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)
8 R0 f6 y. _: H% m3 V( ADim i As Long, y As Double
# m, j$ k5 o# ?8 j: Y4 {/ cFor i = 1 To n
0 ]3 R+ _  t# @    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
& r; g! S0 ~6 DNext
& Y: `4 X' Q1 T5 bArccos = y
End Function
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用的是Arccos的微分来算近似值，怎么不对？
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煎饼

公式对就没有问题拉

游侠无极限

Const pi As Double = 3.14159265359
/ S; }5 N5 M+ d0 a9 f* u# D  BPrivate Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
/ h; O$ P; U" r+ g+ P6 [0 R    Select Case x
% H  g$ M5 e- M6 \: T3 N! {    Case 1
* I' h! B9 Y0 P) K  s, b8 Q  _        Arccos = 0
$ ~( P9 t$ c. ^+ Q, Y    Case -1
        Arccos = pi
1 Y8 R/ m9 S) V! D" O) f$ Q    Case -1 To 1
        Dim dx As Double, y As Double, i As Long
7 e# V$ h! R- N        dx = x / n
        For i = 1 To n
            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
        Next
        Arccos = y + pi / 2
6 C% M# C: l( [9 l& m" u    Case Else
1 U% b' s$ x; E4 B. s7 \8 h        '异常
    End Select
End Function
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% F3 N8 O- z% c' V1 I& m搞定了