用迭代法计算方程 x^3-x-1=0 在x=1.5附近的根,要求误差: s10^-9
时间: 2023-05-21 10:01:30 浏览: 127
答案:使用牛顿迭代法,设初始值为x0=1.5,迭代公式为xn+1 = xn - f(xn)/f'(xn),其中f(x) = x^3 - x - 1,f'(x) = 3x^2 - 1。根据迭代公式,可以得到以下迭代过程:
x1 = x0 - f(x0)/f'(x0) = 1.5 - (1.5^3 - 1.5 - 1)/(3*1.5^2 - 1) ≈ 1.324717957
x2 = x1 - f(x1)/f'(x1) = 1.324717957 - (1.324717957^3 - 1.324717957 - 1)/(3*1.324717957^2 - 1) ≈ 1.220744084
x3 = x2 - f(x2)/f'(x2) = 1.220744084 - (1.220744084^3 - 1.220744084 - 1)/(3*1.220744084^2 - 1) ≈ 1.167303978
x4 = x3 - f(x3)/f'(x3) = 1.167303978 - (1.167303978^3 - 1.167303978 - 1)/(3*1.167303978^2 - 1) ≈ 1.147231724
x5 = x4 - f(x4)/f'(x4) = 1.147231724 - (1.147231724^3 - 1.147231724 - 1)/(3*1.147231724^2 - 1) ≈ 1.145366891
x6 = x5 - f(x5)/f'(x5) = 1.145366891 - (1.145366891^3 - 1.145366891 - 1)/(3*1.145366891^2 - 1) ≈ 1.145365385
x7 = x6 - f(x6)/f'(x6) = 1.145365385 - (1.145365385^3 - 1.145365385 - 1)/(3*1.145365385^2 - 1) ≈ 1.145365385
x8 = x7 - f(x7)/f'(x7) = 1.145365385 - (1.145365385^3 - 1.145365385 - 1)/(3*1.145365385^2 - 1) ≈ 1.145365385
因此,方程 x^3-x-1=0 在x=1.5附近的根约为1.145365385,误差小于s10^-9。