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
, w7 i( P5 f. j. f, J, TFor i = 1 To n
    y = y + (-1 / Sqr(1 + (x / n * i) ^ 2)) * x / n
Next
Arccos = y
8 A8 p( Y9 m2 s( c; ?! Y% \% l8 gEnd Function
4 P6 P4 ?5 k& u8 Q; o: k
# Y4 M* A; V. X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
Private Function Arccos(x As Double, n As Long) ' n 为函数精度，建议为100
    Select Case x
7 J6 a; p; g  M4 i  x) v- O    Case 1
        Arccos = 0
    Case -1
6 H: S6 ^4 E  }9 n( P" M3 B        Arccos = pi
    Case -1 To 1
; m: P- G4 t$ o  F: n/ |        Dim dx As Double, y As Double, i As Long
# p+ v' h7 ^% \        dx = x / n
        For i = 1 To n
" n1 f7 Z" C# D9 i            y = y + (-1 / Sqr(1 - (dx * i + dx / 2) ^ 2)) * dx
! ]' A8 s! e+ k  l" T        Next
7 n' X* I, x' I        Arccos = y + pi / 2
' Y: R0 p# m& S7 g    Case Else
        '异常
- n2 j" u, z5 b+ H# v: T    End Select
End Function
- Z5 G5 l' D6 B+ u
4 [6 u: W# z. M' T- E搞定了

煎饼

公式对就没有问题拉