93 | (1)由题意知\[\int_{C}y\ee^{xy}\cos z\dd x+x\ee^{xy}\cos z\dd y-\ee^{xy}\sin z\dd z=\int_{0}^{2\pi}\ee^{\sin t\cos t}\left(\cos 2t-\sin t\right)\dd t,\]首先由 94 | \[\begin{split} 95 | \int_{0}^{2\pi}\ee^{\sin t\cos t}\sin t\dd t&=\int_{0}^{\pi}\ee^{\sin t\cos t}\sin t\dd t+\int_{\pi}^{2 \pi}\ee^{\sin t\cos t}\sin t\dd t\\&=\int_{0}^{\pi}\ee^{\sin t\cos t}\sin t\dd t-\int_{0}^{\pi}\ee^{\sin x\cos x}\sin x\dd x=0,\end{split}\]其次\[\int_{0}^{\pi}\ee^{\sin t\cos t}\cos 2t\dd t=\left.\ee^{\sin t\cos t}\right|^{2\pi}_0=0,\]于是原积分为0. 96 | 97 | (2)置\(x=\cos^3t,y=\sin^3t\), \(t\)从\(0\)到\(2\pi\), 于是 98 | \[\begin{split} 99 | &\int_C (3x+2y)\dd x+(x^2-y^2)\dd y\\ 100 | =&\int_{0}^{2\pi}\left(3\left(\cos^6t-\sin^6t\right)\sin^2t\cos t-3(3\cos^3t+2\sin^3t)\cos^2t\sin t\right)\dd t\\ 101 | =&-6\int_{0}^{2\pi}\sin^4t\cos^2t\dd t=-\frac{3}{4}\pi.\qedhere 102 | \end{split}\] 103 | \end{solution} 104 | \woe 设\(C\)为\(\mathbb{R}^2\)中\(C^1\)简单曲线. \(P,Q\)为\(C\)上的连续函数, 证明\(\left|\int_C P\dd x+Q\dd y\right|\leqslant \int_C\sqrt{P^2+Q^2}\dd s\). 105 | \begin{proof} 106 | 107 | \[ 108 | \left|\int_{C}P\dd x+Q\dd y\right|=\left|\int_C\left(P,Q\right)\cdot \dd \boldsymbol{s}\right|=\int_C\left|\left(P\cos\alpha+Q\sin\alpha\right)\dd s\right|\leqslant\int_C\left|P\cos\alpha+Q\sin\alpha\right|\dd s,\]由Cauchy不等式知\(\left|P\cos\alpha+Q\sin\alpha\right|\leqslant\sqrt{P^2+Q^2}\cdot\sqrt{\cos^2\alpha+\sin^2\alpha}\), 命题获证. 109 | \end{proof} 110 | \end{quiza} 111 | \begin{quizb} 112 | \woe 对于\(\mathbb{C}\)中的\(C^1\)曲线\(C\)及其上的复值连续函数\(f\), 记\(f(z)\dd z=f(x+\mathrm{i}y)(\dd x+\mathrm{i}\dd y)=f(x+\mathrm{i}y)\dd x+\mathrm{i}f(x+\mathrm{i}y)\dd y\). 若令\(s\)为\(C\)的弧长参数, \(C_{s,\Delta s}\)为\(C\)上对应于弧长从\(s\)到\(s+\Delta s\)的那一段, 则\(\lim_{\Delta s\rightarrow 0}\frac{\left|\int_{C_s,\Delta s}\dd z\right|}{|\Delta s|}=1.\)这样相当于\(|\dd z|=\dd s\). 证明: \(\left|\int_C f(z)\dd z\right|\leqslant \int_C|f(z)|\dd s\). 113 | \begin{proof} 114 | 利用题设条件与Cauchy不等式有:\[\begin{split} 115 | &\left|\int_Cf(z)\dd z\right|=\left|\int_C f(x+\mathrm{i}y)\dd x+\mathrm{i}f(x+\mathrm{i}y)\dd y\right|\\ 116 | \leqslant&\int_C\left|\left(f(x+\mathrm{i}y),\mathrm{i}f(x+\mathrm{i}y)\right)\cdot (\dd x,\dd y)\right|\leqslant\int_C|f(z)|\dd s \qedhere 117 | \end{split}\] 118 | \end{proof} 119 | \end{quizb} 120 | 121 | \section{第二型曲面积分} 122 | \precis{第二型曲面积分,第一、二型曲面积分的关系,通量,曲面的侧} 123 | \begin{quiza} 124 | \woe 计算以下第二型曲面积分: 125 | \begin{quizs} 126 | \item \(\iint_{\varSigma}(x+y)\dd y\dd z+(y+z)\dd z\dd x+(z+x)\dd x\dd y\), 其中\(\varSigma\)为曲面\(x^2+\frac{y^2}{4}+\frac{z^2}{9}=1\)的上半部分, 方向取上侧. 127 | \item \(\iint_{\varSigma}x^2\dd y\dd z+y^2\dd z\dd x+(z+2)^2\dd x\dd y\), 其中\(\varSigma\)为锥面\(z=\sqrt{x^2+y^2}\)对应于\(0\leqslant z\leqslant 1\)的那部分, 方向取上侧. 128 | \end{quizs} 129 | \begin{solution} 130 | (1)为计算\(\iint_{\varSigma}(x+y)\dd y\dd z\), 将\(\varSigma\)投影到\(yz\)平面, 其投影为\[D_1=\left\lbrace(y,z)\big|\frac{y^2}{4}+\frac{z^2}{9}\leqslant 1,z\geqslant 0\right\rbrace.\]曲面分为两部分:\[\begin{split} 131 | \varSigma_1&=\left\lbrace (x,y,z)\big|x=\sqrt{1-\frac{y^2}{4}-\frac{z^2}{9}},(y,z)\in D_1 \right\rbrace,\text{方向为前侧},\\ 132 | \varSigma_2&=\left\lbrace (x,y,z)\big|x=-\sqrt{1-\frac{y^2}{4}-\frac{z^2}{9}},(y,z)\in D_1 \right\rbrace,\text{方向为后侧}. 133 | \end{split}\]于是\[\begin{split} 134 | \iint_{\varSigma}(x+y)\dd y\dd z&=\iint_{\varSigma_1}(x+y)\dd y\dd z+\iint_{\varSigma_2}(x+y)\dd y\dd z\\ 135 | &=\iint_{D_1}\left(\sqrt{1-\frac{y^2}{4}-\frac{z^2}{9}}+y\right)\dd y\dd z+\iint_{D_1}\left(\sqrt{1-\frac{y^2}{4}-\frac{z^2}{9}}-y\right)\dd y\dd z\\&=2\iint_{D_1}\sqrt{1-\frac{y^2}{4}-\frac{z^2}{9}}\dd y\dd z=4\pi. 136 | \end{split}\] 137 | 同理计算可得\(\iint_{\varSigma}(y+z)\dd z\dd x=4\pi,\,\iint_{\varSigma}(z+x)\dd x\dd y=4\pi\). 从而\[\iint_{\varSigma}(x+y)\dd y\dd z+(y+z)\dd z\dd x+(z+x)\dd x\dd y=12\pi.\] 138 | 139 | (2)由对称性\[\iint_{\varSigma}x^2\dd y\dd z=\iint_{\varSigma}y^2\dd z\dd x=0.\]于是仅需计算\[\iint_{\varSigma}(z+2)^2\dd x\dd y=\iint\limits_{x^2+y^2\leqslant 1}\left(\sqrt{x^2+y^2}+2\right)^2\dd x\dd y=\frac{43}{6}\pi.\] 140 | \end{solution} 141 | \woe 设\(\varSigma\)为\(\mathbb{R}^3\)中的有向\(C^1\)曲面, \(P,Q,R\)为\(\varSigma\)上的连续函数. 证明:\[\left|\iint_{\varSigma}P\dd y\dd z+Q\dd z\dd x+R\dd x\dd y\right|\leqslant\iint_{\varSigma}\sqrt{P^2+Q^2+R^2}\dd S.\] 142 | \begin{proof} 143 | 由第一, 第二型曲面积分的关系与Cauchy不等式得到\[\left|\iint_{\varSigma}P\dd y\dd z+Q\dd z\dd x+R\dd x\dd y\right|=\left|\iint_{\varSigma}\left(P,Q,R\right)\cdot\boldsymbol{n}(\boldsymbol{x})\dd S\right|\leqslant\iint_{\varSigma}\sqrt{P^2+Q^2+R^2}\dd S.\]其中\(\boldsymbol{n}\)为\(\varSigma\)对应的单位法向量. 144 | \end{proof} 145 | \end{quiza} 146 | \begin{quizb} 147 | \woe 若请你对\(\mathbb{R}^3\)中的曲面\(\varSigma\)即函数\(P\)定义\(\iint_{\varSigma}P(x,y,z)\dd y\dd z\), 则对于\(\varSigma\)和\(P\)的要求要怎么提? 148 | \begin{solution} 149 | 150 | \end{solution} 151 | \end{quizb} 152 | \section{Green公式,Gauss公式,Stokes公式} 153 | \precis{向量场,单连通域,Ostrogradsky-Gauss定理(散度定理),Green公式,Stokes公式,曲线积分和路径无关性,原函数的存在性,循环常数,场论初步,梯度场(保守场),散度场,向量线,环量,旋度,无源场,无旋场,Hamilton算子,Laplace算子,分部积分公式,Green第一、第二公式} 154 | \begin{quiza} 155 | \woe 计算以下曲线积分绕原点的循环常数:\vspace{8pt}\\\vspace{8pt} 156 | \begin{tabular}{lcccl} 157 | \((1)\int_C\frac{-y\dd x+x\dd y}{x^2+y^2}\)&&&\((2)\int_{C}\frac{x\dd x+y\dd y}{x^2+y^2}\). 158 | \end{tabular} 159 | \begin{solution} 160 | 不妨取\(C=\{(x,y)\big|x^2+y^2=1\}\), 置\(x=\cos\theta,y=\sin\theta,(\theta\in[0,2\pi])\), 于是得到\[\int_C\frac{-y\dd x+x\dd y}{x^2+y^2}=2\pi,\quad\int_{C}\frac{x\dd x+y\dd y}{x^2+y^2}=0.\qedhere\] 161 | \end{solution} 162 | \woe 设曲线\(C\)为\(\begin{cases} 163 | x=\cos t,\\y=2t\sin t,\\z=t 164 | \end{cases}\)对应于\(t\)从\(0\)到\(2\pi\)的第一段. 试计算\[\int_C\ee^{xy}\left[\left(y\cos(x+y)+yz^2-\sin(x+y)\right)\dd x+\left(x\cos(x+y)+xz^2-\sin(x+y)\right)\dd y+2z\dd z \right].\] 165 | \begin{solution} 166 | 方便起见设原积分为\(\int_CP\dd x+Q\dd y+R\dd z\). 令\(L\)为直线段\(x=1,y=0,z\)从\(2\pi\)跑到\(0\), 于是\(C\cup L\)为闭曲线, 任取曲面\(\varSigma\)以\(C\cup L\)为边界, 方向与\(C\cup L\)的方向成右手系, 由Stokes公式得到\[\int_{C\cup L}P\dd x+Q\dd y+R\dd z=\iint_{\varSigma}\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)\dd y\dd z+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)\dd z\dd x+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\dd x\dd y=0.\]从而\[\int_CP\dd x+Q\dd y+R\dd z=-\int_LP\dd x+Q\dd y+R\dd z=\int_{0}^{2\pi}2\ee z\dd z=4\ee\pi^2.\qedhere\] 167 | \end{solution} 168 | \woe 设\(R>0\), 计算积分\(\iint_{\varSigma}\frac{Rx\,\dd y\dd z+(z+R)^2\,\dd x\dd y}{\sqrt{x^2+y^2+z^2}}\), 其中\(\varSigma\)为下半球面\(z=-\sqrt{R^2-x^2-y^2}\)的上侧. 169 | \begin{solution} 170 | 记原积分为\(I_2\), 选取平面\(D=\{(x,y,z)\big|x^2+y^2\leqslant R^2,z=0\}\), 方向为上侧, 可以得到\[I_2:=\iint_{D}\frac{Rx\dd y\dd z+(z+R)^2\dd x\dd y}{\sqrt{x^2+y^2+z^2}}=\iint_{D}\frac{R^2}{\sqrt{x^2+y^2}}\dd x\dd y=2\pi R^3.\]记\(\varOmega\)为\(\varSigma\cup D\)的内部, 于是由Gauss公式\[\begin{split} 171 | -I_1+I_2&=\iiint_{\varOmega}\left(\frac{\partial}{\partial x}\left(\frac{Rx}{\sqrt{x^2+y^2+z^2}}\right)+\frac{\partial}{\partial z}\left(\frac{(z+R)^2}{\sqrt{x^2+y^2+z^2}}\right)\right)\dd x\dd y\dd z\\ 172 | =&\frac{13}{6}\pi R^3+\pi R^2-\frac{2}{3}\pi^2R^3 173 | \end{split}\]于是\(I_1=\frac{2}{3}\pi^2R^3-\pi R^2-\frac{1}{6}\pi R^3.\) 174 | \end{solution} 175 | \woe 设\(S\)为椭球面\(\frac{x^2}{2}+\frac{y^2}{2}+z^2=1\)的上半部分\((z>0)\), 对于\(P=(x,y,z)\in S\), \(\varSigma\)为\(S\)在点\(P\)处的切平面, \(\rho(x,y,z)\)为原点到平面\(\varSigma\)的距离, 求积分\(\iint_{S}\frac{z}{\rho(x,y,z)}\dd S\). 176 | \begin{solution} 177 | 首先, 记\(f=\frac{x^2}{2}+\frac{y^2}{2}+z^2-1\), 对于\(P=(A,B,C)\in S\), \(\varSigma\)的方程为\[\left(f_x\big|_{x=A},f_y\big|_{y=B},f_z\big|_ {z=C}\right)\cdot (x-A,y-B,z-C)=0,\]即\(Ax+By+2Cz-A^2-B^2-2C^2=0.\)由此\[\rho(x,y,z)=\frac{x^2+y^2+2z^2}{\sqrt{x^2+y^2+4z^2}},\]即计算\(\iint_{S}\frac{z\sqrt{x^2+y^2+4z^2}}{x^2+y^2+2z^2}\dd S.\) 注意到\(\varSigma\)上的点\((x,y,z)\)处, 指向曲面外侧的的单位法向量为\(\frac{(x,y,2z)}{\sqrt{x^2+y^2+4z^2}}\), 这样就是计算\[I(S):=\iint_{S}\frac{zx}{x^2+y^2+2z^2}\,\dd y\dd z+\frac{zy}{x^2+y^2+2z^2}\,\dd z\dd x+\frac{2z^2}{x^2+y^2+2z^2}\,\dd x\dd y.\]置\(D=\{(x,y,z)\big| \frac{x^2}{2}+\frac{y^2}{2}=1,z=0\}\), 易见\(I(D)=0\). 令\(\varOmega\)表示\(S\cup D\)的内部, 利用Gauss公式, 得到\[\begin{split} 178 | I&=\iiint_{\varOmega}\frac{6z(x^2+y^2+2z^2)-2x^2z-2y^2z-8z^3}{(x^2+y^2+2z^2)^2}\dd x\dd y\dd z\\&=\int_{0}^{2\pi}\dd\theta\int_{0}^{\pi/2}\dd\varphi\int_{0}^{1}\left(6r\cos\varphi\sin\varphi-2r\cos\varphi\sin^3\varphi-4r\cos^3\varphi\sin\varphi\right)\dd r=\frac{\pi}{2}.\qedhere 179 | \end{split}\] 180 | \end{solution} 181 | \woe 设\(\theta_1<\theta_2\leqslant\theta_1+2\pi\). 证明: 在极坐标下由射线\(\theta=\theta_1,\theta=\theta_2\)和曲线\(C:\rho=\rho(\theta)(\theta_1\leqslant \theta\leqslant\theta_2)\)所围成的曲边扇形(如图\ref{fig:m}所示)的面积为\(\frac{1}{2}\int_{\theta_1}^{\theta_2}\rho^2\left(\theta\right)\dd\theta\). 当\(\rho\)连续可导时, 验证这一结果与公式(11.5.9)给出的结果一致. 思考这一面积为什么不是\(\frac{1}{2}\int_C\rho^2\dd\sigma\). 182 | \begin{figure}[H] 183 | \centering 184 | \begin{tikzpicture}[scale=0.5] 185 | \begin{scope} 186 | \clip (1,10) rectangle (5.1,1); 187 | \filldraw[fill=cyan!20,draw=black,thick] (-5,2) circle [radius=10]; 188 | \fill[color=cyan!60!white](60:7.7)--(75:9.1)--(70:8.6); 189 | \fill[color=cyan!60!white](60:7.7)--(75:9.1)--(65:8.16); 190 | \fill[color=cyan!60!white](60:7.7)--(75:9.1)--(0,0); 191 | \draw[black,thick] (-5,2) circle [radius=10]; 192 | \end{scope} 193 | \begin{scope} 194 | \fill[color=cyan!20](0,0)--(1,10)--(1,3)--(4.9498,1); 195 | \draw[-latex,thick] (0,0)--(10,0); 196 | \draw[thick](0,0)--(1,10);` 197 | \draw[thick](0,0)--(4.9498,1);` 198 | \end{scope} 199 | \begin{scope} 200 | \fill[color=cyan!60!white](60:7.5)--(75:9)--(0,0); 201 | \draw[thick] (75:9.1)--(0,0); 202 | \draw[thick] (60:7.7)--(0,0); 203 | \end{scope} 204 | \draw[thick] (1.5,0) arc (0:60:1.5); 205 | \draw[thick] (2.5,0) arc (0:75:2.5); 206 | \node at (1.5,1.1) {$\theta$}; 207 | \node at (4.2,0.3) {$\theta=\theta_1$}; 208 | \node at (2.8,2) {$\theta+\mathrm{d}\theta$}; 209 | \node at (-0.6,4.5) {$\theta=\theta_2$}; 210 | \node at (3.5,8.2) {$\mathrm{d}s$}; 211 | \node at (5.8,5.7) {$\rho=\rho(\theta)$}; 212 | \node at (-0.3,-0.3) {$O$}; 213 | \node at (9.3,-0.3) {$x$}; 214 | \end{tikzpicture} 215 | \caption{第5题图} 216 | \label{fig:m} 217 | \end{figure} 218 | \begin{solution} 219 | \end{solution} 220 | \woe 试对于足够光滑的\(f\)和\(\boldsymbol{F}\), 化简以下表达式:\\ 221 | \begin{tabular}{lclclclcl} 222 | \((1)\nabla\cdot\nabla f\);&&\((2)\nabla\times\nabla f\);&&\((3)\nabla\left(\nabla\cdot\boldsymbol{F}\right)\);&&\((4)\nabla\cdot\left(\nabla\times\boldsymbol{F}\right)\);&&\((5)\nabla\times\left(\nabla\times\boldsymbol{F}\right)\). 223 | \end{tabular} 224 | \begin{solution} 225 | 不妨设\(f=f(x,y,z),\boldsymbol{F}=X(x,y,z)\boldsymbol{i}+Y(x,y,z)\boldsymbol{j}+Z(x,y,z)\boldsymbol{z}\), 于是\[\nabla\cdot\nabla f=\nabla\cdot\left(\frac{\partial f}{\partial x}\boldsymbol{i}+\frac{\partial f}{\partial y}\boldsymbol{j}+\frac{\partial f}{\partial z}\boldsymbol{k}\right)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}.\]容易验证\(\nabla\times\nabla f=\boldsymbol{0},\nabla\cdot\left(\nabla\times\boldsymbol{F}\right)=0\). 226 | \[\nabla\left(\nabla\cdot\boldsymbol{F}\right)=\left(\frac{\partial}{\partial x}\left(\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}+\frac{\partial Z}{\partial z}\right),\frac{\partial}{\partial x}\left(\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}+\frac{\partial Z}{\partial z}\right),\frac{\partial}{\partial z}\left(\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}+\frac{\partial Z}{\partial z}\right)\right),\] 227 | \[\begin{split} 228 | &\nabla\times\left(\nabla\times\boldsymbol{F}\right)=\nabla\times\left(\frac{\partial Z}{\partial y}-\frac{\partial Y}{\partial z},\frac{\partial X}{\partial z}-\frac{\partial Z}{\partial x},\frac{\partial Y}{\partial x}-\frac{\partial X}{\partial y}\right)\\=&\left(\frac{\partial^2Y}{\partial x\partial y}-\frac{\partial^2X}{\partial y^2}-\frac{\partial^2X}{\partial z^2}+\frac{\partial^2Z}{\partial x\partial z},\frac{\partial^2Z}{\partial y\partial z}-\frac{\partial^2Y}{\partial z^2}-\frac{\partial^2Y}{\partial x^2}+\frac{\partial^2X}{\partial y\partial x},\frac{\partial^2X}{\partial z\partial x}-\frac{\partial^2Z}{\partial x^2}-\frac{\partial^2Z}{\partial y^2}+\frac{\partial^2Y}{\partial z\partial y}\right). 229 | \end{split}\]值得一提的是\(\nabla\left(\nabla\cdot\boldsymbol{F}\right)-\nabla\times\left(\nabla\times\boldsymbol{F}\right)=\Delta\boldsymbol{F}\). 230 | \end{solution} 231 | \woe 保守场, 无源场和无旋场之间有没有什么关系? 232 | \begin{solution} 233 | content... 234 | \end{solution} 235 | \woe 用记号\(\nabla\)重写Gauss公式和Stokes公式. 236 | \begin{solution} 237 | Gauss公式可以表示为\[\iint_{\partial \varOmega}\boldsymbol{F}\cdot\dd\boldsymbol{S}=\iiint_{\varOmega}\nabla\cdot\boldsymbol{F}\dd V,\]Stokes公式可表示为\[\int_{\partial\varSigma}\boldsymbol{F}\cdot\dd\boldsymbol{s}=\iint_{\varSigma}\left(\nabla\cdot\boldsymbol{F}\right)\cdot\dd\boldsymbol{S}.\qedhere\] 238 | \end{solution} 239 | \woe 验证\[\int_{\partial\varOmega}\omega=\int_{\varOmega}\mathrm{d}\omega\]适用于Green公式, Ostrogradski\v{\i}-Gauss公式, Stokes公式以及Newton-Leibniz公式. 240 | \begin{solution} 241 | 242 | \end{solution} 243 | \woe 试利用Stokes公式计算\(\int_Cy\dd x+2z\dd y+3x\dd z\), 其中\(C\)为圆周\(\begin{cases} 244 | x^2+y^2+z^2=1,\\x+y+z=0. 245 | \end{cases}\)从\(z\)轴正向\footnote{表示从上往下看.}看去, 曲线是逆时针方向的. 246 | \begin{solution} 247 | 平面\(x+y+z=0\)的法线的方向余弦为\(\cos\alpha=\cos\beta=\cos\beta=\frac{1}{\sqrt{3}}\), 于是由Stokes公式\[\int_Cy\dd x+2z\dd y+3x\dd z\] 248 | \end{solution} 249 | \end{quiza} 250 | \begin{quizb} 251 | \woe 设\(a,b,c>0,\) \(\varSigma\)为曲面\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1.\) 利用Gauss公式计算曲面积分\[\iint_{\varSigma}\frac{\dd S}{(x^2+y^2+z^2)^{3/2}\sqrt{\displaystyle\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}.\] 252 | \begin{solution} 253 | 记原积分为\(I\), 注意到有\[I=\iint_{\varSigma}\frac{\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{y^2}}{(x^2+y^2+z^2)^{3/2}\sqrt{\displaystyle\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}\dd S.\] 254 | 注意到曲面\(\varSigma\)的单位法向量为\(\frac{\left(x/a^2,y/b^2,z/c^2\right)}{\sqrt{x^2/a^4+y^2/b^4+z^2/c^4}}\). 