VB中定义Arccos函数

游侠无极限

我是这样的：
9 c/ `" M6 T  S, \Private Function Arccos(x As Double, n As Long)
% P" ^+ f+ a# A4 K$ ^$ PDim i As Long, y As Double
: `4 u3 F/ t5 M$ n: x2 l4 g# ^For i = 1 To n
    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
7 W0 A3 q, t* p9 ^/ L! w1 F" TNext
Arccos = y
& H/ B1 e: P$ b% E1 }8 e7 WEnd Function

用的是Arccos的微分来算近似值，怎么不对？
/ u. X2 \9 Y& ?4 P! y! ?( _

煎饼

公式对就没有问题拉

游侠无极限

Const pi As Double = 3.14159265359
Private Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
8 |4 q) H4 J% T- D    Select Case x
    Case 1
        Arccos = 0
  |' i: Y' ~+ L  y    Case -1
8 v7 V3 S4 {& u8 }- b        Arccos = pi
    Case -1 To 1
+ K  [$ O. t  k0 i* V  O6 w( g9 j! b        Dim dx As Double, y As Double, i As Long
) _% |2 x- e% W1 `8 V3 S% W        dx = x / n
, Z, H+ E. w8 H7 E        For i = 1 To n
            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
        Next
        Arccos = y + pi / 2
    Case Else
        '异常
+ o  v7 \& O; v5 b& Q2 ~: v/ U    End Select
0 y8 S' r9 i) Z- KEnd Function
% B# Z5 s. w2 |8 q! E
搞定了