函数

实数与函数

张瑞
中国科学技术大学数学科学学院

函数

定义 1.
设$D\subset\mathbb{R}$为非空数集，若有某种确定的关系（或对应关系）$f$，对于每个$x\in D$，都有唯一的一个实数$y$与其对应，则称这个对应关系$f$定义了从$D$到$\mathbb{R}$中的函数，记为$y=f(x)$，或$f:D\to\mathbb{R}$。

$x$称为自变量，$y$称为因变量，$f(x)$称为函数在$x$处的值。

自变量$x$的变化范围$D$称为函数的定义域，记为$\mathcal{D}(f)$。

当$x$在定义域中任意变动时，函数值$y=f(x)$的变化范围称为函数的值域，记为$\mathcal{R}(f)$。

若集合$A\subset D$，则$f(A)=\{f(x)|x\in A\}$称为集合$A$在函数$f$下的像。

$f^{-1}(B)=\{x\in D | f(x)\in B\}$称为集合$B$关于函数$f$的原像。

函数的运算

  • $(f\pm g)(x)=f(x)\pm g(x)$

  • $(f\cdot g)(x)=f(x)g(x)$

  • $\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$

  • $(f\circ g)(x)=f(g(x))$称为$f$和$g$的复合函数

定义 2.
设$f:A=\mathcal{D}(f)\to B=\mathcal{R}(f)$。若存在函数$g:B\to A$，满足

\[(g\circ f)(x)=x, \quad \forall x\in A \]

和

\[(f\circ g)(x)=y, \quad \forall y\in B \]

则称$g$是$f$的反函数，记为$g(y)=f^{-1}(y)$，简记为$g=f^{-1}$。

  • 反函数是唯一的。
  • 反函数存在的充要条件是函数是一一的。

函数的表示方法

  • 显式， $y=f(x)$
  • 隐式， $f(x,y)=0$
  • 参数式，$\left\{\begin{aligned} x=\phi(t) \\y=\psi(t) \end{aligned}\right. $
  • 极坐标，$r=f(\theta)$

常见函数

  • 常值函数： $f(x)\equiv c$
  • 取整函数： $f(x)=[x]$
  • Dirichlet函数:
    \[f(x)=D(x)=\left\{\begin{aligned} 1 &, x\in Q \\ 0 &, x\notin Q \end{aligned}\right. \]
  • Riemann函数:
    \[f(x)=R(x)=\left\{\begin{aligned} \dfrac{1}{p} &, x=\dfrac{q}{p} \in Q \\ 0 &, x \notin Q \end{aligned}\right. \]

基本初等函数

指数函数: $f(x)=a^x$, $a>0$

对数函数: $f(x)=\log_a(x)$, $a>0$

% 画 e^x 和 log(x) 图像 \begin{tikzpicture}[ scale=1.0] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.8,0) -- (2.8,0) node[right] {$x$}; \draw[->] (0,-2.8) -- (0,2.98) node[left] {$y$}; \draw[domain=-2.6:2.6, color=red, dashed] plot (\x,\x);% node[right] {$f(x) =x$}; \draw[domain=-2.5:1.1, color=blue, thick] plot (\x,{exp(\x)}) node[right] {$e^x$}; \draw[blue] (0,1) node[left] {$1$}; \draw[color=orange, domain=0.1:3.0, thick] plot (\x,{ln(\x)}) node[above] {$\log(x)$}; \draw[dashed, orange] (1,0) node[below] {$1$}; \end{tikzpicture}

