VB中定义Arccos函数

游侠无极限

我是这样的：
Private Function Arccos(x As Double, n As Long)
Dim i As Long, y As Double
# q" q( ]% d" AFor i = 1 To n
: N) D( ?  S) z6 ]' i    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
Next
Arccos = y
End Function

用的是Arccos的微分来算近似值，怎么不对？

煎饼

公式对就没有问题拉

游侠无极限

Const pi As Double = 3.14159265359
  z! E8 |( b/ |Private Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
    Select Case x
    Case 1
0 y7 Q0 K7 @/ ?1 m- o( ]0 A        Arccos = 0
    Case -1
        Arccos = pi
6 E/ Q9 D5 N) f: ^/ V    Case -1 To 1
, j3 X9 r" c1 k) x0 `; u        Dim dx As Double, y As Double, i As Long
: |! k9 ^! t8 f3 n9 a        dx = x / n
0 O8 d) C4 e3 @4 W        For i = 1 To n
. A4 a/ d# V. U% [9 y- K            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
/ B% z. T' [' n; ]* h, D4 ?        Next
        Arccos = y + pi / 2
    Case Else
9 x& }: ?) W. `. ?        '异常
4 p% W  T+ y7 P% O% ^# H* C    End Select
End Function

4 K" i- X- m9 I! o  B; m. N% o" L搞定了