人工智能教程 - 数学基础课程1.1 - 数学分析（一）15-17 微分方程和分离变量,定积分及性质,微积分第一定理

微分方程 differential equation

Ex:

$\frac{dy}{dx}=f(x)$

$y=\int f(x)dx$

solved substitution

Ex2:

$(\frac{d}{dx}+x)$为annihilation operator 湮没算符 in quantum mechnics

$\frac{dy}{dx}=-xy$

$\frac{dy}{y}=-xdx$

$\int \frac{dy}{y}=-\int xdx$

$lny=-x^2/2+C$

$e^{lny}=e^{-x^2/2}+C$

$y=A{e}^{-x^2/2}(A=e^c)$

Solution:

$y=a{e}^{-x^2/2}anya$

$\frac{dy}{dx}=a.\frac{d}{dx}.e^{-x^2/2}$

$=a.(-x).e^{-x^2/2}$

$=-x.y$

by the way,the function is known as the normal distribution 正态分布

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分离变量法

SEPARATION OF VARIABLES

$\frac{dy}{dx}=f(x).g(y)$

$\frac{dy}{g(y)}=f(x)dx$

$H(y)=\int \frac{dy}{g(y)};F(x)=\int f(x)dx$

$H(y)=F(x)+C\to implicit$

$y=H^{-1}(F(x)+C)$

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定积分

Definite Integrals

Find Area under a curve =${\int }_a^{b}f(x)dx$

累积和(cumulative sum)

To compute the area

1. divide into rectangles
2. add up areas
3. take the limit as rectangles get thin

简写 abbreviation

${\sum }_{i=1}^n{a}_i=a_1+a_2+...+a_n$

$\sum$ is called sigma

Notation (Riemann Sums)

General Procedure for definite integrals:

$\Delta x=\frac{b-a}{n}$

Pick any height of f in each interval:

${\sum }_{i=1}^{n}f(c_i)\Delta x\to {\int }_a^{b}f(x)dx$

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微积分第一定理

Fundamental theorem of calculus(FTC1)

If F’(x) = f(x), then

${\int }_a^{b}f(x)dx=F(b)-F(a)=F(x){|}_a^b$

$F=\int f(x)dx$

NOTATION

$F(b)-F(a)=F(x){|}_a^b=F(x){|}_{x=a}^{x=b}$

Ex1:

$F(x)=x^3/3$

$F^{\prime }(x)=x^2(=f(x))$

${\int }_a^b{x}^{2}dx=F(b)-F(a)=\frac{b^3}{3}-\frac{a^3}{3}$

${\int }_0^b{x}^{2}dx=\frac{x^3}{3}{|}_0^b=\frac{b^3}{3}$

True geometric interp of definit integral is that area above x-axis minus the area below the x-axis

定积分的性质

Properties of integrals

$1.{\int }_a^b(f(x)+g(x))dx={\int }_a^{b}f(x)dx+{\int }_a^{b}g(x)dx$

$2.{\int }_a^{b}Cf(x)dx=C{\int }_a^{b}f(x)dx$

$3.{\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx(a<b<c)$

$4.{\int }_a^{a}f(x)dx=0$

$5.{\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

6. (Estimation)If f(x)<=g(x),then:${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx$

change of variables: (=substituion)

${\int }_{u_1}^{u_2}g(u)dx={\int }_{x_1}^{x_2}g(u(x)).u^{\prime }(x)dx$

u = u(x)	u1 = u(x1)
du = u’(x)dx	u2 = u(x2)

Only works if u’ does not change sign