幂函数: $f(x)=x^\mu$, $x^{\mu}=e^{\mu \log(x)}, x>0$

% 画 e^x 和 log(x) 图像 \begin{tikzpicture}[ scale=1.2] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.8,0) -- (2.8,0) node[right] {$x$}; \draw[->] (0,-1.8) -- (0,2.98) node[above] {$y$}; \draw[domain=-0.6:2.6, color=red, dashed] plot (\x,\x) node[above] {$x^1$}; \draw[domain=-1.5:1.5, color=blue, thick] plot (\x,{pow(\x,2)}) node[right] {$x^2$}; \draw[domain=-2.5:2.5, color=blue, dashed, thick,samples=100] plot (\x,{\x/abs(\x)*abs(\x)^(1/3)}) node[right] {$x^{\frac13}$}; \draw[blue] (0,1) node[left] {$1$}; \draw[color=orange, domain=0:3.0, thick] plot (\x,{pow(\x, 0.5)}) node[above] {$x^{\frac12}$}; \draw[color=orange, domain=0:3.0, dashed, thick] plot (\x,{pow(\x, 0.5*sqrt(2))}) node[above] {$x^{\frac{\sqrt 2}2}$}; \draw[color=orange, domain=0.1:3.0, dashed, thick] plot (\x,{pow(\x, -0.5)}) node[above] {$x^{-\frac12}$}; \draw[dashed, orange] (1,0) node[below] {$1$}; \end{tikzpicture}

三角函数与反三角函数

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm \coordinate (A) at (0,0); \coordinate (C) at (35:2); \coordinate (g1) at (0:2); \coordinate (B) at (C|-g1); \draw[name path=ac] (A) node[left] {$A$} -- node[above, sloped] {$1$} (C) node[above] {$C$}; \draw (A) -- (B) node[below] {$B$}; \draw[red] (B) -- node[right] {$x$} (C); \draw ($ (A)+0.2*(g1) $) to[out=90,in=-55] node[right, pos=0.5] {$\alpha=\arcsin(x)$} ($ (A) + 0.2*(C) $); \end{tikzpicture} \begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm \coordinate (A) at (0,0); \coordinate (C) at (35:2); \coordinate (g1) at (0:2); \coordinate (B) at (C|-g1); \draw[name path=ac] (A) node[left] {$A$} -- node[above, sloped] {$1$} (C) node[above] {$C$}; \draw[red] (A) -- node[below] {$x$} (B); \draw (B) node[below] {$B$} -- (C); \draw ($ (A)+0.2*(g1) $) to[out=90,in=-55] node[right, pos=0.5] {$\alpha=\arccos(x)$} ($ (A) + 0.2*(C) $); \end{tikzpicture} \begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm \coordinate (A) at (0,0); \coordinate (C) at (35:2); \coordinate (g1) at (0:2); \coordinate (B) at (C|-g1); \draw[name path=ac] (A) node[left] {$A$} -- (C) node[above] {$C$}; \draw (A) -- node[below] {$1$} (B) node[below] {$B$}; \draw[red] (B) -- node[right] {$x$} (C); \draw ($ (A)+0.2*(g1) $) to[out=90,in=-55] node[right, pos=0.5] {$\alpha=\arctan(x)$} ($ (A) + 0.2*(C) $); \end{tikzpicture}