于是有\[I=\iint_{\varSigma}\frac{x\dd y\dd z+y\dd z\dd x+z\dd x\dd y}{\left(x^2+y^2+z^2\right)^{3/2}}=:\iint_{\varSigma}P \dd z+Q\dd z\dd x+R\dd x\dd y,\]由于\(P,Q,R\)在原点处不可导, 而在别处有\[\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0,\]再由Gauss公式有\[I=\iint_{\varSigma'}\frac{\dd S}{x^2+y^2+z^2}=\iint_{\varSigma'}\frac{\dd S}{\rho^2}=\frac{4\pi\rho^2}{\rho^2}=4\pi,\]其中\(\varSigma'\)是半径\(\rho\)充分小使得含于\(\varSigma\)的球面\(x^2+y^2+z^2=\rho^2\). 255 | \end{solution} 256 | \woe 设\(\boldsymbol{A}\)是\(n\)阶方阵, \(\varOmega\)为\(\mathbb{R}^n\)中具有\(C^1\)边界的有界区域, \(\boldsymbol{\varphi}\in C^2_0(\varOmega;\mathbb{R}^n)\). 证明: \[\int_{\varOmega}\mathrm{det}\left(\boldsymbol{A}+\boldsymbol{\varphi_x}\right)\dd x=\mathrm{det}(\boldsymbol{A})\left|\varOmega\right|.\] 257 | \begin{proof} 258 | 259 | \end{proof} 260 | \woe 设\(n\geqslant 1\), \(R\)是仅在点\(0,0\)为零的\(2n\)次二元多项式, \(P,Q\)是不超过\(2n-2\)次的二元多项式. 若\(\varliminf_{(x,y)\rightarrow\infty}\frac{R(x,y)}{(x^2+y^2)^n}>0\), 且在点\((0,0)\)之外成立\(\frac{\partial}{\partial x}\frac{Q(x,y)}{R(x,y)}=\frac{\partial}{\partial y}\frac{P(x,y)}{R(x,y)}\), 证明: 曲线积分\(\int_{C}\frac{P\dd x+Q\dd y}{R}\)绕点\((0,0)\)的循环常数为零. 261 | \begin{proof} 262 | 263 | \end{proof} 264 | \woe 在上一题中, 去掉条件\(\varliminf_{(x,y)\rightarrow\infty}\frac{R(x,y)}{(x^2+y^2)^n}>0\)后结论是否依然成立. 265 | \begin{solution} 266 | 267 | \end{solution} 268 | \end{quizb} 269 | \section{调和函数与解析函数} 270 | \precis{调和函数,平均值公式,最值原理,Poisson公式,复可导与复解析的等价性,Cauchy定理,最大模原理,Liouville定理,共轭函数,利用解析函数计算} 271 | \begin{quiza} 272 | \woe 利用解析函数的性质证明: \(\int_{0}^{+\infty}\frac{\ee^{-x}-\cos x}{x}\dd x=0\). 273 | \begin{proof} 274 | 考虑积分\(\int_{L}\frac{\ee^{\ii z}}{z}\dd z\), 其中\(L\)是如下图所示的路径: 275 | \begin{figure}[H] 276 | \centering 277 | \begin{tikzpicture}[scale=0.8] 278 | \draw[-latex,thick] (-1,0)--(6,0); 279 | \draw[-latex,thick] (0,-1)--(0,6); 280 | \draw[thick,color=red] (0.8,0) arc (0:45:0.8); 281 | \draw[thick,->,color=red] (0,0.8) arc (90:45:0.8); 282 | \draw[thick,->,color=red] (5,0) arc (0:45:5); 283 | \draw[thick,color=red] (0,5) arc (90:45:5); 284 | \draw[->,thick,color=red] (0.8,0)--(2.5,0); 285 | \draw[thick,color=red] (2,0)--(5,0); 286 | \draw[thick,color=red] (0,0.8)--(0,2.5); 287 | \draw[<-,thick,color=red] (0,2.5)--(0,5); 288 | 289 | \node at (-0.3,-0.3) {$O$}; 290 | \node at (6,-0.3) {$x$}; 291 | \node at (-0.3,6) {$y$}; 292 | \node at (3.5,4) {$\Gamma_R$}; 293 | \node at (0.8,0.7) {$\Gamma_r$}; 294 | \node at (0.8,-0.3) {$A$}; 295 | \node at (5,-0.3) {$B$}; 296 | \node at (-0.3,5) {$C$}; 297 | \node at (-0.3,0.8) {$D$}; 298 | \node at (4.5,5) {$L$}; 299 | \end{tikzpicture} 300 | \caption{} 301 | \end{figure} 302 | 其中\(\Gamma_R\)表图中半径为\(R\)的弧. 由于\(f(z)=\frac{\ee^{\ii z}}{z}\)在\(\Gamma_R\)所围区域内解析, 从而\[\int_{\Gamma_R}\frac{\ee^{\ii z}}{z}\dd z=0.\]又\[\int_{L}f(z)\dd z=\int_{AB}f(z)\dd z+\int_{\Gamma_R}f(z)\dd z+\int_{CD}f(z)\dd z\dd z+\int_{\Gamma_r}f(z)\dd z,\] 303 | 注意到\[\begin{split} 304 | &\int_{AB}f(z)\dd z+\int_{CD}f(z)\dd z =\int_{r}^{R}\frac{\ee^{\ii x}}{x}\dd x+\int_{R}^{r}\frac{\ee^{-y}}{y}\dd y\\=&\int_{r}^{R}\frac{-\ee^{-x}+\ee^{\ii x}}{x}\dd x=-\int_{r}^{R}\frac{\ee^{-x}-\cos x}{x}\dd x+\ii\int_{r}^{R}\frac{\sin x}{x}\dd x, 305 | \end{split}\] 306 | 307 | 下面我们说明\(\int_{\Gamma_r}f(z)\dd z=-\frac{\pi\ii }{2}\), 由于\(\lim_{r\rightarrow 0}zf(z)=1\), 则\(r\)充分小时, 有\[\left|z f(z)-1\right|<\frac{2}{\pi}\varepsilon,\]又\[\left|\int_{\Gamma_r}f(z)\dd z-\ii\frac{\pi}{2}\right|=\left|\int_{\Gamma_r}\frac{zf(z)-1}{z}\dd z\right|<\varepsilon,\]从而结论得证. 又由Jordan引理可知\[\lim_{R\rightarrow+\infty}\int_{\Gamma_R}\frac{\ee^{\ii z}}{z}\dd z=0,\]结合上述结论可知\(\int_{0}^{+\infty}\frac{\ee^{-x}-\cos x}{x}\dd x=0\). 308 | \end{proof} 309 | \woe 设\(1\leqslant p<+\infty\), \(\varphi\)为\(\mathbb{R}^n\)上的调和函数, \(\varphi\in L^p(\mathbb{R}^n)\). 证明\(\varphi\equiv 0\). 310 | \begin{proof} 311 | 312 | \end{proof} 313 | \woe 设\(\varphi\in C^2(\mathbb{R}^n)\)为\textit{上调和函数}, 即满足\(-\Delta\varphi\geqslant 0\). 任取\(\boldsymbol{x}_0\in\mathbb{R}^n\), 证明\(F(r)=\frac{1}{|B_r(\boldsymbol{x}_0)|}\cdot\int_{B_r(\boldsymbol{x}_0)}\varphi(\boldsymbol{x})\dd x\)关于\(r>0\)单调. 314 | \begin{proof} 315 | 316 | \end{proof} 317 | \woe 设\(f\)在有界复区域\(D\)上复解析, 在\(\overline{D}\)上连续. 若存在\(z_0\in D\)满足\(|f(z_0)|\leqslant\min_{z\in\partial D}|f(z)|\). 证明: \(f\)在\(D\)内为常数或有零点. 318 | \begin{proof} 319 | 假设\(f\)不为常数且无零点, 则\(\frac{1}{f(z)}\)在\(D\)上解析, 依条件有\(\left|\frac{1}{f(z_0)}\right|\geqslant\max_{x\in\partial D}\left|\frac{1}{f(z)}\right|\), 这与最大模原理矛盾, 从而\(f\)在\(D\)内有零点. 320 | \end{proof} 321 | \woe 设\(f\)在\(\mathbb{C}\)上复解析, \(\lim_{z\rightarrow\infty}|f(z)|=+\infty\). 证明: \(f\)有零点. 进一步, 若存在\(n\geqslant 1\)使得\(\lim_{z\rightarrow\infty}\frac{|f(z)|}{|z|^n}=a\in\left(0,+\infty\right]\), 则\(f\)至少有\(n\)个零点(含重数). 322 | \begin{proof} 323 | 由条件知\(\lim_{z\rightarrow\infty}\left|\frac{1}{f(z)}\right|=0\), 若\(f\)无零点, 则\(\frac{1}{f(z)}\)解析, 但依极限知\(\exists z_0\), 使得\(\forall |z|>|z_0|\)时有\(\left|1/f(z)\right|>\left|1/f(z_0)\right|\), 与最大模原理矛盾, 从而\(f\)有零点. 进一步, 若\(\lim_{z\rightarrow\infty}\frac{\left|f(z)\right|}{|z|^n}=a\in\left(0,+\infty\right]\), 则\(\lim_{z\rightarrow\infty}\left|\frac{f(z)}{z^{n-1}}\right|=+\infty\), 归纳即得结论. 324 | \end{proof} 325 | \woe 证明: 对任何\(z\in\mathbb{C}\)以及\(\delta\in(0,1)\), 有\(\lim_{s\rightarrow+\infty}\frac{\displaystyle\int_{(0,1-\delta)\cup(1+\delta,+\infty)}x^z\left(x\ee^{-x}\right)^s\dd x}{\displaystyle\int_{0}^{+\infty}\left(x\ee^{-x}\right)^s\dd x}=0.\) 326 | \begin{proof} 327 | 记\(\alpha\)为\(x\ee^{-x}\)在\([0,1-\delta]\cup[1+\delta,+\infty]\)上的最大值, 给定\(\beta\in(\alpha,\ee^{-1})\), 则存在长度为\(l\)的区间\(I\subseteq[1-\delta,1+\delta]\), 使得\(x\ee^{-x}\geqslant \beta\)对\(x\in I\)恒成立. 从而\(\int_{0}^{+\infty}\left(x\ee\right)^s\dd x\geqslant l\cdot\beta^s\), \(s\)足够大时, 存在常数\(C_1\)使得\[\left|\int_{0}^{1-\delta}x^z(x\ee^{-x})^{s}\dd x\right|\leqslant C_1\cdot\alpha^s,\]又存在常数\(C_2\)使得\[\begin{split} 328 | &\left|\int_{1+\delta}^{+\infty}x^z(x\ee^{-x})^s\dd x\right|\leqslant\int_{1+\delta}^{+\infty}x^{\mathrm{Re}\,z}(x\ee^{-x})^s\dd x\\&\leqslant\int_{1+\delta}^{+\infty}x^{\mathrm{Re}\, z}(x\ee^{-x})^{s-1}\cdot\left[s(x-1)-\mathrm{Re}\,z\right]\ee^{-x}\dd x=-\int_{1+\delta}^{+\infty}\dd\left[x^{\mathrm{Re}\, z}(x\ee^{-x})^s\right]\leqslant C_2\cdot\alpha^s.\qedhere 329 | \end{split}\] 330 | \end{proof} 331 | \end{quiza} 332 | \begin{quizb} 333 | \woe 设\(\alpha\in\mathbb{R}\), \(\boldsymbol{\sigma}\in S^{n-1}\). 证明: 在\(\mathbb{R}^n\)的单位球\(B_1(\boldsymbol{x})\)内, \(f(\boldsymbol{x})=|\boldsymbol{\sigma}-\boldsymbol{x}|^\alpha\)可以展开成\(\boldsymbol{x}\)的幂级数, 且对任何\(\delta\in(0,1)\), 该幂级数关于\((\boldsymbol{\sigma},\boldsymbol{x})\in S^{n-1}\times B_{\delta}(\boldsymbol{0})\)一致收敛. 334 | \woe 已知当\(0<\alpha<1\)时, 成立\[\sum_{n=1}^{\infty}\frac{1}{n^2-\alpha^2}=\frac{1}{2\alpha^2}-\frac{\pi\cot (\alpha\pi)}{2\alpha}.\]试利用复解析函数的性质证明: 当\(\alpha>0\)时, 成立\[\sum_{n=1}^{\infty}\frac{1}{n^2+\alpha^2}=-\frac{1}{2\alpha^2}+\frac{\pi}{2\alpha\tanh(\alpha\pi)}.\] 335 | \woe 试用不同的方法证明\[\mathrm{sh}\, x=x\prod_{n=1}^{\infty}\left(1+\frac{x^2}{n^2\pi^2}\right).\] 336 | \end{quizb} 337 | \section{附录: \texorpdfstring{\(C^1\)}{}曲面上的Hausdoff测度} 338 | \precis{Binet-Cauchy公式,\(C^1\)曲面的Hausdorff公式} 339 | \begin{theorem}{}{kwei} 340 | 设\(1\leqslant k\leqslant n,\) \(D_0\)为\(\mathbb{R}^k\)中得区域, 单射\(\boldsymbol{\varphi}:D_0\rightarrow\mathbb{R}^n\)连续可微, 则对于紧包含于\(D_0\)得可测集\(D,\,\varSigma=\boldsymbol{\varphi}(D)\)得\(k\)维测度为\[V_k(\varSigma)=\int_D\sqrt{\mathrm{det}\left(\boldsymbol{\varphi_u}^T\boldsymbol{\varphi_u}\right)}\dd\boldsymbol{u}.\]等价地,\[V_k(\varSigma)=\int_D\sqrt{\boldsymbol{\varphi_u}\text{的所有}k\text{阶子式的平方和}}\,\dd\boldsymbol{u}.\] 341 | \end{theorem} 342 | 343 | \begin{quiza} 344 | \woe 设\(\Omega\subseteq\mathbb{R}^n\)为区域, \(\boldsymbol{\varphi}\in C^1\left(\varOmega;\mathbb{R}^m\right)\). 设\(F\subset \Omega\)为紧集,\[\omega(r)=\underset{0<|\boldsymbol{u}-\boldsymbol{v}|\leqslant r\atop\boldsymbol{u}\in F,\boldsymbol{v}\in\Omega}{\mathrm{sup}}\frac{\left|\boldsymbol{\varphi}(\boldsymbol{u})-\boldsymbol{\varphi}(\boldsymbol{v})\right|}{|\boldsymbol{u}-\boldsymbol{v}|},\quad r>0.\]证明\(\lim_{r\rightarrow 0^+}\omega(r)=0.\) 345 | \begin{proof} 346 | 347 | \end{proof} 348 | \end{quiza} 349 | \begin{quizb} 350 | \woe 试减弱定理\reff{Th:kwei} 得条件使得结论仍然成立. 351 | \begin{solution} 352 | 353 | \end{solution} 354 | \end{quizb} 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\section{三角级数,Fourier级数} 3 | \precis{三角级数,Fourier级数,三角级数的复形式,偶延拓,奇延拓,余弦级数,正弦级数} 4 | 5 | \begin{quiza} 6 | \woe 设\(f\)以\(2\pi\)为周期, 对于\(x\in\left[0,2\pi\right),f(x)=\chi_{[a,b]}(x)\), 其中\(0\leqslant a0\)使得\[\left|\int_{A}^{+\infty}f(x)\sin nx\dd x\right|<\varepsilon,\]令\(F(x)=xf(x)\), 则\(F(0^+)=0\), 于是由Dirichlet引理可得\[\lim_{n\rightarrow+\infty}\int_{0}^{A}\frac{F(x)-F(0^+)}{x}\sin nx=0.\]再由\(\varepsilon\)的任意性即得结论. 130 | \end{proof} 131 | \woe 设\(n\geqslant 1\), 证明: \(\int_{0}^{\pi/2}x\left(\frac{\sin nx}{\sin x}\right)^4\dd x<\frac{n^2\pi^2}{4}.\) 132 | \begin{proof} 133 | \def\r{\frac{\sin nx}{\sin x}} 134 | 只需证\(n\geqslant 2\)的情况, 令\[\int_{0}^{\pi/2}x\left(\r\right)^4\dd x=\int_{0}^{\pi/(2n)}x\left(\r\right)^4\dd x+\int_{\pi/(2n)}^{\pi/2}x\left(\r\right)^4\dd x=:I+J.\] 135 | 对\(I\), 用数学归纳法易知\(\left|\r\right|\leqslant n\), 从而\(I\leqslant\frac{n^2\pi^2}{8}\); 对\(J\), 利用\(|\sin nx|\leqslant 1\)及\(\frac{2}{\pi}t<\sin t\left(0< t< \frac{\pi}{2}\right),\) 可得\(J<\frac{n^2\pi^2}{8}\). 相加即得结论. 136 | \end{proof} 137 | \woe 设\(f\in C(\mathbb{R})\)以1为周期, \(f(x)+f\left(x+\frac{1}{2}\right)=f(2x)\). 若存在\(g\in L^1[0,1]\)使得\(f(x)=f(0)+\int_{0}^{x}g(t)\dd t\), 证明: \(f\equiv 0\). 138 | \begin{proof} 139 | 设\(f\)的Fourier展开式为\[\frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k\cos\left(2k\pi x\right)+b_k\sin\left(2k\pi x\right)\right),\]则\(f(x)+f\left(x+\frac{1}{2}\right)\)的Fourier展开为\[\begin{split} 140 | &a_0+\sum_{k=1}^{\infty}\left(\left(1+(-1)^k\right)a_k\cos(2k\pi x)+\left(1+(-1)^k\right)b_k\sin(2k\pi x)\right)\\=&a_0+\sum_{k=1}^{\infty}\left(2a_{2k}\cos\left(4k\pi x\right)+2b_{2k}\sin\left(4k\pi x\right)\right), 141 | \end{split}\]而\(f(2x)\)的Fourier展开为\[\frac{a_0}{2}+\sum_{k=1}^{\infty}\left(a_k\cos(4k\pi x)+b_k\sin(4k\pi x)\right).\]比较系数得到\[a_0=0,\quad a_k=2a_{2k},\quad b_k=2b_{2k},\quad k\geqslant 1,\]从而\[a_k=2^na_{2^nk},\quad b_k=2^nb_{2^nk},\quad\forall k\geqslant 1,n\geqslant 1,\]由题设\(f(x)=f(0)+\int_{0}^{x}g(t)\dd t\), 于是\[\lim_{n\rightarrow+\infty}na_n=\lim_{n\rightarrow+\infty}2n\int_{0}^{1}f(x)\cos(2n\pi x)\dd x=-\lim_{n\rightarrow+\infty}\frac{1}{\pi}\int_{0}^{1}g(x)\sin\left(2n\pi x\right)\dd x=0,\]同理\(\lim_{n\rightarrow+\infty}nb_n=0\), 所以\[a_k=\lim_{n\rightarrow+\infty}\frac{1}{k}2^nka_{2^nk}=0,\quad b_k=\lim_{n\rightarrow+\infty}\frac{1}{k}2^nkb_{2^nk}=0,\quad\forall k\geqslant 1,\]这样\(f\)的Fourier系数均为\(0\), 所以\(f\equiv 0\). 142 | \end{proof} 143 | \woe 设\(1\leqslant p<+\infty,\,f\in L^p[0,2\pi]\), 证明: \(\lim_{n\rightarrow+\infty}\left\|\sigma_n(f;\cdot)-f(\cdot)\right\|_{L^p[0,2\pi]}=0.\) 144 | \begin{proof} 145 | 由于\[\left\|\sigma_n(f;x)-f(x)\right\|_{L^p[0,2\pi]}=\int_{0}^{2\pi}\left|\frac{1}{2n\pi}\int_{0}^{2\pi}\left(f(x+t)-f(x)\right)\left(\frac{\sin\left(nt/2\right)}{\sin(t/2)}\right)^2\dd t\right|^p\dd x,\] 146 | \end{proof} 147 | \end{quiza} 148 | \begin{quizb} 149 | \woe 设\(f\in L^1_{\#}\left(\mathbb{R}\right),g_n\in L^{\infty}_{\#}(\mathbb{R})(n\geqslant 1)\), 满足\[\int_{-\pi}^{\pi}g_n(x)\dd x=1,\quad\int_{-\pi}^{\pi}|g_n(x)|\dd x\leqslant M,\quad\forall n\geqslant 1,\]其中\(M\)为常数. 又对任何\(\delta>0\), 成立\[\lim_{n\rightarrow+\infty}\int_{0}^{\pi}\left(|g_n(x)|+|g_n(-x)|\right)\dd x=0,\quad \sup_{\delta\leqslant|x|\leqslant\pi \atop n\geqslant 1}|g_n(x)|<+\infty.\]证明: 150 | \begin{quizs} 151 | \item 若\(f\)在点\(x_0\)连续, 则\(\lim_{n\rightarrow+\infty}\int_{-\pi}^{\pi}f(y)g_n(x_0-y)\dd y=f(x_0)\). 152 | \item 若\(f\)在\(\mathbb{R}\)上连续, 则\(\lim_{n\rightarrow+\infty}\sup_{x\in\mathbb{R}}\left|\int_{-\pi}^{\pi}f(y)g_n(x-y)\dd y-f(x)\right|=0\). 153 | \end{quizs} 154 | \woe 推广上一题的结果. 155 | \woe 计算\(\int_{0}^{\pi/2}x\ln(\sin x)\ln(\cos x)\dd x.\) 156 | \begin{proof} 157 | 容易得到\(\int_{0}^{\pi/2}x\ln(\sin x)\ln(\cos x)\dd x=\frac{\pi}{4}\int_{0}^{\pi/2}\ln(\sin x)\ln(\cos x)\dd x\), 利用\[\ln\left(2\sin\frac{x}{2}\right)=-\sum_{n=1}^{\infty}\frac{\cos nx}{n},\quad \forall x\in (0,2\pi).\] 158 | 置\(I=\int_{0}^{\pi/2}x\ln(\sin x)\ln(\cos x)\dd x\)我们有\[\begin{split} 159 | I&=\frac{\pi}{8}\int_{0}^{\pi/2}\left\lbrace\left[\ln(\sin x)+\ln(\cos x)\right]^2-\ln^2(\sin x)-\ln^2(\cos x) \right\rbrace\dd x\\ 160 | &=\frac{\pi}{8}\int_{0}^{\pi/2}\left[\ln^2\frac{\sin 2x}{2}-2\ln^2(\sin x)\right]\dd x\\ 161 | &=\frac{\pi}{8}\int_{0}^{\pi/2}\left\lbrace\left[2\ln 2+\sum_{n=1}^{\infty}\frac{\cos 4nx}{n}\right]^2-2\left[\ln 2+\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right]^2 \right\rbrace\dd x\\ 162 | &=\frac{\pi}{8}\left(\pi\ln^2 2-\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}\right)=\frac{\pi^2}{8}\ln^22-\frac{\pi^4}{192}.\qedhere 163 | \end{split}\] 164 | \end{proof} 165 | \woe 试考察函数\(f(x)=\sum_{n=2}^{\infty}\frac{\cos nx}{\ln x}\)在\([-\pi,\pi]\)上的可积性. 166 | \woe 设\(\alpha\in(0,1)\), 考察函数\(f(x)=\sum_{n=1}^{\infty}\frac{\sin nx}{n^\alpha}\)和\(g(x)=\sum_{n=1}^{\infty}\frac{\cos nx}{n^\alpha}\)当\(x\rightarrow 0^+\)时的阶. 167 | \woe 设\(f\in C_{\#}[0,2\pi]\)的Fourier级数为\(\sum_{n=1}^{\infty}\left(a_n\cos nx+b_n\sin nx\right)\). 问\(\sum_{n=1}^{\infty}\left(b_n\cos nx-a_n\sin nx\right)\)是不是某个\(g\in C_{\#}[0,2\pi]\)的Fourier级数? 168 | \woe 设\(\{b_n\}\)为单调下降的正数列, 证明: \(\sum_{n=1}^{\infty}b_n\sin nx\)在\([-\pi,\pi]\)上一致收敛的充要条件是\(nb_n\rightarrow 0\). 169 | \woe 试讨论如何定义方程(12,2,47)的解, 以及在何种条件下, 方程(12.2.47)有唯一解, 而(12.2.48)-(12.2.49)给出了方程的解. 170 | \woe 对于\(p\in[1,+\infty)\), 证明(12.3.38)式与(12.2.40)式的等价性. 171 | \woe 证明(12.2.39)与(12.2.41)式等价. 172 | \end{quizb} 173 | \section{Fourier变换} 174 | \precis{Fourier变换,速降函数(Schwarz函数),Fourier变换的导数,导数的Fourier变换,Fourier逆变换,卷积的Fourier变换,乘积的Fourier变换,Plancherel定理,Hausdoff-Young不等式,处处连续无处可微函数,处处连续无处H\"{o}lder连续函数,热传导方程求解,Heisenberg不确定性原理,\(L^1(\mathbb{R}^n)+L^2(\mathbb{R}^n)\)上的Fourier变换,Borwein积分} 175 | \begin{theorem}{}{C12ff} 176 | 设\(f,g\in\mathscr{S}\), 则\[\begin{split} 177 | &(f*g)^{\land}(\boldsymbol{x})=\widehat{f}(\boldsymbol{x})\widehat{g}(\boldsymbol{x}),\quad\forall\boldsymbol{x}\in\mathbb{R}^n\\ 178 | &(fg)^{\land}(\boldsymbol{x})=\left(\widehat{f}*\widehat{g}\right)(\boldsymbol{x}),\quad\forall\boldsymbol{x}\in\mathbb{R}^n. 179 | \end{split}\] 180 | \end{theorem} 181 | \begin{theorem}{}{C12fe} 182 | 设设\(f,g\in\mathscr{S}\), 则\[\int_{\mathbb{R}^n}f(\boldsymbol{x})\overline{g(\boldsymbol{x})}\dd\boldsymbol{x}=\int_{\mathbb{R}^n}\widehat{f}(\boldsymbol{x})\overline{\widehat{g}(\boldsymbol{x})}\dd\boldsymbol{x}.