数学要背的内容
数学要背的内容
常用泰勒公式
x->0 时才能使用
\(f(x)=f(0) + f'(0)x +\Large \frac{f''(0)x^2}{2!} + ... + \frac{f^{(n)}(0)x^n}{n!}\)
\(\sin x = x - {\Large\frac{x^3}{6}} + o(x^3)\)
\(\cos x = 1 - {\Large \frac{x^2}{2} + \frac{x^4}{4!}} + o(x^4)\)
\(\sec x=1 + {\Large \frac{x^2}{2} } + o(x^3)\)
\(\tan x = x + {\Large\frac{x^3}{3}} + o(x^3)\)
\(\arcsin x = x + {\Large\frac{x^3}{6}} + o(x^3)\)
\(\arctan x = x - {\Large\frac{x^3}{3}} + o(x^3)\)
\(\ln (1 + x) = x - {\Large \frac{x^2}{2} + \frac{x^3}{3}} + o(x^3)\), 第n项：\((-1)^{n-1}\Large \frac{x^n}{n}\)
\(\ln (x + \sqrt{1 + x^2}) = x - {\Large \frac{x^3}{6}} + o(x^3)\)
\(e^x = 1 + x + {\Large \frac{x^2}{2} + \frac{x^3}{3!}} + o(x^3)\) 。原式： \(e^x=\Large\sum_{n = 0} ^\infin\frac{x^n}{n!}\)
\((1 + x)^a = 1 + ax + {\Large \frac{a(a - 1)}{2}}x^2 + o(x^2)\)
常用等价替换
当 x->0 时，有：
\(\sin x \sim x，\tan x \sim x，\arcsin x \sim x，\arctan x \sim x\)
\(1-\cos x \sim {\Large \frac{x^2}{2}} ， a^x-1 \sim x\ln a ，e^x-1 \sim x\)
\(1-\cos ^a x\sim {\Large \frac{ax^2}{2}}\)
\(\ln(1+x) \sim x ，(1+x)^a-1 \sim ax\)
$\ln(x+\sqrt{1 + x^2}) \sim x $
当 \(x\rightarrow\infty\) 时，有：
重要极限:
\(\underset{x\rightarrow \infty}{\lim}(1 + \frac{1}{x})^x = e\)
\(\underset{x\rightarrow \infty}{\lim}{\Large \frac{\sin x}{x}} = 1\)
\(\Large \underset{x\rightarrow \infty}{\lim} x^{(\frac{1}{x})} = e^{\underset{x\rightarrow \infty}{\lim}\frac{1}{x}\ln x} = 1\) （洛必达法则）
\(\underset{x\rightarrow 0}{\lim}\Large(1 - x)^{\frac{1}{x}} = \frac{1}{e}\)
\(\underset{x\rightarrow 0}{\lim}\Large(1 + x)^{\frac{1}{x}} = e\)
三角函数公式
在直角三角形中：
\(\sin \alpha\) ：对边除以斜边, 音标[saɪn]
\(\cos \alpha\) ：邻边除以斜边, 音标[ˈkəʊsaɪn]
\(\tan \alpha = \Large \frac{\sin \alpha}{\cos \alpha}\) ： 对边除以邻边, 音标[ˈtændʒənt]
\(\cot \alpha = \Large \frac{1}{\tan \alpha} = \frac{\cos \alpha}{\sin \alpha}\) : 邻边除以对边, 音标['kəʊ'tændʒənt]
\(\sec \alpha = \Large \frac{1}{\cos \alpha}\) : 斜边除以邻边, 音标['si:kənt]
\(\csc \alpha = \Large \frac{1}{\sin \alpha}\) : 斜边除以对边, 音标['kəʊ'si:kənt]
三角函数的变换
半径为 1，圆心在 (0，0) 的圆可以推导下面公式：
\(\sin^2 \alpha + \cos^2 \alpha = 1\)
\(\sec^2 \alpha - 1 = \tan^2 \alpha\)
\(\sin({\Large\frac{\pi}{2}}+\alpha)=\cos \alpha\)
\(\sin({\Large\frac{\pi}{2}}-\alpha)=\cos \alpha\)
\(\cos({\Large\frac{\pi}{2}}+\alpha)=-\sin \alpha\)
\(\cos({\Large\frac{\pi}{2}}-\alpha)=\sin \alpha\)
\(\tan({\Large\frac{\pi}{2}}+\alpha)=-\cot \alpha\)
\(\tan({\Large\frac{\pi}{2}}-\alpha)=\cot \alpha\)
\(\cot({\Large\frac{\pi}{2}}+\alpha)=-\tan \alpha\)
\(\cot({\Large\frac{\pi}{2}}-\alpha)=\tan \alpha\)
和差角公式