从左图可以得到$\cos(\alpha)=AB=\sqrt{1-x^2}$。类似有

\[\begin{aligned} \cos(\arcsin(x))=\sqrt{1-x^2}, \sin(\arccos(x))=\sqrt{1-x^2}, \\ \sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}, \cos(\arctan(x))=\frac{1}{\sqrt{1+x^2}}, \end{aligned} \]
% 画 sin 和 arcsin 图像 \begin{tikzpicture}[ scale=1.5] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-1.8,0) -- (1.8,0) node[right] {$x$}; \draw[->] (0,-1.8) -- (0,1.98) node[above] {$y$}; \draw[domain=-1.6:1.6, color=red, dashed] plot (\x,\x);% node[right] {$f(x) =x$}; \draw[domain=-pi/2:pi/2, color=blue, thick] plot (\x,{sin(\x r)}) node[right] {$\sin x$}; \draw[dashed, blue] (pi/2,1) -- (pi/2,0) node[below] {$\frac{\pi}2$}; \draw[dashed, blue] (pi/2,1) -- (0,1) node[left] {$1$}; \draw[dashed, blue] (-pi/2,-1) -- (-pi/2,0) node[above] {$-\frac{\pi}2$}; \draw[dashed, blue] (-pi/2,-1) -- (0,-1) node[right] {$-1$}; \draw[color=orange, domain=-1.0:1.0, thick] plot (\x,{asin(\x)/180.0*pi}) node[above] {$\arcsin(x)$}; \draw[dashed, orange] (1,pi/2) -- (0,pi/2) node[left] {$\frac{\pi}2$}; \draw[dashed, orange] (1,pi/2) -- (1,0) node[below] {$1$}; \draw[dashed, orange] (-1,-pi/2) -- (0,-pi/2) node[right] {$-\frac{\pi}2$}; \draw[dashed, orange] (-1,-pi/2) -- (-1,0) node[above] {$-1$}; \end{tikzpicture}
% 画 cos 和 arccos 图像 \begin{tikzpicture}[ scale=1.2] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-1.2,0) -- (pi+0.2,0) node[right] {$x$}; \draw[->] (0,-0.2) -- (0,pi+0.3) node[above] {$y$}; \draw[domain=-0.2:pi/2+0.2, color=red, dashed] plot (\x,\x);% node[right] {$f(x) =x$}; \draw[domain=0:pi, color=blue, thick] plot (\x,{cos(\x r)}) node[right] {$\cos x$}; \draw[blue] (0,1) node[left] {$1$}; \draw[blue] (pi/2,0) node[below] {$\frac{\pi}2$}; \draw[dashed, blue] (pi,-1) -- (pi,0) node[above] {${\pi}$}; \draw[color=orange, domain=1.0:-1.0, thick] plot (\x,{acos(\x)/180.0*pi}) node[above] {$\arccos(x)$}; \draw[orange] (1,0) node[below] {$1$}; \draw[orange] (0, pi/2) node[left] {$\frac{\pi}2$}; \draw[dashed, orange] (-1,pi) -- (0,pi) node[right] {${\pi}$}; \draw[dashed, orange] (-1,pi) -- (-1,0) node[below] {$-1$}; %\draw[dashed, orange] (-1,-pi/2) -- (0,-pi/2) node[right] {$-\frac{\pi}2$}; \end{tikzpicture}
% 画 tan 和 arctan 图像 \begin{tikzpicture}[ scale=1.0] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.8,0) -- (2.8,0) node[right] {$x$}; \draw[->] (0,-2.8) -- (0,2.98) node[above] {$y$}; \draw[domain=-2.6:2.6, color=red, dashed] plot (\x,\x);% node[right] {$f(x) =x$}; \draw[domain=-1.25:1.25, color=blue, thick] plot (\x,{tan(\x r)}) node[right] {$\tan x$}; \draw[dashed, blue] (pi/2,2.73) -- (pi/2,0) node[below] {$\frac{\pi}2$}; %\draw[dashed, blue] (pi/2,1) -- (0,1) node[left] {$1$}; \draw[dashed, blue] (-pi/2,-2.73) -- (-pi/2,0) node[above] {$-\frac{\pi}2$}; %\draw[dashed, blue] (-pi/2,-1) -- (0,-1) node[right] {$-1$}; \draw[color=orange, domain=-3.0:3.0, thick] plot (\x,{atan(\x)/180.0*pi}) node[below] {$\arctan(x)$}; \draw[dashed, orange] (3,pi/2) -- (0,pi/2) node[left] {$\frac{\pi}2$}; %\draw[dashed, orange] (1,pi/2) -- (1,0) node[below] {$1$}; \draw[dashed, orange] (-3,-pi/2) -- (0,-pi/2) node[right] {$-\frac{\pi}2$}; %\draw[dashed, orange] (-1,-pi/2) -- (-1,0) node[above] {$-1$}; \end{tikzpicture}