\] 183 | \end{theorem} 184 | \begin{quiza} 185 | \woe 设\(L\in L^1(\mathbb{R}^n),g(\boldsymbol{x})=f(r\boldsymbol{x})\), 其中\(r>0\)为给定实数, 试求\(f\)的Fourier变换表示\(g\)的Fourier变换. 186 | \begin{solution} 187 | 188 | \end{solution} 189 | \woe 设\(f\in C_c^2(\mathbb{R})\), 证明\(\lim_{x\rightarrow\infty}\left|x^2\widehat{f}(x)\right|=0\). 进而对于任何\(g\in L^{\infty}(\mathbb{R})\), 有\[\lim_{T\rightarrow +\infty}\sum_{n=-\infty}^{+\infty}\frac{1}{T}\widehat{f}\left(\frac{n}{T}\right)g\left(\frac{n}{T}\right)=\int_{\mathbb{R}}\widehat{f}g(x)\dd x.\] 190 | \begin{proof} 191 | 192 | \end{proof} 193 | \woe 对于\[f\in X_a=\{f\big|f\text{非负, 偶, }\mathrm{supp}\,f=\left[-\frac{a}{2},\frac{a}{2}\right],\,f\text{在}\left(-\frac{a}{2},\frac{a}{2}\right)\text{内Lipschitz连续}\},\]证明: \(\int_{\mathbb{R}}\widehat{f}(x)\dd x=f(0).\) 194 | \begin{proof} 195 | 196 | \end{proof} 197 | \woe 设\(f(x)=\pi\ee^{-2\pi |x|}(x\in\mathbb{R})\). 试求\(f\)的Fourier变换. 198 | \begin{solution} 199 | 200 | \end{solution} 201 | \woe 证明: \(g(z)=\int_{-\infty}^{+\infty}\ee^{-\pi y^2}\ee^{2\pi yz}\dd y\)复可导. 202 | \begin{proof} 203 | 204 | \end{proof} 205 | \woe 试用积分号下求导的方法计算\[F(x)=\int_{\bbr}\ee^{-\pi y^2}\ee^{-2\pi\mathrm{i}xy}\dd y,\qquad x\in\mathbb{R}.\] 206 | \begin{solution} 207 | 首先\(F(0)=\int_{\bbr}\ee^{-\pi y^2}\dd y\), 即\[F(0)=\int_{-\infty}^{+\infty}\ee^{-\pi y^2}\dd y=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}\ee^{-t^2}\dd t=1,\]又\[F'(x)=\int_{\bbr}\ee^{-\pi y^2}\ee^{-2\pi\ii xy}(-2\pi\ii y)\dd y=-2\pi\ii\ee^{-\pi x^2}\int_{\bbr}\ee^{-\pi(y+\ii x)^2}y\dd y,\]由\[\begin{split} 208 | &\int_{\bbr}\ee^{-\pi(y+\ii x)^2}y\dd y=\int_{\bbr}\ee^{-\pi(y+\ii x)^2}\left(\left(y+\ii x\right)-\ii x\right)\dd y\\=&\frac{1}{2}\int_{\bbr}\ee^{-\pi(y+\ii x)^2}\dd\left(y+\ii x\right)^2-\ii x\int_{\bbr}\ee^{-\pi(y+\ii x)^2}\dd y=-\ii x, 209 | \end{split}\]于是\(F'(x)=-2\pi x\ee^{-\pi x^2}\), 可知\(F(x)=\int_{0}^{x}F'(x)\dd x+F(0)=\ee^{-\pi x^2}\). 210 | \end{solution} 211 | \end{quiza} 212 | \begin{quizb} 213 | \woe 试仿下列等式给出一些类似的等式. 214 | \tcbline 215 | 对于正数\(a_1,a_2,\cdots,a_n\), 当且仅当\(a_1+a_2+\cdots+a_n\leqslant 1\)时, 成立\[\begin{split} 216 | &\int_{0}^{+\infty}\frac{\sin x}{x}\cdot\frac{\pi^2\cos a_1x}{\pi^2-4a_1^2x^2}\cdots\frac{\pi^2\cos a_nx}{\pi^2-4a_n^2x^2}\dd x=\frac{\pi}{2},\\ 217 | &\frac{1}{2^n\sin^n(1/2)}\int_{0}^{+\infty}\prod_{k=1}^{n}\left(\frac{\sin\frac{2a_kx+1}{2}}{2a_kx+1}+\frac{\sin\frac{2a_kx-1}{2}}{2a_kx-1}\right)\dd x=\frac{\pi}{2},\\ 218 | &\int_{0}^{+\infty}\frac{\sin x}{x}\frac{\pi^2-2a_1\pi x\sin a_1x}{\pi^2-4a_1^2x^2}\cdots\frac{\pi^2-2a_n\pi x\sin a_nx}{\pi^2-4a_n^2x^2}=\frac{\pi}{2},\\ 219 | &3^n\int_{0}^{+\infty}\frac{\sin x}{x}\frac{\sin a_1x-a_1x\cos a_1x}{a_1^3x^3}\cdots\frac{\sin a_nx-a_nx\cos a_nx}{a_n^3x^3}=\frac{\pi}{2}.\\ 220 | \end{split}\] 221 | \tcbline 222 | \begin{solution} 223 | 224 | \end{solution} 225 | \woe 设\(f\in C^1(\mathbb{R})\), 且\(\int_{\mathbb{R}}\left(f^2(x)+\left(f'(x)\right)^2\right)\dd x=1\). 证明: \(\lim_{x\rightarrow\infty}f(x)=0\), 且\(\left\|f\right\|_{\infty}<\frac{\sqrt{2}}{2}\). 226 | \begin{proof} 227 | 228 | \end{proof} 229 | \woe 设\(\alpha\in\mathbb{R}\), 计算含参变量积分\(\int_{0}^{+\infty}\frac{\cos\left(\alpha\pi x\right)}{1+x^2}\dd x,\,\int_{0}^{+\infty}\frac{x\sin\left(\alpha\pi x\right)}{1+x^2}\dd x\). 230 | \begin{solution} 231 | 记\(J(\alpha)=\int_{0}^{+\infty}\frac{\cos\left(\alpha\pi x\right)}{1+x^2}\dd x.\) 显然有\(J(\alpha)=J(-\alpha)\), 为此我们不妨设\(\alpha\geqslant 0\). 注意到\(\int_{0}^{+\infty}\ee^{-t(1+x^2)}\dd t=\frac{1}{1+x^2}\), 于是\[\int_{0}^{+\infty}\cos\left(\alpha\pi x\right)\left(\int_{0}^{+\infty}\ee^{-t(1+x^2)}\dd t\right)\dd x=\int_{0}^{+\infty}\ee^{-t}\dd t\int_{0}^{+\infty}\ee^{-tx^2}\cos\left(\alpha\pi x\right)\dd x,\]记\(I(\alpha,t)=\int_{0}^{+\infty}\ee^{-tx^2}\cos\left(\alpha\pi x\right)\dd x\), 如果我们置\(x=\sqrt{\frac{\pi}{t}}p,\alpha=2\sqrt{\frac{t}{\pi}}q\), 则\[I(\alpha,t)=\sqrt{\frac{\pi}{t}}\int_{0}^{+\infty}\ee^{-\pi p^2}\cos\left(2\pi pq\right)\dd p,\]注意到在12.3.A第6题中算得\[\int_{\bbr}\ee^{-\pi y^2}\ee^{-2\pi\ii xy}\dd y=\ee^{-\pi x^2},\]这意味着\(I(\alpha,t)=\frac{1}{2}\sqrt{\frac{\pi}{t}}\ee^{-\pi q^2}=\frac{1}{2}\sqrt{\frac{\pi}{t}}\exp\left(-\frac{\pi^2\alpha^2}{4t}\right)\). 于是\[J(\alpha)=\frac{\sqrt{\pi}}{2}\int_{0}^{+\infty}\exp\left(-t-\frac{\pi^2\alpha^2}{4t}\right)\frac{\dd t}{\sqrt{t}},\]置\(t=u^2\), 得\[J(\alpha)=\sqrt{\pi}\int_{0}^{+\infty}\exp\left(-\left(u^2+\frac{\pi^2\alpha^2}{4u^2}\right)\right)\dd u=\sqrt{\pi}\ee^{-\pi\alpha}\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\dd u,\]置\(y=u-\frac{\pi\alpha}{2u}\), 有\(\dd y=\left(1+\frac{\pi\alpha}{2u^2}\right)\dd u\), 并且\(u=0,y=-\infty\) ; \(u=+\infty,y=+\infty\), 于是\[\begin{split} 232 | \int_{-\infty}^{+\infty}\ee^{-y^2}\dd y&=\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\left(1+\frac{\pi\alpha}{2u^2}\right)\dd u\\&=\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\dd u+\frac{\pi\alpha}{2}\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\frac{\dd u}{u^2}, 233 | \end{split}\] 234 | 对于\(\frac{\pi\alpha}{2}\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\frac{\dd u}{u^2}\), 令\(u=-\frac{\pi\alpha}{2w}\), 则\(\dd u=\frac{\pi\alpha}{2}\frac{\dd w}{w^2}\), 于是\[\begin{split} 235 | \frac{\pi\alpha}{2}\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2}\right)^2\right)\frac{\dd u}{u^2}&=\frac{\alpha}{2}\int_{-\infty}^{0}\exp\left(-\left(w-\frac{\pi\alpha}{2w}\right)^2\right)\frac{2}{\pi\alpha}\dd w\\&=\int_{-\infty}^{0}\exp\left(-\left(w-\frac{\pi\alpha}{2w}\right)^2\right)\dd w, 236 | \end{split}\]即有\( \int_{-\infty}^{+\infty}\ee^{-y^2}\dd y=\int_{-\infty}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\dd u\), 从而\[\int_{0}^{+\infty}\exp\left(-\left(u-\frac{\pi\alpha}{2u}\right)^2\right)\dd u=\int_{0}^{+\infty}\ee^{-y^2}\dd y=\frac{\sqrt{\pi}}{2},\] 237 | 即得\(J(\alpha)=\frac{\pi}{2}\ee^{-\pi|\alpha|}\). 对于\(\int_{0}^{+\infty}\frac{x\sin\left(\alpha\pi x\right)}{1+x^2}\dd x\), 注意到\[J'(\alpha)=-\pi\int_{0}^{+\infty}\frac{x\sin(\alpha\pi x)}{1+x^2}\dd x=-\frac{\pi^2}{2}\ee^{-\pi\alpha},\quad \alpha>0,\]类似可得其余结果, 最终\(\int_{0}^{+\infty}\frac{x\sin\left(\alpha\pi x\right)}{1+x^2}\dd x=\mathrm{sgn}(\alpha)\frac{\pi}{2}\ee^{-\pi|\alpha|}\). 238 | \end{solution} 239 | \woe 证明: (12.3.27)式给出了方程(12.3.23)的唯一解. 240 | \begin{proof} 241 | 242 | \end{proof} 243 | \woe 设\(p\in(1,2)\), 速降函数列\(\{\varphi_k\}\)在\(L^p\left(\mathbb{R}^n\right)\)中强收敛于\(f\). 证明: \(\{\widehat{\varphi}_k\}\)在\(L^{p'}(\mathbb{R}^n)\)中强收敛于\(\widehat{f}\), 其中\(p'\)是\(p\)的对偶数. 244 | \begin{proof} 245 | 246 | \end{proof} 247 | \woe 设\(f\in L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)\). 证明: 可取到速降函数列\(\{f_k\}\)同时在\(L^1(\mathbb{R}^n)\)和\(L^2(\mathbb{R}^n)\)中强收敛于\(f\). 248 | \begin{proof} 249 | 250 | \end{proof} 251 | \woe 证明: Plancherel定理在\(L^2(\mathbb{R}^n)\)中成立. 252 | \begin{proof} 253 | 254 | \end{proof} 255 | \woe 设\(p\in[1,2]\), 证明Hausdorff-Young不等式对于\(f\in L^p\left(\mathbb{R}^n\right)\)成立. 256 | \begin{proof} 257 | 258 | \end{proof} 259 | \woe 试推广定理\reff{Th:C12ff} 和\reff{Th:C12fe}. 260 | \begin{solution} 261 | 262 | \end{solution} 263 | \end{quizb} 264 | \section{Fourier级数的唯一性} 265 | \precis{Cantor引理,Riemann第一定理,Riemann第二定理,Cantor-Lebesgue定理,Du Bois-Reymond-de la Vall\'{e}e-Poussin定理} 266 | \begin{lemma}{Cantor引理}{anbn} 267 | 设\(\sum_{n=1}^{\infty}\left(a_n\cos nx+b_n\sin nx\right)\)在区间\([a,b]\)上收敛, 则\(\lim_{n\rightarrow+\infty}a_n=\lim_{n\rightarrow+\infty}b_n=0\). 268 | \end{lemma} 269 | \begin{lemma}{Riemann第二定理}{Riemma} 270 | 设\(\lim_{n\rightarrow+\infty}c_n=0\), 则\[\lim_{h\rightarrow 0}\sum_{n=1}^{\infty}\frac{c_n\sin^2nh}{n^2h}=0.\] 271 | \end{lemma} 272 | \begin{lemma}{}{end} 273 | 设\(\sum_{n=1}^{\infty}c_n=A\), 则\[\lim_{h\rightarrow 0}\sum_{n=1}^{\infty}c_n\left(\frac{\sin nh}{nh}\right)^2=A.\] 274 | \end{lemma} 275 | \begin{quiza} 276 | \woe 试利用闭区间套定理证明引理\reff{le:anbn}. 277 | \begin{proof} 278 | 279 | \end{proof} 280 | \woe 证明引理\reff{le:Riemma} 281 | \begin{proof} 282 | 283 | \end{proof} 284 | \end{quiza} 285 | \begin{quizb} 286 | \woe 推广引理\reff{le:end}. 287 | \begin{solution} 288 | 289 | \end{solution} 290 | \woe 利用Rabbe判别法讨论\(\sum_{n=1}^{\infty}\frac{n!n^{-p}}{q(q+1)\cdots(q+n)}(p,q>0)\)的敛散性. 291 | \begin{solution} 292 | 即计算\(\lim_{n\to+\infty}n\left(\frac{a_n}{a_{n+1}}-1\right)\), 有\[\begin{split} 293 | &\lim_{n\to+\infty}n\left(\frac{n!n^{-p}}{q(q+1)\cdots(q+n)}\cdot\frac{q(q+1)\cdots(q+n)(q+n+1)}{(n+1)!(n+1)^{-p}}-1\right)\\ 294 | =&\lim_{n\to+\infty}n\left(\frac{q+n+1}{n+1}\left(\frac{n+1}{n}\right)^p-1\right)\\ 295 | =&\lim_{n\to+\infty}\left(\frac{q(n+1)^{p-1}}{n^{p-1}}+\frac{(n+1)^p-n^p}{n^{p-1}}\right)=q+p, 296 | \end{split}\] 297 | 则\(p+q>1\)时, 原级数收敛; \(p+q<1\)时, 原级数发散. \textbf{而\(p+q=1\)时, Rabbe判别法失效.} 298 | \end{solution} 299 | \tcblower 300 | 事实上由上式可知\[\begin{split} 301 | \frac{a_n}{a_{n+1}}&=\left(1+\frac{q}{n+1}\right)\left(1+\frac{1}{n}\right)^p=\exp\left(\ln\left(1+\frac{q}{n+1}\right)+p\ln\left(1+\frac{1}{n}\right)\right)\\ 302 | &=\exp\left(\frac{p+q}{n}+O\left(\frac{1}{n^2}\right)\right)=1+\frac{p+q}{n}+o\left(\frac{1}{n\ln n}\right), 303 | \end{split}\] 304 | 由Gauss判别法(定理14.3.8)可知\(p+q=1\), 级数发散. 305 | 306 | 307 | 308 | 309 | 310 | 311 | \end{quizb} -------------------------------------------------------------------------------- /Chapters/chapter13.tex: -------------------------------------------------------------------------------- 1 | \chapter{附录} 2 | 3 | \section*{参考文献} 4 | 5 | \section*{索引} 6 | \begin{multicols}{4} 7 | \newcommand{\infe}[2]{#1, #2\\} 8 | \newcommand{\nle}[1]{{\Large\textbf{#1}}\\} 9 | \noindent \(\eta\)函数,129\newline 10 | \(n\)次方根,43\\ 11 | \(p-\text{范数}\),80\\ 12 | \nle{A} 13 | \infe{Abel(阿贝尔)群}{27}\infe{Abel变换}{119}\infe{Abel判别法}{120} 14 | \infe{Archimedes(阿基米德)公理}{28}\infe{Archimedes性}{34,37}\infe{凹函数}{258} 15 | \nle{B} 16 | \infe{Banach-Tarski(巴拿赫-塔斯基)定理}{20} 17 | \infe{Bernoulli(伯努利)方程}{235} 18 | \infe{Bernstein(伯恩斯坦)定理}{17} 19 | \infe{Bolzano-Weierstrass(波尔查诺-维尔斯特拉斯)定理}{91,95} 20 | \infe{比较判别法}{110} 21 | \infe{闭包}{83}\infe{闭集}{82}\infe{闭集套定理}{96}\infe{闭区域}{152}\infe{闭区域}{152} 22 | \infe{边界}{80}\infe{边界点}{80}\infe{表示矩阵}{175}\infe{补集}{13}\infe{不定积分}{209} 23 | \infe{不定型}{69}\infe{不动点}{159}\infe{部分和}{59}\infe{部分积}{59} 24 | \nle{C} 25 | \infe{Cauchy-Cantor(柯西-康托尔)闭区间套定理}{90}\infe{Cauchy-Hadamard(柯西-阿达马)公式}{122}\infe{Cauchy-Schwarz(柯西-施瓦茨)不等式}{78}\infe{Cauchy乘积}{123}\infe{Cauchy列}{48,92} 26 | \infe{Cauchy判别法}{111,115}\infe{Cauchy中值定理}{241}\infe{Cauchy准则}{92,96}\infe{插值多项式}{292}\infe{常数变易公式}{233}\infe{超越数}{15}\infe{稠密}{83}\infe{纯虚数}{31} 27 | \nle{D} 28 | \infe{D'Alembert(达朗贝尔)判别法}{111}\infe{Darboux(达布)定理}{241}\infe{De Morgan(德摩根)定律}{13}\infe{Dedekind(戴德金)分割}{34}\infe{Desartes(笛卡尔)乘积}{10}\infe{Dirichlet(迪利克雷)函数}{32}\infe{Dirichlet判别法}{120}\infe{大\(O\)记号}{69}\infe{大大大于}{63}\infe{大大小于}{63}\infe{代表元}{50} 29 | \infe{代数差}{77}\infe{代数和}{77}\infe{代数基本定理}{158}\infe{代数数}{15}\infe{单侧极限}{134}\infe{单调收敛定理}{85}\infe{单射}{11}\infe{导函数}{171}\infe{导集}{81}\infe{导数}{171}\infe{到上的}{11} 30 | \infe{道路连通分支}{152}\infe{道路连通集}{152}\infe{等价}{50,69}\infe{等价类}{50}\infe{低阶无穷大量}{69}\infe{低阶无穷小量}{68}\infe{调和级数}{96}\infe{调和平均}{270}\infe{定义域}{11}\infe{度量空间}{78}\infe{对角线法}{96}\infe{对偶数}{267}\infe{对数求导法}{186}\infe{对应}{11}\infe{多元函数}{12}\infe{多重(n重)极限}{164}\infe{多重零点}{195}\infe{多重指标}{194} 31 | \nle{E} 32 | \infe{Euclid(欧几里得)距离}{78}\infe{Euler(欧拉)常数}{100}\infe{Euler乘积公式}{129}\infe{Euler公式}{129,201}\infe{二分法}{347}\infe{二项式系数}{30} 33 | \nle{F} 34 | \infe{Fermat(费马)引理}{239}\infe{发散}{57}\infe{法向量}{278,324}\infe{泛函}{11}\infe{范数}{77}\infe{(轴平行)方体}{80}\infe{方向导数}{173}\infe{方向极限}{164}\infe{分部积分公式}{220}\infe{分离变量法}{230}\infe{复合映射}{12}\infe{复数}{31}\infe{复指数函数}{199}\infe{负元}{37} 35 | \nle{G} 36 | \infe{高阶导数}{191}\infe{高阶无穷大量}{69}\infe{高阶无穷小量}{68}\infe{孤立点}{81}\infe{拐点}{350}\infe{关系}{11}\infe{光滑函数}{194}\infe{广义实数系}{30} 37 | \nle{H} 38 | \infe{H\"{o}lder(赫尔德)条件}{194}\infe{H\"{o}lder不等式}{268}\infe{H\"{o}lder连续}{194}\infe{Heine-Borel(海涅-博雷尔)有限覆盖定理}{94,97}\infe{海涅定理}{135}\infe{Hesse(黑塞)矩阵}{266}\infe{函数}{11}\infe{函数极限}{133}\infe{混合偏导}{191} 39 | \nle{J} 40 | \infe{Jensen(詹森)不等式}{258} 41 | \infe{基本列}{49,92}\infe{基数}{13}\infe{积分因子}{233}\infe{极大化序列(极小化序列)}{74}\infe{极大值点}{156}\infe{极限点}{81}\infe{极小值点}{156}\infe{级数的重排}{125}\infe{集合}{7}\infe{集合平均}{43}\infe{加细}{101}\infe{夹逼准则}{63}\infe{贾宪三角}{30}\infe{间断点}{147}\infe{交错级数}{121}\infe{(可)交换群}{27}\infe{阶乘}{28}\infe{介值定理}{154}\infe{紧包含于}{98}\infe{紧集}{98}\infe{局部}{97}\infe{矩阵指数函数}{336}\infe{聚点}{81}\infe{聚点原则}{91,96}\infe{距离空间}{78}\infe{绝对收敛}{119} 42 | \nle{K} 43 | \infe{开覆盖}{94}\infe{开集}{82}\infe{可公度}{22}\infe{可列集}{13}\infe{可数集}{13}\infe{可微}{174} 44 | \nle{L} 45 | \infe{L'H\^{o}pital(洛必达)法则}{249}\infe{Lagrange(拉格朗日)中值定理}{240}\infe{Lagrange乘子法}{331}\infe{Lagrange型插值多项式}{289}\infe{Lagrange型余项}{284}\infe{Lagrange中值定理}{241}\infe{Landau(兰道)不等式}{311}\infe{Lebesgue(勒贝格)数}{101}\infe{Leibniz(莱布尼茨)公式}{193}\infe{Leibniz判别法}{121}\infe{Liouville(刘维尔)定理}{16}\infe{Liouville数}{16}\infe{Lipschitz(利普希茨)连续}{194} 46 | \infe{Lipschitz条件}{194}\infe{Loewner(勒夫纳)偏序}{95}\infe{累次极限}{164}\infe{累级数}{126}\infe{连通分支}{152}\infe{连通集}{152}\infe{连续}{143}\infe{连续公理}{28}\infe{连续函数}{143}\infe{连续可微}{177}\infe{连续模}{157}\infe{连续势}{14}\infe{连续统}{14}\infe{连续统假设}{19}\infe{链式法则}{183}\infe{列紧集}{99}\infe{邻域}{80} 47 | \nle{M} 48 | \infe{Maclaurin(麦克劳林)展开式}{183}\infe{Mertens(梅尔腾斯)定理}{124}\infe{Minkowski(闵可夫斯基)不等式}{268}\infe{满射}{11}\infe{幂平均}{269} 49 | \nle{N} 50 | \infe{Newton(牛顿)(迭代)法}{347}\infe{内部}{80}\infe{内点}{80}\infe{内积}{77}\infe{内积空间}{84}\infe{逆向}{11}\infe{逆映射}{12}\infe{逆元}{39} 51 | \nle{P} 52 | \infe{Peano(佩亚诺)型余项}{283}\infe{Pythagoras(毕达哥拉斯)恒等式}{201}\infe{偏导数}{173}\infe{偏序集}{27}\infe{平方收敛}{348}\infe{平行四边形法则}{79} 53 | \nle{Q} 54 | \infe{齐次方程}{234}\infe{恰当方程}{210}\infe{强制条件}{326}\infe{切平面}{278,324}\infe{切线法}{347}\infe{求和符号}{28}\infe{求积符号}{28}\infe{球体}{80}\infe{区间}{30}\infe{区域}{152}\infe{曲率}{349}\infe{曲率半径}{349}\infe{曲线}{164}\infe{全导数}{174}\infe{全微分方程}{233}\infe{全序集}{27}\infe{确界存在定理}{85} 55 | \nle{R} 56 | 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\begin{proposition}{}{C31} 5 | 设实函数\(f,g\)在\((a,+\infty)\)内局部有界, 且\begin{compactenum}[(i)] 6 | \item 对于任何\(x>a\), \(g(x+1)>g(x)\); 7 | \item \(\lim_{x\rightarrow+\infty}g(x)=+\infty\), 8 | \end{compactenum}则\[\varliminf_{x\rightarrow+\infty}\frac{f(x+1)-f(x)}{g(x+1)-g(x)}\leqslant\varliminf_{x\rightarrow+\infty}\frac{f(x)}{g(x)}\leqslant\varlimsup_{x\rightarrow+\infty}\frac{f(x)}{g(x)}\leqslant\varlimsup_{x\rightarrow+\infty}\frac{f(x+1)-f(x)}{g(x+1)-g(x)}\] 9 | \end{proposition} 10 | \begin{theorem}{Heine定理}{C33} 11 | 设\(E\subseteq\mathbb{R}^n\), \(\boldsymbol{f}:E\rightarrow\mathbb{R}^m\)为\(E\)上的映射, \(\boldsymbol{x}_0\in E'\), 则 \(\lim_{\substack{\boldsymbol{x}\rightarrow\boldsymbol{x}_0\\\boldsymbol{x}\in E}}\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{A}\)当且仅当对\(E\setminus\{\boldsymbol{x}_0\}\)中任何趋于\(\boldsymbol{x}_0\)的点列\(\{\boldsymbol{x}_k\}\), 成立\(\lim_{k\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{x}_k)=\boldsymbol{A}\). 