\(\sin(\alpha \pm \beta)=\sin \alpha\cos \beta \pm \cos \alpha \sin \beta\)
\(\cos(\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha\sin \beta\)
\(\tan(\alpha \pm \beta)= \Large \frac{\tan \alpha \pm tan \beta}{1 \mp \tan \alpha \cdot \tan \beta}\)
\(\cot(\alpha \pm \beta)= \Large \frac{\cot \alpha \cdot \cot \beta \mp 1}{\cot \beta \pm \cot \alpha}\)
和差化积公式
\(\sin \alpha + \sin \beta = 2\sin{\Large \frac{\alpha + \beta}{2}}\cdot \cos {\Large \frac{\alpha - \beta}{2}}\)
\(\sin \alpha - \sin \beta = 2\cos{\Large \frac{\alpha + \beta}{2}}\cdot \sin {\Large \frac{\alpha - \beta}{2}}\)
\(\cos \alpha + \cos \beta = 2\cos{\Large \frac{\alpha + \beta}{2}}\cdot \cos {\Large \frac{\alpha - \beta}{2}}\)
\(\cos \alpha - \cos \beta = -2\sin{\Large \frac{\alpha + \beta}{2}}\cdot \sin {\Large \frac{\alpha - \beta}{2}}\)
$\sin \alpha + \cos \alpha = \sqrt{2} \cos (\alpha - {\Large\frac{\pi}{4})} = -\sqrt{2}\sin(\alpha + {\Large\frac{\pi}{4})} $
积化和差公式
\(\sin \alpha \cos \beta = {\Large\frac{1}{2}}[\sin(\alpha + \beta)+\sin(\alpha - \beta)]\)
\(\cos \alpha \sin \beta = {\Large\frac{1}{2}}[\sin(\alpha + \beta)-\sin(\alpha - \beta)]\)
\(\cos \alpha \cos \beta = {\Large\frac{1}{2}}[\cos(\alpha + \beta)+\cos(\alpha - \beta)]\)
\(\sin \alpha \sin \beta = {\Large\frac{1}{2}}[\cos(\alpha + \beta)-\cos(\alpha - \beta)]\)
倍角公式
\(\sin 2 \alpha = 2\sin \alpha \cos \alpha=\Large \frac{2\tan \alpha}{1 + \tan^2 \alpha}\)
\(\cos 2 \alpha = \cos^2 \alpha - \sin^2 \alpha \\ \qquad = 2\cos^2 \alpha - 1 \\\qquad = 1 - 2\sin^2 \alpha \\\qquad = \Large \frac{1-\tan^2 \alpha}{1 + \tan^2 \alpha}\)
\(\tan 2 \alpha = {\Large \frac{2\tan \alpha}{1-\tan^2 \alpha}}\cdot \cot 2 \alpha = \Large \frac{\cot^2 \alpha - 1}{2\cot \alpha}\)
\(\sin 3 \alpha = 3\sin \alpha - 4\sin^3 \alpha\)
\(\cos 3 \alpha = 4\cos^3 \alpha - 3\cos \alpha\)
基本求导公式
\((x^a)' = ax^{a - 1}, a为常数\)
\((a^x)' = a^x\ln a,(a>0, a\neq 1)\)
\((e^x)' = e^x\)
\((\log _ax)' = {\Large \frac{1}{x\ln a}},(a>0, a\neq 1）\)
\((\ln|x|)' = \Large \frac{1}{x}\)
\((\sin x)' = \cos x\)
\((\cos x)' = -\sin x\)
\((\arcsin \Large\frac{x}{a})'=\Large \frac{1}{\sqrt{a^2 - x^2}}\)
\((\arccos x)' = -\Large \frac{1}{\sqrt{1 - x^2}}\)
\((\tan x)' = \sec ^2x\)
\((\cot x)' = -\csc ^2x\)
\((\Large \frac{1}{a}\arctan \frac{x}{a})' = \Large \frac{1}{a^2 + x^2}\)
\((arccot x)' = -\Large \frac{1}{1 + x^2}\)
\((\sec x)'= \sec x \tan x\)
\((\csc x)' = -\csc x \cot x\)
\((\ln(x + \sqrt{x^2 + a^2}))' = \Large \frac{1}{\sqrt{x^2 + a^2}}\)