$AC=1$，$CD\perp AD$, $DF\perp AB$, $DE\perp BC$,

则

$AB=\cos(\alpha+\beta)$, $BC=\sin(\alpha+\beta)$

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm \coordinate (A) at (0,0); \coordinate (C) at (50:2); \coordinate (g1) at (0:2); \coordinate (g2) at (30:2); \coordinate (g3) at ($ (C) + (-60:2) $); \draw[name path=ac] (A) node[left] {$A$} -- node[above, sloped] {$1$} (C) node[above] {$C$}; \path[name path=ag1] (A) -- (g1); \path[name path=ag2] (A) -- (g2); \path[name path=cg3] (C) -- (g3); \path[name intersections={of=ag2 and cg3}]; \coordinate (D) at (intersection-1); \coordinate (F) at (D|-A); \coordinate (E) at (D-|C); \coordinate (B) at (C|-g1); \draw (A) -- (B) node[below] {$B$}; \draw[red] (B) -- (C); \draw (A) -- (g2); \draw ($ (D)+(30:0.1) $) -- ++(120:0.1) -- ++(210:0.1); \draw[green] (C) -- (D) node [below right] {$D$}; \draw[green] (B) -- (F); \draw[green] (D) -- (E) node[left] {$E$}; \draw[green] (D) -- (F) node[below] {$F$}; %\draw[blue, name path=func] (-0.2,0.3) .. controls (1,0.3) and (2,0.6) .. (2.5,1.9); \draw ($ (A)+(0:0.5) $) to[out=90,in=-60] node[right, pos=0.5] {$\alpha$} ($ (A) + (30:0.5) $); \draw ($ (C)+(-90:0.2) $) to[out=0,in=-150] node[below, pos=0.7] {$\alpha$} ($ (C) + (-60:0.2) $); \draw ($ (A)+(30:0.52) $) to[out=120,in=-40] node[right, pos=0.8] {$\beta$} ($ (A) + (50:0.52) $); \draw ($ (A)+(30:0.45) $) to[out=120,in=-40] ($ (A) + (50:0.45) $); \end{tikzpicture}
\[\begin{aligned} BC=&CE+DB =CD\cos(\alpha)+AD\sin(\alpha) \\ =&\sin(\beta)\cos(\alpha)+\cos(\beta)\sin(\alpha) \\ \sin(\alpha+\beta)=&\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta) \\ AB=&AF-BF =AD\cos(\alpha)-CD\sin(\alpha) \\ \cos(\alpha+\beta)=&\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \end{aligned} \]
\[\begin{aligned} \sin(\alpha+\beta) =&\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta) \\ \sin(\alpha-\beta) =&\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta) \\ \end{aligned} \]

得到

\[\begin{aligned} \sin(\alpha+\beta)-\sin(\alpha-\beta) =&2\cos(\alpha)\sin(\beta) \\ \sin(\alpha+\beta)+\sin(\alpha-\beta) =&2\sin(\alpha)\cos(\beta) \\ \end{aligned} \]

令$\alpha=\frac{a+b}2$, $\beta=\frac{a-b}2$，得到

\[\begin{aligned} \sin(a)-\sin(b)=&2\cos(\frac{a+b}2)\sin(\frac{a-b}2) \\ \sin(a)+\sin(b)=&2\sin(\frac{a+b}2)\cos(\frac{a-b}2) \\ \end{aligned} \]
\[\begin{aligned} \cos(\alpha+\beta) =&\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \\ \cos(\alpha-\beta) =&\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta) \\ \end{aligned} \]

得到

\[\begin{aligned} \cos(\alpha+\beta)-\cos(\alpha-\beta) =&-2\sin(\alpha)\sin(\beta) \\ \cos(\alpha+\beta)+\cos(\alpha-\beta) =&2\cos(\alpha)\cos(\beta) \\ \end{aligned} \]