12 | \end{theorem} 13 | \begin{quiza} 14 | \woestar 设\(E\subseteq\mathbb{R}^n\), \(\boldsymbol{f}:E\rightarrow\mathbb{R}^m,\,\boldsymbol{x}_0\in E'\), 证明: \(\lim_{\substack{\boldsymbol{x}\rightarrow\boldsymbol{x}_0\\\boldsymbol{x}\in E}}\boldsymbol{f}(\boldsymbol{x})\)存在当且仅当对\(E\setminus\{\boldsymbol{x}_0\}\)中任何趋于\(\boldsymbol{x}_0\)的点列\(\{\boldsymbol{x}_k\}\), 极限\(\lim_{k\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{x}_k)\)存在. 15 | \begin{proof} 16 | 由Heine定理可知必要性成立, 下证充分性. 易见只需证明若对于\(E\setminus\{\boldsymbol{x}_0\}\)中任何趋于\(\boldsymbol{x}_0\)的点列\(\{\boldsymbol{x}_k\}\), 极限\(\lim_{k\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{x}_k)\)存在且相等即可. 假设序列\(\{\boldsymbol{a}_n\}\)与\(\{\boldsymbol{b}_n\}\)均趋于\(\boldsymbol{x}_0\), 但\[\lim_{n\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{a}_n)\ne \lim_{n\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{b}_n),\]构造序列\(\boldsymbol{c}_n:=\{\boldsymbol{a}_1,\boldsymbol{b}_1,\boldsymbol{a}_2,\cdots,\boldsymbol{a}_{(n+1)/2},\boldsymbol{b}_{n/2},\cdots\}\), 易见\(\lim_{k\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{c}_k)\)不存在, 矛盾. 这说明对于任何趋于\(\boldsymbol{x}_0\)的点列\(\{\boldsymbol{x}_n\}\), 若极限\(\lim_{k\rightarrow+\infty}\boldsymbol{f}(\boldsymbol{x}_k)\)存在, 则必相等, 依Heine定理可知充分性成立. 17 | \end{proof} 18 | \woe 证明\(\lim_{n\rightarrow\infty}n\sum_{k=1}^{n}\left|\sin\frac{k}{n^3}-\frac{k}{n^3}\right|=0\)并由此计算\(\lim_{n\rightarrow\infty}n\sum_{k=1}^{n}\sin\frac{k}{n^3}\). 19 | \begin{proof} 20 | 21 | \end{proof} 22 | \woe 证明命题\reff{po:C31}. 23 | \begin{proof} 24 | \end{proof} 25 | \woe 设数列\(\{x_n\},\{y_n\}\)满足\[\begin{cases} 26 | x_n^2+y_n^2+2y_n=1+\sin\frac{1}{n}\\ 27 | x_n+\left(1+\frac{1}{3n}\right)y_n=\frac{1}{\sqrt[n]{n}} 28 | \end{cases}\] 29 | 证明\(\{x_n\},\{y_n\}\)收敛并求其极限. 30 | \begin{proof} 31 | 取\(z'_n=(x_n,y_n+1)\)则\[|z'_n|^2=x_n^2+y_n^2+2y_n+1=2+\sin\frac{1}{n}\leqslant 3.\]从而\(\{z'_n\}\)是有界列, 于是\(z_n=(x_n,y_n)\)也有界, 取\(\{z_n\}\)的一收敛子列\(\{z_{n_k}\}\), 其收敛于\(z=(x,y)\)且满足\[\begin{cases} 32 | x^2+y^2+2y=1\\ 33 | x+y=1 34 | \end{cases}\]解得\(x=1,y=0\). 35 | \end{proof} 36 | \woe 设\(\psi,\varphi\)为定义在\((0,+\infty)\)上的周期函数, 满足\(\lim_{x\rightarrow+\infty}\left(\psi(x)-\varphi(x)\right)=0\). 证明\(\varphi(x)\equiv\psi(x)\). 37 | \begin{proof} 38 | 若\(\exists x_0\in (0,+\infty)\)使得\(\varphi(x_0)\ne\psi(x_0)\), 则\(\lim_{x\rightarrow+\infty}\left(\psi(x)-\varphi(x)\right)\ne 0\)矛盾, 从而\(\varphi\equiv\psi\). 39 | \end{proof} 40 | \end{quiza} 41 | \begin{quizb} 42 | \woe 证明对于\(x\ne 0\), 成立\(\prod_{n=1}^{\infty}\cos\frac{x}{2^n}=\frac{\sin x}{x}\). 特别地, 有\textbf{Vi\`{e}ta公式}: \(\prod_{n=2}^{\infty}\cos\frac{\pi}{2^n}=\frac{2}{\pi}\). 43 | \begin{proof} 44 | 令\(J_n=\prod_{k=1}^{n}\cos\frac{x}{2^k}\), 有\[\sin\frac{x}{2^n}\cdot J_n=\frac{\sin x}{2^n},\qquad\text{即}\qquad J_n=\frac{\sin x}{2^n\cdot\sin\dfrac{x}{2^n}}\]令\(n\rightarrow+\infty\)即得结果. 后者过程类似. 45 | \end{proof} 46 | \woe 仿照习题3.1 \(\boldsymbol{\mathcal{A}}\)第2题编写一个习题. 47 | \woe 抽取下列几个极限类型, 写出相关定义(其中\(a,A\)为实数), Cauchy准则和Heine定理. 48 | \begin{table}[H] 49 | \centering 50 | \begin{tabular}{|c|c|c|c|c|c|c|c|} 51 | \hline 52 | 自变量变换&极限值&&自变量变化&极限值&&自变量变化&极限值\\\cline{1-2}\cline{4-5}\cline{7-8} 53 | \(x\rightarrow a\)&\(A\)&&\(x\rightarrow a^+\)&\(A\)&&\(x\rightarrow a^-\)&\(A\)\\\cline{1-2}\cline{4-5}\cline{7-8} 54 | \(x\rightarrow a\)&\(+\infty\)&&\(x\rightarrow a^+\)&\(+\infty\)&&\(x\rightarrow a^-\)&\(+\infty\)\\\cline{1-2}\cline{4-5}\cline{7-8} 55 | \(x\rightarrow a\)&\(-\infty\)&&\(x\rightarrow a^+\)&\(-\infty\)&&\(x\rightarrow a^-\)&\(-\infty\)\\\cline{1-2}\cline{4-5}\cline{7-8} 56 | \(x\rightarrow a\)&\(\infty\)&&\(x\rightarrow a^+\)&\(\infty\)&&\(x\rightarrow a^-\)&\(\infty\)\\\cline{1-2}\cline{4-5}\cline{7-8} 57 | \(x\rightarrow \infty\)&\(A\)&&\(x\rightarrow +\infty\)&\(A\)&&\(x\rightarrow -\infty\)&\(A\)\\\cline{1-2}\cline{4-5}\cline{7-8} 58 | \(x\rightarrow \infty\)&\(+\infty\)&&\(x\rightarrow +\infty\)&\(+\infty\)&&\(x\rightarrow -\infty\)&\(+\infty\)\\\cline{1-2}\cline{4-5}\cline{7-8} 59 | \(x\rightarrow \infty\)&\(-\infty\)&&\(x\rightarrow +\infty\)&\(-\infty\)&&\(x\rightarrow -\infty\)&\(-\infty\)\\\cline{1-2}\cline{4-5}\cline{7-8} 60 | \(x\rightarrow \infty\)&\(\infty\)&&\(x\rightarrow +\infty\)&\(\infty\)&&\(x\rightarrow -\infty\)&\(\infty\)\\\hline 61 | \end{tabular} 62 | \end{table} 63 | \woe 设\(a_0,a_1,\ell,\alpha\)为正数, \(a_1\ne a_0,\,a_{n+1}=\frac{(\ell+n^\alpha)a_n^2}{\ell a_n+n^\alpha a_{n-1}}(n\geqslant 1)\). 证明: 64 | \begin{quizcs} 65 | \item 若\(\alpha<1\), 则\(\{a_n\}\)有正的极限. 66 | \item 若\(\alpha=1, \ell>1\), 则\(\{a_n\}\)有正的极限. 67 | \item 若\(\alpha>1\), 则\(\{a_n\}\)发散或极限为零. 68 | \item 若\(\alpha=\ell=1\), 则\(\{a_n\}\)发散或极限为零. 69 | \end{quizcs} 70 | \end{quizb} 71 | \section{连续函数} 72 | \precis{连续函数,左连续,右连续,连续函数的四则运算,复合函数的连续性,基本初等函数的连续性,反函数的连续性,紧集上逆映射的连续性,间断点,一些重要的极限,\(\ee^x\)的无穷级数表示} 73 | \begin{quiza} 74 | \woe 证明Dirichlet函数\[D(x)=\begin{cases} 75 | 0,\quad x\text{为有理数},\\ 76 | 1,\quad x\text{为无理数} 77 | \end{cases}\]在\(\mathbb{R}\)上处处不连续. 78 | \begin{proof} 79 | 80 | \end{proof} 81 | \woe 考虑\([0,1]\)上的Riemann函数\[R(x)=\begin{cases} 82 | 1,\quad x=0,\\ 83 | \frac{1}{q},\quad x=\frac{p}{q}\,\text{(其中\(p,q\)为既约整数)},\\ 84 | 0,\quad x\text{为无理数}. 85 | \end{cases}\]证明\(R\)在有理点不连续, 在无理点连续. 86 | \begin{proof} 87 | 对于任意的\(x_0\in[0,1]\)来说, 若任取\(\varepsilon>0\), 则满足不等式\(q<\frac{1}{\varepsilon}\)的正整数\(n\)至多只有有限个, 即在\([0,1]\)中至多只有有限个有理数\(\frac{p}{q}\), 使得\(f\left(\frac{p}{q}\right)=\frac{1}{q}>\varepsilon\). 因而我们可以取\(\delta>0\), 使得\(\mathring{B}_\delta(x_0)\)内不含这样的有理数. 于是, 只要\(0<|x-x_0|<\delta\), 不论\(x\)是否为有理数, 都成立\(|f(x)|<\varepsilon\). 即证明了对于\([0,1]\)中的任意点\(x_0\), 都成立\[f(x_0+0)=f(x-0)=0.\] 88 | 若\(x_0\)为无理数, 则\(f(x_0)=0\), 可见\(f(x)\)在\(x_0\)处连续; 若\(x_0\)是有理数, 则\(f(x_0)\ne 0\), 点\(x_0\)即成为函数\(f(x)\)的可去间断点. 89 | \end{proof} 90 | \woe 求\(\lim_{n\rightarrow+\infty}\left(\frac{n!}{n^n}\right)^{1/n}\). 91 | \begin{solution} 92 | 我们先计算\(\lim_{n\rightarrow+\infty}\left(\frac{n^n}{n!}\right)^{1/n}\), 取对数整理得到\[\ln\frac{n}{\sqrt[n]{n!}}=\frac{n\ln n-\left(\ln 2+\ln 3+\cdots+\ln n\right)}{n}=:\frac{b_n}{n},\]则有\[\lim_{n\rightarrow+\infty}\frac{b_n}{n}=\lim_{n\rightarrow+\infty}\left(b_{n+1}-b_n\right)=\lim_{n\rightarrow+\infty}\ln\left(1+\frac{1}{n}\right)^n=1,\]故\(\lim_{n\rightarrow+\infty}\left(\frac{n!}{n^n}\right)^{1/n}=\ee\). 93 | \end{solution} 94 | \woe 设\(S_n=\frac{1}{n^2}\sum_{k=0}^{n}\ln C_n^k\), 求\(\lim_{n\rightarrow+\infty}S_n.\) 95 | \begin{solution} 96 | 注意到\[\frac{C_{n+1}^1C_{n+1}^2\cdots C_{n+1}^n}{C_n^1C_n^2\cdots C_n^n}=\frac{1}{n!}(n+1)^n,\]于是依Stolz定理\[\lim_{n\rightarrow+\infty }S_n=\lim_{n\rightarrow+\infty}\frac{n\ln(n+1)-\ln(n!)}{\left((n+1)^2-n^2\right)}=\lim_{n\rightarrow+\infty}=\frac{n+1}{2}\ln\left(1+\frac{1}{n+1}\right)=\frac{1}{2}.\qedhere\] 97 | \end{solution} 98 | \woe 当\(\alpha\in\mathbb{R}\)为何值时, 极限\(\lim_{n\rightarrow+\infty}\frac{1^\alpha+2^\alpha+\cdots+n^\alpha}{n^{\alpha+1}}\)存在? 99 | \begin{solution} 100 | 101 | \end{solution} 102 | \woe 设\(a>0,b>0\), 求证: \(\lim_{n\rightarrow+\infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n=\sqrt{ab}\). 103 | \begin{proof} 104 | \[\begin{split} 105 | \lim_{n\rightarrow+\infty}\left(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\right)^n&=\lim_{n\rightarrow+\infty}\exp\left(n\ln\left(\frac{\sqrt[n]{a}-1+\sqrt[n]{b}-1}{2}-1\right)\right)\\&=\lim_{n\rightarrow+\infty}\exp\left(n\left(\frac{\ln a}{2n}+\frac{\ln b}{2n}\right)\right)\\&=\lim_{n\rightarrow+\infty}\exp\left(\ln\sqrt{ab}\right)=\sqrt{ab}. 106 | \end{split}\] 107 | \end{proof} 108 | \woe 设\(\lim_{x\rightarrow 0}f(x)=0,\lim_{x\rightarrow 0}\frac{f(2x)-f(x)}{x}=0\). 证明\(\lim_{x\rightarrow 0}\frac{f(x)}{x}=0.\) 109 | \begin{proof} 110 | 依题设, 对于任给的\(\varepsilon>0\), 存在\(\delta>0\), 使得当\(0<|x|<\delta\)时,\[\left|\frac{f(x)-f(x/2)}{x}\right|<\varepsilon.\]于是(用\(x/2^{k-1}\)代\(x\))\[\left|\frac{f\left(x/2^{k-1}\right)-f\left(x/2^k\right)}{x/2^{k-1}}\right|<\varepsilon\qquad (k=1,\cdots,n+1).\] 111 | 112 | 因为\[f(x)-f\left(\frac{x}{2^{n+1}}\right)=\sum_{k=1}^{n+1}\left(f\left(\frac{x}{2^{k-1}}\right)-f\left(\frac{x}{2^k}\right)\right).\] 113 | 所以\[\left|\frac{f(x)-f\left(x/2^{n+1}\right)}{x}\right|\leqslant\sum_{k=1}^{n+1}\left|\frac{f\left(x/2^{k-1}\right)-f\left(x/2^k\right)}{x}\right|=\sum_{k=1}^{n+1}\frac{1}{2^k}\left|\frac{f\left(x/2^{k-1}\right)-f\left(x/2^k\right)}{x/2^k}\right|.\]应用最初结果, 得到\[\left|\frac{f(x)-f\left(x/2^{n+1}\right)}{x}\right|<\sum_{k=1}^{n+1}\frac{1}{2^k}\cdot 2\varepsilon=\left(1-\frac{1}{2^{n+1}}\right)\cdot2\varepsilon<2\varepsilon.\]在此式中令\(n\rightarrow +\infty\), 注意\(\lim_{x\rightarrow 0}f(x)=0\), 即得\[\left|\frac{f(x)}{x}\right|\leqslant 2\varepsilon.\]于是\(\lim_{x\rightarrow 0}\frac{f(x)}{x}=0.\) 114 | \end{proof} 115 | \woe 设实数列\(\{x_n\}\)满足\[x_n^3+\left(n^{1/n}+2+\frac{1}{n}\right)x_n^2+\left(\frac{1}{n}+2n^{1/n}+n^{(1-n)/n}+1\right)x_n+n^{1/n}+n^{(1-n)/n}+\frac{1}{n^2}=0.\]证明: \(\{x_n\}\)极限存在, 并求其极限. 116 | \begin{proof} 117 | 注意到\[\begin{split} 118 | &x_n^3+\left(n^{1/n}+2+\frac{1}{n}\right)x_n^2+\left(\frac{1}{n}+2n^{1/n}+n^{(1-n)/n}+1\right)x_n+n^{1/n}+n^{(1-n)/n}\\&=(x_n+1)(x_n+n^{1/n})(x_n+1+\frac{1}{n}), 119 | \end{split}\]当\(n\geqslant 3\)时有\(1<1+\frac{1}{n}+n^{1/n}\), 因为\[\lim_{n\rightarrow\infty}(x_n+1)(x_n+n^{1/n})(x_n+1+\frac{1}{n})=-\lim_{n\rightarrow\infty}\frac{1}{n^2}=0,\]考虑到函数的单调性以及三个零点的极限均为\(-1\), 所以\(\lim_{n\rightarrow\infty}x_n=-1.\) 120 | \end{proof} 121 | \woe 称\(\mathbb{R}\)上函数\(f\)满足局部Lipschitz条件, 如果对于任何\(x\in\mathbb{R}\), 存在\(\delta>0\)以及\(M>0\)使得\[|f(y)-f(z)|\leqslant M|y-z|,\qquad\forall y,z\in(x-\delta,x+\delta).\]证明: \(f\)在\(\mathbb{R}\)上满足局部Lipschitz条件等价于对任何\(A>0\), 存在\(M_A>0\)使得\[|f(x)-f(y)|\leqslant M_A|x-y|,\qquad \forall x,y\in[-A,A].\] 122 | \woe 设\(f\)为\(\mathbb{R}\)上函数. 对任何\(x\in\mathbb{R}\), 存在\(\delta>0\)以及\(M>0\)使得\[|f(y)-f(x)|\leqslant M|y-x|,\qquad\forall y\in (x-\delta,x+\delta).\]问: \(f\)是否一定满足局部Lipschitz条件? 123 | \woe 设\(f\)是\([a,b]\)上的单调连续函数, \(\{a_n\}\)是\([a,b]\)中的点列. 124 | \begin{quizs} 125 | \item 若\(f\)单增, 则\(\varlimsup_{n\rightarrow\infty}f(a_n)=f\left(\varlimsup_{n\rightarrow\infty}a_n\right)\), \(\varliminf_{n\rightarrow\infty}f(a_n)=f\left(\varliminf_{n\rightarrow\infty}a_n\right)\). 126 | \item 若\(f\)单减,则\(\varlimsup_{n\rightarrow\infty}f(a_n)=f\left(\varliminf_{n\rightarrow\infty}a_n\right)\), \(\varliminf_{n\rightarrow\infty}f(a_n)=f(\varlimsup_{n\rightarrow\infty}a_n)\). 127 | \end{quizs} 128 | \begin{proof} 129 | (1) 130 | 131 | (2) 132 | \end{proof} 133 | \end{quiza} 134 | \begin{quizb} 135 | \woe 将习题3.2 \(\boldsymbol{\mathcal{A}}\) 第11题推广到广义实数系情形. 136 | \begin{solution} 137 | 138 | \end{solution} 139 | \woe 证明区间\(I\)上的单调函数的间断点均为跳跃间断点, 且间断点的全体至多可列. 140 | \begin{proof} 141 | 不妨设\(f\)是定义在区间\(I\)上的单调增加函数, 设\(x_0\in(a,b)\subset I\)是\(f\)的间断点, 任取\(x_0\)两侧的点\(x,x'(x0\). 任取\(\varepsilon\in(0,1)\), 存在\(N>0\), 使得当\(n\geqslant N\)时, 成立\(a_n\geqslant 1-\varepsilon\)且\(\sqrt[n]{\displaystyle\prod_{n=1}^{n}a_k}\leqslant 1+\varepsilon\). 159 | 160 | 任取\(\delta>\varepsilon\), 我们来估计\(a_{N+1},a_{N+2},\cdots,a_n\)中大于\(1+\delta\)的数的个数\(m_{n,\delta}\):\[\ln(1+\varepsilon)\geqslant\frac{1}{N}\sum_{k=1}^{N}\ln a_k+\frac{n-N-m_{n,\delta}}{n}\ln(1-\varepsilon)+\frac{m_{n,\delta}}{n}\ln(1+\delta),\]由此可得\[\frac{m_{n,\delta}}{n}\leqslant\frac{\displaystyle\ln(1+\varepsilon)-\ln(1-\varepsilon)-\frac{1}{n}\sum_{k=1}^{n}\ln a_k+\frac{N}{n}\ln(1-\varepsilon)}{\ln(1+\delta)-\ln(1-\varepsilon)}.\]因为\[\frac{1}{n}\sum_{k=1}^{n}a_k\leqslant\frac{1}{n}\sum_{k=1}^{N}a_k+(1+\delta)+\frac{m_{n,\delta}}{n}M,\]所以\[\varlimsup_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{n}a_k\leqslant 1+\delta+\frac{\ln(1+\varepsilon)-\ln(1-\varepsilon)}{\ln(1+\delta)-\ln(1-\varepsilon)}M,\]令\(\varepsilon\rightarrow 0^+\)得\(\varlimsup_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{n}a_k\leqslant 1+\delta\). 在令\(\delta\rightarrow 0^+\)得\(\varlimsup_{n\rightarrow+\infty}\frac{1}{n}\sum_{k=1}^{n}a_k\leqslant 1\). 由此结论得证. 161 | \end{proof} 162 | \woe 证明: \(\bbr\)上实连续函数全体的势是\(\aleph\). 163 | \begin{proof} 164 | 我们先来证明这件事情: 若\(f,g\)是定义在\(R\)上的连续函数, 并且对\(x\in\bbq\)有\(f(x)=g(x)\), 则有\(f\equiv g\). 165 | 166 | 只要对\(\bbr\)中每个无理数\(x\)证明\(f(x)=g(x)\)成立即可. 取有理数序列\(\{r_n\}\), 使\(\lim_{n\rightarrow+\infty}r_n=x\). 例如, 取无理数\(x\)的不足近似值\(r_n=\frac{\left[10^nx\right]}{10^n}\), 则有\[r_n=\frac{\left[10^nx\right]}{10^n}=\frac{10^nx-\theta_{x,n}}{10^n},\]其中\(0\leqslant\theta_{x,n}<1\). 因此\(r_n\rightarrow x(n\rightarrow+\infty)\). 167 | 168 | 由于\(f(r_n)=g(r_n),n\in\bbn_+,f,g\)在点\(x\)连续, 由此知\[f(x)=\lim_{n\rightarrow+\infty}f(r_n)=\lim_{n\rightarrow+\infty}g(r_n)=g(x).\] 169 | 由此便得结论. 记所有连续函数构成的集合为\(T\). 易见所有常值函数\(f(x)\equiv c\)与\(\bbr\)等势, 由伯恩斯坦定理, 若能 170 | 171 | \end{proof} 172 | \end{quizb} 173 | \section{连续函数的基本性质} 174 | \precis{道路,连通集,区域,拓扑学视角下的连续性,相对开集,相对闭集,介值定理,最值定理,连续函数的有界性,一致连续性,\(\mathbb{R}^n\)中范数的等价性,代数基本定理,不动点,压缩映射原理,摄动法,利用极限定义指数函数和对数函数} 175 | \begin{proposition}{}{C32} 176 | \(\mathbb{R}\)中的道路连通集必是空集, 单点集或区间. 