AdvancedMath

本文最后更新于 2024年10月28日 晚上

高等数学 - 竞赛向

参考资料:

1 极限

求极限一些方法方法:

  1. 等价无穷小
  2. 洛必达法则
  3. 泰勒展开
  4. 夹逼准则

给出 泰勒展开 的常用公式:

ex=n=01n!xn=1+x+12!x2+(,+)sinx=n=0(1)n(2n+1)!x2n+1=x13!x3+15!x5+,x(,+)cosx=n=0(1)n(2n)!x2n=112!x2+14!x4+,x(,+)tanx=n=1B2n(4)n(14n)(2n)!x2n1=x+13x3+215x5+,x(π2,π2)arcsinx=n=0(2n)!4n(n!)2(2n+1)x2n+1=x+16x3+340x5+5112x7+351152x9++,x(1,1)arctanx=n=0(1)n2n+1x2n+1=x13x3+15x5++x[1,1]ln(1+x)=n=0(1)nn+1xn+1=x12x2+13x3+,x(1,1]11x=n=0xn=1+x+x2+x3+,x(1,1)11+x=n=0(1)nxn=1x+x2x3+,x(1,1)(1+x)α=1+n=1α(α1)(αn+1)n!xn=1+αx+α(α1)2!x2+,x(1,1)\begin{aligned} e^{x}&=\sum_{n=0}^{\infty} \frac{1}{n !} x^{n}=1+x+\frac{1}{2 !} x^{2}+\cdots \in(-\infty,+\infty) \\ \sin x&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} x^{2 n+1}=x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}+\cdots, x \in(-\infty,+\infty) \\ \cos x&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n) !} x^{2 n}=1-\frac{1}{2 !} x^{2}+\frac{1}{4 !} x^{4}+\cdots, x \in(-\infty,+\infty) \\ \tan x&=\sum_{n=1}^{\infty} \frac{B_{2 n}(-4)^{n}\left(1-4^{n}\right)}{(2 n) !} x^{2 n-1}=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\cdots,x\in (-\frac{\pi}{2},\frac{\pi}{2})\\ \arcsin x&=\sum_{n=0}^{\infty} \frac{(2 n) !}{4^{n}(n !)^{2}(2 n+1)} x^{2n+1}=x+\frac{1}{6} x^{3}+\frac{3}{40} x^{5}+\frac{5}{112} x^{7}+\frac{35}{1152} x^{9}+\cdots+, x \in(-1,1)\\ \arctan x&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1} x^{2 n+1}=x-\frac{1}{3} x^{3}+\frac{1}{5} x^{5}+\cdots+ x \in[-1,1] \\ \ln (1+x)&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} x^{n+1}=x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}+\cdots, x \in(-1,1] \\ \frac{1}{1-x}&=\sum_{n=0}^{\infty} x^{n}=1+x+x^{2}+x^{3}+\cdots, x \in(-1,1) \\ \frac{1}{1+x}&=\sum_{n=0}^{\infty}(-1)^{n} x^{n}=1-x+x^{2}-x^{3}+\cdots, x \in(-1,1)\\ (1+x)^{\alpha}&=1+\sum_{n=1}^{\infty} \frac{\alpha(\alpha-1) \cdots(\alpha-n+1)}{n !} x^{n}=1+\alpha x+\frac{\alpha(\alpha-1)}{2 !} x^{2}+\cdots, x \in(-1,1) \\ \end{aligned}

2 微分

基本函数

  • d(C)=0\text{d}(C) = 0
  • d(xn)=nxn1dx\text{d}(x^n) = nx^{n-1} \, \text{d}x
  • d(1x)=1x2dx\text{d}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \, \text{d}x
  • d(ax)=axlnadx\text{d}(a^x) = a^x \ln a \, \text{d}x
  • d(ex)=exdx\text{d}(e^x) = e^x \, \text{d}x
  • d(lnx)=1xdx\text{d}(\ln x) = \frac{1}{x} \, \text{d}x
  • d(logax)=1xlnadx\text{d}(\log_a x) = \frac{1}{x \ln a} \, \text{d}x

