切线法求高次方程x³-3x+1=0的根、初始值设为2
问题描述：
切线法求高次方程x³-3x+1=0的根、初始值设为2
答
f(x) = x^3 - 3x + 1,f'(x) = 3x^2 - 3,x1 = 2
x2 = x1 - f(x1)/f'(x1) = 2 - (2^3 - 3*2 +1)/(3(2)^2 - 3) = 2- 3/9 = 5/3
x3 = x2 - f(x2)/f'(x2) = 5/3 - ((5/3)^3 - 3(5/3) + 1)/(3(5/3)^2 - 3) = -1.548611...