177 | \end{proposition} 178 | \begin{quiza} 179 | \woe 设\(f\)为有界区间\((a,b)\)上的连续函数, \(f(a^+),f(b^-)\)都存在且\(f(a^+)<0,f(b^-)>0\). 证明: 存在\(\xi\in (a,b)\)使得\(f(\xi)=0\). 180 | \begin{proof} 181 | 由函数连续性以及\(f(a^+)<0,f(b^-)>0\), 必存在\(\eta>a\)与\(\mu0\), 由函数的介值定理即得结论. 另外, 可以闭区间套. 182 | \end{proof} 183 | \woe 设\(f\)在\((0,+\infty)\)内连续, 且满足\(f(x^2)=f(x)(\forall x>0)\). 证明: \(f\)在\((0,+\infty)\)内为常数. 184 | \begin{proof} 185 | 对任意\(x\in(0,+\infty)\), 有\[f(x)=f\left(\sqrt{x}\right)=f\left(x^{1/4}\right)=\cdots=f\left(x^{1/(2n)}\right),\]令\(n\rightarrow\infty\), 即有\(f(x)=f(1)\), 从而\(f(x)\)是常数. 186 | \end{proof} 187 | \woe 设\(f\)为区间\((a,+\infty)\)上的连续函数, \(f(a^+)\)存在且为负数, \(f(+\infty)=+\infty\). 证明: 存在\(\xi\in(a,+\infty)\)使得\(f(\xi)=0\). 188 | \begin{proof} 189 | 由\(f(+\infty)=+\infty\)知, 对于任意大得正数\(A\), 存在\(X\)使\(x\geqslant X\)时\(f(x)\geqslant A\), 即\(f(X)\)为正, 于是\(f(a^+)f(X)<0\), 由零点存在定理可知有\(\xi\in(a,X)\)使得\(f(\xi)=0\). 190 | \end{proof} 191 | \woe 设\(f\)是\(\mathbb{R}\)上的连续周期函数. 若\(f\)不为常数, 证明它一定有最小正周期. 192 | \begin{proof} 193 | 194 | \end{proof} 195 | \woestar 设\(E\subset\mathbb{R}^n\)为非空有界集. 证明: 连续函数\(\boldsymbol{f}:E\rightarrow\mathbb{R}^m\)一致连续的充要条件是对任何\(\boldsymbol{x}_0\in E',\,\lim_{\substack{\boldsymbol{x}\rightarrow\boldsymbol{x_0}\\\boldsymbol{x}\in E}}\boldsymbol{f}(\boldsymbol{x})\)存在. 这等价于\(\boldsymbol{f}\)是\(\overline{E}\)上的一个连续函数在\(E\)上的限制. 196 | \begin{proof} 197 | 198 | \end{proof} 199 | \woe 证明命题 \reff{po:C32}. 200 | \begin{proof} 201 | 202 | \end{proof} 203 | \woe 试用闭区间套定理证明介值定理. 204 | \begin{proof} 205 | 设\(f(x)\)是\([a,b]\)上的连续函数, \(f(a)0\). 记\(I_1=[a,b]\). 于是若\(F\left(\frac{a+b}{2}\right)=0\), 则定理证毕. 否则若\(F\left(\frac{a+b}{2}\right)>0\), 则令\(I_2=\left[a,\frac{a+b}{2}\right]\), 否则令\(I_2=\left[\frac{a+b}{2},b\right]\). 记\(a'=\frac{a+b}{2}\), 不妨设\(I_1\)为前者, 继续考察\(F\left(\frac{a+a'}{2}\right)\)的符号, 重复上述步骤得到\(I_3,I_4,\cdots I_n,\cdots\). 设\(I_n\)区间的端点为\(a_n,b_n\), 则或者在有限步得到\(F\left(\frac{a_n+b_n}{2}\right)=0\), 则定理证毕. 206 | 否则由\(I_n\)的构造可以发现\(F(a_n)F(b_n)<0\). 并且\[I_{n+1}\subset I_{n},\quad a_n-b_n\rightarrow 0(n\rightarrow+\infty),\]即\(\{I_n\}\)形成闭区间套. 存在有\(\xi=\lim_{n\rightarrow+\infty}a_n=\lim_{n\rightarrow+\infty}b_n\), 并且\[\lim_{n\rightarrow+\infty}F(a_n)\leqslant 0,\quad\lim_{n\rightarrow+\infty}F(b_n)\geqslant 0,\]从而\(F(\xi)=0\), 即\(f(\xi)=\mu.\) 207 | \end{proof} 208 | \woe 证明道路连通集一定是连通集. 209 | \begin{proof} 210 | 设\(E\subseteq\bbr^n\)是一个道路连通集, 且\(E=A\cup B\), 其中\(A,B\)是两个互不相交的非空集. 在\(A\)中任取一点\(\boldsymbol{p}\), 在\(B\)中任取一点\(\boldsymbol{q}\), 则有一条连续曲线\(\varGamma\subset E\)把\(\boldsymbol{p,q}\)两点连接. 令\[\boldsymbol{\varPhi}(t)=\left(\varphi_1(t),\varphi_2(t),\cdots,\varphi_n(t)\right),\quad (a\leqslant t\leqslant b)\]为\(\varGamma\)的参数方程, 并令\[F=\{t\in[a,b]\big|\boldsymbol{\varPhi}(t)\in A\},\quad G=\{t\in[a,b]\big|\boldsymbol{\varPhi}(t)\in B\}.\]易见\(F\)与\(G\)是互不相交的非空集合, 且\(F\cup F=[a,b]\). 由于区间\([a,b]\)是连通集, \(F'\cap G\)与\(F\cap G'\)这两个集合至少有一个非空. 不妨设\(c\in F\cup G'\), 从\(c\in G'\)可知, 有数列\(\{t_n\}\subset G\)使得\(\lim_{n\rightarrow+\infty}t_n=c\). 由于\(\varphi_1,\varphi_2,\cdots,\varphi_n\)连续, 所以\[\lim_{n\rightarrow+\infty}\boldsymbol{\varPhi}(t_n)=\boldsymbol{\varPhi}(c).\]一方面, 由\(\boldsymbol{\varPhi}(t)\in B(i=1,2,\cdots)\), 可知\(\boldsymbol{\varPhi(c)}\in B'\); 另一方面, 利用\(c\in F\)又知\(\boldsymbol{\varPhi(c)}\in A\). 由此得\(\boldsymbol{\varPhi}(c)\in A\cap B'\), 它不是空集, 所以\(E\)是连通的. 211 | \end{proof} 212 | \woe 证明在\(\mathbb{R}\)中的连通集必为空集, 单点集或区间, 即必为道路连通集. 213 | \begin{proof} 214 | 215 | \end{proof} 216 | \woe 参考开集和闭集的性质, 建立相对开集和相对闭集的性质. 217 | \woe 设\(E\subseteq\mathbb{R}^n\)为连通集, \(f:E\rightarrow\mathbb{R}\)连续. 若\(\boldsymbol{x}_1,\boldsymbol{x}_2\in E,\,f(\boldsymbol{x}_1)<\boldsymbol{\eta}<\boldsymbol{f}(\boldsymbol{x}_2)\). 证明: 存在\(\boldsymbol{\xi}\in E\)使得\(\boldsymbol{f}(\boldsymbol{\xi})=\boldsymbol{\eta}\). 218 | \woe 设\(a_n>-1(n\geqslant 1)\). 下表罗列了级数\(\sum_{n=1}^{\infty}a_n\)与\(\sum_{n=1}^{\infty}a_n^2\)收敛的各种情形. 试确定这些情形是否可能发生, 在可能发生的情况下, 讨论此时无穷乘积\(\prod_{n=1}^{\infty}(1+a_n)\)的收敛性. 若无穷乘积在绝对收敛, 条件收敛和发散三种情形中, 有两种以上的情形可能发生, 请举出相应的例子; 同时, 若至少有一种情形不会发生, 给出证明. 219 | \def\j{绝对收敛}\def\t{条件收敛}\def\f{发散} 220 | \begin{table}[H] 221 | \centering 222 | \begin{tabular}{|c|c|c|c|} 223 | \hline 224 | 情形&\(\sum_{n=1}^{\infty}a_n\)的收敛性&\(\sum_{n=1}^{\infty}a_n^2\)的收敛性&\(\prod_{n=1}^{\infty}(1+a_n)\)的收敛性\\\hline 225 | 1&\j&\j&\\\hline 2&\j&\t&\\\hline 3&\j&\f&\\\hline 4&\t&\j&\\\hline 5&\t&\t&\\\hline 226 | 6&\t&\f&\\\hline 7&\f&\j&\\\hline 8&\f&\t&\\\hline 9&\f&\f&\\\hline 227 | \end{tabular} 228 | \end{table} 229 | \begin{solution} 230 | \textbf{情形 1. }此时\(\prod_{n=1}^{\infty}(1+a_n)\)收敛. 231 | 232 | \textbf{情形 2. } 233 | 234 | \textbf{情形 3. } 235 | 236 | \textbf{情形 4. } 237 | 238 | \textbf{情形 5. } 239 | 240 | \textbf{情形 6. } 241 | 242 | \textbf{情形 7. } 243 | 244 | \textbf{情形 8. } 245 | 246 | \textbf{情形 9. } 247 | 248 | \end{solution} 249 | \woe 设\(f\)在\(\left[0,+\infty\right)\)上连续, \(\lim_{n\rightarrow+\infty}f(\sqrt{n})=0\). 证明: \(\lim_{x\rightarrow+\infty}f(x)\)存在的充要条件是\(f\)在\(\left[0,+\infty\right)\)上一致连续. 250 | \begin{proof} 251 | 252 | \end{proof} 253 | \end{quiza} 254 | \begin{quizb} 255 | \woe 证明区域是道路连通集. 256 | \begin{proof} 257 | 设\(D\)是\(\bbr^n\)中的一个非空连通开集. 取点\(\boldsymbol{x}\in D\), 设\(U\left(\boldsymbol{x}\right)\)为\(D\)中所有与\(\boldsymbol{x}\)有\(D\)中连续曲线相连结的点的集合. 容易看到\(U\left(\boldsymbol{x}\right)\)是一个道路连通集. 我们证明\(U\left(\boldsymbol{x}\right)=D\). 258 | 259 | 设\(\boldsymbol{y}\in U\left(\boldsymbol{x}\right)\), 并取\(\delta>0\)使\(O_{\delta}(\boldsymbol{y})\subset D\). \(\forall\boldsymbol{z}\in O_{\delta}(\boldsymbol{y})\)存在\(O_{\delta}(\boldsymbol{y})\)中的直线段连结\(\boldsymbol{z}\)到\(\boldsymbol{y}\), 从而存在\(D\)中的连续曲线连结\(\boldsymbol{z}\)到\(\boldsymbol{x}\), 所以\(O_{\delta}(\boldsymbol{y})\subset U\left(\boldsymbol{x}\right)\). 因而\(U\left(\boldsymbol{x}\right)\)是包含\(\boldsymbol{x}\)的开集. 如\(D\backslash U\left(\boldsymbol{x}\right)\ne\varnothing\), 则\(D\backslash U\left(\boldsymbol{x}\right)=\bigcup U\left(\boldsymbol{y}'\right)\), 其中\(\boldsymbol{y}'\)取遍\(D\backslash U\left(\boldsymbol{x}\right)\)的所有点. 按前面的证明, 每个\(U\left(\boldsymbol{y}'\right)\)都是开集, 因而\(D\backslash U\left(\boldsymbol{x}\right)\)也是开集. \(D\)有开集分解式\(D=U\left(\boldsymbol{x}\right)\cup \left(D-U\left(\boldsymbol{x}\right)\right)\), 与\(D\)是连通开集矛盾. 这就证明了\(D-U\left(\boldsymbol{x}\right)=\varnothing\), 所以\(D=U\left(\boldsymbol{x}\right)\)是道路连通集. 260 | \end{proof} 261 | \woe 设\(V\subseteq\mathbb{R}^n\)为开集. 证明它可以表示为至多可列个两两不交的区域的并. 在一维情形, 这些两两不交的区域为两两不交的开区间, 即为\(V\)的构成区间. 262 | \begin{proof} 263 | 264 | \end{proof} 265 | \woe 设\(f\)在\((0,+\infty)\)上连续, 且对任何\(a\in (0,+\infty),\lim_{n\rightarrow+\infty}f(na)=0.\) 证明:\[\lim_{x\rightarrow+\infty}f(x)=0.\] 266 | \begin{proof} 267 | 对于任意\(\varepsilon>0\)与\(a\in(0,+\infty)\)存在\(N(a)\in\bbz_+\)使得\(n>N\)时有\(|f(na)|<\frac{\varepsilon}{2}\). 结合\(f(x)\)的连续性, 可知有\(\delta>0\)使得当\(0<|x-x_0|<\delta\)有\(\left|f(x)-f(x_0)\right|<\frac{\varepsilon}{2},\)由\(a\)的任意性可以取到\(a\)与\(n>N(a)\)使得\(0<|x-na|<\delta\)此时\[\left|f(x)-f(na)\right|<\frac{\varepsilon}{2}\Rightarrow |f(x)|<\varepsilon,\]由此即得结论. 268 | \end{proof} 269 | \woe 设\(a>0,\,a^2+4b<0\), 求证: 不存在\(\mathbb{R}\)上具有介值性的函数\(f\left(f(x)\right)=af(x)+bx\). 270 | \begin{proof} 271 | 若这样的函数存在, 注意到\(x=\frac{f\left(f(x)\right)-af(x)}{b}\). 所以\(x\)由\(f(x)\)唯一确定, 即\(f\)是单射. 由介值性, \(f\)必严格单调, 结合介值性和单调性, 可得\(f\)连续. 272 | 273 | 由上可得\(x\mapsto f\left(f(x)\right)\)严格增加, 即\(x\mapsto af(x)+bx\)严格增加. 由于\(a>0\), 因此\(b<-\frac{a^4}{4}\leqslant 0\). 而\(f(x)=\frac{af(x)+bx}{a}-\frac{bx}{a}\), 可知\(f\)是严格单曾函数. 特别地,\[af(x)+bx>af(0),\quad\forall x>0.\]因此\(f(+\infty)=+\infty\). 进而当\(x\gg 1\)时, \(f(x)>0\), 所以\[\frac{f\left(f(x)\right)}{f(x)}=a+b\frac{a}{f(x)},\quad \forall x\gg 1.\]记\(L=\varlimsup_{x\rightarrow+\infty}\frac{f(x)}{x}\), 则\(L\in[0,+\infty]\). 注意到\(b<0\), 我们有\[L=\varlimsup_{x\rightarrow+\infty}\frac{f\left(f(x)\right)}{f(x)}=a+b\varliminf_{x\rightarrow+\infty}\frac{x}{f(x)}=a+\frac{b}{L},\]其中对应于\(L\)取\(0\)和\(+\infty\), 规定\(\frac{b}{L}\)分别为\(-\infty\)和\(0\). 则由上式可知\(L\)有限, 且\(L^2-aL-b=0\). 但\(a^2+4b<0\), 上述方程无解, 矛盾. 因此满足题设条件的函数\(f\)不存在. 274 | \end{proof} 275 | \woe 设\(\mathbb{R}\)上连续函数列\(\{f_n(x)\}\)对于每一个\(x\in\mathbb{R}\)都是有界的(称为“逐点有界”), 证明: 存在区间\((a,b)\)使得\(\{f_n(x)\}\)在\((a,b)\)上一致有界, 即存在与\(n\)无关的常数\(M>0\)使得\[|f_n(x)|\leqslant M,\qquad \forall n\geqslant 1,x\in (a,b).\] 276 | \begin{proof} 277 | 278 | \end{proof} 279 | \woe 设\(\boldsymbol{A},\boldsymbol{B}\)均为\(n\)阶方阵, 且\(\boldsymbol{A+B}\)非奇异, 证明: \(\boldsymbol{A(A+B)^{-1}B=B(A+B)^{-1}A}.\) 280 | \begin{proof} 281 | 当\(\boldsymbol{A,B}\)都可逆时, \(\boldsymbol{A(A+B)^{-1}B}\)与\(\boldsymbol{B(A+B)^{-1}A}\)也都可逆, 注意到\[\begin{split} 282 | \left(\boldsymbol{A(A+B)^{-1}B}\right)^{-1}&=\boldsymbol{B^{-1}}\left(\boldsymbol{A+B}\right)\boldsymbol{A^{-1}}=\boldsymbol{A^{-1}}+\boldsymbol{B^{-1}},\\ 283 | \left(\boldsymbol{B(A+B)^{-1}A}\right)^{-1}&=\boldsymbol{A^{-1}}\left(\boldsymbol{A+B}\right)\boldsymbol{B^{-1}}=\boldsymbol{A^{-1}}+\boldsymbol{B^{-1}},\\ 284 | \end{split}\]从而\(\boldsymbol{A,B}\)都可逆时等式成立, 否则\(\boldsymbol{A}(t)=\boldsymbol{A}+t\boldsymbol{E}\)与\(\boldsymbol{B}(t)=\boldsymbol{B}+t\boldsymbol{E}\)可逆, 令\(t\rightarrow 0\)即得结论. 285 | \end{proof} 286 | \woe 设\(\boldsymbol{A,B}\)为半正定矩阵, 证明: 存在正定矩阵\(\boldsymbol{S}\)使得\(\boldsymbol{ASB=BSA}.\)\\ 287 | \textbf{提示: }存在正交矩阵\(\boldsymbol{P}\)和\(0\leqslant m\leqslant n\)使得\(\boldsymbol{P(A+B)P^{T}=\left(\begin{matrix} 288 | O&O\\O&\varLambda 289 | \end{matrix}\right)}\), 其中\(\boldsymbol{\varLambda}\)是\(m\)阶正交矩阵. 此时必有\(\boldsymbol{PAP^{T}=\left(\begin{matrix} 290 | O&O\\O&\tilde{A} 291 | \end{matrix}\right),PBP^{T}=\left(\begin{matrix} 292 | O&O\\O&\tilde{B} 293 | \end{matrix}\right)}\), 其中\(\tilde{A},\tilde{B}\)都是\(m\)阶半正定矩阵. 294 | \woe 设\(\boldsymbol{A,B}\)是两个\(n\)阶方阵, 满足\(\boldsymbol{AB=BA}\), 证明\(\det\left(\begin{matrix} 295 | \boldsymbol{A}&\boldsymbol{-B}\\\boldsymbol{B}&\boldsymbol{A} 296 | \end{matrix}\right)=\det\left(\boldsymbol{A}^2+\boldsymbol{B}^2\right)\). 297 | \begin{proof} 298 | 将分块矩阵的第二行乘以\(\ii\)加到第一行上, 再将第一列乘以\(-\ii\)加到第二列上, 可得\[\left(\begin{matrix} 299 | \boldsymbol{A}&\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 300 | \end{matrix}\right)\rightarrow\left(\begin{matrix} 301 | \boldsymbol{A}+\ii\boldsymbol{B}&\ii\boldsymbol{A}-\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 302 | \end{matrix}\right)\rightarrow \left(\begin{matrix} 303 | \boldsymbol{A}+\ii\boldsymbol{B}&\boldsymbol{O}\\\boldsymbol{B}&\boldsymbol{A}-\ii\boldsymbol{B} 304 | \end{matrix}\right),\]第三类分块初等变换不改变行列式的值, 因此可得\[\begin{split}&\det\left(\begin{matrix} 305 | \boldsymbol{A}&\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 306 | \end{matrix}\right)=\det\left(\begin{matrix} 307 | \boldsymbol{A}+\ii\boldsymbol{B}&\boldsymbol{O}\\\boldsymbol{B}&\boldsymbol{A}-\ii\boldsymbol{B} 308 | \end{matrix}\right)=\det\left(\boldsymbol{A}+\ii\boldsymbol{B}\right)\det\left(\boldsymbol{A}-\ii\boldsymbol{B}\right),\\=&\det\left(\left(\boldsymbol{A}+\ii\boldsymbol{B}\right)\left(\boldsymbol{A}-\ii\boldsymbol{B}\right)\right)=\det\left(\boldsymbol{A}^2+\boldsymbol{B}^2-\ii\left(\boldsymbol{AB}-\boldsymbol{BA}\right)\right)=\det\left(\boldsymbol{A}^2+\boldsymbol{B}^3\right).\qedhere\end{split}\] 309 | \end{proof} 310 | \woe 设\(\boldsymbol{A},\boldsymbol{B}\)均为\(n\)阶方阵, 证明\(\det\left(\begin{matrix} 311 | \boldsymbol{A}&\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 312 | \end{matrix}\right)=\det\left(\boldsymbol{A}+\boldsymbol{B}\right)\det\left(\boldsymbol{A}-\boldsymbol{B}\right)\). 313 | \begin{proof} 314 | 将分块矩阵的第二行加到第一行上, 再将第二列减去第一列, 可得\[\left(\begin{matrix} 315 | \boldsymbol{A}&\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 316 | \end{matrix}\right)\rightarrow\left(\begin{matrix} 317 | \boldsymbol{A}+\boldsymbol{B}&\boldsymbol{A}+\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 318 | \end{matrix}\right)\rightarrow \left(\begin{matrix} 319 | \boldsymbol{A}+\boldsymbol{B}&\boldsymbol{O}\\\boldsymbol{B}&\boldsymbol{A}-\boldsymbol{B} 320 | \end{matrix}\right),\]第三类分块初等变换不改变行列式的值, 因此可得\[\det\left(\begin{matrix} 321 | \boldsymbol{A}&\boldsymbol{B}\\\boldsymbol{B}&\boldsymbol{A} 322 | \end{matrix}\right)=\det\left(\begin{matrix} 323 | \boldsymbol{A}+\boldsymbol{B}&\boldsymbol{O}\\\boldsymbol{B}&\boldsymbol{A}-\boldsymbol{B} 324 | \end{matrix}\right)=\det\left(\boldsymbol{A}+\boldsymbol{B}\right)\det\left(\boldsymbol{A}-\boldsymbol{B}\right).\qedhere\] 325 | \end{proof} 326 | \woe 设\(\boldsymbol{A,B}\)均为\(n\)阶方阵, \(\boldsymbol{AB}=\boldsymbol{BA}\). 证明: 327 | \begin{compactenum}[(1)] 328 | \item 若\(\boldsymbol{A,B}\)都是正定矩阵, 则\(\boldsymbol{AB}\)也是正定矩阵. 329 | \item 若\(\boldsymbol{A,B}\)都是半正定矩阵, 则\(\boldsymbol{AB}\)也是半正定矩阵. 330 | \end{compactenum} 331 | \begin{proof} 332 | (1)由于\(\boldsymbol{A}\)正定, 则有非异阵\(\boldsymbol{C}\)满足\(\boldsymbol{A}=\boldsymbol{C}^\top\boldsymbol{C}\), 这是因为\(\boldsymbol{A}\)正定当且仅当\(\boldsymbol{A}\)合同于\(\boldsymbol{I}_n\), 由于\(\boldsymbol{AB}=\boldsymbol{BA}\)且\(\boldsymbol{A,B}\)都是正定矩阵, 于是\(\left(\boldsymbol{AB}\right)^\top=\boldsymbol{B}^\top\boldsymbol{A}^\top=\boldsymbol{BA}=\boldsymbol{AB}\), 从而\(\boldsymbol{AB}\)是对称的. 另一方面\[\boldsymbol{AB}=\boldsymbol{BA}=\boldsymbol{BC}^\top\boldsymbol{C}=\boldsymbol{C}^{-1}\boldsymbol{CBC}^\top\boldsymbol{C},\]从而\(\boldsymbol{AB}\)与\(\boldsymbol{CBC}^\top\)相似, 后者显然是正定的, 从而\(\boldsymbol{AB}\)正定. 333 | 334 | (2)我们先证明, \(n\)阶实对称矩阵\(\boldsymbol{A}\)是半正定矩阵的充分必要条件是对于任意正实数\(t\), \(\boldsymbol{A}+t\boldsymbol{I}_n\)都是正定阵. 335 | \end{proof} 336 | \woe 设\(\boldsymbol{A,B,C}\)均为\(n\)阶方阵, \(\boldsymbol{ABC=CBA}\). 证明: 337 | \begin{compactenum}[(1)] 338 | \item 若\(\boldsymbol{A,B,C}\)都是正定矩阵, 则\(\boldsymbol{ABC}\)也是正定矩阵. 339 | \item 若\(\boldsymbol{A,C}\)都是半正定矩阵, \(\boldsymbol{B}\)是正定矩阵, 则\(\boldsymbol{ABC}\)是半正定矩阵. 340 | \item 若\(\boldsymbol{A,B,C}\)都是半正定矩阵, 则\(\boldsymbol{ABC}\)是半正定矩阵. 341 | \end{compactenum} 342 | \begin{proof} 343 | (1) 344 | 345 | (2) 346 | 347 | (3) 348 | \end{proof} 349 | \woe 编写若干与本节内容相关的习题. 350 | \end{quizb} 351 | \section{方向极限与累次极限} 352 | \precis{曲线,方向极限,沿曲线的极限,累次极限,多重极限} 353 | \begin{quiza} 354 | \woe 考虑函数在一点的二重极限和二次极限. 