VB中定义Arccos函数

游侠无极限

我是这样的：
: g8 Z" B: g% A5 U9 c+ IPrivate Function Arccos(x As Double, n As Long)
' y9 b5 `6 M3 z5 k" GDim i As Long, y As Double
For i = 1 To n
    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
6 G$ T5 l2 o$ |4 [$ O( B0 w7 ~Next
% ?6 {3 _; v1 {9 F. s$ {Arccos = y
+ r% d. d9 p" _) @" d' r+ U9 T* gEnd 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
Private Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
0 F0 w# y0 I1 G# J# N# N    Select Case x
1 Q7 y. r% M6 R    Case 1
        Arccos = 0
    Case -1
        Arccos = pi
1 q! N2 ]( V2 A& t4 z1 a1 n    Case -1 To 1
, ?, N7 e6 u: [+ l* E  U' U, e, P        Dim dx As Double, y As Double, i As Long
        dx = x / n
        For i = 1 To n
$ |' q2 }# _5 g7 s- \& a3 Y, |            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
        Next
) U8 J; s6 ?# d, q1 ^. v        Arccos = y + pi / 2
" z% V$ ^1 A* G' u/ P0 V1 I    Case Else
        '异常
    End Select
! w( D) |# x4 D8 l( [8 L8 QEnd Function

搞定了