三角函数

  • d(sinx)=cosxdx\text{d}(\sin x) = \cos x \, \text{d}x
  • d(cosx)=sinxdx\text{d}(\cos x) = -\sin x \, \text{d}x
  • d(tanx)=sec2xdx\text{d}(\tan x) = \sec^2 x \, \text{d}x
  • d(cotx)=csc2xdx\text{d}(\cot x) = -\csc^2 x \, \text{d}x
  • d(secx)=secxtanxdx\text{d}(\sec x) = \sec x \tan x \, \text{d}x
  • d(cscx)=cscxcotxdx\text{d}(\csc x) = -\csc x \cot x \, \text{d}x

反三角函数

  • d(arcsinx)=11x2dx\text{d}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}} \, \text{d}x
  • d(arccosx)=11x2dx\text{d}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}} \, \text{d}x
  • d(arctanx)=11+x2dx\text{d}(\arctan x) = \frac{1}{1 + x^2} \, \text{d}x
  • d(arccot x)=11+x2dx\text{d}(\text{arccot}\ x) = -\frac{1}{1 + x^2} \, \text{d}x

3 积分

基本函数

  • 0dx=C\int 0 \, \text{d}x = C
  • xadx=xa+1a+1+C(a1)\int x^a \, \text{d}x = \frac{x^{a+1}}{a+1} + C \, (a \neq -1)
  • 1xdx=lnx+C(x0)\int \frac{1}{x} \, \text{d}x = \ln |x| + C \, (x \neq 0)
  • axdx=axlna+C(a>0,a1)\int a^x \, \text{d}x = \frac{a^x}{\ln a} + C \, (a > 0, a \neq 1)
  • exdx=ex+C\int e^x \, \text{d}x = e^x + C

三角函数

  • cosxdx=sinx+C\int \cos x \, \text{d}x = \sin x + C
  • sinxdx=cosx+C\int \sin x \, \text{d}x = -\cos x + C
  • tanxdx=lncosx+C\int \tan x \, \text{d}x = \ln |\cos x| + C
  • cotxdx=lnsinx+C\int \cot x \, \text{d}x = \ln |\sin x| + C
  • secxdx=lnsecx+tanx+C=lntanx2+π4+C\int \sec x \, \text{d}x = \ln |\sec x + \tan x| + C = \ln \left| \tan \frac{x}{2} + \frac{\pi}{4} \right| + C
  • cscxdx=lncscxcotx+C=lntanx2+π4+C\int \csc x \, \text{d}x = \ln |\csc x - \cot x| + C = \ln \left| \tan \frac{x}{2} + \frac{\pi}{4} \right| + C

反三角函数

  • 11x2dx=arcsinx+C(arccosx+C)\int \frac{1}{\sqrt{1 - x^2}} \, \text{d}x = \arcsin x + C \, (\text{或} \, \arccos x + C)
  • 11+x2dx=arctanx+C(arccotx+C)\int \frac{1}{1 + x^2} \, \text{d}x = \arctan x + C \, (\text{或} \, \text{arccot} x + C)

其他

  • 1cos2xdx=sec2xdx=tanx+C\int \frac{1}{\cos^2 x} \, \text{d}x = \int \sec^2 x \, \text{d}x = \tan x + C
  • 1sin2xdx=csc2xdx=cotx+C\int \frac{1}{\sin^2 x} \, \text{d}x = \int \csc^2 x \, \text{d}x = -\cot x + C
  • 1a2x2dx=12alna+xax+C\int \frac{1}{a^2 - x^2} \, \text{d}x = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C
  • xa2x2dx=a2x2+C\int \frac{x}{\sqrt{a^2 - x^2}} \, \text{d}x = -\sqrt{a^2 - x^2} + C
  • a2x2dx=x2a2x2+a22arctanxa2x2+C\int \sqrt{a^2 - x^2} \, \text{d}x = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \arctan \frac{x}{\sqrt{a^2 - x^2}} + C
  • dxx2+a2=lnx+x2+a2+C\int \frac{\text{d}x}{\sqrt{x^2 + a^2}} = \ln |x + \sqrt{x^2 + a^2}| + C
  • exsinxdx=12ex(sinxcosx)+C\int e^x \sin x \, \text{d}x = \frac{1}{2} e^x (\sin x - \cos x) + C
  • excosxdx=12ex(sinx+cosx)+C\int e^x \cos x \, \text{d}x = \frac{1}{2} e^x (\sin x + \cos x) + C

4 微分方程

5 应用题

6 证明题

7 无穷级数


AdvancedMath
https://blog.dwj601.cn/GPA/1st-term/AdvancedMath/
作者
Mr_Dwj
发布于
2024年7月12日
更新于
2024年10月28日
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