【斜杠笔记】大学院数学之微积分
微分
导数
导数的性质
\((\dfrac{f}{g})'=\dfrac{f'g-fg'}{g^2}\)
\((fg)'=f'g+fg'\)
\(\dfrac{dy}{dx}=\dfrac{dy}{dt}\cdot \dfrac{dt}{dx}\)
\(\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\)
导数与极限
\(\displaystyle \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\)
重要的极限
\(\displaystyle \lim_{x\to \infty}\frac{\sin x}{x}=1\)
\(\displaystyle \lim_{x\to \pm \infty}(1+\frac{1}{x})^x =e\)
\(\displaystyle \lim_{x\to \infty}\frac{x^{\alpha}}{e^x}=0 (\alpha >0)\) 洛必达
\(\displaystyle \lim_{x \to \infty} \frac{\log(1+x)}{x}=1\)
\(\displaystyle \lim_{x \to +0}x^x=1\)
\(\displaystyle x^x=e^{x\ln x}=e^{\frac{\ln x}{x}}\to e^{\frac{\frac{1}{x}}{-\frac{1}{x^2}}}=e^{-x}\to 0\)
\(\displaystyle \lim_{x\to 0}\frac{\arctan x}{x}=1\)
\(\displaystyle \lim_{x\to 0}\frac{\sin x^3}{(\sin x)^2}=0\)
\(\displaystyle \lim_{x \to \infty}(1+\frac{\alpha}{x})^{\beta}=e^{\alpha \beta}\)
\(\displaystyle \lim_{n\to \infty} \frac{n^k}{a^n}=0 (k>0,a>1)\)
\(\displaystyle \frac{(n^k)'}{(e^{n\ln a})'}=\frac{kn^{k-1}}{\ln ae^{n\ln a}}=\frac{k!}{(\ln a)^k}\cdot e^{\frac{1}{n\ln a}}=0\)
\(\displaystyle \lim_{n\to \infty}\frac{a^n}{n!} (a>1)=0\)
\(\displaystyle \frac{a^n}{n!}=\frac{a}{n}\cdot \frac{a}{n-1}\dots \cdot \frac{a}{1}\)
\(n>a \Rightarrow 0\)
\(\displaystyle \lim_{n\to \infty}\sqrt[n]{n}=\lim_{n\to \infty}n^\frac{1}{n}=e^\frac{1}{n}\ln n=e^0=1\)
常见函数的导数
\((\alpha^x)'=\alpha^x \log x\)
\((\log |x|)'=\dfrac{1}{x}\)
\((\arctan x)'=\dfrac{1}{1+x^2}\)
\((arccot x)'=-\dfrac{1}{1+x^2}\)
\((x^\alpha)^{(n)}=\alpha (\alpha-1)\dots(\alpha-n+1)x^{\alpha-n}\)
\((\log_a|x|)'=(\dfrac{\log |x|}{\log a})'=\dfrac{1}{\log a}\cdot \dfrac{1}{x}\)
\((\sin x)^{(n)}=\sin(x+\dfrac{n\pi}{2})\)
\((\cos x)^{(n)}=\cos(x+\dfrac{n\pi}{2})\)
为了便于计算机计算,将函数用多项式逼近
泰勒级数 テイラー級数
$f(x)=f(a)+(x-a)+(x-a)2+(x-a){n-1}+R_n $ 极小项
\(\displaystyle =\sum_{k=0}^{n-1} \frac{f^{(k)}(a)}{k!}(x-a)^k+R_n\)
a=0のとき、
\(\displaystyle f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f'(0)}{2!}x^2+\dots +\frac{f^{(n-1)(0)}}{(n-1)!}x^{n-1}+R_n\)
\(\displaystyle e^x=1+x+\frac{x^2}{2!}+\dots +\frac{x^{n-1}}{(n-1)!}+R_n\)
\(\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots +(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}+R_{2n+1}\)
\(\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots +(-1)^n\frac{x^{2n}}{(2n)!}+R_{2n+2}\)
\(\displaystyle \log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots +(-1)^n\frac{x^{n-1}}{n-1}+R_n\)
\(\displaystyle (1+x)^n=1+C_n^1x+C_n^2x^2+\dots +C_n^{n-1}x^{n-1}+R_n\) 二项式定理
$=1+x+x2++xn+$ 等比求和
由导数判定极值
\(f'(x)=0\) 判定前后f'(x)正负(单调性) \(f''(x)>0 \leftrightarrow y=f(x)\) 下凸 如\(y=x^2\) \(f''(x)<0 \leftrightarrow y=f(x)\) 上凸
多元函数的极值计算
黑塞矩阵
对于二元函数\(f(x,y)\),若有\(f_x(x_0,y_0)=f_y(x_0,y_0)=0\)
如何判定在点\((x_0,y_0)\)处取极值呢?
