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)
* N8 s+ o2 n+ `/ @; y4 ~. _Dim i As Long, y As Double
3 w4 ^5 n' ]" O+ A# N" F! VFor i = 1 To n
- M- h9 ^( J% E+ }    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
2 e+ |7 l) D5 F4 l& X& tNext
, H( b/ s' h5 F& m0 u) r$ b, gArccos = y
End Function
7 e/ ~$ K- E! o: x
2 b3 P* D) W  v/ r. u$ B% V1 ]用的是Arccos的微分来算近似值，怎么不对？

游侠无极限

Const pi As Double = 3.14159265359
$ T! [$ U+ e' q  C. f4 y% YPrivate Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
    Select Case x
+ k$ z+ ?% Y# Z7 `% m    Case 1
        Arccos = 0
0 Z, H* o4 y6 ^3 ]. H* L    Case -1
        Arccos = pi
    Case -1 To 1
( P4 Z6 n8 f' f, \# B        Dim dx As Double, y As Double, i As Long
& [; b* _2 D( x) d        dx = x / n
        For i = 1 To n
            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
8 n9 Z) u9 I. U$ o        Next
        Arccos = y + pi / 2
    Case Else
$ c2 ^8 |* f0 p- W7 s2 @+ W) w5 }        '异常
    End Select
End Function
/ \. R* {/ H8 I' f
. ?0 B' s& C+ _; v: X+ _搞定了

煎饼

公式对就没有问题拉