人工智能教程 - 数学基础课程1.1 - 数学分析（一）22-24 数值积分,三角函数的积分
Assumption1
hits=ce−r2hits=ce^{-r^2}hits=ce−r2
r1<r=x<r2r_1<r=x<r_2r1<r=x<r2
Part=∫r1r2(2πr)e−r2drPart = \int_{r_1}^{r_2}(2 \pi r)e^{-r^2}drPart=∫r1r2(2πr)e−r2dr
=−πe−r2∣r1r2=- \pi e^{-r^2}|_{r_1}^{r_2}=−πe−r2∣r1r2
=π(e−r12−e−r22)=\pi (e^{-r_1^2}-e^{-r_2^2})=π(e−r12−e−r22)
WHOLE=U≤r<∞WHOLE = U\leq r< \inftyWHOLE=U≤r<∞
=cπ(e−02−e−∞2)=cπ=c\pi (e^{-0^2}-e^{-\infty^2})=c\pi=cπ(e−02−e−∞2)=cπ
P(r1<r<r2)=PartWholeP(r_1<r<r_2)=\frac{Part}{Whole}P(r1<r<r2)=WholePartP(0<r<∞)=1P(0<r<\infty)=1P(0<r<∞)=1
Weight:
W(r)=2πcre−r2W(r)=2\pi cre^{-r^2}W(r)=2πcre−r2
数值积分 NUMERICAL INTEGRATION 1.RIEMANN SUMS (y0+y1+...+yn−1)Δx(left hand)(y_0+y_1+...+y_{n-1})\Delta x (left \ hand)(y0+y1+...+yn−1)Δx(left hand) (y1+y2+...+yn)Δx(right hand)(y_1+y_2+...+y_{n})\Delta x (right \ hand)(y1+y2+...+yn)Δx(right hand) 2.梯形法 Trapezoidal rule Δx(y02+y1+y2+...+yn−1+12yn)=LEFT R−S + RIGHT R−S2\Delta x(\frac{y_0}{2}+y_1+y_2+...+y_{n-1}+\frac{1}{2}y_n)=\frac{LEFT \ R-S \ + \ RIGHT \ R-S}{2}Δx(2y0+y1+y2+...+yn−1+21yn)=2LEFT R−S + RIGHT R−S 3.辛普森法 Simpson’s rule 2Δx (y0+4y1+y26)2\Delta x \ (\frac{y_0+4y_1+y_2}{6})2Δx (6y0+4y1+y2) Simpson’s - Exactans ≈(Δx)4\large \approx (\Delta x )^4≈(Δx)4 Ex: Δx≈110\Delta x \approx \frac{1}{10}Δx≈101 (Δx)4≈10−4(\Delta x )^4\approx 10 ^{-4}(Δx)4≈10−4Simpson’s is derived using the exactanswer for all degree 2 polynomials
二次多项式的精确解 V=∫0∞2πre−r2drV= \int_{0}^{\infty}2\pi re^{-r^2}drV=∫0∞2πre−r2dr =−πe−r2∣0∞=π(1−0)=π=-\pi e^{-r^2}|_0^{ \infty }=\pi (1-0) = \pi=−πe−r2∣0∞=π(1−0)=π 三角函数的积分及三角替换 半角公式 Half-angle formula:sin2Θ+cos2Θ=1\large sin^2\Theta +cos^2\Theta =1sin2Θ+cos2Θ=1
cos(2Θ)=cos2Θ−sin2Θcos(2\Theta) = cos^2\Theta - sin^2\Thetacos(2Θ)=cos2Θ−sin2Θ
sin(2Θ)=2sinΘ.cosΘsin(2\Theta) = 2sin\Theta . cos\Thetasin(2Θ)=2sinΘ.cosΘ
cos(2Θ)=cos2Θ−sin2Θ=cos2Θ−(1−cos2Θ)=2cos2Θ−1cos(2\Theta) = cos^2\Theta - sin^2\Theta= cos^2\Theta - (1-cos^2\Theta)= 2cos^2\Theta - 1cos(2Θ)=cos2Θ−sin2Θ=cos2Θ−(1−cos2Θ)=2cos2Θ−1
⇒cos2Θ=1+cos2Θ2{\color{Red} \Rightarrow cos^2\Theta = \frac{1+cos2\Theta}{2}}⇒cos2Θ=21+cos2Θ sin2Θ=1−cos2Θ2{\color{Red} sin^2\Theta = \frac{1-cos2\Theta}{2}}sin2Θ=21−cos2ΘKNOW:
dsinx=(cosx)dx:∫cosxdx=sinx+cdsinx = (cosx)dx:\int cosxdx =sinx+cdsinx=(cosx)dx:∫cosxdx=sinx+c
dcosx=−(sinx)dx:∫sinxdx=−cosx+cdcosx = -(sinx)dx:\int sinxdx =-cosx+cdcosx=−(sinx)dx:∫sinxdx=−cosx+c
作者：KuFun人工智能
