2018-11-19 回答

用牛顿迭代法求方程'a * x ^ 3 + b * x ^ 2 + c * x + d = 0, 系数a = 1, b = 2, c = 3, d = 4, x在0附近的一个实数根为1.33333333333。算法代码如下：

private sub command1_click() '牛顿迭代法

dim a as double, b as double, c as double, d as double, xx1 as double

dim n as long

a = 1

b = 2

c = 3

d = 4

xx1 = 0 '初始值为0，x在0附近

print nim(a, b, c, d, xx1, n) '方程在xx1附近的根

print n '迭代次数

end sub

'牛顿迭代法newton iteration method

function nim(byval a as double, byval b as double, byval c as double, byval d as double, byval x0 as double, byref n as long) as double

dim x as double, y as double, dy as double, ydy as double, i as long

n = -1 '迭代次数

x = x0

for i = 0 to 1000

y = a * x ^ 3 + b * x ^ 2 + c * x + d

dy = 3 * a * x ^ 2 + 2 * b * x + c

if abs(x - x0) 1000 then n = -2 '该方程无解解

end function