分别以二重极限存在和不存在为前提, 讨论以下各条. 对必然成立或必然不成立的, 给出证明. 对于有时候成立, 有时候不成立的, 给出例子. 355 | \begin{quizcs} 356 | \item 两个二次极限存在且相等. 357 | \item 两个二次极限都存在但不相等. 358 | \item 一个二次极限存在, 另一个不存在. 359 | \item 两个二次极限都不存在. 360 | \end{quizcs} 361 | \begin{solution} 362 | 首先考虑二重极限存在的情况. 363 | 364 | (1)不一定成立. 如考虑\(f(x,y)=y\sin\frac{1}{x}\)在\((x,y)\rightarrow(0,0)\)时的形态, 其二重极限存在, 但二次极限\(\lim_{y\rightarrow 0}\lim_{x\rightarrow 0}f(x,y)\)不存在. 365 | 366 | (2) 367 | \end{solution} 368 | \woe 试就函数在一点的二重极限和方向极限, 仿第1题讨论相关问题. 369 | \woe 试就函数在一点的二次极限和方向极限, 仿第1题讨论相关问题. 370 | \woe 设\(E\subseteq\mathbb{R}^n,\boldsymbol{f}:E\rightarrow\mathbb{R}^m\), 且\(E\)包含\(\boldsymbol{x}_0\)的一个去心邻域. 证明: \(\lim_{\boldsymbol{x}\rightarrow\boldsymbol{x}_0}\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{A}\)当且仅当\[\lim_{t\rightarrow0^+}\sup_{\boldsymbol{\tau}\in S^{n-1}}\left|\boldsymbol{f}(\boldsymbol{x}_0+t\boldsymbol{\tau})-\boldsymbol{A}\right|=0.\] 371 | \end{quiza} 372 | \begin{quizb} 373 | \woe 设有\([a,b]\times[c,d]\)上的函数\(f\), 对固定的\(y\in[c,d],\,f(x,y)\)作为\(x\)的函数在\([a,b]\)上连续, 而\(f(x,y)\)对\(y\)的连续性关于\(x\in [a,b]\)是一致的, 即\(\forall y_0\in[c,d]\),\[\lim_{y\rightarrow y_0}\sup_{x\in [a,b]}\left|f(x,y)-f(x,y_0)\right|=0.\]证明: \(f\)是\(D=[a,b]\times [c,d]\)上的二元连续函数. 374 | \begin{proof} 375 | 376 | \end{proof} 377 | \end{quizb} -------------------------------------------------------------------------------- /Chapters/chapter4.aux: -------------------------------------------------------------------------------- 1 | \relax 2 | \providecommand\hyper@newdestlabel[2]{} 3 | \@writefile{toc}{\contentsline {chapter}{\numberline {第4章\hspace {.3em}}导数与微分}{66}{chapter.4}\protected@file@percent } 4 | 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\setcounter{tcbrastercolumn}{1} 51 | \setcounter{tcbrasterrow}{1} 52 | \setcounter{tcbrasternum}{1} 53 | \setcounter{tcbraster}{0} 54 | \setcounter{mn@abspage}{99} 55 | \setcounter{LT@tables}{1} 56 | \setcounter{LT@chunks}{1} 57 | \setcounter{tcb@cnt@theorem}{0} 58 | \setcounter{tcb@cnt@definition}{0} 59 | \setcounter{tcb@cnt@proposition}{0} 60 | \setcounter{tcb@cnt@lemma}{0} 61 | \setcounter{woes}{2} 62 | \setcounter{lstlisting}{0} 63 | \setcounter{mistake}{7} 64 | } 65 | -------------------------------------------------------------------------------- /Chapters/chapter5.tex: -------------------------------------------------------------------------------- 1 | \chapter{不定积分} 2 | \section{不定积分} 3 | \precis{原函数,不定积分,恰当方程} 4 | \begin{theorem}{}{C51} 5 | 设\(\Omega\)为\(\mathbb{R}^n\)中的区域, \(\boldsymbol{f}:\Omega\rightarrow\mathbb{R}^m\)在\(\Omega\)中可微且微分恒为\(\boldsymbol{0}\), 则\(\boldsymbol{f}\)在\(\Omega\)中恒为常量. 6 | \end{theorem} 7 | \begin{quiza} 8 | \woe 计算\(\int \frac{\sin^2 x}{\cos 2x+1}\dd x\). 9 | \begin{solution} 10 | \(\int \frac{\sin^2 x}{\cos 2x+1}\dd x=\frac{1}{2}\int\frac{\sin^2x}{\cos^2x}\dd x=\frac{1}{2}\int\frac{1-\cos^2x}{\cos^2x}\dd x=\frac{1}{2}\left(\tan x-x\right)+C.\) 11 | \end{solution} 12 | \woe 对于复数\(\lambda\)以及实数\(x\), 定义\(x^\lambda=\ee^{\lambda\ln x}\). 验证: \(\frac{\dd}{\dd x}x^{\lambda}=\lambda x^{\lambda-1}\). 13 | \begin{solution} 14 | \(\frac{\dd}{\dd x}x^\lambda=\frac{\dd}{\dd x}\ee^{\lambda\ln x}=\ee^{\lambda\ln x}\cdot\frac{\lambda}{x}=\lambda x^{\lambda-1}.\) 15 | \end{solution} 16 | \woe 设\(a0\), 则\(y(x)<0\); 若\(x<0\), 则\(y(x)>0\). 并且易见\(y(0)=0\). 若\(x_n>0(n\geqslant 0)\), 则\(x_{n+1}=y(x_n)<0\), 于是\[|x_{n+1}|-|x_{n}|=-y(x_n)-x_n=-y(x_n)-1+y(x_n)+\ee^{y(x_n)}=\ee^{y(x_n)}-1<0,\] 116 | 反之若\(x_n<0\)同理可得\(|x_{n+1}|<|x_n|\). 这意味着如果\(x_0\)取定, 那么无论如何, \(x_n\)都将落在\(\left[-|x_0|,|x_0|\right]\)内, 注意到这时\[\left|\frac{\dd y}{\dd x}\right|=\frac{1}{1+\ee^y}\leqslant\frac{1}{1+\ee^{-|x_0|}}<1,\]于是由Lagrange中值定理与压缩映射原理即得结论, 即\(\lim_{n\rightarrow+\infty}x_n=0\). 117 | \end{solution} 118 | \end{quiza} 119 | \begin{quizb} 120 | \woe 考虑熟知的那些函数的两两组合, 试计算其乘积的不定积分. 121 | \woe 设\(m,n\)为整数, 讨论不同情形下, 什么方法是计算不定积分\(\int\sin^m x\cos^nx\dd x\)是最简单的. 122 | \end{quizb} 123 | \section{有理函数不定积分} 124 | \precis{有理函数,最简分式} 125 | \begin{quiza} 126 | \woe 计算\(\int\frac{1}{3+\cos x}\dd x\). 127 | \begin{solution} 128 | 设\(t=\tan\frac{x}{2}\), 则有\[\int\frac{1}{3+\cos x}\dd x=\int\frac{1}{2+t^2}\dd t=\frac{\sqrt{2}}{2}\arctan\frac{t}{\sqrt{2}}+C=\frac{\sqrt{2}}{2}\arctan\frac{\tan\frac{x}{2}}{\sqrt{2}}+C.\qedhere\] 129 | \end{solution} 130 | \woe 设\(Q(x,y)\)为\(x,y\)的有理函数, \(ad-bc\ne 0,m\geqslant 2\), 试验证变量代换\(t=\sqrt[m]{\frac{ax+b}{cx+d}}\)将积分\(\int Q(x,\sqrt[m]{\frac{ax+b}{cx+d}})\dd x\)转化为有理积分. 试利用这一变量代换计算以下积分. 131 | \begin{quizs} 132 | \item \(\int\frac{1}{x}\sqrt{\frac{x+2}{x-1}}\dd x\); 133 | \item \(\int\frac{1}{x+1}\sqrt[3]{\frac{x-2}{x+2}}\dd x\); 134 | \item \(\int\sqrt{x^2-3x+2}\dd x\);\quad\textcolor{red}{\(\left(\int\sqrt{x^2-3x+2}\dd x=\int\sqrt{(x-1)(x-2)}\dd x=\int (x-2)\sqrt{\frac{x-1}{x-2}}\dd x.\right)\)} 135 | \item \(\int\sqrt{4-3x-x^2}\dd x\). 136 | \end{quizs} 137 | \begin{solution} 138 | 由题设 139 | \[\int Q(x,\sqrt[m]{\frac{ax+b}{cx+d}})\dd x\xlongequal{t=\sqrt[m]{\left( ax+b \right) /\left( cx+d \right)}}\int Q(\frac{b-dt^m}{ct^m-a},t)\cdot\frac{(ad-cb)mt^{m-1}}{(ct^m-a)^2}\dd t.\] 140 | 141 | (1)利用上述变换, 有\[\begin{split} 142 | &\int\frac{1}{x}\sqrt{\frac{x+2}{x-1}}\dd x=\int\frac{6t^2}{(1-t^2)(2+t^2)}\dd t=\int\left(\frac{1}{1-t}+\frac{1}{1+t}-\frac{4}{2+t^2}\right)\dd t\\=&\ln\left|\frac{1+t}{1-t}\right|+2\sqrt{2}\arctan\left(\frac{t}{\sqrt{2}}\right)+C.\\=&\ln\left|\frac{1+\sqrt{\frac{x+2}{x-1}}}{1-\sqrt{\frac{x+2}{x-1}}}\right|+2\sqrt{2}\arctan\sqrt{\frac{x+2}{2x-2}}+C. 143 | \end{split}\] 144 | \end{solution} 145 | \woe 设\(Q(x,y)\)为\(x,y\)的有理函数, \(a>0\), 试讨论变量代换\(t=\sqrt{ax^2+bx+c}+\sqrt{a}x\)与\(t=\sqrt{ax^2+bx+c}-\sqrt{a}x\)在将积分\(\int Q(x,\sqrt{ax^2+bx+c})\dd x\)转化为有理函数积分时的适用性. 进一步, 试利用上述变换计算第2题的(1), (3). 146 | \begin{solution} 147 | 148 | \end{solution} 149 | \woe 设\((Q(x,y))\)为\(x,y\)的有理函数, \(c>0,b^2-4ac\ne 0\). 试讨论变量代换\(t=\frac{\sqrt{ax^2+bx+c}+\sqrt{c}}{x}\)与\(t=\frac{\sqrt{ax^2+bx+c}-\sqrt{c}}{x}\)再将积分\(\int Q(x,\sqrt{ax^2+bx+c})\dd x\)转化为有理函数积分时的适用性. 进一步, 试利用上述变换计算第2题的(3), (4). 150 | \begin{solution} 151 | 152 | \end{solution} 153 | \woe 求\(A,B\)使得\[3\cos x+4\sin x=A(2\cos x+\sin x)+B(2\cos x+\sin x)'.\]由此计算积分\(\int\frac{3\cos x+4\sin x}{2\cos x+\sin x}\dd x.\) 154 | \begin{solution} 155 | 易见\(A=2,B=-1\). 由\[\begin{split} 156 | I_1&:=\int\frac{2\cos x+\sin x}{2\cos x+\sin x}\dd x=x+C,\\ 157 | I_2&:=\int\frac{(2\cos x+\sin x)'}{2\cos x+\sin x}\dd x=\int\frac{\dd\left(2\cos x+\sin x\right)}{2\cos x+\sin x}=\ln\left|2\cos x+\sin x\right|+C 158 | \end{split}\]由线性性有\[\int\frac{3\cos x+4\sin x}{2\cos x+\sin x}\dd x=AI_1+BI_2=2x-\ln\left|2\cos x+\sin x\right|+C.\qedhere\] 159 | \end{solution} 160 | \end{quiza} 161 | \begin{quizb} 162 | \woe 设\(Q,Q_1,Q_2\)为有理函数, 对于使得\(Q(x),Q_1(x),Q_2(x)\)都有意义的实数\(x\), 成立\(Q_1(x)+Q_2(x)=Q(x)\). 证明: 对于使得\(Q(z),Q_1(z),Q_2(z)\)都有意义的复数\(z\), 成立\(Q_1(z)+Q_2(z)=Q(z)\). 163 | \begin{proof} 164 | 165 | \end{proof} 166 | \woe 设\(Q,Q_1,Q_2\)为有理函数, 对于使得\(Q(x),Q_1(x),Q_2(x)\)都有意义的实数\(x\), 成立\(Q_1(x)Q_2(x)=Q(x)\). 证明: 对于使得\(Q(z),Q_1(z),Q_2(z)\)都有意义的复数\(z\), 成立\(Q_1(z)Q_2(z)=Q(z)\). 167 | \begin{proof} 168 | 169 | \end{proof} 170 | \woe 讨论在复平面\(\mathbb{C}\)的什么子区域内, 可以连续地定义\(w=\ln z\)使得\(\ee^w=z\), 进一步, 171 | \begin{quizs} 172 | \item 考察对于复数\(z_0\), 等式\(\frac{\dd x}{x+z_0}=\ln (x+z_0)+C\)的合理性. 173 | \item 考察在不同区域内, 一阶微分形式\(\frac{\dd x+\mathrm{i}\dd y}{x+\mathrm{i}y}\)的原函数的存在性. 174 | \item 设\(n\geqslant 2\), 考察在不同区域内, 一阶微分形式\(\frac{\dd x+\mathrm{i}\dd y}{(x+\mathrm{i}y)^n}\)的原函数的存在性. 175 | \end{quizs} 176 | \end{quizb} 177 | \section{求解简单的微分方程} 178 | \precis{常微分方程,特解,通解,分离变量法,初值问题,解的最大存在区间,一阶线性方程,常数变异法,积分因子法,全微分方程,齐次方程,Bernoulli方程} 179 | \begin{quiza} 180 | \woe 求解下列方程: 181 | \begin{quizs} 182 | \item \(x(1+x^2)y'=(1-x^2)y.\) 183 | \begin{solution} 184 | 分离变量有\(\frac{\dd y}{y}=\frac{(1-x^2)}{x(1+x^2)}\dd x\), 得\(y=\frac{Cx}{1+x^2}.\) 185 | \end{solution} 186 | \item \(y(x^2-1)y'=x(1-y^2).\) 187 | \begin{solution} 188 | 即\(\frac{y}{1-y^2}\dd y=\frac{x}{x^2-1}\dd x\), 得\(y^2=1-C(x^2-1).\) 189 | \end{solution} 190 | \item \(y'-\frac{y}{x+2}=\frac{1}{(x+2)^2}.\) 191 | \begin{solution} 192 | 对应得齐次方程为\(y'=\frac{y}{x+2}\), 即得\(y=c(x+2)\), 使用常数变易法, 令\(c=c(x)\), 代入原方程, 得\[c'(x)\cdot(x+2)=\frac{1}{(x+2)^2},\]得到\(c=-\frac{1}{2(x+2)^2}+C\), 代入\(y\)得到\(y=C(x+2)-\frac{1}{2(x+2)}.\) 193 | \end{solution} 194 | \item \(\frac{\dd y}{\dd x}=\frac{y}{x+y^3}\). 195 | \begin{solution} 196 | 原方程即为\(\frac{\dd x}{\dd y}=\frac{x+y^3}{y}\), 对应齐次方程\(\frac{\dd x}{\dd y}=\frac{x}{y}\), 得到\(x=cy\), 使用常数变易法, 令\(c=c(y)\)代入前述方程得到\(c(y)=\frac{1}{2}y^2+C\), 从而\(x=\frac{1}{2}y^3+Cy\). 197 | \end{solution} 198 | \item \(\frac{\dd y}{\dd x}+\frac{y}{x}-y^2\ln x=0.\) 199 | \begin{solution} 200 | 这正是Bernoulli方程, 令\(z=-y^{-1}\), 原方程变为\[\frac{\dd z}{\dd x}-\frac{z}{x}-\ln x=0,\]对于\(\frac{\dd z}{\dd x}-\frac{z}{x}=0\), 有\(z=cx\), 使用常数变易法, 令\(c=c(x)\)代入前述方程得到\(c'(x)=\frac{\ln x}{x}\), 即\(c=\frac{\ln^2x}{2}+C\), 代回得到原方程得解为\[\frac{1}{y}=Cx-\frac{1}{2}x\ln^2x.\qedhere\] 201 | \end{solution} 202 | \end{quizs} 203 | \woe 试给出方程\(y'=y^{2/3}\)在\(\mathbb{R}\)上的所有解. 204 | \begin{solution} 205 | 当\(y\ne 0\)时, 分离变量得到\(y^{-2/3}y'=1\), 两边积分得到\(3y^{1/3}=x+C\), 因此\(y=\left(\frac{x+C}{3}\right)^3\). 易见此为该方程在\(\mathbb{R}\)上的通解. 特解有\(y \equiv 0\)不在这之中. 206 | \end{solution} 207 | \end{quiza} 208 | 209 | \begin{quizb} 210 | \woe 若\(P,Q\)是满足\[P(tx,ty)=t^{\alpha}P(x,y),\qquad Q(tx,ty)=t^{\alpha}Q(x,y).\qquad\forall t>0,\alpha\in\mathbb{R}.\]的可微函数, 证明\(\frac{1}{xP(x,y)+yQ(x,y)}\)是方程\(P(x,y)\dd x+Q(x,y)\dd y=0\)的一个积分因子. 211 | \begin{proof} 212 | 设\(y=ux\), 则\(\dd y=x\dd u+u\dd x\), 于是\begin{gather*} 213 | P(x,y)=P(x,ux)=x^{\alpha}P(1,u),\\ 214 | Q(x,y)=Q(x,ux)=x^{\alpha}Q(1,u), 215 | \end{gather*}方程\(P\dd x+Q\dd y=0\)两边乘\(\mu:=\frac{1}{xP+yQ}\)得到\[\frac{P}{xP+yQ}\dd x+\frac{Q}{xP+yQ}\dd y=0,\]即\(\frac{P(x,ux)+uQ(x,ux)}{xP(x,ux)+yQ(x,ux)}\dd x+\frac{xQ(x,ux)}{xP(x,ux)+yQ(x,ux)}\dd u=0\), 即\[\frac{1}{x}\dd x+\frac{Q(1,u)}{P(1,u)+uQ(1,u)}\dd u=0,\]这是一个恰当方程, 这就证明了\(\mu\)是题设方程的一个积分因子. 216 | \end{proof} 217 | \woe 求解方程\((x^3y-y^5)\dd x+(-x^4+x^2y^3)\dd y=0.\) 218 | \begin{solution} 219 | 令\(y=tx\), 则\(\dd y=t\dd x+x\dd t\), 原方程化为 220 | \begin{equation}\label{c5fc}\tag{\(\clubsuit\)} 221 | \left(t^4-t^5\right)\dd x+\left(t^3x-1\right)\dd t=0, 222 | \end{equation} 223 | 记上述方程的积分因子为\(\mu(t)\), 则有\[\frac{\partial\left(\mu(t)\left(t^4-t^5\right)\right)}{\partial t}=\mu'(t)(t^4-t^5)+\mu(t)(4t^3-5t^4)=\frac{\partial\left(\mu (t)\left(t^3x-1\right)\right)}{\partial x}=\mu(t)t^3,\]即\[\mu'(t)(t-t^2)+\mu(t)(3-5t)=0,\]容易解得\(\mu(t)=\frac{1}{t^3(1-t)^2}\). 方程\eqref{c5fc}两侧乘以\(\mu\), 得到恰当方程:\[\frac{t}{1-t}\dd x+\frac{t^3x-1}{t^3(1-t)^2}\dd t=:P\dd x+Q\dd t=0.\]则存在\(u(x,t)\)使得\[\frac{\partial u}{\partial x}=\frac{t}{1-t},\quad \frac{\partial u}{\partial t}=\frac{t^3x-1}{t^3(1-t)^2},\]对左侧公式积分得到\(u=\frac{tx}{1-t}+\varphi(t)\), 带入右式得到\[\varphi'(t)=-\frac{1}{t^3(1-t)^2},\]设\[\frac{1}{t^3(1-t)^2}=\frac{t}{t^4(1-t)^2}=\left(\frac{At^2+Bt+C}{t^2(1-t)}\right)'+\frac{Dt^2+Et+F}{t^2(1-t)},\]求得\footnote{这里方程个数将少于未知数个数.}\(A=0,B=\frac{3}{2},C=-\frac{1}{2},D=E=0,F=3\), 即\[\int\frac{1}{t^3(1-t)^2}\dd t=\frac{3t-1}{2t^2(1-t)}+3\int\frac{1}{t^2(1-t)}\dd t,\]容易得到\[\int\frac{1}{t^2(1-t)}\dd t=\int\left(\frac{1}{t}+\frac{1}{t^2}+\frac{1}{1-t}\right)\dd t=\ln|t|-\frac{1}{t}-\ln|1-t|+C,\]即\(\varphi(t)=\frac{1-3t}{2t^2(1-t)}+\frac{3}{t}-3\ln\left|\frac{t}{1-t}\right|+C\), 带回\(y=tx\)与\(u(x,t)\), 得到原方程的解为\[\frac{xy}{x-y}+\frac{(x-3y)x^2}{2y^2(x-y)}+\frac{3x}{y}-3\ln\left|\frac{x}{x-y}\right|=C.\qedhere\] 224 | \end{solution} 225 | \end{quizb} -------------------------------------------------------------------------------- /Chapters/chapter6.aux: 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| \chapter{微分问题} 2 | \section{隐函数存在定理} 3 | \precis{隐函数存在定理,曲面的切平面,法向量} 4 | \begin{quiza} 5 | \woe 证明方程\(\ee^z+\sin z+\ee^y+y-\sin x+x=2\)在点\((0,0,0)\)附近确定一个隐函数\(z=z(x,y)\), 并计算\(\frac{\partial z}{\partial x},\frac{\partial^2 z}{\partial x\partial y}\). 6 | \begin{solution} 7 | 记\(F(x,y,z)=\ee^z+\sin z+\ee^y+y-\sin x+x-2\), 由隐函数定理可得\[\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=\frac{\cos x-1}{\ee^z+\cos z},\quad\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=-\frac{\ee^y+1}{\ee^z+\cos z}\]进一步得到\[\frac{\partial^2z}{\partial x\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{(1-\cos x)(\ee^z+\cos z)}{(\ee^z+\cos z)^2}\frac{\partial z}{\partial y}=\frac{(\cos x-1)(\ee^z+\cos z)(1+\ee^y)}{(\ee^z+\cos z)^3}.\qedhere\] 8 | \end{solution} 9 | \woe 证明方程组\[\begin{cases} 10 | x+y+u+v=0,\\ x^2+y^2+u^4+v^4=2 11 | \end{cases}\]在\((-1,0,1,0)\)附近确定一个隐函数\(\left(\begin{matrix} 12 | u\\ v 13 | \end{matrix}\right)=\left(\begin{matrix} 14 | u(x,y)\\v(x,y) 15 | \end{matrix}\right)\). 进一步, 计算\(\frac{\partial u}{\partial x},\frac{\partial^2u}{\partial x^2}.\) 16 | \begin{proof} 17 | 18 | \end{proof} 19 | \woe 证明方程组\[\begin{cases} 20 | x+y+u+v=0,\\x^2+y^2+u^4+v^4=2 21 | \end{cases}\]在\((-1,0,1,0)\)附近确定一个隐函数\(\left(\begin{matrix} 22 | y\\ u 23 | \end{matrix}\right)=\left(\begin{matrix} 24 | y(x,u)\\u(x,v) 25 | \end{matrix}\right)\). 进一步, 计算\(\frac{\partial u}{\partial x},\frac{\partial^2u}{\partial x^2}.\) 26 | \begin{proof} 27 | 28 | \end{proof} 29 | \end{quiza} 30 | \begin{quizb} 31 | \woe 设\(n\geqslant 2\), \(\varOmega\)是\(\mathbb{R}^n\)中的区域, \(F:\varOmega\rightarrow\mathbb{R}\)连续可微, 且\(F_{\boldsymbol{x}}(\boldsymbol{x})\ne\boldsymbol{0}(\forall\boldsymbol{x}\in\varOmega)\). 又设\(\boldsymbol{x}_0\in\varOmega\)使得\(F(\boldsymbol{x}_0)=0\). 证明存在\(\delta>0\), 使得对于任何满足\(F(\boldsymbol{x}_1)=0\)的\(\boldsymbol{x}_1\in B_\delta(\boldsymbol{x}_0)\), 均存在曲面\(F(\boldsymbol{x})=0\)上连接\(\boldsymbol{x}_0,\boldsymbol{x}_1\)的\(C^1\)曲线\(\boldsymbol{\tau}\). 即存在\(\boldsymbol{\tau}\in C^1\left([0,1];\mathbb{R}^n\right)\)满足\(F\left(\boldsymbol{\tau}(t)\right)=0(\forall t\in[0,1])\)以及\(\boldsymbol{\tau}(0)=\boldsymbol{x}_0,\boldsymbol{\tau}(1)=\boldsymbol{x}_1\). 32 | \begin{proof} 33 | 34 | \end{proof} 35 | \end{quizb} 36 | \section{极值问题} 37 | \precis{强制条件,极值问题,无条件极值,一阶必要条件,二阶必要条件,驻点,二阶必要条件,最小二乘法,线性拟合,条件极值,Lagrange乘子法,矩阵的诱导范数} 38 | \begin{quiza} 39 | \woe 设\(n\geqslant 2\), \(\boldsymbol{\xi,\eta}\in\mathbb{R}^n\). 