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)
Dim i As Long, y As Double
7 v3 t1 Y4 x( a* \% BFor i = 1 To n
    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
$ }% Z1 F' r/ |9 }7 h1 S! bNext
- |! z  M1 v( z/ l7 y- S5 f# Q( r/ X+ EArccos = y
5 s: u$ x4 G1 O) t# U+ TEnd Function

1 L4 S* P$ _" y* ?: l6 t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
* H; T1 d6 w1 E5 K7 KPrivate Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
) j* ^7 S' u$ x4 S+ o4 I- m3 i    Select Case x
    Case 1
        Arccos = 0
0 h/ }5 M* h9 J    Case -1
        Arccos = pi
    Case -1 To 1
        Dim dx As Double, y As Double, i As Long
& B3 r' q2 O# _8 j0 V        dx = x / n
        For i = 1 To n
            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
$ x' v4 }5 |) B% M. S  e5 l        Next
4 ^$ b0 K6 G# t9 a5 n0 v        Arccos = y + pi / 2
; U& v% Y( l. N; v: O, y3 j" u/ k- {    Case Else
0 L& ^+ w1 q! F% h% d        '异常
    End Select
End Function
4 \- Y0 ~5 E0 h  @! H
搞定了