VB中定义Arccos函数

游侠无极限

我是这样的：
  ?# E) q2 C5 z" d' Z1 E1 h7 JPrivate Function Arccos(x As Double, n As Long)
  e# K8 p$ k0 {9 U' ^Dim i As Long, y As Double
For i = 1 To n
* {! @5 S5 S8 g7 G8 M% m, s    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
Next
& _' a0 v  M8 L5 a( R8 ?Arccos = y
9 W" p, D9 \' P2 B/ SEnd Function
1 e1 s0 S% l& C# z: n. L- q
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
  {2 ^3 t5 B/ _( Q! S- KPrivate Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
    Select Case x
+ ?  T, J4 @3 s    Case 1
        Arccos = 0
    Case -1
        Arccos = pi
( Y% p* F# a5 Q: [6 ?1 W7 e    Case -1 To 1
        Dim dx As Double, y As Double, i As Long
( c! v- S5 P: h$ O1 R; \6 u6 d2 M1 q        dx = x / n
        For i = 1 To n
# e( T, A  i' L8 @            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
, Q# H: N+ u, C        Next
        Arccos = y + pi / 2
5 d0 F! [2 n2 `, s  R    Case Else
2 @' R- k) J4 }  P        '异常
    End Select
End Function
4 m3 N7 Q9 O' [* E  [
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