\((\ln(x + \sqrt{x^2 - a^2}))' = \Large \frac{1}{\sqrt{x^2 - a^2}}\)
积分公式
\(\Large\int\frac{dx}{\sin x}=\ln | \csc x- \cot x| + c\)
\(\Large\int\frac{dx}{\cos x}=\ln | \sec x- \tan x| + c\)
\(\Large\int\frac{1}{x^2 - a^2}dx=\frac{1}{2a}\ln |\frac{x-a}{x+a}| + c\)
换元法：
分清下面式子：
\(\{f[g(x)]\}'=\Large\frac{d\{f[g(x)]\}}{dx}\)
\(f'[g(x)]=\Large \frac{d\{f[g(x)]\}}{d[g(x)]}\)
反函数求导：
反函数和原函数关于直线 x=y 对称
\(y=f(x) 单调可导，f'(x) \neq 0, 则存在反函数 x=\varphi(y), \varphi'(y) = \Large\frac{1}{f'(x)}\)
\(若 f(x) 二阶可导，则 \varphi''(y)=-\Large\frac{f''(x)}{[f'(x)]^3}\)
\(f''(x) = -\Large\frac{\varphi ''(y)}{[\varphi '(y)]^3}\)
高数 18 讲
171页 华里士公式
泰勒公式展开为幂级数，84页
$\int e^{ax}\sin bx dx $ 166页
变限积分的奇偶性、周期性
反常积分审敛:
\(\Large \underset{x\rightarrow 0}{\lim}x^a\ln x = 0, a>0\)
\(\Large\underset{x\rightarrow \infty}{\lim} \frac{lnx}{x^a} = 0, a>0\)
\(\Large\int_0^{1}\frac{1}{x^p}dx \lbrace ^{收敛，0<p<1} _{发散， p\geq 1}\)
\(\Large\int_1^{+\infty}\frac{1}{x^p}dx \lbrace ^{收敛，p>1} _{发散， p\leq 1}\)
有理数的不定积分
第10讲，一元函数积分学的公式
几何图形
椭球体
\(\Large \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\)
体积：\(\frac{4}{3}\pi abc\)
椭圆面积: \(ab\pi\)
球体体积：\(\frac{4}{3}\pi r^3\)
球体表面积：\(4\pi r^2\)
圆柱体积：\(\pi r^2h\)
圆锥体积：\(\Large \frac{1}{3}\pi r^2h\)
组合表示：
\(\Large n \choose k\) 表示从 n 个元素中取出 k 个元素的组合数（高中的 \(C_n^k)\)
\(\Large {n \choose k}=\frac{a(a - 1)(a - 2)\cdots (a - k + 1)}{k!} =\frac{(a)_k}{k!}\)
牛顿二项式定理：
\(\Large (x + y)^a=\sum_{k = 0} ^a{a \choose k}x^{a-k}y^k\)
平方和公式：
\(1^2 + 2^2 + ... + n^2 = {\Large\frac{1}{6}}n(n+1)(2n+1)\)
\(a³＋b³=( a＋b)(a²-ab＋b²）\)
\(a³-b³=(a-b)(a²+ab+b²)\)
等比数列求和公式：
\(\Large s_n = a_1\frac{1-q^n}{1-q}\)
等差数列：
第 n 项：\(a_n = a1 + (n-1)d\)
前 n 项和：\(\Large s_n = \frac{n(a_1 + a_n)}{2}、s_n=na_1 + \frac{n(n-1)d}{2}\)
求根公式：
\(\Large \frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\Gamma\) 函数：
\(\Gamma(a)=\Large\int_0^{+\infty}x^{a-1}e^{-x}dx=2\int_0^{+\infty}t^{2a-1}e^{-t^2}dt, (x, t > 0)\)
\(\Gamma(a + 1)=a\Gamma(a)\)
\(\Gamma(1)=1, \Large \Gamma(\frac{1}{2})=\sqrt{\pi}\)
不等式
对于正数：开根号、取对数都不改变不等号的方向
\(\Large\sqrt[3]{abc}\leq\frac{a+b+c}{3}\leq\sqrt{\frac{a^2+b^2+c^2}{3}}, (a,b\geq0)\)
在 \((0,0.86),x<\arctan x\)
\(\Large \frac{2}{\frac{1}{a} + \frac{1}{b}}\leq\sqrt{ab}\leq \frac{a+b}{2}\leq \sqrt{\frac{a^2+b^2}{2}},(a,b\geq 0)\)
\(x>lnx\)
\(x>\sin x, (x>0)\)
\(x<\tan x, (0<x<{\Large\frac{\pi}{2}})\)
\(x<y, a + b < 1时， x<ax+by<y\)
\(a^2 + b^2 \geq \Large \frac{(a+b)^2}{2}\)
中值定理解题技巧
看见 \(f(x) + f'(x)\):
\([e^xf(x)]' = e^x(f'(x)+f(x))\)
物理公式
万有引力公式：\(G\Large\frac{Mm}{r^2}\), （M、m是两物体的质量）