\(H=det\begin{pmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{pmatrix}=f_{xx}f_{yy}-(f_{xy})^2\)
- H>0: 若\(f_{xx}>0\),则\((x_0,y_0)\)处取极小值,若\(f_{xx}<0\),则\((x_0,y_0)\)处取极大值
- H<0:则\((x_0,y_0)\)是鞍点,\((x_0,y_0)\)处不是极值点
- H=0: 则二阶导数无法判断该临界点的性质,需要考察更高阶的导数。
积分
不定积分
\(\int f(x)=F(x)+C\)
定积分
\(\displaystyle \int_a^{a+T}f(x)dx=\int_0^T f(x)dx\)
\(\displaystyle \int_a^b f(x)dx=\int_{\alpha}^{\beta}f\{\Phi(t)\}\Phi'(t)dt\)
部分积分
\(\displaystyle \int_a^b f(x)g'(x)dx=[f(x)g(x)]_a^b-\int_a^b f'(x)g(x)dx\)
广义积分
\(\displaystyle \int_a^{+\infty}f(x)=\lim_{b\to +\infty}\int_a^b f(x)dx\)
\(\displaystyle \int_{-\infty}^b f(x)dx=\lim_{a\to -\infty}^b f(x)dx\)
\(\displaystyle \int_a^b f(x)dx=\lim_{x\to 0+}\int_a^{c-\varepsilon_1}f(x)dx+\lim_{x\to 0+}\int_{c+\varepsilon_2}^b f(x)dx\) c处有断点
三重积分
\(x=r\sin \theta \cos \phi\)
\(y=r\sin \theta \sin \phi\)
\(z=r\cos \phi\)
\(\int_A \int \int f(x,y,z)dxdydz=\iiint_B f(r\sin\theta\cos \phi, r\sin \theta\sin \phi, r\cos \theta)r^2 \sin \theta drd\theta d \phi\)
\(\displaystyle \int f(x)dx=F(x)+C\)
求解方法
1)基本积分
\(\displaystyle \int \tan(\alpha x)dx=-\frac{1}{\alpha}\log |\cos \alpha x|\)
\(\displaystyle \int \cot\alpha x dx=\frac{1}{\alpha}\log |\sin \alpha x|\)
\(\displaystyle \int \frac{1}{x}dx=\log |x|\)
\(\displaystyle \int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan (\frac{x}{a})\)
\(\displaystyle \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log |\frac{x-a}{x+a}|\)
\(\displaystyle \int ax^k dx=a\frac{1}{k+1}x^k+C\)
\(\displaystyle \int \frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}\)
2)分式积分
有理函数
1次多项式
\(\displaystyle \frac{A}{(b_kx+C_k)^n}\Rightarrow \int \frac{A}{b_k^n (x+\frac{c_k}{b_k})^n}=\frac{A}{b_k^n}\cdot \frac{1}{n+1}(x+\frac{c_k}{b_k})^{n+1} (n>1)\)
\(\displaystyle =\frac{A}{b_k}\log (x+\frac{c_k}{b_k}) (n=1)\)
2次多项式
分母凑成平方和
\(\displaystyle \int \frac{x}{(x^2+a^2)^m}dx\)
大式拆为小式
\(\left \{ \begin{array}{cc} \displaystyle \frac{-1}{2(m-1)(x^2+a^2)^{m-1}} &(m>1)\\\displaystyle \frac{1}{2}\log (x^2+a^2) &(m=1)\end{array}\right.\)
例:
\(\displaystyle \frac{4x^4+2x^3+10x^2+3x+9}{(x+1)(x^2+2)^2}=\frac{A}{x+1}+\frac{Bx+C}{x^2+2}+\frac{Dx+E}{(x^2+2)^2}\)
\(\displaystyle \int \frac{1}{(x^2+a^2)^n}dx=I_n\)
\(\displaystyle I_n=\frac{x}{(x^2+a^2)^n}-\int \frac{x(-n)\cdot 2x}{(x^2+a^2)^{n+1}}dx\)
\(\displaystyle =\frac{x}{(x^2+a^2)^n}+2n\int \frac{x^2}{(x^2+a^2)^{n+1}}dx\)
\(\displaystyle =\frac{x}{(x^2+a^2)^n}+2n\int \frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}}dx\)
\(\displaystyle =\frac{x}{(x^2+a^2)^n+2n\int \frac{1}{(x^2+a^2)^n}}dx-2a^2n\int \frac{1}{(x^2+a^2)^{n+1}}dx\)
\(\displaystyle =\frac{x}{(x^2+a^2)^n}+2nI_n-2a^nI_{n+1}\)
\(\displaystyle \therefore I_{n+1}=\frac{1}{2a^2}[\frac{x}{(x^2+a^2)^n}+(2n-1)I_n]\)
无理函数
1) \(\sqrt{ax^2+bx+c}\) 配平方式
2)\(\sqrt[n]{ax+b}\) 换元法
3)\(\displaystyle \sqrt[n]{\frac{ax+b}{cx+d}}\) 换元法
4)\(\displaystyle \sqrt{x^2+a^2}\) \(x=a\tan \theta\)
5)\(\sqrt{a^2-x^2}, x=a\sin \theta\)
6)\(\sqrt{x^2-a^2}, x=a\cos \theta\)
三角函数
1)\(\int f(\sin x)\cos x dx\) 把cos x放到dx中
2)\(\int f(\cos x)\sin x dx\)把sin x放到dx中
3)\(\int f(\sin^2 x, \cos^2x, \tan x)dx\) 令\(t=\tan x\)使用万能公式
4)\(\int f(\sin x,\cos x)dx\) 令\(t=\tan \frac{x}{2}\)
定积分的应用
1.求面积
\(S=\int_a^b \{f(x)-g(x)\}dx, r=f(\theta)\)
\(S=\dfrac{1}{2}\int_{\alpha}^{\beta}r^2d\theta\)
\(=\dfrac{1}{2}\int_{\alpha}^{\beta}f^2(\theta)d\theta\)
2.回转体的表面积
\(y=f(x)\)绕X轴旋转
\(S=2\pi \int_a^b y\cdot \sqrt{(dx)^2+(dy)^2}\)
\(=2\pi \int_a^b f(x)\sqrt{1+f'^2(x)}dx\)
周长\(2\pi r(y) \times 高 \rightarrow 弧长\)
\(V=\pi \int_a^b f^2(x)dx\)
圆盘法向量垂直于旋转轴,切成一个个圆盘,a,b是高度方向上的上下限。
\(\pi f^2(x)=\pi r^2\)
3.弧长
\(l=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt\)
\(=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta\)