令$\alpha=\frac{a+b}2$, $\beta=\frac{a-b}2$，得到

\[\begin{aligned} \cos(a)-\cos(b)=&-2\sin(\frac{a+b}2)\sin(\frac{a-b}2) \\ \cos(a)+\cos(b)=&2\cos(\frac{a+b}2)\cos(\frac{a-b}2) \\ \end{aligned} \]
\[\begin{aligned} \tan(\alpha+\beta) =&\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} =\frac{\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)} {\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)} \\ =&\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)} \end{aligned} \]
\[\tan(\alpha-\beta)=\tan(\alpha+(-\beta)) =\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} \]
\[\tan(\alpha)-\tan(\beta)=(1+\tan(\alpha)\tan(\beta))\tan(\alpha-\beta) \]
\[\tan(\alpha)-\tan(\beta) =\frac{\sin(\alpha)}{\cos(\alpha)}-\frac{\sin(\beta)}{\cos(\beta)} =\frac{\sin(\alpha-\beta)}{\cos(\alpha)\cos(\beta)} \]

公式

\[\tan(\alpha-\beta) =\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)} \]

中，记$\alpha=\arctan(x)$, $\beta=\arctan(y)$，得到

\[\tan(\arctan(x)-\arctan(y)) =\frac{x-y}{1+xy} \]

因此，有

\[\arctan(x)-\arctan(y) =\arctan\frac{x-y}{1+xy} \]

阿基米德螺线(Archimedean spiral)

  • $r=a\theta, \theta\in\mathbb{R}$

    chap1-spiral-1 chap1-spiral-2

伯努利双纽线 (Lemniscate of Bernoulli)

  • $(x^2+y^2)^2=2a^2(x^2-y^2)$
  • $r^2=2a^2\cos(2\theta)$
  • $x=\dfrac{a\sqrt{2}\cos(t)}{\sin^2(t)+1}, y=\dfrac{a\sqrt 2\cos(t)\sin(t)}{\sin^2(t)+1}$
% 画 lemniscate 伯努利双纽线 \begin{tikzpicture}[ scale=3] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-1.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.5) -- (0,0.5) node[above] {$y$}; \draw[domain=-pi:pi, samples=101, color=blue, thick] plot ({cos(\x r)/(sin(\x r)^2+1)},{sin(\x r)*cos(\x r)/(sin(\x r)^2+1)}); \draw[dashed, orange] (1,0.4)--(1,0) node[below] {$1$}; \draw[orange] ({sqrt(2)/2},0) node[below] {$a$}; \draw[orange] (-{sqrt(2)/2},0) node[below] {$-a$}; \node at ({sqrt(2)/2},0) [circle,fill, orange,inner sep=1pt] {};%{$a$}; \node at ({-sqrt(2)/2},0) [circle,fill, orange,inner sep=1pt] {}; \draw[dashed, orange] (1,0.4)--(0,0.4) node[left] {$0.4$}; \end{tikzpicture}

平面上到两个定点距离的乘积等于两个定点距离一半的平方的点的轨迹，是伯努利双纽线。

解. 设定点为$(-a,0)$和$(a,0)$，则两个定点距离一半为$a$。点$(x,y)$在双纽线上，则满足

\[\sqrt{(x-a)^2+y^2}\sqrt{(x+a)^2+y^2} = a^2 \]

两边平方，得到

\[(x^2+y^2+a^2-2ax)(x^2+y^2+a^2+2ax)=a^4 \]
\[(x^2+y^2+a^2)^2-a^4=4a^2x^2 \]
\[(x^2+y^2)^2+2a^2(x^2+y^2)=4a^2x^2 \]
\[(x^2+y^2)^2=2a^2(x^2-y^2) \]

chap1-lemniscate

摆线 (cycloid)

  • $x=a(\theta-\sin\theta), y=a(1-\cos \theta)$

    chap1-cycloid

圆内摆线 (Hypocycloid)