试计算\(\max_{x\in S^{n-1}}\{\inp{\boldsymbol{\xi,x}},\inp{\boldsymbol{\eta,x}}\}\). 40 | \begin{solution} 41 | 42 | \end{solution} 43 | \woe 设\(1\leqslant p,q\leqslant+\infty\), 对\(\boldsymbol{A}\in\mathbb{R}^{m\times n}\), 定义\(\left\|\boldsymbol{A}\right\|_{p,q}=\max_{\left\|x\right\|_q=1}\left\|\boldsymbol{Ax}\right\|_p\). 证明: \(\left\|\cdot\right\|_{p,q}\)定义了一个范数. 44 | \begin{proof} 45 | 我们只需验证由此法的定义满足非负性, 齐次性与三角不等式. 显然满足前两者, 下面证明 46 | \end{proof} 47 | \woe 设\(\boldsymbol{A}\)为\(n\)阶正定矩阵, 证明\(\inp{\boldsymbol{x,y}}_{\boldsymbol{A}}=\boldsymbol{y}^{\mathrm{T}}\boldsymbol{Ax}\)定义了\(\mathbb{R}^n\)中的一个内积. 48 | \begin{proof} 49 | 由于\[\inp{\boldsymbol{x,y}}_{\boldsymbol{A}}=\boldsymbol{y}^{\top}\boldsymbol{Ax}=\left(\boldsymbol{y}^{\top}\boldsymbol{Ax}\right)^\top=\boldsymbol{x}^{\top}\boldsymbol{Ay}=\inp{\boldsymbol{y,x}}_{\boldsymbol{A}},\]也有\(\inp{\boldsymbol{x+y,z}}_{\boldsymbol{A}}=\inp{\boldsymbol{x,z}}_{\boldsymbol{A}}+\inp{\boldsymbol{y,z}}_{\boldsymbol{A}}\), \(\inp{c\boldsymbol{x,y}}_{\boldsymbol{A}}=c\inp{\boldsymbol{x,y}}_{\boldsymbol{A}}\), 其中\(c\)为一常数. 结合\(\boldsymbol{A}\)正定可知对于任意\(\boldsymbol{x}\ne\boldsymbol{0}\), 有\(\inp{\boldsymbol{x,x}}_{\boldsymbol{A}}=\boldsymbol{x}^\top\boldsymbol{Ax}>0\), 从而上述的运算定义了一个内积. 50 | \end{proof} 51 | \woe 设\(n\geqslant 2\), \(\boldsymbol{A}\)为\(n\)阶实对称矩阵, \(\boldsymbol{\xi}_1\in S^{n-1}\)满足\(\lambda_1\equiv \boldsymbol{\xi}_1^{\mathrm{T}}\boldsymbol{A}\boldsymbol{\xi}_1=\max_{x\in S^{n-1}}\boldsymbol{x}^{\mathrm{T}}\boldsymbol{A}\boldsymbol{x}.\)考虑在\(\boldsymbol{x}\in S^{n-1}\)以及\(\boldsymbol{x}\cdot\boldsymbol{\xi}_1=0\)的约束条件下最大化\(\boldsymbol{x}^{\mathrm{T}}\boldsymbol{A}\boldsymbol{x}\)的最大值问题, 证明存在\(\lambda_2\leqslant\lambda_1\)以及\(\boldsymbol{\xi}_2\in S^{n-1}\), 使得\(\boldsymbol{A\xi}_2=\lambda_2\boldsymbol{\xi}_2,\boldsymbol{\xi}_2\cdot\boldsymbol{\xi}_1=0\). 一般地, 证明存在\(\lambda_1\geqslant \lambda_2\geqslant\cdots\geqslant\lambda_n\)以及两两正交的\(\boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_n\in S^{n-1}\)使得\(\boldsymbol{A}\boldsymbol{\xi}_k=\lambda_k\boldsymbol{\xi}_k(1\leqslant k\leqslant n)\). 52 | \begin{proof} 53 | 由于\(\boldsymbol{A}\)实对称, 故存在正交矩阵\(\boldsymbol{P}\)使得\[\boldsymbol{P}^\top\boldsymbol{AP}=\left(\begin{matrix} 54 | \lambda_1'&&&\\ 55 | &\lambda_2'&&\\ 56 | &&\ddots&\\ 57 | &&&\lambda_n' 58 | \end{matrix}\right),\]其中\(\lambda_i'(1\leqslant i\leqslant n)\)是\(\boldsymbol{A}\)的特征值. 59 | \end{proof} 60 | \woe 设\(\boldsymbol{A}\)为\(m\times n\)实矩阵, 证明\(\left\|\boldsymbol{A}\right\|=\max_{\substack{|\boldsymbol{z}|=1\\ \boldsymbol{z}\in\mathbb{C}^{n}}}|\boldsymbol{Az}|\). 61 | \begin{proof} 62 | 63 | \end{proof} 64 | \woe 设\(\boldsymbol{A}\)为\(n\)阶实方阵, \(\lambda\)是它的(复)特征值. 证明: \(|\lambda|\leqslant\left\|\boldsymbol{A}\right\|\). 65 | \begin{proof} 66 | 67 | \end{proof} 68 | \woe 设\(\boldsymbol{x}\in\mathbb{R}^n\), 证明\(|\boldsymbol{x}|=\max_{\substack{\left\|\boldsymbol{A}\right\|\\\boldsymbol{A}\in\mathbb{R}^{n\times n}}}|\boldsymbol{Ax}|.\) 69 | \begin{proof} 70 | 71 | \end{proof} 72 | \woe 是推导两空间直线\[\frac{x-x_i}{l_i}=\frac{y-y_i}{m_i}=\frac{z-z_i}{n_i},\quad i=1,2\]间的距离公式, 其中\[(l_1m_2-l_2m_1)^2+(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2>0.\] 73 | %\begin{solution} 74 | %记题设直线分别为\(L_1,L_2\), 取\(\boldsymbol{x_1}\in L_1\), \(\boldsymbol{x_2}\in L_2\), 则\(L_1\)于\(L_2\)之间的距离\(d = \min\left\|\boldsymbol{x}_1-\boldsymbol{x}_2\right\|\), 于是不妨设\[\boldsymbol{x}_1=\left(l_1t_1+x_1,m_1t_1+y_1,n_1t_1+z_1\right),\quad \boldsymbol{x}_2=\left(l_2t_2+x_2,m_2t_2+y_2,n_2t_2+z_2\right),\]记\(D(t_1,t_2)=\left(l_1t_1+x_1-l_2t_2-x_2\right)^2+\left(m_1t_1+y_1-m_2t_2-y_2\right)^2+\left(n_1t_1+z_1-n_2t_2-z_2\right)^2\), 原问题即为求\(D(t_1,t_2)\)的最小值, 分别对\(t_1,t_2\)求偏导有\[\begin{split} 75 | %\frac{\partial D}{\partial t_1}&=2A_1t_1-2Bt_2+2l_1(x_1-x_2)+2m_1(y_1-y_2)+2n_1(z_1-z_2),\\ 76 | %\frac{\partial D}{\partial t_2}&=2A_2t_2-2Bt_1-2l_2(x_1-x_2)-2m_2(y_1-y_2)-2n_2(z_1-z_2), 77 | %\end{split}\] 78 | %其中\(A_i=l_i^2+m_i^2+n_i^2(i=1,2),B=l_1l_2+m_1m_2+n_1n_2\). 令\(\frac{\partial D}{\partial t_1}=\frac{\partial D}{\partial t_2}=0\), 并且注意到\[A_1A_2-B_2^2=(l_1m_2-l_2m_1)^2+(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2>0,\] 79 | %由Cramer法则可得\[t'_1=\frac{BC_2-A_2C_1}{A_1A_2-B^2},\quad t'_2=\frac{A_1C_2-BC_1}{A_1A_2-B^2}\]其中\(C_i=l_i(x_1-x_2)+m_i(y_1-y_2)+n_i(z_1-z_2)(i=1,2)\). 将\(t'_1,t'_2\)代入\(D(t_1,t_2)\)得到 80 | %\end{solution} 81 | \begin{solution} 82 | 记题设直线分别为\(L_1,L_2\), 取\(\boldsymbol{x_1}\in L_1\), \(\boldsymbol{x_2}\in L_2\), 则\(L_1\)于\(L_2\)之间的距离\(d = \min\left|\boldsymbol{x}_1-\boldsymbol{x}_2\right|\), 不妨设\[\boldsymbol{a}_i=\left(x_i,y_i,z_i\right)^\top,\quad \boldsymbol{d}_i=\left(l_i,m_i,n_i\right)^\top,\quad i=1,2,\]于是\(\boldsymbol{x}_1=\boldsymbol{a}_1+t_1\boldsymbol{d}_1,\boldsymbol{x}_2=\boldsymbol{a}_2+t_2\boldsymbol{d}_2\), 从而设\[\begin{split} 83 | D(t_1,t_2)&=\left|\boldsymbol{x}_1-\boldsymbol{x}_2\right|^2=\left|\boldsymbol{a}_1-\boldsymbol{a}_2+t_1\boldsymbol{d}_1-t_2\boldsymbol{d}_2\right|^2=\inp{\boldsymbol{a}_1-\boldsymbol{a}_2+t_1\boldsymbol{d}_1-t_2\boldsymbol{d}_2,\boldsymbol{a}_1-\boldsymbol{a}_2+t_1\boldsymbol{d}_1-t_2\boldsymbol{d}_2}\\&=\left|\boldsymbol{d}_1\right|^2t_1^2-2\inp{\boldsymbol{d}_1,\boldsymbol{d}_2}t_1t_2+\left|\boldsymbol{d}_2\right|^2t_2^2+2\inp{\boldsymbol{a}_1-\boldsymbol{a}_2,\boldsymbol{d}_1}t_1-2\inp{\boldsymbol{a}_1-\boldsymbol{a}_2,\boldsymbol{d}_2}t_2+\left|\boldsymbol{a}_1-\boldsymbol{a}_2\right|^2, 84 | \end{split}\] 85 | 分别记\(A_i=\left|\boldsymbol{d}_i\right|^2,B=\inp{\boldsymbol{d}_1,\boldsymbol{d}_2},C_i=\inp{\boldsymbol{a}_1-\boldsymbol{a}_2,\boldsymbol{d}_i}(i=1,2)\), 这样就有\[D(t_1,t_2)=A_1t_1^2-2Bt_1t_2+A_2t_2^2+2C_1t_1-2C_2t_2+\left|\boldsymbol{a}_1-\boldsymbol{a}_2\right|^2,\] 86 | 分别对\(t_1,t_2\)求偏导有\[\frac{\partial D}{\partial t_1}=2A_1t_1-2Bt_2+2C_1,\quad 87 | \frac{\partial D}{\partial t_2}=2A_2t_2-2Bt_1-2C_2,\] 88 | 令\(\frac{\partial D}{\partial t_1}=\frac{\partial D}{\partial t_2}=0\), 并且注意到\[A_1A_2-B_2^2=(l_1m_2-l_2m_1)^2+(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2>0,\] 89 | 由Cramer法则可得\[t'_1=\frac{BC_2-A_2C_1}{A_1A_2-B^2},\quad t'_2=\frac{A_1C_2-BC_1}{A_1A_2-B^2},\] 90 | 代入\(D(t1,t2)\)开方即得结果. 91 | \end{solution} 92 | \woe 计算两空间直线\(\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-5}{3},\frac{x-7}{4}=\frac{y-9}{5}=\frac{z-11}{6}\)的公垂线. 93 | \begin{solution} 94 | 95 | \end{solution} 96 | \woe 试推导点\((x_0,y_0,z_0)\)与平面\(Ax+By+Cz+D=0(A^2+B^2+C^2\ne 0)\)的距离公式. 97 | \begin{solution} 98 | 我们先用条件极值的方法. 设\[f(\alpha,\beta,\gamma)=(\alpha-x_0)^2+(\beta-y_0)^2+(\gamma-z_0)^2,\]即求\(f\)在约束条件\(A\alpha+B\beta+C\gamma+D=0\)下的极值. 设\[F(\alpha,\beta,\gamma,\lambda)=(\alpha-x_0)^2+(\beta-y_0)^2+(\gamma-z_0)^2+\lambda\left(A\alpha+B\beta+C\gamma+D\right),\]则令\[\begin{cases} 99 | F_{\alpha}=2\left(\alpha-x_0\right)+A\lambda=0,\\ 100 | F_{\beta}=2\left(\beta-y_0\right)+B\lambda=0,\\ 101 | F_{\gamma}=2\left(\gamma-z_0\right)+C\lambda=0,\\ 102 | F_{\lambda}=A\alpha+B\beta+C\gamma+D=0, 103 | \end{cases}\] 104 | 可以解得\[\lambda=\frac{2(Ax_0+By_0+Cz_0+D)}{A^2+B^2+C^2},\alpha=x_0-\frac{A\lambda}{2},\beta=y_0-\frac{B\lambda}{2},\gamma=z_0-\frac{C\lambda}{2},\]此时得到\[f_{\min}=\frac{\left(Ax_0+By_0+Cz_0+D\right)^2}{A^2+B^2+C^2},\]即得点\((x_0.y_0,z_0)\)到平面\(Ax+By+Cz+D=0\)的距离为\(\sqrt{f_{\min}}=\frac{\left|Ax_0+By_0+Cz_0+D\right|}{\sqrt{A^2+B^2+C^2}}\). 105 | \tcbline 106 | 令一种简单的方法是考虑几何关系, 记\((x_0,y_0,z_0)\)为\(P_1\), 取平面上一点\(P_2(a,b,c)\), 则考虑向量\(\overrightarrow{P_1P_2}=\left(a-x_0.b-y_0,c-z_0\right)\)与平面法向量\((A,B,C)\)夹角的余弦值(记较小的那个角为\(\theta\))可得点到平面距离\[d=\left|\overrightarrow{P_1P_2}\right|\cos\theta=\left|\overrightarrow{P_1P_2}\right|\frac{\left|\overrightarrow{P_1P_2}\cdot\left(A,B,C\right)\right|}{\left|\overrightarrow{P_1P_2}\right|\sqrt{A^2+B^2+C^2}}=\frac{\left|Ax_0+By_0+Cz_0+D\right|}{\sqrt{A^2+B^2+C^2}}.\qedhere\] 107 | \end{solution} 108 | \woe 设\(f(x,y)=xy-x\ln x+x-\ee^y(x>0,y\in\mathbb{R})\). 证明: \(f(x,y)\leqslant 0(\forall x>0,y\in\mathbb{R})\). 109 | \begin{proof} 110 | 111 | \end{proof} 112 | \end{quiza} 113 | \begin{quizb} 114 | \woe 试构造\(\bbr^2\)上的二元实函数\(f\)使得点\(\boldsymbol{0}\)不是它的极小值点, 但对任何\(\alpha\in[0,2\pi],g(t)=f(t\cos\alpha,t\sin\alpha)\)在\(t=0\)取得严格极小值. 进一步, 能否取到这样的一个\(f\)使得它在\(\bbr^2\)上连续? 请证明你的结论. 115 | \begin{solution} 116 | 117 | \end{solution} 118 | \woe 举例说明存在\(n\)阶方阵, 使得其所有特征值的绝对值都严格小于\(\left\|\boldsymbol{A}\right\|\). 119 | \begin{solution} 120 | 121 | \end{solution} 122 | \woe 若\(\boldsymbol{A}\)是\(n\)阶方阵, \(\left\|\boldsymbol{A}\right\|<1\). 证明: \(\left(\boldsymbol{I}-\boldsymbol{A}\right)^{-1}=\sum_{k=0}^{\infty}\boldsymbol{A}^k.\) 123 | \begin{proof} 124 | 125 | \end{proof} 126 | \woe 若\(\boldsymbol{A}\)是\(n\)阶可逆矩阵, 取\(\alpha>0\)足够小使得\(\boldsymbol{I}-\alpha\boldsymbol{AA}^\mathrm{T}\)正定. 令\[\boldsymbol{B}_0=\alpha\boldsymbol{A}^\mathrm{T},\boldsymbol{B}_{k+1}=\boldsymbol{B}_k\left(2\boldsymbol{I}-\boldsymbol{AB}_k\right)(k\geqslant 0).\]证明: \(\{\boldsymbol{AB}_k\}\)是单调增加的正定矩阵, 其极限为\(\boldsymbol{I}\). 特别地, \(\lim_{k\rightarrow+\infty}\boldsymbol{B}_k=\boldsymbol{A}^{-1}\). 127 | \begin{proof} 128 | 129 | \end{proof} 130 | \woe 试对于\(\boldsymbol{A}\)为复矩阵的情形, 定义诱导范数并建立相关性质. 131 | \woe 试改编以往遇到过的一个习题, 将一维情形的结果推广到矩阵情形. 132 | \end{quizb} 133 | \section{常系数线性微分方程} 134 | \precis{一阶常系数线性微分方程,矩阵指数函数,高阶常系数线性微分方程,特征方程,算子法} 135 | \begin{quiza} 136 | \woe 求以下方程的通解:\vspace{8pt}\\ 137 | \begin{tabular}{lcl} 138 | \((1)\,y''(x)-3y'(x)+2y(x)=0\);&\qquad\qquad&\((2)\,y''(x)-3y'(x)+2y(x)=x\);\vspace{0.3cm}\\ 139 | \((3)\,y''(x)-3y'(x)+2y(x)=x\ee^{3x}\);&&\((4)\,y''(x)-3y'(x)+2y(x)=x\ee^{2x}\);\vspace{0.3cm}\\ 140 | \((5)\,y''(x)-3y'(x)+2y(x)=x^2\ee^{x}\cos x\);&&\((6)\,y''(x)-3y'(x)+2y(x)=x^2\ee^{x}\sin 2x\).\vspace{0.3cm}\\ 141 | \end{tabular} 142 | \begin{solution} 143 | 由特征方程\(\lambda^2-3\lambda+2=0\)得\(\lambda_1=1,\lambda_2=2\), 从而齐次方程\((1)\)的通解为\[y=C_1\ee^x+C_2\ee^{2x},\]根据线性方程得通解结构, 为得到后续方程得通解, 只需要得到这些方程得特解. 使用算子法, 即得各个特解如下: 144 | 145 | \noindent (2) \(y^*=\frac{1}{(\DD-1)(\DD-2)}x=\left(\frac{1}{\DD-2}-\frac{1}{\DD-1}\right)x=\left(\frac{1}{2}+\frac{3}{4}\DD\right)x=\frac{1}{2}x+\frac{3}{4}.\)\vspace{0.5em} 146 | 147 | \noindent (3)\(y^*=\frac{1}{(\DD-1)(\DD-2)}x\ee^{3x}=\ee^{3x}\frac{1}{(\DD+2)(\DD+1)}x=\ee^{3x}\left(\frac{1}{2}-\frac{3}{4}\DD\right)x=\ee^{3x}\left(\frac{1}{2}x-\frac{3}{4}\right).\)\vspace{0.5em} 148 | 149 | \noindent (4)\(y^*=\frac{1}{(\DD-1)(\DD-2)}x\ee^{2x}=\ee^{2x}\frac{1}{\DD(\DD+1)}x=\ee^{2x}\frac{1}{\DD}\left(1-\DD\right)x=\ee^{2x}\left(\frac{1}{2}x^2-x\right).\)\vspace{0.5em} 150 | \end{solution} 151 | \woe 求以下方程的通解:\vspace{8pt}\\ 152 | \begin{tabular}{lcl} 153 | \((1)\,y''(x)-4y'(x)+4y(x)=0\);&\qquad\qquad&\((2)\,y''(x)-4y'(x)+4y(x)=x\);\vspace{0.3cm}\\ 154 | \((3)\,y''(x)-4y'(x)+4y(x)=x\ee^{3x}\);&&\((4)\,y''(x)-4y'(x)+4y(x)=x\ee^{2x}\);\vspace{0.3cm}\\ 155 | \((5)\,y''(x)-4y'(x)+4y(x)=x^2\ee^{2x}\cos x\);&&\((6)\,y''(x)-4y'(x)+4y(x)=x^2\ee^{x}\sin x\).\vspace{0.3cm}\\ 156 | \end{tabular} 157 | \begin{solution} 158 | 由特征方程\(\lambda^2-4\lambda+4=0\)得\(\lambda_1=\lambda_2=2\), 从而齐次方程\((1)\)的通解为\[y=C_1\ee^{2x}+C_2x\ee^{2x},\]根据线性方程得通解结构, 为得到后续方程得通解, 只需要得到这些方程得特解. 使用算子法, 即得各个特解如下: 159 | 160 | \noindent (2)\(y^*=\left(\frac{1}{\DD-2}\right)^2x=\frac{1}{4}(1+\DD)x=\frac{1}{4}x+\frac{1}{4}.\)\vspace{0.5em} 161 | 162 | \noindent (3)\(y^*=\left(\frac{1}{\DD-2}\right)^2x\ee^{3x}=\ee^{3x}\left(\frac{1}{\DD+1}\right)^2x=\ee^{3x}(1-2\DD)x=\left(x-2\right)\ee^{3x}.\)\vspace{0.5em} 163 | 164 | \noindent (4)\(y^*=\left(\frac{1}{\DD-2}\right)^2x\ee^{2x}=\ee^{2x}\frac{1}{\DD^2}x=\frac{1}{6}x^3\ee^{2x}.\)\vspace{0.5em} 165 | \end{solution} 166 | 167 | \woe 求以下方程的通解:\vspace{8pt}\\ 168 | \begin{tabular}{lcl} 169 | \((1)\,y''(x)+y'(x)+2y(x)=0\);&\qquad\qquad&\((2)\,y''(x)+y'(x)+2y(x)=x^3+1\);\vspace{0.3cm}\\ 170 | \((3)\,y''(x)+y'(x)+2y(x)=x\ee^{3x}\);&&\((4)\,y''(x)+y'(x)+2y(x)=x\sin 2x\);\vspace{0.3cm}\\ 171 | \((5)\,y''(x)+y'(x)+2y(x)=x^2\ee^{2x}\cos x\);&&\((6)\,y''(x)+y'(x)+2y(x)=x^2\ee^{x}\sin 2x\).\vspace{0.3cm}\\ 172 | \end{tabular} 173 | \begin{solution} 174 | 由特征方程\(\lambda^2+\lambda+2=0\)得\(\lambda_1=\) 175 | \end{solution} 176 | 177 | 178 | \woe 求以下方程的通解:\vspace{8pt}\\ 179 | \begin{tabular}{lcl} 180 | \((1)\,y^{(5)}(x)+2y'''(x)+y'(x)=0\);&\qquad\qquad&\((2)\,y^{(5)}(x)+2y'''(x)+y'(x)=(x+1)^2\);\vspace{0.3cm}\\ 181 | \((3)\,y^{(5)}(x)+2y'''(x)+y'(x)=x\sin 2x\);&&\((4)\,y^{(5)}(x)+2y'''(x)+y'(x)=x\sin x\);\vspace{0.3cm}\\ 182 | \((5)\,y^{(5)}(x)+2y'''(x)+y'(x)=x^2\ee^{2x}\cos x\);&&\((6)\,y^{(5)}(x)+2y'''(x)+y'(x)=x^2\ee^x\sin x\).\vspace{0.3cm}\\ 183 | \end{tabular} 184 | \begin{solution} 185 | 由特征方程\(\lambda^5+2\lambda^3+\lambda=0\)得\(\lambda_1=0,\lambda_2=\ii,\lambda_3=-\ii\), \(\lambda_2,\lambda_3\)均为二重根. 于是齐次方程(1)的通解为\[y=C_1+C_2\cos x+C_3x\cos x+C_4\sin x+C_5x\sin x,\] 186 | \end{solution} 187 | \woe 求方程\(y''(x)-y(x)=1\)满足\(y(0)=1,y'(0)=2\)的特解. 188 | \woe 试研究当\(b\)为何值时, 对所有\(y_0,y_1\in\mathbb{R}\), 方程\(y''(x)+y(x)=0\)满足\(y(0)=y_0,y(b)=y_1\)的特解总是存在. 189 | \woe 若\(n\)阶方阵\(\boldsymbol{A},\boldsymbol{B}\)可交换, 证明: \(\ee^{\boldsymbol{A}+\boldsymbol{B}}=\ee^{\boldsymbol{A}}\ee^{\boldsymbol{B}}\). 190 | \woe 设\(n\geqslant 2,\lambda\in\mathbb{C},n\)阶方阵\(\boldsymbol{A}\)为\[\begin{pmatrix} 191 | \lambda&1&0&\cdots&0&0\\ 192 | 0&\lambda&1&\cdots&0&0\\ 193 | \vdots&\cdots&\vdots&&\vdots&\vdots\\ 194 | 0&0&0&\cdots&\lambda&1\\ 195 | 0&0&0&\cdots&0&\lambda\\ 196 | \end{pmatrix}.\]试计算\(\ee^{x\boldsymbol{A}}\). 197 | \end{quiza} 198 | \begin{quizb} 199 | \woe 设\(P_1,P_2\)是次数依次为\(m,n\geqslant 1\)的多项式, \(Q_1,Q_2\)是次数小于\(m,n\)的非零多项式, 满足\(\frac{1}{P(t)}=\frac{Q_1(t)}{P_1(t)}+\frac{Q_2(t)}{P_2(t)}\), 其中\(P(t)=P_1(t)P_2(t)\). 若\(f\)为\(I\)上的连续函数, \(f_1,f_2\)依次为区间\(I\)上的\(m,n\)次连续可微函数, 满足\(P_1(\mathrm{D})f_1(x)=P_2(\mathrm{D})f_2(x)=f(x)\). 令\(F(x)=P_1(\mathrm{D})f_1(x)+P_2(\mathrm{D})f_2(x)\). 证明: \(F\)是\(m+n\)阶连续可导函数, 进而\(P(\mathrm{D})F(x)=f(x)\). 200 | \woe 举例说明上题中, \(P_1(\mathrm{D})f_1\)可以不是\(m+n\)阶连续可导的. 201 | \end{quizb} 202 | \section{导数的其他应用} 203 | \precis{Newton切线法,平方收敛,平面曲线的曲率和曲率半径,一元实函数的草图,拐点} 204 | \begin{quiza} 205 | \woe 设\(a>0\), 用Newton法通过求解\(f(x)=\ee^x-a\)的零点计算\(\ln a\). 