假设级数表达式为f(i),随i值的变化而减小,则可在循环时利用级数f(i)和f(i-1)的差值来与10^-6比较,当两次计算的结果满足 dp
p = Abs(dc)
b = b * -1
n = n + 1
mv = rad ^ (2 * (n - 1))
For i = 2 * (n - 1) To 2 Step -1
mv = mv / i
Next i
dc = dc + b * mv
p = Abs(p - Abs(dc))
Loop
zcos = dc
End Function
Sub main()
MsgBox Round(zcos(3.1415926 * (135 / 180),0.000001),6)
End Sub
你可以自行在main程序里做个窗体并输入精度值,我的代码为0.000001,你可以试试.