  • $\begin{cases}x=(R-r)\cos\theta+r\cos(\dfrac{R-r}r\theta),\\ y=(R-r)\sin\theta-r\sin(\dfrac{R-r}r\theta)\end{cases}$

    chap1-hypocycloid-1 chap1-hypocycloid-2

圆外摆线 (epicycloid)

  • $\begin{cases}x=(R+r)\cos\theta-r\cos(\dfrac{R+r}r\theta),\\ y=(R+r)\sin\theta-r\sin(\dfrac{R+r}r\theta)\end{cases}$

    chap1-epicycloid-1 chap1-epicycloid-2

箕舌线 (The Witch of Agnesi)

  • $y=\dfrac{8a^3}{x^2+4a^2}$
  • $x=2a\tan\theta, y=2a\cos^2\theta$
    % 画 箕舌线 (The Witch of Agnesi) \begin{tikzpicture}[ scale=2] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.0,0) -- (2.0,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.1) node[above] {$y$}; \draw[domain=-0.35*pi:0.35*pi, samples=101, color=blue, thick] plot ({tan(\x r)},{cos(\x r)^2}); \draw[orange] (0,0.5) circle (0.5); \draw[orange] (-1.9,1)--(1.9,1); \draw[orange, ] (0,0) -- ({tan(pi/3 r)},1); \draw[orange, dashed] ({tan(pi/3 r)},1) -- ({tan(pi/3 r)},0) ; \draw[orange, dashed] ({cos(pi/3 r)*sin(pi/3 r)},{cos(pi/3 r)^2}) -- ({tan(pi/3 r)},{cos(pi/3 r)^2}) ; \node at ({tan(pi/3 r)},{cos(pi/3 r)^2}) [circle,fill,blue,inner sep=1pt] {}; \draw[blue] (0,0.1) node[above right] {$\theta$}; \draw[blue] (30:0.15) arc (30:90:0.15);% node[above right] {$\theta$}; \draw[orange] (0,0.5) node[left] {$a$}; \node at (0,0.5) [circle,fill, orange,inner sep=1pt] {}; \end{tikzpicture}

玫瑰线 (Rose Curve)

  • $r=\cos(k\theta)$
  • $x=\cos(k\theta)\cos(\theta) , y=\cos(k\theta)\sin(\theta)$

    chap1-rose-1 chap1-rose-2

参数曲线$\displaystyle \begin{cases} x=\frac{2a}{1+t^2} \\ y=\frac{2at}{1+t^2} \end{cases}$

可以得到$y=\pm \sqrt{x(x-2a)}$，是一个圆。

下图中$t\in[-10,10]$

% 画 tan 和 arctan 图像 \begin{tikzpicture}[ scale=1.4] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-.1,0) -- (2.1,0) node[right] {$x$}; %\draw[->] (0,-1.1) -- (0,1.1) node[above] {$y$}; \draw[domain=-10:10, samples=1001, color=blue, thick] plot ({2/(1+(\x)*(\x))},{2/(1+(\x)*(\x))*(\x)});% node[right] {$\tan x$}; \draw[dashed, orange] (1,0) node[below] {$1$}; \draw[ orange] (2,0.1)--(2,-0.1) node[below] {$2$}; \end{tikzpicture}

双曲线$\displaystyle y=x+\frac1x$。 下图中$x\in[-5,5]$

% 画 tan 和 arctan 图像 \begin{tikzpicture}[ scale=0.4] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-5.1,0) -- (5.1,0) node[right] {$x$}; \draw[->] (0,-5.1) -- (0,5.1) node[above] {$y$}; \draw[domain=0.2:5, samples=101, color=blue, thick] plot (\x,{\x+1/\x}); \draw[domain=-5:-0.2, samples=101, color=blue, thick] plot (\x,{\x+1/\x}); \draw[dashed, orange] (1,2)--(1,0) node[below] {$1$}; \draw[dashed, orange] (1,2)--(0,2) node[left] {$2$}; \end{tikzpicture}