若取初值为\(1\), 试讨论Newton迭代法的收敛性, 并讨论误差估计. 206 | \begin{solution} 207 | 208 | \end{solution} 209 | \woe 对于\(n\geqslant 2\)以及\(a>0\), 讨论通过Newton法计算\(f(x)=x^{10}-a\)的零点的可行性与误差估计. 210 | \begin{solution} 211 | 212 | \end{solution} 213 | \woe 计算椭圆\(\frac{x^2}{4}+\frac{y^2}{9}=1.\)上各点的曲率. 214 | \begin{solution} 215 | 我们先推到由参数方程确定的平面曲线的曲率计算公式. 对于\(y_0=f(x_0)\), 当\(f''(x_0)\ne 0\)时曲率为\(K=\frac{\left|f''(x_0)\right|}{\left(1+\left|f''(x_0)\right|^2\right)^{3/2}}\).先设平面曲线有如下的参数方程\(\begin{cases} 216 | x=\varphi(t),\\ 217 | y=\phi(t) 218 | \end{cases}\),则有\[y'(x)=\frac{\dd y}{\dd x}=\frac{\dd y/\dd t}{\dd x/\dd t}=\frac{\phi'(t)}{\varphi'(t)},\]进一步有\[y''(x)=\frac{\dd}{\dd x}\left(\frac{\dd y}{\dd x}\right)=\frac{\dd}{\dd t}\left(\frac{\dd y}{\dd x}\right)\cdot\frac{1}{\dd x/\dd t}=\frac{\phi''(t)\varphi'(t)-\phi'(x)\varphi''(x)}{\left(\varphi'(x)\right)^3},\]将其代入\(K\)得到\[K=\frac{\left|\phi''(t)\varphi'(t)-\phi'(x)\varphi''(x)\right|}{\left(\left(\varphi'(t)\right)^2+\left(\phi'(t)\right)^2\right)^{3/2}}.\]一般地, 椭圆\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)的参数方程可设为\(\left(a\cos\theta,b\sin\theta\right)\), 代入上述公式即得\[K=\sqrt{\frac{a^2 b^2}{\left(a^2 \sin ^2(t)+b^2 \cos ^2(t)\right)^3}}.\qedhere\] 219 | \end{solution} 220 | \woe 设\(y=\sqrt[3]{\frac{x^4}{x+1}}\), 试讨论函数的单调性, 极值, 凸性, 拐点, 求出它的渐近线, 并画出它的简图. 221 | \begin{solution} 222 | 首先在\(x=-1\)处\(y(x)\)无定义, 易见其为铅直渐近线. 直接计算得到\[y'(x)=\frac{x^3(3x+4)}{3(1+x)^2}\cdot\left(\frac{x^4}{x+1}\right)^{-2/3},\quad x\ne -1,\]于是\(y(x)\)有驻点\(-\frac{4}{3},0\). 关于函数的单调性与极值等信息如下表: 223 | \begin{center} 224 | \begin{tabular}{c|c|c|c|c|c|c|c} 225 | \Xhline{1.2pt} 226 | \phantom{\Large\(\int\)}&\(\left(-\infty,-\frac{4}{3}\right)\)&\(-\frac{3}{4}\)&\(\left(-\frac{4}{3},-1\right)\)&\(-1\)&\(\left(-1,0\right)\)&\(0\)&\(\left(0,+\infty\right)\)\\\hline 227 | \(f'\)&+&0&\(-\)&不存在&\(-\)&0&+\\\hline 228 | \(f\)&严格单增&极大值点&严格单减&渐近线&严格单减&极小值点&严格单增\\\Xhline{1.2pt} 229 | \end{tabular} 230 | \end{center} 231 | 又有\[y''(x)=\frac{4x^2}{9(1+x)^3}\cdot\left(\frac{x^4}{x+1}\right)^{-2/3},\quad x\ne -1,\]可见\(y(x)\)在\((-\infty,-1)\)上凹而在\((-1,+\infty)\)上凸. 并且\[\lim_{x\rightarrow\infty}\frac{y(x)}{x}=1,\lim_{x\rightarrow\infty}\left(y(x)-x\right)=\frac{1}{3},\]即\(y\)有渐近线\(f(x)=x-\frac{1}{3}\)其简图如下: 232 | \end{solution} 233 | \woe 设\(y=\ln\frac{x^2+x+1}{x^2-x+1}\), 试讨论函数的单调性, 极值, 凸性, 拐点, 求出它的渐近线, 并画出它的简图. 234 | \begin{solution} 235 | 先设\(g(x)=\frac{x^2+x+1}{x^2-x+1}\), 有\(g(0)=1>0\), 当\(x\ne 0\)时, \[g(x)=\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1\frac{2}{x+1/x-1},\]\(x>0\)时\(x+\frac{1}{x}>2\), 从而\(g(x)>0\); \(x<0\)时\(x+\frac{1}{x}\leqslant -2,\frac{2}{x+1/x-1}<-1\), 也有\(g(x)>0\). 从而\(y(x)\)的定义域为\(\bbr\). 直接计算得到\[y'(x)=\frac{-2x^2+2}{x^4+x^2+1},\]于是\(y(x)\)有驻点\(\pm1\). 关于函数的单调性与极值等信息如下表: 236 | \begin{center} 237 | \begin{tabular}{c|c|c|c|c|c} 238 | \Xhline{1.2pt} 239 | &\(\left(-\infty,-1\right)\)&\(-1\)&\(\left(-1,1\right)\)&\(1\)&\(\left(1,+\infty\right)\)\\\hline 240 | \(f'\)&\(-\)&0&\(+\)&0&\(-\)\\\hline 241 | \(f\)&严格单减&极小值点&严格单增&极大值点&严格单减\\\Xhline{1.2pt} 242 | \end{tabular} 243 | \end{center} 244 | \noindent 245 | 又有\[y''(x)=\frac{4x(x^4-2x^2-2)}{(x^4+x^2+1)^2},\]令\(y''(x)=0\), 得到三个实根分别为\(0,\pm\sqrt{1+\sqrt{3}}\). 关于函数凹凸性有下表: 246 | \begin{center} 247 | \begin{tabular}{c|c|c|c|c} 248 | \Xhline{1.2pt} 249 | \phantom{\Large\(\int\)}&\(\left(-\infty,-\sqrt{1+\sqrt{3}}\right)\)&\(\left(-\sqrt{1+\sqrt{3}},0\right)\)&\(\left(0,\sqrt{1+\sqrt{3}}\right)\)&\(\left(\sqrt{1+\sqrt{3}},+\infty\right)\)\\\hline 250 | \(f''\)&\(-\)&\(+\)&\(-\)&\(+\)\\\hline 251 | \(f\)&严格凹&严格凸&严格凹&严格凸\\\Xhline{1.2pt} 252 | \end{tabular} 253 | \end{center} 254 | \noindent 255 | 又有\[\lim_{x\rightarrow\infty}\ln\frac{x^2+x+1}{x^2-x+1}=0,\]从而\(y(x)\)有水平渐近线\(y=0\), 其简图如下.\[?\] 256 | \end{solution} 257 | \end{quiza} 258 | \begin{quizb} 259 | \woe 试给出由参数方程确定的平面曲线曲率的计算公式. 260 | \begin{solution} 261 | 参见本节\(\boldsymbol{\mathcal{A}}\)中的第\textbf{3}小题. 262 | \end{solution} 263 | \woe 设\(\varphi\)为\((0,+\infty)\)内的凹函数, 证明:\[\lim_{x\rightarrow+\infty}\left(\varphi(2x)-2\varphi(x)\right)=-\infty\]当且仅当\(y=\varphi(x)\)当\(x\rightarrow+\infty\)时没有渐近线. 264 | \end{quizb} -------------------------------------------------------------------------------- /Chapters/chapter8.aux: 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64 | \setcounter{mistake}{7} 65 | } 66 | -------------------------------------------------------------------------------- /Fonts/LXGWWenKai-Regular.ttf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Acwuld/LouAnalysis/da9f12974e8a6775a97f4ac7b8247a3d5174a1e0/Fonts/LXGWWenKai-Regular.ttf -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2024 Lorente 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /LouAnalysis.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Acwuld/LouAnalysis/da9f12974e8a6775a97f4ac7b8247a3d5174a1e0/LouAnalysis.pdf -------------------------------------------------------------------------------- /LouAnalysis.tex: -------------------------------------------------------------------------------- 1 | \documentclass{mbook} 2 | %\def\chap#1{Chapters/chapter#1} 3 | \begin{document} 4 | %\include{\chap{1}} 5 | %\include{\chap{2}} 6 | \include{Chapters/chapter0} 7 | \include{Chapters/chapter1} 8 | \include{Chapters/chapter2} 9 | \include{Chapters/chapter3} 10 | \include{Chapters/chapter4} 11 | \include{Chapters/chapter5} 12 | \include{Chapters/chapter6} 13 | \include{Chapters/chapter7} 14 | \include{Chapters/chapter8} 15 | \include{Chapters/chapter9} 16 | \include{Chapters/chapter10} 17 | \include{Chapters/chapter11} 18 | \include{Chapters/chapter12} 19 | \end{document} -------------------------------------------------------------------------------- /Pics/wx.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Acwuld/LouAnalysis/da9f12974e8a6775a97f4ac7b8247a3d5174a1e0/Pics/wx.png -------------------------------------------------------------------------------- /Pics/zfb.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Acwuld/LouAnalysis/da9f12974e8a6775a97f4ac7b8247a3d5174a1e0/Pics/zfb.jpg -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | Hello. 2 | 3 | 这是楼红卫《数学分析》的答案. 4 | 5 | 使用的主要字体是[霞鹜文楷](https://github.com/lxgw/LxgwWenKai) 6 | -------------------------------------------------------------------------------- /clean.pyw: -------------------------------------------------------------------------------- 1 | import os 2 | files = os.listdir('./') 3 | del_suffix = ['toc', 'vrb', 'aux', 'log', 'nav', 'out', 'snm', 'synctex.gz'] 4 | for file in files: 5 | for suffix in del_suffix: 6 | if file.endswith(suffix): 7 | os.remove(file) -------------------------------------------------------------------------------- /mbook.cls: -------------------------------------------------------------------------------- 1 | \NeedsTeXFormat{LaTeX2e}[2007/10/19] 2 | \ProvidesClass{mbook} 3 | [2011/04/05 v0.1 Thesis Template for Doctor] 4 | \LoadClass[oneside]{ctexbook}[2007/10/19] 5 | \RequirePackage{amsmath,amssymb,fancyhdr,xeCJK,tcolorbox,color,paralist,amsthm,tikz,float,caption,turnstile,mathrsfs,anyfontsize,extarrows,makecell,titlesec,geometry,fontawesome,enumitem,calc,wrapfig} 6 | %\RequirePackage{pgfornament} 7 | \RequirePackage[colorlinks,linkcolor=red]{hyperref} 8 | \tcbuselibrary{skins,theorems,breakable,documentation} 9 | 10 | \captionsetup[figure]{labelsep=space} 11 | \everymath{\displaystyle} 12 | \RequirePackage{framed} 13 | 14 | \RequirePackage{extarrows,pgfplots,longtable,multicol,graphicx,amssymb} 15 | %\hypersetup{bookmarks=true,bookmarksopen=false} 16 | 17 | 18 | 19 | \ctexset {chapter = {pagestyle = fancy}} 20 | \setlength{\headheight}{15pt} 21 | 22 | \newenvironment{solution}{\begin{proof}[\kaishu 解]\renewcommand{\qedsymbol}{$\blacksquare$}}{\end{proof}} 23 | \renewcommand{\proofname}{\noindent\kaishu 证明} 24 | \linespread{1.6} 25 | \usetikzlibrary{decorations.markings} 26 | 27 | \RequirePackage{bookmark} 28 | \bookmarksetup{ 29 | open, 30 | numbered, 31 | addtohook={% 32 | \ifnum\bookmarkget{level}=0 % chapter 33 | \bookmarksetup{bold}% 34 | \fi 35 | \ifnum\bookmarkget{level}=-1 % part 36 | \bookmarksetup{color=ocre,bold}% 37 | \fi 38 | } 39 | } 40 | 41 | 42 | \setCJKmainfont[Path=Fonts/]{LXGWWenKai-Regular.ttf} 43 | 44 | \geometry{a4paper,left=2.2cm,right=2.2cm,top=2cm,bottom=2cm} 45 | \pagecolor{yellow!10!white} 46 | \newtcbtheorem[number within=section]{theorem}{Theorem}% 47 | {colback=red!8!white,breakable, 48 | colframe=red!28!white,arc=0mm,breakable,coltitle=purple,fonttitle=\bfseries}{Th} 49 | 50 | \newtcbtheorem[number within=section]{definition}{Definition}% 51 | {colback=orange!8!white,breakable, 52 | colframe=orange!28!white,arc=0mm,breakable,coltitle=magenta,fonttitle=\bfseries}{Def} 53 | 54 | \newtcbtheorem[number within=section]{proposition}{Proposition}% 55 | {colback=cyan!8!white,breakable, 56 | colframe=cyan!28!white,arc=0mm,breakable,coltitle=teal,fonttitle=\bfseries}{po} 57 | 58 | \newtcbtheorem[number within=section]{lemma}{Lemma}% 59 | {colback=cyan!8!white,breakable, 60 | colframe=cyan!28!white,arc=0mm,breakable,coltitle=teal,fonttitle=\bfseries}{le} 61 | 62 | \newcounter{woes} 63 | \renewcommand{\thewoes}{\arabic{woes}} 64 | \newcommand{\woe}{\par\noindent\refstepcounter{woes}\textbf{\thewoes.}\hspace{0.5em}} 65 | \newcommand{\woestar}{\par\noindent\refstepcounter{woes}\(\boldsymbol{\thewoes^*.}\)\hspace{0.5em}} 66 | \newtcolorbox{questionA}{ 67 | colback=green!5!white,colframe=green!30!white, 68 | parbox=false,before upper=\par,before lower=\par, 69 | coltitle=green!25!black,breakable, 70 | title={\thesection \(\,\boldsymbol{\mathcal{A} }\)},fonttitle=\bfseries} 71 | \newenvironment{quiza}{\begin{questionA}\setcounter{woes}{0}}{\end{questionA}} 72 | 73 | \newtcolorbox{questionB}{ 74 | colback=blue!5!white,colframe=blue!30!white, 75 | parbox=false,before upper=\par,before lower=\par, 76 | coltitle=blue!25!black,breakable, 77 | title={\thesection \(\,\boldsymbol{\mathcal{B}}\)},fonttitle=\bfseries} 78 | \newenvironment{quizb}{\begin{questionB}\setcounter{woes}{0}}{\end{questionB}} 79 | \newenvironment{quizs}{\begin{enumerate}[itemindent=0.5em,label=\textcolor{red!50!black}{(\arabic*)}]}{\end{enumerate}} 80 | \newenvironment{quizss}{\vspace{0.2em}\begin{asparaenum}[(1)]}{\end{asparaenum}} 81 | 82 | 83 | \newenvironment{quizcs}{\begin{compactenum}[\quad (1)]}{\end{compactenum}} 84 | \newcommand{\precis}[1]{\addtocontents{toc}{% 85 | \unexpanded{\unexpanded{{\small#1\par}}}}} 86 | 87 | 88 | \newcommand{\inp}[1]{\langle #1 \rangle} 89 | \newcommand{\dd}{\,\mathrm{d}} 90 | \newcommand{\CC}{\,\mathrm{C}} 91 | \newcommand{\ee}{\mathrm{e}} 92 | \newcommand{\DD}{\mathrm{D}} 93 | \newcommand{\ii}{\mathrm{i}} 94 | \newcommand{\bbr}{\mathbb{R}} 95 | \newcommand{\bbc}{\mathbb{C}} 96 | \newcommand{\bbn}{\mathbb{N}} 97 | \newcommand{\bbz}{\mathbb{Z}} 98 | \newcommand{\bbq}{\mathbb{Q}} 99 | \newcommand{\supp}{\mathrm{supp}\,} 100 | \newcommand{\reff}[1]{\,\textbf{\ref{#1}}} 101 | \def\upint{\mathchoice% 102 | {\mkern13mu\overline{\vphantom{\intop}\mkern7mu}\mkern-20mu}% 103 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 104 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 105 | {\mkern7mu\overline{\vphantom{\intop}\mkern7mu}\mkern-14mu}% 106 | \int} 107 | \def\lowint{\mkern3mu\underline{\vphantom{\intop}\mkern7mu}\mkern-10mu\int} 108 | 109 | 110 | 111 | 112 | 113 | 114 | 115 | \setcounter{tocdepth}{1} 116 | \setcounter{secnumdepth}{2} 117 | \pagestyle{fancy}\empty{} 118 | \lhead{\textit{\leftmark}} 119 | \rhead{\textit{\rightmark}} 120 | %\chead{\textcolor{red!88!black}{\pgfornament[scale=0.25]{80}}} 121 | %\chead{\textcolor{red!88!black}{\pgfornament[scale=0.1]{69}}} 122 | %\fancyfoot[L]{\begin{tikzpicture} 123 | % \node (A) at (0,0) {}; 124 | % \node (B) at (5,0) {}; 125 | % \path (A)--(B) coordinate[pos=.5] (c1); 126 | % \draw [orange] (A) to [ornament=84] 127 | % (c1) to [ornament=84] (B); 128 | % \end{tikzpicture}\qquad\textcolor{red!88!black}{\pgfornament[scale=0.3]{17}}} 129 | %\fancyfoot[R]{\textcolor{red!88!black}{\pgfornament[scale=0.3]{18}}\qquad\begin{tikzpicture} 130 | % \node (A) at (0,0) {}; 131 | % \node (B) at (5,0) {}; 132 | % \path (A)--(B) coordinate[pos=.5] (c1); 133 | % \draw [orange] (A) to [ornament=84] 134 | % (c1) to [ornament=84] (B); 135 | % \end{tikzpicture}} 136 | \cfoot{\textcolor{red!88!black}{\thepage}} 137 | 138 | \renewcommand{\maketitle}{ 139 | \newgeometry{left=0.1cm,right=0.1cm,top=0.1cm,bottom=0.1cm} 140 | \vspace*{\fill} 141 | \vspace*{\fill} 142 | \begin{center} 143 | \normalfont 144 | {\Huge\bfseries \href{https://github.com/Acwuld/LouAnalysis}{数学分析}} 145 | \\ 146 | \bigskip 147 | {\Large \kaishu 答案} 148 | \\ 149 | \bigskip 150 | \bigskip 151 | {\Large Le} 152 | \\ 153 | \vspace{1.3cm} 154 | \end{center} 155 | 156 | 157 | 158 | \begin{center} 159 | \large 更新中, 项目已开源至\href{https://github.com/Acwuld/LouAnalysis}{Github}. 点击标题即可获取\LaTeX 源码 160 | \end{center} 161 | 162 | \vspace{1cm} 163 | \usetikzlibrary {datavisualization.formats.functions} 164 | \begin{figure}[h] 165 | \centering 166 | \tikz [scale=2.1]\datavisualization [ 167 | scientific axes=clean, 168 | y axis={ticks={style={ 169 | /pgf/number format/fixed, 170 | /pgf/number format/fixed zerofill, 171 | /pgf/number format/precision=2}}}, 172 | x axis={ticks={tick suffix=${}^\circ$}}, 173 | visualize as smooth line/.list={1,2,3,4,5,6}, 174 | style sheet=vary hue] 175 | data [format=function] { 176 | var set : {1,...,6}; 177 | var x : interval [0:60]; 178 | func y =20* sin(\value x * (\value{set}+10))/(\value{set}+5); 179 | }; 180 | \end{figure} 181 | \vspace{\stretch{3}} 182 | \usetikzlibrary {decorations.fractals,spy} 183 | \begin{tikzpicture} 184 | \draw [decoration=Koch snowflake] 185 | decorate { decorate{ decorate{ decorate{ (0,0) -- (20,0) }}}}; 186 | \draw[black] (0,0)--(20,0); 187 | \end{tikzpicture} 188 | } 189 | \ctexset {chapter={name={第,章},number={\arabic{chapter}},beforeskip={3pt},afterskip={3pt}}} 190 | \endinput 191 | --------------------------------------------------------------------------------