牛頓拉弗森法解方程式

2012-10-31 7:41 pm
以牛頓拉弗森法求解方程式 X三次方=20 在2至3間之根(以X零=3為初始值)

回答 (3)

2012-10-31 9:02 pm
✔ 最佳答案
令f(x)=x^3-20

df(x)/dx=3x^2

△c(0)=c-f(x(0))=0-l 3^3-20 l=-7

[df/dx]=3*3^2=27

△x(0)=c(0)/[df/dx](0)=-7/27=-0.259

x(1)=x(0)+△x(0)=3-0.259=2.741

x(2)=x(1)+△x(1)=2.741-(0.593/22.539)=2.715

x(3)=x(2)+△x(2)=2.715-)0.012/22.113)=2.7144


x值約為2.7144
2012-11-01 7:45 pm
他數學可能比老師強!
2012-10-31 10:32 pm
牛頓拉弗森法

x_n+1 = x_n - f(x_n)/f'(x_n)

f(x) = x^3 - 20, f'(x) = 3x^2

x_n+1 = x_n - [(x_n)^3 - 20]/[3x_n^2]

x_0 = 3

x_1 =3 - 7/27 = 2.740741

x_2 = 2.740741 - [(2.740741)^3 - 20]/[3 * 2.740741^2] = 2.714669625

x_3 = 2.71441764

x_4 = x_5 = x_6 = ... = 2.714417617

因此x^3 = 20 的根為2.714417617


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