牛顿蛇形线$\displaystyle y=\frac{2x}{1+x^2}$。 下图中$x\in[-5,5]$

% 画 tan 和 arctan 图像 \begin{tikzpicture}[ scale=1] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-5.1,0) -- (5.1,0) node[right] {$x$}; \draw[->] (0,-1.1) -- (0,1.1) node[above] {$y$}; \draw[domain=-5:5, samples=101, color=blue, thick] plot (\x,{2*\x/(1+\x*\x)}); \draw[dashed, orange] (1,1)--(1,0) node[below] {$1$}; \draw[dashed, orange] (1,1)--(0,1) node[left] {$1$}; \end{tikzpicture}

牛顿三次曲线$\displaystyle y=x^3+\frac1x$。 下图中$x\in[-2,2]$

% 画 牛顿三次曲线 \begin{tikzpicture}[ scale=0.4] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.1,0) -- (2.1,0) node[right] {$x$}; \draw[->] (0,-5.1) -- (0,5.1) node[above] {$y$}; \draw[domain=0.2:2, samples=101, color=blue, thick] plot (\x,{\x^3+1/\x}); \draw[domain=-2:-0.2, samples=101, color=blue, thick] plot (\x,{\x^3+1/\x}); \draw[dashed, orange] (1,2)--(1,0) node[below] {$1$}; \draw[dashed, orange] (1,2)--(0,2) node[left] {$2$}; \end{tikzpicture}
% 画 lemniscate 伯努利双纽线 \begin{tikzpicture}[ scale=3] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-1.1,0) -- (1.1,0) node[right] {$x$}; \draw[->] (0,-0.5) -- (0,0.5) node[above] {$y$}; \draw[domain=-pi:pi, samples=101, color=blue, thick] plot ({cos(\x r)/(sin(\x r)^2+1)},{sin(\x r)*cos(\x r)/(sin(\x r)^2+1)}); \draw[dashed, orange] (1,0.4)--(1,0) node[below] {$1$}; \draw[orange] ({sqrt(2)/2},0) node[below] {$a$}; \draw[orange] (-{sqrt(2)/2},0) node[below] {$-a$}; \node at ({sqrt(2)/2},0) [circle,fill, orange,inner sep=1pt] {};%{$a$}; \node at ({-sqrt(2)/2},0) [circle,fill, orange,inner sep=1pt] {}; \draw[dashed, orange] (1,0.4)--(0,0.4) node[left] {$0.4$}; \end{tikzpicture}
% 画 lemniscate 伯努利双纽线 \begin{tikzpicture}[ scale=2] %\draw[very thin,color=gray] (-1.5,-1.2) grid (1.5,1.22); \draw[->] (-2.0,0) -- (2.0,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.1) node[above] {$y$}; \draw[domain=-0.35*pi:0.35*pi, samples=101, color=blue, thick] plot ({tan(\x r)},{cos(\x r)^2}); \draw[orange] (0,0.5) circle (0.5); \draw[orange] (-1.9,1)--(1.9,1); \draw[orange, ] (0,0) -- ({tan(pi/3 r)},1); \draw[orange, dashed] ({tan(pi/3 r)},1) -- ({tan(pi/3 r)},0) ; \draw[orange, dashed] ({cos(pi/3 r)*sin(pi/3 r)},{cos(pi/3 r)^2}) -- ({tan(pi/3 r)},{cos(pi/3 r)^2}) ; \node at ({tan(pi/3 r)},{cos(pi/3 r)^2}) [circle,fill,blue,inner sep=1pt] {}; \draw[blue] (0,0.1) node[above right] {$\theta$}; \draw[blue] (30:0.15) arc (30:90:0.15);% node[above right] {$\theta$}; \draw[orange] (0,0.5) node[left] {$a$}; \node at (0,0.5) [circle,fill, orange,inner sep=1pt] {}; \end{tikzpicture}

箕舌线 (The Witch of Agnesi)

chap1-witch-agnesi